the berkeley review - biology part 1

BiologyParti

Physiology

Sections I-VSection I

Nerve and Muscle

Section IIHeart and Lung

Section IIIGastrointestinal Tract

and Kidney

Section IVReproduction

and Development

f Section V

Endocrinologyand Immunology

The

BerkeleyJUr*e*v*i*e*w®

Specializing in MCAT Preparation

ERKELEYR • E • V • ?^ E • W

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BiologySection I

Nerve

and

Muscle

The Nervous System

1. Types of Tissues

2. Membrane Potentials

3. Action Potentials

4. The Neuromuscular Junction5. The Sarcomere

6. Nervous System Components7. Control of Body Activity8. Receptors and Sensory Input

Practice Passages and Answers

BERKELEYSpecializing in MCAT Preparation

•*

Nerve and MuscleTop 10 Section Goals

Be familiar with the different types of tissues.

The four general classes of biological tissues to consider are epithelial tissue, connective tissue,muscle tissue, and nerve tissue.

£) '^ Understand the concept of a membrane potential.Know the relative intracellular and extracellular concentrations of ions like Na®, K®, and Cle. Befamiliar with the Nernst equation and how to calculate a membrane voltage.

Q •jja Know the different phases of the action potential.Understand depolarization, repolarization, hyperpolarization, and refractory period.Be familiar withthe differences oetween action potentials in skeletal muscle and cardiac muscle.

O -im Be familiar with the neuromuscular junction.Understand howa signal ispropagated from thepresynaptic membrane to thepostsynaptic membrane.Know how neurotransmitters are released and how they are degraded.

0 v^ Be familiar with muscle contraction and the sarcomere.Beaware of how the different components of muscle interact in order to contract and relax. Knowthe differences between actin, myosin, troponin, tropomyosin, and their interactions with Ca2®.

@ ^ Be familiar with the components of the nervous system.^The three primary divisions of the brain are the forebrain, midbrain, and hindbrain. Know thecomponents ofeach and how they relate to thebodyas a whole.

q ^^ Beable to distinguish between consciously directed and involuntary muscle activity.^5f Understand the differences between motor and sensory neurons and how they relate to the spinal^ cord. Be able to describe the structure and function of a simple reflex arc, and of smooth versus

skeletal muscle.

9!»Know the different divisions of the nervous system.

The two major divisions are the central nervous system (CNS) and the peripheral nervous system(PNS). Know the subdivisions of the PNS and their actions.

q ^^ Know the types of neurotransmitters that are released from the CNS and PWS.These neurotransmitters include acetylcholine, epinephrine, and norepinephrine. Understand whysome neurotransmitters or hormones are called amines, peptides, or steroids.

Be familiar with the different types of sensory input received by the brain.This sensory input can come from mechanoreceptors, chemoreceptors, thermoreceptors, andphotoreceptors. Be familiar with the action ofeach receptor type.

°p

Biology Nerve fif Muscle

The Nervous SystemTypes Of Tissues

Let's consider a series ofdiscussions on cellular physiology. Forexample, we willconsider how muscle andnerve cells function. How does the chemical energy ofATP (which was generated in glycolysis, the Krebs cycle, and oxidativephosphorylation) become converted into the mechanical movement of, say,muscle cells? How is it that the chemical energy of ATP is converted into anelectrical signal that allows various nerves to communicate with those muscles?

Before we can discuss the cellular mechanisms of muscles and nerves, we firstneed toconsider some of thegeneral characteristics ofcells, tissues, and organs.The general body plan of an animal is fairly simple and can be divided into anumber ofsystems thatrepresent a variety oforgans working in concert withoneanother. Forexample, one body system you are probably quite familiar with isthe skeletal system. Another is the muscular system. Others are the circulatory,integumentary (skin), endocrine, nervous, and digestive systems, to name but afew.

The digestive system is formed byan alimentary canal (gastrointestinal "tube")that begins at the mouth andends at the anus. This system issuspended within abody cavity referred to as the coelom. The coelom is separated into a thoraciccavity (upper) and anabdominal cavity (lower). These two cavities are separatedby the dome-shaped mass ofskeletal muscle called the diaphragm. Within thethoracic cavity, one finds the lungs and the heart. The abdominal cavity containsthe liver, stomach, and intestines.

As we examine these various systems, we will find different levels oforganization. There are individual cells, and then there are cells ofa particulartype which coalesce to form tissue. One example of a tissue is the layer ofepithelial cells that line one of the principal organs of the alimentary canal, thestomach. Some of the simple epithelial cells within the stomach secretehydrochloric acid (pH «1) to aid in the digestion offood. Other epithelial cells ofthe stomach secrete mucus to help prevent that acid from digesting the lining ofthe stomach. Still other epithelial cells secrete enzymes. These epithelial cells arejust one type of tissue that is involved in forming the stomach. The stomach isalso composed of other types of tissue. For example, nervous tissue helps toinnervate thestomach, connective tissuehelps tohold the stomach in its properposition, and muscle tissue helps to propel food through the stomach. Thus,these four groups of primary tissue (epithelial, connective, muscle, and nerve)have the ability to form the various organs of the body. An organ is a structurethat is composed oftwo or more tissues that act in such a way as to perform a specificfunction.

Epithelial TissuesLet's examine the epithelial cells in a little more detail. The epithelial tissue thatconstitutes the various organs of the body can be either simple epithelium(consisting of a single layer of cells) or stratified epithelium (consisting of two ormore layers of cells). Theseepithelial cellscomein a varietyof shapes and sizes.For example, there are squamous (flat), cuboidal, and columnar epithelialcells(refer to Figure 1-1).

Types Of Tissues

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Biology Nerve & Muscle Types Of Tissues

On the lumenal side of the simple epithelial cells are projections calledmicrovilli (singular, microvillus—see Figure 1-1 and Figure 1-2). Theseprojections increase the total absorptive area of the cell (sometimes by as much as25%). Sometimes you find specialized structures called cilia {singular, ciiium)projecting outward on the apical surface of these cells. For example, in therespiratory tract these hair-likeappendages move in a coordinated unidirectionalwave to move foreign particles out of the mucous lining of the lungs andbronchial tubes.

Simple squamousepithelial cells

Cuboidal and columnar

epithelial cells

microvilli

Basal

lamina

Basal

lamina

Simple columnarepithelial cells

Stratified squamousepithelial cells (non-keratinized)

Figure 1-1Types of epithelial cells.

These cells are bounded by a number of specialized junctions. For example, tightjunctions act as a permeability barrier (see Figure 1-2).Not only do they preventthe transport of protein molecules from the lumenal side of the cell towards thebasolateral side of the cell, but they also act to hold neighboring cells together.

Epithelial cells are also held together by structures called desmosomes (seeFigure 1-2)One type of desmosome joins the epithelial cell to a structure on thebasal side of the cell called the basal lamina (or basement membrane). The basallamina is in close contact with connective tissue that helps to anchor the cells inplace.

Gap junctions provide a means for water-soluble molecules to pass from thecytoplasm of one cell to the cytoplasm of another cell (see Figure 1-2). Thesejunctions allow for equilibration within the connected epithelial cells andtherefore allow those cells to function as a unit. For example, the beating ciliaappear to be coordinated by waves of calcium, which flow in the plane of thejuxtaposed epithelial cells.


Biology

Microvilli

Tight junctions'

Gapjunction

Desmosome

Basal lateral IZ^surface

Nerve fie Muscle

LUMENApical surface

BLOOD

^Zl Microvillus

\ I Tight junctions

Gapjunction

Desmosomes

\ I Basal lamina(basement membrane)

Figure 1-2Different components of an epithelial cell.

As we have mentioned, epithelial cells can secrete substances into a lumenalspace. For example, hydrochloric acid can be secreted into the lumen of thestomach. If a cell secretes a substance into the lumen by way of a duct, it isreferred to as an exocrine gland. Endocrine glands secrete substances into theblood. For example, insulin is a protein hormone secreted into the blood byclusters of specialized epithelial cells in the pancreas.

Basement ^

membrane *

Collagen fibers fN

Dead skin cells

S—I (keratinized)

Epidermalcells

Connective

tissue

Figure 1-3Stratified squamous epithelial cells.

Stratified squamous epithelium usually has a protective function. Your skin iscomposed of many layers of stratified squamous epithelial cells. The outer cellsof your skin are dead, and they contain a large amount of the fibrous proteinkeratin (Figure 1-3). These cells are constantly being lost and replaced, as cellsbegin to move toward the surface from below.

Types Of Tissues


Biology Nerve fie Muscle Types Of Tissues

Consider a segment of skin. This organ comprises about 15% of your total bodyweight. The epidermal region contains stratified epithelial cells that act toprotect the deeper layers of the skin. Below the epidermis is the dermis. Withinthe dermis are a variety of structures. Surrounding the hair follicles are erectormuscles, which act to straighten the hair shaft. This causes the skin close to thehair follicle to become depressed and gives the characteristic appearance of"goose pimples." Those erector muscles are innervated by nerves which causethem to contract at specific times (e.g., when it is cold outside). The skin is also ahighly vascularized organ. When it is hot outside, the blood is shunted towardsthe surface of the skin where it can dissipate some of its heat to the outsideenvironment. Below the dermis is the subcutaneous tissue. This is where one

finds adipose deposits.

Connective TissuesThis type of tissue helps to anchor and support the various structures of thebody. There are a variety of types of connective tissues, a few of which arestructural, blood cells, mast cells, adipose cells, and melanocytes.

Many of the proteins that make up structural connective tissue are secreted bycells called fibroblasts. Collagen, reticulin, and elastin are structural proteinswhich are secreted by these cells. Collagen is a triple-stranded, insoluble, fibrousprotein (see Figure 1-3) that is highly cross-linked, a feature that makes thesefibers quite strong and rather flexible. Besides having a very high tensilestrength, collagen is also the most abundant protein found in mammals.Reticulin is a thin fiber found in the spleen and lymph nodes. It is not as highlycoiled as collagen. Elastin is also a highly cross-linked protein found associatedwith organs that require some degree of elasticity (like the lungs, skin, and bloodvessels).

Another type of structural connective tissue, cartilage, is secreted by aspecialized fibroblast cell called a chondrocyte. There are different types ofcartilage, but in general it is found in places where there is a certain amount ofstress placed on the body. For example, cartilage can be found in the nose, on thearticulating surfaces of bones (including the intervertebral discs of the vertebralcolumn), and in the external ear.

Bone is also a structural connective tissue. About one-third of the weight of bonecomes from organic material such as collagen, while the remaining two-thirds isinorganic material such as calcium phosphate and calcium carbonate. Thecollagen found in bone matrix is secreted by specialized fibroblast cells calledosteoblasts. Collagen lends flexibility to bone, while the inorganic crystals lendrigidity. Within the central cavity of bone, we find a spongy marrow where redblood cells and white blood cells are formed. Towards the surface of bone the

cellular arrangement is more compact. [As a comparison, the main structuralcomponent of chitin (found in the exoskeleton of insects) consists of speciallymodified glucose residues linked to one another to form long polymers.Associated with these polymers is calcium carbonate (CaCC>3). This combinationadds rigidity to the exoskeleton,but offers little in the way of flexibility.]

We mentioned that blood cells and mast cells are kinds of connective tissue. We

will discuss blood cells in a separate lecture. Mast cells can be found in therespiratory tract, as well as in the gastrointestinal tract. Mast cells can release

Copyright © byTheBerkeley Review 6 The Berkeley ReviewSpecializing in MCAT Preparation

Biology Nerve & Muscle

histamines in response to an allergic reaction, an infection, or even an injury.Histamine causes an increase in blood flow to the blood vessels of the affected

region.

Other types of connective tissue involve adipose cells and melanocytes. Adiposecells are simply cells that store fat whereas melanocytes are cells which storepigments.

Muscle TissuesWe will be discussing various types of muscle in future lectures. For example,when we examine skeletal muscle we will find that it is voluntary muscle. Thatis, we can generally control its action. Cardiac muscle and smooth muscle areexamples of involuntary muscles.

Nervous TissuesThe nervous systems allow one to adapt rather quickly to external stimuli. Forexample, consider a simple reflex arc. If someone were to tap on your knee with arubber hammer, your lower leg would extend outward. As the hammerimpinged upon the patellar tendon in your knee, an electrical impulse wasgenerated and traveled via a sensory nerve to your spinal cord. That sensoryneuron synapsed with a motor neuron, which returned the impulse to themuscle that was initially stimulated and caused it to contract. We will come backto this example and examine it in a bit more detail later. First, let's consider someterminology.

Cell body

Dendrites

Axon

A

Neurotransmitters

are released from

synaptic bulbs -

A Typical Neuron

Figure 1-4The major components of a neuron.

Nerve cells and associated supporting cells make up the nervous system. Nervecells are also called neurons, and they are the basic structural unit that make upthe nervous system. The major anatomical features of a neuron are the cell body(involved in integration of information), the dendrites (involved in receiving andtransmitting information towards the cell body), and the axon (involved inconducting information away from the cell body). When a neuron becomesexcited and receives electrical information in the form of a stimulus, the cell bodyprocesses that information and transmits it down the axon in the form of a nerveimpulse called an action potential. When that action potential reaches the endjofthe axon (referred to as the synaptic bulb or bouton terminal), it causes the

Types Of Tissues


Biology Nerve fif Muscle Types Of Tissues

release of a chemical substance called a neurotransmitter (see Figure 1-4). Theneurotransmitter diffuses across the synaptic cleft and induces an identicalaction potential in an adjoining neuron, muscle cell, or gland cell. The junctionbetween two such cells is called a synapse.

Copyright © by The Berkeley Review 8 The Berkeley ReviewSpecializing in MCAT Preparation


Membrane Potentials

The generation of electrical signals in the nervous system is concerned with thediffusion of ions from a high to a low concentration (see Figure 1-5). In otherwords, charged ions diffuse down their concentration gradient. In theextracellular space of vertebrates the concentration of Na® is about 150 mM,while that of K® is about 5 mM. The concentration of Cle is about 130 mM,while that of HC03e is about 25 mM. Note that the concentrations of the cations(Na® and K®) and the anions (Cle and HCC>3e) balance one another. In otherwords, we find 155 mM of the cations and 155 mM of the anions. This representselectroneutrality.

Inside Cell Outside Cell

K+ = 5 mM

Na+=150mM

K+ = 120 mM to 140 mM *

Na+=10mMtol5mM

>

• Cations

Cr=5mMto40mM ^ Cr=130mM

HC03-=12mMto25mM > Anions HC03- = 25 mM

Proteins" .;j

Figure 1-5Typical cellular concentrations of the common ions.

Within the cell, we find a high concentration of K® (about 120 mM to 140 mM)and a low concentration of Na® (about 10 mM to 15 mM). We also find a lowerconcentration of Cle (about 5 mM to 40 mM) and usually a lower concentrationofHCC>3e (about12mM to 25 mM). There are manynegatively charged proteinswithin the cell. Electroneutrality can also be found on the inside of the cell aswell.

If you look at the distribution of ions across the cell's membrane, you will find itto be asymmetrical. Let's consider a resting nerve (i.e., a nerve that is waiting togenerate an action potential) that has a permeability to potassium which is muchgreater than its permeability to sodium. In other words, PK® >» PNa® (where"P" refers to permeability). Because the concentration of K® is higher inside thecell than outside the cell, potassium will diffuse down its concentration gradientand leave the cell (see Figure 1-6). As the positive K® cation leaves the cell, thereis correspondingly less positive charge remaining inside the cell. In other words,the inside of the cell is now more negatively charged with respect to the outsideof the cell. This sets up a voltage that is positive on the side of the cell to whichpotassium is trying to diffuse to (i.e., the outside of the cell). As that positivevoltage begins to build up, it tends to push potassium back into the cell(remember, like charges repel each other).

Membrane Potentials


Biology Nerve S£ Muscle Membrane Potentials

Those two forces (chemical and electrical) do not exactly match each other. Itturns out that the force of diffusion is a little larger than the electrical force. Thisresults in a little bit of leakage of K® out of the cell, as well as a little bit ofleakage ofNa® into the cell. The K® thatleaked outof thecell has tobepumpedback into the cell, and theNa® that leaked into the cell has to be pumped out ofthe cell. The pumping action of these two ions is provided for by the Na®/K®ATPase pump. This pump is responsible for the generation of the asymmetricalconcentration gradient of Na® and K® across the cell'smembrane.

Cell

membrane

1Inside K*cell -<-|

Diffusion

Outside

K+ cell

-87mV

Na+

Voltage ^

<^Na+

f(I+ 87mV

^

K+

Where

PK >» PNaand the

anions are

impermeable

Figure 1-6Cellular gradients where PK >» PNa.

We can calculate the membrane voltage (the potential difference) across the cell'smembrane using the Nernst equation as shown in (1-1). In this equation, V is thevoltage in millivolts (1 mV = 10"3 volts), i refers to inside, o refers to outside, R isthe gas constant, T is the temperature in Kelvin, Z is the ion's valance, and F isthe Faraday constant. If we let the cell's membrane be permeable to just K®, wefind that the voltage is -87 mV inside the cell with respect to outside the cell.Remember, this is if potassium is the only permeable ion. It is the gradient ofpotassium alone across the cell's membrane that is able to generate this potential.

Vio = 2.3 RT

ZFlog [K+]0

[K+]i

Vio =60 log [5mM]°[140mM]i

Vi0 = - 87 mV

(M)

(1-2)

(1-3)


Biology Nerve &? Muscle

If we stimulate a nerve, it leaves its resting state and enters an active state inwhich the cell's membrane is more permeable to sodium (Na®) than it is topotassium. In other words, PNa » PK. Since the concentration of sodium ishigher outside the cell than inside the cell, sodium will tend to diffuse down itsconcentration gradient and into the cell.

Once again, we can establish a separation of charge. As the Na® ions enter thecellwith their positive charges, there is that much less positivecharge outside thecell. The outside of the cell becomes more negatively charged than it was beforeNa® started to diffuse into the cell. Similarly, as the Na® ions diffuse into thecell, the inside of the cell begins to accumulate more positive charge. The insideof the cell becomes more positively charged than it was before the Na® ionsstarted to diffuse in (see Figure 1-7). As the Na® ions diffuse into the cell downtheir concentration gradient, a chemical and an electrical equilibrium is beingestablished.

+K

Cell

membrane

^Inside

cell

•<•1

Where |K+

Na

Na"1

i # Diffusion

Voltage 44

Ii ^

Outside

cell

Na+

+ 60mV -60mV

Figure 1-7Cellular gradients, where PNa » PK.

We can calculate the magnitude of the potential across the cell's membrane byusing the Nernst equation. We find that the potential difference is +60mV (insidethe cell with respect to outside the cell).

Vio =601og[150mM]°[15mM]i

Vio = + 60 mV

d-4)

(1-5)

(1-6)

Membrane Potentials


Biology Nerve & Muscle Membrane Potentials

Suppose we were to measure the potential across the plasma membrane of aneuron (nerve cell). We can do this by using two electrodes. A microelectrode isinserted into the neuron itself while another electrode is placed in the fluid thatsurrounds the neuron (see Figure 1-8). If we connect these two electrodes to avoltmeter that can read the potential difference between the two environments,we will find that the inside of neuron registers about -80 mV (with respect to theextracellular space). Note that this voltage is quite close to the voltage that wewould get (-87 mV) if this plasma membrane were permeable only to K®. Thereason that the inside of the neuron is not -87 mV is because the plasmamembrane is not exclusively permeable to K®.

microelectrode

80

\Voltmeter

Extracellular Fluid

Axon

Figure 1-8Measuring electrical potential across a plasma membrane.


Biology Nerve 8? Muscle

Action Potentials

We can reduce the plasma membrane potential by transiently increasing themembrane's permeability to Na®. This can be accomplished by stimulating thenerve. The transient reduction in the membrane potential is referred to as adepolarization of the membrane. If this stimulus is strong enough and the initialdepolarization exceeds a specific minimum value characteristic of the cell (calledthe threshold potential), then within milliseconds a burst of Na® ions will enterthe cell and generate an action potential (see Figure 1-9).As the depolarizationof the membrane continues, its permeability to Na® becomes much greater thanits permeability to K® (i.e., PNa » PK). What we find is that the membranepotential goes from a resting membrane potential of about -80 mV to amembrane potential that approaches that for Na®. In thisexample, at the heightof the wave of depolarization, the membrane potential would be about +40 mV.[Why isn't the membrane potential +60 mV as would be calculated usingequation (1-4)?]

Na+ equilibrium potential

**

(A) Stimulus p p(B) Depolarization * ^K >:> FNa

( }/ V(C) (C) Repolarization **PNa>PK(D) Hyperpolarization(E) Refractory

Action

Potential^Threshold potential

H* / \~f" - - -V - - - -- - -----^ Resting potential(A) (D) \!^^%

K+equilibrium potential+

2 4 6 Time (msec)

Figure 1-9A typical action potential.

Once the membrane potential reaches the equilibrium potential for Na® there isno more influx of Na® ions into the cell. At this point (indicated by the peak ofthe action potential) the membrane potential, which is +40 mV inside the cell, isbalanced by the concentration gradient of Na®. Recall that the concentration ofNa® outside the cellis greater than the concentration of Na® inside the cell.

One millisecond after the burst of Na® into the cell, the ion channels(ionophores) that let Na® into the cell close and become temporarily inactive.When the Na® channels are temporarily inactive they are said to be in arefractory period, which usually lasts for several milliseconds. During this time aneuron will not be able to generate another action potential, because the Na®channels cannot open to allow the cell membrane to depolarize.

Action Potentials


Biology Nerve &Muscle Action Potentials

Once the Na® channels have closed, the K® channels begin to open and K®begins to leave the cell. This phase of the action potential is referred to asrepolarization. As the cellmembrane's permeability to K® increases, a massiveamount ofK® flows outofthecell and themembrane potential passes itsoriginalresting state of about -80 mV. This event is called hyperpolarization (see Figure1-9). Eventually the cell will reach equilibrium, and once again an appropriatestimulus will generate an action potential.

The generation of an action potential is said to be an all-or-none phenomenon.What does this mean? If we apply a stimulus to a nerve (see point (A) in Figure1-9), the plasma membrane of the neuron will become depolarized. Themembrane potential will become less negatively charged during the increase inthe permeabilityof the plasma membrane to Na® ions. Remember,Na® wants toflow down its concentration gradient and into the cell. As Na® enters the cell, themembrane becomes more depolarized. This increase in depolarization tends toinduce neighboring Na® channels to open, thus lettingmore Na® ions into thecell. However, we know that the cell is dynamic rather than static; K® ions areleaving the celland flowingdown their concentration gradient. This action has atendency to repolarize the plasma membrane. In other words, the generation ofan action potential depends on the ratio of membrane permeability to these twoions. If there is not a sufficient depolarization (i.e., influx of Na® ions) of themembrane, then there will be no action potential. However, if there is a sufficientdepolarization of the membrane because there are more Na® ions entering thecellthan K® ions leaving the cell, then anaction potential will begenerated ifthatdepolarization exceeds some threshold value. Once an action potential isgenerated in a nerve cell it will always have the same magnitude. Thus, an actionpotential is either all-or none. It either happens or it doesn't.

In Figure 1-9 we have indicated how an action potential is generated. Wementioned that in order to generate an action potential there must be adepolarization of the membrane abovesome threshold value, a repolarizationofthe membrane, a hyperpolarization, and a refractory period that allows themembrane potential to return to its resting value before a new action potentialcan be generated. Now let's examine how an actionpotentialpropagatesalonganerve.

Consider the chargeseparation across the plasma membrane (see Figure 1-10) ofa nerve as an action potential moves from, say, the center of the axon outwards.Suppose we have a stimulus that exceeds a given threshold value. As the actionpotential is generated a separation of charge will be established. Because Na®ions are rapidly entering the cell, the inside of the cell will become morepositively charged (+) with respect to the outside of the cell. Since there is adeficit of positive charge on the outside of the cell (compared to the originalresting condition), we say that the outside of the cell is now more negativelycharged (-) with respect to the inside of the cell. In other words, at that particularplace on the axon, where the action potential has been generated, there is areversal in the membrane's polarity. This is shown in Figure l-10a.

The depolarization of the central region of an axon spreads outward in bothdirections as moreNa® ionsenter thecell. In Figure l-10a the arrows representalocal flow of current between the area of the plasma membrane that has beendepolarized and the adjacent areas of the membrane which are still at their


Biology Nerve Si Muscle

resting potential. The flow of positive charge within the cell causes the adjacentareas to become more depolarized. In other words, the polarity of the membraneis being decreased in front of the propagating action potentials. Because themembrane potential in these adjacent areas is becoming depolarized, more Na®channels begin to open. This lets more Na® into the cell, which increases thedepolarization until an action potential is generated.

+ + + + + + + - -^+^++ + + + + + + + +

Action Potentials

Axon

• = Na+ Channels

+ + + + + + + + + + + + +

Figure 1-10Action potential propagation.

Note that as the action potentials spread outward, they create a region which isrefractory (see Figure l-10b). Recall that the refractory period is due toinactivation of the Na® channels. If the plasma membrane has been depolarizedfrom an action potential, then another action potential cannot be generated untilthat membrane has once again reached its resting membrane potential.

During the state ofhyperpolarization, the K® channels arewide open and K® ispouring out of the cell. Even though an action potential could be generatedduring this time (of hyperpolarization), it is rather difficult unless a strongenough stimulus is applied. Also, the action potentials in Figure l-10b appearbackwards compared to the action potential in Figure 1-9, because the actionpotential in Figure 1-9 shows a fixed point on the axon and the membrane'spotential variation with time at that point. The actionpotentials in Figure l-10bshow a distribution of the membrane potential along an axon (not at a fixedpoint) at a particular instant in time.

There are a many kinds of nerve fibers in the body, and they do not all conductaction potentials at the same rate. For example, large diameter neurons will


Biology

JV

Nerve & Muscle Action Potentials

conduct a depolarization or action potential further and faster than a smalldiameter neuron. What this means is that the ability of a given neuron to conducta current (the flow of charge represents a current) depends on the cross-sectionalarea of that neuron. The larger the cross-sectional area of a neuron, the morecytosol in that neuron, and the more ions there will be to conduct that current.The more ions available to conduct the current, the faster and farther that currentwill spread.

The nerves which we have been discussing so far are referred to asunmyelinated nerves. Myelinated nerves have the ability to greatly increase therate at which action potentials are conducted. Myelin is a specialized membranethat is composed of just a few types of proteins and a large proportion ofphospholipids. A specialized type of cell (called a glial cell) attaches itself to asection of an unmyelinated axon and begins to rotate itself around that axon anumber of times. During this process a myelin sheath is deposited on the axon.Sometimes this myelin sheath is thicker than the axon itself. The deposition ofmyelin along the axon is interrupted by areas where there is no myelin. Theseareas are called the nodes of Ranvier. See Figure 1-11.

Myelin sheath(membranes)

•0-

tfPlasma membrane

(containsa high densityof Na+ channels)

Action potential

Node of

Ranvier

Axon

Figure 1-11Action potential propagation from node to node.

Later we will learn that the body's nervous system is separated into two divisions(based on function). The central nervous system is made up of the brain andspinal cord while the peripheral nervous system represents the nerves in theperiphery. The point here is that the glial cells in the central nervous system arecalled oligodendrocytes, while the glial cells in the peripheral nervous systemare called Schwann cells.

We mentioned that myelinated nerves increase the rate at which action potentialsare conducted. The myelin that surrounds the axon acts as an electrical insulatorand prevents the transfer of ions across the plasma membrane of the axon. Theonly region of the axon where ions can pass across the plasma membrane is atthe nodes of Ranvier. Thesenodes contain a high density of Na® channels. As anaction potential is generated, there is a flow of current through the cytoplasm ofthe axon (called the axoplasm) and the extracellular fluid from node to node. Theplasma membrane at each of these nodes is in turn depolarized enough to


Biology Nerve &Muscle

generate an action potential. The nerve impulse seems to "jump" from node tonode along the axon. Thus, this type of impulse transmission is referred to assaltatory conduction (from the Latin saltatorius—pertaining to dancing orleaping). See Figure 1-11.

Nerve fibers can be grouped into different classes. The nerve fibers in some ofthese classes are myelinated, while those in other classes are unmyelinated.Some of the myelinated fibers that impinge on skeletal muscle can transmitmotor impulses quite rapidly, about 60 to 120 m/sec. In contrast, some of theunmyelinated fibers which transmit sensory impulses of certain types of pain, doso rather slowly (relative to a myelinated fiber), about 0.6 to 2 m/sec. As you cansee, myelinated nerve fibers greatly increase the rate at which information, in theform of action potentials, is conducted.

So far, we have been considering the propagation of an action potential within asingle nerve cell. Let's see how this information is transmitted to other cells.

Action Potentials


Biology

Action

potential

Terminal

bouton

Nerve & Muscle The Neuromuscular Junction

The Neuromuscular Junction

At the end of the axon is a specialized area called the terminal bouton (orsynaptic bulb, synaptic knob, or even terminal foot). This specialized end regionof the axon can impinge or synapse on another axon of a different neuron, on adendrite, or even on some type of cell body (like a muscle cell). The spacebetween this synapticjunctionis called the synaptic cleft. The plasma membraneat the terminal bouton is called the pre-synaptic membrane while the plasmamembrane that the synapse is being made to is called the post-synapticmembrane. See Figure 1-12.

• Choline

Presynaptic Synaptic cleftmembrane *\ a.

Y

ACh

ACh

ACh

ACh

Acetlycholinestrase

Postsynapticmembrane

ACh receptorcomplex

Figure 1-12Presynaptic and postsynaptic membranes.

Let's consider a neuromuscular junction. This is the synapse between theterminal bouton of an axon and a muscle fiber. Within the terminal bouton of this

junction are hundreds of thousands of synaptic vesicles that contain theexcitatory neurotransmitter acetylcholine (abbreviated ACh). There may bethousands of molecules of acetylcholine in each synaptic vesicle. Acetylcholineitself is synthesized in the cytosol of the neuron from the molecules acetyl-CoA(which caries an acetate moiety) and choline.

As the action potential reaches the terminal bouton, it triggers the opening ofcalcium (Ca^®) channels. A transient influx of Ca^® into the terminal regioncauses the synaptic vesicles to fuse with the presynaptic membrane and releasetheir neurotransmitters into the synaptic cleft (via exocytosis). Calcium ispumped out of the cytosol and back into the extracellular fluid. The membranesof the synaptic vesicles that fused with the presynaptic membrane are recycled


Biology Nerve & Muscle The Neuromuscular Junction

(via endocytosis). The released neurotransmitters diffuse through the synapticcleft and bind to specific postsynaptic membrane receptors. Binding ofacetylcholine to this postsynaptic membrane receptor conformational^ changesthat receptor into a channel (ionophore) that is large enough to allow cationssuch as Na® through. [Note that this channel is not a voltage-activated channellike the channels we examined in the axon itself. This channel is a ligand-activated channel. In this case, the ligand (a molecule or ion bound to a protein)is acetylcholine.] As Na® enters the postsynaptic membrane, the muscle fiberbegins to depolarize and an action potential is eventually generated.

If acetylcholine were to remain in the synaptic cleft, then it would continue tobind to the postsynaptic membrane and action potentials would continually begenerated. This could result in prolonged muscle spasms. However, the enzymeacetylcholinesterase, which is bound to the surface of the postsynapticmembrane, hydrolyzes acetylcholine to acetate and choline. These two productsare then recycled as they are transported back into the presynaptic terminalwhere they are used in the synthesis of acetylcholine (Figure 1-12).

There are many different types of neurotransmitters. For example, the amino acidglutamate and the amino acid derivatives epinephrine, norepinephrine,dopamine, and serotonin can act as neurotransmitters. There are also drugs thatcan act at the level of these postsynaptic receptors.

EPSPs and IPSPs

We said that the binding of acetylcholine to its postsynaptic membrane receptorelicits an excitatory response that generates an action potential. These types ofpotentials are referred to as excitatory postsynaptic potentials (EPSPs). There canalso be inhibitory postsynaptic potentials (IPSPs) as well.

(b)

Excitatory PostSynaptic Potcntia

Inhibitory PostSynaptic Potential

ftp Na

Depolarization

The result of

synapse with < *A and B !

ftp CI

Hyperpolarization

Figure 1-13EPSPs and IPSPs.

Suppose neuron A synapses with neuron C, as shown in Figure l-13a. If thesynaptic connection between these two neurons is excitatory, then the same


Biology

Refractoryperiod

Nerve & Muscle The neuromuscular Junction

number of action potentials that pass down the axon of neuron A will also passdown the axon of neuron C. In the case of an excitatory postsynaptic potential,there is an increase in the permeability of the postsynaptic membrane to Na®(i.e., neuron C will depolarize).

Suppose neuron B, which is inhibitory, also synapses with neuron C, as shown inFigure l-13b. In the case of an inhibitory postsynaptic potential, there is anincrease in the postsynaptic membrane to K® and to CI" ions. Neuron C willhyperpolarize. Remember, becoming more hyperpolarized does not generate anactionpotential. In fact, it will keep neuron C quiescent. In other words, if we justexcited neuron B alone, then nothing would happen to neuron C.

However, if we integrated the action potentials that neuron C receives from bothneuron A and neuron B,we find that neuron C only generates 2 action potentials(Figure l-13b). In this case,neuron B will decrease the excitability of neuron C tothe stimulation it is receiving from neuron A.

Consider two action potentials coming toward one another, as shown in Figure1-14. When those two action potentials meet what will happen? Will theycontinue past one another? Will their amplitudes add together? The region infront of each actionpotentialis continually being depolarized, while the regionbehind the action potential is in its refractory period for a few milliseconds andcannot immediately be depolarized. Therefore, when the two action potentialsmeet one another they will stop their propagation along the axon.

+ + + + + + + + + + + + + + +

+ + + + + + + + + + + + + + +

Refractoryperiod

Figure 1-14Propagation of action potentials.



The Sarcomere

We mentioned that the synapse in Figure 1-12 involved a neuromuscularjunction. We have considered the neuronal portion of that junction already. Let'snow examine the muscular portion and follow the fate of the action potential thatwas generated.

If you were to examine a typical muscle, you would find that in most cases it isattached to a specific area of the skeleton by tendons at either end of the muscle.The region between the tendons is referred to as the belly of the muscle. A cross-section of the belly of the muscle shows a great number of multinucleatedmuscle fibers (cells), as shown in Figure 1-15. These muscle cells do not divide toproduce new muscle cells. However, if you were to exercise your muscles, thosecells would just get larger.

Myofibril

Actin

J~l Myosin

•^ Heads

Z-line H-zonc Z-line

Sarcomere

Figure 1-15Muscle anatomy.

Closer examination of skeletal muscle fibers shows striations in both thetransverse and longitudinal directions. The striations in the longitudinal

The Sarcomere


Biology

Figure 1-16Sarcomere details.

Nerve fif Muscle The Sarcomere

direction are due to myofibrils. The myofibrils contain the contractile units ofthe muscle. Each contractile unit (Figure 1-16),bounded by a structure referred toas the Z-line, is referred to as a sarcomere and each sarcomere contains two typesof contractile proteins. The thin contractile protein is called actin and the thickcontractile protein is called myosin. The myosin thick filament comprises the A-band and the region between two A-bands, where there is just the thin actinfilaments, is the I-band. The H-zone is that region in the center of the A-bandand between the ends of the actin filaments.

Sarcomere Sarcomere

Myosin filaments arearranged toward thecenter of thesarcomere. They give riseto the characteristic dark bands one sees when examining muscle tissue. Actinfilaments are attached to the Z-lines. The actin and myosin filaments interactwith eachother throughprojections stemmingfrom the myosin filaments. Thoseprojections are sometimes referred to as myosin heads. The myosin thickfilament does not haveany of these head groups in its central regionbut ratherconcentrates thosehead groups towards its terminal regions.

The actin thin filament is composed of a protein subunit called G actin ("G" forglobular), which is roughly spherical in shape. The actin filaments can grow bythe addition of G actin to the ends of the already existing filament. Each actinfilament is composed of two rows of G actin monomers wound around eachother to form a helix.

Actin and MyosinLet's consider how myosin and actin interact with each other at the molecularlevel. When a muscle is in its relaxed state, ATP is bound to the myosin headgroups. The myosin head is not bound to the actin filament, because ATPreduces myosin's affinity for actin. When ATP is bound to the myosin head, themyosin head itself is at about a 45° angle with respect to the actin filament. Sinceactin is not bound to myosin, the ATP on the myosin heads is hydrolyzed toADP and Pi (inorganic phosphate). These hydrolysis products remain on themyosin head. More importantly, the myosin head now undergoes aconformational change, so that it is situated at a 90° angle with respect to theactin filament. This high energy, stable, myosin-ADP-Pi head complex binds tothe actin filament. As wewill see, this step is dependent on Ca2® being present.



The interaction between the actin filaments and myosin head groups causes therelease of Pj and then ADP from the myosin heads. This process causes aconformational change in the myosin heads, so that they shift by about 45° in adirection that is away from the Z-lines. This step, in which the actin filamentsmove relative to the myosin filaments, is called the power stroke, and theproduct of this step is referred to as the rigor complex or rigor state. See Figure1-17.

ATP binds and

causes myosin headto release actin

Actin filament

moves

Actin

ATP is

hydrolyzed

Actin and

myosin bind

Figure 1-17The contraction cycle.

After the myosin heads have swiveled and have pulled the actin filament pastthe myosin filament, the myosin heads remain bound to the actin filaments. Thisis the so-called rigor state. In order for the myosin heads to dissociate from theactin filaments, ATP needs to bind to those myosin heads. Remember, ATPreduces myosin's affinity for actin. Without ATP, muscles remain in a state ofrigor for a given period of time. This is what happens to muscles in your bodywhen you die (i.e., rigor mortis). The myosin head groups can no longer bindATP, because the metabolic pathways that generate this energy-rich nucleotidehave ceased to function.

Troponin, Tropomyosin, & CalciumWhen we examined actin, we saw that it was composed of two long rows ofroughly spherical protein subunits, which polymerized together and then woundabout each other in a helical fashion. If we look closely at the helical structure ofactin, we notice two grooves. Running the length of these grooves is a coiledprotein called tropomyosin, composed of two helical polymers wound abouteach other (see Figure 1-18). When tropomyosin resides in the actin groove, itcovers up the binding sites for the myosin heads and prevents those head groupsfrom attaching to the actin filament.

The Sarcomere


Biology Nerve & Muscle The Sarcomere

In the case of striated (skeletal) muscle, this is often referred to as the actin-basedregulation of muscle contraction. Cardiac muscle and smooth muscle are bothcontrolled by myosin-based regulation.

Troponin ProticnComplex

Tropomyosin

Removal of

Calcium

Myosin Head Binding Sites

^2+Ca-+ BindingSite

Addition of

Calcium

Actin

Relaxation

Contraction

Figure 1-18Sarcomere detail.

Troponin, a multi-subunit binding protein, interacts with tropomyosin, actin,and Ca2®. When Ca2® isbound toa particular subunitof the troponin complex,it causes tropomyosin to shift its position and expose the myosin head bindingsites. The myosin heads then bind to the actin filaments (see Figure 1-17) andmuscle contraction follows. [In Figure 1-17, Ca2® would bind at Step 3.] IfCa2®is not in the medium, there will be none to bind to the troponin complex andtropomyosin will remain in the actin groves and cover the myosin head bindingsites. This is the relaxed state.

Surrounding each myofibril is a membranous structure, a modified version of theendoplasmic reticulum called the sarcoplasmic reticulum. Calcium issequestered within this smooth membranous structure. Also surrounding eachmyofibril is an invagination of the sarcolemma (i.e., the plasma membrane) calledthe transverse tubule (abbreviated as T-tubules). These T-tubules follow the Z-lines of each myofibril. Selected anatomical features of the structures associatedwith the sarcoplasmic reticulum are shown in Figure 1-19.

After an action potential crosses the last synaptic junction on its way to a musclecell, it passes down each T-tubule and somehow stimulates the release ofCa2®from the sarcoplasmic reticulum. [At the present time, it is thought that thelumen of the T-tubules and the lumen of the sarcoplasmic reticulum are not



continuous.] As the Ca2® flows from a region of high concentration (thesarcoplasmic reticulum) to a region of low concentration (the cytosol), it binds toits binding site on the troponin complex. Each sarcomere contractssimultaneously because of the way in which the nerve impulse is carried alongthe sarcolemma and into the T-tubules.

Sarcolemma

Myofibrils

Transverse tubules

Figure 1-19The sarcoplasmic reticulum.

Once contraction has taken placeand the nerve impulse has ceased, the Ca2® inthe cytosol is pumped back into the sarcoplasmic reticulum by a Ca2®-ATPasepump. In the sarcoplasmic reticulum Ca2® associates with a specific bindingprotein. When the next action potential passes down the T-tubules, the cycle willbegin again and another muscle contraction will take place.

There are a number of ways in which the strength of muscle contraction can bevaried. The strength of contraction can be varied by (1) the size of the motor unit(to be defined in a moment), (2) the number of available motor units, and (3) theamount of actin and myosin contained with each cell.

A motor unit is simply a motor neuron and the muscle fibers that it innervates.We have already examined the muscle fibers. What is a motor neuron? Motorneurons are nerve cells whose cell bodies are located in the central nervous

system (e.g., the spinal cord or brain stem), and whose myelinated axonsinnervate skeletal muscle. Recall that myelinated axons allow action potentials tobe transmitted rapidly to the desired effector organ (in this case, a muscle).

If you wanted to precisely move a muscle (e.g., the muscles associated with youreyes or fingers), then you would need motor units which were rather small insize. The smaller the size of the motor unit, the smaller the strength ofcontraction. In contrast the postural muscles of the back or the legs require largemotor units to be effective. Not only can strength be controlled but precision canbe controlled as well.

The Sarcomere


Biology Nerve fie Muscle The Sarcomere

The number of motor units also controls the strength of contraction. For example,if you wantedtopickup a lightobject (e.g., a pencil), you would need to employonly a few motor units. However, if you wanted to pick up heavier objects (e.g.,gym weights), you would need to utilize more motor units. When you liftweights regularly, the size of each muscle cell increases, because the amount ofactin and myosin in eachmusclecellhas increased.The more actin and myosin ina muscle cell, the greater the strength of contraction that can be generated.

Even though the strength of muscle contraction can vary with the size of themotor units, the number of motor units, and the amount of actin and myosin ineach muscle cell, the ultimate determinant of muscle contraction is theconcentration of ATP in your muscle cells. The energy for contraction comesfrom the hydrolysis of ATP to ADP and Pi (inorganic phosphate). See equation(1-7):

ATP • > ADP + Pj + Energy

For muscle contraction to continue, ATP must be constantly regenerated. Recallfrom basic biology that under aerobic conditions (when O2 is not limited)glucosewill be oxidized to CO2 and H2O. Energy can be extracted (in the form of36 molecules of ATP~if we use the glycerol-phosphate shuttle) during thiscatabolic process from glycolysis, the Krebs cycle, and oxidativephosphorylation/electron transport. See equation (1-8):

Aerobic

(Slow)

Glucose • > C02 + H20 + 36 ATP(1-8)

During anaerobic conditions (when O2 is limited), glucose is metabolized tolactate (the anion of lactic acid). Only 2 ATP molecules are extracted for the useof energy in this process. See equation (1-9). [This equation is not balanced.]

Anaerobic

(Fast)Glucose ' > Lactate + 2 ATP

(1-9)

Even though the efficiency of ATP production is greater for aerobic metabolismthan for anaerobic metabolism (36 ATPs compared to 2 ATPs), ATP can beproduced much more quickly through anaerobic metabolism than throughaerobic metabolism, owing to enzyme regulation in the glycolytic pathway.

During anaerobic metabolism, the concentration of lactic acid begins to increase.This means that the pH of the medium will decrease (i.e., become more acidic).One of the key regulatory enzymes in the glycolytic pathway cannot functionwell below a certain pH value. This enzyme has some optimum pH range atwhich it functions and if the pH falls below (or above) that range, the enzyme isinhibited. Glycolysis comes to a halt and the ATP yield becomes insufficient tocarry out normal metabolic processes. In other words, fatigue sets in.


Biology Nerve & Muscle Nervous System Components

Nervous System Components

The nervous system in its simplest form can be found among the cnidarians. Allthe neurons in these creatures are linked together in a nerve net (much like aweb). Stimulation of the nerve net causes muscle contraction. This simpleprocedure is referred to as a reflex arc. In the case of the cnidarians there is noassociative activity, meaning that the transmission of the action potential is notinfluenced by other neuronal activity.

Associative neurons are found in higher animals (Platyhelminthes and above).Here, different neurons can interact to produce a given response. When anumber of these associative neurons are grouped together, the nervous systemexpands. A grouping of nerve cells is called a ganglion (or a nucleus or nuclei).Groupings ofneurons in higher animals also lead tomore elaborate sense organs,differentiation into a central region of cells (thecentral nervoussystem, or CNS)and a peripheral region of cells (the peripheral nervous system,or PNS), variousassociative areas, and a brain.

The three basic anatomical divisions of thevertebrate brain can beseen in Figure1-20. Those divisions are: (1) the forebrain (prosencephalon), (2) the midbrain(mesencephalon), and (3) the hindbrain (rhombencephalon).

ProsencephalonA

Mesencephalon

Rhombencephalon

nal cord

Figure 1-20Anatomical divisions of the vertebrate brain.

Inferior to (meaning "below") the brain is the spinal cord. Recall that the brainand the spinal cord together are the central nervous system (CNS). From the spinalcord, nerves extend into the limbs and extremities (the periphery) of the body.These nerves represent the peripheral nervous system (PNS). If a neuron carriesinformation into the spinal cord and brain, that neuron is said to be an afferent(sensory) neuron. If a neuron carries the information away from the brain orspinal cord, that neuron is said to be an efferent (motor) neuron.

The forebrain has several anatomical subdivisions, including the cerebrum,thalamus, and hypothalamus (see Figure l-21a). The cerebrum is divided intoright and left cerebral hemispheres, joined by the corpus callosum. The cerebralhemispheres are divided into the frontal lobes (associated with movement andpersonality), parietal lobes (associated with touch and stretch sensation),occipital lobes (associated with vision), and temporal lobes (associated withhearing) as shown in Figure l-21b.


Biology

(a)

CorpusCallosum

Nerve St Muscle

Thalamus

Cerebral

Cortex

nervous System Components

(b)

Frontal

Temporal

The Cerebrum and its Four Lobes

Central Sulcus

O

Hypothalamus ^

Pituitary^/Gland

MedullaCord

Figure 1-21Anatomical divisions of the vertebrate brain.

\\ Spinal Cerebellum

The outermost layer of the cerebrum is called the cerebral cortex. It consists ofgray matter, which is simply nerve cell bodies and their dendrites. Beneath thegray matter is the white matter, or myelinated axons of the nerve cells. In thespinal cord, the situation is reversed: The gray matter is more centralized, whilethe white matter is more peripheral.

The cerebral cortex has many important landmarks, one being the central sulcus,a prominent groove that separates the frontal lobes and the parietal lobes.Anterior to this sulcus is the motor cortex, which controls the movement ofindividual muscles. Posterior to this sulcus is the (somatic) sensory cortex, whichdetects sensations in various parts of the body.

Somatic receptors send their information into the spinal cord via afferent nervefibers, which either cross over to the opposite side of the spinal cord and thenascend to the sensory cortex in the brain, or ascend in the spinal cord and thencross over in the brainstem before reaching the sensory cortex.

HomunculusWilder Penfield, a Canadian neurosurgeon, was able to map the sensory andmotor areas of the brain by electrically stimulating certain regions of the brains ofhis patients during surgery and observing their actions. Throughout thisprocedure his patients wereconscious but were unable to feel any pain becausethereareno sensory receptors forpain in the cerebrum. He was able to show thatthe largest number of cortical neurons found in the sensory cortex registersensation in the fingers, hands, lips, and tongue. This is represented as thesensory homunculus (a schematic model of a human beingmapped out on thesensory cortex) shown in Figure 1-22.

Penfieldwas also able to show that the largest number of corticalneurons foundin the motor cortex control individual movement of the fingers, hands, andspeech. Groups of muscles are controlled by an area just anterior to the motorcortex called the premotor cortex. This is an association area. Stemming from

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Biology Nerve fie Muscle Nervous System Components

this association area are neuronal connections with the thalamus and cerebellum.Together these structures plan specific movements that the motor cortex thenexecutes. It is thought that cognitive functioning like speech, writing, andreading are localized in the lefthemisphere of thebrain, while intuitive functionsare localized in the right hemisphere of the brain. This is not proven, onlytheoretical.

l*H Tongue

Sensory receptors on the right sideof the body project to the

somatosensory cortex on the left hemisphereof the brain and vice versa

Left

HemisphereL Right

Hemisphere

Figure 1-22The homunculus.

The thalamus is a relay station for much of the visual and auditory informationthat we receive from our environment. The hypothalamus is concerned with thevisceralactivities of the body. The pituitary gland is the master endocrine glandof the body. It receives information from the hypothalamus and sends outinformation to regulate different parts of the body.

The brainstem contains such anatomical features as the midbrain, cerebellum,pons, medulla, and the reticular formation. These areas coordinate motor andvisceral activities. The midbrain detects movement and can direct the head and

eyes towards that movement. The midbrain can also sense pleasure and pain.The cerebellum is responsible for the bulk of regulation and coordination ofmuscular activity. The pons and medulla coordinate visceral activities. Thereticular formation, which is the core of the brainstem, is essentially an activatingsystem designed to alert the brain. The reticular formation also inhibits motorand sensory impulses and can induce sleep. Below the medulla is the spinal cord.


Biology Nerve & Muscle Control Of Body Activity

Control Of Body Activity

Recall that when we first mentioned the nervous system, we looked at a simplereflex arc. If someone were to tap you on your knee with a rubber-headedhammer, your lower leg would extend outward. The mechanism behind this isquite simple. As the hammer impinges on the patellar tendon, sensory neuronslocated in the tendon of the quadriceps muscle are excited. These impulses travelalong an axon that enters the spinal cord through the dorsal root ganglion andsynapses with two neurons (Figure 1-23).

Biceps ('hamstring")muscle is a flexor

O

Figure 1-23The reflex arc.

Tibia

This area, even though it is supposedto be gray matter, is shown as being

i clear so you can see the synapses, j

Polysynaptic T .„ J ' Interncuron

Reflex Arc

Gray Matter

Spinal Cord

One of the synapses is to a motor neuron that immediately leaves the spinal cordand returns to the quadriceps muscle, causing a contraction. As this musclecontracts, the leg straightens (extends) at the knee joint. Because it elicits thiskind of action, the quadriceps muscle is termed an extensor muscle. The type ofsynapse just described (making just one synaptic connection), is referred to as amonosynaptic reflex arc.

The other synapse is made to an interneuron which, in this case, is inhibitory.This interneuron in turn synapses with a motor neuron that innervates the bicep("hamstring") muscle in the back of the leg. When the bicep muscle contracts, thelower portion of the leg bends or flexes at the knee joint. We call this kind ofmuscle a flexor muscle, and this type of synapse (because it makes at least twosynaptic connections) is referred to as a polysynaptic reflex arc.

If contraction of the biceps muscle is inhibited as the quadriceps musclecontracts, then one observes a smooth and coordinated movement at the kneejoint, as the lower portion of the leg extends outward after stimulation by thetapping of the hammer on the patellar tendon.


Biology Nerve fit Muscle

Neurovisceral Control

The autonomic nervous system, which is part of the efferent division of theperipheral nervous system, can be subdivided into the sympathetic and theparasympathetic systems. Nerve fibers from the autonomic nervous system leavethe spinal cord to innervate various glands, smooth muscle, and cardiac muscle.Let's examine these two subdivisions.

Parasympathetic DivisionThe parasympathetic division of the autonomic nervous system has nerve fiberswhich leave from the sacral portion of the spinal cord and from the midbrain(mesencephalon) and medulla (part of the rhombencephalon), as shown inFigure 1-24. Parasympathetic nerve impulses tend to increase the rate ofdigestion and lower the heart rate. The blood pressure is also lowered, and thepupils constrict (contract). In general, the parasympathetic division conservesenergy and helps in the restoration of various bodily functions (e.g., by aiding inthe digestion of food for later use in metabolic processes).

Control Of Body Activity

Preganglionic ^-Ganglionnerve fiber f Neurotransmitter is

acetylcholine

Midbrain 1Medulla J Lacrimal glands

Lungs andBronchi

i Large intestines

Figure 1-24Parasympathetic nerves.

The parasympathetic division has both preganglionic and postganglionic nervefibers. Thecellbodies of the preganglionic neurons are found in the sacral region


Biology Nerve &Muscle Control Of Body Activity

of the spinal cord and in the brainstem. The ganglia of the parasympatheticdivision lie near or in the organs that are to be innervated. Therefore, thepreganglionic nerve fibers are rather long, while the postganglionic nerve fibersare rather short. Both the preganglionic and the postganglionic nerve fibers inthe parasympathetic division release acetylcholine (ACh) as theneurotransmitter. [Nerve fibers that release acetylcholine as theirneurotransmitter are called cholinergic nerve fibers.]

The most prominent nerve in the parasympathetic division is the vagus nerve[also called the (tenth) X cranial nerve]. Roughly three-quarters of all the neuronsin the parasympathetic division can be found in the vagus nerve. The reason forthis should become obvious, if you look at Figure 1-24 The vagus nerveinnervates the heart, lungs, stomach, liver (not shown), small intestine, largeintestine, and kidneys (not shown), among other organs.

Sympathetic DivisionThe sympathetic division of the autonomic nervous system has nerve fibersbranching off from the thoracic and lumbar regions of the spinal cord, as shownin Figure 1-25. Sympatheticnerve fibers tend to condition the body for a "fight-or-flight" response (a term first proposed by the Harvard physiologist WalterCannon in the early 1900s). The heart rate increases, blood pressure elevates,pupils dilate (openwider), and the digestivefunctions decrease, all as a result ofsympathetic nervous innervation.

The sympathetic nerves that leave the thoracic and lumbar regions of the spinalcord first enter chains of ganglia connected by nerve fibers on either side of thespinal column. These chains of ganglia are collectively called the sympathetictrunk. As these spinal nerves, called preganglionic nerve fibers, enter thesympathetic trunk, they can either (a) pass through this collection of ganglia tosynapse with other ganglia outside the sympathetic trunk, (b) pass into thesympathetic trunk and ascend or descend to synapse with ganglia at other levels,or (c) pass into the sympathetic trunk and directly synapse with a givenganglion.

The nerve fibers leaving a synapse in a given ganglion are referred to as postganglionic nerve fibers. In the caseof the sympatheticdivision the preganglionicnerve fibers tend to be short, while the postganglionic nerve fibers tend to belonger. The preganglionic fibers release acetylcholine, while the postganglionicfibers release norepinephrine as their neurotransmitters. [Nerve fibers thatrelease norepinephrine (or epinephrine (adrenaline)) are called adrenergic nervefibers.]

One set of spinalnerves that originates in the lower thoracic regionof the spinalcord send long preganglionic nerve fibers to the adrenal medulla, a regionwithin the adrenal glands (locatedon the superior surface of the kidneys). Thesenerve fiberssynapse directlyon the ganglia in the adrenal medulla. There are nopostganglionic nerve fibers. When the adrenal medulla is stimulated, bothnorepinephrine and epinephrine are released directly into the bloodstream.Because these substances are released into the bloodstream, we can refer to themas hormones. Therefore, the adrenal medulla can be considered as an endocrinegland. These hormones, distributed throughout the body by the circulatorysystem,can increase the heart rate and cause the pupils to dilate.

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Biology Nerve &? Muscle Control Of Body Activity

Postganglionicnerve fiber

Neurotransmitters are

acetylcholine andnorepinephrine

Preganglionic,nerve fiber

Thoracic

LumbarLarge

intestine

Figure 1-25Sympathetic nerves.

Somatic vs Autonomic Nervous SystemLet's review the basic differences between the somatic nervous system and theautonomic nervous system. In the somatic nervous system, we find that (a) oncethe nerve fibers leave the central nervous system, they do not make a synapseuntil they have reached their effector organ. When the synapse is made at theeffector organ, (b) the neurotransmitter that is released is acetylcholine. Thesomatic nervous system (c) innervates skeletal muscle. Innervation of thatskeletal muscle (d) leads to excitation of the muscle itself.

In the autonomic nervous system, we find that (a) once the nerve fibers leave thecentral nervous system, they synapse with a ganglion before they make the finalsynapse with their effector organ. The preganglionic fibers in both theparasympathetic and sympathetic divisions release (b) acetylcholine as theneurotransmitter. The postganglionic fibers in the parasympathetic divisionrelease acetylcholine; in the sympathetic division, norepinephrine is released.The autonomic nervous system (c) innervates glands, and smooth and cardiacmuscle. The cells innervated by the autonomic nervous system can (d) be eitherexcitatory or inhibitory.


Biology

Somatic

Nervous

System

Autonomic

Nervous -^System

CNS

r

^.

Figure 1-26CNS and PNS review.

Nerve St Muscle Control Of Body Activity

The adrenal medulla, a specialized ganglion in the sympathetic division of theautonomic nervous system, is directly stimulated by a preganglionic fiber. Thisnerve fiber releasesacetylcholine, which causes the cells of the adrenal medullato release two typesofhormones into the blood. The majorhormone released isepinephrine. The minor hormone released into the blood is norepinephrine.These differences are illustrated in Figure 1-26.

PNS

Acetylcholine

Postganglionic nerve fiber

Acetylcholine

Skeletal muscle• ^F •& V V V •

Glands, cardiacand smooth muscle

<P!l*\% Jd^ Hormone C=^ Blood CZ^ Organ^ ' (Epineprhine)

Norepinephrine

COV-

3

Sra.o

Adrenal Medulla Ganglion

Preganglionic nerve fiber

Glands, cardiac ^and smooth muscle JS

Acetylcholine

X"A A A A A A »v


Biology Nerve St Muscle Receptors and Sensory Input

Receptors and Sensory Input

There are many different types of receptors receiving sensory information fromthe environment and passing that information to the nervous system. Forexample, there are mechanoreceptors, which are concerned with pressure,hearing, balance and blood pressure. Nociceptors sense pain. Mechanoreceptorsrespond to a change in their configuration. Some canrespond toa lightpressure,while others respond to a deeper pressure. There are specialized regions on thesides offish called lateral lineorgans that respond toa change in thepressure ofthe water. There are pressure receptors in the walls of the aorta that are able tosense an increase in blood pressure.

Thermoreceptors detect cold and warmth, while chemoreceptors can detecttaste, smell, oxygen, carbon dioxide, hydrogen ions, and blood glucose levels.The taste receptors in the tongue can distinguish between food which is bitter,sour, salty,or sweet. Olfactory cells in the nasalcavity can distinguish hundredsof different odors. Photoreceptors in the retina of the eye can respond to thepresence of a single photon of light. There are also receptors concerned withelectricity and magnetism. For example, catfish have electrical receptors thathelp them detect prey, and birds have magnetic receptors that help them tonavigate.

The sensory information that a receptor receives is specific for that particularreceptor. The stimulus that is received by that receptor changes (converts ortransduces) the potential (Vm) across the receptor's membrane. This change inmembrane potential is called a receptor potential. If this receptor potential wereto exceed a specific threshold, an action potential would be generated. We alsofind that if the pressure on this receptor is great and the change in membranepotential is large, then the receptor potential will exceed the threshold level. Theresult is an increase in the frequency of the action potentials being generated.Theamplitude of the action potentials will not be change,only the frequency.

Receptors and TransducersLet's consider the receptor potential for a special type of mechanoreceptor calleda pacinian corpuscle. A pacinian corpuscle has an unmyelinated nerve endingwhich is encapsulated in layers of connective tissue. However, the afferent nervethat leaves this encapsulated ending is myelinated.

Figure 1-27Pacinian corpuscle.

Pressure

Copyright © by The Berkeley Review

SaltatoryConduction

35 The Berkeley ReviewSpecializing in MCAT Preparation


Whenthe nerveendingis depressed, a localdeformation causessodium channelsto open and Na® ions to rush into the nerve ending. An action potential is notgenerated in the receptor region. Instead, the establishment of this receptorpotential causes a localized flow of current to be propagated along the nervefiber. If the threshold has been reached, then this local current flow will initiatean action potential at the first node of Ranvier, located at the outer edge of thepacinian corpuscle. The action potential canspread down the afferent nerve andtowards the central nervous system by saltatory conduction, as shown in Figure1-27.

Supposewe were to depress the paciniancorpuscle just a small amount. If thethreshold is not reached, an action potential will not be generated. This meansone would not feel the pressure that is being applied. If we apply a secondstimulus, we might depolarize the membrane even more. If we were to apply athird stimulus that was even stronger, we could exceed threshold and an actionpotential would be generated. If we maintain this pressure such that we are justabove threshold by a certain amount, we will continue to generate actionpotentials at the rate of,say, 2 actionpotentials every 10 milliseconds.

+40

>

ft1st

Weak 2nd

2 action potentialsper 10 milliseconds

Action

Potential

Potential

H

3 action potentialsper 10 milliseconds

rh

WWN-WWW—

/\

4th

^ Strength ofstimulus

Strong

Time (ms)

Figure 1-28Action potential frequencies.

What would happen if we apply a fourth stimulus, much stronger than theprevious stimuli? The membrane would become more depolarized, and wewould move further above the threshold level. Once this happens, the frequencyof action potentials generated would increase, say, to 3 action potentials every 10milliseconds. This is shown in Figure 1-28.Note that in each case the amplitudeof the action potentials remains the same. What changes is the frequency of theaction potentials propagating along the axon. It is the frequency of actionpotentials that signals to the central nervous system the magnitude of thestimulus being received.All receptors function on this basic principle.



AdaptationWhen you wake up in the morning and take a coolshower, you feel the coolnessof the water on your skin. Whenyou get dressed, you feel the clothes touchingyour skin. In each case, though, after a brief period of time you get used to thewater or the clothes touching your body. These are examples of sensoryadaptation. One type of receptor that undergoes sensory adaptation is thepressure receptor. If we plot the frequency of action potentials (as they travelback to the central nervous system along a given axon) as a function of the timeof a sustained stimulus, we would find that pressure receptors adapt rapidly.However, pain receptors do not adapt rapidly, as is shown in Figure 1-29. If thepressure receptors did not adapt to the touch of the clothes that we wear, itwould prove to be rather inconvenient. [On the other hand, the body does notadaptas readily to the sensation of pain-a good thing, evolutionarily speaking,since pain is a warning of potential damage to the body's tissue and notsomething to be responded to as a matter of convenience, but as a matter ofsurvival.]

pain

pressure

10

Time of sustained stimulus (sec)

Figure 1-29Adaptation to pressure and to pain.

Tactile DiscriminationRecall that the end of a neuron can be divided into many branches. Each of thesebranches in turn can end at a receptor (such as a pacinian corpuscle). Theseaxonal branches and their corresponding receptors constitute a receptive field.Depending on which area of the body one is describing, there can be manyreceptive fields, some of which overlap, or there can be a few receptive fields,some of which do not overlap. Let's consider a set of receptive fields that overlapin order to determine how two points in space can be distinguished from oneanother by the same touch.

Suppose we have a set of receptive fields like the ones shown in Figure l-30a.What one generally finds is that a greater frequency of action potentials will begenerated (assuming threshold has been reached) if the central region rather thanthe periphery of a given receptive field is stimulated. In other words, there seemto be more receptors in the central region of a receptive field than in itsperiphery.

What would happen if receptive fields overlapped, as shown in Figure 1-30? Ifwe were to stimulate the receptive field represented by (b) in Figure 1-30vigorously enough, then because of the field overlap, we would also stimulatethe receptive fields in (a) and (c). It would feel to a subject as if three differentareas of the body were being stimulated when, in fact, just the receptive field in(b) is being stimulated. How can the information being received from thesesensory afferent neurons be fine-tuned to let us know that only field (b) is being


Biology Nerve & Muscle Receptors and Sensory Input

stimulated, and not field (a) or field (c)? This is accomplished by a process calledlateral inhibition, mediated through interneurons within the spinal cord.

In Figure 1-30, the axon that leads away from field (b) has lateral connections tointerneurons that inhibit the impulses being sent down the axons from fields (a)and (c). In other words, because of lateral inhibition the impulses generated inreceptive fields (a) and (c) are suppressed, allowing the impulses from thereceptive field in (b) reach the proper spinal tracts that ascend to the brain.

ReceptiveField

Action

Potentials

A.Ganglion

19 Axon

9 Axon

9 Axon

Lateral Inhibition

(interneuron)

To Brain

Figure 1-30Receptive fields.

Somatic Sensory PathwaysOnce the sensory afferent nerve fibers enter the spinal cord they cross to the sideopposite from the side they entered, either in the spinal cord or in the brainstem.In other words, the sensory input from the right side of the body will berepresented on the somatosensory cortex of the left cerebral hemisphere, andsensory input from the left side of the body will be represented on thesomatosensory cortex of the right cerebral hemisphere. [In Figure 1-30, whichside of the body are the receptive fields located, the right side or the left side?Will the sensory input go to the somatosensory cortex of the right or left cerebralhemisphere? How do you know?]

Where do these sensory neurons synapse as they ascend to the brain? As ageneral rule, there are three neurons involved in sensory pathways. In thecase ofpressure, we find that the first-order neurons carrying information from thereceptive field(s) enters the spinal cord and synapse with second-order neuronsthat ascend on the oppositeside of the spinal cord to the thalamus. Here, anothersynapse is made with third-order neurons that continue to ascend, until theyreach a specific region of the somatosensory cortex of the cerebral hemisphereopposite to the side of the body in which the sensation was perceived. Thecerebral cortex itself contains cells that are organized into 6 horizontal layers.The sensations of pressure would be registered in cells arranged in columns thatcross these different layers.


Nerve and Muscle

15 Passages

100 Questions

Passage Titles QuestionsI. Types of Transport 1 -8

II. Autonomic Nervous System 9- 13III. Action Potentials 14-20IV. Local Anesthetic 21 -26

V. The Lens, the Iris, and Associated Muscles 27-33VI. Resting Membrane Potential 34-40

VII. Nicotine Replacement Therapy 41 -46VIII. Retinal Projection 47-52

IX. Axonal Transport 53-58X. Huntington's Disease 59-64

XL Photoreceptors 65-71

XII. Sound Transmission in the Ear 72-79XIII. Tryptophan and Serotonin Experiment 80-87

XIV. Frog Muscle Experiment 88-94XV. Skeletal Muscle Groups 95- 100

Ilk

BERKELEY1 ^ i*J® JLrR-E-V-I-E-W®

zing in MCAT Preparation<Speciali

••

Suggestions

Thepassages that follow are designed to getyou to think in a conceptual manner about the processesof physiology at the organismal level. If you have a solid foundation in physiology, many of theseanswers will be straightforward. If you have not had a pleasantexperience with the topic,some of theseanswers might appear to comefrom the void past theOortfield of the solar system.

Picka few passage topicsat random. For theseinitialfew passages,do not worry about the time. Justfocus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, orgraphs. Third, read each question and the accompanying answers carefully. Fourth, answer thequestions the best youcan. Check the solutions and seehow youdid. Whether you got the answersrightor wrong, it is important to read the explanations and see if you understand (and agree with) what isbeing explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block ofpassages and try them. Beaware that time is going to become important. Generally, you will have about1 minute and 15 seconds to complete a question. Be a little more creative in how you approach this nextgroup. If you feel comfortable with the outline presented above, fine. If not, then try differentapproaches to a passage. For example, you might feel well versed enough to read the questions first andthen try to answer some of them, without ever having read the passage. Maybe you can answer some ofthe questions by just looking at the diagrams, charts, or graphs that are presented in a particular passage.Remember, we are not clones of one another. You need to begin to develop a format that works best foryou. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where thelevel of distraction is at a minimum. Get out your watch and time yourself on these passages, eitherindividually or as a group. It is important to have a feel for time, and how much is passing as you try toanswer each question. Never let a question get you flustered. If you cannot figure out what the answer isfrom information given to you in the passage, or from your own knowledge-base, dump it and move onto the next question. As you do this, make a note of that pesky question and come back to it at the end,when you have more time. When you are finished, check your answers and make sure you understandthe solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tendsto stay with you longer. (Forexample, what is the Oort field?)

The estimated score conversions for 100 questions are shown below. At best, these are roughapproximations and should be used only to give one a feelfor which ballpark they are sitting in.

Section I

Estimated Score Conversions

Scaled Score Raw Score

>12 86 - 100

10-11 79-85

8-9 65-78

7 59-64

6 54-58

5 48-53

<4 0-47

Biology Types of Transport Passage I

Passage I (Questions 1-8)

Animal cells have membranes containing a vast arrayof lipids and proteins. Some of the transmembraneproteins in these membranes act as channels that allow forthe simple diffusion {passive transport) of water-solublemolecules or ions of low molecular weight between theintracellular and extracellular regions of the cell. Theionic components of these two regions differ considerably(see Table 1).

Selected Cellular Ionic Components

Ion [Intracellular] [Extracellular] +Ei0nNa+ 15 mM 140 mM +58 mV

K+ 135 mM 4mM -92 mV

cr 4mM 120mM -89 mV

Ca2+ 2 x lO'4 mM 4mM +129 mV

Mg2+ 40 mM 2mM

*Ejon =Equilibrium potential for the ion

Table 1

Other transmembrane proteins act as carriers totransport ions and molecules across the cell membrane. Ifthe substance is too large (e.g., glucose), it will require acarrier protein. This type of carrier-mediated transport iscalled facilitated diffusion, and the rate of diffusion willincrease until all of the carrier binding sites on themembrane transport protein are filled.

If energy is required to move a substance across amembrane, the process is called active transport. Primaryactive transport uses the energy of ATP directly to move asubstance against its concentration gradient (Figure 1).The carrier molecule is the membrane-bound proteinATPase. Only three primary active transport systemshave been identified: one for Na® and K® ions, one forCa2® ions,and one for H® ions.

Secondary active transport systems do not use theenergy of ATP directly, but rather use the energy stored ina concentration gradient established across a cellularmembrane.

Transmembrane proteins that act as channels in nerveor muscle cells can be electrically or chemically activated.The resting membrane potential (Em) in these cells isabout -80 mV. The driving force that moves an ionthrough one of these channels is the difference betweenEm and Ejon, as shown in the following equation:

Force Moving Ion = E m

Copyright © by The Berkeley Review 41

The transport of substances across a cellularmembrane by a carrier protein occurs either through auniport, a symport, or an antiport mechanism.

>»

«

'a.

ATPase,

ATP Ca2+ ADP + P

2K+

Ca

Ca2+ tmq+3Na+ Glucose Na+

Figure 1

1. Compared to the extracellular fluid, the intracellularfluid will contain higher concentrations of all of thefollowing substances EXCEPT:

A. proteinB. HCO3-C. PO43-D. K®

2. All of the following will lead to inhibition of theNa+/K+-ATPase EXCEPT a low:

A. intracellular [ATP].B. intracellular [Na®].C. extracellular [K®].D. extracellular [Na®].

Glucose can be carried across the cell membrane byall of the following EXCEPT:

A. facilitated diffusion.

B. primary active transport.C. a symport mechanism.D. secondary active transport.

The Berkeley ReviewSpecializing in MCAT Preparation

Biology Types of Transport Passage I

4.

5.

6.

Glucose, despite its relatively high intracellularconcentration, can dissociate from its carrier proteinon the cytosolic side of the membrane and into thecytoplasm, because the:

A. low intracellular [Na®] allows sodium to bereleased from the carrier protein.

B. high extracellular [Na®] allows glucose to bereleased from the carrier protein.

C. affinity of the carrier protein for intracellularsodium is high.

D. affinity of the carrier protein for glucosedecreases, once sodium binds to the carrierprotein on the extracellular side of themembrane.

Which of the following graphs best represents therelationship between simple diffusion (SD) andfacilitated diffusion (FD)?

A. B.

[Gradient] [Gradient]

C. D.

[Gradient] [Gradient]

The acetylcholine receptor within the postsynapticmembrane of a neuromuscular junction binds twomolecules of acetylcholine and allows Na® and K®to flow through the membrane. The acetylcholinereceptor can best be described by its ability to allow:

A. Na® to enterthecell, leading to repolarization.B. more Na® to enter the cell than K® to leave

the cell.

C. more K® to enter the cell than Na® to leavethe cell.

D. a change in the membrane potential to lead toits activation.


7.

8.

42

All of the following characteristics are common togap junctions that exist between cells EXCEPT:

A. cytoplasmic continuity.B. rapid transmission of action potentials.C. coordination of muscle contraction in the

heart.

D. a high resistance electrical pathway.

Which of the following will NOT occur across acell's membrane, if the electron transport chain isinhibited?

A. Decreased primary active transport.B. Decreased secondary active transport.C. Increased sodium concentration gradient.D. Increased intracellular calcium levels.


Biology Autonomic Nervous System Passage II

Passage II (Questions 9-13)

The autonomic nervous system (ANS) is concernedprimarily with the regulation of visceral function inresponse to external and internal stimuli. The twodivisions of the ANS are the parasympathetic nervoussystem (PNS) and sympathetic nervous system (SNS).

The SNS is associated with readying the body forstressful situations, while the PNS is involved inmaintaining bodily functions at basal levels. Thesesubgroups use a two-neuron system. A primary neuronwill extend from the central nervous system (CNS) whilea secondary neuron will extend from a ganglion to thenerve's target organ.

The primary and secondary neurons for the PNSsecrete the neurotransmitter acetylcholine (ACh). TheSNS, however, uses two different neurotransmitters. AChis secreted from the primary neuron, while norepinephrine(NE) is secreted from the secondary neuron.

The SNS has large collections of ganglia that form anintricate network. The most prominent of these gangliaare the paravertebral ganglia, which are adjacent to thevertebral column. The prevertebral ganglia lie along theanterior surface of the great vessels of the abdominalcavity and on the adrenal medulla. The PNS has gangliathat are immediately adjacent to the organs that are beinginnervated.

9. The control of individual organs is considered animportant function of the ANS. However, the PNScomponent is thought to be more specific in itsregulation than the SNS. Which statement(s) bestexplain this theory ?

I. A response to stressful stimuli needs a generalstimulation of the body, thus the SNS is lessspecific.

II. The SNS developed much later in evolutionand thus is not as refined as the PNS.

III. The regulation needed to maintain the bodilysystems at equilibrium and after stress must bemore specialized to each organ innervated.

IV. The location of the ganglia of the PNS allowsit to regulate organ function more closely.

A. I onlyB. II, III, and IV onlyC. Ill and IV onlyD. I, III, and IV only


10. Which of these situations would stimulate the SNS?

I. Extreme heat.

n. Extreme cold.

m. Sleeping.IV. Running a marathon on a sunny day

A. I and II onlyB. I, II, and III onlyc III and IV onlyD. I, II, and IV only

11. Which of the following symptoms would beconsidered a PNS response?

A. BlushingB. Gastrointestinal immotiltiyC. Piloerection

D. Penile erection

12. The adrenal gland is considered an extension of theSNS. Stimulation of the SNS will initiate the release

of the hormones epinephrine and norepinephrinefrom the adrenal medulla. Cortisol and aldosterone,produced in the adrenal cortex, will also be released.Which of the effects below can be attributed to the

hormones released by SNS stimulation of the adrenalcortex ?

I. GluconeogenesisII. Retention of ions and water in the kidney.III. ProteolysisIV. Storage of glucose in the liver.

A. I and II onlyB. I, II, and in onlyC. Ill and IV onlyD. I, II, m, and IV

13. The vagus (10th cranial) nerve is a component of thePNS. Which of the following effects is LEASTlikely to be associated with the PNS?

A. Constriction of the pupils.B. Increased heart rate.

C. Increased gastric motility.D. Increased secretion of HC1 in the stomach.


Biology Action Potentials Passage III

Passage III (Questions 14-20)

Many animal cells have an intracellular fluid (ICF)that is electrically negative relative to the extracellularfluid (ECF), creating a voltage difference across the cell'smembrane called the resting membrane potential (Em).The Em in nerve cells and muscle cells is about -80 mV,while in epithelial and red blood cells it is about -30 mV.

In nerve and muscle cells the concentration gradientsand the permeability of the membrane to both Na® andK® determine the Em. Theconcentrations of these ions inthe ICF and ECF are shown in Table 1. They aremaintained by the Na®/K®-ATPase pump, a pump thattransports 3 Na® ions to the ECF for every 2 K® ionstransported to the ICF. At rest, the relative permeabilityof themembrane to K® is roughly 10times greater than itis to Na®.

Table 1Ion [Intracellular] [Extracellular] tEi(

Na+

K+

15 mM

135 mM

140 mM

4mM

'Ejon = Equilibrium potential for the ion

+58 mV

-92 mV

If the Em transiently changes from its resting value,an electrical signal is generated in the nerve cell. Thereare two types of electrical signals: graded potentials andaction potentials (AP). Graded potentials are signals thatoperate over short distances, while action potentials aresignals that operate over long distances.

Along the plasma membrane of a nerve cell arevoltage-sensitive Na® and K® channels that containgates. TheNa® channel has anmgate that opens quicklyand an h gate that closes slowly upon depolarization. TheK® channel has an n gate that opens slowly upondepolarization. The Na® gate can be blocked bytetrodotoxin (TTX), and the K® gate can be blocked bytetraethylammonium (TEA). In the resting state the Na®and K® channels have their gates arranged as shown inFigure 1:

m ** ECF n

I j membrane |1/\l icf 7

K+

Figure 1


If a stimulus is strong enough so that the thresholdpotential of a cell is reached (about 15 mV from the Em),an action potential (Figure 2) will be generated that willpropagate along the nerve cell until it reaches thepresynaptic ending. A neurotransmitter is released anddiffuses across the synaptic cleft, where it binds to thepostsynaptic membrane. A postsynaptic potential of equalmagnitude is generated in the postsynaptic cell and thesignal continues. The response of the action potential isall-or-none. The phases are shown in Table 2.

Time (msec)

Figure 2

Table 2

1. Resting State: Thecell at itsEm.2. Threshold: Potential that generates an AP.

3. Depolarization: Due to influx of Na+.

4. Overshoot: AP with a positive membrane potential.

5. Repolarization: Due to efflux of K+.

6. Hyperpolarization: Membrane potential is morenegative thantheE^

7. Absolute Refractory Period: AP cannot be generated.

8. Relative Refractory Period: AP can be generated if thestimulus is strong enough.

14. The threshold potential for the generation of anaction potential can be found at that value of the Emwhere the:

A. influx of K® balances the efflux of Na®.B. efflux of K® is 10times greater than the influx

ofNa®.C. efflux of K® balances the influx of Na®.D. influx of Na® is 10 times greater than the

efflux of K®.


Biology Action Potentials Passage m

15.

16.

17.

18.

19.

The threshold potential for the action potentialshown in Figure 1 is:

A.

B.

C.

D.

-50 mV.

-65 mV.

-80 mV.

-92 mV.

During depolarization, the quick opening of thevoltage dependent m gates results in all of thefollowing EXCEPT:

A. an influx of Na® ions.

B. an increase in the opening of fast m gates inneighboring Na® channels.

C. an increase in the opening of fast n gates inneighboring K® channels.

D. an absolute refractory period.

Which of the following voltage gated positions bestrepresents the repolarization phase?

A. m gate open; h gate closed; n gate open.B. m gate closed; h gate closed; n gate open.C. m gate open; h gate open; n gate open.D. m gate closed; h gate open; n gate closed.

During the last half of the relative refractory periodshown in Figure 1, the membrane potential isbeginning to:

A. repolarize.B. hyperpolarize.C. depolarize.D. hypopolarize.

If TTX and TEA are added to a nerve cellpreparation and a depolarizing stimulus is applied tothe presynaptic ending, a postsynaptic potential willbe produced. This shows that:

A. Na® and K® ions are needed for the release ofthe neurotransmitter to occur.

B. Na® and K® ions are not responsible for therelease of the neurotransmitter.

C. TTX and TEA can amplify the postsynapticpotential by allowing neurotransmitter release.

D. an increase in ICF K® ions and an increase inECF Na® ions inhibit the release of theneurotransmitter, unless a depolarizingstimulus is applied.


20. Gamma-aminobutyric acid (GABA) is a majorinhibitory neurotransmitter in the central nervoussystem. It is synthesized from glutamate, a majorexcitatory neurotransmitter in the brain.

COOH© i

HjN-C-H

CH,I

CH,I

COOH

Glutamic acid

Enzyme

© iHjN-C-H

CH2

CH,I

COOH

GABA

The enzyme that catalyzes this reaction is bestdescribed as:

A. GABA decarboxylase.B. glutamic acid decarboxylase.C. GABA carboxylase.D. glutamic acid carboxylase.


Biology Local Anesthetic Passage IV

Passage IV (Questions 21-26)

Local anesthetics provide a reversible block in theconduction of impulses in nerve fibers. All localanesthetic drugs contain a lipophilic group, anintermediate chain, and a hydrophilic group. Theintermediate chain may be either an ester (as in procaine)or an amide (as in lidocaine). The activity and potency ofany local anesthetic dependson severalaspects, includingits chemical structure, and the pKa of the substance.

Local anesthetics are weak bases, so they are kept inan acid solution in which they form a water-soluble salt.The base is lipid-soluble and can penetrate various tissuebarriers. The concentration of base or cation in the

solution depends on the pKa of the local anesthetic. Adecrease in the amount of base facilitates removal of the

local anesthetics, resulting in a shorter duration of action.The duration of action of a local anesthetic is also affected

by the local vasodilation which most of them produce.Finally, experiments have shown that a nerve which hasbeen recently and repetitively stimulated is more sensitiveto a local anesthetic than a resting nerve.

21.

22.

C,Hc

\ /r~ v~ c~ CH2~ -N-~qH<i

CH,

Lidocaine at pH = 9.0

Figure 1

Local anesthetics penetrate the blood-brain barrierwith great ease, most likely because of their:

A. large molecular size and high lipid solubility.B. large molecular size and low lipid solubility.C. small molecular size and high lipid solubility.D. small molecular size and low lipid solubility.

Based on the passage, which of the followingstatements is most likely true?

A. The higher the pKa of the local anesthetic, thelower the concentration of base in the tissue.

B. The lower the pKa of the local anesthetic, thehigher the concentration of acid in the tissue.

C. The higher the Ka of the local anesthetic, thelower the concentration of base in the tissue.

D. The lower the Ka of the local anesthetic, thelower the concentration of acid in the tissue.


23. Which of the following statements is LEASTcompatible with the effects of local anesthetics?

A. Local anesthetics slow the rise of the action

potential in nerve fibers.B. Local anesthetics slow the propagation of

nerve impulses.C. Local anesthetics decrease threshold for

electrical stimulation in nerve fibers.

D. Local anesthetics prevent the depolarization ofcell membranes.

24. The pKa of lidocaine is somewhere between 7.6 and7.8. The pKa of procaine is between 8.1 and 8.6.Based on this information, which of the followingstatements is the MOST likely conclusion?

A. The onset of lidocaine is more rapid, becausemore exists in its base form at body pH.

B. The onset of lidocaine is more rapid, becausemore exists in its acidic form at body pH.

C. The onset of lidocaine is slower, because moreexists in its base form at body pH.

D. The onset of lidocaine is slower, because moreexists in its acid form at body pH.

25. If acidosis develops after a local anesthetic haspenetrated a tissue barrier, it is most likely that:

A. the amount of ionized form will be decreased,and the local anesthetic will freely cross thetissue barrier.

B. the amount of ionized form will be decreased,and the local anesthetic will be trapped in thetissue.

C. the amount of ionized form will be increased,and the local anesthetic will freely cross thetissue barrier.

D. the amount of ionized form will be increased,and the local anesthetic will be trapped in thetissue.

26. The structure of lidocaine is given in Figure 1.Lidocaine is administered through injection, ratherthan orally. This is because lidocaine:

A. will be deactivated by the acidic conditions ofthe stomach.

B. will not be absorbed across the intestinal wall.C. is biotransformed by reactions occurring in the

liver.

D. is biotransformed by reactions occurring in thespleen.


Biology The Lens, the Iris, & Associated Muscles Passage V

Passage V (Questions 27-33)

Figure 1 shows the lens, the iris, and the muscles ofthe eye.

OpticNerve

Pupil

Cornea

SuspensoryLigaments

Figure 1

The opening of the pupil of the eye is controlled bytwo sets of muscles. The circularly arranged smoothmuscle is under parasympathetic control, and the radiallyarranged smooth muscle is under sympathetic control.Figure 2 shows the relation between these muscles and thepupil of the eye. The size of the pupil reacts to the amountof light present.

Radial Smooth

Muscle Fibers

Dim Light

Figure 2

Circular Smooth

Muscle Fibers

Bright Light

The lens of the eye is suspended within andsurrounded by a ring of tissue called the ciliary muscle.Suspensory ligaments connect the lens to the ciliarymuscle. When the ciliary muscle is relaxed, thesuspensory ligaments pull the lens taut and Hat. Since wespend most of our time in far-away vision (20 feet ormore), this is ideal. For closer work, the lens mustaccommodate and become thicker for focusing. In theprocess, the ciliary muscle tightens, the suspensoryligaments become slack, and the lens thickens andbecomes more convex.


27. For a person watching a meteor shower at midnight,what are the contraction states of the circular muscle

of the iris, the radial muscle of the iris, and theciliary muscle?

A. The circular muscle of the iris is contracted,the radial muscle of the iris is relaxed, and theciliary muscle is relaxed.

B. The circular muscle of the iris is relaxed, theradial muscle of the iris is contracted, and theciliary muscle is contracted.

C. The circular muscle of the iris is relaxed, theradial muscle of the iris is relaxed, and theciliary muscle is relaxed.

D. The circular muscle of the iris is relaxed, theradial muscle of the iris is contracted, and the

ciliary muscle is relaxed.

28.

29.

As people age, the lens becomes less llexible andtherefore less able to change shape duringaccommodation. What happens to a person's visionwith these age changes?

A. The lens is less able to focus the light raysfrom near objects.

B. The lens is more able to focus the light raysfrom near objects.

C. The lens is more able to focus the light raysfrom far objects.

D. The lens is less able to focus the light raysfrom far objects.

Special solutions are used to dilate the pupil during aretinal exam, so that the retina can be viewed. Thesolution causes an inability to focus on near objects,reduces clear distance vision, and enlarges the pupil.What is the stimulation or inhibition required toproduce these changes in the eye?

A. Sympathetic nerves arc inhibited, andparasympathetic nerves are stimulated.

B. Sympathetic nerves are stimulated, andparasympathetic nerves are inhibited.

C. Both are stimulated.D. Both are inhibited.


Biology The Lens, the Iris, & Associated Nusdes Passage V

30. When light rays travel through the pupil and thelens, upon which part of the eye does the imagefocus?

A. Retina

B. Cornea

C. Optic nerveD. Vitreous body

31. One eye exercise involves focusing on an objectwhile it moves from arm's length to the closest pointupon which you can focus. This causes the lens toundergo accommodation, that is, the lens mustchange shape to keep the object in focus as it movescloser. As the object moves nearer, what changesoccur in the eye?

I. The pupil gradually contracts.II. The ciliary muscle gradually contracts.III. The suspensory ligaments gradually contract.

A. I onlyB. II onlyC. IH onlyD. II and III only

32. The iris is pigmented epithelial tissue. The color ofthe eye is determined by the amount of pigment.Blue eyes have the least pigment, brown eyes havemore, and black eyes have the greatest amount ofpigment. What color are the eyes of a person whohas no pigment in her iris?

A. Black

B. White

C. Pink

D. Green

33. Astigmatism is the condition of having a non-uniformly shaped lens or cornea. This means thatparallel light rays do not focus, so that a sharp imageis not formed. How could this eye condition becorrected?

A. By providing a convex corrective lens.B. By providing a concave corrective lens.C. By providing a non-uniform corrective lens.D. By providing a uniform corrective lens.


Biology Resting Membrane Potential Passage VI

Passage VI (Questions 34-40)

Separated electric charges of opposite sign have thecapacity to do work if they are allowed to come together.This electric potential is more commonly known assimply the potential. All cells in a resting state have adifference in potential across their cell membranes, theinside being negative with respect to the outside.

The magnitude of the resting membrane potential isdetermined primarily by two factors. One is the typicalconcentration of three kinds of ions within the

intracellular and extracellular fluids (Table 1):

Ion

Extracellular

(mmol/L)Intracellular

(mmol/L)

Na®

Cle

K®

150

110

5

15

10

150

Table 1

The second factor is the ability of each kind of ion topenetrate the plasma membrane, an ability that fluctuatesfrom one moment to the next. In Table 1, notice that thesodium and chloride ion concentrations are generallylower inside in the cell, while potassium ion concentrationis lower outside the cell. The concentration differencesbetween sodium and potassium are produced by a plasmamembrane active transport system that simultaneouslypumps sodium out of the cell and potassium into the cell.

As an ion moves down its concentration gradient, anelectric force is established that opposes the movement ofthat ion. The membrane potential that exists when theelectric force is equal in magnitude but opposite indirection to the concentration force is the equilibriumpotential for that ion. The equilibrium potential for eachion species is different in magnitude-sometimes even indirection—and can be calculated using the Nernstequation:

c =.RT ln CoZF d

In a nerve cell at rest, the cell membrane is 50 to 75times more permeable to potassium than to sodium.During the resting potential, there is a net diffusion ofions into and out of the cell. The concentration gradients,which would eventually dissipate, are maintained throughthe Nae/Ke pump. If the concentration gradients remainfixed and the permeabilities of ions do not change, theresting membrane potential remains constant.


34. The cell membrane acts as an insulator, because the:

A. proteins making up the cell membrane arecharged.

B. proteins making up the cell membrane areuncharged.

C. lipids making up the cell membrane arecharged.

D. lipids making up the cell membrane areuncharged.

35. The resting membrane potential of a typical neuronis*

36.

A.

B.

C.

D.

positive, with the excess charge representing avery large fraction of the total number of ionsinside and outside the cell.

positive, with the excess charge representing avery small fraction of the total number of ionsinside and outside the cell.

negative, with the excess charge representing avery large fraction of the total number of ionsinside and outside the cell.

negative, with the excess charge representing avery small fraction of the total number of ionsinside and outside the cell.

Which of the following diagrams BEST representsthe equilibrium potential for potassium across anerve cell membrane?

B.

High

K®

Low

K®


Biology Resting Membrane Potential Passage VI

37.

38.

The resting membrane potential lies above theequilibrium potential for potassium, because a smallnumber of:

A. chloride ions diffuse into the cell.

B. chloride ions diffuse out of the cell.

C. sodium ions diffuse into the cell.

D. sodium ions diffuse out of the cell.

According to information in the passage, theconcentration gradients of ions are maintainedthrough the Nae/ K® pump. The ultimate sourceofenergy for this pump is the:

A. formation of ATP.

B. breakdown of ATP.

C. oxidation of nutrients.

D. reduction of nutrients.

39. The constancy of the resting membrane potential isan example of:

A. an equilibrium state.B. a dynamic state.C. a steady state.D. a static state.

40. The plasma membrane of many cells is permeable tochloride ions and does not contain chloride ionpumps. The membrane potential set up by other ionsthus acts on chloride ions. According to thisinformation and information in the passage, the newresting membrane potential is:

A. changed only slightly by the chloride ion.B. changed significantly by the chloride ion.C. changed greatly by the chloride ion.D. not affected at all by the chloride ion.


Biology Nicotine Replacement Therapy Passage VII

Passage VII (Questions 41-46)

Acetylcholine is one of the body's most importantneurotransmitters, responsible for the transmission ofnerve impulses across synaptic junctions. There are twomain classes of acetylcholine receptors. The nicotinicacetylcholine receptor responds to nicotine as an agonistand to curare as an antagonist. The muscarinicacetylcholine receptor responds to muscarine as anagonist and to scopolamine as an antagonist.

Nicotine is a psychoactive alkaloid extracted fromtobacco plants. It is a toxic substance, one that places astress on the heart and the entire cardiovascular system.Because this drug is quite addictive, several techniqueshave been designed to cure nicotine dependence.

In a novel nicotine replacement therapy, acombination of physostigmine and scopolamine isadministered. Physostigmine is an acetylcholinesteraseinhibitor at both muscarinic and nicotinic acetylcholinereceptors. It must be noted that scopolamine is functionalin the central nervous system, but receptors in heart tissueare insensitive to scopolamine. Figure 1 schematicallydepicts the action of these substances in the autonomicnervous system.

Parasympathetic division

1*1

/Muscarinic

receptor

1 n :

Effectororgan•••• "1 • w

/Nicotinicreceptor

Sympathetic division

PSl^/

Nicotinicreceptor

Effectororgan1 o :

/Adrenergic

receptor

CNS = Central nervous system

• = Acetycholine o = Norepinephrine

Figure 1

41. Which of the following ions is responsible for therelease of acetylcholine-filled vesicles from apresynaptic nerve terminal?

A. Na®

B. K®

C. Cle

D. Ca2®


42.

43.

51

The following graph shows the results of a bindingassay on the muscarinic acetylcholine receptor:

[Muscarine]

Scopolamine, a competitive inhibitor of muscarine,is added to the solution and the assay is repeated.Which of the following graphs BEST represents theresults of the second assay?

[Muscarine] [Muscarine]

C.

[Muscarine] [Muscarine]

The nicotinic acetylcholine receptor, which is madeof a pentamer of subunits, is essentially a sodiumchannel. Which of the following occurs directlyafter acetylcholine binds to a nicotinic acetylcholinereceptor?

A. An action potential is generated in thepostsynaptic cell.

B. An inhibitory postsynaptic potential isgenerated.

C. An excitatory postsynaptic potential isgenerated.

D. There is no change in the resting membranepotential of the postsynaptic cell.


Biology Nicotine Replacement Therapy Passage VII

44.

45.

46.

The novel nicotine treatment described in the

passage theoretically should be superior totraditional nicotine addiction treatment, because itoffers patients:

A. no net nicotine excitation.

B. a net nicotinic and sympathetic nervous systemexcitation.

C. a net nicotinic and parasympathetic nervoussystem excitation.

D. a net nicotinic and neuromuscular excitation.

Cardiovascular stress is one common result of

ingesting nicotine by smoking cigarettes. Thissupports the idea that:

A. nicotine stimulation of the nicotinic

acetylcholine receptor is dominant in thesympathetic nervous system.

B. nicotine stimulation of the nicotinic

acetylcholine receptor is weaker in thesympathetic nervous system.

C. nicotine acts directly on cardiac muscle topromote overstimulation and stress.

D. nicotine acts indirectly through the centralnervous system to produce cardiac stress.

As part of an experiment, a molecule is introduced atthe synapses between heart muscle and nerve fibersof both the sympathetic and parasympatheticnervous systems. The molecule moves in aretrograde fashion through both of these divisions.By the time it reaches the next synapse, themolecule will have traveled through:

A. a dendrite, moving a greater distance in the.parasympathetic nervous system than in thesympathetic nervous system.

B. a dendrite, moving a shorter distance in theparasympathetic nervous system than in thesympathetic nervous system.

C. an axon, moving a shorter distance in thesympathetic nervous system than in theparasympathetic nervous system.

D. an axon, moving a greater distance in thesympathetic nervous system than in theparasympathetic nervous system.


Biology Retinal Projections Passage VIII

Passage VIII (Questions 47-52)

The optic nerve is formed from the axons of allretinal ganglion cells. The optic nerves from each eyejoin at the optic chiasm and eventually enters either theleft or right optic tract. The optic tract projects to threesubcortical areas. One is the lateral geniculate nucleus,which is responsible for processing visual information.One is the pretectal area, which produces pupillar reflexesbased on information from the retina. Finally, thesuperior colliculus uses the information from the retina togenerate eye movement.

When light is shone upon one eye, it causesconstriction of the pupil in both eyes. Constriction of theeye in which the light is shone is the direct response whileconstriction of the other is known as the consensual

response. The pupillary reflexes are mediated throughretinal ganglion neurons that project to the pretectal areawhich lies anterior to the superior colliculus. The cells inthe pretectal area project bilaterally to preganglionicparasympathetic neurons in the Edinger-Westphalnucleus. This is also known as the accessory occulomotornucleus. The preganglionic parasympathetic neurons inthe Edinger-Westphal nucleus send axons through theocculomotor nerve to innervate the ciliary ganglion. Theciliary ganglion's postganglionic neuron innervates thesmooth muscle of the pupillary sphincter.

47. It has been determined that the frequency of actionpotentials increases dramatically in axons once theyhave left the optic nerve. The most likelyexplanation for this increase is:

A. a higher density of sodium channels are foundin the axons leaving the optic disc.

B. a lower density of sodium channels are foundin the axons leaving the optic disc.

C. the axons are myelinated by Schwann cells.D. the axons are myelinated by oligodendrocytes.

48. The right optic tract can be described as an:

A. efferent pathway, containing nerve axons fromonly one eye.

B. efferent pathway, containing nerve axons fromboth ways.

C. afferent pathway, containing nerve axons fromonly one eye.

D. afferent pathway, containing nerve axons fromboth eyes.


49. The blind spot, located at the optic disc, is called sobecause:

A. this is the region where ganglion cells leavethe retina.

B. color vision is not available at this retinallocation.

C. this area is free of photoreceptors.D. light is unable to reach this small area of the

retina.

50. The neurotransmitter released by the axons in theEdinger-Westphal neurons is most likely:

A. epinephrine.B. norepinephrine.C. acetylcholine.D. glutamine.

51. As part of a routine eye exam, the following isnoticed: If light is shone directly into the patient'sleft eye, the patient exhibits a consensual but not adirect response. Which of the following is a likelyexplanation?

A. The optic nerve of the left eye is intact, but theefferent limb of the left eye is damaged.

B. The optic nerve of the left eye is damaged, butthe efferent of the left eye is intact.

C. The optic nerve of the right eye is intact, butthe efferent limb of the right eye is damaged.

D. The optic nerve of the right eye is damaged,but the efferent limb of the right eye is intact.

52. In another patient, the following is observed: Lightshone into the right eye does not elicit a response ineither pupil. Light shone directly into the left eyecauses a direct and consensual response. It can beconcluded that there is a lesion in the:

A. left optic nerve.B. right optic nerve.C. left preganglionic parasympathetic neuron.D. right preganglionic parasympathetic neuron.


Biology Axonal Transport Passage IX

Passage IX (Questions 53-58)

Movement of substances from the soma to thesynaptic endings of a nerve cell by simple diffusion is aninefficient process, because axons are often very long.However, components which originate in the soma mustbe distributed along the axon. Axonal transport, a processwhich costs metabolic energy, is a special transportmechanism that accomplishes this function. Membrane-bound organelles are transported rapidly by fast axonaltransport, while substances dissolved in cytoplasm aremoved by slow axonal transport. Microtubules provide apathway along which membrane-bound organelles canmove. These organelles may interact with themicrotubules through a linkage similar to that betweenthick and thin filaments of skeletal muscle cells.

Transport from the soma toward the axonal terminalis known as anterograde axonal transport. Retrogradeaxonal transport is transport in the opposite direction. Inorder to trace neural pathways experimentally, markersubstances can be used that are transported eitherretrogressively or anterogressively.

53. A neurotransmitter traveling from the soma to thepresynaptic terminal will travel:

A. anterogressively, via fast axonal transport.B. anterogressively, via slow axonal transport.C. retrogressively, via fast axonal transport.D. retrogressively, via slow axonal transport.

54. Which of the following ions most likely triggers aninteraction between organelles and microtubules?

A. Na®B. K®C. Ca2®D. Mg2®

55. Neuropeptide packaging into vesicles will includeall of the following EXCEPT the:

A. signal recognition particle receptor.B. signal peptide.C. complete translation of the peptide in the

cytoplasm.D. transport through the Golgi apparatus.


56. Glucose does not simply diffuse over the innermitochondrial membrane. Glucose enters the

mitochondrial matrix as:

A. glucose-6-phosphate.B. phosphoenolpyruvate.C. pyruvate.D. oxaloacetate.

57. Phaseolus vulgaris leucoagglutinin is taken up byneurons and is transported anterogressively. Thismarker substance is most likely injected at the:

A. synapse to trace the pathway of neuronalaxons.

B. synapse to trace the location of cell bodies.C. soma to trace the pathway of neuronal

dendrites.

D. soma to trace the synaptic endings of neurons.

58. The graph shown below represents the response of apostsynaptic cell to acetylcholine in the synapticcleft. Each vertical line (I) represents an actionpotential.

time L $

Which of the following graphs BEST represents theresponse following the addition of the enzymeacetylcholinesterase?

A.

B.

C.

D.

time $

time ^>

illtime ^>

time L $


Biology Huntington's Disease Passage X

Passage X (Questions 59-64)

Huntington's disease is a relatively rare disease thatcauses progressive degeneration of the cerebral cortex andthe basal ganglia of the brain. The basal ganglia aremasses of neuron cell bodies (gray matter) located deepwithin the cerebrum. Within the basal ganglia, certaintypes of neurons are destroyed, while others remain intact.

Huntington's disease is a genetic disorder caused by anautosomal dominant genetic defect on the tip ofchromosome 4. The symptoms of the disease arechoreiform movements (rapid, uncontrolled, jerkymovements), mental deterioration, and emotionaldisturbances. The disease is relentless and usually leadsto complete dehabilitation and death within 15 years of itsonset. The age of onset is usually from 35-50 years,usually after the patient has had children.

The proposed mechanism of the defect lies withcertain neurotransmitters and is twofold. First of all, thereis a decreased level of y-aminobutyric acid, an inhibitoryneurotransmitter, and a decreased level of the enzyme thatsynthesizes it, glutamic acid decarboxylase. Secondly, adeficiency of the enzyme choline acetylase leads todecreased levels of the excitatory neurotransmitter,acetylcholine. The delicate homeostatic balance betweenthe two neurotransmitters is disrupted. These deficienciesare thought to lead to the observed characteristicmovements and mental symptoms of Huntington'sdisease.

59.

60.

Based on information in the passage, Huntington'sdisease is:

A. more common in men than in women.B. more common in women than in men.

C. equally common in both men and women.D. more common in children.

Based on this passage, the basal ganglia most likelycontrol:

A. pituitary gland secretion.B. speech.C. the parasympathetic nervous system.D. voluntary movements.


61. The following molecule is referred to as GABA.What is its IUPAC name?

© o

H3N— CH2- CH2- CH2- COO

A. y-amino butyrateB. 3-amino butanoic acid

C. 1-amino butyrateD. 4-aminobutanoic acid

62. Which amino acid is the precursor to GABA?

63.

64.

A. Histidine

B. Leucine

C. TyrosineD. Glutamate

After having one child, a 40-year old woman with apaternal family history of Huntington's disease isdiagnosed as having the disease. Assuming that thefather is not affected, what is the MOST probablelikelihood that the child will also develop thedisease?

A. 100%

B. 75%

C. 50%

D. 25%

In what structure of the neuron are neurotransmitters

stored?

A. The synaptic cleft.B. The vesicles of presynaptic neurons.C. The receptors of postsynaptic neurons.D. The vesicles of postsynaptic neurons.


Biology Photoreceptors Passage XI

Passage XI (Questions 65-71)

The retina of the human eye contains two types ofphotoreceptors known as rods and cones, both located atthe back of the retina. Upon stimulation by a photon, thesodium channels of these photoreceptor cells close. Thehighest density of cones found on the retina will give thegreatest visual acuity, or highest visual precision. Inaddition to the photoreceptor cells, the retina houses fourtypes of neurons. These cells are the bipolar, ganglion,horizontal, and amacrine cells. The rods and conessynapse with bipolar cells, which then go on to synapsewith ganglion cells. The axons of the ganglion cellsconverge and leave the eyes as the optic nerve.

The visual field is the view seen by the two eyeswithout movement of the head. As shown in Figure 1, theleft visual hemifield projects to the right side of the brain,while the right visual hemifield projects to the left half ofthe brain.

LeftVisual

Hemifield

Left Optic Tract

Figure 1


RightVisual

Hemifield

Right Optic Tract

56

65.

66.

67.

68.

Which of the following retinal cells is LEASTresponsible for vision in the dark?

A. Cone cells.B. Rod cells.

C. Ganglion cells.D. Bipolar cells.

Stimulation of a photoreceptor by a single photonwill result in a:

A. monopolarization of the photoreceptor.B. depolarization of the photoreceptor.C. hyperpolarization of the photoreceptor.D. micropolarization of the photoreceptor.

To view an object with greatest acutiy, one willfocus which of the following structures on thatobject?

A. Fovea centralis

B. Cornea

C. Optic discD. Choroid

An experiment involved the removal of the lateralgeniculate nucleus of the thalamus. To determine theeffects on processing of visual information,electrodes were placed in which of the followinglobes of the brain?

A. Parietal

B. Frontal

C. TemporalD. Occipital

69. Light entering the eye will pass which of these cellsfirst?

A. Rod cells

B. Cone cells

C. Bipolar cellsD. Ganglion cells

The Berkeley ReviewSpecializing in NCAT Preparation

Biology Photoreceptors

70. The region of the retina where the axons of theganglion cells converge and leave as the optic nerveis best described as the region:

I. of highest visual precision, because of thedense axonal population.

II. of the retina most sensitive to stereovision,because of a high density of photoreceptors.

III. where no vision is possible, because of a lackof photoreceptors.

A. I onlyB. I and II onlyC. Ill onlyD. II only

71. According to Figure 1, light originating from the lefthemivisual field will strike the nasal hemiretina of

the:

A. right eye and the temporal hemiretina of theleft eye.

B. right eye and the temporal hemiretina of theright eye.

C. left eye and the temporal hemiretina of theright eye.

D. left eye and temporal hemiretinal of the lefteye.


Passage XI


Biology Sound Transmission In The Ear Passage XII

Passage XII (Questions 72-79)

Sound is created by disturbances within a mediumresulting in the production of pressure waves detected bythe ear. These pressure waves consist of alternatingcompressions and rarefactions of the surroundingmedium. The loudness of a sound is determined by theamplitude of these pressure waves and is measured on thedecibel (db) scale. The relationship is as follows:

Sound Pressure Level (SPL) = 20 logio ?t / Pr

where Pt is the test pressure and Pr is the referencepressure. Pr has the value of20 micro-Newtons/m2. Thepressure waves travel through the external ear canal andwill cause the tympanic membrane of the middle ear tovibrate.

A set of small bones is responsible for transmittingthis vibration throughout the middle ear, and the last ofthese bones (the stapes) is attached to the oval window.The oval window is the connection between the middleand internal ear, which is composed of the cochlea and thevestibular apparatus (Figure 1).

Scala Media

6. i " o' ,-. Incus

Malleus ~Z*WTwH4 \'• StaPes.„ ™ , . 1°- \ Scala VestibulL

TympanicMembrane

AuditoryCanal

Middle Ear

Figure 1

The transduction of pressure waves into electricalsignals occurs in the internal ear. The vibration of thestapes produces pressure waves in the fluid of the scalavestibuli and scala tympani. This wave motion in theparalymph naturally sets up oscillations in the endolymph,located in the scala media. The transduction organ (organof Corti) is located on the basement membrane of thescala media. Oscillating movements of this organ exciteand inhibit sensory transduction cells, which transmitimpulses to the brain.

Inducing a vibration in the temporal bone will causesounds to reach the cochlea. This sound transmission isunique in that the middleear is bypassed. While this is aninefficient method of energy transfer, such a method isclinically useful for diagnosing auditory problems.


. , Tympaniindow

Internal Ear

58

72.

73.

74.

75.

The human ear normally responds to a range ofsound covering about 120 dB. The loudest soundpressure level that can be heardover this range is:

A. ten thousand times less than the reference

pressure.

B. one million times less than the referencepressure.

C. ten thousand times greater than the referencepressure.

D. one million times greater than the referencepressure.

According to Figure 1, the tympanic membrane hasan area greater than the oval window. A result ofthis design is that the total:

A. force acting on the oval window is increased.B. force/unit area acting on the oval window is

decreased.

C. force acting on the oval window is decreased.D. force/unit area acting on the oval window is

increased.

The relationship between the speed of sound (C),wavelength (k), and frequency (v) is:

C = (X)(v)

where C = 340 m/s. What is the period of a soundwave with a wavelength of3.4x 10"4 km?

A. Ixl0"6s.B. 1 x 10"3 s.C. 1 x 103 s.D. 1 x 106 s.

As a result of movement in the organ of Corti, thesensory transducing cells most likely repond with apotential that is:

A. depolarizing.B. hyperpolarizing.C. oscillating depolarizing-hyperpolarizing.D. returning to its resting membrane potential.


Biology Sound Transmission In The Ear Passage XII

76.

77.

The following picture represents themembrane uncoiled and stretched out flat.

basilar

m

Flexible

region

According to the resonance theory, different areas ofthe basilar membrane are affected by varying soundfrequencies. High frequency sounds resonate best inwhich region?

A. Region 1.B. Region 2.C. Region 3.D. Region 4.

The following circuit model attempts to explainaway the problem that sound arrives at the two earsat different times. Cells will fire with maximum

output when bilateral inputs arrive at the same time.

Input fromleft ear

Cells

<2>

<Z>

Input fromright ear

If sound to the right ear were delayed relative to theleft, which of the following cell(s) would most likelyfire?

A. Cells 1 and 2.

B. Cell 3.

C. Cells 3 and 4.

D. Cell 5.


78. A patient complains of auditory problems and so atuning fork is set into vibration and placed at thepatient's ear. When the patient indicates the sound isinaudible, the fork is placed on the temporal bone.If the sound becomes audible to the patient, thedamage can be narrowed down to the:

A. external ear.

B. middle ear.

C. inner ear.

D. auditory cortex.

79. It has been discovered that every hair cell respondsmaximally to a particular frequency. Which of thefollowing graphs best represents this idea?

A.

C.

CL

E<

Frequency (kHz)

Frequency (kHz)

B.

E<

D.

Frequency (kHz)

Frequency (kHz)


Biology Tryptophan and Serotonin Experiment Passage XIII

Passage XIII (Questions 80-87)

The change in tryptophan levels in the brain and inthe synthesis rate of serotonin (a neurotransmitter) fromtryptophan in thebrain after the ingestion of two meals isexamined in several experimental animals. The synthesisand release of serotonin by brain neurons is known to bestrongly influenced by the local tryptophan concentrationnear these neurons.

Brain tryptophan (Trp) concentration reflects theuptake of tryptophan from the blood into the brain. Thisuptake is accomplished by a transporter in the blood-brainbarrier that is shared by the large, neutral amino acids(LNAAs): leucine, isoleucine, tyrosine, phenylalanine,and tryptophan. Food ingestion can alter levels ofserotonin in the brain by altering blood concentrations oftryptophan and its LNAA competitors.

The design of the experiment requires that two mealsgiven two hours apart be fed to laboratory rats that havefasted overnight. The five experimental diets fed to therats contain varying ratios of protein and carbohydrate,with fat held at a low and constant percentage in all of thediets. The diets are designated as CHO (all carbohydrate,no protein), 6% protein, 12% protein, 24% protein, and40% protein. For the sake of reference, standard rat chowis about 10% protein, and the typical human diet is about12-15% protein.

Rats in Trial 1 are given the all-carbohydrate diet,followed by one of the protein diets two hours later.Those in Trial 2 receive one of the protein diets, followedby the CHO diet two hours later. The rats are sacrificedtwo hours after the second meal, and their brain tissue isexamined for tryptophan concentration.

Table 1 shows the study results for the animalsreceiving CHO first (Trial 1). Serum tryptophan andserum LNAA are in umol/1, and cortex tryptophan is inumol/g.

GroupSerum

TrpSerumLNAA

CortexTrp

No Food 96 482 28

CHO-CHO 140* 376 36*

CHO - 6% Protein 150* 385 36*

CHO-12% Protein 156* 437 35*

CHO - 24% Protein 160* 675 28

CHO - 40% Protein 162* 976 23

*p = 0.05 versus no food

Table 1.


Table 2 shows the study results for the animals receivingprotein first (Trial 2). Serum tryptophan and serumLNAA are in umol/1, and cortex tryptophan is in umol/g.

80.

81.

GroupSerum

TrpSerum

LNAA

CortexTrp

No Food 142 511 26

CHO-CHO 162* 497 33*

6% Protein - CHO 165* 533 31*

12% Protein-CHO 190* 621 27*

24% Protein - CHO 184* 743 28

40% Protein - CHO 194* 814 25

*p = 0.05 versus no food

Table 2.

The diagram below shows the molecular structure oftryptophan and the chemical reaction by which itbecomes serotonin.

H

Tryptophan

COOH

H

Serotonin

Tryptophan is converted into serotonin by firstundergoing:

A. reduction, followed by carboxylation.B. hydroxylation, followed by decarboxylation.C. amidation, followed by hydroxylation.D. amination, followed by decarboxylation.

Which of the following statements about the data inTable 1 and Table 2 is FALSE?

A. Serum LNAA concentration increases as

dietary protein content increases.B. A 24% protein meal at two hours following a

CHO meal attenuates the increase in serotonin

due to tryptophan.C. At higher levels of protein, the serum LNAAs

are converted to serotonin.

D. Serum tryptophan concentrations increase asdietary protein content increases.


Biology Tryptophan and Serotonin Experiment Passage XIII

82. Which of these statements can be inferred from thedata in Table 1 and Table 2?

I. In Trial 1, the higher protein levels (24% and40%) cause competition between tryptophanand the other LNAAs.

II. In Trial 2, the higher protein levels (12%,24%, and 40%) cause competition betweentryptophan and the other LNAAs.

III. In Trials 1 and 2, serum tryptophan increasewith feeding.

A. I onlyB. II onlyC. II and III onlyD. I, II, and III

83. It is accurate to describe tryptophan as:

I. a neurotransmitter.

II. an essential amino acid.

III. an acidic amino acid.

A. I onlyB. II onlyC. I and II onlyD. I, II, and III

84. As part of this experiment, researchers administer aninhibitor of one of the enzymes that participates inthe conversion pathway of tryptophan to serotonin inthe brain. Examining which factor would providethe BEST estimate of the new rate of serotoninsynthesis caused by this inhibitor?

A. Accumulation of an intermediate metabolite inthe brain

B. Concentration of tryptophan in the bloodC. Concentration of the inhibitor in the brainD. Accumulation of the inhibitor in the blood


85.

86.

In gauging the metabolic effect of eating a largeamount of pure tryptophan as a dietary supplement,we should conclude that dietary tryptophan alonewould:

A. not be absorbed readily in the small intestine,so it would have no metabolic effect.

B. increase brain serotonin concentrations, sincefew LNAAs would be competing.

C. increase the synthesis of all classes ofneurotransmitters.

D. cause more alert behavior, due to theproduction of serotonin.

The LNAAs described in the passage are allessential amino acids. If the diet of an adult humanbeing is deficient in phenylalanine, which of thefollowing non-essential amino acids now becomesrequired in the diet?

A. AsparagineB. CysteineC. Serine

D. Tyrosine

87. The selectivity of the blood-brain barrier is due to:

A.

B.

C.

D.

tight junctions,gap junctions,desmosomes.

adherens.


Biology Frog Muscle Experiment Passage XIV

Passage XIV (Questions 88-94)

The gastrocnemius muscle along with the sciaticnerve from a frog's leg were dissected and attached to adevice to measure action potentials and record them onchart paper. Figure 1 is a diagram of the experimentalsetup:

Figure

Weights (1, 2, 5, 10, 20, and 50 gm) were hungopposite the muscle, so that a known set of tensions wereapplied to it. The action potentials generated by eachweight were recorded on the chart paper. This isillustrated in Figure 2. Each vertical ([) line represents anaction potential.

1 gm

5 gm

10 gm

20 gm

50 gm

time :>

Figure 2

The purpose of this experiment was to learn moreabout the interaction between muscle and nerve, and how

muscle stretch is communicated.


88. The muscle is stretched more and more as heavierweights are added to the experimental apparatus.How is the intensity of the muscle stretch indicatedby the activity of the nerve fiber?

A. A larger amplitude of the action potentialsindicates greater intensity.

B. A larger amplitude of the action potentialsindicates lower intensity.

C. A greater frequency of the action potentialsindicates greater intensity.

D. A greater frequency of the action potentialsindicates lower intensity.

89. Two more trials were performed on the same musclepreparation. A 75-g and a 100-g weight were usedto stretch the muscle. The printouts of the actionpotentials looked exactly like the 50-g trial. What isthe explanation for this?

A. The action potentials had increased to amaximum with the 50-g weight and increasedno further, due to the refractory period of theneuron.

B. The nerve and muscle were damaged by theweights, and the nerve therefore slowed itstransmission of action potentials.

C. The speed of the chart recorder could not keepup with the increased action potentials.

D. The action potentials were identical, due to aneurotransmitter defect.

90. Figure 3 indicates the output of action potentialsafter a 5-mg weight was added to the musclepreparation. What is the explanation for this effect?

A.

B.

C.

D.

time :>

Figure 3

The weight was not heavy enough to exceedthe action-potential threshold.The weight was too heavy and suppressed theaction-potential threshold.The weight must produce an action potential,so the recording equipment must be damaged.The size of the action potential was too smallto read on the printout.


Biology Frog Muscle Experiment Passage XIV

91. Actions potentials are conducted by the nerve cellthrough which structure?

A. CytoplasmB. Schwann cell membraneC. Endoplasmic reticulumD. Plasma membrane

92. When an action potential is communicated, what isthe sequence and direction of movement of sodiumand potassium?

A. Potassium moves from the intracellular fluid tothe outside of the cell, and then sodium movesfrom the extracellular fluid into the cell.

B. Sodium moves from the extracellular fluid intothe cell, and then potassium moves from theintracellular fluid to the outside of the cell.

C. Sodium moves from the intracellular fluid tothe outside of the cell, and then potassiummoves from the extracellular fluid into the cell.

D. Potassium moves from the extracellular fluidinto the cell, and then sodium moves from theintracellular fluid to the outside of the cell.

93. What type of nerve function is studied using thisexperimental design?

A. Efferent function

B. Afferent function

C. Sympathetic functionD. Parasympathetic function

94. The sodium-potassium pump, an active transportsystem, restores the proper balance of intracellularand extracellular ions. When an organism ispoisoned with cyanide, which halts ATP production,what happens to neural action potentials?

A. Action potentials stop immediately.B. Action potentials will continue, even though

ATP production stops.C. Action potentials continue briefly but stop

eventually, due to a lack of ATP.D. Action potentials will generate ATP during

cyanide administration.


Biology Skeletal Muscle Groups Passage XV

Passage XV (Questions 95-100)

Muscles of the lower extremities aid in the movementof bones such as the femur, tibia, and fibula. They helpmaintain an erect skeletal system and keep the body inbalance by acting as an antagonist to gravity. Thesemuscles provide the pumping action that helps circulatebloodand lymphthrough the lowerextremities.

A diagram of the major bones found in the lowerextremity is shown in Figure 1.

ilium

<^U Posterior

medialcondyle

ftSuperior

Anterior C^>

pubic bone

femur

Figure 1

A list of the major muscle groups of the thigh regionof the lower extremity is provided below:

Muscles of the Thigh

I. Quadriceps Group

Made of four muscles, three of which originatefrom the upper, anterior portion of the femur andone that originates from the anterior inferior iliacspine of the ilium. They all descend and insert intothe broad patellar tendon which in turn crosses theknee joint and inserts into the tibial tuberosity ofthe upper, anterior tibia. These muscles areinnervated by branches of the femoral nerve.


II. Hamstring Group

Made up of three muscles. All originate from theischium and run along the posterior side of thefemur. Two of the muscles of the hamstring groupinsert onto the medial condyle of the tibia, whilethe third inserts onto the lateral condyle of thetibia and the head of the fibula. The hamstringsare innervated by branches of the sciatic nerve.

III. Adductor Group

Several muscles comprise this group. All of thesemuscles originate from the pubic bone region andextend to insert along the medial and posteriorinferior shaft of the femur. The adductors are

innervated mainly by the obturator nerve.

95. Judging by where they attach, which of the followingmuscle groups would play the strongest role inflexing the hip joint (i.e., pulling the knee upwardstowards the chest)?

A. Quadriceps.B. Hamstrings.C. Adductors.

D. Quadriceps and hamstrings together.

96. The pumping action of muscles in the lowerextremity is crucial in order for blood to properlyflow back to the heart. Which of the followingstatements must be true if blood is to be returned to

the heart effectively while a person is standing?

A. The veins involved must have inelastic walls.

B. The arteries involved must have elastic walls.

C. The veins involved must have valves.

D. The arteries involved must contract under

sympathetic control.


Biology Skeletal Muscle Groups Passage XV

97. If an individual contracts the hamstring muscles, the:

A. quadriceps must contract.B. quadriceps must relax.C. quadriceps are unaffected.D. obturator nerve is responsible.

98. Paralysis of the femoral nerve would MOSTSTRONGLY affect the ability:

A. to stand on tip-toes.B. to extend the hip posteriorly.C. to kick a ball.

D. of the adductor muscles to contract.

99. Often when a person suffers a herniated vertebraldisk in the lower back, the sciatic nerve iscompressed and damaged. Which of the followingwould most likely NOT be a symptom of such acondition?

A. Difficulty walking.B. Difficulty moving the lower leg posteriorly.C. Pain radiating down the back of the leg.D. Weakness in the adductor muscles.

100. Claudication is a medical condition in which arteries

which feed the muscles of the lower extremitybecome partially occluded. In these patients,moderate exercise might cause all of the following inthe affected muscles EXCEPT:

A. ATP deprivation.B. lactic acid buildup.C. lower-than-normal CO2 levels.D. O2 deprivation.


Biology Nerve & Muscle Section I Answers

Passage 1(1 - 8) Types of Transport

B iscorrect, HC03e. Roughly 60% of the body by weight is due to water, and all of the water can be divided intotwo compartments separated by a cellular membrane. One compartment contains the intracellular fluid (ICF) whilethe other compartment contains the extracellular fluid (ECF). The larger of the two compartments is the ICFcompartment, and itcontains about 2/j ofall the water in the body. The ECF compartment is itself divided into twotypes of fluids: interstitial fluid and blood plasma. Table 1in the passage lists the major ions of the ICF and ECF.The first thing to note is that the ICF has a higher concentration of K® than the ECF. This allows us to eliminatechoice D.

The remaining three choices are not mentioned in the passage and require a little thought. Phosphate (P043e) isrequired not only in cellular DNA and RNA, but it is also required by a vast array of proteins that participate inphosphorylation reactions within the cell. Simply consider phosphorylating ADP with Pj to make ATP. Therefore,we would expect the levels of phosphate to be higher in the ICF than in the ECF. and they are. Phosphate levels inthe ICF are about 20 mM, while in the ECF they are about 4 mM. Eliminate choice C. We can use the samereasoning for the relative levels of proteins. All of our cells are carrying out metabolic reactions, and to do themajority of these reactions enzymes (proteins) mustbe employed. This means that there will be more protein insidea cell than outside a cell. Eliminate choice A. Cells which are catabolizing molecules are producing COt gas (e.g.,

via the Krebs cycle). Carbon dioxide can combine with water to form carbonic acid, which can then dissociate toform the bicarbonate ion (HC03e). Since CO2 is a waste and it needs to be eliminated via the blood, the cell wantsto get rid of it. We find that the levels of bicarbonate are higher in the ECF (i.e., the blood plasma) than in the ICF.Bicarbonate levels in the ICF are about 10 mM, while in the ECF they are about 24 mM. The correct choice is B.

D is correct, extracellular [Na®]. Extracellular refers to what is happening outside the cell and intracellular refersto what is happening inside the cell. The Na®/K®-ATPase requires ATP in order to function. If the levels of ATPinside the cell are low, the ATPase activity will decrease and eventually be inhibited. The remaining ATP leftwithin the cell will be diverted to reactions that are more important to the cell's survival. We can eliminate choiceA. A low intracellular concentration of Na® will also lead to inhibition of the ATPase, because if there is no Na® topump to the extracellular space, then the enzyme cannot catalyze the transport of sodium out of the cell andpotassium into the cell. We can eliminate choice B. The same reasoning applies to low extracellular concentrationsof K®. This allows us to eliminate choice C. A low extracellular concentration of Na® will not directly affect theATPase, because the ATPase is located within the cell. It is not found on the outside of the cell. The correct

choice is D.

B is correct, primary active transport. In the second paragraph of the passage, we learned that glucose can cross thecell's membrane by facilitated diffusion. At the bottom of Figure 1 In the passage, we see glucose and Na® beingtransported across the cell's membrane in the same direction. The protein carrier is acting as a symport. If thecarrier were a uniport, only one molecule would be crossing the membrane. An antiport allows two molecules tocross the membrane, but each is crossing opposite to the other. Glucose is entering the cell along with sodiumthrough a symport mechanism. The sodium was actively transported out of the cell. Sodium can re-enter the celldown its concentration gradient, allowing glucose to be transported with it by a secondary active transport process.A primary active transport system uses the energy of ATP directly to move a substance across a cell's membrane.We do not observe this happening in the case of glucose, especially since we read in the third paragraph that only

three primary active transport systems have been identified: one for Na® and K® ions, one for Ca-® ions, and onefor H® ions. The correct choice is B.

A is correct, low intracellular |Na®) allows sodium to be released from the carrier protein. A high extracellularNa® concentration is whatallowsglucose to bind to the symport on the extracellularside of the cell's membrane andbe transported into the cytosol. We can eliminate choice Band D. Ifthe affinity of the carrier protein for Na® werehigh on the cytosolic side of the membrane, then sodium and glucose could not be released into the cytosol. WhenNa® is released into the cytosol from the carrier protein, thecarrier protein's affinity for glucose is reduced. We cannow eliminate choice C. The correct choice is A.

A is correct. Simple diffusion (SD) involves a substance moving from a high concentration to a low concentration.Diffusion of lipid-solublc substances can occur through a lipid bilayer, while diffusion of water-soluble substanceswill occur through a transmembrane channel. The lamer the gradient, the more a substance will diffuse across the



membrane. We would expect to see a straight line showing that as the gradient of a substance increases, the flux ofthe substance across the membrane increases as well. Facilitated diffusion (FD) requires a carrier protein. Thesecarrier proteins have a limited number of carrier sites for the substance that is to be transported across themembrane. Therefore, at a higher concentration gradient all of these carrier sites will be filled. The carrier will thenbe saturated (think of Michaelis-Menten kinetics). At this point, the rate of diffusion of the substance across themembrane will no longer increase with an increasing concentration gradient. We would expect to see a plateau inthe curve as soon as all of the carrier sites become saturated. The correct choice is A.

B is correct, more Na® to enter the cell than K® to leave the cell. The resting membrane potential of aneuromuscular cell is about -HO mV. If a positively charged Na® ion enters the cell, it will not make the inside ofthe cell more negative, but rather make it more positive. The resting membrane potential will be depolarized, notrepolarized. Eliminate choice A. Table 1 in the passage tells us that the concentration of K® is higher in the cellthan it is outside the cell. It also tells us that the concentration of Na® is higher outside the cell than it is in the cell.Therefore, when the acetylcholine receptors open and allow for the flow of K® and Na® ions, we find that Na® ionsenterthe cell, while K® ions leave the cell down their concentration gradients. Thisallows us toeliminate choice C.The question tells us that two molecules of acetylcholine will bind to the acetylcholine receptor. This is achemically mediated and activated response. The channel in the acetylcholine receptor is not activated (opened) byan electrical stimulus (i.e., a depolarization). If that were the case, acetylcholine would not need to bind to thereceptor. We can eliminate choice D.

By the process of elimination, we arrive at choice B. But how do we know that more Na® will enter the cell thanK® will leave the cell? The electrochemical gradient for Na® is greater than the electrochemical gradient for K®.The electrochemical gradient that allows an ion to pass through a membrane is simply the difference between themembrane potential (Em) and the equilibrium potentials for the ion in question (Ejon). The resting membrane

potential for our cell is about -80 mV. The resting membrane potential for K® is about -92 mV, while the restingmembrane potential for Na® is about +58 mV. Clearly, the electrochemical gradient for Na® is greater than that forK®. The correct choice is B.

D is correct, a high-resistance electrical pathway. Gap junctions join the cytoplasm of one cell to the cytoplasm of aneighboring cell through intramembrane proteins. These proteins (connexons) contain 6 subunits that form a centralpore that allows molecules with molecular weight up to 1500 to pass through. Molecules like ATP and ions likeNa® can easily pass through these channels. This helps to establish a cytoplasmic continuity between the cells.When an action potential reaches a gap junction, electrical coupling between the two cells occurs, and the actionpropagates to the next cell. This type of coupling is important in heart tissue, where entire collections of cells mustcontract in a coordinated fashion. Therefore, a gap junction must provide a low-resistance (not a high-resistance)electrical pathway that enables current to pass from one cell to the next. The correct choice is D.

C is correct, increased sodium concentration gradient. Inhibition of the electron transport chain means that theamount of ATP being synthesized is substantially reduced. If the amount of ATP is reduced, then the amount ofenergy transferred from ATP to the membrane ATPases is reduced. The primary active transport systems (thoseinvolving the Na®/K®-ATPase, Ca2®-ATPase, and H®-ATPase) all show a decrease in their activity. In secondaryactive transport, the energy that was stored in the Na® concentration gradient across the cell's membrane (highextracellular [Na® 1and low intracellular [Na®]) is used to transport molecules like glucose and amino acids intothe cell and calcium out of the cell. If the synthesis of ATPdecreases, the Na® concentration gradient across thecell's membrane is not as great (i.e., it decreases, not increases), and the rate of secondary active transportdecreases.Low levels of ATP also mean that the Ca-®-ATPase cannot adequately pump calcium out of the cell. Since calciumis also transported out of the cell (against its concentration gradient) by a secondary active transport system thatutilizes the sodium concentration gradient (which is now rather low), the levels of intracellular calcium begin toincrease. Calcium is able to diffuse back into the cell though calcium channels. The correct choice is C.

Passage II (9 - 13) Autonomic Nervous System

9. D is correct, I, III, and IV only. The sympathetic system is important in getting the body ready for a perceivedstressful situation. This response is a general response, affecting all parts of the body almost simultaneously. Thisis essential to the survival of the animal. You wouldn't want to have to wait for each organ or muscle to determinethat this is a stressful situation while a lion is running at you! But the parasympathetic system has the luxury ofdetermining specifically for each structure it innervates whether the job is done during this time of stress and when



it's time to return to basal levels. The parasympathetic system can do this type of individual monitoring, partiallybecause of the proximal location of its ganglia next to the structure in question. The correct choice is D.

10. D is correct I, II, and IV only. All three situations are interpreted by the body as demands to increase particularfunctions beyond their normal basal levels. In other words, they represent a stress to the body. Sweating on a warmday is a response to release heat and cool down the external body. On cold days shivering and piloerection arestimulated to increase heat in the body. Both of these reactions are instigated by the sympathetic system.Furthermore, running is a stressor on the body. The correct choice is D.

11. D is correct, penile erection. Penile erection can be caused by direct stimulation involving penilemechanoreceptors, or it can be caused by central nervous system activity stemming from sights and smells or eventhoughts and emotions. In either case, input reaches the neurons of the penis. Parasympathetic nervous systemactivity is increased while sympathetic nervous system activity is decreased. When parasympathetic activityincreases and sympathetic activity decreases, vascular dilation of the arterioles occurs so that the tissues of the penisbecome engorged with blood. As erection continues, the veins leaving the penis become compressed and little bloodis allowed to leave, thus maintaining the penis in the erect state. The correct choice is D.

12. B is correct, I, II, and III only. The adrenal gland sits on top of the kidney. Cortisol, a glucocorticoid produced bythe adrenal cortex, controls various aspects of metabolism. An increase in plasma concentrations of Cortisol willlead to an increase in gluconeogenesis (i.e., the synthesis of glucose from precursor molecules like lactate), adecreased uptake of glucose by cells, an increase in protein catabolism, and an increase in triacylglycerol breakdown(i.e., the release of free fatty acids). Aldosterone, a mineralocorticoid produced by the adrenal cortex, controlselectrolyte balance by stimulating sodium reabsorption and potassium secretion in the kidney. Sodium reabsorptionalso leads to water reabsoiption in the kidney. The correct choice is B.

13. B is correct, increased heart rate. The vagus (10th cranial) nerve is a component of the parasympathetic nervoussystem. Think of the parasympathetic system as being passive. The iris muscle of the eye will contract, making thepupil smaller. The motility and secretions (e.g., HC1 from the parietal (oxyntic) cells) of the stomach will increase,especially after a meal. An increased heart rate and contractility is due to the sympathetic nervous system (think ofthefight-or-flight response). The correct choice is B.

Passage III (14-20) Action Potentials

14. C is correct, efflux of K® balances the influx of Na®. A small depolarization in the membrane will allow someNa® to enter the cytoplasm of the cell. At the end of the second paragraph in the passage we see that the relativepermeability of the membrane to K® is roughly 10 times greater than it is to Na®. Since there is more K® in theICF than Na® (from Table 1), we see that more K® will flow out of the cell than Na® into the cell. Therefore, ifthere is a small depolarization, then the influx of Na® will be more than balanced by the efflux of K®. If the effluxof K® is greater than the influx of Na®, theexplosive nature of an action potential will not result. What this meansis that the threshold potential for the generation of an action potential is that point where the influx of Na® exactlymatches the efflux of K®. Remember, there is a higher concentration of Na® in the ECF and a higher concentrationof K® in the ICF. The nanosecond that there is more of an influx of Na® than there is an efflux of K®, an actionpotential will be generated. The correct choice is C.

15. B is correct, -65 mV. As stated in the fourth paragraph of the passage, the threshold is about 15 mV away from theEm. In Figure 2 of the passage, the Em is at -80 mV. The threshold is given by the dashed line labeled with thenumber 2. Therefore, the threshold potential must be -65 mV. The correct choice is B.

16. C is correct, an increase in the opening of fast /; gates in neighboring K® channels. As depolarization takes place,Na® rapidly enters the cell (Table 2 in the passage). In order for Na® to enter the cell, the fast acting /;/ gate mustopen to allow Na® to rush into the cell down its concentration gradient. As Na® enters the cell, it causes furtherdepolarization and allows more m gates to open. This is the explosive nature of the action potential. The absoluterefractory period begins at about the time the depolarization of the membrane takes place. Another action potentialcannot be generated at this time because of the opening of the /;/ gates from the first action potential. This allows usto eliminate choices A, B, and D. Depolarization of the membrane will lead to an eventual opening of the // gates.The problem is that the n gates are slow and not fast. The correct choice is C.

17. A is correct, m gate open; h gate closed; n gate open. Table 2 in the passage says that during the repolarization

phase K® is flowing out of the cell and down its concentration gradient. If K® is flowing out of the cell, the n gate


Biology Nerve 8t Muscle Section I Answers

must be open. Looking through our choice of answers, we see that the n gate is closed in choice D. Therefore, weeliminate choice D.

We know that during repolarization, we do not want Na® entering the cell. If Na® enters the cell as K® leaves, wewill not be able to obtain the resting membrane potential of the cell as quickly as we would like. Which one of theNa® gates is closed, or are they both closed? The last paragraph of the passage tells us that during depolarizationthe h gate begins toslowly close. Let's assume that the h gate has closed as repolarization begins. If this is the case,then we can eliminate choice C.

However, all through depolarization Na® is entering the cell, which means that the in gate is remaining open. Bythe time the peak of the action potential has occurred, the /; gate has closed, the mgate is still open but beginning toclose, and the // gate is open. The correct choice is A.

18. A is correct repolarize. An action potential will initially depolarize the resting membrane potential (Em) and causethat potential to be less negative. As the Na® gates close and the K® gates open, the membrane potential is beingrepolarized toward the Em of the cell. Because of the high permeability to K® during repolarization, the membranepotential becomes even more negative than the Em. This is called hyperpolarization. Once the cell has completedhyperpolarized, it will return to the Em value of the cell. This is what is occurring during the last half of the relativerefractory period. Whenever a membrane potential returns to its true Em, the process is referred to as arepolarization. The correct choice is A.

19. B iscorrect, Na® and K® ions are not responsible for the release of the neurotransmitter. The passage states thatTTX blocks the Na® channels, while TEA blocks the K® channels. If this is the case, then the depolarization of thenerve cell membrane will not occur. However, if the presynaptic ending is stimulated with a depolarizing signal, apostsynaptic potential is produced. The last paragraph of the passage says that a neurotransmitter is released fromthe presynaptic terminal, diffuses across the synaptic cleft, binds to the postsynaptic terminal, and produces apostsynaptic potential. Since we are using a stimulus (signal) of depolarization to cause the release of a

neurotransmitter, it must mean that Na® and K® are not responsible for the release of the neurotransmitter. If theywere needed, then the stimulus that we applied would have no effect on the generation of a postsynaptic potential.We can eliminate choice A.

As mentioned above, TTX and TEA block Na® and K® ion channels, respectively. They do not allowneurotransmitter release. If they did. we would not have needed the depolarizing stimulus. We can eliminate choiceC. Because the channels for Na® and K® are blocked, it does not (necessarily) mean that the concentrations of ICFK® and ECF Na® are going to increase. Remember, the membrane is permeable to both Na® and K® (see thesecond paragraph in the passage). These ions will tend to diffuse down their concentration gradients and theNa®/K®-ATPase pump will redistribute them again. Blocking the Na® and K® channels is analogous to not havinga stimulus. The nerve cell will be at its resting state. We can eliminate choice D. The correct choice is B.

20. B is correct, glutamic acid decarboxylase. The a-carboxyl group of glutamic acid is being removed. When CCb isremoved from a molecule, the reaction is called a decarboxylation and is carried out by a decarboxylase enzyme. IfCCb were added to a molecule, it would be a carboxylation reaction, which would be carried out by a carboxylaseenzyme. Since the CCb is being removed from glutamic acid, the enzyme that catalyzes this reaction is a glutamicacid decarboxylase. Note that once GABA is formed, the numbering of the carbon atoms changes. What was the a-position becomes the y-position. There is no carboxyl group at the y-position. Instead, there is an amino group. Ifwe were to remove the amino group of GABA, the four carbon compound would be butyric acid. Hence the name,y-aminobutyric acid. The correct choice is B.

Passage IV (21 - 26) Local Anesthetic

21. C is correct, small molecular size and high lipid solubility. The passage tells us that the base form of the localanesthetic, being lipid-soluble, can penetrate various tissue barriers. The blood brain barrier is certainly one of thesetissues. The reason lipid soluble substances are able to penetrate the blood brain barrier is that the leaflets of the cellmembrane, including those of the capillary endothelium, are composed of lipid molecules. Lipid soluble moleculescan thus diffuse through the two-dimensional liquid and through the membrane. Furthermore, it is logical that thesmaller the molecule, the greater its ease in crossina a membrane. The correct choice is C.


Biology Nerve & Muscle Section 1 Answers

22. A is correct, the higher the pKa of the local anesthetic, the lower the concentration of base in the tissue. Thepassage tells us that the concentration of base or cation in the solution depends on the pKaof the local anesthetic.Remember that local anesthetics are weak bases. Look at the structure of lidocaine. When it is protonated, it carriesa positive charge. In other words, the cation is theacid form. Therefore, when we havea high pKa, we have a smallKa. This, of course, means the reaction does not go very far to the right. In other words, we have a largeconcentration of acid and a small concentration of base in the tissue. The correct choice is A.

23. C is correct, local anesthetics decrease threshold for electrical stimulation in nerve fibers. We are looking for astatement that is least compatible with the way in which local anesthetics act. Remember that local anesthetics actto produce a reversible, local nerve conduction block. Therefore, weare looking for a statement that contradicts thisgoal. A decrease in the threshold value indicates it is now easier to initiate a nerve action potential. Creatingconditions where action potentials are formed more easily does certainly contradict the goal of the local anesthetic.The correct choice is C.

24. A is correct, the onset of lidocaine is more rapid, because more exists in its base form at body pH. We know fromthe passage that the base form is lipid-soluble, and therefore it penetrates the tissue. One can logically conclude thatthe local anesthetic that exists more in its base form (as a percentage) at tissue pH will have a quicker onset of actionbecause more penetration of tissue will occur. That is why the pKas of these local anesthetics are important. Thelocal with the lower pKa will have a higher percentage of molecules in the base form and thus have a more rapidonset. Lidocaine has a more rapid onset because more of it exists in its base form. The correct choice is A.

25. D is correct, the amount of ionized form will be increased, and the local anesthetic will be trapped in the tissue.Again, the local anesthetic penetrates the tissue in its neutral, base form. If acidosis (lowering of the pH) occurred intissue after penetration of the local anesthetic, this would drive the base form back into its acidic form. In otherwords, a lower percentage of the molecule would exist in the base form. The problem with this is that the acidicform is charged and thus cannot escape the tissue barrier. The trapping of the charged form of the local anestheticcan contribute to an increase in the degree of toxicity. The correct choice is D.

26. C is correct, is biotransformed by reactions occurring in the liver. Lidocaine is injected (rather than taken orally) toavoid biotransformation reactions which occur in the liver. If the drug is taken orally, it is not destroyed by theharsh conditions of the stomach. For example, the drug is not a protein, so there is no concern over the proteasecontent of the stomach. After absorption across the intestinal wall occurs (most will probably occur in the smallintestine), the drug enters into the hepatic portal system and is transported to the liver. Remember that the livercontains enzymes (we do not need to concern ourselves with the exact reactions) which are involved in modificationof substances. Such modification can render the local anesthetic inactive. Furthermore, we wish to avoid having alocal anesthetic circulating in the blood stream. The correct choice is C.

Passage V (27 - 33) The Lens, the Ins, St Associated Muscles

27. D is correct, The circular muscle of the iris is relaxed, the radial muscle of the iris is contracted, and the ciliarymuscle is relaxed. Although this answer seems complicated, we can figure it out. First, figure out what the iris isdoing. In the dark, the pupil will be large to let in light. The radial muscle will contract, and the circular musclewill relax. This immediately eliminates choices A and C. We must now figure out what the ciliary muscle is doingto choose between choices B and D. The ciliary muscle, as we are told in the passage, is relaxed for vision beyondabout 20 feet. Since the night sky extends beyond 20 feet, the ciliary muscle is relaxed. Eliminate choice B. Thecorrect choice is D.

28. A is correct, the lens is less able to focus the light rays from near objects. Remember, the "default" shape of thelens in the relaxed state is flatter, for far vision. This really does not change with age. Choices C and D areincorrect. Near vision requires a flexible lens that can plump up into a convex shape to allow focusing on nearobjects. A less flexible lens is less able to focus on near objects. Choice B is incorrect. The correct choice is A

29.

30.

B is correct, sympathetic nerves are stimulated, and parasympathetic nerves are inhibited. What situation leads tolarge pupils? The radial muscles must contract under sympathetic stimulation. The circular muscles must berelaxed under parasympathetic inhibition. Choice A is backwards and thus incorrect. Two different stimuli arerequired, so choices C and D are incorrect. The correct choice is B.

A is correct, retina. Rays converge and focus on the retina. The cornea is a clear, tough membrane covering thefront of the eyeball. Choice B is incorrect. The optic nerve is responsible for transmitting visual information to thebrain, but it does not focus the image. Choice C is incorrect. The vitreous body is the inert stuff inside the eyeball.Choice D is incorrect. The correct choice is A.



31. B is correct, II only. As the object moves closer, the lens must become more convex. This is accomplished bycontraction of the ciliary muscle. Choice II is correct. The ligaments do not contract, they merely pull the lens tautby default. Choice III is incorrect. The pupil responds to changes in light, not in focusing. Choice I is incorrect.Since choice III is false, choice D is incorrect as well. The correct choice is B.

32. C is correct, pink. The condition of not having epithelial pigment is call albinism. A person or animal withalbinism is called an albino. Most strains of laboratory rats are albinos. They have pink eyes, due to the color of theblood vessels visible in their iris. Also, you may have seen white bunnies with pink eyes. Pigment covers up thecolor of the blood vessels in people and animals who don't have albinism. Black eyes contain the most pigment.Choice A is incorrect. Green eyes contain some pigment, as well. Choice D is incorrect. You may not know aboutthe lab rats, but a person would not have white eyes due to the presence of blood vessels in the iris. Choice B isincorrect. The correct choice is C.

33. C is correct, by providing a non-uniform corrective lens. Since the person has an irregularly shaped lens or cornea,the corrective lenses used should be shaped to correct these problems and allow focusing of the image on the retina.This would require that the corrective lens have a compatible non-uniform shape. Convex lenses correctfarsightedness. Choice A is incorrect. Concave lenses correct nearsightedness. Choice B is incorrect. A uniformlens would not help the person with astigmatism. Choice D is incorrect. The correct choice is C.

Passage VI (34 - 40) Resting Membrane Potential

34. D is correct, lipids making up the cell membrane are uncharged. Cell membranes are made up primarily of lipids.Lipids contain very few charged groups, cannot carry current, and have a high electrical resistance. Materials withhigh electrical resistance are known as insulators. The lipid layers of the plasma membrane are regions of highelectrical resistance separating two water compartments of low resistance. The correct choice is D.

35. D is correct, negative, with the excess charge representing a very small fraction of the total number of ions insideand outside the cell. This problem requires outside knowledge about the resting membrane potential of a typicalneuron. The typical resting membrane potential lies somewhere between -40 mV to -75 mV. The minus signindicates that the potential is negative with reference to the inside of the cell. In other words, there is an excess ofnegative charge inside the cell relative to the outside. This eliminates choices A and B. Now, do these excesscharges represent a small or large fraction of the total ions in the cell? The excess charges are but a very smallfraction of the total number of ions inside and outside the cell. The correct choice is D.

36. B is correct. We are looking for a picture that represents the potassium equilibrium potential. The equilibriumpotential will exist where the force due to the concentration gradient is equal to the electrical force gradient.Potassium ions have a higher concentration inside the cell, eliminating choices C and D. To discriminate betweenchoice A and choice B, we simply look at the magnitude of the arrows. Again, at the equilibrium potential, the twoforces will be equal. Answer choice A represents the resting membrane potential. The correct choice is B.

37. C is correct, sodium ions diffuse into the cell. One can arrive at this answer by thinking about the membranepotential and the direction in which it is going. We are told the resting membrane potential lies above that for thepotassium equilibrium potential. This means that some positive charge is coming back into the cell, making thepotential higher. Of the possible answers, the only way this is going to happen (given the true ion concentrations) isto have sodium coming into the cell. This is indeed the case. There is a permeability to sodium ions over theplasma membrane, and this contribution causes a rise in the membrane potential. Chloride ions have a higherconcentration outside of the cell, so if they move down their gradient, they will decrease the membrane potential.The correct choice is C.

38. C is correct, oxidation of nutrients. The pump that transports sodium and potassium ions does breakdown ATP tocouple the energy of that reaction to the work it must perform in transporting the ions. However, the questionalludes to the ultimate source of the energy. Therefore, we need to think about how we are getting the ATP that isused by this pump. This answer comes from our knowledge of metabolism and an understanding that food isoxidized in the course of glycolysis and the Krebs cycle. The electrons released are eventually transported down theelectron transport chain and are coupled to a proton gradient, which results in the formation of ATP. The correctchoice is C.

39. C is correct, a steady state. The constancy of the resting membrane potential is best described as a steady-statepotential. The answer really comes down to two choices. Is it an equilibrium, or is it a steady state? With anequilibrium, no energy input is needed to maintain the state. Is this the case? No. We use the energy of ATP to runthe Na®/K® ATPase pump that maintains the concentration gradients of the these ions. Since there is energy used,this cannot be described as an equilibrium. When a state is constant, but energy is used to maintain that state, this istermed a steady-state system. The correct choice is C.


Biology Nerve St Muscle Section I Answers

40. D is correct, the new resting membrane potential is not affected at all by the chloride ion. We are told that theplasma membrane of many cells are permeable to chloride ions and do not contain chloride-ion pumps. Therefore,in these cells, the membrane potential set up by other ions will act on chloride ions. The inside negativity moveschloride out of the cell until a concentration gradient develops. Looking at Table 1. we see the concentration ofchloride is larger extracellular!}'. This concentration gradient will force chloride ions into the cell. However, thediffusion force will be exactly equal to the electrical force pushing chloride ions out of the cell. The result is that theequilibrium potential for chloride ions is equal to the resting membrane potential, and the chloride ion makes nocontribution to the magnitude of the membrane potential. The correct choice is D.

Passage VII (41 -46) Nicotine Replacement

41. D is correct, Ca-®. Calcium ions are the link between depolarization of the presynaptic membrane andneurotransmitter release. Depolarization of the terminal causes voltage-sensitive calcium channels in the membraneto open, and calcium diffuses into the axon terminal from the extracellular fluid, including that in the synaptic cleft.The increase in the calcium level in the terminal causes vesicles filled with neurotransmitter to fuse with thepresynaptic membrane and release their contents into the synaptic cleft. The correct choice is D.

42. C is correct. In the question, we are given a graph of a binding assay for the muscarinic acetylcholine receptor.where the substrate is muscarine. The question tells us that scopolamine, a competitive inhibitor, is added to thesolution and the assay is run again. A competitive inhibitor will compete for the same site on the receptor asmuscarine. It will not remove or incapacitate any of the receptors (this would be a non-competitive inhibitor). Sincethe total number of receptors remain in place, the level of maximum binding will not change. This eliminateschoices A and B. What will change is the apparent affinity of the receptor for muscarine, which can be representedby half-maximum binding. The apparent affinity of the receptor for muscarine will decrease, because the inhibitor iscompeting for the same site. Therefore, we will see an increase in the half-maximum binding site of muscarine,because it now takes a higher concentration of muscarine to reach the half-maximum binding site. Remember, thetell-tale sign of a competitive inhibitor is that if we increase the concentration of either substrate, one of them shouldbe able to out-compete the other and reach maximum binding. The correct choice is C.

43. C is correct, an excitatory post-synaptic potential is generated. We arc told from the question that the nicotinicacetylcholine receptor is essentially a sodium channel. Therefore, when acetylcholine binds to a nicotinicacetylcholine receptor, we are essentially activating a sodium channel. When a sodium channel opens, whathappens? Since sodium always has an extracellular concentration higher than its intracellular concentration, sodiumcomes into the cell and causes a net movement of positive ions in the cell. At this point, we can eliminate choice D.This causesa slightdepolarization and is known as a excitatory post-synaptic potential (EPSP). Eliminate choice B.This is a graded potential, and it is not an all-or-nothing action potential. In order to generate an action potential, thedepolarization must be beyond a threshold value set by that particular cell. Usually, several EPSPs added togetherwill eventually generate an action potential at an area known as the axon hillock, where a high concentration ofsodium channels are found. The correct choice is C.

44. C is correct, a net nicotinic and parasympathetic nervous system excitation. We want to offer the person who isaddicted to nicotine the "high." without the cardiovascular risks associated with nicotine. Therefore, we will want tostimulate the nicotinic receptors in the central nervous system to achieve this sensation. The problem is thatnicotinic receptors arc located in the pre-ganglionic synapse of both the sympathetic and parasympathetic nervoussystems. We might think that since both are stimulated, they simply cancel each other out. However, we know thatnicotine causes stress on the heart, so the nicotinic receptor in the sympathetic nervous system must be dominant.With this in mind, let us look at the two drugsadded. The first is physostigmine (eserine). This acetylcholinesteraseinhibitor will cause increased levels of acetylcholine in both nicotinic and muscarinic receptors. This will give usthe nicotine "high" that we want. Yet wc do not want to stimulate all the muscarinic receptors, so we addscopolamine. This acts to block all muscarinic receptors. This is what we want. We get the familiar feeling causedby nicotine and have blocked all muscarinic receptors. We are not done, though. We are still left to deal with thecardiovascular stress. The physostigmine will increase acetylcholine (ACh) levels at the first synapse in thesympathetic nervous system. But in the parasympathetic system, it increases ACh levels at both the nicotinic andthe muscarinic receptors (remember that scopolaminedoes not work directly upon heart tissue). Therefore, we havetwo stimulations in the parasympathetic system while we only have one in the sympathetic system. Therefore, weare left with a net nicotinic stimulation with a parasympathetic excitation. The correct choice is C.

45. A is correct, nicotine stimulation of the nicotinic acetylcholine receptor is dominant in the sympathetic nervoussystem. We would think that because the first synapse in both the sympathetic and parasympathetic are nicotinic thatnicotine would stimulate both of them equally, and therefore there would be no stress on the heart. However, we



know this not to be the case, as nicotine places considerable stress on the cardiovascular system. Therefore, nicotinestimulation of the nicotinic acetylcholine receptor in the sympathetic nervous system must be dominant over the onein the parasympathetic nervous system. There is no evidence in the passage for the claim made in answer choice C.Cardiovascular stress is not brought about by activity in the central nervous system, but by activity in the peripheralnervous system. We can eliminate choice D. The correct choice is A.

46. D is correct, axons, moving a greater distance in the sympathetic nervous system. The question tells us we have amolecule that will be moving in a retrograde (backward) fashion from the synapses to the heart between bothsympathetic and parasympathetic nervous systems. The molecule will thus be moving through axons (eliminatingchoices A and B), because the axons, not the dendrites will be closest to the organ. In other words, an electricalsignal reaching the organ will travel through a nerve's axon to reach the synapse between the nerve and the organ.So the next question becomes one of distance. Remember that in the sympathetic nervous system, the preganglionicfiber is short while the postganglionic fiber is long. The molecule will be traveling through the postganglionic fiberto reach the first synapse (the first synapse is the junction between the pre- and postganglionic fibers). Therefore,the molecule will travel a longer distance in the sympathetic nervous system. The correct choice is D.

Passage VIII (47 - 52) Retinal Projections

47. D is correct, the axons are myelinated by oligodendrocytes. This question calls on our knowledge of the nervoussystem outside of what is stated in the passage. We are looking for the most likely explanation for the increase in thefrequency of the action potential. Myelinated nerves have the ability to increase the frequency of action potentialconduction. Therefore, we can narrow the options down to choices C or D. The question then becomes: Whichcells are responsible for the myelinalion? In both cases, glial cells are responsible for laying down the myelinsheath. In the central nervous system, these cells are called oligodendrocytes, while in the PNS they are calledSchwann cells. Since we are talking about nerves located in the CNS, the best answer becomes choice D. Thecorrect choice is D.

48. D is correct, afferent pathway, containing nerve axons from both eyes. Again, this question requires us to draw onour knowledge of the nervous system, and in particular, the eye. One must remember that the information leadingtowards the brain is found in the afferent nerves. Information leading away from the CNS is found in the efferentnerves. Therefore, answer choices A and B can be eliminated. Now. the question becomes whether the optic tractcontains information from one eye or two. The answer is two. Information from the nasal hemiretina (medial halfof the retina) of the left eye cross the optic chiasm and enters the right optic tract. The right optic tract is also madeup of nerve fibers originating from the temporal hemiretina of the right eye. In that way, the right side of the brainprocesses information from the left side of the visual world. Because the tract contains information from both eyes,D is the correct answer. The correct choice is D.

49. C is correct, this area is free of photoreceptors. The question is fairly straightforward. It asks for the bestexplanation of the "blind spot." The blind spot is the area on the retina where the optic nerve bundle leaves the eye.That is not the reason why the area is termed the blind spot. The region is blind because it has no photoreceptors. Ifthere are no photoreceptors to be found, there can be no transduction of light into a visual image. The correctchoice is C.

50. C is correct, acetylcholine. We are told from the passage that the neurons which make up the Edinger-Westphalnucleus are parasympathetic neurons. Therefore, this question is really testing one's knowledge of theneurotransmitter used by parasympathetic neurons. We cannot be expected to know from the question alone whichneurotransmitter these neurons use. However, we are supposed to be aware that neurons that are parasympatheticuse the neurotransmitter acetylcholine. The correct choice is C.

51. A is correct, the optic nerve of the left eye is intact, but the efferent limb of the left eye is damaged. The questiontells us we see a consensual but not a direct response. We can therefore conclude that the optic nerve of the left eyeis intact because the optic nerve of the right eye is not involved in the response. The response involves informationgoing down the left optic nerve to the pretectal area. Prom the pretectal area, neurons project bilaterally to theEdinger-Westphal nucleus. Axons from neurons in the nucleus innervate the ciliary ganglion. We see a consensualresponse in the right eye. We can therefore conclude that the bilateral projection and the efferent pathway to theright eye arc unharmed. In addition, we can conclude that there is some problem with the efferent pathway to theleft eye. We are not seeing a constriction of the pupil in response to the light being shone. We are left with thefollowing conclusion: The left optic nerve is intact, but the efferent pathway of the left eye is somewhere andsomehow damaged. The correct choice is A.

52. B is correct, right optic nerve. The light shone in the right eye elicits neither a direct or consensual response. Thismost likely means that there is damage to the right optic nerve. The result of the damage is that the information


Biology Nerve 8t Muscle Section I Answers

leading away from the eye is unable to elicit a reflex which results in the constriction of both eyes. This can be theonly explanation for the information given in the question, because light shone in the other eye elicits both a directand a consensual response. The correct choice is B.

Passage IX (53 - 58) Axonal Transport

53. A is correct, anterogressively, via fast axonal transport. Recall from the passage that transport from the somatowards the axon terminal is known as anterograde axonal transport. Therefore, this eliminates choices C and D.Next, is the neurotransmitter moving via fast or slow axonal transport? Recall that neuropeptides are packaged intomembrane bound vesicles as they are formed near the soma. From the passage, we know that membrane-boundorganelles are transported rapidly by fast axonal transport. While the vesicle may not be technically considered anorganelle, given the choice between membrane-bound organelle and cytoplasm, a vesicle is best aligned with themembrane-bound organelle system. Our best choice includes fast axonal transport. The correct choice is A.

54. C is correct, Ca-®. We are told from the passage that organelles interact with microtubules through a linkagesimilar to that between thick and thin filaments of skeletal muscle cells. Recall that thick filaments refer to myosin,while thin filaments refer to actin. Recall that calcium is needed for actin and myosin interaction, as calciumremoves the troponin-tropomyosin complex which covers the actin binding sites. Since the interaction of theorganelles with the microtubules is similar to that of actin/myosin, one can best conclude that calcium triggers theinteraction. The correct choice is C.

55. C is correct, complete translation of the peptide in the cytoplasm. Recall that peptides that are produced by a cellfor the purpose of secretion (neuropeptides are certainly an example) are processed through the rough endoplasmicreticulum. The peptide being translated will have a signal peptide that attracts a cytoplasmic signal receptor particle(SRP). When the SRP attaches to the signal peptide, a pause in translation will occur. The translation does notbegin again until the SRP complex (SRP/signal peptide/ribosome) binds to an SRP receptor located on the roughER. When the SRP complex binds to the SRP receptor, translation begins along with transport of the new proteininto the lumen of the RER. At that point, the signal peptide is cleaved off and the proteins are modified in both theRER and the lumen of the Golgi apparatus. From the trans face of the Golgi, the peptide buds off as a vesicle.According to this hypothesis, there is no complete translation of the neuropeptide in the cytoplasm. The correctchoice is C.

56. C is correct, pyruvate. Glucose will go through the glycolytic pathway and the end product of glycolysis ispyruvate. Recall that glycolysis occurs in the cytosol of cells. This is a very straightforward question asking us toremember that pyruvate is transported over both the outer and inner mitochondrial membranes to enter the matrix.The correct choice is C.

57. D is correct, soma to trace the synaptic endings of neurons. The marker substance moves anterogressively.According to the passage, the marker substance will move from the soma toward the synaptic ending. If we areinterested in tracing a neural pathway, we should use this marker to identify synaptic endings. In other words, weshould inject this marker in a given cell body, and it will trace the axon and the synaptic endings of the neuron. Ifwe inject the marker into the soma, it will not trace the dendrites because of its direction of movement. Therefore,we can eliminate choice C. Furthermore, we do not want to inject the marker by the synapse, because there isnothing left to trace. The correct choice is D.

58. D is correct. We are told from the graph in the question that action potentials arise in a postsynaptic cell after theaddition of acetylcholine. We then add acetylcholinesterase, which breaks down ACh. We must look at the changein the pattern of the action potential response. Why? Recall that action potentials are all-or-nothing events, so theywill not change in amplitude. We can easily eliminate two of the choices. We then look at the pattern, and in thiscase, the frequency of response. If we are going to have lower levels of ACh due to the addition ofacetylcholinesterase, we should see a less frequent train of action potentials in the postsynaptic cell. The correctchoice is D.

Passage X (59 - 64) Huntington's Disease

59.

60.

C is correct, equally common in both men and women. The passage tells you the defect is an autosomal dominantone. This means it is not sex-linked, so eliminate choices A and B as incorrect. We also learn from the passage thatthe age of onset is from 35 years of age. Eliminate choice D. The correct choice is C.

D is correct, voluntary movements. The hypothalamus produces releasing hormones that affect the pituitary.Choice A is incorrect. The control of speech is localized in the cerebral cortex. Choice B is incorrect. The



parasympathetic nervous system (PNS) is not controlled by the basal ganglia. In the second paragraph, we learn thatsymptoms of the disease include rapid, uncontrolled, jerky movements. The correct choice is D.

61. D is correct, 4-aminobutanoic acid. This molecule is a 4-carbon organic acid. The skeletal structure is calledbutanoic acid. The carbon of the carboxyl group is designated 1. This means the amino group is attached to the 4thcarbon. This gives the name 4-aminobutanoic acid. The correct choice is D.

62. D is correct, glutamate. This information is given in the second paragraph of the passage. A key missing enzymethat leads to GABA deficiency is glutamate decarboxylase. If we correctly named the previous molecule, GABA,then we can imagine adding on a carboxyl group on the C-4 carbon to make glutamate. Choice A is incorrect:histidine is the precursor of histamine. Choice B is incorrect. Choice C is incorrect: tyrosine is the precursor ofepinephrine and norepinephrine. The correct choice is D.

63. C is correct, 50%. Since the gene is autosomal dominant and the father is not affected, he is homozygous recessivefor this disease. The gene is rare, and only one of the woman's parents shows a familial history for the disease. Thewoman is probably heterozygous. Therefore, she either passed on the dominant, disease-causing gene or therecessive gene. The chance is 50/50. since she has 2 X-chromosomes and only one is affected. The correct choiceisC.

64. B is correct, the vesicles of presynaptic neurons. Neurotransmitters are synthesized in presynaptic neurons, storedin vesicles there, and released by the arrival of a neural impulse. The released neurotransmitter crosses the synapticcleft and interacts with the receptors on the postsynaptic neuron's membrane. The correct choice is B.

Passage XI (65 - 71) Photoreceptors

65. A is correct, cone cells. This answer can be arrived at using previous knowledge, and not information specificallyfrom the passage. The fact is that cones are used for color vision, while rods are used for black-and-white vision.The question asks about seeing in the dark, which definitely qualifies for a lack of color. Of course, one may seesome color in the dark and the cones would be responsible, but the question asks which cell is most likely notinvolved. Consider the other answers. We have already discussed that rods play a role in black-and-white vision,making their role in dark vision likely. The other cells are involved in taking information from the photoreceptors tothe brain. It is stated in the passage that the photoreceptors synapse with the bipolar cells, which then synapse withthe ganglion cells. Therefore, for both color and black/white vision, these cell types will be involved in thetransmission of visual information. The correct choice is A.

66. C is corrrect, hyperpolarization of the photoreceptor. It is stated in the passage that upon stimulation of aphotoreceptor cell by a photon, the sodium channels close. This implies that when there is no light stimulation, thesodium channels are open. When sodium channels are open, the sodium ion comes in (following its gradient) andcauses the cell to be depolarized. In an unstimualted cell, a very active Na®/K® pump is constantly restoring thesodium gradient. Again, the net result of this is that a photoreceptor cell in the dark is normally depolarized. When aphoton hits, the sodium channels close and the pump continues to work. Both of the these events cause ahyperpolarization of the photoreceptor cell. This is a very special case for photoreceptors, and it is important to beaware of this phenomonon. Based on this information, the other answers can easily be eliminated. The correctchoice is C.

67. A is correct, fovea centralis. The area where there is the highest density of cones on the retina will give the greatestacuity, or precision. At this point, one must recall the anatomy of the eye. The area of highest cone density is thefovea centralis, and when we look at objects, we move our eyes to focus light onto this region of the retina. Thisanswer could not be obtained from the passage, but we thought it would be nice to review some eye anatomy. Lookat the other answers. The cornea is the transparent structure located at the front of the eye and is the structurethrough which light rays enter. The optic disc is where the optic nerve leaves the eye and the retinal blood vesselsenter. Since there are no photoreceptors overlying this disk, it is known as the blind spot. The choroid is a pigmentedlayer behind the retina, which contains many of the blood vessels that nourish structures in the eye. It also absorbslight not taken up by photoreceptors. It absorbs rather than reflects, so as not to alter the photoreceptor responses toother light. The correct choice is A.

68. D is correct, occipital. We need to recall our knowledge of the cortices of the brain. The experiment removed thethalamus. The thalamus, in particular the lateral geniculate nucleus, is a relay station for visual information. Fromthis nucleus, information then goes on to the visual cortex, located in the occipital lobe of the brain. The occipitallobe is located above the cerebellum. Based on this, the only possible answer choice is D. As a very brief review,one can associate the frontal lobe with association processes, the temporal lobe with hearing, and the parietal lobewith high level sensory and motor control. The correct choice is D.



69. D is correct, ganglion cells. The answer can be obtained by either previous knowledge, or by reading the passagecarefully. It is stated that both photoreceptorsare located at the back of the retina. It clearly states that the four othertypes of cells (bipolar, etc.) are part of the retina. Since the photoreceptors synapse with the bipolar cells, and thebipolar cells synapse with the ganglia, one could create a picture in their head where the ganglion cells are furthestaway from the photoreceptor cells, or the first layer in the retina. Based on this thought process, light entering theeye will first pass over the ganglion cells. The correct choice is D.

70. C is correct, III only. The place where the axons of the ganglion cells leave as the optic nerve is called the opticdisc. There are no photoreceptors at this region of the retina, because the axons are leaving and blood vessels areentering. If there are no photoreceptors, then vision is not possible. This is of course one's "blind spot." Consider theother possibilities. The region of highest acuity is where one finds the highest density of cones. This is certainly notthe case. In the case of stereovision, it is not the structure of one eye that allows for depth perception, but the factthat we have two eyes looking at one object from different angles. This gives us stereovision. Therefore, statement IIdoes not apply. Statement III is the only applicable statement. The correct choice is C.

71. C is correct, left eye and the temporal hemiretina of the right eye. If light originates in the left visual hemifield, thenwe known that our right side of the brain will be processing this information. In order to get to the right side of thebrain, we need to have light strike the temporal hemiretina of the right eye. If you follow the optic nerve, you willsee it goes to the right side of the brain. In addition, light from the left side will strike the nasal hemiretina of the lefteye. The optic nerve from this part of the retina will cross over at the optic chiasm and move on to the right side ofthe brain. Draw some lines of light from the left visual hemifield; it may augment your understanding of how lightoriginating from one side of the visual field becomes processed in the opposite side of your brain. The correctchoice is C.

Passage XII (72 - 79) Sound Transmission in the Ear

72. D is correct, one million times greater than the reference pressure. The question asks for the relationship betweenthe loudest sound pressure level and the reference pressure. We know from the question that the upper available dBrange is 120. Inorder to arrive at this figure, we must have a test pressure that is 106 times as large as the referencepressure, because 20 log 10" = 120. The correct choice is D.

73. D is correct, force/unit area acting on the oval window is increased. We know the sound wave is conducted throughthe set of small bones in the middle ear. Since the passagedoes not mention any significant loss of force during thistransmission, we can assume the force associated with the wave does not increase or decrease. Well, how about thepressure? Pressure is the force/unit area. The force does not change. However, the area becomes smaller.Therefore, the total force/unit area acting on the oval window is increased. The correct choice is D.

74. Biscorrect, 1x 10"3 s. The question isasking us to carry out some calculations. First, letus convert the 3.4 x 10"4kilometers into meters. This gives us 0.34 m. Based on this value and the speed of sound, we get a frequency of1000 Hz. Recall that the period of a wave is the inverse of its frequency. Therefore, the period of this wave is1/1000 Hz, or 1x 10"3 seconds. The correct choiceis B.

75. C is correct, oscillating depolarizing-hyperpolarizing. The passage tells us that oscillating movements of the organof Corti both excite and inhibit sensory transducing cells, because the hair cells attached to these cells are movingback and forth. In one direction, these hair cells cause cation channels to open. This will cause a depolarization.However, in the other direction, the hair cells cause a closure of these channels. This is associated with ahyperpolarization. Therefore, the cell undergoes an oscillating depolarizing-hyperpolarizing potential. The correctchoice is C.

76. A is correct, region 1. The question is asking which region of the basilar membrane would be associated with high-frequency sounds. High-frequency sounds are associated with short wavelengths and will be more energetic thanlower-frequency sound waves. We can assume that the broad flexible region of the basilar membrane will not beable to resonate with these short energetic movements. In other words, the broad flexible region will not be able toreceive high frequency sound as well. Such movements are achieved by the short, stiff region of the membrane.The correct choice is A.

77. D is correct, cell 5. Let us look at the circuit model. We know that information is arriving at the left ear at a timebefore that of the right ear. We also know that these cells fire maximally when the inputs from both sides arrivesimultaneously. For example, if sound arrived at both ears at the same time, cell 3 would fire maximally, becauseinput would arrive at that neuron simultaneously. Based on that logic, if information is arriving at the left ear beforethe right, it will travel up the entire circuit before the information from the right ear will arrive. If the two inputsarrive at any cell at the same time, that cell must be located in the upper region of the model. Therefore, cell 5 is thebest answer. The correct choice is D.



78. B is correct, middle ear. The passage tells us that conducting sound through the temporal bone is one way to bypassthe middle ear. The passage also tells us that this means of transmitting sound is quite inefficient. Therefore, whenthe fork is placed on the patient's temporal bone and the patient hears the sound (after they no longer heard thesound with the fork at their ear), we can narrow down the problem to conduction of air in the middle ear. Thecorrect choice is B.

79. C is correct, graph C. The Y-axis is amplitude and the X-axis is frequency. If cells have a characteristic frequency,the data should reach only one point on the X-axis. Some graphs show two. Choices A and D can be eliminated.Now that we know that only one characteristic frequency point should be reached, which direction will the curvego? At a cell's best frequency, does the loudness of a sound need to be high or low to elicit a response? It is low.The amplitude of the sound does not need to be high, if we are at the cell's characteristic frequency. However, if wemove away from that frequency in either direction, the amplitude of the sound needed to elicit a response increases.Based on this information, graph C is best. The correct choice is C.

Passage XIII (80 - 87) Tryptophan St Serotonin Experiment

80.

81.

82.

B is correct. Choice A can be eliminated, because a carboxyl group is not being added to serotonin. Choice C canbe eliminated, because amidation is the process of forming an amide (CONH). There are no amides (think of apeptide bond) associated with serotonin. Choice D can be eliminated, because amination involves the addition of anamino (NH2) group to a molecule.

COOH

Tryptophan 5-1lydroxy tryptophanH

Serotonin

Comparing the two molecules shown in the question, we see that a hydroxyl (OH) group has been added to the C-5carbon of the indole ring system, and a carboxyl (COOH) group has been removed from the side chain. As shown inthe reaction sequence above, a hydroxylase enzyme adds a hydroxyl group to make the molecule 5-hydroxytryptophan. A decarboxylase enzyme next removes the carboxyl group to give serotonin, making choice Bthe best answer. The correct choice is B.

C is correct. We are looking for the false answer. Choices A and D appear to be the easiest choices to eliminatefirst. Serum large, neutral amino acid (LNAA) concentration increases in both trials as dietary protein contentincreases. This can be seen by reading the values in the Serum LNAA columns in Table 1 and Table 2. Therefore,choice A is a true statement and can be eliminated as the best answer. Serum tryptophan levels also show anincreasing trend as dietary protein content increases. This can be seen by reading the values in the Scrum Trpcolumns in Table 1 and Table 2. Therefore, choice D is a true statement and can be eliminated as the best answer. In

choice B, a 24% protein meal is given to the rats two hours after they were fed an all-carbohydrate meal. This meanswe are just considering the information in Tabic 1, because that reflects Trial 1. To attenuate means to reduce,weaken, or diminish. The 24% protein meal in Trial 1 docs attenuate the increase in serotonin due to tryptophan. Weknow from the first paragraph that serotonin is synthesized from tryptophan. If the diet is higher in protein (readdown the Group column in Table 1), there will be more competition from LNAAs at the transporters in the blood-brain barrier. If there is more competition at the transporters, there should be less tryptophan crossing the blood-brain barrier. If there is less tryptophan near the neurons (see the column Cortex Trp in Table 1), there will be lesssynthesis of serotonin. In other words, the levels of serotonin are attenuated. Therefore, choice B is true statementand can be eliminated as the best answer. Choice C is false, because there is no indication in the passage that higherlevels of serum LNAAs lead to an increase in the synthesis of serotonin. In fact, based on the last two columns ofeach table, it appears that an increase in LNAAs actually leads to a decrease in serotonin synthesis. The correctchoice is C.

D is correct. The question is asking us to draw a conclusion from the data in Table 1 and Table 2. One aspect ofboth tables that stands out is that the serum LNAAs increase with higher protein diets. If the LNAAs other thantryptophan (e.g., leucine, isoleucine, tyrosine, and phenylalanine) were competing with tryptophan for the LNAAreceptors associated with the blood-brain barrier, then the tryptophan levels in the brain should decrease as serumprotein levels increase. This is especially true for the higher protein diets in both tables. In Table 1, the levels of24% protein and 40% protein cause this competition. In Table 2, the levels of 12% protein, 24% protein, and 40%



protein cause thiscompetition. In bothtrials, thelevel of serum tryptophan increased for the group of laboratory ratsthat fasted overnight. Therefore, because statements I, II, and III are correct, we can eliminate all of the answerchoices except choice D. The correct choice is D.

83. B is correct. In the first paragraph we read that serotonin, a neurotransmitter, is synthesized from tryptophan. If weassume that one neurotransmitter is not synthesized from an existing neurotransmitter, we could conclude thattryptophan is not a neurotransmitter. Also, paragraph two states that all of the LNAAs, including tryptophan, areamino acids. Statement I is eliminated, which means that choices A, C, and D are all eliminated. Support for theelimination of choice D comes from paragraph two where we read that tryptophan is a neutral amino acid, not anacidic amino acid. By the process of elimination, this leaves us with choice B as the best answer. Essential aminoacids are amino acids that cannot be synthesized de novo by human beings. Even though tryptophan is indeed anessential amino acid, there is no indication in the passage as to whether it is an essential amino acid. Tryptophan isfound in most protein-based foods, like red meat, poultry, fish, milk, and eggs. It is also found in chocolate. Thecorrect choice is B.

84. A is correct. If one of the enzymes in a pathway (see below) is blocked by an inhibitor, then some intermediatemetabolite is not processed through to the final product. This means the intermediate form may remain andaccumulate in the tissue.

Tryptophan • 5-Hydroxytryptophan • SerotoninHydroxylase Decarboxylase

If an inhibitor of the tryptophan hydroxylase enzyme were added to rat brain tissue, the intermediate (i.e., 5-hydroxytryptophan) between tryptophan and serotonin accumulates. The factor to be examined in order to estimatethe new rate of serotonin synthesis is not the concentration of the inhibitor itself, either in the blood or the brain, justwhether it is working or not, so choices C and D are incorrect. Exclude choice B, also: The tryptophan concentrationin the blood would not provide any information about the rate of the conversion of tryptophan to serotonin in thebrain. Remember, the LNAAs, including tryptophan, have to cross the blood-brain barrier to reach the brain. Thecorrect choice is A.

85. B is correct. The small intestine, which includes the duodenum, jejunum, and ileum, is where the majority ofdigestion and absorption of food takes place. Whether tryptophan is obtained from the food we eat or in the form ofa dietary supplement, makes no difference in terms of its metabolic effect. Either way, it is easily absorbed in thesmall intestine by a carrier-mediated process, so eliminate choice A. Some common neurotransmitters includeaspartate, glutamate, gamma-aminobutyric acid, acetylcholine, dopamine, norepinephrine, and serotonin, to namebut a few. Even though tryptophan leads to the synthesis of the neurotransmitter serotonin, it does not lead to thesynthesis of the other neurotransmitters just mentioned. For example, acetylcholine is an ester of acetic acid andcholine. It does not involve the indole ring structure like that found in tryptophan and serotonin, so choice C iseliminated.

We are now left with choices B and D. To determine whether choice D is the best answer, we would need to bring insome outside information. About 10% of the human body's serotonin concentration is located in the central nervoussystem (CNS); the reminder can be found in the gastrointestinal system. As a neurotransmitter, serotonin is believedto be involved with a wide variety of body functions. It is thought to be involved with some social behaviors, as wellas memory, learning, mood, sleep, appetite, sexual desire, and temperature regulation. A deficiency in serotonin isthought to be involved in depression, whereas an increase in serotonin above normal levels can lead to anxiety,insomnia, and gastrointestinal disturbances. An excess of serotonin, usually brought about by a drug-inducedaugmentation of the neurotransmitter, includes an elevated body temperature and sweating, nausea, vomiting, arapid heart beat, blood pressure fluctuations, loss of coordination, and hallucinations. This hyperserotonergic state isreferred to as serotonin syndrome and is a potentially life-threatening condition. These conditions do not mimic oneof alertness, so we can rule out choice D as the best answer. A careful reading of the passage does not help us learnwhether dietary tryptophan, when converted to serotonin, causes more alert behavior. However, based on theinformation in paragraph two about the uptake of LNAAs, we could conclude that if a large dietary supplement oftryptophan alone were ingested, without the presence of any other LNAAs, then tryptophan would be the primaryamino acid being transported across the blood-brain barrier. In this case, the brain concentrations of serotonin wouldincrease, making choice B the best answer. The correct choice is B.

86. D is correct. The nine essential amino acids for adult human beings are Valine, Methionine, Histidine, Leucine,Phenylalanine, Threonine, Isoleucine, Lysine, and Tryptophan. A helpful mnemonic is: Very Many Happy LittlePigs Take Iced Lemon Tea. The degradation and catabolism of amino acids occurs as a result of protein turnover.We can assume from the question that the metabolism of phenylalanine leads to the production of one of the amino



acids listed in the answer choices, because if there were a deficiency in phenylalanine in the diet, then one of thoseamino acids would need to be supplemented in the diet. The easiest way to arrive at the best answer is by structuralanalogy between the amino acids. Phenylalanine has a phenyl group (i.e., a ring structure) attached to an alaninecomponent. The only amino acid among the answerchoices that has a ring structure is tyrosine. The correct choiceisD.

87. A is correct. There are four types of cellular junctions in vertebrates: tight junctions, gap junctions, desmosomes,and adherens. Tight junctions help to prevent the passage of molecules and ions between cells. Gap junctionsconnect the cytoplasm of two cells, and allows the passage of selected molecules and ions between those cells.Desmosomes are specialized structures that allow for cell-to-cell adhesion on the lateral sides of plasma membranes.They help resist the shearing forces associated with cells. Adherens appear as actin bands that can encircle a cell,helping to create a strong attachment between cells. Since these junctions are all associated with a variety of organsystems in the body, we need to know a little more about the blood-brain-barrier in order to answer the question.The primary function of the blood-brain barrier is to separate circulating blood from the cerebrospinal fluid of thecentral nervous system. This barrier to the central nervous system is due to the selectivity of the tight junctionsfound between the endothelial cells that make up the capillaries supplying nutrients to the brain. The correctchoice is A.

Passage XIV (B8 - 994) Frog Muscle Expenment

88. C is correct, a greater frequency of the action potentials indicates greater intensity. Action potentials act in an "all-or-none" fashion. There is no amplitude modulation. Choices A and B are incorrect. Identical action potentials areproduced more frequently to indicate a stronger stimulus, such as the stretch caused by a set of increasing weights.The correct choice is C.

89. A is correct, the action potentials had increased to a maximum with the 50-g weight and increased no further, due tothe refractory period of the neuron. The action potentials were limited by the nonconducting refractory period of theneuron. Increasing the weight could not speed up the "recovery time." The nerve was not damaged by the weight.Choice B is incorrect. Do not assume equipment failure when a biological mechanism is being studied. Choice C isincorrect. A neurotransmitter defect would probably mean no action potentials rather than slowed ones, exactlyidentical to another weight. Choice D is incorrect. The correct choice is A.

90. A is correct, The weight was not heavy enough to exceed the action-potential threshold. 5 mg (1/200) of the 1 gramweight caused only a few action potentials. Since there was not an action potential generated, this weight must havebeen too small to overcome the threshold for generating action potentials. The weight was not too heavy, so choiceB is incorrect. The weight does not have to produce an action potential, if its stimulus is lower than the threshold.Choice C is incorrect. All the action potentials were identical in size, so frequency was modulated, not theamplitude. Choice D is incorrect. The correct choice is A.

91. D is correct, plasma membrane. The cytoplasm of a neuron is a poor conductor. Choice A is incorrect. TheSchwann cell acts to insulate nerve fibers and to provide gaps between the conducting regions. Choice B isincorrect. The endoplasmic reticulum does not function as a conductor. Choice C is incorrect. The plasmamembrane is the site of conduction of action potentials. The correct choice is D.

92. B is correct, sodium moves from the extracellular fluid into the cell, and then potassium moves from theintracellular fluid to the outside of the cell. This is an easy question, but all the words make it easy to make amistake. First, eliminate wrong answers based on ion location. Sodium is the principle extracellular ion, whilepotassium is the principle intracellular ion. Choices C and D are incorrect based on this fact. Then eliminatechoices based on order of action. The sodium channels open first, followed by the potassium channels. Choice A isincorrect. The correct choice is B.

93. B is correct, afferent function. Afferent pathways carry information from the tissues to the nervous system.Information of muscle stretch was being conveyed back toward the (no-longer-present) nervous system. Efferentfunction would be tested, if a nerve were stimulated with electrodes to make a muscle twitch. Choice B is incorrect.Since the nerve and muscle under study are isolated from the intact frog, there is no parasympathetic or sympatheticfunction. The correct choice is B.

94. C is correct, action potentials continue briefly but stop eventually, due to a lack of ATP. The Na®/K® pumprequires ATP to perform active transport. The cell has a small reserve of ATP that would provide brief function ofthe pump if ATP production stopped. Choice A is incorrect. Action potentials stop when all the ATP is gone.Choice B is incorrect. Action potentials require ATP, they do not generate it. Choice D is incorrect. The correctchoice is C.



Passage XV (95 -100) Skeletal Muscle Groups

95. A is correct, quadriceps. According to the passage, the quadriceps group has at least one muscle that crosses the hipjoint anteriorly. It attaches above the hip joint at the anterior inferior iliac spine and below the hip joint at the tibialtuberosity. Thus, as the muscle shortens it will aid in flexion of the hip by pulling the femur and tibia upwardstowards the chest. Refer to the skeleton diagram and try to visualize exactly how this works. The other answerchoices can be eliminated after consideration. The hamstrings cross the hip joint behind the femur. When theyshorten, they would pull the femur backwards. This is hip extension, not flexion; eliminate choice B. Choice D canalso be ruled out on the basis of this information. The adductors mainly insert along the medial and posterior shaft ofthe femur. When they contract, they pull the thigh inwards. This is adduction; eliminate choice C. The correctchoice is A.

96. C is correct, the veins involved must have valves. How can blood be returned from the lower extremity againstgravity and under the low pressure that is normally found in the venous system? Muscles of the thigh and lower leg"massage" the elastic veins, pumping blood in one direction: upwards. Blood can only go in this direction becauseveins in the lower extremity have one-way valves which prevent backflow of blood away from the heart. These arevery important in maintaining proper circulation. Varicose veins result from defects in these valves. Let's considerthe other answer choices. Eliminate choices B and D because arteries do not return blood to the heart, veins do.

Eliminate choice A because if veins had inelastic walls, they couldn't be squeezed by muscles and the process wouldnot work. The correct choice is C.

97. B is correct, the quadriceps must relax. The quadriceps and the hamstrings are opposing muscle groups withopposite actions. The hamstrings extend the hip posteriorly while the quadriceps flex the hip anteriorly. In order forone of these antagonistic muscle groups to work, the other must relax. If both contract simultaneously, nothingwould get accomplished because the two actions would cancel each other out. Knowing these should lead us toeliminate choices B and C. Choice D can be ruled out because the obturator nerve, according to the passage,innervates the adductor muscles, which aren't involved here. The correct choice is B.

98. C is correct, to kick a ball. The passage states that the femoral nerve innervates the quadriceps muscles. Thequadriceps are responsible for flexing the hip anteriorly (see prior question) and also for extending the knee. In otherwords, since the quadriceps crosses the knee joint, contraction pulls the lower leg and foot anteriorly (a kickingaction, known as extension of the knee joint). Paralysis of the femoral nerve might therefore affect the ability to kicka ball, among other things. Let's consider the other answer choices. Choice A can be eliminated because thequadriceps muscle doesn't cross the ankle joint and hence can't plantar flex the foot (i.e., cause a person to stand ontip-toes). Choice B can be ruled out because extending the hip posteriorly is the job of the hamstrings, not thequadriceps. Choice D can be eliminated because the adductor muscles aren't innervated by the femoral nerve; theobturator innervates them. The correct choice is C.

99. D is correct, weakness in the adductor muscles. What would happen if an intervertebral disk herniated (squeezedout of its sheath) and smashed the adjacent sciatic nerve? From the passage, we learn that the sciatic nerveinnervates the hamstring muscles. These muscles are normally responsible for extending the hip posteriorly andflexing the knee (i.e., bringing the heel upwards posteriorly). These actions of the hamstring group can be explainedby looking at where they attach. The hamstrings attach to the ischium posteriorly, cross the hip joint, and attachinteriorly to the femur, tibia, and fibula. They therefore cross both the hip joint and the knee joint. Shortening of thehamstrings results in flexion of the knee and extension of the hip posteriorly. Since the hamstrings are innervated bythe sciatic nerve, damage to this nerve would affect walking (i.e., hip extension), difficulty moving the legposteriorly (knee flexion), and sensory stimulus along the back of the thigh and lower leg (which manifests as pain).The combined conditions are known as sciatica. Choice D is our answer because the adductor muscles are not

innervated by the sciatic nerve. According to the passage, they are innervated by the obturator nerve. We would notexpect them to be affected by damage to the sciatic. The correct choice is D.

100. C is correct, lower-than-normal CCb levels. What effect would reduced blood flow have on exercising muscle

cells? To start, Cb (which would be in high demand in active muscle) would not reach muscle fast enough due topoor circulation. This would lead to a decline in aerobic respiration (oxidative phosphorylation, etc.) resulting inATP deprivation. In order to partially compensate, anaerobic respiration (glycolysis and lactic acid fermentation)would be stepped up. resulting in a buildup of the by-product lactic acid (possibly causing severe pain). CCb levelswould actually rise because it could not be carried away effectively by the blood. The correct choice is C.


BiologySection II

Heart

and

Lung

A. The Heart

1. Cardiovascular Anatomy2. Cardiac Output3. Poiseuille's Law

4. Diffusion

5. Osmosis

6. Lymphatics7. Blood Clotting

B. The Lung

1. Gases

2. Gas Exchange


f?EBKELEYSpecializing in MCAT Preparation

Heart and LungTop 10 Section Goals

.^ Be familiar with the basics of cardiac anatomy.

Know how to describe the heart in the anatomical position, and understand its relationship to theother major organs of the body.

Qlftfc Be able to trace the path of blood flow through the heart.

? Starting with thevenacava, bloodflows intotherightatrium, to the rightventricle, to the rightandleft lungs, to the left atrium, to the leftventricle, and then out to the systemiccirculation.

Q&* Understand how blood flows through the systemic circulation.jar Blood leaving the heart passes through the aorta, toarteries, arterioles, capillaries, venules, and backJ to the veins beforeentering the heart againby way of the vena cavas.

O^i Be familiar withthe electrical activity of the heart.]Br Know the actions of the sympathetic and parasympathetic nervous systems on the heart. Be able to* trace thewaves ofelectrical activity through theheart, andknowhow theyrelate toheart function.

©'fm Understand the cardiac cycle in relation to pressure and volume curves.jfir Understand the relationship between the opening and closing of the valves of the heart and how

•» theyrespondto pressure and volume changes.

®%% Have a basic understanding of the lymphaticsystem.Understand the relationship ofthelymphatic system to thesystemic andpulmonary circulations.Befamiliar with the conceptof edema.

o*m Bef;^5r Have

J infra

amiliar with the basics of respiratory anatomy.Have amental picture of how the thoracic cage, diaphragm, visceral pleura, parietal pleura, andintrapleural space relate to one another. Review the mechanics of inspiration and expiration.

(*)-£* Understand the processof gas exchange at the level of the alveoli.]5r Have a general feel for the differentpartial pressures ofO2 and CO2, both at the level ofthe lung* and tissues. Be familiar withthegradients orexchange foreach ofthese gases.

(§) -§% Understand the mechanics behind the oxygen-hemoglobin saturation curve.Know why the siemoidal saturation curve shifts to therightor to the left. Be ableto relate thisshiftto theeffects ofpH,[H®], pO^ PCO2, temperature, and[2,3-BPG].

Know the basis of the Bohr effect and how this relates to ventilation.

This is tied into the concept ofmetabolic acidosis, metabolic alkalosis, respiratory acidosis, andrespiratory alkalosis. Beable to read and interpret graphs that describethese events.

Biology Heart St Lungs

The HeartCardiovascular Anatomy

Within living creatures there are two types of circulatory systems. In an opencirculatory system the circulating fluid within the body is mixed with the bodyfluid itself. In other words, the blood in these creatures does not circulate entirelywithin the confines of vessels like arteries and veins. Most mollusks and

arthropods have an open circulatory system. Conversely, in a closed circulatorysystem the blood that flows throughout the body is confined to vessels likearteries and veins. It does not freely mix with the fluid of the body. Annelids andmammals like ourselves have a closed circulatory system.

The function of the circulatory system is to bring nutrients and oxygen to thetissues of the body while simultaneously removing waste products from thosevery tissues. Because the circulatory system is continually flowing it helps tomaintain body temperature. Also, the circulatory system can act as a means totransport hormones to various locations within the body. Ultimately, though, asthe blood enters into the smallest vessels, the capillaries, there will be diffusionbetween those capillaries and the cells in the immediate environment.

General Anatomical FeaturesLet's examine the general anatomical features of the mammalian cardiovascularsystem. The foundation of the circulatory system is the heart. The heart can bethought of as being divided into two halves. These two halves are often referredto as the right heart and the left heart. When we look at a picture of a heart wemust remember that we are viewing that heart in its anatomical position. In theanatomical position the body is standing upright, the arms to the sides, handswith plams up, and the face positioned forward. Thus, when we look at adiagram of a heart on paper, the right side of that paper really represents the leftside of the heart and vice versa.

The right heart pumps blood to the lungs and back to the left heart. The left heartpumps that blood to the remaining tissues of the body and back to the rightheart. The blood that is pumped from the right heart to the lungs and back to theleft heart is called the pulmonary circulation while the blood that is pumpedfrom the left heart to the rest of the tissues and back to the right heart is called thesystemic circulation. Both the pulmonary and systemic circulations lie in serieswith one another.

We can pick a starting point in the circulatory system and follow a red blood cellas it migrates through the pulmonary and systemic circulations. Let's start with ared blood cell in the right heart. The right heart is composed of the right atriumand right ventricle. As deoxygenated blood passes from the right atrium into theright ventricle it is pumped into the pulmonary artery and to the lungs where itis oxygenated. The oxygenated blood returns to the left heart by way of the (leftand right) pulmonary veins. The blood enters the left atrium and passes into theleft ventricle where it is pumped out the aorta and to the branching arteries,arterioles, and capillaries. It is at the level of the capillaries that the bloodexchanges nutrients and oxygen for waste products created by metabolism.Deoxygenated blood passes from the capillaries to venules and then to largerveins and eventually to the superior and inferior vena cava which enter the right

Cardiovascular Anatomy


Biology Heart St Lungs Cardiovascular Anatomy

atrium. The circulation of our red blood cell can now start again. As you arestudying this circuit it is important to note that arteries carry blood away fromthe heart while veins carry blood from the tissues and lungs towards the heart.This can be seen in the diagram shown in Figure 2-1.

cs

Superior VenaCava

Inferior Vena

Cava

Capillaries of Head and Arms

Capillariesof Abdomen and Legs

PulmonaryArtery

PulmonaryVein

Left

Atrium

Left

Ventricle

Figure 2-1Blood flow through the cardiovascular system.

Aorta and ArteriesAs shown in Figure 2-1, blood leaves the left ventricle of the heart and exitsthrough the aorta and passes to both the superior and inferior portions of thebody. The total cross-sectional area of the aorta is about 5 cm2. Under restingconditions the velocity of blood flow in the aorta is about 30 cm/sec. The bloodpressure fluctuates between 120 and 80 mmHg with an average being about 100mmHg. The volume of blood found in the aorta and arteries at any given time isabout 16% to 20%.

As the left ventricle contracts (systole) it propels blood out the aorta and into thearteries with a pressure of about 120 mmHg. At sea level this force is enough toraise a column of mercury 120 mm above the ground. After contraction takes


Biology Heart & Lungs

placeand the ventricles begin to relax (diastole) and fill with blood, the pressurein the arteries is about 80 mmHg.

Measurement of Blood Pressure: When you have your blood pressure taken apressure cuff is placed around your upper arm and a stethoscope is placed overthe antecubital artery (the artery at the bend between your upper and lowerarm). The cuff is inflated above arterial systolic pressure. This causes the artery inyour arm to collapse thus stopping the flow of blood. The pressure in the cuff isdecreased and the pressure needle monitored. As soon as the pressure in the cuffis below the systolic pressure, blood will jet through the small opening in theartery. The flow through the artery is turbulent because of the great pressurepushing the blood through such a narrow opening. The "tapping" sounds that areheard in the stethoscope approximately correspond to the systolic pressure in theheart. The cuff is continually deflated and the artery slowly regains its originalshape. The blood has an easier time passing through the vessel and the flowbegins to change from turbulent to laminar (smooth). Whenthe turbulent soundsdisappear it means that you are recording the diastolic pressure of the heart.Blood pressure readings are given as systolic/diastolic (e.g., 120 mmHg/80mmHg). These are the approximate readings that the baroreceptors are sensingas they monitor your blood pressure.

The blood that moves through the arteries is under a lot of pressure. Therefore,those arteries have to be rather durable. They have thick walls and are composedof smooth muscle and connective tissue that contain both collagenous andelastic fibers. The elasticity of the arteries prevents the blood pressure frombecoming too high when it is ejected out of the heart and it alsomaintains a higharterial pressure between the systolic and diastolic phases of the heart. Thisallows blood to flow to the rest of the circulatory system without a sudden loss ofpressure.

The lumen of all the blood vessels in the body are lined with epithelial cells andbecause these cells are inside the cardiovascular system they are referred to asendothelial cells. Damage to these endothelial cells by the pulsating arterialpressure or even by abrasive substances in the blood can lead to the diseaseknown as atherosclerosis. Once these cells are damaged cholesterol can build upat the site of the lesion and a plaque will form. During the later stages of thedisease the arteries become "hardened" from layer upon layer of deposit. This isreferred to as "hardening of the arteries"or arteriosclerosis.

Regulation of the circulatory system is controlled (in part) by the sympatheticand parasympathetic divisions of the autonomic nervous system. Thesympathetic division is the more important ofthe two. Besides nervous control ofblood flow there is also humoral control from the action of ions or hormones andlocal control at the level of the individual tissues from various metabolites.

How are metabolites important? If you were to start exercising a particularmuscle, you would notice that the blood flow to that muscle would increase.Recall that when muscles contract ATP is hydrolyzed to ADP and Pi, lacticacidis formed, and other such metabolites circulate in the tissue. When thesemetabolites diffuse from the tissue out to the arterioles the smooth muscle dilatesand increases the flow of blood to that particular area. The blood brings not onlynutrients and oxygen to the working tissue, but it carries away the wasteproducts of metabolism as well.



Biology Heart St Lungs Cardiovascular Anatomy

ArteriolesAfter the aorta branches to form a variety of arteries, the arteries themselvesbranch to form arterioles. The arterioles are important because they represent themajor area of resistance in the cardiovascular system. Arterioles have strongmuscular walls and are regulated as mentioned above.

CapillariesDiffusion takes place at the level of the capillaries. The total cross sectional areaof thecapillaries is about 3000 cm2. The velocity ofblood flow has beenreducedto about 2 cm/sec. The pressure is roughly 25 mmHg and the total volume ofblood is about 5% to 7%.The wall of the capillaries are composed of a unicellularlayer of endothelial cells. Surrounding these cells is a basement membrane.However, there is no connective tissue or smooth muscle. The capillary itself isjust large enough for a red blood cell to squeeze through. At the entrance to thecapillary bed is a precapillary sphincter composed of smooth muscle whichhelps to regulates the flow of blood to the area.

VeinsOnce the blood flows through the capillaries and reaches the veins the bloodpressure has been reduced to a value between 0 mmHg and 25 mmHg. Thevelocity of blood flow through the veins is about 20 cm/sec and the total crosssectional area isroughly 7cm2. Roughly 50% ofthe total blood volume ispresentin the veins at any given time.

Figure 2-2A valve in a vein showing directionality.

Since there is not much pressure in the veins the amount of smooth muscle andelastic tissue surrounding the veins is reduced. However, they are under controlof the sympathetic nerves. Note that the rate of blood flow in the veins is similarto that in the arteries. This is due to specialized valves that allow the blood toflow in only one direction (see Figure 2-2). As the muscles that surround theveins in your body contract they squeeze the blood back towards the heart. Sincethese valves are "one-way valves" blood is prevented from flowing backward. Ifthe valves become damaged and blood is allowed to flow backwards, pressure inthe veins can increase. This has a tendency to cause varicose veins, which areprotrusions of the dilated veins beneath the skin.

Blood Flow in the HeartWe have mentioned that deoxygenated blood returns from the tissues and entersthe right atrium via the superior and inferior vena cava. As the right ventriclebegins to relax blood from the right atrium is pumped into the right ventricle.The right ventricle contracts and forces blood out the pulmonary artery and tothe lungs where it is oxygenated. Blood returning to the heart enters the leftatrium via the right and left pulmonary veins. As the left ventricle begins to relax



blood from the left atrium is pumped into the left ventricle. Contraction forcesthe blood in the left ventricle out the aorta and to the tissues of the body. SeeFigure 2-3.

Pulmonary i!*^««•••Artery ^r3!iM


PulmonaryArtery

PulmonaryVeins 1"I

SuperiorVena Cava g^

Pulmonary Valves(semilunar) -y

RightAtrium

R. Atrioventricular

Valves (tricuspid)

Inferior

Vena Cava

PulmonaryVeins

Left

Atrium

Aortic Valves

(semilunar)

L. Atrioventricular

Valves (mitral)

Left

Ventricle

Bundle of Hisgives rise to

L. fir R.

Bundle Branches

(further divides intoChordae tendineae i/am(m«Ia ~ ... >i irunner aiviaes inVentricle Descending Papillary purkmje Fibers)Aorta

Muscle

Figure 2-3Anatomical landmarks of the heart.

Within the heart there are valves between the right atrium and right ventricle(the right atrioventricular valve), between theright ventricle and thepulmonaryartery (the pulmonary valve, or tricuspid), between the left atrium and the leftventricle (theleft atrioventricular valve, or mitral), and between the leftventricleand the aorta (theaorticvalve). Oncethe ventricles are filled withbloodand theybegin to contract the valves between the atria and the ventricles close. Thisprevents any backflow ofblood into the atria and ensures that the blood will bedirected to the pulmonary and systemic systems of the body. The closing of theatrioventricular valves between the atria and the ventricles as the ventricles arecontracting gives the characteristic "lub" sound when listening to theheart. Astheblood flows out of the pulmonary arteryand theaorta, the pulmonary valveand the aortic valve close, giving the characteristic "dub" sound. Theclosing ofthese valves as blood is pumped either from theatria intothe ventricles or fromtheventricles to thepulmonary or systemic tissues prevents backflow intoeither


Biology Heart & Lungs Cardiovascular Anatomy

the atria or the ventricles, respectively. The valves between the atria and theventricles themselves do not invert because of tendinous cords, called chordaetendineae, which hold them in place.

Electrical Activity of the HeartLocated near the junction of the superior vena cava and the right atrium is aspecialized region of myocardium called the sinoatrial node (SA node) or thepacemaker of the heart (Figure 2-4).

Figure 2-4The S A and A V nodes of the heart.

The SA node is the point of origin for the electrical impulse that propagatesthrough the rest of the heart. This electrical impulse spreads out over the atriacausing them to contract and fill the ventricles. Located in the lower portion ofthe right atrium and near the right ventricle is the atrioventricular node (AVnode). Impulses from the SA node also spread to the AV node and then from theAV node through a collection of fibers called the bundle of His. Branches of thebundle of His surround the ventricles, and when this bundle receives animpulses it causes the ventricles to contract and eject blood to the pulmonary andsystemic systems. Anatomically speaking, ventricular contraction is from theapex of the heart towards the base of the heart. It is interesting to note that if theSA node is damaged, the AV node takes over and slows the heart down to about40 beats per minute.

Cardiovascular ControlThe average blood pressure leaving the aorta is about 100 mmHg. This bloodpressure is monitored by baroreceptors and chemoreceptors located in thecarotid arteries and in the aortic arch. The baroreceptors (pressure receptors) arecontinually monitoring how much the aortic and carotid arteries are expanding



and contracting.Suppose we were to decrease the arterial pressure. There wouldbe less stretch on the aortic and carotid arteries. The baroreceptors would sensethis and send impulses to the medulla in the brain stem. The medulla respondsbyactivating thesympathetic nerves of the autonomic nervous system. Impulsesare sent via the sympathetic nerves and norepinephrine is released at the SAnode to increase the rate of the heart. Thisactionhelps to increasethe contractionof the heart, which in turn will elevate the blood pressure. Sympathetic nervefibers will also stimulate the adrenal medulla to release epinephrine (adrenaline)into the blood. This hormone acts to increase the rate of the heart, thereforeincreasing the contraction of the heart. Both of these actions act in a negativefeedback manner to "negate" the initial loss of pressure due to, say,hemorrhaging.

The sympathetic nerves will also cause constriction of the blood vessels that leadto the gastrointestinal system and the kidneys. During hemorrhaging the brainand heart receive firstpriority in termsofblood. Therestof the organsystems ofthebody see a decreasein the flow of blooduntil the problemis corrected.



Biology Heart St Lungs Cardiac Output

Cardiac Output

Every time your heart beats, a certain volume of blood is pushed out into thecirculatory system. The cardiac outputis thatamount ofblood which is pumpedper minute by each of the two individual ventricles of the heart. We candefinethe cardiac output (for either ventricle in liters/minute) as being equal to theheart rate (in beats/minute) times the stroke volume (in liters/beat). This isshownin equation (2-1). The stroke volume is simply the amountofbloodejectedby eachventricle during onebeatof the heart.

If the average heart rate is 72 beatsper minuteand the average strokevolume is70 milliliters (or 0.07 liters) per beat, then the cardiac output would be about 5liters perminute. This value is for theaverage resting adult male. Aswe will seelater, the volume of the cardiacoutput can be influenced by the diameter of thebloodvessels in the periphery, the amountof blood returned to the heart by thesuperior and inferior vena cava (i.e., return of the venous blood), and the heartrate and force of ventricular contraction.

Cardiac Output = (Stroke Volume) (Heart Rate)(2-1)



Poiseuille's Law

Not all of the arteries and veins in the circulatory system have the same diameter.Jean Poiseuille, a French physician studying the flow of blood in blood vessels,established a relationship between the radius and length of a tube, the change inpressure between the two ends of the tube, the viscosity, and the flow rate of afluid in that tube. This relationship, known as Poiseuille's law, is given inequation (2-2).

Flow =AP £Ri= (P, - P2 )££i8ilL to\L (2-2)

In this equation (2-2), AP is the pressure drop between the two ends of the tube(i.e., Pi - P2), R is the radius of the tube, eta (r|) is the coefficient of viscosity, L isthe length of the tube, and rc/8 is a proportionality constant adjusting for thecross-sectional area of the tube. What can we say about this equation?

(a) Notice that the flow rate is proportional to R4. This tells us that the rate ofblood flow is extremely dependent on the radius of the vessel. If the radiusof the vessel were reduced by a factor of 2, then the rate of blood flow wouldbe reduced by a factor of 16. Similarly, if the radius of the vessel wereincreased by a factor of 1.5, then the flow rate would increase by a factor of5.1.

(b) The flow rate is also inversely proportional to the length of the vessel. Inother words, the longer the vessel, the slower the rate of flow. The shorterthe vessel, the faster the rate of flow.

(c) The flow rate is also inversely proportional to the viscosity of the solution.This tells us that a high viscosity gives a low flow rate.

(d) The value of APis provided for by the strength of the heart's contraction. Inother words, the difference in the pressure is what drives the blood in thecardiovascular system.

Poiseuille's Law


Biology Heart St Lungs Diffusion

Diffusion is simply the process by which molecules randomly move from oneplace to the next. The molecular weightofa molecule and the temperature of themedium have a lot to do with the velocity at which a molecule moves. Smallermolecules tend to move faster than larger molecules. Similarly, a highertemperature provides more energy to a system and therefore imparts morethermal motion to the various molecules in that system. You might think thatbecause smaller molecules move relatively fast they would have no problemtraversing their environment. However, don't forget that there are millions andmillions of molecules within a given system and that each of those molecules,even thought they are moving at a given velocity, collide with their neighbors.These collisions tend to alter the path of the molecules, thus confining them to arandom walk through their medium.

Consider an imaginary sphere of water with a given radius. Suppose we placesome dye outside this sphere of water and ask how long it will take to reach thecenter. The answer clearly depends on the radius of that sphere. If the radiuswere 7 microns, it would take the dye about 5.4 seconds to reach the center.However, if the radius were 1 centimeter, it would take about 11,000 seconds or alittle more than 3 hours to reach the center. [As a reference a red blood cell isabout 7 microns in length. There are about 1,400 red blood cells end to end in 1centimeter.] What this is telling us is that simple diffusion is a rather poor wayfor a molecule to trek across long distances.

The law that governs diffusionis given in equation (2-3) where J is the net flux ornet rate of diffusion (in molesper unit time, usually in seconds; that is, mol/sec),D is the proportionality constantcalled the diffusioncoefficient, A is the area oftheplane ofinterest (in cm2), and AC/Ax is theconcentration gradient across thatplane (in mol/cm4, because concentration isinunits ofmol/cm3 and distance isinunits of cm). Since the net flux always proceeds down a concentration gradient,from a high concentration to a low concentration,we need to add a minus sign infront of this equation. [The minus sign indicates the direction of the flux ordiffusion.] Thisequation is sometimesreferred to as Fick's law (after the Germanphysiologist who postulated it in the 19th century). In equation (2-3), what arethe units of the diffusion coefficient, D?

J = - (D)(A) AC _ (D)(A)(Cout - Cin)Ax Ax (2-3)

Copyright© by The BerkeleyReview 92 The Berkeley ReviewSpecializing in MCAT Preparation


Osmosis

In order to understand hydrostatic pressure we must first review osmosis. Recallthat osmosis is simply the net movement of water from a region of highconcentrationto a region of low concentration. Generalchemistrytellsus that theconcentration of pure water is 55.5 moles/liter. If we had a beakerof pure waterand we added a solute to that beaker (say glucose), then we would decrease theconcentration of the water in the beaker. Remember, when we add a solutemolecule (or molecules) we are occupying a volume of space that was onceoccupied by a water molecule. The more solute molecules we add to our beakerof pure water, the more water molecules we will displace and the lower will bethe concentration of pure water.

Glucose, when added to a solution of pure water, does not ionize. It stays asglucose. However, ifweadd a molecule ofsodium chloride toa solution ofpurewater, it will ionize into a Na® ion and a Cle ion. Because we now have added asodium chloride molecule which has dissociated into two ions in solution, wehave displaced (lowered the concentration of) the water molecules twice as muchas we would if we had added a glucose molecule. What this is telling us is thatthe concentration of water in a given solution depends on the number of soluteparticles (e.g., glucose, Na®, or Cle) in that solution. We can define the totalsolute concentration in our solution as the osmolality, where one osmol issimply one mole of a molecule that does not ionize. If we had a IM solution ofglucose, it would have a concentration of 1 osmol per liter. If we have a 1Msolution of sodium chloride, we would find that it would have a concentration of2 osmolsperliter (one from the Na® ionand onefrom theCle ion). Ifwe have a150mM concentration of NaCl, then after ionization we would have 150 mM ofNa® ions and 150 mM of Cle ions ora total of 300 milliosmols (mosmols) perliter. Therefore, the osmolarity refers to the concentration ofsolute particles thatwe have in our solution.

(a) (b)

> iJih # •

• ••

• •• ••

H2 i2o H2(1 ^ • • •• ••°<_ —>- 1 • •

•• • •

Permeable Membrane Semipermeable Membrane

Figure 2-5Water movement across a permeable and semipermeable membrane.

Suppose we have a U-tube with water and a permeable membrane as shown inFigure 2-5a. Water is free to pass back and forth across this membrane andbecause of this the height of the water in each of the columns of the U-tube willbe the same. However, suppose we now replace the permeable membrane with asemipermeable membrane and then add some protein to the right side of the U-tube as shown in Figure 2-5b. What willhappen? The level of water in the right

Osmosis


Biology Heart St Lungs Osmosis

side of the tube will rise and the level of water in the left side of the tube willdrop. Why?

In order to understand this we must examine the semipermeable membrane andthe solutes which we have added to the right side of the tube in a little moredetail. First, the concentration of solute on the right side of the tube is greaterthan the concentration of the solute on the left side of the tube. This established asolute concentration gradientin which the solutes on the right side of the U-tubewant to diffuse down their concentration gradient to the solution on the left sideof the U-tube. This cannot happen, because we have said that the membrane issemipermeable. In other words, the membrane will not allow these solutes topass through only water. What aboub the concentration of the water? Theconcentration of the water on the left side of the U-tube is greater than theconcentration of water on the right side of the U-tube. Again, a concentrationgradient has been established that will allow water todiffuse from theleft side ofthe U-tube to the right side of the U-tube. Since the membrane is permeable towater we find that water diffuses down its concentration gradient (from left toright). This leads toan increase in the volume in therightsideof theU-tube anda decrease in the volume in the left side of the U-tube. The effect is to decreasethe solute concentration in the right side of the U-tube. An equilibrium willeventually be established when the concentrations of both the water and thesoluteare equalon bothsidesof the semipermeable membrane.

Once equilibrium isreached nomore water will flow from theleft side ofthe U-tube to theright side of the U-tube. This is because thepressure has increased inthe right side of the U-tube (because there is now a larger volume of solutionpushing onthe semipermeable membrane). The amount ofpressure thatstoppedosmosis is referred to as the osmotic pressure (abbreviated as rtosm). A directmeasure of theosmotic pressure is thedifference in the levels of waterin the leftandright sides ofthe U-tube. This difference, Ah, is referred toas thehydrostaticpressure (orfluid pressure) which canbe abbreviated as PH2O.

The osmotic pressure is proportional to thenumber of dissolved molecules in asolution and isrepresented in Figure 2-6. As we increase theconcentration of theprotein in solution (i.e., solute in solution) we find that the osmotic pressureincreases as well.

<uh-

3t/3

<D '"i,

0 J>•s KO "-'

ECfl

O

[Protein]

Figure 2-6Relationship between osmotic pressure and dissolved particles.

Now, let's return tothe hydrostatic pressure generated bytheheart which forcesfluid out of the capillaries and into the interstitial space. Since the hydrostaticpressure in the capillaries turns out to be abit greater than the osmotic "pullingpressure" of the solutes in the blood, there is a net movement of fluid from thecapillaries to the interstitial space and eventually into the lymphatic capillaries.


Biology Heart & Lungs

Lymphatic System

The lymphatic system lies parallel to the systemic and pulmonary circulations.This can be seen in Figure 2-7. The lymphatic system collects the excess fluid(about 4 liters per day) that leaks into the interstitial space from the capillariesand returns it by way of the vena cava back to the circulatory system. Lymphnodes locatedalong the lymphatic system help to filter out foreign particles thatcouldpotentiallylead to disease. If the lymph flow through the lymphatic systemwere blocked, edema would result. This is simply an increase in the interstitialfluid (because it cannot be reabsorbed by the lymphatic system). Patients whohave heart surgery usually have swollen legs and ankles. This is because theheart cannot pump the blood out fast enough, and as a result blood within theheartbegins to back up. This translates to a back-up in theveins and eventuallyin the lymphatic system. Since gravity pools fluid toward the lower extremities,edema results in the legs and ankles.

Valve IZ^

Lymphnode

Figure 2-7The lymphatic system.

^D Lymphatic capillaries

Pulmonarycapillaries

Artery

Systemiccapillaries

^3 Lymphatic capillaries

Lymphatic System


Biology Heart St Lungs Blood Clotting

Blood Clotting

Blood clotting occurs via a cascade process. Thrombin, which is involved in bloodclotting, is a serine protease. We can achieve amplification of a very weak signalby a cascade process. If we consider blood clotting, we will find that there is anintrinsic route (due to contact with some abnormal surface) and an extrinsicroute (due to trauma to the tissue), both shown in Figure 2-8.

In the outline shown in Figure 2-8 we will use the Roman numerals to representthe clotting factors. The subscript "a" means that we are dealing with the activeform of the molecule. Again, this would be some type of conversion of aproenzyme to an enzyme. Factor IX will be converted to Factor IXa by someintrinsicfactor. Factor IXa will be the trigger which will convert Factor X toFactor Xa. From the extrinsic portion of this scheme we start with some tissuefactor which will convert Factor VII to the active form, Factor Vila. Factor Vilacan also bring about the conversion of FactorXto FactorXa. It is FactorXawhichis involved in the conversion of prothrombin (II) to thrombin (Ha). Thrombin (Ila)will convertfibrinogen (I) tofibrin (la). Thesefibrin fibers are then crossed-linkedby FactorXHIa (which is an enzymecalled transglutaminase) to form the maturecross-linked fibrin clot.

Some of the serine proteases mentioned in Figure 2-8 are Factors VII, Vila, Xa,II, and Ila. In the blood clotting process there are more than 15 different factorsinvolved, and about 8 or so of them are serine proteases.

Intrinsic Pathway(Damaged Surface)

Factor IX ' > FactorIXa

FactorX ' > Factor Xa <

VIIL

Prothrombin (II) t

Figure 2-8The intrinsic and extrinsic pathways.

Extrinsic Pathway(Trauma)

FactorVIIa <=Jl

=> Factor VII

^ Factor X

> Thrombin (IIa)

uFibrinogen (I) i > Fibrin (Ia)

JI XIIIa (transglutaminase)

Crossed-linked Fibrin Clot

A lot is known about the biochemistry of blood clotting. In the conversion ofFactor Xto Factor Xa by Factor IXa, we find that we need an additional factorcalled Factor Villa. Individuals lacking Factor Villa havehemophilia (asex-linkedrecessive characteristic). Factor Villa is sometimes called the antihemophiliacfactor.


Biology Heart 8e Lungs

0 HO Carboxylase enzyme 0 HO

II 1 IIC—N — C—C—

1 1H CH2 a

|

requiring Vitamin K

and HCO3-.II 1 IIC— N — C— C

1 1H CH2

1

• /

CH2 P

Cv YY-Carboxyglutamate

residue

1C-H

o=c' Nc-01 1

0° °©Preprothrombin fragment +Ca+

Prothrombin fragmentchelated with calcium.

Figure 2-9Chelating action.

Vitamin K is one of the fat-soluble vitamins that is found in green leafyvegetables. In addition, our intestinal flora can make a form of Vitamin K. Whatis the importance of this vitamin? Prothrombin exists in even an earlier formcalled preprothrombin. In preprothrombin there are certain Glu residues whichare carboxylated by a carboxylase enzyme. This carboxylase enzyme has anabsolute requirement for Vitamin K. The carboxylase enzyme adds anothercarboxyl group to the gamma position of the first ten Glu residues located in theamino terminal region of preprothrombin. Thus, in the presence of Vitamin K,HCO3 and the carboxylase enzyme, we willbe able toconvert preprothrombin toprothrombin. This structure now has a great affinity (a good chelating agent) fordivalent ions such as Ca2®. This is shown in Figure 2-9. (Calcium ions areessential for blood clotting.)

A section of prothrombin

Platelet Membrane

following injury.

After cut is made,thrombin is released.

A

Figure 2-10Binding of prothrombin to a membrane.

Prothrombin is next converted to thrombin in the presence of Factor Xa. Theconversionof prothrombin to thrombin in the presence of FactorXacan be picturedas shown in Figure 2-10. Blood platelets have phospholipid molecules in theirmembrane. The head of the phospholipid is negatively charged. This will allow

Blood Clotting


Biology Heart St Lungs Blood Clotting

the y-carboxyglutamate residues ofprothrombin tobind via Ca2®. Factor Xa alsohas y-carboxyglutamate residues on it, and can also bind to the membrane viainteraction withCa2e. The enzyme Factor Xa willmake cuts (atArg-X residues)in the prothrombin molecule as shown. The portion of the prothrombin moleculethat is cut away is called thrombin. Thrombin will drift away and convertfibrinogen into fibrin in the vicinity of the damaged area. It is fibrin that willform the blood clot. There is also an auxiliary factor, Factor Va, which isinvolved in this process. So, what you need for this clotting process to take placeis the (a) platelet membrane, (b) enzyme, (c) Ca2® ions, (d) an auxiliary factor,and (e) the substrate prothrombin.

Thrombin can now act as a proteolytic enzyme that converts fibrinogen to fibrin.Fibrinogen is a large soluble protein. Its solubility is due to an excess ofnegatively charged amino acids (Glu, Asp, Tyr-SC^), particularly in the centraldomain of the molecule. The net charge in the central domain of the fibrinogenmolecule is -8, while the net charge at the terminal ends is -4. If very largemolecules have a net charge of zero, they will tend to come together. However, ifwe have an excess of negative charges or positive charges, the molecules willrepel one another.

Fibrinogen © TfrTP»mrm«

V

Fibrinopeptides

Aggregation offibrin monomers

f . \ssssf +Wis/ - j f - Wsh/+ \pm¥md - j

O—0—00=0—GFibrin Clot

Figure 2-11Excess charge allows for aggregation.

Thus, one way to make thingssoluble is to have an excess of charge. In order toconverta soluble proteinto an insoluble proteinone must remove that portionofthe molecule contributing all those negative (or positive) charges. Whenfibrinogen is converted to fibrin we find that 4 Arg-Gly bonds are brokenin thecentral domain. This releases four peptides containing an excess of negativecharge calledfibrinopeptides.



Once these fibrinopeptides are released, the overall net charge in the centraldomain now becomes +5. The fibrin monomer that is now formed (due to therelease of the fibrinopeptides) has the ability to interact with other fibrinmonomers through electrostatic interactions between the terminal and centraldomain regions of the polypeptide. This is shown in Figure 2-12. Thisaggregation of fibrin monomers leads to the formation of the fibrin clot.

The clot that is initially formed is called a soft clot. The hard clot involves afurther step, in which there is a cross-linking via the enzyme transglutaminase(or Factor XHIa). Cross-links are formed between the individual subunits thatmakeup the aggregate. Thosecross-links occurbetweenGin and Lys residues asshown in Figure 2-12. This type of reaction is referred to as a transaminationreaction.

O

IIFibrin— CH2 CH2 — C— NH2

Gin

©+ NH3-(CH2)4 Fibrin

Lys

Xllla (transglutaminase)

O

Fibrin — CH2 CH2 — C— N- (CH2)4- Fibrin

Cross-linked

Fibrin Clot

H

Figure 2-12Cross-linking leads to a hard clot.

Once the damaged area has been repaired, a serine protease called plasminhydrolyzes specific regions in the fibrin clot in order to dissolve it into smallerpeptide fragments (to remove the clot). Tissue plasminogen activator (TPA)converts plasminogen into this active protease.

Let's return to vitamin K for a moment. Vitamin K is one of the fat-soluble

vitamins that is found in green leafy vegetables. In addition, out intestinal floracan make a certain form of vitamin K. What is the importance of this vitamin?We have already discussed the importance of this vitamin in connection withblood clotting. If you were to have a vitamin K deficiency, you could suffer frombleeding disorders. One way to have a vitamin K deficiency is a failure to absorblipids properly. Another way is to take a lot of antibiotics. Antibiotics have atendency to kill off the intestinal flora that help synthesize vitamin K. Finally,there may not be enough of this vitamin in the diet.

There are also antagonists to vitamin K, like dicoumarol. In the 1920's manyfarmers in Wisconsin were having problems with their cows. They were feedingtheir cows various feeds, some of which were causing hemorrhaging in the cows.It turns out that dicoumarol is a naturally occurring material in the rotting cloverof the feeds that the cows consumed. Cows that ate the dicoumarol had an

abnormal prothrombin that did not bind Ca2®.

Blood Clotting


Biology Heart & Lungs Blood Clotting

It occurred to some chemists that they might be able to find an inhibitor similarto dicoumarol which would not be toxic to cows, chickens, or even people, butmight have an effect on rats or mice. One of the things that farmers didn't likewas rats and mice running around eating their chicken feed. The chemists at theWisconsin AgriculturalResearch Stationdeveloped a vitamin K antagonist calledWarfarin. If you mixed Warfarin with chicken feed, and that feed was eaten bythe rats and mice, then eventually they would die of hemorrhaging. It would takemuch greater levels of warfarin to be toxic to humans. It turns out that warfarinis much more active in mammals than in birds, so the chickens were not affected.Warfarin is a Vitamin K antagonist that is used as rat poison. The structures ofvitamin K, Warfarin, and dicoumarol are shown in Figure 2-13.

H CH3 \11CH2 - C= C- CH2 -f H

Vitamin K2

Dicoumarol

O O0. // \\ .0

CH2

OH OH

Figure 2-13Chemical structures of vitamin K, Warfarin, and dicoumarol.

O,

OH CH

Warfarin

O

?H^\\ //2

C=0

ICH,



The LungsGases

Glucose can be oxidized to carbon dioxide and water as shown by equation (2-4).The majority ofcarbondioxide is generated in theKrebs cycle. Oxygen is used asthe final electron acceptor in the electron transport chain. Water is generated inthat reaction. Out of this metabolic pathway weare able to extract energy in theform of ATPand that ATPis ultimately used to keep the organismalive.

6 02 + C6H1206 i==> 6C02 + 6 H20 + Energy(2-4)

In general, respiration is the process by which oxygen is brought to the cells ofthe tissues and carbon dioxide is removed as a waste product. As we breathe airinto our lungs it first enters our system by way of the nose or mouth. The airpasses form the oral cavity to the pharynx, into the larynx, and then down thetrachea. At the end of the trachea the air passes into two tubular passagewayscalled bronchi. One bronchus enters into each lung and continues to divide intosmaller passageways called bronchioles, ending eventually in the functionalunits of the lungs which are the alveoli. Each alveolus consists of a single layer ofepithelial cells juxtaposed to a very thin basement membrane. Surroundingeach single alveolus is a capillary network. The epithelial cell layer of eachalveolus and the endothelial layer of the capillaries are separated from eachother by a very narrow interstitial space (if they are separated by an interstitialspace at all). This means that the air in each alveolus and the blood in thecapillaries are separated by a very small distance (about 0.2 to 0.3 |im, comparedto the average diameter of an erythrocyte, which is about 7 (im). The two lungsare composed of millions of alveoli, and if they were all laid flat on a surface,their combined total surface area would be between 70 m2 to 100 m2 (which isabout the size of a tennis court). Therefore, large quantities of oxygen in thealveoli of the lungs can quickly be equilibrated with the blood in the capillariesbecause of the large surface area and the thin barrier to diffusion for gasexchange.

The air that we breathe on a normal day is composed of roughly 78% N2, 21%02, 0.3% C02, and 0.7% H20. [These values vary slight from textbook totextbook.] Because all of these gas molecules are (normally) quite far apart, theytend not to interfere with one another. This means that the pressure exerted byone gas is independent of the pressure exerted by all of the other gases. In otherwords, the sum of each of these individual gas pressures equals the total pressureof this mixture of gas. The partial pressure of a gas (e.g., PN2, PO2, PCO2, etc.) istherefore a measure of the concentration of a given gas (such as O2) in a mixtureof gases (i.e., the air). If we add up all the partial pressures of the individualgases in the atmosphere, we will come up with the atmospheric pressure. At sealevel the atmospheric pressure is 760 mmHg.

What is the partial pressure of oxygen gas (PO2) at sea level? Roughly 21% of thetotal atmospheric pressure at sea level is oxygen gas. Therefore, the partialpressure of oxygen gas at sea level is about 160 mmHg (from 0.21 x 760 mmHg =160 mmHg). What happens to the partial pressure of oxygen as we ascend to the

Gases


Biology Heart & Lungs Gases

top of a high mountain? The concentration of oxygen in the air on this mountaintop will still be about 21%. However, as we increase our altitude the atmosphericpressure begins to decrease. The result is a decrease in the partial pressure ofoxygen.

How does a gas behave in a liquid? Consider an open glass of water sitting on atable as shown in Figure 2-14. The surface of that water is constantly beingbombarded with gas molecules from the air.

Figure 2-14Gas movement across a water surface.

When a gas molecule such as O2 comes in contact with that water surface it candissolve in the liquid. The number of O2 molecules which dissolve in the water isdirectly proportional to the partial pressure of the O2 gas. Just as O2 molecules inthe air can hit the surface of the water and dissolve in the liquid so too can O2molecules in the liquid hit the surface of the water and escape into the air.Therefore, at equilibrium the number of O2 molecules dissolving in the waterwill equal the number of O2 molecules leaving the water. In other words, thepartial pressure of oxygen in the gas phase (air) is equal to the partial pressure ofoxygen in the liquid phase (water), as shown in equation (2-5).

(Po2)gas - (Po2)liquid (2-5)

We can say that the pressure of the air acting on the membranes of the epithelialcells in the alveoli of the lungs is the sum partial pressures of all the gases in theair. Thus, the total pressure for any given gas molecule is directly proportional tothe concentration of that gas molecule in the air. For example, we can account forthe many factors that affect the rate of diffusion of a gas into a liquid byconsidering equation (2-3). In this case J is the flux or diffusion rate of the gasmolecule across the membrane, D is the diffusion coefficient, A is the area of theplane the molecule is diffusing across, C is the concentration of the molecule, andx is the distance of diffusion. Since the concentration of a gas is proportional tothe pressure of the gas, we can write that as shown in equation (2-6).

j_ (DXAXCft -C2°2) a (DXAXP!02 -P202)Ax Ax (2-6)



The respiratory passages have a unique anatomical structure. Let's consider across-section of one of these passageways and work our way from the lumenoutward. The epithelial cells which line the lumen of the passagewaysto the endof the bronchioles have cilia which are continually beating towards the pharynx.Scattered among these epithelial cells are glands which secrete mucus. As thecilia beat towards the pharynx the mucus is moved upward towards the oralcavity. Any foreign matter that has become trapped in the mucus is eventuallytransported to the oral cavity where it is swallowed in a normal reflex action.Oneof the reasons that smokingis bad foryou (besides itscarcinogenic nature) isthat it decreases the activity of the cilia and lowers the body's defenses againstlung infection by bacteria that can enter the respiratory tract on airborne dustparticles. Immediately beneath the layer of mucus is a layer of fluid in which thecilia operate. Air that flows within the passageways of the respiratory tract iswarmed and moistened.

The upper passageways of the respiratory tract maintain their opening by meansof cartilage rings that surround most of the diameter of the passageways. By thetime the bronchioles are reached the cartilage has disappeared. Smooth muscle isfound in almost all areas of the respiratory tract where there is no cartilage. Forexample, the walls of the bronchioles are mainly smooth muscle. The bronchiolesof the lungs are innervated by nerve fibers from parasympathetic nerves (whichtravel in the vagus nerve).

Asthma is usually caused by an allergic hypersensitivity to airborne antigenswhich have entered the respiratory tract. The direct result is to cause the mastcells within the bronchioles to release a number of different substances which

cause the smooth muscle surrounding bronchioles to spasm and constrict.

Mechanics of BreathingThe thoracic cage which contains the lungs is separated from the abdomen by asheet of skeletal muscle and connective tissue called the diaphragm (see Figure2-15). The lungs themselves are encased in a pleural membrane. The visceralpleura covers the lungs, while the parietal pleura adheres to the diaphragm andthe chest wall. Between the visceral pleural and parietal pleura is the intrapleuralspace which contains a watery fluid. As the muscles of the diaphragm contractthey pull the diaphragm itself downward. Simultaneously the muscles of the ribcage contract and cause the rib cage to move upward and outward. Both of theseactions enlarge the area of the thoracic cage that contain the lungs.

Since the diaphragm is attached to the parietal pleura this pleura is also pulleddownward. The watery fluid in the intrapleural space is rather indistensible andthe pulling of the parietal pleura translates through the intrapleural fluid. It alsopulls the visceral pleura downward as well. Enlargement of the thoracic cageexpands the lungs and creates a subatmospheric pressure in the alveoli. Airrushes down its pressure gradient from 760 mmHg at sea level to whatever lowerpressure is found in the lungs at inspiration.

Once the muscles of inspiration stop contracting, the elastic tissue found withinthe thoracic cage and lungs returns to its normal length. Air within the alveoli iscompressed and forced out through the passageways of the respiratory tractduring expiration.

Gases


Biology Heart & Lungs Gases

If the lungs were to become separated from the visceral pleura, they wouldcollapse. This is because they have no anatomical structures to maintain rigidity.One way to separate the lungs from the visceral pleura is to receive a very strongblow to the chest area (as often happens during football games).

Parietal

pleural

Figure 2-15The thoracic cage.

To the pharynx and oral cavity

Larynx ^^

Cartilagerings

Intrapleuralspace

Ribs



Gas Exchange

It is important to remember that molecules will diffuse from a site of highconcentration to a site of low concentration. As deoxygenated blood (a dark redcolor) is coming back from the tissues it enters the right atrium via the superiorand inferior vena cava and then passes into the right ventricle, where it ispumped out to the lungs. The PO2 is about 40 mmHg and the PCO2 is about 46mmHg as this deoxygenated blood enters the capillaries surrounding the alveoliof the lungs. The PO2 and PCO2 within the alveoli are about 105 mmHg and 40mmHg, respectively. Passage of the blood through the capillaries is relativelyslow and an equilibration can be reached between the gas exchange in alveoliand the capillaries. Oxygen will diffuse down its concentration gradient from thealveoli to the capillaries, and carbon dioxide will diffuse down its concentrationgradient from the capillaries to the alveoli. As the oxygenated blood (a bright redcolor) leaves the capillaries of the lungs and enters the left atrium of the heart, thePO2 is about 100 mmHg and the PCO2 is about 40 mmHg. See Figure 2-16.

All values are in

mmHg

P02 = 160

P02 = 105

P02 = 40PC02 = 46

PCO2 = 0.3

PC02 =40 JAlveoli

P02 = 100PCO2 = 40

Capillaries at level of alveoli

Right Heart Left Heart

Capillaries at level of tissues

P02 = 40PC02 = 46

PC02 > 46

P02 = 100PC02 = 40

Cells

Figure 2-16Gas exchange at the level of the lungs and tissues.

Oxygenated blood will pass from the left atrium to the left ventricleof the heartand then be pumped to the tissue capillaries. At the levelof the tissuesthe PO2inthe cells (depending on which cells you are considering) is less than 40 mmHg,while the PCO2 in those cells is greater than 46 mmHg. Oxygen will diffuse fromthe blood in the capillaries to the cells while carbon dioxide will diffuse from thecells to the blood in the capillaries. The deoxygenated blood that leaves the tissuecapillaries returns to the right atrium of the heart via the venous system.

Gas Exchange


Biology

100-r

O

O

aX 60o

co

-a 402

80-

{£20-

Heart & Lungs Gas Exchange

Oxygen can travel in the blood by being dissolved in the blood itself or by beingbound to a transport protein in the red blood cells (erythrocytes) calledhemoglobin (abbreviated as Hb). Since oxygen is rather insoluble in water, notmuch is actually dissolved in the blood and transported in that manner.However, since hemoglobin has such a high affinity for oxygen, more than 98%of the oxygen in contact with hemoglobin is transported by this protein.

Hemoglobin is a protein that is composed of four polypeptide subunits. Whenthese subunits interact with each other to form the hemoglobin molecule, theygive hemoglobin a quaternary structure. Located within each of the fourpolypeptide subunits is a heme prosthetic group that has an iron atom in thecenter which is in the ferrous (Fe2®) oxidation state. Since one hemoglobinmolecule has four binding sites for oxygen, there is a potential for 4 O2 moleculesto bind to one hemoglobin molecule. Every time hemoglobin takes up O2 fromthe blood more O2 can leave the gas phase in the alveoli and enter into the liquidphase in the blood. This increases oxygen's solubility in the blood.

Oxygen Saturation Curve for HemoglobinIn Figure 2-17 we have a plot of the percent saturation of hemoglobin withoxygen as a function of the partial pressure of oxygen at various places in thebody. Recall that when the PO2 is about 100 mmHg, we are at the level of thealveoli and when the PO2is about 40 mmHg, we are at the level of the tissues. [Itis best to read this curve from right to left and from top to bottom.]

Venous

Blood

40

pH7.4

Shift due to:

Decrease in pHIncrease in temperature

Increase in 2,3-DPG

Resting Muscle

Working Muscle

P02 (mmHg)

Arterial

Blood

100

„ O2 releasedto tissues

IExtra O2released

to tissues

Figure 2-17Oxygen-hemoglobin dissociation curve.

At the level of the capillaries (PO2 ~ 100 mmHg) surrounding the alveoli roughly98% of the hemoglobin is saturated with oxygen. As the blood passes to thetissues (PO2 = 40 mmHg) roughly 25% of the oxygen that was bound tohemoglobin is given up to the tissues. This can be seen in Figure 2-17.



Effects of pH, Temperature, and 2,3-BPGThe upper oxygen dissociation curve shown in Figure 2-17 is for normalconditions in which the body temperature is about 37 "C and the blood pH isabout 7.4. If we were to decrease the pH of the blood (to a pH of 7.2) or increasethe temperature, then the oxygen dissociation curve would shift to the right anddownward as shown in the lower curve in Figure 2-17. This happens when youexercise. The same type of shift occurs when a by product of glycolysis, 2,3-bisphosphoglycerate (abbreviated as 2,3-BPG), binds to hemoglobin. 2,3-BPG isusually synthesized in increased amounts when the body has been deprived ofoxygen for an extended period of time (e.g., when you visit high altitudes).

All three of these interactions with hemoglobin (the lowering of the pH, theincrease in temperature, and the binding of 2,3-BPG) cause hemoglobin to releasemore oxygen to the tissues at the same partial pressure of oxygen as in thestandard case. The oxygen dissociation curve can be shifted to the left andupward by reversing these interactions.

Carbon DioxideHow is CO2 carried in the blood? It can be carried by (a) dissolving in the plasmaand the red blood cells, (b) binding to a specific site on the hemoglobin molecule,or (c) in the form of bicarbonate ions (HC03e). About 70% of the carbon dioxideis carried in the blood in the form of bicarbonate ions, roughly 20% is carried bythe hemoglobin itself, and about 10% is dissolved in the plasma and red bloodcells. See Figure 2-18.

Plasma

/ \^^Red Blood Cell

70%

^y^ ""\ HC03e

/ Carbonic[ anhydrase( C02 + H20 • H2C03 H® +HC03e)\ Nl 20% Hb-C02 5% C02 J

>^^dissolved/

x\>^ 5% CO2 dissolved

P ^ CO2 from tissues

Figure 2-18Gas exchange at the level of the tissues.

At the level of the tissues the PCO2 is greater than 46 mmHg. However, in theblood of the capillariesit is about 40 mmHg. Therefore, CO2 will diffusedown itsconcentration gradient and into the blood. Some of the carbon dioxide willdissolve in the blood plasma, some will dissolve in the red blood cell, and somewill bind to hemoglobin. The remaining CO2 will react with water and beconverted to carbonic acid (H2CO3) by the enzyme carbonic anhydrase. Carbonicacidwill ionizeto the bicarbonate ion and a proton (He). Bicarbonate willdiffuseinto the blood plasma and be carried by the circulatorysystem to the capillariesof the lungs. See Figure 2-18.

Gas Exchange

Copyright © by The Berkeley Review 107 The Berkeley ReviewSpecializing in NCAT Preparation

Biology Heart St Lungs Gas Exchange

At the level of the capillaries in the lungs, the PCO2 in the blood is about 46mmHg, while the pC02 in the alveoli of the lungs is about 40 mmHg. Onceagain, CO2 will diffuse down its concentration gradient and into the alveolarspace. The CO2 that is dissolved in the plasma and in the red blood cell diffusesinto the alveoli as does the CO2 that was bound to the hemoglobin. Bicarbonateion in the plasma will diffuse into the red blood cell and combine with a protonto become carbonic acid. Carbonic anhydrasewill convert carbonic acid to waterand carbon dioxide. The CO2 diffuses into the lungs. This is shown in Figure 2-19.

Plasma HC03<

CapillaryTo the alveoli

Figure 2-19Gas exchange at the level of the alveoli.

Control of BreathingThe coordinated rhythm for breathing isgenerated by the medulla and the ponsin the brainstem. Nerve impulses from these centers cause the respiratorymuscles to contract. What is the stimulus that causes contraction? The majorstimulus is a change in the concentration of CO2 and H® that affects chemo-sensitive areas in themedulla. The concentration of O2 in the bloodis sensed bychemoreceptors in the carotid arteries and in the arch of the aorta. Thesechemoreceptors transmit signals to the respiratory center in the brainstem,informing it of the levels of O2 in the blood.

If you were to breathe into a container that collected your exhaled CO2, youwould soon find yourself breathing deeper and faster. The increase of CO2 inyour blood indicates to your respiratory center that C02 is building up in thebody. One way to eliminate CO2 would be to breathe faster and deeper. Whatwould happen if you were to breathe pure O2? The chemoreceptorsfor O2 wouldrelay a message to the respiratory centers indicating that there is too muchoxygen in the blood. Impulses from the brain stem would be sent to therespiratory muscles in order to slow their action.


Heart and Lungs

15 Passages

100 Questions

Passage Titles Questions

I. Mechanics ofBlood Plow 1 -7

II. Enzymes in Blood Clotting 8- 14

III. Cardiac Output and Venous Return 15-21

IV. Circulatory Pressure, Area, Velocity, and Volume 22-28

V. Thoracic Cavity 29-34

VI. Electrocardiogram 35-40

VII. Folate Experiment 41 -47

VIII. Measurement ofBlood Pressure 48-54

IX. Aortic Compliance 55-60

X. Heart Muscle Action Potentials 61 -66

XI. Capillary Filtration 67 - 72

XII. Respiratory Calculations 73-78

XIII. Aspirin 79-85

XIV. Sickle Cell Anemia 86-92

XV. Ventilation Regulation 93- 100

fgERKELEYSpecializing in MCAT Preparation

Suggestions

The passages that follow aredesigned togetyouto think in a conceptual manner about theprocessesof physiology at the organismal level. If you have a solid foundation in physiology, many of theseanswerswill be straightforward. If youhavenot had a pleasant experience with the topic, some of theseanswers mightappear to come from thevoidpast theOortfield of the solarsystem.

Picka fewpassagetopics at random. Forthese initial few passages, do not worry about the time. Justfocus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, orgraphs. Third, read each question and the accompanying answers carefully. Fourth, answer thequestions the best you can. Check thesolutions and seehow youdid. Whether you got the answersrightor wrong, it is important to read the explanations and see if you understand (and agree with) what isbeing explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block ofpassages and try them. Beaware that time is going to become important. Generally, you will have about1 minute and 15 seconds to complete a question. Be a little more creative in how you approach this nextgroup. If you feel comfortable with the outline presented above, fine. If not, then try differentapproaches to a passage. For example, you might feel well versed enough to read the questions first andthen try to answer some of them, without ever haying read the passage. Maybe you can answer some ofthe questions by just looking at the diagrams, charts, or graphs that are presented in a particular passage.Remember, we are not clones of one another. You need to begin to develop a format that works best foryou. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where thelevel of distraction is at a minimum. Get out your watch and time yourself on these passages, eitherindividually or as a group. It is important to have a feel for time, and how much is passing as you try toanswer each question. Never let a question get you flustered. If you cannot figure out what the answer isfrom information given to you in the passage, or from your own knowledge-base, dump it and move onto the next question. As you do this, make a note of that pesky question and come back to it at the end,when you have more time. When you are finished, check your answers and make sure you understandthe solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tendsto stay with you longer. (For example, what is the Oort field?)

The estimated score conversions for 100 questions are shown below. At best, these are roughapproximations and should be used only to give one a feel for which ballpark they are sitting in.

Section II



>12 86 - 100

10-11 79-85

8-9 65-78

7 59-64

6 54-58

5 48-53

<4 0-47

Biology Mechanics Of Blood Flow Passage I


Flow can be used in three different contexts: (1) asflow, (2) as average velocity, and (3) as the linear velocityof a small element of fluid. The total volume of fluid

passing a given point per time is known as the flow rate.The flow at a given point divided by the cross-sectionalarea is known as the average velocity offlow. The linearvelocity is the distance traveled by small volume of bloodper unit time. The flow rate (Q) must remain constantthroughout the entire cardiovascular system.

Blood flow through a blood vessel results from adriving force in the form of a pressure gradient. Theperfusion pressurealong the length of any blood vessel isgiven by the differences between the pressure at theproximal and distal end of the vessel. The followingrelationship between blood flow and perfusion pressurecan be established:

BloodFlow = Perfusion PressureFlow Resistance

The meanarterialpressure (MAP) is a time-weightedaverage of the arterial pressureover the entire heartcycle.The net flow resistance of the systemic loop can bedefined as the total peripheral resistance (TPR) while theentire venous pressure (entire vascular system acting asone unit) is termed the central venous pressure (CVP).

The viscosity of a fluid is a measure of the internalwork necessary to make a fluid flow. The viscosity ofblood is determined by the composition of blood, thenature of the vessel in which it is flowing, and the meanflow velocity. Plasma has a viscosity of about 3, twicethat of water, while whole blood has a viscosityapproximately three times that of water.

1. Capillary walls provide the minimum possiblebarrier to diffusion. Relative to the arteries andveins, the blood in capillaries has a:

A. low average velocity because of a high totalcross-sectional area.

B. low average velocity because of a low totalcross-sectional area.

C. high average velocity because of a high totalcross-sectional area.

D. high average velocity because of a low totalcross-sectional area.

2. The diameter of a uniform blood vessel is reducedby a factor of 2. The resistanceof this blood vesselincreases by a factor of:

A. 2.

B. 4.

C. 8.

D. 16.

Copyright © by The Berkeley Review HI

Which of the following equations BEST representsblood flow for the entire vascular system acting as aunit?

A. Q x TPR = (CVP-MAP)B. MAP x Q = (TPR - CVP)C. Q x TPR = (MAP - CVP)D. CVPxMAP = (Q-TPR)

The increased viscosity of whole blood relative toplasma is best explained by the contribution of:

A. albumin.

B. glucose.C. white blood cells.

D. red blood cells.

The structure of systemic veins differ from thestructure of systemic arteries in that only:

A. the inner surface of veins is lined with a singleendothelial layer.

B. arteries contain a component of elastic tissue.C. veins contain valves that favor one-way blood

flow.

D. arteries contain a layer of smooth muscle.

The most likely reason that the central venouspressure (CVP) is not an average value similar to theMAP is because the:

A. cross-sectional area of the veins is larger thanthe arteries.

B. venous pressure does not change significantlyover the heart cycle.

C. arterial pressure does change significantlyoverthe heart cycle.

D. arterial pressure rises rapidly with increasedvolume while the venous pressure remainswithin a few mmHg with the same increase involume.

In the classical method of measuring blood pressure,pressure from a cuff is applied to the brachial arteryin the arm to collapse the artery and prevent bloodflow. The pressure of the cuff in this initial stepmust exceed:

A. CVP.

B. MAP.

C. arterial diastolic pressure.D. arterial systolic pressure.


Biology Enzymes In Blood Clotting Passage II


Figure 1. Blood Clotting Cascade

Intrinsic Pathway

Kallikrein <^ I Prekallikrein

Extrinsic Pathwayt

—, Thrombin

y nFactor^-^-^ <J?XITJ > S~~^

Xffla f FibrinogenFibrin \ ' Fibrin

(hard clot) (soft clot)

Table 1. Blood Coagulation Factors

Common Name

FibrinogenProthrombin"!"*

Tissue factor (thromboplastin)Calcium

Proaccelerin

Proconvertin**

Antihemophilic factorChristmas factor t*Stuart-Prower factor t*

Plasma thromboplastin antecedent*Hageman factor*Fibrin-stabilizing factor

Factor Half-life*

I

II

III

IV

V

VII

VIII

IX

X

XI

XII

XIII

Fletcher factor* Prekallikrein

90-120

50-120

N/A

N/A

12-24

2-6

10-12

18-30

25-60

45-80

40-70

70-200

48-52

§ Ca2+, Phospholipid Membranef Dependent on Vitamin K

$ Undergoes activation to Serine Protease

A Reported in hours


Hemostasis is the process by which bleeding isarrested. At the time of injury, enucleated blood cellscalled platelets bind to the exposed subendothelium of thedamaged blood vessel. As platelets begin to aggregate toone another, a platelet plug is formed. Vasoconstrictionof the blood vessel is stimulated by the release ofthromboxane A2 and serotonin from the platelets. Thepurpose of the platelet plug is to provide a surface thatallows for the activation of blood coagulation factors(Table 1), many of which are synthesized in the liver.Activation of these coagulation factors initiates the bloodclotting cascade (Figure 1). The end result of this cascadeis to form a blood clot (thrombus) that will seal thedamaged blood vessel.

Many of the coagulation factors are glycoproteinsthat can be classified as being pre-enzymatic. They existin their inactive form as zymogens of serine proteases,proteolytic enzymes that have a common catalyticmechanism based on a highly reactive serine residue atthe active site. Once activated these serine proteasesenhance the coagulation cascade.

The blood clotting cascade can follow either theintrinsic pathway, initiated by contact of an appropriatesurface, or the extrinsic pathway, initiated by tissuetrauma. Both pathways converge at factor Xa, theactivated form of factor X. Factor Xa initiates the finalsequence of events in the common pathway that will leadto the formation of a clot.

Vitamin K is essential for the synthesis of many ofthe coagulation factors, including prothrombin. Thereduced form of this cofactor, vitamin K hydroquinone, isinvolved in a carboxylation reaction that converts many ofthe glutamate residues located in the N-terminal segmentof these clotting factors into y-carboxyglutamate. VitaminK hydroquinone is regenerated in a two-step reaction thatcan be inhibited by antagonists like dicoumarol andwarfarin.

y-carboxyglutamate is an excellentchelatorof Ca2+and, following injury, acts as an anchoring mechanismbetween calcium-dependent coagulation factors and thephospholipid membranes of platelets. The functionalsignificance of this event is that it brings specific factorstogether that aid in the formation of a clot, therebyaccelerating clot formation many fold.

Defects in the blood clotting cascade may be assessedusing screening tests. Partial thromboplastin time (PTT)screens the intrinsic and common pathways whileprothrombin time (PT) screens the extrinsic and commonpathways. A deficiency in factor XIII cannot be detectedby using these two tests.


Biology Enzymes In Blood Clotting Passage II

8. Mature red blood cells obtain their energy from:

A. electron transport.B. oxidative phosphorylation.C. Krebs cycle.D. glycolysis.

9.

10.

11.

Glutamate is converted to y-carboxyglutamate in thevitamin K dependent reaction shown below:

h oI n

n-c-c-n —

H

I

CH,I

CH2

coo©

H

Glutamate

Vitamin K ,,Hydroquinone

Vitamin K _^Quinone "

y-Carboxyglutamate

ooc coo

The enzyme which BEST regenerates vitamin Khydroquinone is called a:

A. decarboxylase.B. carboxylase.C. reductase.

D. protease.

The only coagulation factor listed below that DOESNOT require vitamin K is:

A. Factor II.

B. thromboplastin.C. Christmas Factor.

D. Factor X.

All of the following mechanisms will limit clotgrowth EXCEPT:

A. consumption of leafy green vegetables.B. removal from circulation by the liver.C. inhibitors of serine proteases.D. blood flow dilution.


12. Hemophilia A results from a deficiency of factorVIII while hemophilia B results from a deficiencyoffactor IX. Genes coding for both hemophilias resideon the X chromosome and are geneticallytransmitted as a sex-linked recessive trait. What isthe probability that a woman whose father hashemophilia, and who marries a normal man, willhave an affected son?

A. 1.00

B. 0.75

C. 0.50

D. 0.25

13. Which of the following compounds, when added tostored whole blood, BEST prevents clotting?

14.

A.

C.

coo

CH,

I 0HO - C- COO

CH,

COO

Citrate

coo

CH,

CH2I ©coo

Succinate

B.

~ H O© I ll ©

HjN-C-C- O

CH,I *

CH2I ©coo

Glutamate

D.

H O© I II ©

HjN-C-C" O

CH2

0 M 0OOC COO

Y-Carboxyglutamate

The most common oral anticoagulant in use at thepresent time is warfarin. Administration of thiscompound begins with a daily dose of about 5 mgand adjustments are made 3 days later to maintain aPT between 1.2-1.5 times the control. This is mostlikely attributed to a decrease in:

A. prothrombin.B. proconvertin.C. Christmas factor.

D. Stuart-Prower factor.


Biology Cardiac Output St Venous Return Passage III

Passage DI (Questions 15-21)

The blood returning to the right heart from the venoussystem must equal the blood leaving the left heart to thearterial system over any extended period of time. In anormal individual the venous return and the cardiac outputis about 5 L/min.

The vascular function curve in Figure 1 depicts arelationship between venous pressure and venous return.As the heart contracts, blood is removed from the vesselsof the venous system and pumped into the vessels of thearterial system.

-4 0 4 8 12 16

Venous pressure (mmHg)

Figure 1

If the cardiac output and the venous return is increased,the pressure in the arterial system will increase while thepressure in the venous system will decrease. This is dueto the increased transfer of blood from the venous to thearterial system.

If the heart were stopped and the cardiac output wasreduced to zero, the pressures in the venous and arterialsystems would soon equilibrate. However, the pressurein the venous system will be at its highest value, simplybecause the veins now contain more blood than when theheart was contracting. This pressure is referred to as themean circulatory pressure (MCP) and varies withsympathetic stimulation and blood volume.

The cardiac function curve in Figure 2 depicts arelationship between venous pressure and cardiac output.

-4 0 4 8 12 16


Figure 2


If there is an increase in the venous pressure, then duringdiastole there will be more filling of the ventricles. Thisstretches the muscle fibers in the heart tissue and leads to

a greater force of contraction during systole. This isknown as Starling's law of the heart, and simply statedsays that an increase in venous pressure leads to anincrease in cardiac output.

15. The component of the circulatory system with theLEAST total percentage of blood volume is thenetwork of:

A. capillaries.B. arterioles.

C. arteries.

D. small veins and venules.

16. The pH of the blood passing through the pulmonaryartery, compared to the pH of the blood passingthrough the renal artery, is:

A. lower due to the loss of O2.B. higher due to the loss of O2.C. lower due to the accumulationof CO2.D. higherdue to the accumulationof CO2.

17. Which of the following curves BEST represents anincrease in only arteriolar resistance to blood flow?

A.

9 10

C.

Normal

-4 0 4 8 12 16


-? 10.,

-4 0 4 8 12 16


B.

s 10

D.

Normal

-4 0 4 8 12 16


-=• 10

E

E

8-

6

4

2-1

0NormaP

-4 0 4 8 12 16



Biology Cardiac Output St Venous Return Passage ID

18. Changes in blood pressure or blood volume aredetected by stretch receptors in the atria and arterialbaroreceptors. Which of the following pairs ofhormones act directly to increase blood volume?

A. Renin and vasopressinB. Angiotensin II and aldosteroneC. Renin and angiotensin ID. Vasopressin and aldosterone

19. Both the cardiac output (CO.) from the heart and thevenous return (V.R.) to the heart depend on venouspressure. This allows the curves in Figure 1 andFigure 2 of the passage to be combined into onegraph as shown below:

-4 0 4 8 12 16


Which of the following graphs BEST representsactivation of the sympathetic nervous system? (Solidlines represent normal curves; dashed lines representchanges.)

A.

C.

-4 0 4 8 12 16


"s 10 -i-

8 12 16


B.

D.

-4 0 4 8 12 16


10 •

8 12 16



20. Sympathetic nerve fibers are supplied to all parts ofthe heart while parasympathetic nerve fibers arelocated primarily at the sinoatrial (SA) node and theatrioventricular (AV) node. The neurotransmitterreleased by sympathetic heart nerves is:

A. norepinephrine, because it opens Na® andCa^® channels in myocardial cells.

B. acetylcholine, because it opensK® channels inmyocardial cells.

C. norepinephrine, because it opens K® channelsin myocardial cells.

D. acetylcholine, because it opens Na® and Ca^®channels in myocardial cells.

21. All of the following compensatory mechanismswould be expected to occur during a hemorrhageEXCEPT:

A. increased sympathetic discharge to thearterioles.

B. decreased parasympathetic discharge to theheart.

C. increased cardiac output.D. decreased arterial pressure.


Biology Circulatory Pressure, Area, Velocity, St Volume Passage IV


The total blood volume in an average human being isabout 5.5 L. Blood flows through the circulatory system,which is divided into the systemic and pulmonarycirculatory systems. The systemic system supplies bloodto all of the tissues of the body except the lungs. Thepulmonary system supplies blood to the lungs.

The passage of blood through the circulatory systemcan be examined in terms of differences in the pressure,cross-sectional area, velocity, and total blood volume indifferent segments of this closed system.

00

X66

100 -

80 -

60 -

40 -

20 -

5000 -

<n 4000 -

3000 -

< 2000

1000 -

i i

20 40 60 80

Total Blood Volume (%)

20 40 60 80


20 40 60 80


Blood returns to the right atrium of the heart from thesystemic system under low pressure and is passed to theright ventricle. Contraction of this ventricle pushes theblood into the pulmonary arteries where it flows to thelungs.

Copyright © The Berkeley Review

100

100

116

The pulmonary veins return the oxygenated blood tothe left atrium of the heart which then passes it to the leftventricle. Contraction of the left ventricle sends the blood

into the aorta under high pressure.

Blood passes down the descending aorta and into thesystemic arteries, arterioles, capillaries, venules, and veinsbefore returning to the right atrium by way of the superiorand inferior vena cava (the great veins).

22. The flow of blood throughout the body is primarilyregulated at the level of the:

A.

B.

C.

veins.

capillariesarterioles.

D. aorta.

23. Which point on the graph of pressure versus totalblood volume BEST represents blood flowingthrough the pulmonary artery?

A. I

B. n

C. m

D. IV

24. Which point on the graph of area versus total bloodvolume BEST represents blood flowing through thelungs?

A. I

B. n

C. in

D. rv

25. Which point on the graph of velocity versus totalblood volume BEST represents blood flowingthrough the vena cava?

A. I

B. II

C. m

D. IV


BlOlOgy Circulatory Pressure, Area, Velocity, fif Volume Passage IV

26. During circulation of the blood, turbulent flow couldincrease with a decrease in the:

A. viscosity of the blood.B. velocity of the blood.C. diameter of the blood vessels.

D. Reynold's number.

27. In the circulatory system, the vascular distensibility isthe LEAST for:

A. systemic veins.B. pulmonary veins.C. systemic arteries.D. pulmonary arteries.

28. Which of the following is NOT characteristic of thesystemic and pulmonary capillaries of an averageadult human?

A. The diameter of a capillary is the smallest ofany of the vessels.

B. The thickness of a capillary wall is the thinnestof any of the vessels.

C. The capillaries contain between 5% to 10% ofthe total blood volume.

D. The capillaries contain about 20% of the totalblood volume.

Copyright © The Berkeley Review 117 The Berkeley ReviewSpecializing in MCAT Preparation

Biology Thoracic Cavity Passage V


The thoracic cavity of the body protects the heart andlungs and is important in the mechanics of respiration.The thoracic cage is composed of the ribs, costalcartilages, sternum, vertebral column, primary andassociated musculature.

The walls of the thoracic cage are constructed of theribs. These are curved bones that articulate with the

vertebral column posteriorly and extend anteriorly andmedially to articulate with the sternum. Out of the 12pairs of ribs, only the upper six to seven pairs of ribs willarticulate directly to the sternum. The remaining ribs willattach by costal cartilages.

Spanning the distance between each rib are the(external and internal) intercostal muscles. Attaching tothe most inferior ribs and costal cartilages, and extendingposterior to the vertebral column, is a large, flat, dome-shaped muscle called the diaphragm. These muscles,along with several of the abdominal muscles, areimportant in the expansion and compression of thethoracic cage, which is necessary for the function ofrespiration.

Lining the internal wall of the thoracic cage is a sheetof tissue referred to as the parietal pleura. Lining theexternal surface of the lungs is a similar sheet of tissuecalled the visceral pleura. Between these two layers oftissue is the pleural cavity, a space containing a smallamount of fluid and registering a negative atmosphericpressure.

The functional unit of the lungs is referred to as thealveolus, a small sac lined with epithelial cells. Eachalveolus is surrounded by a network of capillaries. Thewalls of the alveoli and the capillaries are quite thin,allowing for maximal efficiency in gas exchange betweenthe two structures.

Red blood cells are the primary means of gastransport between the lungs and the tissues of the body.The protein hemoglobin, a component of red blood cells,binds oxygen at the level of the alveoli and transports thisgas by way of the circulatory system to the tissues of thebody. Carbon dioxide, a waste product of metabolizingtissues is transported back to the lungs by the circulatorysystem.

The transport of both gases is dependent not only ontheir partial pressures but also on the temperature and pHof their environment. Figure 1 and Figure 2 represent arelationship between the partial pressure of oxygen andthe percent oxygen saturation of hemoglobin.


o

o

So

X

03

c

60

X

O

uu-

80-

Curve II //^S/^

60- /40-

1 1 Curve I

20-

0- • 1 1 1 1

20 40 60 80

Partial Pressure of Oxygen

Figure 1

100

100-1

"soo

S<L>

X

80-

60-

3 40-3

0800

c0)60>>X

o

20-

* 0-

Curve I

20 40 60 80

Partial Pressure of Oxygen

Figure 2

100

29. As the intercostal muscles lift the thoracic cage, howwill pressure in the pleural cavity change?

A. No change to pleural pressure.B. Pleural pressure will become more negative.C. Pleural pressure will become more positive.D. None of the above.


Biology Thoracic Cavity Passage V

30. A pneumothorax is the presence of gas in the pleuralspace of the thoracic cavity. This condition leads to acollapsed lung. In order for a collapsed lung tooccur, which structure associated with the thoraciccavity needs to be punctured?

A. Internal intercostal muscles

B. Parietal pleuraC. Ribs

D. Heart

31. Several muscles are involved in respiratory function.If the external intercostal muscles are non-functional,which muscle group will take over the primaryfunctions of normal respiration?

I. Abdominal muscles

II. DiaphragmHI. Internal intercostals

A. I and LQ onlyB. II onlyC. in onlyD. Normal respiration cannot occur.

32. The main gases that are exchanged in the lungs arecarbon dioxide and oxygen. This exchange dependsto a large extent on the partial pressures of thesegases (pC02 and p02) in the capillaries and thealveoli. Which situation is ideal for normal gasexchange between the capillaries and the alveoli ofthe lungs as blood enters the lungs from thepulmonary arteries?

A. High pC02 in the capillaries; low p02 in thecapillaries.

B. Low pC>2 in the capillaries; low pC02 in thecapillaries.

C. HighpC02 pressure in the lungs; highpC>2 inthe lungs.

D. Low pC>2 in the lungs; high pC02 in the lungs.


33. In the graph shown in Figure 1, curve I represents arelationship between the partial pressure of oxygenand the percent oxygen saturation of hemoglobin at apH of 7.40 and a temperature of 38 °C. If the pH ofthe blood is increased, curve II depicts the result.Which of the statements below explains thisphenomenon?

I. A rise in pH requires a lower partial pressure ofO2 to bind a given amount of O2.

II. A rise in pH requires a greater partial pressureof O2 to bind a given amount of O2.

HI. A rise in pH requires a lower partial pressure ofCO2 to bind a given amount of O2.

A. I onlyB. II onlyC. Ill onlyD. I and ffl only

34. In the graph shown in Figure 2, curve I represents therelationship between the partial pressure of oxygenand the percent oxygen saturation of hemoglobin at apH of 7.40 and a temperature of 38 °C. If thetemperature of the blood is increased, curve II depictsthe result. Which of the statements below explainsthis phenomenon?

I. A rise in temperature requires a lower partialpressure of O2 to bind a given amount O2.

II. A rise in temperature requires a greater partialpressure of O2 to bind a given amount of O2.

III. A rise in temperature requires a lower partialpressure of CO2 to bind a given amount of O2.

A. I onlyB. II onlyC. Ill onlyD. II and III only


Biology Electrocardiogram Passage VI


Potential differences within the heart tissue areconducted to the surface of the skin because the body is agood conductor of electricity. A reading of these patternsprovides a graph representing the electrical activity of theheart, called an electrocardiogram (ECG).

During each heartbeat cycle, 3 distinct wave patternsare produced: P, QRS, and T waves. The waves representchanges in potential occurring across heart tissue as theresult of many action potentials inside the myocardialcells.

The P wave represents the spread of depolarizationthrough the atria. The QRS wave represents the spread ofdepolarization into the ventricles. The T wave representsthe repolarization of the ventricles. The graph shown inFigure 1 indicates a standard cycle of these waves.

1.0 -

R

0.5

P

T

0.0

qV0.5

•

s

1 » '

200 400

Milliseconds

Figure 1

600

35. Which of the following conditions could bemonitored with an ECG?

A. Blood pressureB. Cardiac outputC. ArrhythmiasD. Hematocrit


36. The phase of contraction of the ventricles is calledsystole. Based on the passage, to which part of theECG does systole best correspond?

A. P

B. QRSC. T

D. T-P phase

37. Fibrillation involves random contraction of

myocardial fibers at different times. It can bemonitored with an ECG. What is the effect on an

individual who experiences ventricular fibrillation?

A. More heartbeats than usual.

B. Fewer heartbeats than usual.

C. Unaffected.

D. Death, if a normal rhythm is not introduced.

38. Which group of pacemaker cells provides the triggerfor depolarization in the right atrium and initiates theP wave?

A. Sinoatrial node

B. Atrioventricular node

C. Bundle of His

D. Purkinje fibers

39. Why is the body a good conductor of electricity?

A. Tissue fluids contain ions that participate inconduction.

B. Tissue fluids are free of ions that hamperconduction.

C. Tissue fluids contain ions that hamperconduction.

D. Tissue fluids are free of ions that participate inconduction.

40. The ECG can also be used to count the number ofheartbeats per minute (the cardiac rate). Whichperson would most likely have the slowest cardiacrate at rest?

A. An infant

B. A marathon runner

C. A weight lifterD. A sedentary adult


Biology Folate Experiment Passage VII


Currently, the recommended dietary allowance for theB-vitamin folate is 200 |ig/day for non-pregnant women.One group of scientists explored the adequacy of the 200Hg/day recommendation in the following experiment.

Experiment 1

30 non-pregnant women were assigned to 3 differentdietary groups: either a diet containing 200 Jig/day, 300Hg/day, or 400 Jig/day of folate. They followed theprescribed diet for 10 weeks. Levels of folate in theplasma and in the red blood cells (RBCs) were measuredat the beginning of the experiment and at the conclusion.Results of these measurements are shown in Table 1.

Dosageper Day

(M*>

Initial

Plasma

Folate

(ng/ml)

Initial

RBC

Folate

(ng/ml)

Final

Plasma

Folate

(ng/ml)

Final

RBC

Folate

(ng/ml)

200 3.6 444 3.1 387

300 3.5 450 3.3 423

400 3.6 449 6.7 610

Table 1: Folate concentrations in plasma and RBCs before andafter folate diets.

If plasma folate drops below 3.0 ng/ml, then a folatedeficiency is indicated. Further deficiency will depleteRBC stores and could lead to anemia.

41. In this study, the outcome variables of plasma folateand RBC folate were chosen to reflect the status of

two different body pools of folate. Which body pooldoes each reflect?

A. Plasma folate represents folate intake over thepast year, while RBC folate represents folateintake over the past day.

B. RBC folate represents folate intake over thepast year, while plasma folate represents folateintake over the past day.

C. Plasma folate represents folate intake over thepast day, while RBC folate represents folateintake over the past few months.

D. RBC folate represents folate intake over thepast day, while plasma folate represents folateintake over the past few months.


42.

43.

44.

45.

A microbiological assay using a folate-dependentstrain of Lactobacillus casei is used to determinelevels of folate in the plasma and in the red bloodcells. What type of laboratory technique is used toseparate the plasma and the red blood cells?

A. Extraction with an organic solventB. CentrifugationC. Thin layer chromatographyD. Dialysis

How would a microbiological assay for folate beperformed using Lactobacillus casei ?

A. The subject's sample is added to a knownamount of bacteria, incubated for a timeperiod, and the turbidity of the medium ismeasured.

B. The subject's sample is added to a knownamount of bacteria along with a known amountof folate, incubated for a time period, and theturbidity of the medium is measured.

C. The subject's sample is added to a knownamount of bacteria, incubated for a timeperiod, and the amount of folate produced ismeasured by an enzyme activity assay.

D. The subject's sample is added to a knownamount of bacteria, incubated for a timeperiod, and the number of mutated colonies iscounted.

Based on the results in Table 1, what conclusionwould you make about the levels of folate thesewomen were consuming in their diets before thestudy began?

A. They were consuming between 300 and 400Hg/day.

B. They were consuming more than 400 Jig/day.C. They were consuming between 200 and 300

M-g/day.D. They were consuming less than 200 (ig/day.

When folate is deficient, red blood cell synthesis isdramatically affected. Folate is required forsynthesis of the nucleotide thymidylate. Developingcells will enlarge too much before they are matureand are released from the bone marrow. What typeof anemia is this called?

A. Microcytic anemiaB. Megacytic anemiaC. Millicytic anemiaD. Macrocytic anemia


Biology Folate Experiment

46. What could you infer about the folate needs inpregnant women?

A. Folate needs are still about 200 Jig/day inpregnancy.

B. Folate needs increase during pregnancy due tosynthesisof many newcells.

C. Folate needs decrease during pregnancy due tosynthesis of folate by the placenta.

D. Folate needs of the mother are met by liverstores of folate, since folate is a fat-solublevitamin.

47. Which of the following structures indicates thepyrimidine nitrogenous base thymine?

A.

H,N N

C. D.

0' N


Passage VII


Biology Measurement of Blood Pressure Passage VIII

Passage VIII (Questions 48-54)

The first measurement of blood pressure, in the early1700s by Stephen Hales, used a cannula inserted into theartery of a horse. The pressure of the blood wasdetermined by the height it achieved in a vertical tubeattached to the cannula. Although this is a directtechnique, we currently use an indirect technique based onarterial sounds to measure blood pressure.

A pressure cuff is wrapped around the arm andinflated, compressing and closing the brachial artery.Using a stethoscope, the reader listens for the beginningof blood flow as the pressure is released from the cuff.The tapping sounds she hears are spurts of blood, pushingthrough the partially obstructed artery with each heartbeat.She also listens for the final sound and notes at whatpressure the first and last sounds occur. As the arterybecomes completely opened, no sounds are heard. Thefirst sound is recorded as the systolic pressure, and thefinal sound is recorded as the diastolic pressure. Anattached meter called a sphygmomanometer is used toread the pressures.

In Table 1 are some average arterial(mm/Hg) at various ages for men and women.

pressures

AgeSystolic

Men

SystolicWomen

Diastolic

Men

Diastolic

Women

20-24 123 116 76 72

25-29 125 117 78 74

30-34 126 120 78 75

35-39 127 124 80 78

40-44 129 127 81 80

45-49 130 131 82 82

50-54 135 137 83 84

55-59 138 139 84 84

60-64 142 144 85 85

Table 1

48. As the pressure in the cuff is released, what does thefirst sound, read as the systolic pressure, indicate?

A. The minimum pressure of the blood followinga heartbeat.

B. The beginning of laminar flow in the artery.C. The maximum pressure of the blood following

a heartbeat.

D. The heartbeat itself.


49. If a person has a large amount of fat tissue on theirhow will this change the blood pressurearms,

reading?

A.. The reading will be falsely high.B. The reading will be falsely low.C. The reading will be accurate.D. One cannot tell from this passage.

50. What is laminar flow?

A. A fluid moves as a group of particles, eachwith its own random course.

B. A fluid moves as a series of individual layers,each of the same velocity.

C. A fluid moves as single particles, each with itsown random course.

D. A fluid moves as a series of individual layers,each of a different velocity.

51. Gravity plays a role in reading blood pressure. Forthis reason, blood pressure measurements areconventionally made with the cuff approximatelylevel to the heart. For a 46-year-old woman, to whatvalue would the systolic pressure change if it weremeasured at the level of the ankle? Assume theankle is 100 cm below the heart, and gravityincreases blood pressure 0.8 mm/Hg per cm.

A. 162

B. 210

C. 211

D. 313

52. Astronauts on the space shuttle routinely participatein research experiments. Where should the bloodpressure cuff be placed for a reading of bloodpressure on an astronaut in space?

A. The cuff must be located in the conventional

position at the level of the heart.B. The cuff should be located at the ankle, as far

as possible from the level of the heart.C. The location of the cuff does not matter.

D. The cuff will not work in an anti-gravitysituation.


Biology Measurement of Blood Pressure Passage vm

53. When during the blood pressure measurement is theblood flow turbulent?

I. Before the systolic measurementII. Between the systolic and diastolic

measurements

III. After the diastolic measurement

A. I onlyB. II onlyC. II and m onlyD. I, II, and IH

54. The pulse pressure is the difference between thediastolic pressure andthesystolic pressure. It canbeused to calculate the mean arterial pressure (MAP)during the cardiac cycle.

MAP = diastolic pressure + (pulse pressure/3)

What is the MAP, on average, for a 27-year-oldman?

A. 51

B. 62

C. 78

D. 94


Biology Aortic Compliance Passage IX


To obtain curves shown in Figure 1, aortas wereobtained at autopsy from individuals in different agegroups. Successive volumes of liquid were injected intothis closed elastic system and after each increment ofvolume the internal pressure was measured. The graphcan beused toextract information about theeffects of ageon aortic compliance.

50 100 150

Pressure (mmHg)

Figure 1

The effects of aging on the elastic modulus (Ep) isshown in Figure 2. The elastic modulus is defined as AP/(AD/D). AP is the aortic pulse pressure, D is the meanaortic diameter during the cardiac cycle, and AD is themaximum change in aortic diameter during the cardiaccycle. The fractional change in diameter (AD/D) of theaorta during the cardiac cycle is a reflection of the changein volumeduring left ventricularcontraction. Finally, theheart is unable to eject its stroke volume into a rigidarterial system as rapidly as into a more compliant system.

0 20 40 60 80 100

Age (years)

Figure 2


55. Compliance is best represented by which of thefollowing relationships?

A. AD/D

B. AV/AP

C. AP/ADD. D/AD

56. Theelastic modulus is most likely:

A. constant over a lifetime.B. not related to aortic compliance.C. directly proportional to aortic compliance.D. inversely proportional to aortic compliance.

57. According to Figure 1, which curve most likelyrepresents the oldest age group?

A. Curve A.

B. Curve B.

C. Curve D.

D. Curve E.

58. As compliance diminishes, it is most likely thatpeakarterial pressure occurs progressively:

A. earlier in systole.B. later in systole.C. earlier in diastole.D. later in diastole.

59. According to Figure 1, aortic compliance in theyoungest individuals is:

A. smaller over all pressures when compared toother age groups.

B. greater over all pressures when compared toother age groups.

C. greatest at very high and low pressures andleast over the usual range of pressurevariations.

D. least at very high and low pressures andgreatest over the usual range of pressurevariations.

60. According to Figure 1, it can most likely beconcluded that compliance:

A. increases with age, aincreased arterial rigidity.

B. increases with age, adecreased arterial rigidity.

C. decreases with age, aincreased arterial rigidity.

D. decreases with age, adecreased arterial rigidity.

manifestation of

manifestation of

manifestation of

manifestation of


Biology Heart Muscle Action Potentials Passage X


Contraction of cardiac muscle is triggered bydepolarization of the plasma membrane. A typicalventricular action potential from a contractile myocardialcell is illustrated in Figure 1. In order to bring about thisaction potential, there must be inherent changes in thepermeabilities of ions flowing into and outof themusclecell.

0 0.15 0.30

Time (seconds)

Figure 1

Figure 2 illustrates the membrane permeabilitychanges which are depicted by the action potential.

J5 &

10.0 1 i1.0

0.1 -) /VB ^-^C

•a u

0 0.15 0.30

Time (seconds)

Figure 2

Special cells located in the heart itself are responsiblefor the heart's spontaneous and rhythmic self-contraction.These specialized cells of the sinoatrial (SA) node havesignificantdifferences in their membranepotentialswhencompared to contractile myocardial cells described inFigure 1.

£

Time (seconds)

Figure 3

Copyright© by The Berkeley Review

Figure 3 represents the membrane potentials of a cellfrom the sinoatrial node, which is located in the rightatrium of the heart.

126

61. In passing from the right atrium to the rightventricle, blood will have passed through the:

A. mitral valve.

B. semilunar valve.

C. tricuspid valve.D. pulmonary valve.

62. Figure 2 depicts the permeabilities of ions during theaction potential in a myocardial contractile cell.Graph C most likely depicts the permeability ofwhich of the following ions?

A. Na®B. K®C. Ca2®D. Cle

63. The capacity for spontaneous and rhythmic self-excitation is best manifested by which region of theaction potential shown in Figure 3?

A. Region A.B. Region B.C. Region C.D. Region D.

64. A membrane potential plateau region, similar to theone shown in Figure 1, is best explained by:

A. inactivation of Na® channels.B. a significant increase in the permeabilityof the

membrane to Ca2®.C. the flow of positive ions out of the cell

equaling the flowof positiveions into the cell.D. both Na® and K® channels becoming

inactivated.


Biology Heart Muscle Action Potentials Passage X

65.

66.

The action potential initiated in the SA node spreadsthroughout the right atrium from contractile cell tocontractile cell. The means by which this spreadtakes place is most likely through:

A. tight junctions between cells.B. neurotransmitter communication between

cells.

C. gap junctions between cells.D. desmosomes, which hold adjacent cardiac cells

together.

The following graph depicts three different actionpotentials from a SA nodal cell. Which of thefollowing statements is most likely true of thegraph?

A.

B.

C.

D.

A = control

-50

Time (seconds)

Graph B is the result of acetylcholine releaseonto the heart muscle.

Graph B is the result of vagus nervestimulation.

Graph C is the result of norepinephrine releaseonto the heart muscle.

Graph C is the result of vagus nervestimulation.


Biology Capillary Filtration Passage XI


The direction and magnitude of water movementacross a capillary wall are determined by the algebraicsum of the hydrostatic pressure and osmotic pressure thatexist across that membrane (Figure 1).

Filtration

jttMSSr33 mmHg

\Arteriole

End

Reabsorption

Lumen of Capillary

Interstitial Space

Figure 1

13 mmHg

VenuleEnd

The hydrostatic pressuregradientacross the wall of acapillary, Pc, dependson the arterial pressure (35 mmHg),the venous pressure (15 mmHg), and the interstitial fluidpressure, Pj, which is about 2 mmHg. Precapillary andpost capillary resistances are also important. A reductionin the arterial or venous pressure reduces capillaryhydrostatic pressure. The interstitial fluid pressureopposes capillary filtration, and the difference betweenthe two provides the driving force for filtration.

Important factors responsible for fluid retention in thecapillariesare the osmotic pressureof the plasmaproteins,7Cp, and the osmotic pressure ofthe interstitial proteins, 7tj.The osmotic pressure exerted by these proteins is alsocalled the colloid osmotic pressure or the oncoticpressure.

While this oncotic pressure is low compared to thetotal plasma osmotic pressure, it plays a major role influid reabsorption. This is due to the fact that theelectrolytes that are responsible for the major fraction ofplasma osmotic pressure are equal in concentration onboth sides of the endothelium.

The protein albumin is dominant in determiningplasma oncotic pressure. While albumin is relativelyimpermeable to thecapillary membrane, small amounts ofthe protein do leak into the interstitial fluid and create avery small osmotic force.


67.

68.

An increase in intracapillary hydrostatic pressurewould:

A. increase the concentration of osmoticallyactive particles within the vessel.

B. result from a decrease in the arterial pressure.C. favor movement of fluid from vessel to

interstitial space.D. increase the venous resistance.

Studies using plastic capsules implanted in thesubcutaneous tissue indicate a Pj of -1 to -7 mmHg.If such values are accurate, which of the followingcould most likely be concluded?

A. Pc-Pi>Pc.B. Pc-Pj<Pc.C. Pj-Pc>Pi.D. Pc-Pi<Pi-

69. A reduction in the diameter of a precapillary vesselwould:

A. favor movement of fluid from the vessel to the

interstitial space.B. result in a reduction of capillary hydrostatic

pressure.

C. be equivalent to an increase in venousresistance.

D. significantly increase the concentration ofosmotically active particles in the interstitialspace.

128

70. Fluid movement is most likely described by whichof the following equations? Note: k is the filtrationconstant for the capillary membrane.

A. Fluid movement = k[(Pc + iz\) - (Pj + 7tp)].B. Fluid movement = k[(Pc + Pj) - (7Cp - n\)].C. Fluid movement = k[(Pc - Pi) + (rci + rcp)].D. Fluid movement = k[(n\ + np) - (Pj + Pc)l-


Biology Capillary Filtration

71. Albumin exerts a greater osmotic force than can beaccounted for solely on the basis of the number ofmolecules dissolved in plasma. Based on thisinformation, it can BEST be described that albumin:

A. may be replaced by inert substances with noeffect.

B. dissolves into more than one protein moleculein plasma.

C. carries electrical charges at blood pH whichattracts various electrolytes.

D. is permeable to the capillary endothelium.

72. Only a small percentage (2%) of the plasma flowingthrough the vascular system is filtered, and of this,about 85% is reabsorbed in the capillaries and thevenules. The remaining 15% of fluid:

A. acts to increase intracapillary hydrostaticpressure.

B. remains in the interstitial fluid.

C. returns to the arterial circulation via thelymphatic system.

D. returns to the venous circulation via thelymphatic system.


Passage XI


Biology Respiratory Calculations Passage XII


The bronchial circulation is the nutritive supply to allof the lung support structures. Such structures includeairways, connective tissue, and pleura. The pulmonarycirculation is the nutritive supply to the alveolar walls.The pulmonary capillaries form an extensive networkwithin the alveolar walls. The maximum capillaryvolume is about 200 ml, while the normal capillaryvolume at rest is about 70 ml. This capillary volume canbe increased by openingcompressed or closedcapillaries.Such opening is callled recruitment.

For a given volume, the total gas pressure of allmolecular species is the sum of the individual pressures.This law of partial pressures makes the assumption thatgas molecules do not interact with each other. The dryroom atmospheric pressure is given the value 760 mmHg.This air contains 21% O2, 79% N2, and 0% CO2. Bloodthat leaves the. pulmonary capillaries has come intoequilibrium with all of the alveolar gases. In addition,small quantities of venous blood from bronchial venulesand vessels from the heart join the pulmonary venousoutflow.

When air is inspired, it becomes saturated with watervapor at 37 degrees Celcius in the nose, throat, andtrachea. The source of the heat and water vapor are thepulmonary and bronchial blood flows. The water vaporexerts a mandatory partial pressure, with the PH2O = 47mmHg. The total quantitiy of water in expired gas over a24-hour period accounts for nearly one-half of theobligatory daily water loss from the body.

In the alveoli, some oxygen diffuses directly into thepulmonary capillary blood. In fact, 3.0 ml of O2 aredissolved in 1L of blood for a p02 = 90 mmHg. Thisreduces the fraction of O2 in alveolar gas to 0.143. Mostoxygen binds to the protein hemoglobin, which is able tobind 1.34 ml 02/g, with the normal hemoglobinconcentration in blood being 150 g/L of blood.

73. Besides recruitment, capillary volume can beincreased by enlarging open capillaries as theirinternal pressure rises. Such distension is mostlikely caused by:

A. increased cardiac output.B. increased left atrial pressure.C. decreased left ventricular pressure.D. right heart failure.


74. A respiration physiologist measures a volume of airto be 1L at 0 degrees Celcius and 1 ATM. What willbe the volume of this air at 25 degrees Celcius and1ATM?

A. 0.82 L

B. 0.91 L

C. 1.0 L

D. 1.09 L

75. The partial pressure of O2 in systemic arterial bloodis normally 5-10 mmHg less than that in alveolargas. The difference in partial pressure existsbecause:

A. a fraction of O2 is directly dissolved into thepulmonary blood flow.

B. a fraction of O2 is replaced by the partialpressure of water.

C. the venous blood from bronchial venules andheart vessels contaminate the pulmonaryvenous outflow.

D. the partial pressure of gases dissolved in liquidis not equal to the partial pressure of the gasphase at equilibrium.

76.

77.

78.

130

Pulmonary physiologists are often interested in onlyanhydrous gas volumes and partial pressures. Takingthis into consideration, the partial pressure ofoxygen in dry inspired air at the trachea is:

A.

B.

C.

D.

102 mmHg.115 mmHg.150 mmHg.170 mmHg.

Based on the information in the passage, theanhydrous partial pressure of oxygen in alveolar gasis:

A.

B.

C.

D.

102 mmHg.115 mmHg.150 mmHg.170 mmHg.

At an oxygen partial pressure of 90 mmHg, 97% ofhemoglobin is fully saturated with oxygen. At 90mmHg, which of the following is the BESTcalculation of the TOTAL concentration of oxygen(ml O2/L) in the systemic arterial blood?

A. 145 ml 02/L.B. 195 ml O2/L.C. 197ml02/L.D. 200 ml O2/L.


Biology Aspirin


Aspirin is a non-steroidal anti-inflammatory agent thatworks by inhibiting the initial enzyme in prostaglandinsynthesis pathway, cyclooxygenase (CO). Aspirinacetylates a serine residue in the active site of CO,inactivating it. This makes aspirin a useful drug formodifying prostaglandin synthesis.

The diagram in Figure 1 shows the pathways ofprostaglandin synthesis from the fatty acid, arachidonicacid.

Leukotrienes

Arachidonate

Cyclooxygenase

Prostaglandins Thromboxanes

Figure 1

Low-dose aspirin treatment is often used to reduce riskof heart attack or stroke in people with a history ofcardiovascular disease. Aspirin lowers the chance ofblood clots traveling through the vessels and obstructingblood flow through the coronary arteries of the heart orthrough the carotid arteries leading to the brain. Aspirinworks by altering the ratio of various prostaglandins toreduce clot formation.

Specifically, the important alteration is in the ratio ofprostacyclin to thromboxane A2. Platelets, which have alife span of four days, are important in the clottingprocess. Platelets secrete thromboxane A2, whichpromotes platelet aggregation and vasoconstriction. Theendothelial cells of the blood vessels secrete prostacyclin,which inhibits platelet aggregation and causesvasodilation. The balance between the two causes

appropriate clotting: a strong clot that does not cover anexcessive area, and that permits unrestricted blood flowaround the clot.

Aspirin inactivates CO permanently in both plateletsand endothelial cells. Endothelial cells rapidly producemore CO enzymatically, but platelets cannot manufactureCO.

The model shown in Figure 2 indicates CO levels inplatelets and vascular endothelial cells after the ingestionof 300 mg of aspirin:


79.

80.

131

Passage XIII

1 2 3

Days Following Aspirin Dose

Figure 2

Which of the following statements BEST describeswhat is happening in Figure 1?

B.

D.

After aspirin treatment, platelets produce newCO, but endothelial cells do not.After aspirin treatment, endothelial cellsproduce new CO, but platelets do not.After aspirin treatment, both platelets andendothelial cells produce new CO.After aspirin treatment, neither platelets norendothelial cells produce new CO.

How many degrees of unsaturation are present in thefollowing structure, thromboxane B2?

A.

B.

C.

D.

One degree of unsaturation.Two degrees of unsaturation.Three degrees of unsaturation.Four degrees of unsaturation.


Biology Aspirin

81. When aspirin modifies CO, inactivating it, whichstructure indicates the product amino acid residue?

A.

C.

H

NH,-C-C-0-I

CH,I

O

'AO CH,

B.

D.

NH,

H OI II

_ c-c-o-

CH,I

SI

'A0 CH,

H OI II

NH,-C-C-0

CH,I

O

0 = P-0(

00

— —NH,-C-C-0 —I

CH,

S

0 = P-0*I

©o

82. What symptoms would accompany an intake ofaspirin that was too high?

I. Easy bruising.II. Prolonged bleeding time.

HI. Rapid clotting.

A. I onlyB. I and II onlyC. I and HI onlyD. I, H, and HI

83. The fatty acid molecule that is the precursor toprostaglandins and the substrate for cyclooxygenasehas the common name of arachidonic acid. What is

the IUPAC name?

A. 5, 8,11, 14 arachidonic acid.B. 6,9, 12, 15-arachidonic acid.C. 6, 9, 12, 15-eicosatetraenoic acid.D. 5, 8, 11, 14 eicosatetraenoic acid.


84.

85.

132

Passage XIII

Low-dose aspirin therapy is used every other day toprevent heart attacks in some people with heartdisease. Why does this dosing regimen help?

A. The dosing regimen keeps platelet aggregationlow.

B. The dosing regimen keeps platelet aggregationhigh.

C. The dosing regimen keeps endothelial cellsproliferation low.

D. The dosing regimen keeps endothelial cellsproliferation high.

Although the mechanisms behind the birth processare not well understood, it is clear thatprostaglandins play an important role in initiatingand promoting parturition. What does this indicatefor late pregnancy (months 7-9)?

A. Aspirin would probably stimulate an earlylabor.

B. Aspirin would have no effect on pregnancy.C. Aspirin should not be used in late pregnancy.D. Aspirin would probably cause birth defects.


Biology Sickle Cell Anemia and HbF Passage XIV

Passage XTV (Questions 86-92)

Sickle cell anemia is caused by a single substitution ofone amino acid on the P chain of the adult hemoglobin(HbA) molecule. This change produces the form ofhemoglobin called HbS. Under conditions of low bloodpartial pressure of oxygen (p02), HbS comes out ofsolution and forms a cross-linked crystalline structureinside the red blood cell. This leads to the characteristicsickle shape of the red blood cells. Complications canarise because the sickled red blood cells cannot properlyfold up enough to pass through small capillaries of thecirculatory system. Tissues are deprived of some of theirblood supply by blockages caused by sickled cells.

Homozygotes are severely affected by the sicklingcells, but heterozygotes rarely experience sickling undernormal oxygen levels. Heterozygotes may experienceproblems at high altitudes (such as in a depressurizedplane) or under some types of anesthesia.

Certain drug treatments for sickle cell anemia focus onincreasing levels of fetal hemoglobin (HbF), which doesnot sickle. The following chart shows types of Hb presentat various ages:

a chain

3 6

Gestation

(months)

Birth

Figure 1

3 6

Age(months)

86. Sickle cell anemia is usually diagnosed by 3 monthsof age, but not always at birth. This is most likelydue to:

A. a decrease of HbF and an increase of HbA

with age.B. persistent maternal hemoglobin until age 3

months.

C. a decrease in HbF and an increase in HbS with

age.

D. persistent maternal antibodies to HbS until age3 months.


87.

88.

89.

In the blood cells of an adult heterozygote with thesickle cell trait, how are the hemoglobin typesdistributed?

A.

B.

C.

D.

All cells contain both HbS and HbA.Some cells contain exclusively HbS, whileothers contain exclusively HbA.All cells contain HbA only.All cells contain HbS only.

Treatment with hydroxyurea in homozygotespromotes fewer crisis events. HbF is increased toabout 20% of the cell's hemoglobin content bytreatment with hydroxyurea. What beneficial roledoes HbF play inside the cell?

A. HbF binds and inactivates HbS inside the cell.B. HbF promotes a lower oxygenation state inside

the cell.

C. HbF promotes a higher oxygenation stateinside the cell.

D. HbF binds and activates HbA inside the cell.

Sodium metabisulfate (Na20sS2) functions as anantioxidant in many medical preparations. It istransformed to sodium persulfate (Na20gS2) in itsrole as an antioxidant. How could sodium

metabisulfate be used to test for sickle cell trait

when combined with hemoglobin from red bloodcells?

A. Sodium metabisulfate would promotereduction of the hemoglobin to a rusty color,which would identify affected individuals.

B. Sodium metabisulfate would oxygenatehemoglobin from red blood cells, and thesickling pattern would be noted in affectedindividuals.

C. Sodium metabisulfate would promoteoxidation of the hemoglobin to a rusty color,which would identify affected individuals.

D. Sodium metabisulfate would deoxygenatehemoglobin from red blood cells, and thesickling pattern would be noted in affectedindividuals.


Biology Sickle Cell Anemia and HbF Passage XIV

90. What would happen if an adult had too much HbF,as is the case in some of the thalassemias, in whichtoo much of the y chain of hemoglobin is produced?

A. Decreased delivery of oxygen to the tissues,compared to other adults.

B. Increased delivery of oxygen to the tissues,compared to other adults.

C. The red blood cells would sickle, but at highp02 levels.

D. The red blood cells would sickle, but at lowp02 levels.

91. Which diagnostic technique would distinguishheterozygotes, homozygotes with sicklecell anemia,and wild-type homozygotes?

I. Restriction fragment length polymorphisms(RFLPs) on red blood cell DNA.

II. Gel electrophoresis of hemoglobin samplesfrom the red blood cells.

in. Centrifugation of hemoglobin samples fromthe red blood cells.

A. I onlyB. II onlyC. I and n onlyD. I, n, and HI

92. Which of the following statements is TRUE of HbF?

I. HbF does not bind 2,3-BPG.II. HbF has a higher oxygen saturation at a given

p02 than HbA.III. HbF contains two a and two P chains.

A. I onlyB. I and II onlyC. n and HI onlyD. I, H, and m


Biology Ventilation Regulation Passage XV


A group of inspiratory neurons located in the medullais responsible for controlling ventilation. Inputs to theseneurons from both the peripheral and centralchemoreceptors are important in regulating involuntarycontrol of ventilation. The peripheral chemoreceptorsconsist of two different bodies of receptors located in thecarotid artery of the neck and in the arch of the aorta. Thenerve terminals are intimate with the arterial blood and

respond to changes in pC02 , p02, and H®. Impulsesfrom these nerve terminals travel up afferent fibers towardthe brainstem and eventually synapse with inspiratoryneurons of the breathing center. The centralchemoreceptors, located in the medulla, monitor thebrain's extracellular fluid for changes in the concentrationof protons. Fibers from this receptor synapse withinspiratory neurons.

~ 30c

IS 20

c

>

Bsc

i

10

0 20 40 60 80 100

Arterialp02 (mmHg)

Figure 1

Figure 1 shows the relationship between arterial p02and the ventilation rate. The graph indicates that alowering of the arterial p02 will stimulate the rate ofventilation. This reflex is mediated by the peripheralchemoreceptors which respond to a lower arterial p02 byincreasing their rate of discharge to the breathing center.

c

6-1

16 r

12

i 8c

>

3 4 NormalRestingLevel

40 44 48

ArterialpC02 (mmHg)

Figure 2


Figure 2 shows the results of the same experiment withthe variable being arterial pC02- The reflex involves theperipheral and central chemoreceptors and is closely tiedto changes in the proton concentration.

An increased arterial pC02 will cause an increase inthe arterial concentration of protons. The peripheralchemoreceptors are stimulated by this lowered pH. At thesame time, the elevated arterial pC02 causes an increasein the pC02 of extracellular brain fluid. This causes a risein the proton concentration of the fluid, stimulating thecentral chemoreceptors. It is the role of the centralchemoreceptor which is most important in mediating thisventilation reflex.

93. To assess the effects of a changing arterial p02 onthe ventilation rate accurately, which of thefollowing conditions should exist?

A. The arterial pC02 should decrease at aconstant rate.

B. The arterial pC02 should increase at a constantrate.

C. The arterial pC02 should be maintained at 40mmHg.

D. The arterial pC02 should be maintained at 46mmHg.

94. According to Figure 1, a drop of 30 mmHg from thenormal resting value of p02 does NOT radicallyincrease the ventilation rate. This occurs most likelybecause the:

A. pC02 is increasing.B. total amount of oxygen transported is

relatively unaffected.C. body responds with a decreased use of oxygen.D. pC02 is decreasing.

95. The inspiration of carbon monoxide gas will resultin:

A.

B.

C.

D.

an unchanged arterial p02-a lowered arterial p02-an increased arterial p02-an increased percentage of hemoglobinsaturated with O2.


Biology Ventilation Regulation Passage XV

96.

97.

98.

99.

According to Figure 1 and Figure 2, changes inarterial pC02 are:

A. augmented by reflexes regulating ventilation toa greater degree than are equivalentchangesinarterial p02-

B. augmented by reflexes regulating ventilation toa lesser degree than are equivalent changes inarterial p02-

C. resisted by reflexes regulating ventilation to agreater degree than are equivalent changes inarterial p02-

D. resisted by reflexes regulating ventilation to alesser degree than are equivalent changes inarterial p02-

Which of the following statements provides theBEST support for the type of ventilation found earlyon in metabolic acidosis?

A. Hypoventilation is caused by increased neuraloutput from the peripheral chemoreceptors.

B. Hypoventilation is caused by increased neuraloutput from the central chemoreceptors.

C. Hyperventilation is caused by increased neuraloutput from the peripheral chemoreceptors.

D. Hyperventilation is caused by increased neuraloutput from the central chemoreceptors.

Which of the statements below is FALSE regardingpC02 during exercise?

A. The alveolar pC02 determines arterial pC02-B. Alveolar ventilation increases.

C. Venous pC02 increases.D. Arterial pC02 increases.

In order to hold their breath for a longer duration,swimmers often hyperventilate immediately beforecompeting in a swimming event. This activity can bedangerous because even though the:

A.

B.

D.

low pC02 is permitting one to hold theirbreath, the exercise may lower the p02 tolevels which may induce unconsciousness,low pC02 is permitting one to hold theirbreath, the exercise may raise the p02 to levelswhich may induce unconsciousness,high pC02 is permitting one to hold theirbreath, the exercise may lower the p02 tolevels which may induce unconsciousness,high pC02 is permitting one to hold theirbreath, the exercise may raise the p02 to levelswhich may induce unconsciousness.


100. During times of sleep, the body's ventilation ratedecreases more than the consumption of oxygen bycells. The result of this situation is:

A. an increased arterial pC02 and a increasedarterial p02-

B. an increased arterial pC02 and a decreasedarterial p02-

C. a decreased arterial pC02 and an increasedarterial p02-

D. a decreased arterial pC02 and a decreasedarterial p02-


Biology Heart St Lungs Section II Answers

Passage 1(1 - 7) Mechanics of Blood Flow

1. A is correct, low average velocity because ofa high total cross-sectional area. This is according to the continuityequation describing fluid How. This equation states that since blood flow must be equal throughout an entire closedsystem, the cross-sectional area times the average velocity equals the cross-sectional area times the velocity. Thetotal cross-sectional area of all the capillaries is very large, the largest of all types of vessels seen in thecardiovascular system. Therefore, the blood flow through the capillaries will have a low average velocity because oftheir high total cross-sectional area. The correct choice is A.

2. D is correct, 16. The problem requires that we be familiar with the relationship between resistance of a vessel andthe radius. The resistance of the vessel varies inversely with the radius to the fourth power. Therefore, decreasingthe radius by a factor of 2 leads to an increase in the resistance by a factorof 16. The correct choice is D.

3. C is correct, Q x TPR = (MAP - CVP). We are looking for an expression to represent blood How. The generalequation is given in the passage. Blood flow (Q) = Perfusion Pressure/Flow Resistance. We are asked about theentire vascular unit. The perfusion pressure is therefore going to be the mean arterial pressure minus the centralvenous pressure. The resistance for theentire unit is given by the total peripheral resistance. Placing these values inthe formula and multiplying both sides by TPR will give the solution. The correct choice is C.

4. D is correct, red blood cells. One can arrive at this answer by thinking about the composition of blood. Blood ismade up of plasma and cells. The question is asking about the increase in viscosity seen in whole blood versusplasma. Knowing the composition of blood leads one to believe that increase must come from the cellularcontribution. It then becomes a matter of WBCs or RBCs. At this point, one must be aware that there are millionsof red blood cells in whole blood. In other words, there are many more red cell than white cells. Therefore, themajor contribution must come from the presence of red blood cells. The correct choice is D.

5. C is correct, veins contain valves that favor one-way blood flow. We are required to draw on knowledge about thestructure of blood vessels. Both arteries and veins have a layer of endothelial cells, a layer of elastic tissue, and alayer of smooth muscle. Therefore, we can eliminate answer choices A, B, and D. However, only veins containone-way valves which prevent back flow of blood. One must understand that the pressure in the veins is not verylarge, and so there is not a tremendous amount of force to propel blood. When the blood is propelled (skeletalmuscle contraction is one method of providing the force), we wish the blood to remain moving forward. Thepresence of these one-way valves carry out this function. The correct choice is C.

6. B is correct, venous pressure does not change significantly over the heart cycle. The MAP is an average figurebecause the pressure in the arterial system changes significantly over the heart cycle. It is very high right out of theaorta, but becomes smaller and smaller toward the capillaries. Therefore, we need to take an average value. Thepressure in the veins does not change very dramatically over the entire heart cycle. The total pressure difference inthe peripheral veins is usually 5 to 10 mmHg. Therefore, we do not need to lake an average value. The fact that thearterial pressure does change dramatically is true, but it is not proper rationale for trying to explain why we do notneed to take an average for the veins. Furthermore the cross-sectional area and the pressure responses to increases involume offer us no explanation as to why no average is taken. The average is not taken because the value remainsrelatively constant. The correct choice is B.

7. D is correct, arterial systolic pressure. We are told in the question that we need to use the pressure in the cuff tocollapse the artery totally and prevent blood flow. Therefore, the pressure in the cuff must counter the maximumpressure offered by the artery. The maximum pressure in the artery will be represented by the arterial systolicpressure. Recall that systolic pressure is the maximum arterial blood pressure during cardiac cycle. To collapse theartery, the pressure in the cuff must exceed this value. The correct choice is D.

Passage II (8-14) Enzymes in Blood Clotting

8. D is correct, glycolysis. It is important to know which cells have organelles and which do not. It will also beimportant to know certain metabolic processes and where they take place. Even though we have not formallydiscussed metabolism yet, this question is designed to get you to think about cellular function. Mature red bloodcells are enucleated eukaryotic cells. They do not have any organelles and they have only one membrane, theplasma membrane. As we will see in future discussions, the Krcbs cycle, electron transport, and oxidativephosphorylation are all associated (in eukaryotic cells) with mitochondria. Even though these three systems areresponsible for the majority of energy generated in metabolism, they arc not found in red blood cells. However, redblood cells do require energy and they obtain that energy from glycolysis. The correct choice is D.


Biology Heart & Lungs Section II Answers

9. C is correct, reductase. The vitamin K dependent reaction, as written from glutamate to y-carboxyglutamate, is acarboxylation and would require a carboxylaseenzyme. If we wanted to go from y-carboxyglutamate to glutamate,it would be a decarboxylation reaction, involving a decarboxylase enzyme. Proteases are proteolytic enzymes thatcatalyze the hydrolysis of peptide bonds.

Vitamin K Vitamin KHydroquinone Quinone

— N-

H

H

C-C-N —

CH, HI

CH,

' ©coo

Glutamate

rC02

H 0II

:> —N-C-C-N —II I

H CH, HI

r> CH /-vOOC COO

y-Carboxyglutamate

In this reaction we are not degrading a protein. We are regenerating vitamin K hydroquinone, which is the reducedform of vitamin K. In order to go from the oxidized form (vitamin K quinone) to the reduced form, a reductionneeds to occur. This reaction will be catalyzed by a reductase. The correct choice is C.

10. B is correct, thromboplastin (Factor HI or Tissue factor). This question can be answered by looking directly atTable 1 and Figure 1 in the passage. In Table 1 the symbol "f" denotes dependency on vitamin K. All that isneeded is a correlation between the common names and the factor numbers. The correct choice is B.

11. A is correct, consumption of leafy green vegetables. The major source of vitamin K is from one's diet, especially inleafy green vegetables. There are also intestinal bacteria that can synthesize vitamin K. Since vitamin K is beingmade available to be a cofactor in the conversion of glutamate to y-carboxyglutamate, we would expect an increasein clot growth.

Since bloodclots can break free frombloodvessels and wreak havoc within the circulatory system (by causing heartattacks and strokes), their growth must be limited. Clot growth can be limited by inhibiting the serine proteaseswhich act at various steps in the cascade. Since blood is dynamic (constantly moving), the areas near a clot arealways being diluted with fresh blood. This helps to prevent a build up of cascade intermediates involved in theclottingprocess. The intermediates that are removed by the flowof bloodare transported to the liver where they areremoved and degraded. The correct choice is A.

12. C is correct, 0.50 (or 50%). Men have an X and a Y chromosome; women have two X chromosomes. We are toldin the question that the defect for both types of hemophilia (h) resides on the X chromosome and that it is recessive.The woman's father has hemophilia, which we can designatedas XnY. Since her mother is normal (XX) the womanmust be a carrier (XXn) because she received one good X chromosome from her mother and one defective Xnchromosome from her father. This is shown in the Punnett square in Figure 1 below:

X" xxh XXh

XY XY

Figure 1

X

X" X

xxh XX

XhY XY

Figure 2

The woman (XXn) now marries a man who is normal (XY). The mode of inheritance of their offspring is indicatedin Figure 2 above. There are two sons and two daughters. However, we care about only the sons. One will benormal, while the other will express the trait. Therefore, the probability that this woman will have an affected son is0.50 or 50%. Even though we have not formally discussed simple Mendelian genetics yet, this question is designedto get you to think along those lines. The correct choice is C.



13. A is correct, citrate. Calcium is an important cofactor in thecascade mechanism outlined in Figure 1 in the passage.This ion bears a +2 charge and can be chelated by certain compounds that bear negative charges. As stated in thepassage the modified amino acid y-carboxyglutamate is an excellent chelatorof Ca2®. When this amino acid is tiedup in a protein, it presents two carboxyl groups to the calcium ion. Based on this we could assume that the morecarboxyl groups on a molecule, the better the chelator. That is, the better the molecule is at sequestering calcium.

If we follow this assumption, we can eliminate choice B first. Glutamate has two carboxylate groups. However,note the positively-charged amino group. This would tend to repel a positively-charged calcium ion. Not only that,but the positively-charged amino group will be tied up with the a-carboxyl group through electrostatic interactions.This will diminish the chelating ability of that molecule. We can eliminate choice D based on similar reasoning.The major difference between the last two choices is that choice C (succinate) has just two full carboxylate groupswhereas choice A (citrate) has three full carboxylate groups. In this case, the more carboxyl groups, the better thechelator, and the better the molecule is at preventing clotting in stored blood. The correct choice is A.

14. B is correct, proconvertin (Factor VII). This question is designed to get you to think about a number of items at thesame time. As stated in the passage, warfarin is an antagonist (i.e., a competitive inhibitor)of vitamin K. Now, youneed to ask yourself, "Which coagulation factors in the cascade are affected by inhibition of the use of vitamin K?"The factors which are affected (from Table 1) are those which arc dependent on vitamin K. Those factors turn out tobe choices A-D in the answers (i.e., prothrombin, factor II; proconvertin, factor VII; Christmas factor, factor IX;Stuart-Prower factor, factor X). Since there is just one answer (and not four), we need to consider some moreinformation.

In the question it states that 3 days after oral administration, the PT (prothrombin time) is maintained at 1.2-1.5limes the control. The PT screens the extrinsic and common pathways. Note that in the cascade shown in Figure 1,the extrinsic pathway includes factor VII and the common pathway includes factor X and factor II. The extrinsicpathway does not screen factor IX as it is part of the intrinsic pathway. This allows us to eliminate choice C(Christmas factor, factor IX) as a possible solution.

How do we distinguish between the remaining choices? Another way to look at this is to ask yourself, "Whichsingle factor decreases so much after 3 days as to maintain a PT between 1.2-1.5 limes the control?" You shouldnow be think about the half-life of these factors (see Table 1 in the passage).

The one factor which has the shortest half-life is factor VII (between 2-6 hours). Let's assume that its half-life isabout 4 hours. In 3 days (72 hours) factor VII will have been reduced by 18 half-lives. The half-life of factor X isbetween 25-60 hours. Let's assume that its half-life is about 43 hours. Factor X will have been reduced by about 2half-lives. Finally, factor II has a half-life between 50-120 hours. Let's assume that its half-life is about 85 hours.Factor II will have been reduced by about 1 half-life. [All values have been rounded up for simplicity.! Based onthis analysis, we see that factor VII will have been decreased the most, because it has such a short life span. Thecorrect choice is B.

Passage III (15- 21) Cardiac Output & Venous Return

15. B is correct, arterioles. The answer to this queslion is not found in the passage but rather comes from ourknowledge of the flow of blood through the circulatory system.

Roughly 64% of blood in the circulatory system is contained in the veins while about 15% is contained in thearteries. The veins include the large veins, small veins, and venules. The small veins and venules together containabout 25% of the total blood volume, leaving about 39% of the total blood volume found in the large veins. Thearteries include the large arteries, small arteries, and arterioles. The large arteries and small arteries together containabout 13%of the total blood volume, leaving about 2%of the total blood volume found in the arterioles. About 5%of the blood is contained in the capillaries. The remaining 16% of the blood is contained in the heat and pulmonaryvessels (associated with the lungs). This allows us to pick the arterioles as having the LEAST blood volume (at anygiven time) in the circulatory system. The correct choice is B.

16. C is correct, lower due to the accumulation of CCb- Recall that as blood passes through the tissues it gives up Cbas a nutrient and picks up CO2 as a waste product. The blood that is leaving the pulmonary artery is coming fromthe right ventricle, which in turn came from the right atrium, which in turn came from the venous system.

CCb + H<>0 f-hCO


0 ©

HCO-, + H


Biology Heart St Lungs Section n Answers

Blood in the venous system is, for all practical purposes,deoxygenatedblood. The oxygen had been given up at thelevel of the respiring tissues. The CO2 that was picked up by the venous system is either dissolved in the blood asfree CO2 (about 5%), bound to hemoglobin as the carbamate (about 5%), or in the form of the bicarbonate ion(about 90%). When the bicarbonate ion is formed (Bohr reaction), hydrogen ions are produced which tend to lowerthe pH of the blood. The correct choice is C.

17. A is correct, (see the graph below). When there is a constriction of a blood vessel there is more resistance to bloodflow. This leads to a greater pressure drop (i.e., a greater change in pressure, AP) between the veins and arteries.The result is a lower venous pressure. Therefore, the curve would be expected to drop along the y-axis.

The curve intersects at the samepointon the x-axis because if the heart were to stop and the cardiacoutput reducedto zero, the pressures in the venous andarterial systems would soonequilibrate. In otherwords, the venous pressurewould not change. This makes choice A a good answer.

18.

19.

*3 10 i

-4 0-4 8 12 16


Choice B indicates a decrease in the blood volume. When the blood volume decreases there is less blood to fill theventricles. This leads to less of a contractile force and a lower blood pressure. Choice C indicates decreasedarteriole resistance while choice D indicates and increase in blood volume. The correct choice is A.

D is correct, vasopressin and aldosterone. When the blood volume is low, neural impulses form the atria andarterial baroreceptors aresent to the hypothalamus and vasopressin (antidiuretic hormone, ADH) is released into theblood. Also, if the blood volume is low, chances are that the blood osmolarity has increased. An increasedosmolarity is also detected at the level of the hypothalamus and this allows for the releaseof ADH as well. ADHacts at the level of the late distal tubules and the collecting ducts in the kidney and directly promotes waterreabsorption (back into the blood). This increases the blood volume.

A low blood volume is usually associated with a low blood pressure. This is detected by the juxtaglomerularapparatus in the kidney and granular cells in that complex release renin. This enzyme converts angiotensinogen toangiotensin I, and then angiotensin I isconverted to angiotensin IIby the angiotensin-converting enzyme. Not onlydoes angiotensin II cause vasoconstriction of the arterioles (tohelp increase blood pressure), but it alsostimulatesthe release aldosterone from the cortex of the adrenal glands. Aldosterone stimulates the reabsorption ofsodium atthe level of the kidney. Water follows down its concentration gradient into the blood and leads to an increase inblood volume. The correct choice is D.

C is correct, (see the graph below). Thesympathetic nervous system is a component of the autonomic nervoussystem. Recall that the sympathetic nervous system is involved in thefight-or-flight response. Activation of thesympathetic nervous system will stimulate the heart and cause it to bea stronger pump. In other words, theactivityof theheart increases. This will cause the cardiac output (CO.)curve toshift toa higher level. Wesee this responsein choices A and C.

Oca

-4 0 4 8 12 16


Choice A

Oca

MCP

0 4 8 12 16


Choice C



Activation of the sympathetic nervous system will also cause vasoconstriction of arteries, arterioles, and veins.Contraction of venous smooth muscle will lead to a decrease in the diameter and increase in the pressure within thevessel. This will allow more blood to bereturned to the right heart (i.e., the right atrium and right ventricle) fromthe veins. The result isan increase invenous return and an increase inthe mean circulatory pressure.

Since cardiac output must equal venous return (see the first paragraph in the passage), we can equate the y-axis ofour curve in the question not only tocardiac output but also tovenous return. Therefore, we would expect the curveto move upwards along the y-axis for venous return (V.R.) and outwards along the x-axis for an increase in meancirculatory pressure (see above). The correct choice is C.

[Note: In the thirdparagraphof the passageit states that if the cardiac output andthevenous return is increased, thepressure in the arterial system will increase while the pressure in the venous system willdecrease. Be careful ofwhat it means when it says "the pressure in the venous system will decrease." If we are at a venous return of 0L/min, the venous pressureis about 8 mmHg (see Figure 1 in the passage, or below). If we now move to a venousreturn of 5 L/min, the venous pressure is about -4 mmHg. It is a decrease form the mean circulatory pressure(MCP).

0 4 8 12 16


In question 5 we have increased the MCP by constricting vessels. An increase in cardiac output due to sympatheticstimulation will decrease the venous pressure relative to the new MCP.]

20. A is correct, norepinephrine, because it opens Na® and Ca^® channels in myocardial cells. This question ties inaconcept we examined when we discussed the nervous system. Recall that the preganglionic and postganglionicnerve fibers of the parasympathetic nervous system release the neurotransmitter acetylcholine (ACh). Thepreganglionic fibers of the sympathetic division release ACh while the postganglionic fibers release theneurotransmitter norepinephrine. We can now eliminate choices B and D.

Next, we need to recall that sympathetic fibers will stimulate the heart (think of the fight-orflight response). If theheart is to be stimulated, we would expect to see rapid depolarization of the myocardial cells. Depolarization of acell's membrane is due to an influx of Na® into the cell. This allows a negative resting membrane potential tobecome more positive. An influx of both Na® and Ca^® into a cell will allow the resting membrane potential tobecome depolarized. Choice A looks likea good answer at the moment. What would happen if the K® channelswere to open up? Potassium would leave the cell, making the resting membrane potential more negative. The cellwould have a higher threshold potential and would not be depolarized as readily. We can eliminate choice C. Thecorrect choice is A.

21. D is correct, decreased arterial pressure. The key word here is compensatory. How does the body handle ahemorrhage? The first thing that must be done is to prevent more loss of blood. Constriction of blood vessels wouldhelp, which means that an increased sympathetic discharge to the arterioles (choice A) is true. We still want theheart to pump blood. Therefore, we do not want to inhibit its action. We can remove inhibitory signals to the heartby decreasing the amount of parasympathetic discharge at the heart (choice B). After a blood loss the body stillneeds to maintain a reasonable cardiac output. Therefore, we would expect to see an increase in the cardiac output(choice C). What we would not expect to see is a decreased arterial pressure (choice D). Why? A decreasedarterial pressure is theRESULT of a hemorrhage. Wewant theCOMPENSATION, which would be to increase thearterial pressure. We can reason this out as outlined below.

When a hemorrhage takes place there is a blood loss and a decrease in the blood volume. The venous pressure,venous return, and atrial pressure will all decrease accordingly. The volume of blood ejected by a ventricle in theheart (i.e., the stroke volume)during one heartbeat is decreasedas well. This leads to a decreased cardiac output anda decreased arterial blood pressure.



Located in the upper neck region are small arteries which branch off of the common carotid arteries. The commoncarotid arteries stem from the aortic arch of the aorta, the main vessel that leaves the left ventricle of the heart. Atthe fork where the small arteries leave the carotids is a region called the carotid sinus. The carotid sinus containsnerve endings which are sensitive to the size of those small arteries. The carotid sinus acts as a baroreceptor,sensing the pressure differences within the small arteries. There are also baroreceptors in the aortic arch.

If there is a decrease in arterial blood pressure, the baroreceptors will sense this and compensate accordingly. As theblood pressure drops there is less expansion of the arterial walls and a decreased discharge of the nerve endings atthe baroreceptors. The result is an increased sympathetic discharge to the arterioles (veins and heart) and adecreased parasympathetic discharge to the heart (because we do not want to inhibit its function). Theparasympathetic and sympathetic divisions will act to increase heart rate and therefore increase cardiac output.

Sympathetic discharge to the arterioles and veins will constrict those vessels. Constriction of the arterioles will leadto an increase in the total peripheral resistance of that system, thereby increasing arterial pressure. This is the endresult of the compensatory mechanism. Constriction of the veins will lead to an increase in the venous pressure,venous return, and cardiac output. An increased cardiac output leads to an increase in the arterial pressure. Again,this is the end result of the compensatory mechanism. The correct choice is D.

Passage IV (22 - 28) Circulatory Pressure, Area, Velocity, and Volume

22.

23.

24.

C is correct, arterioles. Arterioles have strong muscular walls. When stimulated by the sympathetic nervoussystem, they constrict. Metabolic products at the level of the arterioles, such as CCb, H®, and Cb. regulate thedegree of constriction. The correct choice is C.

C is correct, III. It is important to pay attention to the labels on the x-axis and y-axis. If we look at position I onthe graph we note that the total blood volume is quite low, even though the pressure is quite high. Where is bloodexperiencing the highest pressure? As it is being ejected from the left ventricle. The pressure is going to decreaseas the blood reaches the arteries and arterioles. Once the blood reaches the capillaries and the venules, the pressureof the blood will be approaching an even lower value as indicated by position II. As the blood begins to enter theveins, the pressurecontinues to decrease. As blood enters the right atrium, it is coming from the vena cava. This isrepresented by position III. Note that once the blood leaves the vena cava and enters the right atrium, it will thenflow into the right ventricle. Contraction of the right ventricle increases the blood pressure and thus increases thevelocity of the blood leaving the right ventricle as it flows into the pulmonary artery. Blood from the pulmonaryartery will enter the lungs where the pressure drops to a rather low value. The correct choice is C.

D is correct. IV. Consider the x-axis (total blood volume) and the y-axis (area). The total area of the arteries willbe small (point I), and it will carry a small amount of blood volume. The total area of the venules and veins will bethe largest (point II), and it will carry the largest blood volume. As we move from the veins to the vena cava and tothe right atrium, right ventricle, and pulmonary artery, the total blood volume will again be minimal while the areawill also be minimal (point III). However, as the blood leaves the pulmonary artery and enters the capillaries of thelungs, the total area of the vessels begins to increase as does the total blood volume (point IV). Note that the totalblood volume in the lungs will be less than the total blood volume in the veins simply because (a) there are moreveins in the systemic circulation than in the pulmonary circulation, and (b) the veins are quite distensible. Thecorrect choice is D.

25. C iscorrect, III. If we look at position I on the graph, we note that the total blood volume isquite low even thoughthe velocity is quite high. Where is blood experiencing the highest velocity? As it is being ejected from the leftventricle. The velocity is going to decrease as the blood reaches the arteries and arterioles. Once the blood reachesthe capillaries and the venules the velocity of the blood will be at its minimum value as indicated by position II.Blood then begins to enter the veins and the velocity begins to increase again due to muscular contraction squeezingthe veins to help return the blood to the right atrium. As blood enters the right atrium it is coming from the venacava. This is represented by position III. Note that once the blood leaves the vena cava and enters the right atrium,it will then flow into the right ventricle. Contraction of the right ventricle increases the blood pressure and thusincreases the velocity of the blood leaving the right ventricle as it flows into the pulmonary artery. Thusposition IVwill be the velocity of the blood as it leaves the pulmonary artery andenters the capillary system of the lungs. Thecorrect choice is C.

26. A is correct, viscosity of blood. Laminar blood flow is blood which flows at a steady rate through a blood vessel.Blood flow is said to be streamlined. Turbulent flow involves blood which is not flowing in a steady stream through



the blood vessel. Instead, turbulent flow creates eddy currents. These currents result from the blood passing over arough surface in the vessel or encountering obstructions in the vessel such as a bifurcation in the path of blood flow(i.e., a splitting of the vessel itself). Turbulent flow, measured by the Reynold's number (Re), tends to increase asthe velocity of the blood increases and as the diameter of the vessel increases. Turbulent flow is inverselyproportional to the viscosity of blood. By using the equation Re = (v)(d)/(n/p), where v = velocity, d = diameter ofvessel, n = viscosity, and p = density, we find that as the Reynold's number increases, the turbulence of blood flowcan increase. Usually turbulence will occur when the Reynold's number increases above 2000. The correct choiceis A.

27. C is correct, systemic arteries. Arterial walls are stronger than the walls of veins. Blood vessel diameter increasesas the internal pressure within the blood vessel increases. This is because the vessels are distensible. Becausearterial walls are much stronger than the walls of veins, the arterial walls are about 6 to 10 times less distensible thanthe walls of veins. What this is saying is that for an increase in pressure there will be 6 to 10 times more blood in avein as there will be in a comparable artery. Even though pulmonary arteries still have stronger walls than those ofthe pulmonary veins, pulmonary arteries are under less pressure (about 1/6 less) than systemic arteries. It turns outthat pulmonary arteries have distensibilities about 1/2 that of veins. Even though the pulmonary arteries are stilldistensible, they are not as distensible as the pulmonary veins or systemic veins. The vascular distensibility wouldbe least for the systemic arteries. The correct choice is C.

28. D is correct, The capillaries contain about 20% of the total blood volume. Capillaries are about 1 mm in length and,on the average, most cells are about 0.01 cm away from any given capillary. This allows for a highly efficient formof diffusion of 02 from a red blood cell within a capillary to the surrounding cells.

100

60 -

40 -

20 -

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\ :

The value between the twodashed lines represents thetotal blood volume within

/ the capillaries.

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0 20 40 60 80


100

In order for this diffusion to be optimal, capillaries must have not only a very small diameter (about 8 urn, which isjust large enough for the passage of a red blood cell), but they must also have a very thin wall (about 0.5 urn). Thethickness of the capillary wall is just one endothelial cell layer thick. Even though these values were not presentedin the passage, it should be realized that gas exchange (for all practical purposes) occurs at the level of the capillary.It would make sense for them to have the thinnest walls and the smallest diameter, to allow for maximal andefficient diffusion. We can eliminate choices A and B, because they are characteristic of the systemic andpulmonary capillaries.

Capillaries in the systemic system contain about 5% of the total blood volume, while capillaries in the pulmonarysystem contain about 4% of the entire blood volume. This adds up to between 5% and 10% of the total bloodvolume, making choice C characteristic of the systemic and pulmonary capillaries and not a candidate for the bestanswer. Even though this value is not mentioned in the text of the passage, it can be obtained from the first graph inFigure 1 of the passage. In Figure 1 note that the dashed lines border the value of 20%. At first you might beinclined to think that choice D, which states that total blood volume in the capillaries is about 20%, is a truestatement, and therefore not the best answer. But think about what the x-axis is telling us. If we move out to 100%total blood volume, does that mean that at that point in the curve a vessel contains 100% of the total blood volume?No, it does not. What the area between the borders of the dashed lines is telling us is that the total blood volume inthe capillaries is represented by what we see between those two dashed lines. In other words, the total blood volumein the capillaries is less than 10%of the total blood volume in the system. The correct choice is D.

Passage V (29 - 34) Thoracic Cavity

29. B is correct, pleural pressure will become more negative. The two lungs in the chest cavity are individuallysurrounded by a thin layer of cells called pleura. The visceral pleura attaches to the lung while the parietal pleura



attaches to the diaphragm and interior region of the thoracic cage. Between the two pleural surfaces is a very thinlayer of fluid called the intrapleural fluid. The volume of the intrapleural fluid is constant. Pressures in therespiratory system are listed relative to atmospheric pressure, which is 760 mmHg at sea level. Between inspirationand expiration the alveolar pressure is the same as atmospheric pressure. Since pressures are measured relative toatmospheric pressure, the alveolar pressure is said to be 0 mmHg. As stated in the passage the intrapleural pressureis negative relative to atmospheric pressure (about -4 mmHg).

As an individual inspires the thoracic cavity begins to enlarge. The rib cage moves upward and outward. This isdue to contraction of the external intercostal muscles. The diaphragm descends towards the abdomen. Both actionsmove the wall of the thoracic cage away from the surface of the lungs. However, the parietal pleura is in contactwith the intrapleural fluid which is in turn in contact with the visceral pleura. This intrapleural space increases involume by a very small amount. As the volume of the intrapleural space increases (ever so slightly), the pressurewithin that space must decrease (Boyle's Law). In other words, the pressure within the intrapleural cavity willbecome more negative (relative to atmospheric pressure).

The same reasoning applies to the lungs. As the thoracic cage moves away from the walls of the lungs, the volumeof the lungs increases. As the volume of the lungs increases, the pressure within the lungs must decrease. In otherwords, the alveolar pressure becomes subatmospheric and air flows from the outside into the lungs. The correctchoice is B.

30. B is correct, parietal pleura. The parietal pleura and/or the visceral pleura are the most important layers of thethoracic wall for maintaining the pressure component of the thoracic cage. A puncture of the intercostal muscles,ribs, or heart is not going to lead to a collapsed lung because these structures are outside the pleura. The correctchoice is B.

31. B is correct, II only (diaphragm). The diaphragm is the primary muscle involved in normal respiration. Theintercostal and abdominal muscles are execssory muscles and aid in respiration during times of forced expiration andinhalation. The correct choice is B.

A iscorrect, high pCCb in the capillaries; low Cb in the capillaries. As blood returns to the capillaries of the lungsfrom the systemic circulation via the pulmonary arteries, it will have a low pC>2 (deoxygenated blood) and a highpCCb. The partial pressures of these gases in the alveoli of the lungs will be reversed. This sets upa concentrationgradient that will allow oxygen to How from the alveoli to the capillaries and carbon dioxide to flow from thecapillaries to the alveoli. The correct choice is A.

A is correct, a rise in pH requires a lower partial pressure of Cb to bind a given amount of O2. The graph showsthat as the pH rises, the curve shifts to the left. This shift indicates that less of a partial pressure ofoxygen is neededin order to reach half saturation of hemoglobin with oxygen. As the blood becomes more alkaline (due tohyperventilation), hemoglobin shows a tendency to retain more oxygen (at lower partial pressures of oxygen). Thecorrect choice is A.

32.

33.

34. B is correct, a rise in temperature requires a greater partial pressure ofO2 to bind a given amount ofO2. The graphshows that as the temperature increases, the curve moves to the right. The shift to the right indicates that a higherpartial pressure of oxygen is needed to reach halfsaturation of hemoglobin with oxygen. The correct choice is B.

Passage VI (35 - 40) Electrocardiogram

35. C is correct,arrhythmias. Arrhythmia means an irregular heart beat or having an irregular rhythm. Blood pressureis measured with a blood pressurecuff. ChoiceA is incorrect. Cardiac output is a measure of the volume of bloodpumped with each heartbeat. Choice B is incorrect. Hematocrit is a measure of the %RBCs in whole blood. It isused for checking on anemia. Choice D is incorrect. The correct choice is C.

36. B is correct, QRS. P represents the depolarization of the atria. During depolarization, they arc contracting. ChoiceA is incorrect. QRS represents spreadof depolarization through the ventricles. During this depolarization, they arecontracting. Choice B is correct. During the T wave, the ventricles are relaxed and repolarizing. Choice C isincorrect. There is no T-P phase, it is a fake answer. Choice D is incorrect. The correct choice is B.

37. D is correct, death, if a normal rhythm is not introduced. Ventricular fibrillation means the ventricles are notpumping together. The cells are contracting in their own unique patterns, producing an unproductive twitching.



This can be fatal pretty quickly—within minutes. The individual is definitely affected, therefore choice C isincorrect. If there are no heartbeats at all, choices A and B arc incorrect. The correct choice is D.

38. A is correct, sinoatrial node. All the choices are involved in conduction of action potentials in the heart, but onlythe sinoatrial node is actually in the right atrium, as the question asks. The atrioventricular node is located on theseptum between the atria and the ventricles. Choice B is incorrect. The bundle of His travels though theinterventricular septum, and breaks into branches, forming the Purkinje fibers. Choices C and D are incorrect. Thecorrect choice is A.

39. A is correct, tissue fluids contain ions that participate in conduction. Tissue fluids contain ions, such as sodium,potassium, bicarbonate, and chloride. Choices B and D are incorrect. Ions arc required for conduction. That is,dcioni/.cd water does not conduct electricity. Choice C is incorrect. The correct choice is A.

40. B is correct, a marathon runner. Exercise conditions the heart, so that greater volumes can be pumped per beat.This means the heart rate can slow and still maintain the same cardiac output as an untrained person. Infants havefaster heart rales than adults. Choice A is incorrect. Weight lifting does not provide aerobic conditioning tostrengthen the heart. Choice C is incorrect. A sedentary adult is also less conditioned than a trained athlete. ChoiceD is incorrect. The correct choice is B.

Passage VII (41 -47) Folate Experiment

41. C is correct, plasma folate represents folate intake over the past day, while RBC lolate represent folate intake overthe past few months. As stated in the passage, folate is a water-soluble vitamin. It is excreted by the kidney, so it isrequired every day. The serum contains folate eaten in the diet on that day. Choices A and D arc incorrect. RBCslive about 120 days, so their supply of folate represents the composite intake of folate over their lifetime of a fewmonths, not a year. Choice B is incorrect. The correct choice is C.

42. B is correct, ccntrifugation. Both extraction with an organic solvent and thin-layer chromatography would breakthe membrane of the RBC, mixing the plasma and the cellular contents. This would not be helpful. Choices A andC are incorrect. Dialysis tubing would contain the RBCs and probably the folate, too. This would not lead to aseparation. Choice D is incorrect. Finally, ccntrifugation is the perfect way to separate the heavy cells from theplasma. Following a short spin (10 minutes at 3000 rpm) the cells would sink to the bottom of the centrifugationtube and the plasma would move to the top, allowing separation for independent analysis of the two components.The correct choice is B.

43. A is correct, The subject's sample is added to a known amount of bacteria, incubated for a time period, and theturbidity of the medium is measured. In this assay, the growth of the folate-dependent bacteria provide theindication of the folate level present. The serum or RBC sample is added to a known amount of the bacteria in aliquid medium, incubated, and the growth is measured by the turbidity (cloudiness) of the solution. The moregrowth, the more turbid the solution. No extra folate is added to confuse the results. This makes choice B incorrect.No further enzyme activity assay is required, the bacteria provide the enzymatic assay. And they do notproducefolate. Choice C is incorrect on both thesepoints. Choice D is trying to remind us of the Ames's test for mutagens.It is also incorrect. The correct answer is A.

44. Aiscorrect, they were consuming between 300 and 400 jig/day. Compare the initial levels with the final levels foreach dosage group. Although all the women started with nearly identical plasma and RBC folate levels in eachgroup, the 200 and 300 groups's levels fell. This means they were consuming less than their usual levels of folate.This meanschoices C and D arc incorrect. However, the400 group had a rise in both plasma and RBC folate levels.This means the experiment provided more folate than their usual levels. Choice B is therefore incorrect. Thecorrect answer is A.

45. D is correct, macrocytic anemia. The question tells you that the cells become loo large before they are releasedfrom the bone marrow. The prefix "macro" refers lo large, while "micro" refers to small. Choice A is incorrect.Choices B and C are made up words. The correct choice is D.

46. B is correct, folate needs increase during pregnancy due to synthesis of many new cells. Since the experimentindicated 200 |ig/day was not adequate for nonpregnant women, and pregnant women are synthesizing morenucleotides for the fetus, it makes sense that folate needs are increased during pregnancy. Choice A is incorrect.The placenta docs not synthesize folate. Choice C is incorrect. Folate is a water-soluble vitamin, and is not storedin the liver. Choice D is incorrect. The correct choice is B.



47. D is correct, thymine. First, make a distinction between purines and pyrimidines. Purines have 2 rings, whilepyrimidines have 1. Choices A is the purine guanine while choice B is the purine adenine. This leaves us with thestructures shown in choice C and D. Is there any information the passage that will allow us to distinguish betweenthese two nitrogenous bases? No. This question is designed to get us to eliminate as many of the possible choices aswe can, and then to make an educated guess at the correct answer. The correct choice is D.

Passage VIII (48 - 54) Measurement of Blood Pressure

48. C is correct, the maximum pressure of the blood following a heartbeat. When the first sound is heard, the arterialpressure is just overcoming the pressure of the inflated cuff. This is the maximum pressure the artery reaches.Choice A is incorrect. The laminar flow of blood is interrupted by the cuff. It does not return until the cuff is notcompressing the arm at all. Choice B is incorrect. The heartbeat is not represented by taking blood pressure.Pressure is measured. Choice D is incorrect. The correct choice is C.

49. A is correct, the reading will be falsely high. If a cushion of fat absorbs someof the pressure from the cuff, then thecuffmust be inflated more tocompress the artery completely. This would mean that the reading of pressure neededto close the artery is falsely high. This would lead toan inflated blood pressure reading. The correct choice is A.

50. D is correct, a fluid moves as a scries of individual layers, each of a different velocity. If you have notseen this inphysics, try to figure out the answer based on vocabulary. A laminated countertop consists of a layer of plastic orother material glued onto a different material. Laminar refers to layers. Eliminate choice A. Since the question asksabout layers (laminar) eliminate choice C as well, although atoms do have their own random activity. To get thecorrect answer, think of water moving in a stream. It is slowest closest to the bank, where there is a lot of frictionwith dirt and rocks. It is fastest in the center. Each layer moves with a different velocity. Choice B is incorrect.The correct choice is D.

51. C is correct, 211. Read the table of blood pressures. The systolic pressure in a 46-ycar-old woman is 131 at thelevel of the heart. Add in the gravity factor (100 x0.8 = 80) and you get 131 + 80 = 211. If you read the column formen, 210 is your answer and is incorrect. Choice B is incorrect. If you read the diastolic column, 162 is youranswer. Choice A is incorrect. Choice D is just incorrect. The correct choice is C.

52. C iscorrect, the location of the cuff does not matter. On Earth, gravity changes the reading from site to site on thebody. In space, since there is no gravity, this change will not occur. Choices A and B are incorrect. The cuff willfunction just fine, as will the heart, in an anti-gravity situation. Choice D is incorrect. The correct choice is C.

53. B is correct, II only. Normally, arterial blood flow is laminar and smooth. It makes no sounds or vibrations.Turbulent flow is noisy, just like in a stream. When the artery is slightly constricted by the cuff, the flow becomesturbulent and noisy. Before the systolic sound is heard, no blood is flowing through thecuff. Choice I is incorrect.After the diastolic sound, flow is silent and therefore laminar. Choice III is incorrect. Only during the intervalbetween the systolic and the diastolic sound can noises be detected, so that flow must be turbulent. Choice II iscorrect. The correct choice is B.

54. D is correct, 94. The calculation uses data from the table. A 27-year-old man. on average, would have a systolicpressure of 125 and a diastolic pressure of 78. The difference between the two readings is 125 - 78 = 47. The pulsepressure is 47. MAP = 78 + (47/3) = 78 + 16 = 94. The correct choice is D.

Passage IX (55 - 60) Aortic Compliance

55. B is correct, AV/AP. Wc are looking for the best relationship for aortic compliance. Think about what"compliance" means. Compliance is the act of conforming. In this case, it is the conforming (stretch) of the aorta inresponse to successive injections of liquid. In the body blood is injected into the aorta from the heart which willcause a rise in pressure. The volume of the aorta will change, depending on the pressure. The compliance of theaorta is the ratio of the change in volume in response to a change in pressure. This relationship is denoted on thegraph by the slopes of the lines, as the y-axisrepresents the volume and the x-axis representspressure. Complianceis the change in volume in response to a change in pressure. The correct choice is B.

56. D is correct, inversely proportional to aortic compliance. Wc can arrive at this answer by the following logic: Weare told from the passage that the fractional change (AD/D) is a reflection of the change in volume of the aortaduring left ventricular contraction. Therefore, one can say that (AD/D) can be replaced by AV. Making this



substitution in the equation for the elastic modulus, one sees that Ep = AP/AV. This is, of cours,c the inverse ofcompliance, so the elastic modulus is inversely related to the compliance. As the compliance increases, the elasticmodulus decreases. Looking at Figure 2, one sees that with age, the elastic modulus increases. This should clueone to the idea that with age, compliance decreases. Nevertheless, the relationship is one of an inverse proportion.The correct choice is D.

57. D is correct, Curve E. In order to answer this question correctly, one must have made the conclusion thatcompliance decreases with age. From Figure 2 and the ideas in the passage, we can come to this conclusion. Wehave already established that in Figure 1; the compliance will be represented by the slopes of the lines. Each linerepresents a particular age group. Let us look at the normal pressure range of 80-120 mmHg. It is clear that CurveA has the highest slope, which represents the highest compliance. Wc arc looking for the oldest age group, whichshould have the lowest compliance (according to our conclusion). This is represented by the curve with the lowestslope which is best represented by curve E. The correct choice is D.

58. B is correct, later in systole. This question requires a careful reading of the passage. First, think about when thepeak arterial pressure is going to occur. Is it going to occur during diastole or systole? The heart is contractingduring systole, and that is when we shall see the most force and hence the peak arterial pressure. Therefore, we caneliminate the answers which claim that the peak arterial pressure will occur during diastole. Then the questionbecomes when will we sec the peak arterial pressure. The last sentence in the passage tells us that as compliancedecreases, the heart is unable to eject its stroke volume as rapidly when compared to a more compliant system. If theheart is slowed at ejecting its stroke volume, it can be logically concluded that it will take progressively longer forthe peak arterial pressure to occur. The correct choice is B.

59. D is correct, least at very high and low pressures and greatest over the usual range of pressure variations. We cangather this information from looking at Figure 1. We should already be aware that Curve A represents the youngestage group. We simply need to look at slopes. It is clear that at certain points in the curve, the compliance of theyoungest individuals is smaller than those of older individuals (this does not invalidate our general conclusion thatcompliance decreases with age). On the other hand, it is clear that for the youngest individuals, their compliance ishigher for most of the range of pressures. Therefore, when compared to the rest of the age groups, one cannot saythat the compliance for the youngest individuals is either greater or less than for the entire range of pressures. Onecan then eliminate choices A and B. We then look at choices C and D. Curve A is sigmoidal. The slopes are leastat the extreme pressures, and greatest during the normal range of pressures. This translates into the compliance forthe youngest individuals is least at very high and low pressures, and greatest during the normal range of pressures.The correct choice is D.

60. C is correct, decreases with age, a manifestation of increased arterial rigidity. One should easily be able toeliminate choices A and B. These claim that compliance increases with age, and Figure 1 clearly docs not supportthis claim. Therefore, we now look for a more likely manifestation of the decreased compliance. If the aorta doesnotchange its volume as well in response to a given change in pressure (less compliant), the arterial system is morelikely to be rigid. The more rigid the aorta, the less it can conform, or comply with a given change in pressure. Theincreased rigidity is a result caused by progressive changes in the contents of collagen and elastin in the arterialwalls. The increased rigidity with age results in a decreased compliance withage. The correct choice is C.

Passage X (61 - 66) Heart Muscle Action Potentials

61. C is correct, tricuspid valve. This question draws not on information from the passage, but from our ownknowledge of the anatomy of the heart. There are valves between the atria and the ventricles known as theatrioventricular valves. The valve between the right atrium and the right ventricles has three cusps, making it thetricuspid valve. When the right ventricles begins to contract, the valve between the two chambers closes andprohibits the back flow of blood into the right atrium. The correct choice is C.

62. B is correct, K®. We are looking at the permeability of potassium. As the cardiac action potential begins, thepermeability of potassium decreases as potassium channels close. This is the key to answering this question.Decreasing the permeability ofpotassium while increasing the permeability of sodium will result in depolarizationof the membrane. During the plateau of the action potential, the potassium permeability stays below the restingvalue. Ultimately, repolarization does occur when the permeabilities ofcalcium, sodium, and potassium all return totheir original state. As a note. Graph A represents the permeability of sodium, while Graph B represents thepermeability of calcium. Because Graph C dips right as the depolarization begins, one can safely conclude thatGraph C represents the permeability of potassium. The correct choice is B.



63.

64.

65.

66.

A is correct, Region A. This region of the action potential is unlike a normal resting potential. Note that the restingpotential of the SA node cell is not steady but instead manifests a slow depolarization. This gradual depolarizationis known as a pacemaker potential. The pacemaker potential brings the membrane potential to threshold, at whichpoint an action potential occurs. After the peak of the action potential, the potential repolarizes, and the gradualrepolarization begins over again. Thus, the capacity for spontaneous rhythmic self excitation is best manifested byRegion A of the action potential. The correct choice is A.

C is correct, the flow of positive ionsout of thecell equaling the flow of positive ions into the cell. In myocardialcells, the original membrane depolarization causes voltage-gated calcium channels in the plasma membrane toopen.This results in a flow of calcium ions down their electrochemical gradient into the cell. Because there is a delay intheir opening, these are called slow channels. The flow of positive calcium ions into the cell, along with somesodium alsoentering through the slow channels just balances the flow of positive potassium charge out of the cell.This keeps the membrane depolarized at the plateau value. In other words, the flow of positive ions out of the cellequals the flow of positive ions into the cell. Do not be fooled by the other answers. While they may be true forcertain aspects of the cardiac action potential, we are looking for the best reason to explain the fact that themembrane potential is essentially notchanging throughout theplateau region. The correct choice is C.

C is correct, gapjunctions between cells. The wave of depolarization spreads from contractile cell to contractilecell via gap junctions. Recall that gap junctions are channels which allow cells to share cytoplasm. In this way, thecurrent responsible for depolarization in one cell can spread to another cell in a very quick manner. There are noneurotransmitters involved, which greatly reduces the amount of time needed to cause contraction over the entirechamber. The correct choice is C.

Dis correct, Graph Cis the result of vagus nerve stimulation. We know from the diagram that all three graphs arepacemaker potentials. The difference between the three graphs obviously lies in the slope at the beginning part ofthe potential. In Graph B, we see a large slope. In other words, we reach threshold very quickly. On The other hand,Graph Creaches threshold after quite some time. One could conclude that Graph Bwould most likely representsympathetic stimulation, while Graph C represents parasympathetic stimulation. Therefore, Graph C most likelyrepresents stimulation of the vagus nerve (a major parasympathetic fiber). The correct choice is D.

Passage XI (67 - 72) Capillary Filtration

67.

68.

69.

Cis correct, favor movement offluid from vessel to interstitial space. One needs to make the following conclusionfrom the information in the passage. There are two forces which favor movement of fluid out of the vessel. Theseforces are the intracapillary hydrostatic pressure and the interstitial fluid oncotic pressure. There are two forceswhich oppose movement of fluid out ofthe vessel. These are the plasma protein oncotic pressure and the interstitialfluid hydrostatic pressure. With this understanding, we can see that an increase in the intracapillary hydrostatic-pressure would most likely favor movement of fluid from the vessel to the interstitial space. Such action will notaffect the concentration ofosmotically active molecules, and therefore we can eliminate choice A. Also, such actionwill not affect the venous resistance, and therefore we can eliminate choice D. From the passage we can eliminatechoice B because we are told a decrease in the arterial pressure results in a decrease in the intracapillary pressure.The correct choice is C.

A is correct, Pc - Pj > Pc. According to the figure in the passage, filtration is the movement of fluid out of thevessel. In addition, wc are told from the passage that the force of filtration comes from the difference between thePcand Pj. The force of filtration is thus Pc - Pj. According to the diagram, the Pc always has a positive value and isthus larger than the Pj. If Pj is a negative number, then the force of filtration will actually become larger than thecapillary hydrostatic pressure. This is indicated by the equation in choice A. None of the other equations followsthis conclusion, and all of them can be eliminated. The correct choice is A.

B is correct, result in a reduction of capillary hydrostatic pressure. We are reducing the diameterof a precapillaryvessel. This will increase the resistance of that vessel so blood will not flow through that vessel as well. The resultof this increased resistance is that the capillary will not receive as much blood volume. In turn, this will result in adecrease in the capillary hydrostatic (blood) pressure. A reduction in blood volume will decrease the amount offorce/ unit area placed on the walls of the capillary. A reduction in Pc will not favor movement of fluid from thevessel to the interstitial space,eliminating choiceA. The resultof this action will not be equivalent to an increaseinvenous resistance, but is exactly the opposite. Think about it! Finally, resistance changes will not affect theconcentration of osmotically active particles in the interstitial space, so choice D can be eliminated. The correctchoice is B.


Biology Heart & Lungs Section II Answers

70.

71.

72.

A is correct, Fluid movement = k[(Pc + K\) - (Pj + np)]. This question requires us to put together knowledgegathered in the passage into an equation form. It brings us back to our discussion of forces. The forces movingwater out of the vessel are the capillary hydrostatic pressure and the interstitial oncotic pressure. Therefore, weshould see these two forces added together. This eliminates choices C and D. The two opposing forces, Pj and theplasma protein oncotic pressure, should be added together because they move fluid in the same direction. Thiseliminates choice B and leaves us with choice A as the correct answer. We see the opposing forces subtracted fromeach other will result in fluid movement. If the algebraic sum is positive, filtration will occur, and if the sum isnegative, absorption will occur. The correct choice is A.

C is correct, carries electrical charges at blood pH which attracts various electrolytes. The answer can be arrived atthrough a process of elimination. We are told that albumin exerts a greater osmotic force than can be accounted forsolely on the basis of the number of molecules dissolved in the plasma. Therefore, there is something special aboutalbumin. It is not very likely that it could be replaced by an inert molecule and have noeffect on theplasma proteinoncotic pressure. Therefore, we can eliminate choice A. Consider choice B. How many proteins do you know ofthatdissolve intomore thanone protein? Proteins are notsimple ionic salts. We are talking about a largeprotein. Itwill not dissolve into more than one molecule, making choice B an unlikely choice (be aware that this wouldincrease its osmotic activity, if it happened). Choice D provides no explanation for the claim in the question. Infact, choice D would allow us to predict a smaller plasma protein oncotic force. Therefore, the best answer is choiceC. Nothing in the passage tells us that it does carry a charge, but that is not the point. One should be able to reasonthat this is the best answer given our knowledge. This is indeed the case. The protein carries negative charges,which attract primarily sodium ions. In addition, chloride ions attach themselves to the protein, which attract evenmore sodium ions. The additional electrolytes provide the increased osmotic strength not accounted for by theconcentration of plasma albumin. The correct choice is C.

D is correct, returns to the venous circulation via the lymphatic system. Since the remaining fluid is picked up bythe lymphatic system, we can eliminate choices A and B. The fluid will not remain in the interstitial space. If it did,edema would result. Furthermore, the fluid in the interstitial space is certainly not going to increase theintracapillary hydrostatic pressure. Having decided that the lymphatic system is going to pick up the fluid, one hasto ask where is the fluid being returned to. Nothing in the passage tells us this, so one must draw on their ownknowledge. The lymphatic system returns fluid into the venous system, not the arterial system. In fact, the lymphcapillaries drain into larger vessels that finally enter the right and left subclavian veins at their junctions with therespective internal jugular veins. The correct choice is D.

Passage XII (73 - 78) Respiratory Calculations

73. B is correct, increased left atrial pressure. We are looking for a situation which causes an increased volume ofblood in the pulmonary capillaries. The increase in volume will cause the distension of the vessels. Increased leftatrial pressure indicates that blood is filling the left atrium, but it is not being injected into the left ventricle. Thismay be due to left heart failure. If blood is not being ejected from the pulmonary atrium, the pressure will increaseand blood from the pulmonary capillaries will not be able to move into the atrium. This will cause congestion andwill increase the volume in the pulmonary capillaries, causing distension. All of the other answers act to keep bloodaway from or out of the pulmonary capillaries. The correct choice is B.

74. D is corrrect, 1.09 L. According to the ideal gas law, which states that PV= nRT, we find that:

P1V1 P2V-and that: V2 =Vip.

T^

Pj/P2 = 1 and T9/T1 (in Kelvin) > 1. The only possible answer choice is D, because we are looking for an answergreater than one. The answer then becomes (1L)(298/273) = 1.09 L. The correct choice is D.

75. C is correct, the venous blood from bronchial venules and heart vessels contaminate the pulmonary venous outflow.We are looking for an explanation to the decrease in O2 partial pressure. From the passage, we know that thebronchial venous flow along with vessels from the heart join with pulmonary venules. In effect, by increasing thevolume of blood without changing the amount of oxygen, we are decreasing the concentration of oxygen in blood.The correct choice is C.



76. C is correct, 150 mmHg. The question asks for the partial pressure of oxygen in dry inspired air at the trachea.First, we need to know the fraction of oxygen in the air. We get this from the passage, and it is 21%. Note that inthe alveolus the fraction of oxygen dips to 14.3%, but in the trachea it still is 21%. Next, we must get the totalpressure of the dry air. The total pressure of the air is 760 mmHg, but 47 mmHg is due to the vapor pressure ofH2O. The total dry air presssure is 760 - 47 = 713 mmHg. The partial pressure of oxygen in anhydrous air becomes0.21 x 713 mmHg « 150 mmHg. The correct choice is C.

77. A is correct, 102 mmHg. The total anhydrous pressure is 760 - 47 mmHg. The only difference is the fraction ofoxygen found in the air. In alveolar gas, the value is 0.143 (from the passage). To find the partial pressure ofoxygen, we simply multiply 0.143 x 713 mmHg ~ 102 mmHg. The correct choice is A.

78. C is correct, 197 ml ChfL. We are looking for the total concentration of oxygen in the systemic arterial blood. Thisincludes oxygen found in hemoglobin and oxygen dissolved in the blood itself. Let us start with the latter. Weknow this value straight from the passage. The value is 3 ml O2/L. Now, let us determine the oxygen content forhemoglobin. From the passage, we see that hemoglobin binds 1.34 ml 02/g. The concentration of hemoglobin is150 g/L. We need to multiply these values. We get 1.34ml 02/g x 150g/L = 201 ml O2/L. The question states thatonly 97% of hemoglobin is saturated with oxygen, so let us do an approximate calculation. We multiply 0.97 x 200ml O2/L. This gives us 194 ml O2/L. Adding the 3.0 ml O2/L that are dissolved in the plasma, we obtain 197 ml0->/L. The correct choice is C.

Passage XIII (79 - 85) Aspirin

79. B is correct, after aspirin treatment, endothelial cells produce new CO, but platelets do not. From the passage, welearned that the turnovertime of platelets is four days. Since the platelets in Figure 1 have the most active CO after4 days, this means they cannot produce more when aspirin blocks CO. This eliminates choices A and C. However,endothelial cells show a rapid rise in CO after the aspirin dose. These cells simply produce more COenzyme sincethey have the machinery to do so. Choice D is incorrect. The correct choice is B.

80. D is correct, four degrees of unsaturation. Alkanes, which are saturated with hydrogen atoms, have the empiricalformula CnH2n+2, where n represents the number of carbon atoms in the molecule. This formula tells us that forevery carbon atom in an alkane there will be twice as many hydrogen atoms plus two more. Hydrocarbons that arecomposed of a single ring or a single carbon-carbon double bond (i.e., an alkene) are missing two hydrogen atomsand have the empirical formula CnH2n. These molecules are said to have one degree of unsaturation. If ahydrocarbon contains a single carbon-carbon triple bond (i.e., an alkyne), then that molecule is said to have twodegrees of unsaturation and is represented by the empirical formula CnH2n-2. Atoms which are not part of ahydrocarbon can be treated as follows: halides like F, CI, Br, and I can be treated as hydrogen atoms; oxygen can betreated as a carbon atom.

81.

82.

83.

In thromboxane B2 we see that there are two carbon-carbon double bonds and one ring. Thisgives three degrees ofunsaturation, making choice C the apparent answer. However, the carboxylate group (COO") has a carbon-oxygendouble bond (C=0) which adds another unit of unsaturation, giving a total of four units of unsaturation. The correctchoice is D.

A is correct, acetylated serine residue. The passage tells us that a serine is acetylated to inactivate CO. Choices Band D indicate a cysteine residue, and are incorrect. Choice C is a phosphorylated serine residue, and is incorrect.The correct choice is A.

B is correct, I and II only. Too much aspirin would lead to poor clotting. This means that minor accidents couldmore easily lead to bruising, and bleeding time would be prolonged. Choices I and II are correct. A prolongedbleeding time means that the wound is slow to clot, not rapid. Choice III is incorrect. The correct choice is B.

D is correct, 5, 8 II, 14-eicosatetraenoic acid. Start counting from the carboxyl end. There are twenty carbons,which requires the prefix "eicosa". There are 4 double bonds and the carboxyl group, giving the "tetraenoic acid"



84.

85.

part. Choices A and B are incorrect. The carbons arecounted from the carboxyl end, giving 5, 8, 11, and 14 as thepositions for the double bonds. The correct choice is D.

A is correct, the dosing regimen keeps platelet aggregation low. The passage said nothing about the proliferation(growth) of endothelial cells. Eliminate choices C and D right away. To prevent heart attacks in people withnarrowed arteries from cardiovascular disease, it is often desirable to keep the blood from excessive clotting. Thismeans that platelet aggregation (clumping together) should be low. Choice B is incorrect. The correct choice is A.

C is correct, aspirin should not be used in late pregnancy. Sinceaspirin inhibits the synthesis of prostaglandins byinactivating the first enzymatic step, it could possibly interfere with the normal prostaglandin cycles involved inlabor and delivery. This would probably not stimulate labor. Choices A and B are incorrect. Birth defects areinitiated in the first few weeks of pregnancy. Choice D is incorrect. The correct choice is C.

Passage XIV (86 - 92) Sickle Cell Anemia

86. C is correct, a decrease in HbF and an increase in HbS with age. HbA is normal, nonsickling hemoglobin. Anincrease in HbA would not promote sickling of cells. Choice A is incorrect. The maternal hemoglobin is separatefrom the fetal hemoglobin, and they do not mix. Choice B is incorrect. There are no antibodies to HbS, or thatwould lead to an even bigger obstruction problem. Choice D is incorrect. The correct choice is C.

87. A is correct, all cells contain both HbS and HbA. The hemoglobin in mixed in each blood cell. There are noexclusive HbS-containing cells or HbA-containing cells. Choice B is incorrect. In a heterozygote, both HbS andHbA are present. Choices C and D are incorrect. The correct choice is A.

88. C is correct, HbF promotes a higher oxygenation state inside the cell. Adult (HbA) and fetal (HbF) hemoglobins donot bind to each other. This is seen in question 1, in which sickle cell anemia is diagnosed only after fetal Hb levelsfall. We also know that adults who have only HbA do not sickle. Choices A and D are incorrect. The cells sickleunder lower oxygen conditions. This means that a beneficial action is to raise oxygen levels inside the cell, so thatthe HbS does not fall out of solution and cause sickling. Since HbF is beneficial, it must promote a higher oxygenstate inside the cell. Also, we know from our reading that HbF binds oxygen more avidly then HbA. This is howthe oxygen gets transferred from the mother to the fetus. Choice B is incorrect. The correct choice is C.

89. D is correct, sodium metabisulfate would deoxygenate hemoglobin from red blood cells and the sickling patternwould be noted in affected individuals. Forget about the rusty color answers. We are interested in the sicklingphenomenon itself. Choices A and C are incorrect. The conversion of sodium metabisulfate to sodium persulfateinvolves the removal of oxygen from the other substrate. Since HbS crystallizes and causes sickling under loweredoxygen tension, this removal of oxygen would allow cells to sickle if the individual has the sickle cell trait.Hemoglobin would not be oxygenated under these conditions. Choice B is incorrect. The correct choice is D.

90. A is correct, decreased delivery of oxygen to the tissues compared to other adults. HbF binds oxygen more tightlyand therefore releases less oxygen to the tissues. This would be a problem in disease states in which HbF is reallyhigh, such as the thalassemias. Choice B is incorrect. This question has nothing to do with HbS, but rather it asksabout HbF. HbF, as is stated in the passage, does not sickle. Eliminate choices C and D. The correct choice is A.

91. C is correct, I and II only. A single base pair is changed to produce the HbS product. Since the DNA differs atleast in this spot, specific restriction enzymes can used to cleave the DNA to produce a unique pattern of fragments.When fragments are separated by gel electrophoresis, the pattern of homozygotes (both sickle cell affected and wild-type) and the heterozygotes can be easily distinguished. Choice I is correct. The hemoglobin proteins will movedifferently during gel electrophoresis. The pattern of bands can also be used to distinguish the groups. Choice II iscorrect. Centrifugation is used to separated things of different masses or viscosities. The hemoglobin proteins aretoo similar to be separated this way for diagnosis. Choice III is incorrect. The correct choice is C.

92. B is correct, I and II only. HbF does not bind 2,3-BPG. This is one feature that increases its oxygen saturationcapacity. Choice I is correct. HbF does have a higher oxygen saturation at a given p02 than HbA. This is amechanism that ensures the fetus gets oxygen preferentially from the mother. Choice II is true. HbF contains twoalpha and two gamma chains. Choice III is incorrect. The correct choice is B.

Passage XV (93 - 100) Ventilation Regulation

93. C is correct, the arterial pC02 should be maintained at 40 mmHg. We want to assess the effects of a changingarterial pOo on the ventilation rate. To do this, we should keep the pC02 at a constant value. This eliminates


Biology Heart &Lungs Section II Answers

choices A and B. The question becomes whether the pC02 is maintained at 40 or at 46 mmHg. We are interested inthe effects of the arterial p02, where the pC02 is 40 mmHg. Therefore, to obtain the most accurate results, weshould keep constant the pC02 at the value which can be found in the arterial circulation. The correct choice is C.

94. B is correct, total amount of oxygen transported is relatively unaffected. The question tells us that a drop of 30mmHg from the normal resting value of arterial p02 does not change the rate of ventilation. Recall the oxygendissociation curve, which tells of the percent saturation of hemoglobin with oxygen depending on the partialpressure of oxygen. Remember that the curve is sigmoidal, with the top reaching an asymptotic value. One candescend from 100 mmHg of oxygen to 60 mmHg of oxygen and the percent of hemoglobin saturated with oxygendoes not change dramatically. In other words, the amount of oxygen delivered to the body is relatively unaffected.The correct choice is B.

95. A is correct, an unchanged arterial p02- Carbon monoxide has a very strong affinity for the iron on the heme ofhemoglobin which is responsible for binding oxygen. For this reason, the number of hemoglobins able to transportoxygen is decreased. Does this decrease the arterial partial pressure of oxygen? No. The reason is that only theoxygen dissolved in the plasma contributes to the partial pressure. The oxygen on hemoglobin is bound and is thusnot soluble in the fluid. For that reason, carbon monoxide will not affect the partial pressure of oxygen. Thecorrect choice is A.

96. C is correct, resisted by reflexes regulating ventilation to a greater degree than are equivalent changes in arterialp02- This question is asking us to look at the graphs and ask ourselves the following question: Is our body moresensitive to changes in the partial pressure of oxygen or to changes in carbon dioxide? Looking at the curves, theanswer is clearly changes in the partial pressure of carbon dioxide. We also know that the reflexes discussed in thepassage are involved in resisting changes that occur to the partial pressure, in hopes of restoring values to normal.For that reason, changes in the arterial partial pressure of carbon dioxide are resisted by reflexes regulatingventilation to a greaterdegree thanare equivalent changes in arterial p02- The correct choice is C.

97. C is correct, hyperventilation is caused by increased neural output from the peripheral chemoreceptors. In ametabolic acidosis, we have an overproduction of protons from some metabolic condition, like excessive exercise.These extra protons will stimulate both the central chemoreceptors and the peripheral chemoreceptors to increasetheir output, and cause a rise in the rate of ventilation. Therefore, we can eliminatechoices A and B, which call fora hypoventilation. Now, while the rise in protons will stimulate both central and peripheral receptors over time, askyourself which one will be the primary receptor. Can protons simply cross over the blood-brain barrier to haveaccess to the central receptors? The answer is no, because they are a charged species. Therefore, we can deducethat it is the peripheral receptors which are primarily responsible for the rise in the rate of ventilation. The correctchoice is C.

98. D is correct, arterial pC02 increases. During exercise, it is clear that the cells will be using more oxygen and thusproducing morecarbon dioxide. For that reason, the levels of venous pC02 will increase. Will this lead to a rise inthe arterial partial pressure of carbon dioxide? The answer is no, because the rate of ventilation will increase. Weknow that we breathe "harder" when we are exercising. In other words, the excess carbon dioxide:is released as aresult of the higher rate of ventilation. The increase in the carbon dioxide production is equivalent to the increase inbreathing rate. The result of this is that the arterial partial pressure of carbon dioxide does not change, making Dafalse statement. The correct choice is D.

99. A is correct, low pC02 is permitting one to hold their breath, the exercise may lower the p02 to levels which mayinduce unconsciousness. Theswimmer's hyperventilation will result in a lowered partial pressure of carbon dioxide,because they are eliminating a good deal of gas. During the race, that lowered level of carbon dioxide will allowthem to hold theirbreath. However, theexercise will lower the level of oxygen. Recall the body is more sensitiveto levels of carbon dioxide relative tooxygen. Forthat reason, the body does not know to breathe, butoxygen levelsmay be reaching dangerous lows. The level may become low enough to induce unconsciousness, as not enoughoxygen is reaching the brain. The correct choice is A.

100. B iscorrect, an increased arterial pC02 and a decreased arterial p02- This question is very straightforward. Duringtimes of sleep, the body is not breathing as often. The end result of this is that we are bringing in less amount ofoxygen, and eliminating lesser amounts of carbon dioxide. Thequestion tells us that thebody's usage of oxygen perproduction of carbon dioxide is unchanged during this dormant period. Therefore, we should expect to see anincreased pC02 and a decreased p02- The correct choice is B.

Copyright © byThe Berkeley Review 152 The Berkeley ReviewSpecializing in MCAT Preparation

BiologySection III

GastrointestinalTract

and

Kidney

A. The Gastrointestinal Tract

1. Nutrients and Digestion

B. The Kidney1. Renal Function

2. Renal Physiology3. Homeostatic Mechanisms


fEEKKELEYSpecializing in MCAT Preparation

Gastrointestinal Tract and KidneyTop 10 Section Goals

°^

©|m

Be familiar with the anatomy of the gastrointestinal tract.Know how food passes from the mouth, down the esophagus, into the stomach, andthrough theintestinalsystem, and how it is eliminated as feces.

Understand the interactions between the various gastrointestinal secretions.Thefourgastrointestinal peptidehormones ofimportance aregastrin, cholecystokinin, secretin, andglucose-dependent insulinotropic peptide. Be aware oftheir functions.

Be familiar with the process of digestion and the areas of nutrient absorption.

Know what types ofnutrients are broken down inthe stomach andwhattypes are degraded in theintestine. Understandhow peristalsis is relatedto the movementof chymem the system.

Ofm Be familiar with the anatomyof the kidney.jfir Understand how molecules are filtered at the glomerulus and which ones are reabsorbed from the^ different parts ofthe nephron tubular system.

Be aware of the effect of the autonomic nervous system on the kidney.Know how the kidney responds tosympathetic andparasympathetic stimulation. Know theeffectsbetween vasoconstriction and vasodilation at the level of the glomerular arterioles.

Understand how the renin-angiotensin-aldosterone system functions.Thissystemstimulates the reabsorption of Na® in the distalconvoluted tubule and in the collectingduct. Understand why this is important to the body

^|ft* Be aware of the different types of transport systems that line the tubules.^Br Understand the differences between passive reabsorption (no energy required) and active reabsorption^ (energy required) and generally what isreabsorbed atwhich points along the tubules.

Understand how the kidney responds to a decrease in arterial blood pressure.

Beaware that there is a short-term adjustment and a long-term adjustment for a decrease in arterialblood pressure. Understand how they work interms ofthe glomerular filtration rate.

(§)<!§** Be familiar with diabetes mellitus and diabetes insipidus.

? Akey feature ofdiabetes mellitus iselevated blood glucose levels due toinsulin problems, while akeyfeature ofdiabetes insipidusis a vasopressin deficiency. Both lead to a largeurine loss.

Be familiar with some of the conditions brought about by renal failure.The twomost life-threatening consequences ofrenal failure involve the retention ofK®, due to poortubular secretion ofK®, and metabolic acidosis, due to impropertubularsecretion ofH®.

Biology Gastrointestinal Tract St Kidney

Gastrointestinal TractNutrients & DigestionLet's examine the gastrointestinal system. Digestion is the processby which foodthat is eaten is broken down into progressively smaller particles and ultimatelyabsorbed by the intestinal tract. The gastrointestinal system includes the mouthand associated salivary glands, esophagus, stomach, small intestine, largeintestine, and certain aspects of the liver and pancreas. In the average adult thistract, running from mouth to anus, is about 30 feet in length and is actually anextension of the external environment.

The food that we eat has a diverse composition. A typical meal might includemacromolecules such as starch in the form of bread or potatoes, cellulose in theform of a salad or other greens, protein in the form of meat or cheese, and fat inthe form butter or ice cream. Each of these dietary categories needs to bedegraded before they can be absorbed into the body.

Starch and glycogen are both important polysaccharides in nature. Starch,characteristic of plant cells, and glycogen, characteristic of animal cells, arehydrolyzed within the digestive tract by enzymes called amylases. Uponhydrolysis both polymers release the monosaccharide glucose. Cellulose isfound within the cell walls of plants and is a polysaccharide consisting of glucoseresidues linked together. The enzyme cellulase can hydrolyze cellulose into itsconstituent glucose residues. See Figure 3-1.

Amylase

Starch r. => Glucose CelluloseCellulose

^ GlucoseAn example of

bacterial symbiosis

Figure 3-1Enzymatic hydrolysis of starch and cellulose.

It is important to note that even though starch, glycogen, and cellulose all containglucose residues linked together to form long polymers, we as humans cannotdigest cellulose. The reason that we cannot digest cellulose is because the glucoseresidues in cellulose are linked together in a different configuration than theglucose residues in starch or glycogen. This special linkage can only behydrolyzed only by the enzyme cellulase—an enzyme that we do not have. Why,then, can cows eat grass? The only vertebrates that can utilize cellulose are theruminant animals (like the cow and goat). The rumen of these vertebratescontains microorganisms which manufacture and secrete the cellulase that canbreakdown the cellulose in the vegetation being consumed. This is an example ofa symbiotic relationship between the host ruminant and the microorganismsliving within the rumen. The microorganisms get a warm place to live and theruminant gets energy in the form of digested cellulose.

As we have mentioned in previous sections, proteins are composed of aminoacids. Enzymes called proteases can hydrolyze proteins to their constituentamino acid residues. Fat cells or adipocytes store triacylglycerols (also called a

Nutrients & Digestion


Biology Gastrointestinal Tract St Kidney Nutrients &Digestion

triglyceride but better known as just plain fat). Fats are hydrolyzed into fattyacids and glycerol by the enzyme lipase. See Figure 3-2.

Protease _ Lipase Fnrtv arirlsProteins I [> Amino Acids Fats I [> Glycerol

Figure 3-2Enzymatic hydrolysis of proteins and fats.

Vitamins are needed in only small amounts in the diet yet they are essential inorder to prevent certain nutritional deficiencies. [For example, the diseaseberiberi is due to a deficiency of vitamin B\ or thiamine.] As a group, vitaminsparticipate in many different chemical reactions in the body.

Inorganic minerals such as iron, potassium, calcium, and zinc are also importantfor proper development. Iron is the atom in the center of the heme group thatcarries oxygen in both hemoglobin and myoglobin, two proteins that transportoxygen within the body.

All of the proteins that we are concerned with in the biological sciences arecomposed of a basic set of 20 different amino acids joined together in peptidelinkages. Out of these 20 different amino acids we need to obtain 9 of them in ourdiet. The rest we can synthesize. In other words, we cannot synthesize certainamino acids from our metabolic reactions and must therefore obtain them from

the food we eat. The amino acids that we cannot synthesize are referred to asbeing essential amino acids. Thus, the five general components to a completediet are carbohydrates in the form of sugars like glucose, proteins, fats, vitamins,and minerals.

The Gastrointestinal TractThe gastrointestinal tract includes the mouth, esophagus, stomach, smallintestines and large intestines. Let's consider a cross section of the small intestineand examine the tissue layers as we work outward from the lumen. The firstbarrier we encounter is a convoluted layer of epithelial cells (Figure 3-3a).

The convolution of this layer helps to increase the surface area of thegastrointestinal tract for absorption of nutrients. Scattered throughout thisepithelial layer are ducts from external exocrine glands like the pancreas andliver (and the salivary glands in the oral cavity). Juxtaposed to many of theepithelial cells are endocrine cells which contain hormones that can be releasedinto the blood. These hormones influence other cells in the gastrointestinalsystem. Both the parasympathetic and sympathetic nerves (of the autonomicnervous system) innervate the gastrointestinal system. The more importantnerves stem from the parasympathetic system. Recall that the parasympatheticnerves are more active during times of relaxation and digestion while thesympathetic nerves are more active during times of "fight or flight."

As food is passed from the mouth to the lower portions of the gastrointestinaltract the smooth musculature surrounding the epithelial cells begin to contract inperistaltic waves. These waves (peristalsis) are not controlled by consciousthought but rather by the action of the parasympathetic division and by theaction of hormones. Peristaltic action is rather quick because of the electrical



continuityimposed by the gap junctionsin the circular and longitudinalmusclesof the gastrointestinal tract.

(a)

EpithelialCells

Circular

Muscle

LongitudinalMuscle

Blood Vessels

and/or Lymphatics

Gastric Pits

Gland outside the

gastrointestinal tract

Nerves to and

from the CNSSmooth Muscle

Figure 3-3Cross section through the small intestines.

Let's follow a bolus of food as it passes its way through the gastrointestinal tract.In the mouth the muscles of mastication move the jaw and the food is groundbetween the teeth. Salivary amylase is secreted into theoralcavity and begins todigest the starch and glycogen. This secretion is controlled by theparasympathetic nerves.

Asthe food is swallowed and passedinto the pharynx access to the nasal cavityis closed. A flap of tissue called the epiglottis covers the opening to the larynxand prevents food from entering into this passageway (to protect the airway).Instead, the food passes into the esophagus and then down into the stomach.This swallowing reflex is controlled by centers in the medulla.

Once the food reaches the stomach a sphincter (circular muscle) called thegastroesophageal sphincter contracts and prevents regurgitation of the foodback into the esophagus. If this sphincter were not closed off, stomach acidwould enter the esophagus and irritate the nerve endings in the smooth muscle.This burning sensation is referred to as heartburn.

As the foodenters into the stomacha littlemore digestive action takesplace. Twofunctions of the stomach is to break the food down into smaller particlesand todetoxify it (by acidic secretions). There are four major types of secretion withinthe stomach. Mucus is secretedby surface cells and acts to protect the lining ofthe stomach and lubricate the food. Gastrin, located in endocrine cells in the



Biology Gastrointestinal Tract fir: Kidney Nutrients et Digestion

lowerportion of the stomach, is secreted in response to protein enteringinto thestomach. This hormone stimulates the secretion of HCl and pepsinogen. HCl, ata pH ofabout 1,is secreted by parietal cells. Pepsinogen is secreted by thechiefcells. Parietal cells and chief cells are located in the gastric pits the line theepithelium of the stomach (see Figure 3-3b). Pepsinogen is the inactive form ofthe peptidase enzyme pepsin. Peptidases, like pepsin, cleave peptide bonds.Since proteinis composed ofamino acids and amino acidsare linkedvia peptidebonds, pepsin hydrolyzes proteins (at specific places in their amino acidsequence). Pepsinogen is converted to pepsin by the action of HCl. However,once pepsin is formed it can autocatalytically act on pepsinogen to form morepepsin.

Besides secretingHCl the parietal cellsalso secrete a glycoprotein called intrinsicfactor. Thisglycoprotein is important because it complexeswith vitamin B12 andis then absorbed by the intestinal epithelial cells and transported by thebloodstream. If there is a defect in the synthesis of intrinsic factor, vitamin B12cannot be bound and therefore cannot be absorbed by the epithelial cells in theintestine. Vitamin B12 is important in erythrocyte (red blood cell) formation.

If too much acid is secreted into the stomach, ulcers can occur in the stomach andin the small intestine (which is directly connected to the stomach). Ulcers aresimply erosions of the walls of these two organs, and if they are extensiveenough can cause bleeding. One of the most powerful stimulants that causes HClto be released into the lumen of the stomach is histamine. It turns out that a

compound called cimetidine (trade name is Tagamet) inhibits the binding ofhistamine to its receptor on the parietal cells. This reduces the amount of HClsecreted in the lumen of the stomach. Cimetidine and its analogs are quitecommonly prescribed by doctors as a non-surgical treatment for ulcers.

As the dissolved food (referred to as chyme) passes from the lower stomach andinto the small intestine more enzymatic activity takes place. This is partly due todistension of the stomach and small intestine and the generation of nerveimpulses that stimulate enzymatic secretions. Roughly 90% of the digestion andabsorption that takes place in the gastrointestinal tract occurs in the smallintestine. Not only do fluids enter the small intestine from the stomach but thepancreas and liver also add secretions as well. The small intestine acts toneutralize the acid which has been secreted by the stomach and to further digestand absorb food particles.

Distension of the small intestine also causes the hormone cholecystokinin(abbreviated as CCK) to be released from the intestinal mucosa. CCK diffuses bythe way of the bloodstream to the pancreas where it causes the pancreas torelease digestive enzymes. The hormone secretin is released from the smallintestine in response to the entering chyme from the stomach. Secretin is alsoabsorbed by the blood and is transported to the pancreas where it causes therelease of bicarbonate ion and other fluids.

Thepancreas not only hasendocrine cells which secrete insulin and glucagon butit alsocontains secreting structurescalledacini that secretes a fluid into the smallintestine which has a high bicarbonate content that is rather alkaline. Thebicarbonate ions combine with the protons from HCl to produce carbonic acidwhich is then converted into carbon dioxide and water. Carbon dioxide isabsorbed into the blood and transported to the lungs where it is expired. This

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mechanism increases the pH in the small intestine to a more alkaline value. SeeFigure 3-4.

Na+ + HC(V + H+ + CI"

H2C03 c => COo + H90

HCO3"(bicarbonate ion)

H2CO3 (carbonic acid)

Figure 3-4Removal of the bicarbonate ion.

Theliverhas a diverseset of functions, oneof which is tosynthesize a compoundcalled bile which it concentrates and stores in the gallbladder. The majorpigment in bile is a compoundcalledbilirubin, which is a breakdown product ofhemoglobin. Bile also contains bile salts which are important in the digestionand absorption of fats. Whenbileis released fromthe gallbladder it passesdowna duct that joins with the pancreatic duct, through a constriction called thesphincterof Oddi, and emptiesinto the smallintestine. Bile acts toemulsify fats(i.e., decrease their surface tension in order to break them up into smallersizes)and it also helps the epithelial cells of the small intestine absorb those fats. Thepresence of fats in the small intestine releases CCK which then acts on thegallbladder and causes contraction, and on the sphincter of Oddi and causesrelaxation, so the bile can pass into the lumen of the small intestine. [Thegreenish colorof the gallbladder is due to the variousbreakdownproducts foundin the concentrated bile.]

Na+

Glucose

Lumen

Apicalside

Basolateral

^

Na+

Glucose

siae

V +

^ ^

^- Na/Kpump

J Blood

Figure 3-5Transport systems.

Aswe havementioned, the smallintestine is where themajority ofabsorption offluid and nutrients takes place. Located on the apical and basolateral regions ofthe epithelial cells are specialized transport proteins ("carriers") responsible forthe absorption of sodium, chloride, sugars, amino acids, vitamins, and other suchcompounds. For example, consider the absorption of glucose into the epithelialcell. Glucose is only able to enter the cellif sodium is cotransported along with it..Assodium enters the celldown its concentration gradient, it "drags" glucose intothe cytoplasm of the cell. Because the intracellular concentration of glucoseincreases it is able to diffuse out the basolateral side and eventually into theblood. As the concentration of sodium increases in the cell it is also pumped out



Biology Gastrointestinal Tract St Kidney Nutrients 8t Digestion

on the basolateral side in exchange for potassium via the sodium-potassiumpump. See Figure 3-5.

As the ions and nutrients are being absorbed by the epithelial cells of the smallintestine, water is diffusing through the membrane. Water is trying to equilibrateon both sides of the membrane by osmosis. Recall that cholera toxin causesmassive loss of fluid by diarrhea. In the majority of cases this can be treated bygiving the individual large quantities of glucose and saline solutions.

How much fluid is being absorbed by the small intestine? Roughly 1.2 liters ofwater per day are taken in from the foods we eat. Another 7.0 liters are secretedby the stomach, pancreas, liver, small intestines and so on. Roughly 8.1 liters ofthe total 8.2 liters is reabsorbed in the small intestine. Similarly, out of theroughly 800 grams of solid food consumed, roughly 570 grams are absorbed inthe small intestine. The remaining fluid and solid material is excreted as waste.

Recall that fats are degraded to fatty acids and glycerol by the action of theenzyme lipase. These compounds can diffuse into the intestinal epithelial cellswhere they are resynthesized into triglycerides (fats) and aggregate intostructures called chylomicrons. These aggregates are released at the basolateralmembrane and into the extracellular space (Figure 3-6). The chylomicrons enterinto the lymph and are transported to the veins and eventually the tissues.

Fattyacids

Glycerol

^ Fatty^ acids

•^^ Glycerol

r-FATS

To the lymphand tissues

Figure 3-6Release of chylomicrons.

Thelarge intestine absorbs mostof the water and ions that are left in the chymeas it passes from the small intestine. As the sodium and chloride ions areabsorbed an osmotic gradient is established that allows for the absorption ofwater. What is not absorbed is passed out the body in the feces.


Biology Gastrointestinal Tract St Kidneys

The KidneysRenal Function

Roughly 60% (a little less than 2/3) of an average 70-kg (154-lb) man's bodyweight is water. Most of this fluid is located inside the cells (called intracellularfluid) and about 1/3 of it is located in the interstitial spaces outside the cells(called the extracellular fluid). If you think about the organ systems which wehave previously discussed, you will realize that water can be lost from theintegument (theskin) by evaporation, thegastrointestinal tract, and the lungs.Water can also be lost from the urinary system. However, the water that is lostfrom the body by these four avenues exactly matches the water gained by thebody through consumption offluids and regulation at the level of the kidneys.

Recall that we have mentioned that osmolarity refers to the total soluteconcentration of a solution. Remember, the higher the osmolarity of a givensolution, the lower will be the concentration of water in that solution. If anorganism can change the internal ionic concentration of its body fluids to meetthat of the surrounding environment, they are referred to as osmoconformers.Many marine invertebrates are osmoconformers. However, most vertebrates donot change the internal ionic concentration of their body fluids to meet that of asurrounding environment. Organisms like this are referred to as osmoregulators.The difference between osmoconformers and osmoregulators can be seen on thegraph in Figure 3-7.

>>

Osmoregulator

Limited

Osmoregulator

Environmental osmolarity

Figure 3-7The difference between osmoconformers and osmoregulators.

The vertebrate kidney presumably arose from freshwater teleosts (fishes). Theblood osmolarity of a freshwater teleost is about 300 milliosmols/liter, which isgreater than that of the surrounding aqueous environment. This means that theconcentration of water is less in the freshwater fish than it is in the water in

which the fish is swimming. Water from the environment will diffuse down itsconcentration gradient and into the fish. The excess water that is continuallydiffusing into the fish is filtered and removed by the kidneys and excreted in the

Renal Function


Biology Gastrointestinal Tract &Kidneys Renal Function

form of a dilute urine.Since copiousamounts of urine are excreted on an hourlybasis the freshwater fish also loses important ions like sodium and chloride ions.These ions are in rather low concentration in freshwater habitats and so thefreshwater teleost has developed (via evolution) gills for active transport ofsodium and chloride into its circulatory system. See Figure 3-8.

Sallt loss

though gills

Water enters

by osmosis

Copiousurine

(salt and water loss)

Active transportof salt through gills

Figure 3-8Salt regulation.

Generalized Kidney FunctionThe kidneys in vertebrates like the freshwater teleost have three main functions.The first function is that of filtration. The functional unit in the kidney is thenephron which consists of a glomerulus (Latin, meaning "little ball"), Bowman'scapsule, and a tubular system. The glomerulus is a collection of capillaries thatreceives blood from an artery terminating in the renal system. Blood is pumpedinto the glomerulus by the hydrostatic pressure of the heart and that pressureforces the blood through the capillary walls and into Bowman's capsule. The cell-free ultrafiltrate found in Bowman's capsule lacks many of the plasma proteinsfound in the blood. Blood plasma is simply a solution which is about 90% waterthat is composed of organic (e.g., proteins, sugars, amino acids, etc.) andinorganic substances (e.g., various ions like sodium and chloride). Within theblood plasma one also finds red blood cells, white blood cells, and platelets. Thefiltrate in Bowman's capsule is essentially the plasma minus the (large molecularweight) proteins.

The second function of the kidney is reabsorption of important organic andinorganic compounds from the filtrate in Bowman's capsule. Reabsorption occursthrough many of the epithelial cellswhich line the tubular lumen of the nephron.The cilia of these epithelial cells move in such a way as to help propel the filtratethrough these renal tubes. Glucose,small proteins, amino acids, salts, bicarbonateions, and water are reabsorbed along this tubular system and transported backinto the blood. The epithelial cells also have the ability to secrete protons,potassium, urea, uric acid, and ammonia. The third function of the kidney isexcretion of waste, salts, and excess water.


Biology Gastrointestinal Tract 6* Kidney

Renal Physiology

Both human kidneys sit on the dorsal side of the abdominal cavity and are bean-shaped organs about the size of a fist. Blood from the descending aorta entersinto the renal artery and eventually into the glomeruli of the nephrons. Bloodleaves the kidney by way of the renal vein which itselfempties into the inferiorvena cava. Leaving each kidney is a ureter which transports the urine to thebladder. Urine exits the bladder by way of the urethra. These anatomicalconsiderations are shown in Figure 3-9.

Inferior Vena Cava'

Adrenal Gland ^

Renal Vein

Urethra

Abdominal Aorta

Renal Artery

Left

Kidney

Bladder

Figure 3-9Anatomical position of the kidneys.

Each kidney contains more than a million nephrons. The glomerulus andBowman's capsule of each nephron is located in the cortex and gives this portionof the kidney a granular appearance. In contrast, the striated appearance of themedulla is primarily due to a portion of the tubular system of the nephron calledthe loop of Henle and the collecting duct. The collecting ducts, which collect theurine, empty into the renal pelvis of the kidney and then into the ureter. This isshown in Figure 3-10.

Renal Physiology



Renal

Pelvis

Ureter

Renal Physiology

Nephron

Figure 3-10Kidney anatomy.

The NephronWe mentioned that the functional unit of the kidney is the nephron. As the renalartery branches into smaller divisions and enters the kidney it becomes theafferent arteriole. It is the afferent arteriole which enters into Bowman's capsuleand forms the capillary bed called the glomerulus. An efferent arteriole leavesthe glomerulus and forms a capillary network which surrounds the renal tubules.The blood that leaves this capillary network does so by a venule which laterempties into the renal vein that leaves the kidney. Extending from Bowman'scapsule is a long tubular structure which is divided into the proximal convolutedtubule (PCT), loop of Henle, distal convoluted tubule (DCT), and the collectingduct. Many different nephrons can attach to a single collecting duct. Thesestructures are shown in Figure 3-11.

The cardiac output of the heart is roughly 5 liters of blood every minute. Out ofthis 5 liters of blood roughly 20% of it (about 1/5) passes through the kidneyseach minute. Thus, the volume of filtrate passing into Bowman's capsule will berather large. Roughly 180 liters (47 gallons) of filtrate per day passes into allthese capsules and of that about 1.0 to 1.5 liters enters into the urine. Dependingon how much fluid you drank, the upper end of urine production might be about3 liters per day.

The osmolarity of the filtrate in Bowman's capsule and in the proximalconvoluted tubule is about 300 milliosmols per liter. Recall that this is essentiallythe same osmolarity we found in the plasma. The concentration of the urine thatis collected varies depending on the circumstance. If you wanted to excrete adilute urine, then the concentration of the fluid leaving the collecting ductsshould be lower than 300 milliosmols per liter. For example, a urineconcentration which is low might have an osmolarity which is 0.7 times that of


Biology Gastrointestinal Tract & Kidney

theosmolarity of theplasma. In otherwords, theosmolarity of this "dilute" urinemight be 200milliosmols per liter. In contrast, a concentrated urine would have ahigh osmolarity, something in the range of 4.2 times that of the osmolarity ofplasma (or about 1200 milliosmolsper liter).

Bowman's

Capillary . kNetwork *

Glomerulus Proximal

Convoluted

Tubule A

Dustal V CortexConvoluted

Tubule

J

> Medulla

x-_ Collecting

J

Figure 3-11The nephron and its related capillary system.

Consider the proximal convoluted tubule. The PCT is the obligatory section ofthe nephron because roughly 65% of all reabsorption and secretion occurs here.The epithelial cells in the PCT are quite metabolically active and support avariety of active transport processes. Glucose, small molecular weight proteins,amino acids, and vitamins are completely (100%) reabsorbed in the PCT.Roughly 80% of the Na®, CI®, and water are alsoreabsorbed in the PCT. Manyof the mechanisms for reabsorption involve sodium. For example, glucose andsodium ions are reabsorbed together as are the amino acids and sodium ions.Sodium ions are also reabsorbed with chloride ions. Little regulation occurs inthe PCT.

Renal Physiology


Biology Gastrointestinal Tract St Kidney Renal Physiology

As the PCTbegins to descend from the cortexinto the medulla it forms the loopof Henle. The loop of Henle has a thin portion and a thick portion (see Figure 3-11).The descending thin portion of the loop of Henle is very permeable to waterbut only relatively permeable to ions like sodium and molecules like urea. Theascendingthin portionof the loop of Henle is much more permeable to urea butmuch less permeable to water than the thin descending portion of the loop.

As the loop of Henle ascendsit becomes thicker.The epithelial cells in this regionof the loop actively transport ions like sodium and potassium from the lumen ofthe loop into the interstitial fluid. However, this region is impermeable to ureaand water. This means that as the filtrate is passing up the thick portion of theloop of Henle it is becoming more dilute because of the fact that ions are beingtransported out into the interstitial fluid. Remember, the lower the osmolarity,the higher will be the concentrationof water (i.e., a more dilute fluid).

The filtrate from the thick portion of the loop of Henle passes into the distalconvoluted tubule. The segment of the DCT closer to the loop of Henle is alsoquite impermeable to urea and water but rather permeable to ions like sodium.Therefore, as the filtrate passes through this segment of the DCT it will become abit more dilute.

The epithelial cells of the segment of the DCT closest to the collecting duct andthat portion of the collecting duct located in the cortical region of the kidney arestill rather impermeable to urea. However, these two areas of the nephron arequite sensitive to the hormone aldosterone (secreted by the cortex of the adrenalglands which sit on top of the kidneys) which regulates sodium absorption. Anincrease in the concentration of aldosterone causes sodium to be reabsorbed bythe epithelial cells. When sodium ions are pumped out into the interstitial space,potassium ions are simultaneously transported into the lumen of this region ofthe nephron. Epithelial cells in this region of the nephron also respond toantidiuretic hormone (ADH, produced by hypothalamus and released by theposterior pituitary) which regulates water absorption. If the concentration of ADHincreases, then water will be reabsorbed from the epithelial cells in the collectingduct and passed to the interstitial space. Removing water from the lumen of thecollecting duct acts to concentrate the urine. The cells of the collecting duct canalso secrete hydrogen ions (as can the cells of PCT and DCT).


Biology Gastrointestinal Tract & Kidney

Homeostatic Mechanisms ,

Suppose we want to maintain the levels of sodium in the body. Certain cells inthe cortex of the adrenal gland are sensitive to the levels of sodium in the blood.When the concentration of sodium decreases in the blood these cells release thehormone aldosterone. Aldosterone actsat thelevel of theDCT and the collectingducts and stimulates the epithelial cells in those regions to reabsorb sodium. As aresult the blood concentrationof sodium begins to rise.Thisacts as a signal andfeeds back on the cortical cells of the adrenal gland telling it to reduce thesecretion of aldosterone.

Suppose we want to maintain the proper levels of water in the body. If there hasbeen a decrease in the plasma volume of the body, there will be a tendency tohave a lower than normal blood pressure (detected by baroreceptors) but ahigher than normal osmotic pressure (detected by osmoreceptors). Receptorsreceiving this information stimulate specific cells in the hypothalamus tosynthesize and transmit ADH to the posterior pituitary where it is released intothe blood. ADH stimulates the epithelial cells in the latter portion of the DCTand the collecting ducts to reabsorb water. This leads to less water excretion anda higher plasma volume. If you were to drink too much liquid (e.g., pitchers ofbeer), then the opposite process would happen and you would excrete copiousamount of urine. It turns out that the release of ADH is inhibited by alcohol.

Excretion of Nitrogenous WasteProteins and nucleic acids are the two primary metabolic sources of nitrogenouswastes. There are essentially three ways to get rid of the nitrogen that isproduced during the metabolism of these compounds. Nitrogen can be removedas ammonia, as urea, or as uric acid. The structures of these compounds areshown in Figure 3-12.

NH3

Ammonia

(toxic, soluble)

H2N-C-NH2

Urea

(less-toxic, soluble)

Oii

H-N V

I JU0=0

H H

Uric acid

(toxic, insoluble)

Figure 3-12Nitrogenous waste compounds.

As glutamine is metabolized near the gills of fish, the nitrogen in this amino acidcan be converted to ammonia which can then combine with a proton and, inexchange for a sodium ion, be carried away with the passing water. Theammonium ion can also be excreted by mammals. One of the by-products of theamino acid glutamine is urea which is carried by the blood to the kidney where itis excreted in the urine. Urea is excreted by most mammals, amphibians andsome reptiles and birds. The ammonia that is produced from the metabolismfrom the amino acids glycine, aspartate, and glutamate can be converted to uricacid which is the excretory product of birds and reptiles.

Homeostatic Mechanisms


Biology Gastrointestinal Tract St Kidney Homeostatic Mechanisms

Control of ptl and tl®The hydrogen ion concentration of the extracellular fluid is closely monitored.The normal pH of the body is 7.4. This value can fluctuate between a pH of 7.35and 7.45. If the pH of the body decreases to values below 7.35 for an extendedperiod of time, the individual becomes lethargic and death can result. If the pH isabove 7.45 for extended periods of time, the individual becomes hyper and deathcan result. The regulation of blood pH is by means of a buffering system inwhich hydrogen ions are reversibly bound to a buffer like the bicarbonate ion(HCC>3e), the hemoglobin molecule, or even certain plasma proteins. Forexample, if there are hydrogen ions in the plasma, they can be taken up by thebicarbonate ion as shown in equation (3-1) to form carbonic acid. This is a weakacid and decarboxylates to form carbon dioxide and water. Carbon dioxide iseliminated by respiration. In fact, the PCO2 levels in the alveoli of the lungs arequite important in regulating the plasma pH. This can be seen if you examine the(Henderson-Hasselbalch) equation shown in (3-2). The levels of bicarbonate andhydrogen ion are controlled by the kidney whereas the levels of carbon dioxidelevels are controlled by respiration.

H+ + HCO,

pH = pKa + log

— H2C03 ^=^ C02 + H20

[HCO3]

[C02]

(3-D

(3-2)

Cholera Toxin

Suppose you had a severe case of diarrhea caused by the bacterial cholera toxin(Vibrio cholera) that we have previously discussed. This would result in adecrease in the pH and an increase in the PCO2. Why? Recall that thegastrointestinal tract secretes a large amount of sodium bicarbonate. Whathappens if you lose a lot of sodium bicarbonate by way of the gastrointestinaltract? Metabolic acidosis results. Look at equation (3-1). If you decrease theconcentration of HC03e, then by LeChateller's principle (from generalchemistry) CO2 will combine with H2O to make H2CO3. This will dissociate intoHC03, and H®. In otherwords, the hydrogen ion concentration has increasedor,if you like, the pH has decreased. This same conclusion can be derived fromequation (3-2).

A decrease in pH will stimulate the DCT to secrete hydrogen ions and reabsorbas much bicarbonate as possible. By a negative feedback loop this tends tocounter the initial decrease in pH. Simultaneously, an increase in PCO2stimulates the central chemoreceptors in the medulla and pons to increase thecontraction of the diaphragm and chest muscles in order to increase respiration.An increase in the respiratory rate will eliminate more CO2 from the lungs.


Gastrointestinal Tractand Kidney

15 Passages

100 Questions

Passage TitlesI. Gastrointestinal Smooth Muscle

II. Protein TurnoverIII. Liver, Pancreas, and IntestinesIV. Digestion SafeguardsV. Digestion of Fats

VI. Diabetic Diet ExperimentVII. The Kidney

VIII. Kidney and pHIX. Dialysis and UltrafiltrationX. Caffeine.

XI. Kidney and CalciumXII. Lymph

XIII. Renal ClearanceXIV. Gastrointestinal JunctionXV. Cholera Toxin

BERKELEYUr-e-v-i^e-w®


Questions1 -5

6- 12

13-2021 -27

28-3334-40

41 -48

49-53

54 - 60

61 -66

67-72

73 - 79

80-86

87-9394- 100

Suggestions

The passages that follow are designed to get you to thinkin a conceptual manner about the processesof physiology at the organismal level. If you have a solid foundation in physiology, many of theseanswers will be straightforward. If you have not had a pleasant experience with the topic, some of theseanswers might appear to comefrom the void past the Oort fieldof the solar system.

Pick a few passage topics at random. For these initial few passages, do not worry about the time. Justfocus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, orgraphs. Third, read each question and the accompanying answers carefully. Fourth, answer thequestions the best you can. Checkthe solutionsand seehow you did. Whether you got the answers rightor wrong, it is important to read the explanations and see if you understand (and agree with) what isbeing explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block ofpassages and try them. Be aware that time is going to become important. Generally, you will have about1 minute and 15 seconds to complete a question. Be a little more creative in how you approach this nextgroup. If you feel comfortable with the outline presented above, fine. If not, then try differentapproaches to a passage. For example, you might feel well versed enough to read the questions first andthen try to answer some of them, without ever having read the passage. Maybe you can answer some ofthe questions by just looking at the diagrams, charts, or graphs that are presented in a particular passage.Remember, we are not clones of one another. You need to begin to develop a format that works best foryou. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where thelevel of distraction is at a minimum. Get out your watch and time yourself on these passages, eitherindividually or as a group. It is important to have a feel for time, and how much is passing as you try toanswer each question. Never let a question get you flustered. If you cannot figure out what the answer isfrom information given to you in the passage, or from your own knowledge-base, dump it and move onto the next question. As you do this, make a note of that pesky question and come back to it at the end,when you have more time. When you are finished, check your answers and make sure you understandthe solutions. Beinquisitive. If you do not know the answer to something, look it up. The solution tendsto stay with you longer. (For example, what is the Oort field?)


Section ID



>12 86-100

10-11 79-85

8-9 65-78

7 59-64

6 54-58

5 48-53

£4 0-47

Biology Gastrointestinal Smooth Muscle Passage I


The smooth muscle cells of the gastrointestinal (GI)tract are typically of the single unit type, roughly 50 to100 microns long and 2-5 microns wide. The actionpotentials found in smooth muscle are smaller inamplitude, but longer lasting than those in neurons. Themuscle cells can be arranged in sheets and can be excitedby action potentials from neighboring cells. These sheetsof cells also exhibit an intrinsic pattern of periodicdepolarization known as pacemaker activity. Slow wavesof depolarization followed by repolarization make up thebasic electrical rhythm. In the presence of excitatoryneural or hormonal input, a slow wave may exceed thethreshold for action potential generation and initiatestronger contractions.

Motility in the GI tract falls into three majorcategories. The first is peristalsis, which involvescoordinated waves of contraction and relaxation. The

second is segmentation, which involves alternatingcontraction and relaxation of the circular muscle layer.The third is the mixing motions of villi and microvilliwhich reduce the effect of the unstirred layer of fluid thatlies adjacent to the apical surface of cells.

1.

2.

Two types of smooth muscle, single-unit and multi-unit, are distinguishable based on their electricalactivity. Which of the following statements aboutthis distinction is most likely true?

A. Single-unit fibers are coupled, with manyindividual cells containing gap junctions.

B. Single-unit fibers are uncoupled, with fewindividual cells containing gap junctions.

C. Multi-unit fibers are coupled, with fewindividual cells containing gap junctions.

D. Multi-unit fibers are uncoupled, with manyindividual cells containing gap junctions.

The conduction velocity of action potentials alongsmooth muscle fiber is low because activation of:

A. potassium channels is slow.B. sodium channels is slow.

C. calcium channels is slow.

D. magnesium channels is slow.


3.

171

According to the diagram shown below, the force ofsmooth muscle contraction is most likely dependentupon the:

A.

B.

C.

D.

Action Potentials

I jv_Time

size of the action potential.interval between successive sets of action

potentials.number of action potentials within a set.time duration of individual action potentials.

Segmentation, in contrast to peristalsis, is primarilyresponsible for:

A. absorption of nutrients from the lumen of theGI tract.

B. movement of intestinal contents through the GItract.

C. mixing of intestinal contents, breaking largerparticles into smaller ones.

D. stimulating the secretion of digestive enzymesfrom the intestinal exocrine glands.

The mixing of intestinal contents most aids indigestion because:

A. smaller particles move faster through the GItract.

B. smaller particles are able to cross the epitheliallayer with increased efficiency.

C. smaller particles carry less charge, enablingmore efficient passage of particles overbiological membranes.

D. breaking larger particles into smaller particlesincreases the surface area available to digestiveenzymes.


Biology Protein Turnover Passage n


The diagram shown in Figure 1 is a summary of dailyprotein intake and output—turnover—in a 70 kg human.[Note: The gastrointestinal (GI) system is equivalent tothe gut.]

Protein Intake

(100 gm)

Fecal

(10 gm protein)

Body Protein(10,000 gm)

Liver

(25 gm)

Muscle

(50 gm)

Turnover of

Body Protein(250 gm)

♦

if̂White Cells"V* (20 gm)

"^i HhHb

(8gm)

Free

Amino Acids I

(100 gm)

*Urinary

(80 gm proteinequivalent)

Figure 1

Skin

(2gm)+

6. You can see from the diagram that the gut secretesmore protein than the person ingests. Which of thefollowing gut secretions contain recyclable protein?

I. Digestive enzymesII. Mucus

m. Bicarbonate

A. I onlyB. I and II onlyC. II and m onlyD. I, n, and in

7. Given the following equation, what is the efficiencyof protein digestion and absorption in the gut?

Efficiency =

A. 63%

B. 77%

C. 90%D. 94%

Protein Absorbed

Total Protein Available


8.

9.

10.

172

The urine carries 80 gm of protein equivalents perday. The term protein equivalent is used becauseprotein is degraded and another molecule is used asthe excretory route for nitrogen. Which of thefollowing is the chemical formula for this excretoryproduct?

A.

B. NH3

C.

on

H2N— C- NH2

D. H2N- NH2

Rank in increasing order the following organs ortissues according to their amount of daily proteinturnover.

A. White cells, gut, liver, muscle.B. Gut, muscle, liver, white cells.C. Muscle, liver, gut, white cells.D. White cells, liver, muscle, gut.

Albumin is the most abundant plasma protein. Whattype of experiment could be used to quantify thealbumin turnover rate?

A. Give a known dose of albumin and sample theblood periodically to measure albuminconcentration.

B. Give isotopically labeled albumin, sample theblood periodically, and calculate the decay rateof labeled albumin.

C. Give a known dose of albumin and sample theblood periodically to measure albumin decayrate.

D. Combine a blood sample with isotopicallylabeled albumin in a test tube and calculate a

decay rate.


Biology Protein Turnover

11. During a fast, which tissue is the first to break downits protein to provide amino acids for other tissues?

A. Smooth muscle.

B. Cardiac muscle.

C. Enzymes of energy metabolism.D. Skeletal muscle.

12. The following experiment, a nitrogen balance study,was performed to determine nitrogen needs inhuman subjects. To be in nutrient balance meansthat intake - output = zero for that nutrient. Fornitrogen, this means that the nitrogen contained inprotein in food eaten and secreted by the gut equalsthe amount of nitrogen excreted in urine, feces, andthrough skin losses.

Subjects were placed on 3 different dietssequentially. The results of their nitrogen balanceare noted in Table 1.

Protein Intake

(gm/kg)Nitrogen Balance

(mg/kg)

0.1 -42

0.6 -21

1.5 + 12

Table 1

What is the amount of protein needed by thesesubjects to maintain a nitrogen balance?

A. Less than 0.1 gm/kg.B. Between 0.1 and 0.6 gm/kg.C. Between 0.6 and 1.5 gm/kg.D. Greater than 1.5 gm/kg.


Passage U


Biology Liver, Pancreas, and Small Intestines Passage in

Passage HI (Questions 13- 20)

In controlling the processes of digestion, the pancreasand liver work in tandem to prepare food that exits thestomach for absorption in the small intestine. Secretionofbile salts from the liver (sodium and potassium saltsconjugated to glycine and taurine) is important in theformation of micelles, water soluble complexes fromwhich lipids can be more easily absorbed by the intestinallining. Secretions from the pancreas contain a highlyalkaline bicarbonate solution that is important in theneutralization of the hydrochloric acid (HCl) secreted bythe parietal cells of the stomach. Secretion of HCl can beinhibited by gastrin inhibitory peptide (GIP), anenterogastrone released by the duodenum. HCl can leavethe stomach and enter into the duodenum through thepyloric sphincter.

The pancreas also secretes enzymes needed tobreakdown proteins entering the duodenum. All of thesepancreatic enzymes are secreted as inactive proenzymes.Trypsin inhibitors are secreted by the same pancreaticcells to prevent activation of these enzymes. Theseproenzymes travel through the ductal system to theduodenum were the enzyme enteropeptidase(enterokinase) is secreted from the brush border of thesmall intestine. Enteropeptidase initiates the conversionof the pancreatic enzymes to their active state.

Bile salts and exocrine secretions of the pancreas arecontrolled by secretin and cholecystokinin (CCK), twoenterogastrones released from endocrine cells in the upperportion of the small intestine.

Secretin is released from the S cells in the intestinal

mucosa when the pH of the duodenum falls below 4.5.Secretin release is increased by gastric acid bathing thelining of the small intestine and products of proteindigestion. Secretin stimulates the ductal cells of thepancreas to release a fluid containing a high concentrationof bicarbonate ion but a low concentration of chloride ion.

Secretin, also increases the production of pancreaticenzymes, decreases acid secretion in the stomach, andmay cause contraction of the pyloric sphincter.

CCK secretion is stimulated by peptide, amino acid,and fatty acid products coming in contact with theintestinal lining. CCK stimulates the contraction of thegall bladder, a storage organ that concentrates bile saltsproduced in the liver. Furthermore, CCK increases thesecretion of pancreatic fluids rich in enzymes and alkalinephosphates, inhibits gastric emptying, increasesenteropeptidase production, and increases motility of thesmall intestine and colon.


13. Which secretion from the small intestine would be

most appropriate for processing a meal containing ahigh fat content?

A. CCK

B. Bile Salts

C. SecretinD. Pancreatic Fluids

14. Which of the pancreatic enzymes activated in thesmall intestine is important for the initiation of thecascade shown in the diagram below?

15.

16.

Proenzyme Enzyme

t * * _, ^~„, Enteropeptidase ' „,Trypsinogen •— ^> Trypsin

Chymotrypsinogens —ryps'"^_ , Trypsin^Proelastase ^»

Procarboxypeptidases —ryPsm^.

A. Bile salts

B. Secretin

C. CCK

D. Enterokinase

Chymotrypsins

Elastase

Carboxypeptidases

Deficiency of the enzyme enteropeptidase results ina disorder displaying which of the symptoms listedbelow?

A. Fat deficiencyB. Increases in fat absorptionC. Protein deficiencyD. Vitamin B \2 deficiency

Which hormone is important in initiating theneutralization of gastric secretions in the smallintestine?

A. Secretin

B. Enterokinase

C. ProenzymesD. CCK


Biology Liver, Pancreas, and Small Intestines Passage III

17.

18.

19.

The enzyme phospholipase A can remove a fattyacid residue from lecithin (a component of bile) toform lysolecithin, a compound that damages cellmembranes. Activation of phospholipase A in thepancreatic ducts causes damage to the surroundingtissue and can lead to acute pancreatitis, a severe andoften fatal disease. Which of the following catalyticenzyme is most likely involved with this disorder?

A. TrypsinB. EndopeptidaseC. Bile salts

D. Secretin

As acidic chyme enters the small intestine, secretinstimulates the release of the bicarbonate ion from thepancreas. The following reaction will result in theduodenum:

HCl + NaHC03 -> NaCl + H2C03

Carbonic acid (H2CO3) will immediately:

A. begin to increase the acidity of the duodenumand activate CCK in order to decrease motilityof the small intestine and increase absorption.

B. decrease the pH of the intestinal lumen so theentering chyme can easily be hydrolyzed bythe acidic medium.

C. dissociate into CO2 and H2O in order to allowthe NaCl solution in the small intestine to

remain neutral.

D. dissociate into H® and HCC«3e in order toincrease the acidity of the intestinal lumen andprepare it for nutrient absorption.

Severe duodenal ulcers, leading to the removal ofthe duodenum, will cause an increase in all of thefollowing EXCEPT:

A. gallbladder relaxation.B. parietal cell activity.C. pancreatic ductal cell activity.D. passage of chyme into the small intestine.


20. Antacid tablets, which are mixtures of magnesiumhydroxide and aluminum hydroxide, have beenshown to heal duodenal ulcerations.

Mg(OH)2 + 2 HCl -> MgCl2 + 2 H20

Al(OH)3 + 3 HCl -> AICI3 + 3 H20

Based on the diagram shown below, antacids shouldbe taken 1 and 3 hours after a meal because they:

A.

B.

C.

D.

'•3

1 3 5

Hours After Eating a Meal

A = No Antacid

B = Antacid taken 1 hour after meal

C = Antacid taken 1 and 3 hours after meal

will allow for additional release of acid to aid

in food digestion.delay the appearance of acid in the stomach.combine with either magnesium hydroxide oraluminum hydroxide to form an acid which ismore concentrated.

produce water as a by-product of the reactionand reduce the acidity of the stomach.


Biology Digestion Safeguards Passage IV

Passage TV (Questions 21-27)

The body is about 95-98% efficient in digestingprotein of animal origin. Since 20% of the human body isprotein of animal origin, we require some sophisticatedsafeguards to avoid digesting our own tissue.

First, mucopolysaccharides line the epithelial tissue ofthe digestive tract. Mucus is indigestible and protects thetissues from the actions of acids and enzymes ofdigestion. Second, the pancreas contains proteolyticenzymes in an inactive form (zymogens) to avoiddigestion of the pancreatic cells. These zymogens arereleased unchanged through the pancreatic duct into theduodenum.

Trypsin, a pancreatic protease, is activated in theduodenum by the enzyme enterokinase. Trypsin, in turn,activates the other zymogens by cleaving a portion of themolecule so that the active site is exposed. A furthermechanism for protection of the GI tract is the rapidturnover of cells. The epithelial lining of the GI tract isreplaced every 5-7 days. Finally, although the stomachacidifies the bolus of food considerably, the pancreassecretes bicarbonate to neutralize the pH of the duodenalcontents.

21. Raw soybeans contain a protein that acts as a trypsininhibitor. What would be the digestive effect ofeating a large quantity of raw soybeans?

A. Trypsin would be more active, but the otherenzymes would remain as zymogens.

B. Trypsin would be inactive, and the otherpancreatic enzymes would function normally.

C. Trypsin would be inactive, and the otherpancreatic enzymes would remain aszymogens.

D. Trypsin and other pancreatic enzymes wouldfunction normally.

22. Chemotherapeutic agents target rapidly growing anddividing cells. This is beneficial for attacking tumorcells, but may be detrimental to other body cells.Which cells would be most adversely affected?

A. neurons

B. cardiac muscle cells

C. gastric epitheliumD. skeletal muscle cells


23. Which parts of the digestive tract are lined withmucus?

I. Mouth

II. Colon

III. Small intestine

A. I onlyB. I and II

C. II and HI

D. I, H, and in

24. What is the approximate pH of the stomach?

25.

26.

A.

B.

C.

D.

6.8

0.2

9.0

2.0

What could happen if enterokinase entered thepancreatic duct and went to the pancreas?

I. The zymogens would be activated.II. Pancreatic enzymes would appear in the

bloodstream.

III. Insulinsecretion wouldbe impaired.

A. I onlyB. I and II onlyC. II and HI onlyD. I, II, and ffl

What type of acid(s) do the parietal cells of thestomach secrete?

I. Hydrochloric acidII. Sulfuric acid

III. Carbonic acid

A. I onlyB. I and II onlyC. II and m onlyD. I, n, and ffl

27. To what major class of moleculesmucopolysaccharides belong?

A. Protein

B. LipidC. CarbohydrateD. None of the above

do


Biology Digestion of Fats Passage V


The digestion of fats and cholesterol by the body isprimarily carried out in the duodenum of the smallintestine. Triglycerides are the most common fat productsin the diet. The initial process in the digestion of fats isthe emulsification of fat droplets by bile salts releasedfrom the liver and gall bladder. Bile salt release isstimulated by increased levels of long-chained fatty acids(containing more than 10 carbon atoms) within theduodenum. Bile salt emulsification of fat is not essential

for digestion and absorption since as much as 60% of alltriglycerides can be absorbed without the aid of bile salts.Cholesterol processing, however, is dependent on theactions of bile salt.

Cells within the duodenum sensitive to fatty acids willrelease cholecystokinin (CCK), a hormone which willstimulate contraction of the gall bladder and increase fluidsecretion from the pancreas. Once the fats have beenemulsified, pancreatic lipase, an enzyme produced fromthe exocrine pancreas and activated by the enzymetrypsin, will hydrolyze the fats into monoglycerides, freefatty acids, and glycerol. The hydrolyzed products of fatdroplets are then packaged by bile salts into water solublemicelles and then transported to the cellular membrane ofthe intestinal microvilli. The micelles cross the cell

membrane and are incorporated into vesicular packages(chylomicrons) in the endoplasmic reticulum. Thechylomicrons are then exocytosed into the intercellularspaceand taken up by the lactealsof the villi.

Chylomicrons also contain triglycerides, cholesterol,and phospholipids. They are transported by way of thelymphatics and the capillary system to the liver. In routeto the liver, lipoprotein lipase, an enzyme located in thecapillaries, strips triglycerides from the chylomicrons forstorage in adipose tissue. The remaining components ofthe chylomicrons are degraded in the liver and stored forlaterproduction of steroid hormones, VLDLs, or excretedwith bile salts into the intestine.

28. A trypsin antagonist is present during fatmetabolism. Which of the following processes willNOT occur?

A. Emulsification of fats.

B. Increase in fat content in the small intestine.C. Hydrolyzation of fat droplets in the small

intestine.

D. Decrease in CCK production.


29. Which of the components of fat digestion is mostclosely associated with principles utilized byhousehold detergents to remove grease?

A. Emulsification of fats.

B. Micelle endocytosis.C. Exocytosis of fatty acids.D. Packaging of fat products in the endoplasmic

reticulum.

30. Sprue, a disorder of malabsorption in the smallintestine, is characterized initially by the increasedsoap-like appearance of fat in the stools. Which ofthe following reasons could lead to this symptom?

A. Decrease in chylomicron production.B. Decrease in bile salt secretion.

C. Decrease in intestinal cell microvilli.

D. Increase in pancreatic lipase production.

31. What will occur if bile salt secretion is decreased?

A. Increases in cholesterol absorption in the smallintestine.

B. Complete blockage of fatty acid digestion inthe small intestine.

C. Inhibition of cholesterol absorption in thesmall intestine.

D. Decreased levels of CCK secretion from the

small intestine.

32. Chylomicrons are degraded in the liver. Which ofthe following compounds is a component of achylomicron as it enters into the liver to bedegraded?

A. Fatty acids.B. Bile salts.

C. Lipoprotein lipase.D. Cholesterol.

33. Disorders characterized by increases in fatdeposition in the liver may be produced by one ofthe abnormalities below.

A. Decreased levels of pancreatic lipase.B. Decreased levels of CCK.

C. Decreased levels of lipoprotein lipase.D. Decreased levels of chylomicrons.


Biology Diabetic Diet Experiment Passage VI


Diet control is important in non-insulin-dependentdiabetes mellitus (NIDDM), not only to improve controlof hyperglycemia but also to reduce the risk of coronaryheart disease by improving plasma lipid levels.

The American Diabetes Association (ADA) currentlyrecommends the following diet for patients with NIDDM:

• 30% of energy as fat (with 10% as saturated fat)• 40-50% complex carbohydrates (starches)• less than 10% simple carbohydrates (sugars)

Lastly, protein makes up 10-20% of the diet's energy.

Some studies indicate that diets like the ADA diet, thatare high in carbohydrates, lead to increased triglyceridelevels, increased very low density lipoprotein levels, andreduced high density lipoprotein levels: These would notbe beneficial in a diabetic population concerned withmaintaining optimal plasma lipid levels.

The following experiment was designed to test theeffect of the ADA diet versus a higher monounsaturatedfat diet (HF) on plasma glucose and lipid levels.

ExperimentJ

After a baseline measurement of 1 week, 10 NIDDMpatients ate the ADA diet and the HF diet for 5 weeks inrandom order during 2 hospital admissions. Metabolicmeasurements regarding glucose (Glc) are outlined inTable 1. [Note: Hb refers to hemoglobin.]

Table 1: Metabolic variables during thestudy.

Variable

Base

line

ADA

Diet

HF

Diet

Plasma Glc (mg/dL) 129 117 101$

Urinary Glc (mg/day) 550 142 ot

Insulin Requirements(units/day)

84 81 70*

Glycosylated Hb (%) 11.3§ 7.8 8.1

t P < 0.02 for HF versus ADA and baseline.

t P < 0.01 for baseline versus HF values.

§ P < 0.005 for baseline versus HF and ADA values.

Metabolic measurements on plasma lipids andlipoproteins are outlined in Table 2.


Table 2: Plasma lipidand lipoproteinlevelsduring the study.

Measurement

(mg/dL)Base

line

ADA

Diet

HF

Diet

Total Cholesterol 225* 205 196

Triglycerides 285 218 163f

VLDL Cholesterol 58 43 28§

LDL Cholesterol 135 131 134

HDL Cholesterol 32 30 34§

$ P < 0.02 for baseline versus ADA and HF values,

t P < 0.01 for ADA versus HF values.

§ P < 0.005 for ADA versus HF values.

34. Which of the following statements are supported bythe data in Table 1?

I. The HF diet lowered plasma glucose in thesesubjects.

II. Patients required more insulin on the HF dietthan at the baseline measurement.

HI. Less glucose was excreted in the urine duringthe HF diet compared to the baselinemeasurements.

A. I and III onlyB. I and II onlyC. II and ffl onlyD. I, II, and ffl

35. In Table 1, glycosylated hemoglobin (G-Hb) isreported as an indicator of glucose metabolism in thebody. Glycosylatedmeans that glucose residues areattached. G-Hb decreases on both the HF and theADA diets during the study. Is this decreasebeneficial or detrimental?

A. This decrease is detrimental, becauseglycosylated hemoglobin indicates persistentlevels of low plasma glucose.

B. This decrease is detrimental, becauseglycosylated hemoglobin indicates persistentlevels of high plasma glucose.

C. This decrease is beneficial, becauseglycosylated hemoglobin indicates persistentlevels of low plasma glucose.

D. This decrease is beneficial, becauseglycosylated hemoglobin indicates persistentlevels of high plasma glucose.


Biology Diabetic Diet Experiment Passage VI

36.

37.

38.

The table shown below indicates the fatty acidcomposition of both ADA diet and HF diet in %energy intake/day.

FattyAcid

ADA

Diet

HF

Diet

12:0 0.2 0.2

14:0 0.3 0.2

16:0 6.8 7.6

16:1 0.2 0.4

18:0 2.4 2.2

18:1 7.4 29.4

18:2 7.2 8.0

18:3 0.0 0.2

20:4 0.2 0.2

22:0 0.3 1.5

Total: 25.0 49.9

Table 3

Which of the fatty acids was used to supplement themonounsaturated portion of the HF diet ?

A. Palmitoleic acidB. Stearic acid

C. Oleic acid

D. Palmitic acid

Which of the following oils or fats could be used toincrease the amount of monounsaturated fat in a

diet?

A. Olive oil

B. Lard

C. ButterD. Beef tallow

Which of the following statements is/are supportedby the data in Table 2?

I. Both VLDL and LDL were lower in the HFdiet compared to the ADA diet.

II. Triglycerides were significantly lowered bythe HF diet compared to baseline.

III. HDL cholesterol was increased by the HF dietcompared to the ADA diet.

A. I and ffl onlyB. I and II onlyC. n and III onlyD. I, II, and ffl


39. Some patients with NIDDM maintain good glucosecontrol using drugs that increase insulin secretion.What is the target of these drugs that affect insulinsecretion?

A. The alpha cells of the pancreas.B. The exocrine cells of the pancreas.C. The beta cells of the pancreas.D. Both the exocrine and beta cells of the

pancreas.

40. Insulin is secreted from the pancreas into the hepaticportal vein. Which is the first major organ insulinencounters after entering this portal system?

A. Small intestines

B. Stomach

C. Brain

D. Liver


Biology The Kidney Passage VII

Passage VOL (Questions 41-48)

The functional unit of the human kidney is the nephron(Figure 1), and each kidney contains approximately 1million of these tubular structures.

Efferent _, .arteriole Glomerulus

Afferentarteriole

Figure 1

Blood is supplied to each kidney by a renal artery. Thesmallest branch of this artery is the afferent arteriole, avessel influenced by sympathetic stimulation. A capillarybed (glomerulus) is formed from the afferent arteriole andis contained within Bowman's capsule. The efferentarteriole that leaves the glomerulus is further divided intoa capillary system that surrounds different portions of thenephron.

The plasma ultrafiltrate that passes through theglomerulus and into Bowman's capsule is essentiallyprotein-free and devoid of cellular structure. This filtrateflows into the proximal convoluted tubule (PCT), throughthe loop of Henle and into the distal convoluted tubule(DCT), through the collecting ducts and into the renalpelvis before making its way to the ureter and urinarybladder.

The PCT reabsorbs essentially all of the glucose andamino acids filtered by the glomerulus and roughly 70%of the filtered K®, Na®, Cle, and water. Approximately20% of the K®, Na®, Cle, and water thatpasses from thePCT into the loop of Henle is reabsorbed at the loop. Theremaining 10% of the K®, Na®, andCle is reabsorbed atthe DCT. Variable amounts of water also reabsorbed at

the DCT.


Antidiuretic hormone (ADH) and aldosterone are bothregulators of urine composition. Both hormones promoteNaCl reabsorption in the ascending limb of the loop ofHenle and in the DCT and collecting ducts. ADHpromotes water reabsorption in the DCT and collectingducts while aldosterone promotes K® secretion in theDCT.

41. Constriction of the efferent arterial will lead to:

A. a decrease in glomerular pressure, followedfirst by a decrease and then by an increase inglomerular filtration rate.

B. an increase in glomerular pressure, followedfirst by an increase and then by a decrease inglomerular filtration rate.

C. a decrease in glomerular pressure, followedfirst by an increase and then by a decrease inglomerular filtration rate.

D. an increase in glomerular pressure, followedfirst by a decrease and then by an increase inglomerular filtration rate.

42. The reabsorption of sodium across the epithelialcells of the proximal convoluted tubule and into theperitubular capillary would be significantly reducedby:

A. reduced secretion of ADH from the posteriorpituitary.

B. an increased rate of secretion of aldosteronefrom the adrenal cortex.

C. reduced plasma glucose concentrations.D. an increase in the tubular colloid osmotic

pressure.

43. Sympathetic stimulation of the kidneys primarilyaffects the:

A. afferent arterioles and decreases the volume of

urine flow.

B. afferent arterioles and increases the volume ofurine flow.

C. efferent arterioles and decreases the volume of

urine flow.

D. efferent arterioles and increases the volume of

urine flow.


Biology The Kidney Passage VH

44. Which of the following graphs BEST represents theplasma osmolarity (X-axis) of the blood and theplasma concentration of antidiuretic hormone (Y-axis)?

A. B.

X-Axis X-Axis

C. D.

X-Axis X-Axis

45. Which of the following graphs best represents theplasma concentration of glucose (X-axis) and therate of glucose excretion in the urine (Y-axis)?

A. B.

X-Axis X-Axis

C. D.

X-Axis X-Axis

Copyright© by The Berkeley Review 181

46. All of the following statements about aldosterone aretrue EXCEPT it:

A. promotes sodium reabsorption in the distalconvoluted tubule.

B. promotes the secretion of potassium into thelumen of the distal convoluted tubule.

C. is a steroid hormone secreted by the adrenalmedulla.

D. is the major mineralocorticoid in humans.

47. Excessive aldosterone secretion will result in all of

the following EXCEPT:

A. hyperpolarization of nerve and musclemembranes.

B. increased muscle contraction.

C. an excessive loss of potassium ions from theextracellular fluid.

D. hypokalemia.

48. Which of the following substances is filtered by theglomerulus and into Bowman's capsule?

A. Platelets

B. Proteins

C. ElectrolytesD. Erythrocytes


Biology Kidney & pH Passage VHI

Passage VHI (Questions 49-53)

Most metabolic reactions are highly sensitive to the pHof the fluid in which they occur. Due to this sensitivity,the hydrogen ion concentration of body fluids is closelyregulated.

Buffering systems, both intracellular and extracellular,act to minimize changes in the hydrogen ionconcentration. However, these buffering systems do nothave the ability to eliminate or retain hydrogen ions fromthe body. This role is left for the kidneys, whichaccomplish this regulation in two ways. The first is byaltering the secretion of hydrogen ions and the second isby altering the reabsorption of bicarbonate. Hydrogen ionexcretion and bicarbonate reabsorption are both achievedby the secretion of hydrogen ions.

Tubular

LumenEpithelial Cells

Urine CA = Carbonic Anhydrase

Figure 1

Interstitial

Fluid

According to Figure 1, the hydrogen ion to be secretedinto the lumen is generated in the tubular cell rather thancoming from the blood. Once in the lumen, the hydrogencombines with a buffer and is secreted in that form. In

Figure 1, the bicarbonate is transported to the interstitialfluid.

The hydrogen ion in the lumen can also bind withfiltered bicarbonate to form CO2 and H2O according toFigure 2. This is in essence bicarbonate reabsorption.The net result is that filtered bicarbonate disappears whilebicarbonate (the result of intracellular events) appears inthe blood. This situation is equivalent to one wherefiltered bicarbonate is reabsorbed back into the blood.


Tubular

LumenEpithelial Cells

CA = Carbonic Anhydrase

Figure 2

Interstitial

Fluid

In the body, the events shown in Figure 1 and Figure 2occur together, producing a situation where a high rate ofhydrogen ion secretion achieves complete reabsorption offiltered bicarbonate.

49.

50.

Which of the following situations will NOT producea gain of hydrogen ions?

A. An increase in the concentration of CO2.B. Production of lactic acid during severe

exercise.

C. Excessive vomiting.D. Loss of bicarbonate due to diarrhea.

The events shown in Figure 1 and Figure 2 MOSTlikely illustrate renal compensation for:

A. a decreased blood hydrogen ion concentration.B. an increased blood hydrogen ion concentration.C. a decreased sodium ion concentration.

D. an increased sodium ion concentration.


Biology Kidney & pH

51. The transport of a hydrogen ion from a tubular cellto the tubular lumen shown in Figure 1 represents:

A. simple diffusion.B. facilitated diffusion.

C. primary active transport.D. secondary active transport.

52. Respiratory alkalosis is best characterized by :

A. an increase in the hydrogen ion concentration,resulting from an increase in the level of CO2.

B. a decrease in the hydrogen ion concentration,resulting from a decrease in the level of CO2.

C. an increase in the hydrogen ion concentration,resulting from a decrease in the level of CO2.

D. a decrease in the hydrogen ion concentration,resulting from an increase in the level of CO2.

53. During metabolic acidosis, the partial pressure ofCO2 can be expected to:

A. increase, due to increased ventilation.B. increase, due to decreased ventilation.C. decrease, due to increased ventilation.D. decrease, due to decreased ventilation.


Passage VIII


Biology Dialysis and Ultrafiltration Passage IX


Dialysis is a process used to separate molecules insolution based on their size. Semipermeable membranesthat have pores smaller than the macromoleculeof interestare used to contain one or more large molecules. Dialysistubing is filled with the solution of interest, and the endsof tubing are tied. The package is immersed in a largequantity of solvent. Diffusion of molecules smaller thanthe pores of the tubing allows exchange of solvents, salts,and small metabolites between the solution in the tubingand the surrounding solution. This can be used to removeexcess salts, for example, from a protein to be purified.

Ultrafiltration is a related technique that is used toconcentrate macromolecules. Pressure is used to force the

solution through the semipermeable membrane. Thesolvent and small molecules pass through the membrane,while a more concentrated solution of macromolecules is

left behind. Normal kidneys perform ultrafiltrationconstantly in the removal of waste products from theblood. The entire blood supply of 5-6 liters is filteredapproximately every 45 minutes.

Diseased kidneys require exterior dialysis support toprevent dangerous increases in the plasma solute load,including, salts, electrolytes, and urea. Uremia is thecondition of having an increased urea plasmaconcentration.

Dialysis support is conventionally performed in one oftwo ways. The first involves passing the blood through amachine containing semipermeable membranes that allowwaste products to diffuse out. An in vivo technique usesthe patient's peritoneal membranes (in the abdominalcavity) to form an intraabdominal dialysis system. Asterile fluid is instilled through an abdominal catheter,allowed to remain from 15 minutes to 1 hour, and drainedthrough a second abdominal catheter. This process isrepeated many times. Both methods are time-consuming,but they allow the necessary removal of waste productsfrom the blood.

54. What complications could result from the peritonealdialysis procedure?

I. Increased risk of abdominal cavity infections.II. Changes in electrolyte balance.III. Kidney damage by ultrafiltration.

A.

B.

C.

D.

I onlyI and II onlyII and HI onlyI, H, and HI


55. Which structure of the kidney performsultrafiltration?

A. Peritubular capillariesB. Loop of HenleC. Afferent arterioles

D. Glomerular capillaries

56.

57.

A student is purifying an enzyme in the laboratoryby overnight dialysis. The student tests forsuccessful dialysis using an enzyme activity assay.What would the student find if the dialysis tubingwas punctured with a pin before the experiment?

A.

B.

C.

D.

No enzyme activity in the bag.Increased enzyme activity in the bag comparedto the activity measured before the experiment.There is no effect caused by the pinhole, andthe enzyme activity in the bag is the same asbefore.

Slightly decreased enzyme activity in the bagcompared to the activity measured before theexperiment.

Which of the following hormones would most likelybe supplementedduring chronic kidney disease?

A. ErythropoetinB. Antidiuretic hormone (ADH)C. Epinephrine and norepinephrineD. Insulin

58. What would be a physiological effect of uremia?

A. Increased blood pHB. Edema

C. Increased urination

D. Alkalosis


Biology Dialysis and Ultrafiltration Passage IX

59. When a dialysis machine is used to clean blood,what precautions are taken to avoid loss of glucoseand salt?

A. Use of a dialysis membrane that retainsglucose and salts.

B. Use a dialysis solution that is isotonic to theblood in glucose and salts.

C. The patients drinks an electrolyte and glucosedrink during dialysis.

D. Nothing can be done to avoid loss, so thepatient has an IV of saline and glucose at thesame time.

60. If a person has 5 liters of blood and it is passedcompletely through the kidneys every 45 minutes,what is the glomerular filtration rate?

A.

B.

C.

D.

225 ml/min

111 ml/min

66 ml/min

9 ml/min


Biology Caffeine Passage X


Caffeine is a commonly used beverage in the U.S.The average individual consumes 2-3 cups per day, withsome people drinking 10-15cups. Coffee containsabout150mg caffeine/cup for drip-style makers. Caffeine is thephysiologically active agent in coffee. The followingexperiments were carried out to examine the effects ofcaffeine in coffee on free fatty acid (FFA) metabolism:

All subjects followed on overnight fast. At 8:00 AM,each subject drank one of the following beverages:

Experiment1:500 mL hot water containing 5 g instant coffee (250mg caffeine) and 3 tablets of sodium saccharin (15mg).

Experiment2:

500 mL hot water containing 5 g decaffeinated coffee(20 mg caffeine) and 3 tablets of sodium saccharin(15 mg).

Experiment 3:500 mL hot water containing 3 tablets of sodiumsaccharin (15 mg)

Experiment 4:

500 mL hot water containing 5 g instant coffee (250mg caffeine) and 25 g sucrose

Blood samples for all experiments were taken beforethe start of the experiment and at 1, 2, and 3 hoursfollowing ingestion of the test beverage. The followingtable follows the FFA concentrations throughout thestudy:

Expt 0 hours 1 hour 2 hours 3 hours

1 0 +182* +245* +384$

2 0 -16* +18* +93*

3 0 -19 -59 +17

4 0 -96t -90t +120t

* Not significant compared to Experiment 3

XSignificant compared to Experiment 2 and Experiment 3

t Significant compared to Experiment 1

Table 1: Mean Changes in Plasma FFAs (^EQ/L)

61. Which of the following mechanisms is the mostimportant in raising plasma FFA concentrations?

A. LipolysisB. LipogenesisC. Increased fat oxidation

D. Increased dietary fat


62. Which of the following hormones would mostincrease plasma FFA concentration?

A. CCK

B. Thyroid hormoneC. EpinephrineD. Insulin

63. In Experiment 4, how did the sucrose lead to lowerFFA concentrations?

A. Sucrose was digested to glucose and fructose;fructose mediated the effects on FFAs.

B. Sucrose stimulated glucagon, which decreasedrelease of fatty acids from adipocytes.

C. Sucrose acted as a competitive inhibitor ofcaffeine.

D. Sucrose stimulated insulin, which decreasedrelease of fatty acids from adipocytes.

64. The action of caffeine is mediated through increasedcAMP levels in target cells. What is the enzyme thatis activated by second messengers to make cAMP?

A. EpinephrineB. G proteinC. Adenylate cyclaseD. Protein phosphorylase

65. From Experiment 2, what would you conclude aboutthe effects of the caffeine present in decaffeinatedcoffee on plasma FFA concentration?

A. There are slight increases in FFA concentration due to the 20 mg of caffeine present

B. The effects are the same as drinking coffeewith 25 g sucrose.

C. There are slight decreases in FFA concentration due to the 20 mg of caffeine present.

D. The effects are the same as drinking water.

66. What was the point of Experiment 3?

I. To determine the effects of hot water

n. To determine the effects of sodium saccharin

III. To establish control data

A.

B.

C.

D.

I onlyIlonlyI and II onlyI, II, and m


Biology Kidney St Calcium Passage XI


Kidney function can be characterized by threeprocesses. The first involves movement of an essentiallyprotein free plasma from renal glomerular capillaries intoBowman's capsule. This is referred to as glomerularfiltration. Tubular reabsorption refers to the movement ofsubstances from the renal tubular lumen to the peritubularcapillary plasma. Movement in the opposite direction istermed tubular secretion.

The kidney, among its many other functions, isinvolved with regulating levels of calcium ion. Theextracellular calcium concentration is normally heldrelatively constant. However, a low calcium ionconcentration increases the excitability of nerve andmuscle cell membranes.

Calcium homeostasis is achieved through thefunctioning of three major effector sites. These sites arethe kidney, bone, and gastrointestinal (GI) tract. Almost99% of total-body calcium is contained in bone, which isa framework of organic molecules upon which calciumphosphate crystals are deposited. Bone is a dynamic tissuewhich can either remove or deposit calcium into theextracellular fluid. In the GI tract, control of activetransport systems that move calcium into the blood canresult in large fluctuations in the amount of calciumabsorbed. The kidney modulates calcium levels byreabsorption of both calcium and phosphate ions.

All three effector sites are under control of parathyroidhormone, whose plasma concentration fluctuates directlyin response to changes in plasma calcium levels.

67. All of the following statements about glomerularfiltration are true EXCEPT:

A. fluid pressure in Bowman's capsule opposesfiltration.

B. osmotic force due to protein in plasma favorsfiltration.

C. glomerular capillary blood pressure favorsfiltration.

D. forces favoring filtration are larger than thoseopposing filtration.


68.

69.

187

According to the diagrams in Figure 1, all of thefollowing statements concerning substances X, Y,and Z are false EXCEPT:

Renal

Artery

Urine

Afferent

Arteriole

Urine

Figure 1

Efferent

Arteriole

Peritubular

Capillary

A. Z is filtered and partially reabsorbed.B. Y is totally reabsorbed.C. X is filtered and secreted, but not reabsorbed.D. X, Y, and Z are all secreted.

In response to a decrease in plasma calcium levels,one would most likely expect to see:

A. increased plasma parathyroid levels, withincreased urinary calcium excretion.

B. increased plasma parathyroid levels, withincreased removal of bone.

C. decreased plasma parathyroid levels, withdecreased urinary calcium excretion.

D. decreased plasma parathyroid levels, withdecreased removal of bone.


Biology Kidney St Calcium

70. According to information in the passage, it can beconcluded that:

A. the concentration of extracellular calcium in nowayeffectsneuromuscular excitability.

B. persons with low extracellular calciumconcentrations may suffer total skeletal musclespasms.

C. persons with increased extracellular calciumconcentrations may suffer from frequentmuscle spasms.

D. the effect of extracellular calciumconcentration on membranes is distinct fromits role as a mediator of muscle contraction.

71. A decrease in calcium will most likely cause:

A. an increase in the extracellular concentrationof phosphate.

B. no change in the extracellularconcentration ofphosphate.

C. a decrease in the urinary excretion ofphosphate.

D. a reduction in the tubular reabsorption ofphosphate.

72. Parathyroid hormone is secreted by the:

A. anterior pituitary.B. posterior pituitary.C. parathyroid gland.D. thyroid gland.

Copyright© by The BerkeleyReview 188

Passage XI


Biology Lymph Passage XII


The lymphatic system has 3 functions: (1) itslymphocytes provide immunological defenses, (2) ittransports absorbed fat from the small intestine to theblood, and (3) it transports interstitial fluid back to theblood.

Both lymph nodes and lymphocytes of the immunesystem play important roles in immunological defense.Lymph nodes contain phagocytic cells through whichlymph is filtered before returning to the blood supply.Lymph nodes also contains germinal centers that producelymphocytes, which are present in the lymph and in theplasma.

The lymphatic system also carries dietary fat from thesmall intestine to the blood supply. Absorbed fat,including long chain fatty acids and cholesterol, alongwith fat-soluble vitamins, are packaged into chylomicronsby the cells of the small intestinal mucosa. Thesechylomicrons transport fat from the intestine to thegeneral blood circulation. Other nutrients from digestiveprocesses in the small intestine travel first to the liver viathe hepatic portal vein.

The liquid part of lymph is an ultrafiltrate of plasma,called interstitial fluid. This is the fluid that is squeezedout of capillaries under pressure. Interstitial fluid entersthe lymph capillaries through its endothelial cells whichhave porous junctions. This fluid, along withchylomicrons, microorganisms, and lymphocytes, is thencalled lymph. From the lymph capillaries, the lymphmoves into the lymphatic vessels, which have a 3-layerstructure similar to veins. The lymphatic systemultimately returns the fluid to the cardiovascular systemvia the right and left subclavian veins.

73. What is the consequence of the chylomicron systemof fatty acid packaging?

A. Muscle and adipose tissue have first access todietary fatty acids from chylomicrons.

B. Liver tissue has first access to dietary fattyacids from chylomicrons.

C. Microorganisms are packaged intochylomicrons.

D. Interstitial fluid carries chylomicrons.

74. What is the major component of lymph?

A. Fat-soluble vitaminsB. ChylomicronsC. Water

D. Triglycerides


75. Fat-soluble vitamins are carried from the cells of thesmall intestine into the bloodsystem via the lymph.They are packaged into chylomicrons and with fattyacids. Which of the following vitamins are fat-soluble vitamins?

I. Vitamin A

n. Vitamin K

m. Vitamin E

A.

B.

C.

D.

I onlyII onlyII and III onlyI, n, and IE

76. What is the function of the lymph node?

A. To destroy microorganisms.B. To filter chylomicronsout of the lymph.C. To produce chylomicrons.D. To add intestinal nutrients to the lymph.

77. What force moves the lymph inside the lymphaticvessels?

I. Squeezing by the muscles.n. Contraction of the muscular layer of the lymph

vessels.

III. Blood pressure.

A. I onlyB. I and II onlyC. II and m onlyD. I, II, and HI

78. The term ultrafiltrate means fluid formed by thehydrostatic pressure of the blood acting against asemi-permeable membrane. Interstitial fluid is oneexample of an ultrafiltrate. Which of the followingorgans also produces another ultrafiltrate?

A. Liver

B. KidneyC. Pancreas

D. Salivary gland

79. Which of the following statements are TRUE ofboth lymph and plasma?

I. Both contain chylomicrons after a meal.II. Both contain lymphocytes.m. Both contain erythrocytes.

A. I onlyB. I and II onlyC. II and HI onlyD. I, n, and HI


Biology Renal Clearance Passage xm

Passage XHI (Questions 80-86)

Theconcept of renal clearance, based on theprincipleof mass balance, is a means to quantify the excretoryfunction of the kidney. The relationship of clearance (C)to the concentration of a substance in the urine ([ ]u), theurine flow rate (Fu), and the plasma concentration of asubstance ([ ]p)is

C = ([]U)X(FU)/ []p

The glomerular filtration rate (GFR) is the volume perunit time that enters the kidney from the plasma. Undercertain conditions, the amount of a substance filtered isequal to the amount excreted in the urine. To calculatethe amount filtered, one can multiply the GFR by the[substance]p.

Similar to the plasma, not all of a substance travelingthrough the renal artery is filtered at the glomerulus. Thefraction of plasma filtered can be calculated as the ratio ofGFR to renal plasma flow (RPF). The fraction normallytakes on values of 15% to 20%.

The renal handling of metabolites may be instrumentalin diagnosing certain clinical disorders such as diabetes.When determining the amount of glucose reabsorbed inthe nephron, one must subtract the amount excreted fromthe amount filtered at the glomerulus. The amount ofglucose reabsorbed by epithelial cells of the tubule can bethought of as having a maximum transport rate, known asthe tubular transport maximum. The [plasma glucose]will ultimately determine whether glucose is excreted or is100% reabsorbed. The level of plasma glucose whichfirst gives rise to glucose in the urine is termed the plasmathreshold.

80. Mass balance describes the principle that the mass ofa substance entering the kidneys must equal themass of that substance leaving the kidneys. Usingthe following key, and information from the passage,which relationship BEST describes this principle?

Key[ ]A = concentration of substance inrenal artery

[ ]v = concentration of substance in renal vein

PFA = plasma flow rate in renal artery

PFV = plasma flow rate in renal vein

A. []A x PFV = (Hu x Fu) + ([]v x PFA)B. ([]A x PFV) + ([]u x Fu) = []v x PFAC. ([]A x PFA) + Q]ux Fu) = []v xPFvD. []A x PFA = ([]u x Fu) + (Uv x PFV)


81.

82.

83.

84.

190

What is the volume of urine that could be collectedin 30 seconds if the [ ]u of substance Y is 50 mg/ml,the [ ]p ofsubstance Yis 1mg/ml, and the clearancefor substance Y is 50 ml/min?

A. 0.00025 L

B. 0.0005 L

C. 0.001 L

D. 0.0015 L

Which of the following equations represents theglomerular filtration rate (GFR)?

A. GFR = [ substance ]p x []u x FuB. (GFR) x (Fu) = ([ substance ]p) x ([]u)C. GFR = ([ ]u x Fu)/ ([ substance ]p)D. (GFR) x ([]u) x ([]p) x (Fu)=l

According to the passage, the amount of a substancefiltered is equal to the amount excreted in the urineonly under certain conditions. Which of thefollowing conditions CANNOT exist for thisstatement to be TRUE?

A. The substance has no barrier to filtration at the

glomerulus.B. The substance must not be reabsorbed and

must be secreted by the nephron.C. The substance must not be metabolized or

produced by the kidney.D. The rate of glomerular filtration cannot be

altered by the substance.

Glucose is reabsorbed through the epithelial cells ofthe:

A. proximal convoluted tubule.B. loop of Henle.C. distal convoluted tubule.D. collecting duct.


Biology Renal Clearance

85.

86.

According to the graph below, the plasma thresholdis:

3

-J c

1000 •

LevelsOfExcretion,Filtered AndReabsorption(mg/m800

600

400

200

0

1

f Reabsorption

3 2 4 6 8 10

Plasma Glucose (mg/ml)

A. 3.5 mg /ml

B. 7.6 mg /ml

C. 380 mg / mlD. 400 mg / ml

The following graph illustrates the influence of sizeand charge on a substance's filterability at theglomerulus. Which of the following statements ismost likely true based on this graph?

A.

B.

C.

D.

22 26 30 34 38 42

Effective Molecular Radius (A)

The cationic species has the highestfilterability, because cationic atoms are smallerthan anionic atoms.

The cationic species has the highestfilterability, because of the presence of anionicglycoproteins on the surface of all glomerularfiltration components.The anionic species has the highestfilterability, because cationic atoms are largerthan anionic atoms.

The anionic species has the highestfilterability, because of the presence ofcationic glycoproteins on the surface of allglomerular filtration components.


Passage XIII


Biology Gastrointestinal Junction Passage XIV


The pyloric sphincter separates the terminal portion ofthe stomach (gastric antrum) from the beginning segmentof the small intestine (duodenum). This sphincter, whichconsists of a ring of connective tissue preceded by twodistinct rings of circular smooth muscle, carries out twoimportant functions. First, it helps to regulate the rate ofgastric emptying into the duodenum. The transfer of thestomach's contents to the duodenum is dependent on theduodenum's ability to process chyme. Second, it preventsthe reflux of small intestinal contents into the antrum.These two functions are controlled by both hormones andneural input. Norepinephrine and cholecystokinin (CCK),for example, both cause a constriction of this sphincter.

It has been discovered that several distinct mechanismscan lead to an alteration in the rate of gastric emptying. Alowering of the pH in the duodenum has been shown toslow gastric emptying. This decrease in pH has beenshown to not only affect motility, but also increase therelease of bicarbonate ion into the small intestine via thehormone secretin. In addition to pH, the presence of fatresults in a retarded gastric emptying. This response isbelieved to be carried out mainly by the hormone CCK.The hormone gastric inhibitory peptide (GIP) is alsobelieved to be a component of this mechanism, as GIP hasbeen shown to decrease the rate of gastric emptying.Finally, the secretion of gastrin lowers the rate of gastricemptying. Gastrin is secreted in response to the presenceof amino acids and peptides in the stomach.

87. Based on information in the passage, which of thefollowing statements is TRUE?

A. A low rate of gastric emptying may lead toduodenal ulcers.

B. A high rate of gastric emptying may lead tostomach ulcers.

C. Regurgitation of duodenal contents may leadto duodenal ulcers.

D. Regurgitation of duodenal contents may leadto stomach ulcers.

88. Norepinephrine is released in response tosympathetic stimulation and acts to increase pyloricsphinctercontraction. Norepinephrine is an exampleof a:

A. mineralocorticoid.B. catecholamine.

C. steroid.

D. peptide.


89.

90.

192

It is known that a hypertonic solution in theduodenum decreases the rate of gastric emptying.This mechanism prevents a:

A. rise in the blood pH.B. decrease in the blood pH.C. rise in the blood volume.D. decrease in the blood volume.

100 mM HCl was injected into the duodenum of adog while the contractile activities of both theantrum (AC) and duodenum (DC) were measured.Which of the following graphs BESTS representsthe expected results of the experiment?

A.

B.

D.

iiiiiiiiiiiiiii inn minimi

iniiiiiiiiiiiiiiiiiiiiiiimit t

HCL Infusion HCL Infusion

Starts Ends

lllllllllllllllllllll

lllllllllllllllllllllt t


Starts Ends

IIIIIIIIIIIIIII!

iiiiiiiiiiiiiii mil minimit t


Starts Ends

imiiimiiiiimiii

iiiiiiiiiiiiiiiiiiiiiiiiiiiiiit

HCL Infusion

Starts

tHCL Infusion

Ends

AC

DC

AC

DC

AC

DC

AC

DC


Biology Gastrointestinal Junction Passage XIV

91. As stated in the passage, GIP is involved indecreasing the rate of gastric emptying in responseto fats in the small intestine. This hormone is mostlikely to increase the secretion of which of thefollowing hormones?

A. GlucagonB. Bile

C. Insulin

D. Cortisol

92. Cholecystokinin is known to stimulate contraction ofthe antrum and promote constriction of the pyloricsphincter. The overall effect of CCK is to:

A. increase the rate of gastric emptying, andcontribute to the mixing of stomach contents.

B. increase the rate of gastric emptying andinhibit the mixing of stomach contents.

C. decrease the rate of gastric emptying andcontribute to the mixing of stomach contents.

D. decrease the rate of gastric emptying andinhibit the mixing of stomach contents.

93. Which of the following statements is FALSEregarding parietal cell secretion of protons andpancreatic release of bicarbonate ion?

A. The pH of the stomach decreases.B. The pH of the blood increases.C. The acid entering the small intestine is

neutralized.

D. The bicarbonate ion is derived from carbon

dioxide.


Biology Cholera Toxin Passage XV


Cholera is a potentially lethal disease caused by thebacterium Vibrio cholerae. Individuals afflicted withcholera suffer from severe dehydration due to diarrhea,which left untreated can be fatal.

V. cholerae bacteria replicate in the small intestine ofthe host after infection (usually from contaminateddrinking water), producing a protein known as choleratoxin (CT). CT then binds to the surface of intestinalepithelial cells, and through a complex series of eventsirreversibly alters membrane-associated G-proteins suchthat they are constitutively active, allowing for theconstant overproduction of the second messenger cyclic-AMP. This triggers a sodium and chloride efflux that isresponsible for the water loss associated with diarrhea.Rehydration therapy for cholera sufferers takes advantageof an inwardly-pumping sodium/glucose symport presentin the apical membranes of intestinal cells.

CT is frequently used in research to help elucidate thefunctions of G-protein-related signal transductionalmechanisms. Recently, lines of mice have beenengineered which have a transgene encoding the CTprotein linked to a promoter which allows proteinexpression only in cells of the pituitary gland whichproduce growth hormone. These mice exhibit gigantism,an effect of excessive growth hormone.

94. Cholera toxin causes intestinal cells to secrete largeamounts of Na®/Cle into the intestinal lumen. Howdoes this cause dehydration?

A. Salt loss triggers the opening of water-permeable pores in the plasma membrane.

B. Salt loss creates an osmotic gradient, whichwater follows out of the cell.

C. Salt loss triggers the active transport of waterout of the cell.

D. The distal convoluted tubules of the kidneyreabsorb more water.

95. What should be given to help rehydrate a cholerapatient?

A. Distilled water.

B. Water and Na® ions.C. Water and glucose.D. Water, glucose, andNa® ions.


96. Cholera toxin affects a second-messenger signalcascade. The toxin would NOT affect the actions of:

A. epinephrine.B. gastrin.C. norepinephrine.D. testosterone.

97. Why is rehydration therapy alone not sufficient tocure severe cholera?

A. The body cannot establish normal bloodpressure after severe dehydration.

B. Pores in the plasma membrane will allowwater to eventually leak back into the intestinallumen.

C. Chloride ion must be ingested to replace thatwhich is secreted.

D. The underlying bacterial infection must betreated so that dehydrating diarrhea ceases.

98.

99.

Humans are one of the few mammals affected bycholera. Unaffected mammals:

A. lack G-protein mediated signaling.B. always produce large amounts of cyclic-AMP.C. lack membrane receptors for cholera toxin.D. normally have low cyclic-AMP levels.

Assuming they could be applied specifically to thepituitary, which of the following would be aneffective inhibitor of the gigantism seen in thetransgenic mice described above?

I. An enzyme which degrades cyclic-AMPII. Excessive normal G-proteinsHI. Excessive cyclic-AMP

A.

B.

C.

D.

I onlyII onlyIII onlyII and III only

100. A recent hypothesis suggests that individuals whoare heterozygous for the recessive gene that causescystic fibrosis (CF) may exhibit some resistance tocholera (i.e., the CF gene confers some selectiveadvantage). Which of the following statementsprovides evidence in support of this theory?

A. The frequency of the CF gene is low inpopulations chronically exposed to cholera.

B. The frequency of the CF gene is high inpopulations chronically exposed to cholera.

C. The frequency of the CF gene is high inpopulations rarely exposed to cholera.

D. Individuals with cholera are oftenheterozygous for the CF gene.


Biology Gastrointestinal St Renal Section III Answers

Passage 1(1 - 5) Gastrointestinal Smooth Muscle

1. A is correct, single-unit fibers are coupled, with many individual cells containing gap junctions. This questiondraws on our knowledge ofsmooth muscle types. Even ifwe did not know the answer right from the question, theanswer can be arrived at through some common sense. If something is acting as a single unit, then it is likely thatthe muscle fibers which make up the unit will be coupled to each other. Therefore we can eliminate choice B.Furthermore, in order for the muscle fibers to be electrically coupled to each other, many individual cells shouldhave gap junctions. Remember, gap junctions offer connections between two different cellular cytoplasms, whichwill allow current to move from cell to cell in a rapid fashion. The correct choice is A.

2. C is correct, calcium channels is slow. As in cardiac muscle, an inward calcium current is an important componentof the action potential in smooth muscle. We are told that the conduction velocity is low along smooth muscle fiber.It is logical to assume that the conduction velocity of action potentials along GI smooth muscle fibers is slowbecause activation of the calcium channels is slow. This question wasjust designed to makeone thinkaboutthe roleof calcium in the conduction velocity of an action potential of smooth muscle. The correct choice is C.

3. C iscorrect, number of action potentials within a set. We are looking at the diagrams togetan understanding of therelationship between force and action potentials. In the graphs we see three action potentials giving rise to a largerforce of contraction when compared to two action potentials within the set. Therefore, we can say that the size ofthe force depends on the number of action potentials within a burst. There is simply noevidence or support from thegraph to consider the other answers as viable possibilities. We cannotcompare amplitudes as the action potentialsare all of the same size. Thus, the force of contraction depends on the number of action potentials. The correctchoice is C.

4. C is correct, mixing of intestinal contents, breaking larger particles into smaller ones. As stated in the passage,segmentation allows alternating contraction and relaxation within the circular muscle. This type of movement issimilar to opening and closing your hand. If you had a tomato in your hand, it would soon turn to mush. The samehappens to food within the intestines. It is continually beingbrokendown into smaller and smaller pieces. There isno indication that absorption of nutrients occurs because of segmentation, nor is there any direct evidence thatsegmentation causes movement of the intestinal contents through the GI tract. Even though peristalsis andsegmentation may stimulate (indirectly) the secretion of digestive enzymes, this is not the best answer choice. Thecorrect choice is C.

5. D is correct, breaking larger particles into smaller particles increases the surface area available to digestiveenzymes. Wc are told from the question that mixing aids in digestion. We know that mixing will break largerparticles into smaller ones. This aids in digestion because the surface area exposed to digestive enzymes will haveincreased. These enzymes will break down complex molecules into those molecules that can be absorbed across thelayer and eventually be used by the body. While the other answers may seem attractive, one has to ask themselves,what would this do to aid in digestion. Moving through the gut faster does not seem to aid in digestion, therebyeliminating choice A. Furthermore, there is simply no evidence from the passage that smaller molecules will carryless charge or cross over the epithelial layer of the lumen any easier. Absorption will most likely be occurringthrough channels in the cell layer. Therefore, we can say that increasing the surface area available to digestiveenzymes will aid in digestion. The correct choice is D.

Passage II (6 - 12) Protein Turnover

6. A is correct, I only. Enzymes are proteins, and their amino acids, can themselves be digested into amino acids andtaken back up to feed the body. Choice I is correct. Mucus, a protective coating on the GI tract, is made ofpolysaccharides. Mucus is not digestible. How could it protect tissue if it were digested by our enzymes? Choice IIis incorrect. Bicarbonate is not protein. Choice III is incorrect. The correct choice is A.

7. D is correct, 94%. If we merely looked at food intake and fecal output, wc would have calculated 90/100 = 90%.This did not consider secretion of proteins from the gut that were recycled. Choice C is incorrect. If we did notinclude food intake, we would have calculated 63%. Choice A is incorrect. Choice B is also incorrect. The totalprotein available was 100 gm from the diet and 70 gm from gut secretions to make a total of 170 gm of proteinavailable. 10 gm was excreted in feces, so 160 gm was digested and absorbed. 160/170 = 94%. Notice that the skinis not included in this calculation The correct choice is D.

8. C is correct, urea. Choice A is the chemical structure for uric acid. Choice B is ammonia. Choice C is urea.Choice D is hydrazine. The correct choice is C.



9. D is correct, white cells, liver, muscle, gut. FromFigure 1, lookat the amount of proteineach tissue turns over eachday: The correct choice is D.

10. B is correct,give isotopically labeled albumin, sample the blood periodically, and calculate the decay rate of labeledalbumin. The concentration of albumin will not give information about the turnover rate. Choice A is incorrect. Anisotopic label will allow quantification of the decay rate of the label is the blood as the albumin is recycled. Anonlabeledalbumin will not help. Choice C is incorrect. Since a turnover rate in the body involves many differentinteractions, studying this is a test-tube would not work. ChoiceD is incorrect. The correct choice is B.

11. D is correct, skeletal muscle. The first tissue to break down would be the most expendable one. Smooth muscle isthe muscle of the diaphragm and many other internal organs. This is not expendable. Choice A is incorrect.Cardiac muscle is heart tissue. It is not the first to be broken down either. Choice B is incorrect. Enzymes ofenergy metabolism are required even during a fast, especially the enzymes controlling gluconeogcnesis. They arenot used first, cither. Choice C is incorrect. Skeletal muscle serves as a reserve of amino acids for use during a fast.It is the first tissue to break down. D is correct.

12. C is correct, between 0.6 and 1.5 gm/kg. The nitrogen balance will be zero mg/kg at the appropriate protein intake.The chart moves from negative to positive between 0.6 and 1.5 gm/kg. This means zero is contained somewhere inthat intake range. The correct choice is C.

Passage III (13-20) Liver, Pancreas, St Intestines

13. A is correct, CCK. CCK from the small intestine is stimulated by free fatty acids entering the duodenum. CCKsecretion will stimulate bile salt and pancreatic juice secretion. Secretin will be stimulated by gastric acids and highprotein content in the small intestine. Hie correct choice is A.

14. D is correct, enterokinase. Enterokinase (enteropeptidase) is the enzyme located within the mucosal layer of thesmall intestine. It catalyzes the conversion of trypsinogen to trypsin. Trypsin will catalyze the remainingconversions. Secretin and CCK are both hormones that will stimulate release of these proenzymes from thepancreas, while bile salts, which are not enzymes, are produced in the liver. The correct choice is D.

15. C is correct, protein deficiency. The lack of enteropeptidase will result in decreases in the breakdown of proteinproducts entering the small intestine. Breakdown of these proteins is essential for absorption by the small intestine.Enteropeptidase does not affect fat or B12 metabolism. The correct choice is C.

16. A is correct, secretin. Secretin is stimulated by gastric acids entering the small intestine. It will stimulate secretionof bicarbonate from the pancreas to neutralize the acidity of the gastric juices. Enterokinase and proenzymes areenzymes that do not affect gastric acids. CCK will stimulate the secretion of bile salts needed for the emulsificationof fats in the small intestine. The correct choice is A.

17. A is correct, trypsin. Acute pancreatitis is produced by abnormal levels of trypsin, which is produced as aproenzyme (trypsinogen) in the pancreas. Trypsin is a powerful enzyme that is not normally active in the pancreas.Activation of this enzyme within the pancreas will cause the breakdown/digestion of the pancreas itself. One type ofdamage to the pancreas involves lysolecithin (as outlined in the question). Secretin is a hormone in the smallintestine. Bile salts are not found in the pancreas. Endopeptidase is a fictitious enzyme. It sounds likeenteropeptidase (also known as enterokinase), which is the enzyme responsible for converting trypsinogen totrypsin. The correct choice is A.

18. C is correct, dissociate into CCb and H2O in order to allow the NaCl solution in the small intestine to remainneutral. The whole point behind the function of the small intestine is to neutralize the acidic chyme that enters fromthe stomach. The bicarbonate ion (HCC>3e) that is released from the pancreas neutralizes the HCl from the stomachby forming the neutral salt, NaCl. If we want to form more of this neutral salt, then we must pull the reaction givenin the question to the right (via LeChatelier's principle). We can do that by removing carbonic acid (H2CO3). How?By allowing it to dissociate to CO2 and H2O. The CO2 is absorbed into the blood and carried back to the lungswhere it is blown off as a gas. The H2O becomes part of the intestinal fluid. One reason that we do not want toincrease the acidity of the intestinal lumen is because peptic ulcers can result. The correct choice is C.

19. C is correct, pancreatic ductal cells. Severe duodenal ulcers, leading to the removal of the duodenum, will result inthe removal of cell-types associated with the upper small intestine. Two important cell types are those that secreteCCK and secretin. CCK stimulates gallbladder contraction. The gallbladder stores and concentrates bile. If the



20.

cells that secrete CCK are removed, then contraction of the gallbladder will not increase. The gallbladder tends tostay in the relaxed state for longer periods of time. Secretion of HCl from the parietal cells in the stomach isinhibited by GIP, an enterogasterone released by cells in the duodenum. Removal of GIP-secreting cells willremove the inhibition on the parietal cells and lead to an increase in parietal cell secretion (of HCl). Both CCK andsecretin act to inhibit gastric emptying into the small intestine. Removal of the cells that secrete these two hormoneswill result in an increase in the passage of chyme into the small intestine. Secretin also acts to increase theproductionof bicarbonate form the pancreatic ductal cells. Removal of the cells which secrete secretin will lead to aDECREASE in the production of pancreatic bicarbonate and a DECREASE in pancreatic ductal cell activity. Thecorrect choice is C.

B is correct, delay the appearance of acid in the stomach. The name antacid tells us what it does. The antacidtablets act to neutralize the increase of gastrointestinal acid. This is exactly what the two equations in the questionare addressing. We do not see acid being formed in either equation. Instead, we see the neutralization of HCl. Wecan eliminate choices A and C. Even though it is true that water is a by-product of the reaction with the antacid andHCl, it is not addressing the question as to why antacids should be taken 1 to 3 hours after a meal. The delay intaking antacids after a meal is simply to delay the appearance of acid in the stomach. This is seen from the graph inthe question. If the antacid is taken 1hours after a meal, the appearanceof acid in the stomach is delayed for about2 hours. We can eliminate choice D. The correct choice is B.

Passage IV (21 -27) Digestion Safeguards

21. C is correct, trypsin would be inactive, and theotherpancreatic enzymes would remain as zymogens. This is to testour understanding of the passage. Trypsin is required to activate all other secreted pancreatic digestive enzymes.Raw soybeans contain a trypsin inhibitor, so trypsin would be inactivated. Choices A and D are incorrect. Since theother enzymes require cleaving by trypsin to achieve activity, they will remain inactive as zymogens. Choice B isincorrect. The correct choice is C.

22. C is correct, gastric epithelium. The cells that are most rapidly turning over of those listed arc in the gastricepithelium. Neurons and muscle cells turnover minimally, if at all, in adults. The correct choice is C.

23. D is correct, I, II, and III. Since the passage tells you mucus lines the digestive tract, wc could just pick choice Dand be done. However, mucus is present in the mouth to protect the tissue and to lubricate food. Choice I is correct.Mucus is present in the colon. Choice II is correct. Mucus is very important in protecting the small intestine,particularly the duodenum, from auto-digestion. Choice III is correct. The correct choice is D.

24. D is correct, 2.0. The stomach is acidic, choice C, a basic pH, is incorrect. A pH of 6.8 is almost neutral, so choiceA is incorrect. A pH of 0.2 is 10 times more acidic than a pH of 2.0, and is much too acidic. A pH of 2.0 is correct.The correct choice is D.

25. D is correct, I, II, and III. Enterokinase activates trypsin. If enterokinase entered the pancreas via the pancreaticduct, then trypsin would be activated, and it would activate the other zymogens. They would all go to work on thepancreatic tissue, causing damage so that pancreatic enzymes leak into the blood. Extensive damage could damagethe islet cells that secrete insulin, leading to impaired insulin secretion. Actually, a person would probably be verysick or have some sort of surgery before the damage reached this level. It is important that we think of both theexocrine and the endocrine parts of the pancreas. The correct choice is D.

26. A is correct, I only. The stomach's parietal cells arc the acid-secreting cells. HCl, hydrochloric acid, is the acid instomach contents. The correct choice is A.

27. C is correct, carbohydrate. The give-away is in the name polysaccharide. This term refers to sugar units, so it is acarbohydrate derivative. The correct choice is C.

Passage V (28 - 33) Digestion Of Fats

28. C is correct, hydrolyzation of fat droplets in the small intestine. The hydrolyzation of fat droplets is dependent onthe enzyme pancreatic lipase. This enzyme is activated by the pancreatic enzyme trypsin. Emulsification of fats isdependent on bile salts secreted by the gall bladder. Stimulation of bile secretion is induced by increase in fatcontent in the small intestine and CCK stimulation. The correct choice is C.

29. A is correct, emulsification of fats. The removal of household grease docs not involve micelle endocytosis,exocytosis of fatty acids, or packing of fat products in the endoplasmic reticulum. The correct choice is A.



30. C is correct, decrease in intestinal cell microvilli. Because of the soap-like appearance of the stools, bile saltsecretion is probably normal. Therefore, the problem most likely lies in the absorption of the fat droplets.Decreases in the surface area of the small intestine due to decreases in the microvilli will initiate this response.Increases in pancreatic lipase alone will notcause this problem because any increase in free fatty acids andmicelleswill usually be absorbed by the intestines. Chylomicron production is not involved with this process. The correctchoice is C.

31. C is correct, inhibition of cholesterol absorption in the small intestine. Cholesterol is dependent on bile salts forabsorption in the small intestine. Fatty acids, eventhough aided by bile salts,are still able to be absorbed. Withtheincrease in fatty acid and cholesterol content in the small intestine, CCK secretion will be elevated. The correctchoice is C.

32. D is correct, cholesterol. By the time chylomicrons reach the liver, lipoprotein lipase will have cleaved all the fattyacids from the molecule. Lipoprotein lipase is found in the arterial system. The bile salts remaining in thecirculatory system are not attached to the chylomicrons. The correct choice is D.

33. C is correct, decreased levels of lipoprotein lipase. If levels of lipoprotein lipase are decreased, chylomicrondeposition of fats in the liver will increase. Lipoprotein lipase, while in the peripheral circulation, will cleavetriglycerides from the chylomicrons for deposition in fat stores in the periphery. Without lipase the liver will cleaveand store these same fats. Decreased levels of CCK and pancreatic lipase will only allow fatty acids to be lost in thestools due to malabsorption. The correct choice is C.

Passage VI (34 - 40) Diabetic Diet Experiment

34. A is correct, I and III only. Table 1 indicates a drop in the plasma glucose levels from the baseline to the HF group.Choice I is correct. Insulin requirements were lower in the HF diet phase compared to baseline. Choice II isincorrect. Less glucose was excreted in the urine in the HF group compared to baseline. Choice III is correct. Thecorrect choice is A.

35. D is correct, this decrease is beneficial because glycosylated hemoglobin indicates persistent levels of high plasmaglucose. When blood glucose is elevated persistently, then plasma proteins become glycosylated (glucose residuesare attached to the proteins). A decrease in glycosylated hemoglobin means that, on the average, blood glucoselevels were lower during the past few months. (The life span for a RBC is about 120 days). This would be abeneficial change. Choices A and B are incorrect. Glycosylation occurs when plasma glucose is high. Choice C isincorrect. The correct choice is D.

36. C is correct, oleic acid. The notation for the fatty acids is (# of carbons: # of double bonds). In the HF diet, theamount of 18:1 is dramatically increased. In the passage, you are told that the HF diet is specifically higher inmonounsaturated fatty acids. Monounsaturated means having one double bond. Palmitic acid (16:0) nor stearic acid(18:0) have double bonds. Choices B and D are incorrect. Palmitoleic acid is 16:1, and oleic acid is 18:1. Choice Ais incorrect. The correct choice is C.

37. A is correct, olive oil. Even if you have no clue about nutrition, use your test-taking skills to eliminate wronganswers or see patterns. One of these things is not like the others. Olive oil is from plants and the others are fromanimal fats. Anyway, olive oil is a good source of monounsaturated fatty acids (oleic acid to be specific). The otherfats are more saturated (they are solid at room temperature). The correct choice is A.

38. C is correct, II and III only. Although VLDL was lower in Table 2, LDL was not. Choice I is incorrect.Triglycerides are lower on the HF diet. (Use the significance symbols to help interpret the data. Researchers mustapply statistical methods to see if a finding is real or not.) ChoiceII is correct. HDL did increase on the HF diet.Choice IQ is correct. The correct choice is C.

39. C is correct, the beta cells of the pancreas. The alpha cells of the pancreas secrete glucagon. Choice A is incorrect.The exocrine portion of the pancreas secretes digestive enzymes, while the endocrine portion secretes hormones.Choice B is incorrect, as is choice D. The correct choice is C.

40. D is correct, the liver. The word hepatic refers to the liver. The hepatic portal vein drains the blood supply of mostof the organs of digestion and passes it to the liver first for processing. Insulin released from the pancreas firstreaches the liver. Choices A, B, and C are incorrect. The correct choice is D.



Passage VII (41 -48) The Kidney

41. B is correct, an increase in glomerular pressure, followed first by an increase and then by a decrease in glomerularfiltration rate. If the efferent arteriole is constricted, blood cannot flow pass the point of constriction. This leads to anincrease in the glomerular pressure, and allows us to immediately pick choice B or D. Constriction of the efferentarteriole also means that the blood flow in the glomerulus will decrease, and more plasma will begin to filter out intoBowman's capsule. With this information we can pick choice B. The more the plasma filtered out into Bowman'scapsule, the more the solute concentration begins to increase in the glomerulus. Eventually this will decrease theglomerular filtration rate. The filtrate simply will not be able to flow out of the glomerulus and into the capsulebecause of the change in the concentration gradient. The correct choice is B.

42. C is correct, reduced plasma glucose concentrations. Recall that in the proximal convoluted tubule the transport ofsodium across the apical membrane (from the lumen of the proximal convoluted tubule to the cytoplasm of thelumenal epithelial cell) is mediated by proteins called symports (a cotransportcr). For example, these symports allowfor the passage of both sodium and glucose together into the cell. Another example is the symport that allows for thepassage of both sodium and various amino acids into the cell. If the concentration of glucose is decreased in thefiltrate, the amount of sodium that can be reabsorbed at the level of the proximal convoluted tubule is reduced. Thecorrect choice is C.

43. A is correct, afferent arterioles and decreases the volume of urine flow. Sympathetic innervation of the kidneysprimarily affects the afferent arterioles. This innervation constricts the afferent arterioles and reduces the glomerularfiltration rate. The urine output will be dramatically reduced. The correct choice is A.

44. A is correct, (see the diagram below). Notice that along the X-axis the plasma osmolarity is steadily increasing. Atsome particular point the solute concentration in the plasma will be so great that osmoreceptors, located in thesupraoptic nuclei of the anterior hypothalamus quickly respond to a change in the osmolarity of the extracellularfluid (especially to the sodium ion).

45.

X-Axis

Since the osmolarity of the plasma is becoming higher (hyper-osmotic), water will flow down its concentrationgradient (osmosis) and out of the osmoreceptors and cause them to shrink. As the receptors shrink their rate ofdischarge increases. If the rate of discharge increases, the posterior pituitary gland releases more ADH into theplasma. This is exactly what we see as the plasma ADH concentration along the Y-axis begins to increase. If there ismore ADH in the system, then there is more ADH to act at the baso-lateral membrane of the late distal tubules,collecting tubules, and collecting ducts. In other words, the cells in those locations are more permeable to water.Water is reabsorbed into the blood and the urine becomes more concentrated. Notice that as ADH is being released,the osmolarity does not change that much. The correct choice is A.

D is correct, (see the diagram below). Again, along the X-axis the plasma concentration of glucose begins toincrease. As soon as the concentration of glucose reaches a particular point where it can no longer be reabsorbedfrom the lumen of the proximal convoluted tubules, it is passed through the lumen of the kidney tubules in an everincreasing rate.

X-Axis



This is shown along the Y-axis where we observe a continual increase in the rate of excretion of glucose in the urineof the kidney. We do not see this type of curve in the other three answers. The correct choice is D.

46. C is correct, is a steroid hormone secreted by the adrenal medulla. Aldosterone is a cholesterol-derived steroidwhich is synthesized in the zona glomerulosa (the outermost region) of the adrenal cortex. Aldosterone is referred toas a mineralo-corticoid because of its affect on electrolytes such as Na® and K®. This hormone has a half-life ofabout 30 minutes. Aldosterone promotes Na® reabsorption to a great extent at the late distal tubular region, thecortical collecting tubules, and the collecting ducts. Aldosterone also controls the secretion of K® ions into thelumen of the distal convoluted tubule. The correct choice is C.

47. B is correct, increased muscle contraction. Excessive aldosterone secretion results in an excessive loss of K® fromthe extracellular fluid and into the urine. This lowers the plasma K® concentration, a condition referred to ashypokalemia. Loss of K® ions from a cell tends to decrease the K® concentration in that cell, making the restingmembrane potential for K® more negative that it would have been had aldosterone levels been normal. Recall that asK® rushes out of the nerve cell, thatcell tends to hyperpolarize. Hyperpolarization leads to a decrease in the musclecontraction, not an increase. The correct choice is B.

48. C is correct, electrolytes. If we were to centrifuge down a sample of blood, we would find "formed elements" suchas red blood cells (erythrocytes), white blood cells (leukocytes), platelets, and proteins (high molecular weight) atthe bottom of the testtube. In the plasma wc would find such substances as sugars, amino acids, and the electrolytes,to name but a few. Electrolytes are rather small molecules that can easily be filtered by the glomerulus. The correctchoice is C.

Passage VIII (49 - 53) Kidney & pli

49. C is correct, excessive vomiting. We are looking for a situation which will not result in the gain of hydrogen ions.One must realize that a potential source of net bodily gain or loss of hydrogen ions is gastrointestinal secretionsleaving the body. Recall that vomiting has its origins in the stomach, which contains acidic secretions. This acidicvomitus leaving the body will result in a net loss, not a net gain of hydrogen ions. The correct choice is C.

50. B is correct, a increased blood hydrogen ion concentration. It is clear from figures one and two that two events areoccurring. One event is the secretion of hydrogen ions from the tubular cells of the kidney into the renal tubule.These hydrogen ions, after reacting with a buffering system, will leave the body in the urine. The body is removinghydrogen ions. This tells us that the hydrogen ion concentration must have been elevated, and the kidney iscompensating for this elevation. Second, Figure 2 is indicating the reabsorption of bicarbonate. One must understandthat reabsorption of a bicarbonate ion is equivalent to the loss of a hydrogen ion. Therefore, this reabsorption is alsoacting as a compensatory mechanism to relieve an elevated hydrogen ion concentration. The correct choice is B.

51. D is correct, secondary active transport. This question draws on your previous knowledge of transport systems. Asone can see, a hydrogen ion is transported out of the tubular cell while a sodium ion is moved from the renal tubulelumen to the cell. Sodium moves into the cell because it is falling down its concentration gradient. Even though thisis not stated in the passage, one should know that the sodium ion concentration is always higher cxtracellularly whencompared to intracellularly. The reason sodium can continue to fall down its concentration gradient is because of asodium/potassium ATPase pump which uses the energy of ATP to pump sodium out of the cell. One can refer tosodium moving down its gradient as primary active transport. To transport the hydrogen ion across the membrane,one can harness the energy found in the sodium concentration gradient (really a form of potential energy). Since thehydrogen ion is cotransported along with the sodium ion, it is given the term secondary active transport. The correctchoice is D.

52. B is correct, a decrease in the hydrogen ion concentration, with a decrease in the level of CO2. It should be clearthat an alkalosis is a situation where the pH is above that of neutral, and there is a deficiency of hydrogen ions. Inthis case, we have an alkalosis that is brought on by respiratory troubles. Recall the equation that is shown in Figure1. In order to produce a situation where there is a deficiency in the hydrogen ion, we must first create a situationwhere there is a deficiency in C02- A loss of carbon dioxide will pull the equation towards CO2 to compensate forthe loss. The result of that shift is that there is a decrease in the amount of hydrogen ions. A reduction in the level ofhydrogen ions clearly results in alkalosis. The correct choice is B.

53. C is correct, decrease, due to increased ventilation. We have a situation here where there is simply too muchhydrogen ion. We want to restore the original level of hydrogen ion. Again, refer to the equation shown in Figure 1.



One way we can reduce the hydrogen ion concentration is to lower the level of carbon dioxide. This should soundvery familiar, like the previous question. How do we lower the partial pressure of carbon dioxide? We breathe airout with an increased frequency. In other words, we increase the ventilation rate. This will cause a decrease in thepartial pressure of carbon dioxide, and shift the equation in such a manner so we experience a decrease in the levelof hydrogen ion. This will help restore the normal level of hydrogen ion from its current elevated level. The correctchoice is C.

Passage IX (54 - 60) Dialysis St Ultrafiltration

54. B is correct, I and II only. Since the solution is introduced into the abdominal cavity through catheters, there isalways a possibility of infection. Choice I is correct. Depending on the person's state of hydration, dialysis can leadto electrolyte imbalance. Choice II is correct. Since there is no pressure involved in this technique, there is noultrafiltration occurring. Choice III is incorrect. The correct choice is B.

55. D is correct, glomerular capillaries. The glomerular capillaries are fenestrated, that is, they contain large pores.Although large proteins cannot pass, small molecules and water are free to pass. The pressure of the blood acts toforce the molecules through the semipermeable capillaries. Choice D is correct. The ultrafiltrate collects in theglomerular capsule. The afferent arterioles are not permeable to these molecules. Choice C is incorrect. Theperitubular capillaries are involved in resorption. Choice A is incorrect. The loop of Henle is involved inconcentrating the urine. Choice B is incorrect. The correct choice is D.

56. A is correct, no enzyme activity in the bag. Although enzymes are large compared to atoms, they are smallcompared to a pinhole. This is why you test your dialysis tubing for holes before you begin enzyme purification.Since there is no way the amount of enzyme can increase, choice B is incorrect. Since there is an effect caused bythe pinhole, choiceC is incorrect. Choice D is incorrect since theenzyme would all (or almost all) moveout duringan overnight dialysis. The correct choice is A.

57. A is correct, erythropoetin. The kidney produces erythropoetin in response to low blood volume. This hormonesignals the bone marrow to make more red blood cells. If a kidney were diseased, this hormone may not beproduced. Choice Aiscorrect. Choices B, C and Dare incorrect because the kidney does not make these hormones.ADH is released from the posterior pituitary where it is stored after its manufacture in the hypothalamus. Theadrenal gland produces epinephrine and norepinephrine. Insulin is produced by the beta cells of the pancreas. Thecorrect choice is A.

58. B is correct, edema. You are given 2 choices that give the same answer, choices A and D. Eliminate them. Ifuremia is present, the kidney is probably diseased, so urination is probably decreased. Choice C is incorrect. Edemameans an accumulation of fluid in the intracellular spaces, leading to tissue swelling. Uremia indicates kidneyfunction is declining. This would lead to an increased solute load in the blood and retention of fluid. The fluidincrease would mainly move to the extracellular spaces, since cells have only a little room for expansion. Thecorrect choice is B.

59. B is correct, use of a dialysis solution that is isotonic to the blood in glucose and salts. The pores of dialysismembrane are not small enough to retain the small molecules of glucose and salts. Also, if glucose and salts wereretained in the dialysis membrane, urea would be, as well. Choice A is incorrect. Choices C and D are incorrect,because even through glucose and salts will enter the blood, they would be lost to a hypotonic dialysis solution.Finally, to prevent movement ofglucose and salts, the dialysis solution should be isotonic in its concentration ofthese molecules. This will prevent loss. The correct choice is B.

60. B iscorrect, 111 ml/min. 5 liters = 5000 ml. 5000 ml/45 min = 111 ml/min. Thecorrectchoice is B.

Passage X (61 - 66) Caffeine

61. Aiscorrect, lipolysis. Lipolysis is the mechanism ofbreakdown ofstored triglyceride to release FFA and glycerolinto the blood. This would raise FFA concentrations. Lipogenesis is the synthesis of fat. Thiswould not raise FFAlevels. Choice B is incorrect. Increased fat oxidation would decrease plasma FFA levels. Choice C is incorrect.Increasing dietary fat would increase chylomicrons, but not FFAs. Choice Dis incorrect. The correct choice is A.

62. C is correct, epinephrine. CCK is a hormone secreted from the duodenum in response to dietary fat. It causes gallbladder contraction and release of pancreatic enzymes. Choice A is incorrect. Thyroid hormone does not increase



plasma FFA concentration. Choice B is incorrect. Insulin decreases lipolysis, and increases fat storage. Choice Disincorrect. Finally, epinephrinepromotes lipolysis to increase FFA concentration in the blood in response to stresses.Think of the fight or flight mechanism. The muscles will need fuel and the elevated FFAs will provide that. Thecorrect choice is C.

63. D is correct, sucrose stimulated insulin, which decreased release of FFAs from adipocytes. Insulin turns downlipolysis. If you havejust eaten, and insulin is secreted, your body does not need theextra fuel thatcirculating FFAsprovide. You haveswitched from withdrawal from stores to depositing into yourstores. Sucrose docs not stimulateglucagon, so choice B is incorrect. Fructose does not mediate the effect on FFAs, so choice A is incorrect. Sucroseis not an inhibitor of the action of caffeine, so choice C is incorrect. The correct choice is D.

64. C is correct, adenylate cyclase. The wrong answers in this question are all part of the cAMP activation pathway.Be careful! A hormone such as epinephrine stimulates a receptor on the cell. A second messenger, the G protein,communicates the message internally to the enzyme adenylate cyclase. Adenylate cyclase produces cAMP. cAMPactivates protein phosphorylases. The question is for the enzyme, so adenylate cyclase is your answer. The correctchoice is C.

65. D is correct, the effects are the same as drinking water. The answer to this question in the table. The *, t, and tindicate if differences between groups are significant. This is a very common way for data to be summarized.Experiment 2 has the symbol "*", which indicates that the results were not significantly different from Experiment3, water. Although the numbers look like a slight increase, the * indicates there is actually no difference. Thecorrect choice is D.

66. D is correct, I, II, and III only. This is an experiment to determine the control data for plasma FFA concentrationsover time. To simulate the real experiment as much as possible, hot water and sodium saccharin were both included.This means I, II, and III are all valid choices. The correct choice is D.

Passage XI (67 - 72) Kidney & Calcium

67. B is correct, osmotic force due to protein in plasma favors filtration. This is a false statement. Filtration is themovement of substances from the glomerular capillaries into Bowman's capsule. The protein that is unable to befiltered creates an osmotic force which pulls the movement of fluid away from Bowman's capsule and into theglomerular capillaries. Therefore, the protein in the plasma creates an osmotic force which opposes, not favors,filtration. We are looking for a false statement, and choiceB certainly fits the description. The correct choice is B.

68. C is correct, X is filtered andsecreted, butnotreabsorbed. This question is requiring one to interpret Figure 1. It isclear that substance X leaves the glomerular capillaries and enters the renal tubule at Bowman's capsule. Thisclassifies as filtration. Also, it is apparent that substance X leaves the peritubular capillaries and enters the renaltubule after Bowman's capsule. This classifies as secretion. The only event that does not occur is movement ofsubstance X back into the peritubularcapillaries. In other words, there is no reabsorption. Therefore, X is filteredand secreted, but not reabsorbed is a true statement. The correct choice is C.

69. B is correct, increased levels of parathyroid hormone, with increased removal of bone. This question is requiringone to think about the role of parathyroid hormone. Wc have a decrease in the level of calcium, and so we willtherefore want to act to increase the levels of calcium. This results in an increase in the levels of parathyroidhormone. Why? One of the functions of the parathyroid hormone is to remove bone. Removal of bone releasescalcium and phosphate from the organic molecules. The release of calcium acts to counteract the original decrease.Therefore, we will see an increase in the level of parathyroid hormone, and an increase in the removal of bone. Allof the other choices are either false, or lead to a further decrease in calcium levels. The correct choice is B.

70. D is correct, the effect of the extracellular calcium concentration on membrane is distinct from its role as a mediatorof muscle contraction. Wc are told from the passage that a low extracellular calcium concentration leads to anincrease in nerve and muscle cell excitability. In fact, a person with a low calcium concentration may suffer frommuscle spasms. We often think of calcium as being involved with the troponin-tropomyosin complex, wherecalcium is needed to initiate contraction. In that situation, a low calcium concentration would lead to decreasedexcitability or contraction. We are told that in this case, a low extracellular concentration leads to increased nerveand muscle excitability. Based on this apparent contradiction, we can conclude that the effect of calciumconcentration on the membrane is distinct from its role as a mediator of contraction. In fact, these effects reflectcalcium's ability to bind to plasma membrane proteins that function as ion channels. The binding alters the state ofthe these channels. The correct choice is D.



71. Dis correct, a reduction in the tubular reabsorption ofphosphate. This question is not easy. The following logiccan be used to arrive at the answer: We have a low calcium concentration. We will want to counteract that decreasewith an increase. We will see an increase in the removal of bone due to increased levels of parathyroid hormone.This not only releases calcium, but from the passage, one should beable toconclude that it releases phosphate ion aswell. Now, if the extracellular concentration of phosphate were to increase as a result of this release, furthermovement of calcium from bone would be hampered because a high extracellular concentration of phosphate wouldbind with calcium and cause the salt to be re-deposited on bone and other tissues. The point is that we want togetrid of the phosphate ion. One way to do this is to decrease the amount reabsorbed in the kidney. In that way,phosphate ion is released with the urine and calcium levels can return to normal. Again, we will most likely see areduction in the tubular reabsorption of phosphate ions. The correct choice is D.

72. C is correct, the parathyroid gland. This question is simplyasking us to recall the function of our endocrine glands.Parathyroid hormone, a protein hormone, is released by the parathyroid gland. While the parathyroid gland isembedded in the thyroid gland, the glands arc distinct. The correct choice is C.

Passage XII (73 - 79) Lymph

73. A is correct, muscle and adipose tissue have first access to dietary fatty acids from chylomicrons. The chylomicronpackaging system avoids direct delivery of fatty acids to the liver. This is stated in the text of the passage. Choice Bis incorrect. Although microorganisms may be present in the lymph, they are not packaged into the chylomicrons.Choice C is incorrect. Lymph, not interstitial fluid, carries chylomicrons. Choice D is incorrect. The correct choiceis A.

74. C is correct, water. Water is the main component of our body fluids. Since lymph is a derivative of interstitial fluid,it is mostly water. Fat soluble vitamins are usually present in very small amounts (jag), triglycerides may be presentin larger amounts (grams), but water is the most abundant. There must be more water than chylomicrons so that theyremain soluble. Although chylomicrons carry fat, neither they nor the fat they carry is the major component oflymph. Choices A , B, and D are incorrect. The correct choice is C.

75. D is correct, I, II, and III. All three vitamins are fat-soluble vitamins. The correct choice is D.

76. A is correct, to destroy microorganisms. Chylomicrons are produced and added by the intestinal cells. Choices Band C are incorrect. Intestinal nutrients are either packaged in chylomicrons and put into the lymph at the smallintestine, or nutrients from the small intestine directly enter the hepatic portal vein and travel to the liver. The lymphnode does not add intestinal nutrients. Choice D is incorrect. Lymph nodes filter microorganisms through thephagocytic cells, as mentioned in the passage. The correct choice is A.

77. A is correct, I only. In the passage, the structure of the lymph vessel is described as similar to the vein. The veindoes not have a muscular layer like the artery. Choice II is incorrect. Blood in the vein is moved by squeezing by theskeletal muscles. Choice I is correct. Blood pressure leads to the creation of interstitial fluid, it does not movelymph. Choice III is incorrect. The correct choice is A.

78. B is correct, kidney. Urine is formed as an ultrafiltrateof blood in the glomerular capsule of the kidney. Choice B iscorrect. Neither the liver, the pancreas, nor the salivary gland produces an ultrafiltrate. Choices A, C, and D areincorrect. The correct choice is B.

79. B is correct, I and II only. After a meal, chylomicrons first enter the lymphatic vessels, then the blood vessels.Choice I is correct. Lymph does not contain erythrocytes (red bloodcells). Choice III is incorrect. Lymphocytesarethe second most abundant type of white blood cells in the plasma. They are also present in the lymph. Choice II iscorrect. The correct choice is B.

Passage XIII (80 - 86) Renal Clearance

80. D is correct, [ U X PFa = ([ ]u x fu) + ([ Iv x pfv)- The principleof mass balance is the idea of what enters thekidney must leave the kidney. That which enters the kidney will be from only the plasma of the renal artery. Thiscan be quantified as [ ]A X PFA. There are two ways substances can leave the kidney. One is through the urine,quantified as Fu X (' ]u. The other is through the renal vein, quantified as PFU X [ |v. Therefore, mass balance isbest represented as [ ]AX PFA = ( [ ]u X Fu) + ([ ]v X PFV). The correct choice is D.



81. B is correct, 0.0005 L. We are interested in knowing the volume of urine collected in 0.5 minutes. We need toknow the rate of urine flow to figure out the volume. The rate of urine flow can be calculated using the relationshipestablished in the passage. In other words, Fu = (C) X ([ ]p)/[ ]u. Putting in the values, weget Fu = 50 ml/min X 1mg/ml divided by 50 mg/ml, which turns out to be 1 ml/min. We arc asked for 0.5 minutes, so we have 0.5 ml asour volume. 0.5 ml X 1 L/1000 ml = 0.0005 L. The correct choice is B.

82.

83.

84.

85.

86.

C is correct, GFR = [ ]u X Fu/[ substance]p. From the passage, we know that under certain conditions, the amountof a substance filtered is equal to the amount excreted in the urine. The amount of a substance excreted in the urinewill be [ ]u X Fu. To calculate the amount filtered, we can multiply the GFR by the concentration of the substancein the plasma. Equating the two, we have GFR X [ ]p = [ ]u X Fu. Therefore, the GFR = [ ]u XFU = [ ]«. Thecorrect choice is C.

B is correct, the substance must not be reabsorbed, and must be secreted by the nephron. From the passage, weknow that only under certain conditions can the amount filtered at the glomerulus equal the amount excreted in theurine. It should make logical sense that one of thoseconditions should be that nothing is added or taken away fromthe filtrate as it passes through the nephron. Statement B claims that the substance must be secreted into thenephron. If this was the case, the amount excreted will be more than likely greater than the amount filtered.Therefore, condition B cannot exist if our claim is to be valid. The correct choice is B.

A is correct, Proximal Convoluted Tubule. This question is simply calling on our previous knowledge of kidneyfunction. Recall that most reabsorption occurs in the proximal convoluted tubule. This is the first part of thenephron. The substances reabsorbed are returned to the venous system through the peritubular capillaries. Thereabsorption of glucose takes place via a secondary active transport, and under normal conditions, 100% of glucoseis reabsorbed. The correct choice is A.

A is correct, 3.5 mg/ml. The plasma threshold is defined as the level of plasma glucose which first gives rise toglucose in the urine. Looking at the curve, we are interested in the line labeled excretion. When this line firstbecomes positive, we need to look at the value of the blood glucose which gave rise to this first appearance ofglucose in the urine. This line becomes positive between the values of 2 and 4. Looking carefully, the value of theblood glucose is around 3.5 mg/ml. The correct choice is A.

Bis correct, the cationic species has the highest filterability because of the presence of anionic glycoproteins on thesurface of all glomerular filtration components. Let us read the curve given in the question. Draw an imaginaryvertical line in the graph. With this line, we can say that ifa molecule has the same radius, the cationic species willhave the higher filterability. Therefore, any choice that uses the argument that cationic or anionic species havedifferent radii is invalid. In other words, we cancompare a positive and negative molecule that has the same radii,and know from the graph that the positive cationic species will have the greater filterability. Thatmeans there mustbe some other reason for the enhanced filterability. The most likely reason is the presence of anionic glycoproteinson the surface of all filtration components. The opposite signs will create an electrical attraction that can lead to anenhanced filterability. The correct choice is B.

Passage XIV (87 - 93) Gastrointestinal Junction

87. D is correct, regurgitation of duodenal contents may lead to stomach ulcers. We know from the passage that thepyloric sphincter carries out two essential functions. One of these functions is toprevent the reflux of small intestinalcontents into the antrum. This is the case because the lining of the stomach is not protected from the bile that issecreted into the small intestine. The reflux of this bile containing content may contribute to the development ofstomach ulcers. So again, while the stomach is protected from the acid it secretes, it is vulnerable to the bile secretedby the small intestine. The opposite holds truefor thesmall intestine. The small intestine is protected from the bile,but is vulnerable to the acid from the stomach. Therefore, regurgitation of duodenal contents may lead to stomachulcers. The correct choice is D.

88. B is correct, catecholamine. Thisquestion cannot beanswered from any of the information in the passage, but mustbe answered solely on our own knowledge of hormones and hormone classification. Norepinephrine, along withepinephrine and dopamine, are catecholamines which fall under the major category of amine hormones. The correctchoice is B.

89. D is correct, a decrease in blood volume. If a hypertonic solution is introduced into the lumen of the small intestine,this creates a situation where water will enter into the lumen from the vascular system. In other words, the water


Biology Gastrointestinal & Renal Section III Answers

from the vascular system will cross the intestinal wall and diffuse to where there is a higher concentration of solute.This will certainly impair the vascular system. In particular, it will cause a decrease in blood volume which couldlead to a dangerous drop in arterial blood pressure. The correct choice is D.

90. A is correct. We are told from the passage that a decrease in the pH of the duodenum will cause a decrease in therateof gastric emptying. One way to achieve a decrease in gastric emptying would be todecrease thecontractility ofthe antrum and to increase the contractility of the duodenum. In this way, wc could achieveour goal of a decreasedflow of material from the antrum into the duodenum. Looking at Graph A, the injection of HCl into the duodenumcertainly does cause a decrease in thecontractility rate of the antrum. In addition, the injection of the HCl certainlydoes cause a significant increase in the rate of contraction for the duodenum. Therefore, Graph A BEST representsour expected results. The correct choice is A.

91. C is correct, insulin. We are looking for a hormone that is most likely to be released in response to a rise in GIP.Since we are looking for a hormone, bile can automatically be eliminated. We know that GIP is itself released as aresult of fat in the small intestine. The fat will soon find its way into the vascular system by first going through thelymphatic system. The fat will want to enter into the adipose tissue. Insulin is the hormone responsible for taking fatfrom the vascular system and facilitating its uptake into adipose tissue. Think in very general terms. We are in anabsorbing state for the body, as food is being absorbed from the small intestine. For this reason, insulin is thehormone that is necessary, and GIP is stimulating the release of insulin in anticipation of the arrival of fat into thevascular system. The correct choice is C.

92. C is correct, decrease the rate of gastric emptying and contribute to the mixing of stomach contents. Wc know fromthe passage that CCK does retard the rate of gastric emptying. For that reason alone, we can eliminate choices A andB. We now must concentrate our efforts into understanding whether the actions of CCK contribute or inhibit themixing of stomach contents. The question tells us that CCK causes a constriction of the pyloric sphincter. However,it also causes an increase in antrum contractility. Think about this. If the antrum has an increased contractility andthe contents of the stomach cannot enter into the small intestine, they will simply be pushed back into the body ofthe stomach. The contents of the stomach have no where to go but backwards. Therefore, there will be an increase inthe mixing of stomach contents and an overall decrease in the rate of gastric emptying (remember, the pyloricsphincter is closed). The correct choice is C.

93. B is correct, the pH of the blood increases. Wc are looking for afalse statement surrounding the notion of parietaland pancreatic cell secretion. We know that a parietal cell secretes protons into the lumen of the stomach. First, howis this proton obtained. The parietal cell converts a carbon dioxide molecule into a bicarbonate ion and a proton. Thecell secretes the proton into the lumen. What does it do with the bicarbonate ion? It transports the ion into the blood.In that way, the parietal cell itself does not become basic. However, that does present a problem for the pH of theblood, right? Well, not exactly. A pancreatic cell does just the opposite. The cell secretes a bicarbonate ion into thelumen of the small intestine and is left with a proton. The proton is released into the blood. If the secretions of theparietal cell and pancreatic cells are equivalent, the pH of the blood does not change throughout the digestiveprocess. The correct choice is B.

Passage XV (94 - 100) Cholera Toxin

94. B is correct, salt loss creates an osmotic gradient, which water follows out of the cell. The epithelial cells lining theintestine form a selectively permeable boundary between the blood and the lumen, or cavity, of the intestines. Incholera, these epithelial cells secrete sodium and chlorine ions into the lumen. This changes the relativeconcentrations of solutes, meaning that there is a lower concentration of solutes inside the cell in comparison tooutside the cell. Water diffuses across membranes by osmosis, following a gradient from low dissolved solutes tohigh dissolved solutes (i.e., high water concentrations to low water concentrations; the concentration of water is lessin a solution with a lot of ions in it.). Therefore, water leaves the epithelial cells and enters the lumen of the intestine,following the sodium and chloride ions.This causes diarrhea. Choice A is wrong because the membrane is alreadyhighly permeable to water, and extra pores won't make a difference. Choice C is wrong because water cannot beactively transported. Choice D involves reabsorption of waterand is completely incorrect. The correct choice is B.

95. D is correct, Water, glucose, and Na® ions. From the passage, we learn that rehydration therapy involves aninwardly-pumping glucose/sodium symport. A symport is a protein which transports two ions at a time across amembrane, in this case, into the cell. If a cholera patient is given both glucose and Na® with water, theglucose/sodium symport will transport both of these ions into the intestinal cells, increasing the solute concentrationinside. This will cause water to follow osmotically and reenter the blood (via the epithelial cells). All of the other


Biology Gastrointestinal St Renal Section ffl Answers

answer choices are wrong because none of them allows solute to enter the cell, favorably changing the osmoticbalance. The correct choice is D.

96. D is correct, testosterone. This problem requires that we know that mechanisms of the hormones that are mentioned.Epinephrine, gastrin, and norepinephrine are peptide hormones which operate by binding to receptors on the outersurface of cell membranes. This then triggers a signal cascade involving second-messengers like cyclic-AMP. Assuch, the cholera toxin would affect their actions by increasing cyclic-AMP levels. Testosterone, on the other hand,is a steroid hormone. Steroid hormones can pass through the plasma membranes of cells, directly (or through abound protein complex) affecting DNA transcription. Steroid hormones therefore do not use second-messengersystems. The correct choice is D.

97. D is correct, the underlying bacterial infection must be treated so that dehydrating diarrhea ceases. Rehydrationtherapy is just that, a therapy, not a cure. It does not address the underlying problem, which is the bacterial infectionthat produces the cholera toxin in the first place. In severe infections, antimicrobial agents must be administered todestroy the bacteria. Otherwise, dehydration will simply reoccur after each rehydration attempt due to thepersistence of thecholera toxin which causes saltsecretion and water loss. Choice A is wrong because the body canreestablish a normal blood volume and pressure. Choice B is incorrect because pores won't affect the existingosmotic gradient(see Question 1).Choice C is notright because morechloride willjust make the osmoticgradientsteeper and more water will leave the body. The correct choice is D.

98. C is correct, lack membrane receptors for cholera toxin. The question is basically asking why some mammals (aclassof animal closely related to humans, relatively speaking) aren'taffected adversely by cholera in the way thathumans are. Choices A and B are wrong because mammals, like humans, useG-protein mediated signaling (in fact,most studies are done in mice) and generally only have high cellular cyclic-AMP levels when cells are stimulated(i.e., by hormones). Choice D is wrong because if cyclic-AMP levels are normally low, then cholera toxin wouldincrease them. Choice C is correct because thecholera toxin binds to specific receptors on thesurface of intestinalcells. If these receptors are absent, as in many mammal species, the toxin can't attach and exert its effects. Thecorrect choice is C.

99. A is correct, I only. The transgenic mice described in the passage are engineered to have the gene which encodesthecholera toxin. This gene is attached toa promoter (a DNA sequence that allows DNA to be transcribed) which isonly active in the growth-hormone producing cells of the pituitary gland. Therefore, these mice produce choleratoxin specifically in theirpituitary cells, causing a sustained increase in cyclic-AMP levels in these cells.Thesemicesubsequently exhibit gigantism, which results from excess growth hormone. Therefore, we can assume that theincreased cyclic-AMP levels are somehow signaling these cells to produce more growth hormone. The question askswhat would inhibit this effect. Excessive normal G-proteins wouldn't because they would simply bealtered by thecholera toxin and would become constantly active. Excessive cyclic-AMP would add to the gigantism because itseems tobehigh levels ofcyclic-AMP which iscausing it inthe first place. An enzyme which degrades cyclic-AMPwould reduce the effect. The correct choice is A.

100. B is correct, the frequency of the CF gene is high in populations chronically exposed to cholera. The theorymentioned in the question states that the cystic fibrosis gene, when present inone copy (i.e., heterozygously) mayconfer a selectiveadvantage becauseit provides a degree of resistance to cholera. This is a similarcase to sickle-cellanemia heterozygotes, which are somewhat resistant tomalaria. If the CFgene did confer a selective advantage onheterozygotes, then wewould expect to seea high frequency of this gene in areas of high cholera incidence. This isbecause people with the gene would bemore likely tosurvive epidemics and would evolutionarily bemore "fit" thanpeople without thegene. All of theother answers are wrong because they do notprovide evidence thatsupports thistheory, even though they may not contradict it. The correct choice is B.

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BiologySection IV

Reproductionand

Development

A. Reproduction

1. Male Reproduction2. Female Reproduction

B. Development

1. Developmental Stages2. Developmental Mechanisms3. Human Embryo Development


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Reproduction and DevelopmentTop 10 Section Goals

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amiliar with the male reproductive anatomy.

Know howtotrace thepassage ofsperm from thetestis, through theepididymis, intothevasdeferens,through theejaculatory ductandprostate, andoutofthebodythrough theurethra.

@^f^ Know how sperm cells are produced during spermatogenesis.^Sr Be familiar with the actions of LH and FSH on developing sperm. Know the different cell types^» involved in sperm production. Make sure you understand both mitosis and meiosis.

®#* Be familiar with the female reproductive anatomy.Jjr Know the general steps ofhow anovum develops within the ovary. Be aware ofthe different types* of cellswithin the ovary and their relationship to the developing egg.

QIHI Know how eggs are producedduring oogenesis.jSr Be familiar with the actions of LH and FSH on developing ova. Know the different cell types involvedJ in egg production. Make sure you understand both mitosis and meiosis.

€)<§•* Be able to compare and contrast male and female gamete production.jar Understand why the male Leydig cells are analogous to the female theca cells, andwhy the male* Sertoli cells areanalogous to thefemale follicle cells.

(D'fm Understand how to read the hormonalcurves for a woman's monthly cycle.jlf Understand the relationship between LH, FSH, estrogen, and progesterone during both the follicular* phase and the luteal phase of ova production.

Olfe Be familiar with the different hormonal feedback mechanisms in the brain.jSf In the male and female, gonadal hormones circulate to the central nervous system and exhibit•» feedback control onhypothalamic-pituitary secretions. Understand these general feedback mechanisms.

©#* Understand what happens after a sperm fertilizes an egg.jBr Be familiar with the process of fertilization, where itis most likely to occur, and where the fertilized* egg is most likely to implant itself in the uterus.

©#* Understand the hormonal relationships associated with fetal development.j5r Be familiar with how placental hCG affects the corpus luteum andhow estrogen andprogesterone

»

fromthe corpusluteum affects the anterior pituitaryand the mammary glands.

([l)lrm Be familiar with the different stages of embryonic development.Developmentbegins with fertilization and proceeds through cleavage, gastrulation, neurulation,neural crest formation, organogenesis, and eventuallyparturition. Befamiliar with.these stages.

Biology Reproduction St Development

ReproductionMale Reproduction

General Reproductive EndocrinologyDuring the endocrinology of reproduction a hormone called gonadotropinreleasing hormone (GnRH) is released from the hypothalamus. GnRH signalsthe anterior pituitary to release the gonadotropins luteinizing hormone (LH)and follicle stimulating hormone (FSH). Both LH and FSH travel to the gonads.In the male the gonads are the testes while in the female they are the ovaries.Each of these gonadotropins elicitsa number of different responsesat their targettissues.There are two general types of responses that we will be considering. Thefirst is the production of steroid hormones. The second is the production of germcells. In the male the germ cells are called spermatozoa while in the female theyare tlie ova.

Male Reproductive AnatomyThe organs which carry out the reproductive functions in the male are the testes,epididymis, vas deferens, ejaculatory ducts, seminal vesicles, prostate,bulbourethral glands, and penis. These structures are shown in Figure 4-1.

Prostate

"land

Urethra

Corpuscavcrnosum

Corpusspongiosum

Testis

Seminal

vesicle

Ejaculatoryduct

Bulbourethral

gland

Figure 4-1Male reproductive anatomy.

All of these structures are important in the synthesis and delivery of sperm to thefemale vagina. The actual production of sperm is referred to as spermatogenesis.Because the formation of sperm requires a temperature lower than that of bodytemperature, they are produced in the testes which lie in the scrotum outside ofthe body cavity. Within each testis are a series of convoluted tubules called theseminiferous tubules. It is within these tubules that we find the spermatogenic

Male Reproduction


Biology Reproduction &Development Male Reproduction

cells. Not only are sperm produced within the testes but the hormonetestosterone is synthesized within specialized interstitial cells called Leydig cellsthat lie outside the seminiferous tubules. A third cell type found within the areaof the seminiferous tubule is the Sertoli cell. As we will see, these cells act topromote spermatogenesis and they also produce the protein hormone inhibin.

Sperm ProductionThe target tissue for hormones like FSH and LH are the gonads. In the male thegonads are the testes. Within the testes are a series of convoluted tubulesreferred to as the seminiferous tubules. There are several different cell types thatinteract within the seminiferous tubules to allow for the development ofspermatozoa. Between adjacent seminiferous tubules are the Leydig cells (alsocalled the interstitial cells). If we were to take a cross-section of a seminiferoustubule, we would note a lumenal space in the center. Adjacent to the lumenalspace are fairly large cells called Sertoli cells. In direct contact with the Sertolicells are a series of spermatogenic cells. As you proceed from the basementmembrane of the seminiferous tubules toward the lumen, the cells become moreand more differentiated. In other words, those cells are becoming more and moredeveloped. An important aspect of this development is that the Sertoli cells are inconstant contact with the spermatogenic cells. You could view this as the Sertolicells regulating the development of the spermatogenic cells.

Let's consider the development of the spermatogenic cells. If we consider aportion of theseminiferous tubule, we willseea basement membrane separatingthe Sertoli cells from the interstitial cells. The Sertoli cells are in contact with oneanother and essentially form a ring around the inside of the basementmembrane. Different developmental events take place on different sides of theSertoli cells. For example, spermatogenic cells closer to the basement membraneturn out to dividein a completely different way than theydo further awayfromthe basement membrane.The spermatogenic cells that are closer to the basementmembrane are calledspermatogonia. They have 46chromosomes.

The way that spermatogonia divide is by mitosis. The resulting "daughter" cellsare exactly the same as the parental cell. In the area closest to the basementmembrane the spermatogonia are undergoing constant division. Recall that thefirst thing that happens in mitosis is duplication of the chromosomes. Thesechromosomes eventually migrate towards the cell's equator and the cellundergoes cytokinesis to produce two daughter cells which are exactly like theparental cells. The spermatogonia that undergo mitosis by the basementmembrane are referred to as primary spermatocytes. As the primaryspermatocytes form they begin to squeeze by the juxtaposed Sertoli cells andmove towards the lumen of the seminiferous tubule. In the area near the lumen acompletely different set of events takes place. The primary spermatocytesundergo meiosisand will reducetheir genetic complement to 23chromosomes.

Meiosis is simply one duplication with two divisions. Recall that during themeiosis the first thing that happens is duplication of the chromosomes. Thechromosomes lineup at thecell's equatorand the cell undergoes cytokinesis. Thecells that are formed after the first meiotic division are called secondaryspermatocytes. At this point a second meiotic division takes place. Eachsecondary spermatocyte will again divide to give two daughter cells. Thedaughter cells of the second meiotic division are referred to as spermatids andthey have half the chromosomal complement (i.e., 23 chromosomes) as the



original spermatogonia. Therefore, every primary spermatocyte that undergoesmitosis and meiosis ends up producing four spermatids.

The very next event that takes place is a transformation event. Each of thespermatids is transformed into a spermatozoa. The transformation process hasan interesting characteristic to it. When the spermatids are initially formed, theyare all connected to each other by cytoplasmic bridges. During thetransformation process the spermatozoa essentiallybud out from the spermatids.

At the very tip of a spermatozoan there is a structure called the acrosome whichcontains a lot of digestive enzymes. These digestive enzymes help thespermatozoa gain access to the interior of the egg once fertilization has takenplace. Inside the head of the spermatozoa is a nucleus which contains DNA. Inthe midsection are mitochondria, providing energy for the whipping movementof the tail. This enables the sperm to swim towards their destination (the egg).

Hormonal Control of Sperm ProductionIt is important to remember that the spermatogenic cells are in constant contactwith the Sertoli cells. The Leydig cells have a rather special function. Theyconvert cholesterol to testosterone (which we can abbreviate as T). Testosteronehas several roles. Testosterone can diffuse out of the Leydig cells and move toother target tissues of the body. It can also leave the Leydig cells and diffuse intothe Sertoli cells. Within the Sertoli cell testosterone binds to a specific receptorand is converted to a compound called dihydrotestosterone (which we canabbreviate as dHT). This complex diffuses into the nucleus of the Sertoli cell andinstructs the DNA to synthesize RNA. It is the products of the RNA synthesisthat affect the spermatogenic cells. See Figure 4-2.

Recall that we mentioned that the hypothalamus releases GnRH which then actson the anterior pituitary and causes it to release FSH and LH. These twogonadotropins have a direct effect on the Sertoli cells and Leydig cells. LH bindsto a specific receptor on the membrane of the Leydig cell. The secondarymessenger produced from this binding interaction increases the conversion ofcholesterol into testosterone. FSH has a different function. FSH will bind to a

surface receptor on the Sertoli cell and induce a secondary response within theSertoli cell. This secondary response helps convert testosterone intodihydrotestosterone. It also induces the synthesis of the receptors. [A thoughtquestion. If you had to make a male contraceptive, what events would youdisrupt in this pathway? You would disrupt the production ofdihydrotestosterone or FSH. One of the problems with producing a malecontraceptiveis that it is rather difficult to stop the synthesis of FSH.]

As with any hormonal system there is always the consideration of a feedbackmechanism. In males there is a constant production of LH and FSH. The reason isdue to a feedback mechanism. One of the consequences of testosterone diffusingout of the Leydig cells is that it has a negative feedback at the anterior pituitaryand at the hypothalamus. Testosterone is preventingthe synthesis of GnRH andLH. The Sertoli cells are involved in a different feedback mechanism. The Sertolicells also secrete a compound called inhibin. Inhibin acts as a negativemodulator of the anterior pituitary. The reason that the gonadotropin levels arerelatively constant is that you have a constant synthesis of the spermatogeniccells. If testosterone levels are too high, the feedback mechanism will decreasethe levels of LH. If the levels of dihydrotestosterone are too high, there is anincrease in inhibin synthesis which results in a decrease in the levelsof FSH.

Male Reproduction


Biology Reproduction & Development Male Reproduction

(-) Hypothalamusi

GnRH

I..(•).> (-)Anterior Pituitary <• ;

Testosterone LH

iCholesterol

Testosterone

Leydig CellBasement

membrane

FSH

dHT

T*i>nucleus

ReceptorA

Sertoli Cell

Inhibin

TestesSpermatogenic

cell

Figure 4-2Hormonal regulation in the testes.

Secondary Sex Characteristics in the MaleAlmost all of the secondary sex characteristics in the male are due to testosterone.Some of those characteristics are development of a beard, pubic hair, deepeningof the voice, texture of the skin, muscle distribution, bone development, sexualdrive, and, if you are genetically predispositioned to it, baldness.

Sperm DeliverySperm leave the seminiferous tubules and enter into the epididymis and then thevas deferens. Sperm are stored in the epididymis and vas deferens for about 14days before ejaculation. Contraction of smooth muscle lining the walls of thesestructures ejects the sperm down the vas deferens and into tlie ejaculatory duct.The seminal vesicles, prostate gland, and bulbourethral gland secretecomponents that make of the remainder of the fluid that is ejaculated (now calledsemen). Some of the components secreted by these glands are fructose vitamins,bicarbonate zinc, prostaglandins, and mucus. The total volume of the fluidejaculated with the sperm is about 3 to 5 milliliters.



Female Reproduction

Female Reproductive AnatomyTheessentialanatomicalfeatures of the female reproductive systemincludes thevagina, uterus, Fallopian tubes, and ovaries. Once sperm have been ejaculatedinto the vagina they can live for about 48 hours. However, about a half an hourafter ejaculation the leading sperm arrive at the site of fertilization which isusually the oviduct (also called the Fallopian tubes). The fertilized egg continuesto move down the oviduct towards the uterus where it will implant in the uterinelining. About a week after ovulation the fertilized egg, now called a blastocyst,implants in the lining of the uterus where it will continue to grow and developuntil parturition (delivery). A few of the anatomical structures mentioned hereare shown in Figure 4-3.

Bladde

Urethra

Fallopian 0varytube

Vagina

Figure 4-3Female reproductive anatomy.

Ova Production

The production of the female germ cells occurs via a process called oogenesis. Incertain aspects this process is similar to that of spermatogenesis in males. Theimportant similarity in the female is that when cell division is complete the ovumwill have half the DNA as the original parent cell. The regulation of oogenesis,however, is quite different. First, there are a series of events in which the oogoniamitoticallydivide to form primary oocytes. The primary oocytes,which have 46chromosomes, are analogous to the primary spermatocytes in the male. In thefemale all of these divisions occur within the first two to three months of fetaldevelopment. About the third month of gestation all the mitotic divisions cease. Inother words, all the oogonia that a female will produce will be produced withinthe first three months of fetal development. In fact, at birth a female will haveabout 400,000 primary oocytes in both of her ovaries. However, by the time shehas finished her reproductive life only about 400 will have ever matured. Theremaining primary oocytes will have degenerated at various stages during thedevelopmental process. This process of degeneration is referred to as atresia.

Female Reproduction


Biology Reproduction & Development Female Reproduction

The primary oocytes will eventually undergo the first meiotic division, causedby a surge in the gonadotropin LH. The first meiotic division happens inmonthly cycles. Some of these primary oocytes could take as long as 50 years tohave their first meiotic division. Once the first meiotic division is complete thesecondary oocyte, which has 23 chromosomes, is formed. During this divisionone of the two daughter cells obtains all the cytoplasm while the other cell is justa small sphere of DNA called the first polar body. The secondary oocyteundergoes the second meiotic division onlyafterfertilization has taken place. Theproduct of that division is the ovum and a second polar body. [The first polarbody also divides to give two second polar bodies as well.] The ovum now has 23chromosomes. When sperm and ovum unite in fertilization, the complement of46 chromosomes will be restored (23 from the male and 23 from the female).

If we take a cross-section of the ovary, we will see a series of specific cell typescalled primary follicles which are in different developmental states. The primaryfollicle is actually a primary oocyte surrounded by a layer of follicle cells. Thesefollicle cells are in constant contact with the primary oocyte. Eventually one ofthese primary follicles will start to develop.[An exception arises with fraternaltwins in which case two follicles will form and develop at the same time.]

ZonaPellucida

GranulosaCells

Theca GrowingCeIls Follicle

PrimaryFollicleOvum

CorpusOvulation Luteum

Regressing CorpusLuteum

Figure 4-4The ovarian cycle.

In order to mature the primary oocyte estrogen, LH, and FSH are needed. Thisdeveloping time period is referred to as the follicular phase and lasts up to aboutthe 14th day of the woman's monthly cycle. During this period the primaryfollicle gradually develops. Surrounding the primary oocyte will be a membranecalled the zona pellucida. The zona pellucida is surrounded by more follicle cellscalled granulosa cells and finally by theca cells. See Figure 4-4. These cell typesare a direct result of the estrogen, LH, and FSH that are present. The theca cellsare analogous to the Leydig cells in the male while the granulosa cells areanalogous to the Sertoli cells in the male.


Biology

ill d

|£ll Follicular Phase c^

Reproduction St Development

14 28 Days

Figure 4-5The hormones of the ovarian cycle.

Within the primary follicle a fluid starts to build up forming the antrum. Thesystem is now primed for ovulation. The next event that takes place is the LHsurge. This surge causes the primary oocyte to undergo the first meioticdivision, forming a polar body and the secondary oocyte. The LH surge alsocauses the production of a series of enzymes which break down the membrane inthe primary follicle. Once the secondary oocyte is released, ovulation hasoccurred. Ovulation usually comes at about the 14th day in the woman's monthlycycle. As we have mentioned, the process leading up to ovulationis referred to asthe follicular phase. The left over follicle (after ovulation) is transformed into agland-like structure called the corpus luteum. One of the main functions of thecorpus luteum is to produce estrogen and progesterone. If fertilizationdoes notoccur and there is no pregnancy, the corpus luteum will degenerate and thewhole cyclewill start again. From the point of ovulation, at about the 14th day,until the beginning of the menstrual flow is the luteal phase. SeeFigure 4-5.

Hormonal Control of Ova ProductionDuring a woman's monthly cycle her gonadotropic hormones fluctuate ratherdramatically. LH remains relatively constant until just before ovulation at whichtime there is a sudden surge. FSH will also have a surge, but not as great as theLH surge. There will be a surge of estrogen before ovulation and then a smallerone after ovulation. The increase in estrogen before ovulation actually leads tothe increase in LH. Another hormone called progesterone will surge afterovulation as well. How are these hormones regulated?

Female Reproduction



..«..> Hypothalamusi

GnRH

(-)"> Anterior Pituitary

o

cc oO '—

o to

PJ P

LH

|::;::::.,; ;j^

Cholesterol

ITestosterone

\ Theca cell

CorpusLuteum

Basement

membrane

FSH

t i

Testosterone

Estrogen *""

Follicle cell

OvariesPrimaryoocyte

Female Reproduction

^^ Low level

>Highlevel

Figure 4-6The hormones of the ovarian cycle.

In the female the theca cells convert cholesterol into testosterone. The

testosterone diffuses into the folliclecells where it is converted into estrogen. Oneof the effects of estrogen is to help in the development of the primary follicle (i.e.,the primary oocyte). We had mentioned that the primary oocyte reaches maturityby undergoing reactions with estrogen, LH, and FSH. Recall that the anteriorpituitary secretes LH and FSH. LH affects the theca cells while FSH affects thefollicle cells. As estrogen is being synthesized, the primary oocyte is developing.At the same time we have a proliferation of the follicle cells. If we have morefollicle cells, then we will be able to synthesize more estrogen. This is where thefirst estrogen surge comes from. It turns out that at low concentrations ofestrogen there is a negative feedback on FSH production. See Figure 4-6.

However, the follicle cells are growing and at a certain point in time there is athreshold level in estrogen. This means that relative to the system we no longerhave a low concentration of estrogen but instead have a high concentration ofestrogen. At high concentrations we see that estrogen has a positive feedback on


Biology Reproduction & Development

the anterior pituitary and LH production. Thus, with a high level of estrogenproduction we now have the LH surge. This is important. The increasedconcentration of estrogen is having a positive feedback and therefore causingmore LH production-giving the LH surge. The reason that one sees these surgesis because of the rapid growth of the follicle cells during development of theprimary follicle.

Immediately following the LH surge the levels ofLH drop down to very lowlevels. What causes the drop in LH and the increase inestrogen and progesteroneduring the luteal phase? Ifwe no longer have a high concentration ofestrogen,we will lose the positive feedback on the anterior pituitary. Recall that atovulation all the otherfollicle cells become transformed into the corpus luteum.They are no longer follicle cells. During that transformation they transiently losethe ability to produce estrogen. This ability toproduce estrogen is lost just longenough so the LH levels decrease. However, the levels of both estrogen andprogesterone begin to increase. Whereis thiscomingfrom?

The corpus luteum becomes an endocrine gland and begins to synthesizeestrogen and progesterone. Estrogen and progesterone, in combination, have anegative feedbackon LHand FSHproduction in theanterior pituitaryas wellasa negative feedback on the hypothalamus and the synthesis of GnRH. What isthe result of having a negative feedback on LH and FSH? Essentially what ishappening is that the primary follicle is prevented from developing. As long asthecorpus luteum is producingestrogen and progesterone, the primaryfollicle isunable to develop. But this is what wewant to happen. At this point we do not needanother primary follicle being made. [With this information, what is the basis forfemale oral contraception? Birth control pills are essentially a combination ofestrogen and progesterone in concentrations that represent the needs of thewoman.] Remember, estrogen alone has a positive feedback. Estrogen andprogesterone in combination have a negative feedback.

A very different set of events takes place during pregnancy. What happens whenthe spermatozoan comes in contact with the ovum? The acrosome of thespermatozoan allows for the digestion of the membrane of the secondary oocyte.Within the membrane of the ovum (zona pellucida) are receptor sites for thesperm. These receptor sites prevent cross-species fertilization. The only thing thatenters the secondary oocyte is the nucleus of the sperm. After the nucleus of thesperm enters the secondary oocyte the zona pellucida changes and prevents anyother spermatozoa from entering. This process is referred to as fertilization. Atthis point the secondary oocyte undergoes the second meiotic division to formthe ovum and the polar body. The nucleus of the sperm and the nucleus of theegg fuse together to form the zygote, which now has a complement of 46chromosomes.

The zygote rapidly begins to develop and in about 7 days attaches itself (as ablastocyst) to the uterine lining. A blastocyst is essentially a small ball of cellswith a central fluid-filled cavity. When the implantation takes place the placenta,made up of maternal and fetal cell types, begins to form.

What one does not want to happen during pregnancy is the development ofanother primary follicle. Throughout pregnancy there are high levels of estrogenand progesterone. The reasons for maintaining high levels of estrogen andprogesterone are different during the first three months of pregnancy than they

Female Reproduction


Biology

(£

Chorionic

Gonadotropin

0 12 345 67 89

Months after start of last

menstrual period

Figure 4-8

Placental synthesis ofestrogen and progesterone.

Reproduction fie Development Female Reproduction

areduring the last sixmonths ofpregnancy. Shown in Figure 4-7 isadiagram ofthe hormonal considerations during the first three months of pregnancy. Theplacenta itself is an endocrine gland. During the first three months the corpusluteum is still a viable gland and it secretes estrogen and progesterone. Theplacenta synthesizes chorionic gonadotropin (CG) which stimulates the corpusluteum to make estrogen and progesterone. One of the best pregnancy tests is todetect CG in the blood-stream. The only time CG is made is during the first threemonths of pregnancy.

CorpusLuteum

/V

CG (+)

Placenta

Estrogen &Progesterone (-)

Anterior

Pituitary

r°*e (+) (+)

cs

(+)

Prolactin

V

MammaryGlands

Figure 4-7The placenta and its relationships.

What are the functions of estrogen and progesterone? Estrogen and progesteronehave a negative feed-back on the anterior pituitary, thus inhibiting LH and FSHproduction. This prevents the formation of the primary follicles duringpregnancy. It also means that ovulation and menstrual cycles will be eliminatedfor the duration of the pregnancy. Another important gland during pregnancy isthe mammary gland. There are a series of hormones that positively feedback onthe mammary glands. One hormone, prolactin, comes from the anteriorpituitary. The placenta forms another hormone called chorionic somatomammotropin (CS) which also acts on the mammary glands. They help themammary glands to grow. Estrogen and progesterone have positive feedback onthe mammary glands as well.

During the last six months of pregnancy there in only a slight variation on thistheme. CG is no longer made. This results in a loss of feedback to the corpusluteum. After three months the corpus luteum breaks down and the supply ofestrogen and progesterone from the corpus luteum comes to a halt. However, atthat point in time the placenta itself starts to synthesize estrogen andprogesterone. In fact, the placenta makes much more estrogen and progesteronethat the corpus luteum did. This can be seen in Figure 4-8.

Secondary Sex Characteristics in the FemaleSome of the effects of the female sex steroid estrogen involve development offemale body configuration, growth of breasts, growth of external genitalia,pattern of pubic hair, fluid retention, influence on the cardiovascular system, andit may be involved in inhibiting atherosclerosis.



DevelopmentDevelopmental StagesRecall that when we discussed meiosis we mentioned that the haploid sperm andthe haploid egg can unite to form a zygote. The zygote will undergo a series ofmitotic cell divisions which will form a clone of cells. All of those mitoticallydividing cells were derived from a single precursor cell and eventually they willgive rise to an embryo. The embryo will continue to develop and later forms asexually immature juvenile. The juvenile undergoes a maturation process andbecomes the adult.

Brain

Thoracic Cavity

Spinal CorAbdominal Cavity

Figure 4-9The body outline.

In our discussion on development we will focus on the vertebrate. In particular,we will want to examine the arrangement of the body plan in vertebrates. As wewill see, human beings, as all other vertebrates, have essentially the same bodyplan. For example, the skeletal system in vertebrates helps to support the body.One aspect of the skeletal system is a dorsally located vertebral column. Withinthis vertebral column is the spinal cord. Anterior to the spinal cord is the brain.The brain and the spinal cord comprise the central nervous system. Nervoustissue outside the central nervous system is referred to as being part of theperipheral nervous system. Ventral to the vertebral column is the body cavity, orcoelom. The coelom is divided into the upper thoracic cavity and the lowerabdominal cavity. The thoracic cavity contains the heart and lungs while theabdominal cavity contains a variety of organs like the stomach and the intestinalsystem. These basic landmarks are shown in Figure 4-9.

The body is composed of a variety of highly differentiated cell types. Forexample, there are stratified epithelial cells found in the epidermis. Cuboidal

Developmental Stages


Biology Reproduction & Development Developmental Stages

epithelium can be found in thekidney. Smooth muscle cells canbe found in thewalls of the intestine. Skeletal muscle cells can be found in the voluntary musclesand cardiac muscle cells can be found in the heart. This is just a sample of someof the cell types fond in the body. Cells of similar type form tissuesand tissuescan come together to form organs like the heart and liver. During developmentthe cells, which initially all start out alike, begin to grow and differentiate. Thishappens eventhough those cells all contain the samegenetic constitution. Theyall have the samegenes. Not onlyis there differentiation but some of thesecellsundergo morphogenesis (i.e., theybegin to take on differentshapes).

There are a number of stages that one can use to characterize development (seeTable 4-1). The first developmental stage involves the union of the male andfemale gametes. This process is called fertilization. The gamete producingorgans are the gonads, m the male theseare the testesand in the female they arethe ovaries. When the gametesunite a zygote is formed. The next developmentalstage involves cleavage of the zygote. During cleavage the zygote rapidlydivided into many smaller cells without an overall increase in size. Gastrulationis the third developmental stage. Many embryologists consider this to be themost important part of an organism's development. During this stage the cells ofthe zygote move to form the three primary germ layers (ectoderm, mesoderm,and endoderm) of the organism. The next developmental stage is neurulation.During neurulation we begin to see the formation of the nervous system. This isthe first organ system to begin differentiation. Neural crest formation is the fifthdevelopmental stage. The neural crest cells help to form parts of the nervoussystem, skull, and sensory organs. The last stage of development isorganogenesis. It is during this stage that the different organs of the body areformed.

1. Fertilization

2. Cleavage3. Gastrulation

4. Neurulation

5. Neural Crest Formation

6. Organogenesis

Table 4-1

Development In The FrogWe can use the frog as an example of vertebrate development because of thelarge size of its eggs (about1.5millimeters in diameter). The life cycle of the frogis shown in Figure 4-10. As we have mentioned, males give rise to haploid sperm(IN) while females eventually give rise to haploid eggs (IN). Initially the femalefrog will give rise to a primary oocyte which then undergoes the first meioticdivision. The secondary oocyte is formed but it is arrested at metaphase of thesecond meiotic division. This second meiotic division is not completed until thesperm from the male frog fertilize the secondary oocyte. Fertilization initiates aprocess called egg activation. This is a signal for development to begin. Thesecond meiotic division in the secondary oocyte is completed and the zygotenucleus is formed. This nucleus is now diploid or 2N. The zygote undergoes aseries of cleavages and an embryo is formed. Further cell division leads to the


Biology Reproduction & Development

formation of the tadpole. The tadpole, being an aquatic organism, eventuallyloses its tail and forms legs so itcan begin its terrestrial life as an adult frog. Thisis a process of metamorphosis.

\

Female 2n

V y//® bodies / Nj^^Vs^-^//® expelled / Secondary>nd | / / / S y oocyteion •' i / „*' /

Egg's secondmeiotic divis

1/

o

Sperm n

S,Zygote 2«

Fertilization

Figure 4-10The life cycle of the frog.

In the unfertilized egg there is a large amount of yolk that resides in the vegetalpole (i.e., the lower hemisphere). This yolk will act as food for the developingembryo.The animalpole (i.e., the upper hemisphere) contains mainly cytoplasm.In a fertilized egg an interesting phenomenon takes place. On the side of the eggopposite to where the sperm penetrates the egg's membrane, a structure calledthe gray crescent forms. The gray crescent is located on the dorsal aspect (thefuture back) of the animal. This is where we will eventually find the spinal cordand brain. The side opposite the dorsal aspect is the ventralaspect. This is where



Biology Reproduction St Development Developmental Stages

the belly ofthe organism will form. An imaginary line which connects the twopoles is called the meridian. Imaginary lines which run around the circumferenceof the cell are equatorial innature. The middle of the gray crescent defines thebody midline of the future organism. By the time the gray crescent has appearedthemainaxes ofthebodyof theorganism has beenestablished.

CleavageOnce the zygote isformed it undergoes a special cell division to increase itsmassbut not its overall size. The first cellular division occur along the body midlinewhich bisects the gray crescent. This division gives two cells. The next divisionalso occurs along themeridian, but this time at right angles to the last division.We now have four cells.The third cellular division occurs equatorially and givesfour cells in the animal pole and four cells in the vegetal pole. The individualcells involved in this growth arecalled blastomeres. Eventually a small, solid ballof cells will be formed called a morula. Further cellular division results in theformation of a hollow ball of cells called the blastula. Within the blastula is afluid-filled cavity called theblastocoel. These structures are outlinedin Figure4-11.

Animal Pole

Vegetal Pole

Unfertilized

Egg

GreyCrescent

Cleavage Down MeridianBody Midline Cleavage

(2 cells) (4 cells)

EquatorialCleavage(8 cells)

Meridian

Cleavage(16 cells)

Blastocoel

Morula

(32 cells)Blastula Cross Section

of Blastula

Figure 4-11Different stages in embryonic development.

GastrulationDuringgastrulation rearrangement ofcells occurs. Not far from the gray crescentan open develops intheblastula called theblastopore. Cells from theanimal polebegin to migrate inwards through the dorsal lip of the blastopore. As this outerlayer of cells migrates inward they form a second layer of cells immediatelybelow that outer layer. The blastocoel is reduced in size and is eventuallyeliminated. In its place a new cavity is formed called the archenteron. At thisstage the embryo is referred to as a gastrula (because gastrulation has takenplace). Invagination of this outer cell layer produces two cell layers. The outercell layer is called the ectoderm. The inner cell layer is called the endoderm. Alayer ofmesoderm will laterform between these twocell layers. These structuresare shown from different viewpoints in Figure 4-12.



The events that lead up to gastrulation differ widely in the animal world.However, gastrulation iscommon to all of them. It seems that the generation ofthe threecelllayers(ectoderm, mesoderm, and endoderm) has beenconserved inthroughout the evolutionary process. An animal cannot be made if gastrulationdoes not occur.

Blastocoel

Blastocoel

Lateral Marginof Endoderm

Archenteron

Remains of

Blastocoel

Ectoderm

Dorsal Lipof Blastopore

Blastopore

Archenteron


Mesoderm


Yolk Plug



Yolk Plug

Figure 4-12The process of gastrulation.

The ectoderm, mesoderm, and endoderm cell layers play different roles duringthe developmental process. They have different developmental fates. Forexample, the ectoderm will eventually differentiate into structures like the skin,the lens of the eye, and the brain and nervous system. Mesoderm willdifferentiate into structures like the notochord, heart, skeleton, muscle, the outercoverings of internal organs, and the reproductive organs. Endoderm willdifferentiate into the inner lining of the digestive tract and the respiratory tract,and major glands of the body like the liver and pancreas. These are just a few ofthe developmental fates of these cell layers.

NeurulationDuring this stage of development the ectoderm, mesoderm, and endoderm beginto form the structures that will eventually define the embryo and later the adult.



Biology Reproduction & Development Developmental Stages

During neurulation the formation of the notochord takes place along the bodymidline. This structure is derived from mesoderm. Superior to the notochord is amass of ectodermal cells called the neural plate. The neural plate will begin tofold in on itself and form the neural groove. As the edges of this folding fuse withone another the neural tube is formed. Within the neural tube will form thespinal cord (encased in the spinal column) and anterior to the spinal cord willform the brain. In other words, the neural plate, which is composed ofectodermal cell, gives rise to the nervous system. This tissue is referred to asprimordium (from the Latin primus, first, + ordior, to begin) and it is the earlieststage of development of a structure. During neurulation the embryo is sometimescalled a neurula.

As the neural plate begins to fold in on itself it will form the neural groove. Twoviews are shown in Figure 4-13.Below the neural groove is the notochord and oneither side of the notochord is mesoderm. As the neural groove begins to formthe mesoderm is split and forms a coelom (i.e., a body cavity). The lungs willeventually develop within this coelom. When the edges of the neural groove fusetogether the neural tube will be formed. Within the neural tube forms the spinalcord and anterior to the spinal cord forms the brain.

Neural Groove

EctodermNeural Crest

Cells

Neural

Plate |Notochord

Notochord

Ectoderm

Neural

Plate

Mesoderm SomitSCoelom

Figure 4-13Neural tube formation.

Neural

Groove

Neural Crest

Cells

Neural Tube

Neural

Tube

Neural Crest FormationAs the edges of the neural groove fused together and became the neural tube,specialized ectodermal cells were left in a more dorsal position on the neuraltube. These specialized cells are called the neural crest cells (Figure 4-13). Asthese cells begin to move to the sides of the developing embryo, they begin tofunctionalize. For example, ectodermal cells in the anterior portion of thedeveloping embryo associate with the neural crest cells and forms placodes.These structures will eventually form the sense organs located in the head. Someneural crest cells will help form sensory cells (e.g., olfaction and touch). Otherneural crest cellswill form the adrenal medulla. During times of stress the animal



will experience the "fight or flight" syndrome. In order to respond to this stresstheadrenal medulla releases adrenaline. This hormone prepares theanimal torespond byincreasing blood sugarlevels, heart rate, andblood pressure.

OrganogenesisDuring the initial stages of organogenesis there is an interaction betweenectoderm and mesoderm. The neural tubebecomes longer and thinner and hasan anterior to posterior developmental gradient. Neural crest cells begin tomigrate and take up positions in the vicinityof the neural tube. Mesodermalcellsmigrate towards the neural tube. Eventually they will form the vertebral column.The brain begins to form at theanterior portion oftheneural tube. Optic vesiclesbegin to form. As the neural tube continues to form segments of mesodermaltissue called somites beginto appear. Thesomites will eventually give rise to thevertebrae, connective tissue, and the muscles of thebody.

In the frog embryogenesis is complete with the appearance of the sexuallyimmature tadpole. Thesecreatures have gillsand livein an aquaticenvironment.They have no limbs. Instead, they have a tail which is used to propel themthrough the water. The last critical phase in the life of a tadpole is one ofmorphogenesis. Thetadpolewillchange from a sexually immature organism intoa sexually active organism, which is the frog. During this process ofmorphogenesis the tadpole grows limbs, develops lungs, loses its tail and itsgills. This change is initiated by a hormone which is released during a specifictime of development. The hormone which is released is thyroxin and it isreleased from the thyroidgland.

Many organisms pass through a larval stage (sexually immature) like thetadpole. When the frog eggs hatch the tadpoles are left to fend for themselves. Ifthe post-embryonic organism is left on its own, then the development is termedindirect. In contrast, direct development involves care being given to the post-embryonic organismby the mother. For example, a human fetus developswithinthe female of the species. Even after birth the female (and male) care for theyoung until they are able to take care of themselves.



Biology Reproduction fie Development Developmental Mechanisms

Developmental Mechanisms

If the cells of the body all contain the same genetic information, how can theybecome so differentiated?There are two general classes of interaction associatedwith the differentiation of cells that we need to consider. There are intracellularinteractionswhich involve interactions between the components within the cellsthemselves. There are also intercellular interactions in which the interactions arebetween cells. The intracellular interactions usually result in the setting up of aprepattern. The intercellular interactions usually undergo developmentalinduction.

Intracellular InteractionsWe have mentioned that the unfertilized egg has an animal pole, which containsthe cytoplasm, and a vegetal pole, which contains the yolk. The egg is nothomogeneous. However, at this point in the egg's existence a longitudinal axis isalready present. In the frog the animal pole will give rise to the head while thevegetal pole will give rise to the tail. This is true for the frog but it is not a truthset in stone. Remember, there are always variations on a given theme.

Recall that after fertilization and induction of development the gray crescentforms at a point opposite to where the sperm penetrated the egg. The formationof the gray crescent is due to an intracellular interaction. The formation of thegray crescent is not a prepatterned phenomenon that took place before the sperminteracted with the egg. This is clearly shown by the fact that the gray crescentdeveloped after the penetration of the egg by the sperm. The entry of the spermin the animal half of the cell is random. However, one entry is made and the graycrescent forms, the dorsal midline is established. This allows us to definedirections. In other words, prior to the first cellular division the axes of theorganism is established. In this case a prepattern is laid down which is adheredto during the rest of development.

Hans SpemannDuring the 1920's Hans Spemann was able to demonstrate that the gray crescentis an important landmark in the future development of the embryo. In hisexperiment, Spemann took a newly fertilized frog egg and tied a string around itsuch that the string bisected the gray crescent (see Figure 4-14a). He slowly tiedthe string tighter and separated the egg into two halves, each with half a graycrescent. Two separate blastomeres developed and two separate embryos wereformed. Since these embryos were twins, this procedure was called twinning.This experiment demonstrated that each blastomere was equivalent in itspotential to form a complete embryo. Even though two complete embryos wereformed, they were smaller than the normal frog embryos. Why? Because whenthe fertilized egg was cut in half, each blastomere received half of the originalamount of yolk.

There is one other important point to make note of in this experiment. During thelast lecture we said that the body midline bisected the gray crescent and that thiswas what eventually formed the neural tube. In the Spemann experiment thebody midline was itself bisected. In other words, instead of the midline being inthe middle it was to the side of the bisected gray crescent. However, the embryosdeveloped normally. What this means is that the midline had to migrate to aposition which was in the middle of the bisected gray crescent. Hence, there was


Biology Reproduction St Development Developmental Mechanisms

a developmental regulation inside the blastomere which allowed for themovement of the body midline.

Spemann did thesame experiment on another fertilized frog eggbut this time hecleaved the eggat right angles to the cleavage in the previous experiment (seeFigure 4-14b). Two blastomeres were formed. One blastomere had thecompletegraycrescent while the other blastomere did not contain any piece of the graycrescent. As development proceeded a complete embryo was from theblastomere with the complete gray crescent.The other blastomereformed a massofcells thatdid not differentiate into an embryo. It did not gastrulate. Again, theembryo thatwas formed was smaller due to the reduction in theamount ofyolkreceived. These two experiments proved for the first time that there was a clearprepattern during the development of an egg.

(a)

GrayCrescent.

Bisection

e? £

Complete Embryoswill Develop

Bisection

<* s

(b)

GrayCrescent

Undifferentiated

Mass of Cells

CompleteEmbryo

Figure 4-14Hans Spemann's gray crescent experiment.

Hilde Mangold and Hans SpemannIn 1924Hilde Mangold performed an experiment in which she removed a sectionof the dorsal lip from one species of salamander embryo (before gastrulationstarted) and transplanted it into the belly of another species of salamanderembryo. These two species differed in their pigmentation (the donor being oflighter pigmentation that the recipient). Recall that the dorsal lip is that areawhere cells from the animal pole begin to invaginate into the blastula. Thisprocess leads to the formation of the blastopore and the development of thearchenteron. The blastopore, and hence the dorsal lip of the blastopore, formclose to the boundary of the vegetal hemisphere and the gray crescent. [In Figure4-14b one blastula did not receive the gray crescent. Therefore, it could notdevelop the dorsal lip and ectodermal cells could not invaginate. This resulted inthe blastula never developing into an embryo.]



The recipient salamander embryo now has two dorsal lips. When it developed itformed a second embryo (of a lighter pigmentation) at the location of thetransplant (Figure 4-15).

The transplanted dorsal lip developed into a notochord. In turn the notochordinduced the formation of a neural plate and eventually a neural tube. What thisexperiment is saying is that the ectoderm of the recipient salamander wasinduced by the dorsal lip of the donor to take on a fate that it was not normallydesigned to fulfill. In other words, the dorsal lip acts as an organizer andorganizes the cells of the recipient such that a second embryo is formed. Thisprocess later became known as embryonic (or developmental) induction.

The mesoderm is removed

opposite the dorsal lip ofrecepient embryo.

Mesoderm is removed from a donor embryo near thedorsal lip and transplanted into the recepient embryo.

c=>

Recepient Embryo

Primary

Neural Fold

SecondaryNeural Fold

Recepient Embryo

Primary NeuralDevelopment

Secondary NeuralDevelopment

Donor Embryo

Development of a double embryo.

Figure 4-15Transplant of the dorsal lip.

Hilde Mangold was a student of Hans Spemann when they did this experiment.She later died while cooking in her kitchen when her gas stove blew up. Anumber of years later Spemann got the Nobel Prize for his participation in thiswork.

In another experiment Spemann examined a chain of successive inductions thatleads to the formation of the eye. In the anterior most portion of the neural tube(the forebrain) there are two symmetrical protrusions from the developing brain.These protrusions extend to the ectoderm. Upon contact the ectoderm begins toinvaginate and pushes into the growing optic stalk. This creates the optic cup.The optic stalk will eventuallybecome the optic nerve. The lens (placode)of theeye develops from ectoderm which is in contact with the edges of the optic cup.As the developing lens pulls away, the remaining ectoderm develops into the



cornea. The remainder of the optic cup develops into the retina, which containsthe photoreceptors necessaryfor vision (Figure4-16).

EctodermRetina

Optic StalkInvagination

Figure 4-16Development of the eye.

Experiments like these indicate that there is a hierarchy of developmental stages.During the 1930'sthis gave rise to the field of chemicalembryology.Researchersat the time thought that induction was due to some kind of biochemicalagent. Itturns out that this line of thought was a waste of time and effort. Why?Molecular biology had not yet been invented. The selfish gene needed to beexplored.


BlOlOgy Reproduction &Development Human Embryo Development

Iiuinan EmDii^o D •.tii&ti'.

Recall that fertilization of the egg by the sperm usually takes place in the oviduct(Fallopian tube). Within the first 36 hours following fertilization the zygote willundergo its first mitoticdivision. It will continue to divide as it travels down theoviduct toward the uterus. By the fifth or sixth day the embryo will reach theuterus. At this point the embryo is called a blastocyst. The blastocyst is a hollowball of cells with a mass of cells on one side.

The surrounding cells which form the ball are called trophoblast; the others arecalled the inner cell mass. If the lining of the uterus is prepared to receive theembryo, then the embryo will implant. [Otherwise, the embryo is rejected andsloughed off during the next menstrual period.] After implantation thetrophoblastic cells grow into the lining of the uterus. In fact, the cells of thetrophoblast grow little finger-like projections into the uterus.

As the fetus gets bigger it will need to get nourishment from the mother's blood.The blood vessels from the umbilical cord will grow into the projections oftrophoblast. There is a layer of cells between the trophoblast and the fetal bloodvessels called the chorion. The chorionic villi are the projections offetal/umbilical blood vesselsand the chorion that covers them. The trophoblastabuts the uterine cells of the mother, but allows arteries in the uterine lining todrain into sinuses around the chorionic villi. The chorion preserves the barrierbetween the mother's and fetus's blood (but diffusion of nutrients and wasteproducts may still occur). This whole exchange apparatus is called the placenta.

The inner cell mass will undergo changes similar to those in the frog such thatthe three basic cell layers are formed. First, the inner cell mass forms a cavitywithin itself. This is called the amniotic cavity and the cells which line it are theectoderm. Below the amniotic cavity and its ectoderm, on the hollow side of theblastocyst, a layer of cellswill form along the ectoderm and will eventually coverthe entire hollow blastocyst. This layer is the endoderm. So what we have at thisstage is a hollow ball of cells, the blastocyst, which is surrounded by trophoblast.

Inside this ball there are two compartments; an amniotic cavity defined byectoderm and yolk cavity defined by endoderm. It is the two-layered sheet ofcells formed by the ectoderm and endoderm that will develop into the fetus. Sounlike the frog which developed from the whole ball of cells, the human onlydevelops from the sheet of cells suspended in middle of the blastocyst. The restof the cells of the blastocyst will be important in forming the placenta andchorion which protect the fetus inside the womb.

The primitive streak, equivalent to the neural plate in frogs, forms in theectoderm above the endoderm. Cells from the primitive streak migrate downbetween the ectoderm and endoderm to become mesoderm. Further folding ofthe primitive streak gives rise to a neural groove and then a neural tube, etc. Sothe formation of the primitive streak in mammals marks the beginning ofgastrulation. It is quickly followed by neurulation. The fate of cells from the threebasic cell layers will be the same as in frogs: endoderm will become the GI lining;mesoderm will become connective tissue, bones, muscle, blood; ectoderm willbecome skin and nervous tissue (from the neural grove and neural crest cells),etc.


BlOlOgy Reproduction fir: Development Human Embryo Development

Hormones and PregnancyThe embryo cannot implant if the uterus is not receptive, that is, if it is notquiescent. Furthermore, pregnancy will not be maintained by the uterus evenafter implantation unless it remainsquiescent. So to maintain the uterine liningthrough the first part of pregnancy the trophoblastic cells secrete chorionicgonadotropin. Chorionic gonadotropin (CG) is a hormone which causes thecorpus luteum to continue to produce estrogen and progesterone. As long as thelevels of estrogen and progesterone, particularly progesterone, are sufficient,they keep the uterine lining quiescent and pregnancycontinues. Eventually, theplacenta is able to take over the production of estrogen and progesterone itselfand the corpus luteum is not needed.

The placenta secretes increasing amounts of estrogen and progesterone.However, the levels of estrogen increase faster than the levels in progesterone.And at some point near the end of pregnancy the levels of progesterone plateau.At a crucial ratio of [progesterone]:[estrogen] the uterus is no longer quiescent; itbegins to have contractions.The smooth muscle of the uterus contracts,putting itunder tension. Also the size of the fetus at this stage can put the smooth musclein the walls of the uterus under tension. When the walls of the uterus are putunder tension they send nervous impulses to the hypothalamus. Thehypothalamus sends signals to the posterior pituitary (via nerves) to releaseoxytocin. Oxytocin is released into the blood and is a strong inducer of morecontractions of the smooth muscle of the uterus. Oxytocin also stimulates theproduction/secretion of prostaglandins which further induce contractions.

All of these hormones and nerves form a positive feedback system. Once started(i.e., contraction) each step stimulates the next and eventually stimulates the firstprocess more (stronger contractions). Eventually the contractions force the fetusthrough the vagina and into the hard, cold world. We call the birth processparturition.

Lactation

There are two different processes that go on in lactation. First, there must be milkproduction, and second, there must be milk ejection. Breast milk is a fairlycomplex fluid containing proteins, fats, vitamins, and other goodies (antibodies,etc.). There are epithelial cells within the breast that make up the glands thatproduce the milk. Around these epithelial cells are myoepithelial cells which cancontract around the milk glands, ejecting the milk that has been produced.

Production of milk is stimulated by the suckling of the infant on the mother'snipple. The nipple sends a nervous impulse to the hypothalamus. Thehypothalamus releases PRH, prolactin releasing hormone, which acts on theanterior pituitary. PRH stimulates the anterior pituitary to release prolactin.Prolactin goes into the bloodstream and stimulates the epithelial cells in thebreasts which comprise the milk glands to produce more milk.

Milk ejection uses a similar process, but different hormones. The suckling on thebreast sends a nervous signal to the hypothalamus. The hypothalamus send anervous signal to the posterior pituitary. The posterior pituitary responds byreleasing oxytocin. Oxytocin causes contraction of the myoepithelial cells in thebreast. Milk gets squeezed out.


Reproductionand Development

15 Passages

100 Questions

Passage Titles Questions

I. Spermatogenesis, Oogenesis, and RU486 1-8

II. Male Contraception 9- 15

III. Fertilization of the Ovum 16-21

IV. Pregnancy Fuel Utilization 22-29

V. Female Reproductive Hormones 30-35

VI. Endocrine Control of the Ovarian Cycle 36-43

VII. Gestation 44-49

VIII. Female Birth Control Vaccine 50-56

IX. Estrogen 57-63

X. lsoflavone Experiment 64-69

XL Ovulation 70-76

XII. Vertebrate Gastrulation 77-83

XIII. Oxytocinand Labor 84-90

XIV. Puberty 91 -95

XV. Testicular Cancer 96- 100

The

BERKELEY1 ^ dbJ® 1 X-TR-E-V-I-E-W®


Suggestions

The passages that follow are designed to get you to think in a conceptual manner about the processesof physiology at the organismal level. If you have a solid foundation in physiology, many of theseanswers will be straightforward. If you have not had a pleasant experience with the topic, some of theseanswers might appear to come from the void past the Oort field of the solar system.

Pick a few passage topics at random. For these initial few passages, do not worry about the time. Justfocus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, orgraphs. Third, read each question and the accompanying answers carefully. Fourth, answer thequestions the best you can. Check the solutions and see how you did. Whether you got the answers rightor wrong, it is important to read the explanations and see if you understand (and agree with) what isbeing explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block ofpassages and try them. Be aware that time is going to become important. Generally, you will have about1 minute and 15 seconds to complete a question. Bea little more creative in how you approach this nextgroup. If you feel comfortable with the outline presented above, fine. If not, then try differentapproaches to a passage. For example, you might feel well versed enough to read the questions first andthen try to answer some of them, without ever having read the passage. Maybe you can answer some ofthe questionsby just lookingat the diagrams, charts,or graphs mat are presented in a particular passage.Remember, we are not clones of one another. Youneed to begin to develop a format that works best foryou. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where thelevel of distraction is at a minimum. Get out your watch and time yourself on these passages, eitherindividually or as a group. It is important to have a feelfor time, and how much is passing as you try toanswer each question. Never let a question get you flustered. If you cannot figure out what the answer isfrom information given to you in the passage, or from your own knowledge-base, dump it and move onto the next question. As you do this, make a note of that pesky question and come back to it at the end,when you have more time. When you are finished, checkyour answers and make sure you understandthe solutions. Beinquisitive. If you do not know the answer to something, look it up. The solution tendsto stay with you longer. (For example, what is the Oort field?)


Section IV



>12 86 -100

10-11 79-85

8-9 65-78

7 59-64

6 54-58

5 48-53

<4 0-47

Biology Spermatogenesis, Oogenesis, & RU 486 Passage I


Human development begins with the union of thesperm (male gamete) and the ovum (female gamete)during fertilization. The unicellular organism which isformed is called the zygote.

The formation of the male and female gametes occursduring gametogenesis, a process in which there are twosuccessive meiotic divisions to reduce the chromosome

number from a diploid state to a haploid state.

During spermatogenesis the male germ cells (calledspermatogonia) grow and are transformed into primaryspermatocytes. Each primary spermatocyte forms ahaploid secondary spermatocyte after the first meioticdivision. After the second meiotic division, secondaryspermatocytes form spermatids which are then graduallytransformed by differentiation (spermiogenesis) intomature sperm. Spermatogenesis takes about 64 days.

During oogenesis the female germ cells (calledoogonia) grow and are transformed into primary oocytes.These oocytes begin their first meiotic division beforeparturition but are arrested in prophase until after puberty.This reductive division is not completed until shortlybefore ovulation. As the primary oocyte grows duringpuberty, follicular epithelial cells begin to surround theoocyte, forming a primary follicle. As soon as a secondlayer of epithelial cells is formed the primary folliclebecomes a secondary (or mature) follicle. The secondmeiotic division begins at ovulation but is arrestedmidway through the cycle unless the secondary oocyte isfertilized by a sperm. Once fertilization takes place themature oocyte is referred to as the ovum.

The monthly reproductive cycles of the female isregulated by the gonadotropins, follicle-stimulatinghormone (FSH) and luteinizing hormone (LH). FSHinduces the development of an ovarian follicle. However,both FSH and LH are required for follicular maturation.As the mature follicle grows, it begins to produceestrogen. High estrogen levels produce a surge in thesynthesis of LH and this surge triggers ovulation of thesecondary oocyte. The mature follicle develops into aglandular structure called the corpus luteum and begins tosecrete both progesterone and estrogen. These hormonesbegin to prepare the endometrium of the uterus forimplantation of the blastocyst.

If fertilization occurs, the corpus luteum becomes thecorpus luteum of pregnancy and is maintained for about20 weeks until the placenta can assume production ofprogesterone and estrogen. If fertilization does not occur,the corpus luteum begins to degenerate about 12 daysafter ovulation and becomes the corpus luteum ofmenstruation.


Cellular division does NOT occur in which of the

following cell types?

A. SpermatogoniaB. OogoniaC. SpermatidsD. Primary oocytes

Before fertilization the secondary oocyte is arrestedin meiosis II at:

A. interphase.B. prophase.C. metaphase.D. anaphase.

Roughly 3 days after fertilization in the ampulla ofthe human fallopian tube, the morula:

I. develops a blastocoel.II. contains about 16 blastomeres.

III. passes into the uterus.IV. forms an outer cell layer called the tropoblast.

A.

B.

C.

D.

I and II onlyII and III onlyHI and IV onlyI and IV only

As the mature ovarian follicle begins to swell, asmall oval protrusion called a stigma forms on theside of the follicle juxtaposed to the peritonealcavity. Rupture of the stigma:

I. releases progesterone.II. is due to increased follicular pressure.m. releases a secondary oocyte.IV. results from stimulation by prostaglandins.

A. I onlyB. II and HI onlyC. IV onlyD. I, n, m, and IV

Dispermy is an abnormal fertilization process inwhich two sperm fertilize an ovum. The resultingembryo will:

A. show trisomy 21.B. be triploid and therefore exhibit aneuploidy.C. show monosomy.D. will not exhibit polyploidy.


Biology Spermatogenesis, Oogenesis, & RU 486 Passage I

The molecular structure that BEST representsoxytocin is:

A.

B.

c.

D.

HO-P*HO

CH,I

- CH, — NH2©

H 0© I II

H OI II

H3N-C-C-N-C-C-O-CH3

CH2

coo( ""•NQ>

s—s-

HjN- Cys- Tyr- He— Gin- Asn- Cys

H,N- C— Gly - Leu- ProII

7. RU 486, used widely in France to terminateunwanted pregnancies, was originally tested as anantagonist for glucocorticoid receptors. Duringtesting this compound was also shown to have ahigh affinity for other steroid receptors as well.Administered alone, RU 486 has about an 80%success rate in expulsion of the embryo from theendometrial lining of the uterus. However, when adrug which causes uterine contractions isadministered about 36 to 48 hours after receivingRU 486, the success rate increases by about 26%.RU 486 has a high success rate because it blockssteroid receptors from binding:

A. glucocorticoids and allows administeredprogesterone to cause uterine contractions.

B. testosterone and allows administered oxytocinto cause uterine contractions.

C progesterone and allows administered prostaglandins to cause uterine contractions.

D. estradiol and allows administered glucocorticoids to cause uterine contraction.


8. The molecular structure of the synthetic steroid RU486 is shown in Figure 1.

H,C

RU486:•"/ c= c- CH,

Figure 1

The molecular structures of four naturallyoccurring steroids derived from cholesterol areshown below.

h3c oh

Progesterone Testosterone

Based on the structures of the four steroids shown

above, which receptors will RU 486 bind to withthe strongest affinity?

A. Progesterone and testosterone receptors.B. Estradiol and Cortisol receptors.C Testosterone and estradiol receptors.D. Cortisol and progesterone receptors.


Biology Male Contraception Passage D


Sterilization is a contraceptive option for people whoare positive they wish to have no children. About 500,000men in the U.S. choose to have vasectomies per year.Vasectomy is the surgical ligation of the vas deferens,resulting in sterilization. Although sperm are producednormally in men with vasectomies, the sperm cells are notable to exit the testis. In a local phenomenon, sperm cellsare phagocytosed in the testis by macrophages and otherimmune system cells in men with vasectomies. As a resultof this immune attack, antibodies to sperm are elevated inmen with vasectomies.

9. What is the effect of vasectomy on sperm productionand secretion after several months?

A. Both sperm production and sperm secretionincrease.

B. Both sperm production and sperm secretiondecrease.

C. Neither sperm production nor sperm secretionchange.

D. Sperm production is unaffected and spermsecretion stops.

10. What is a potential consequence of reversing avasectomy?

A. Antibodies to sperm may reduce fertility.B. Antibodies to sperm may increase fertility.C. Birth defects are more likely in subsequent

children.

D. Birth defects are less likely in subsequentchildren.

11. If a man were diagnosed with oligospermia, whatimplications would this have for his fertility status?

A. He could not ever father a child.B. He has reduced fertility status compared to the

population average.C. He has increased fertility status compared to

the population average.D. He has the same fertility status as the

population average.


12. What changes does a vasectomy cause in thehypothalamus, particularly involving the secretion ofFSH and LH?

A. Both FSH and LH are increased.B. FSH is increased, and LH is decreased.C. FSH is decreased, and LH is increased.D. Both FSH and LH are unchanged.

13. What is the anatomical location of spermproduction?

A. Vas deferens

B. Seminiferous tubules

C. Prostate glandD. Testicle interstitium

14. Which of the following statements are TRUE?

I. The sperm count ultimately drops to zerofollowing a vasectomy.

II. The female counterpart of vasectomyhysterectomy, the removal of the uterus.

in. Vasectomy decreases testosterone levels.

A. I onlyB. I and II onlyC. II and HI onlyD. I, II, and III

is

15. In an experiment on dogs, researchers used aninjection of a stable polymer to block the vasdeferens. What results could be expected from thisexperiment?

A. The dogs undergoing this treatment wouldremain fertile.

B. The dogs undergoing this treatment woulddevelop enlarged testicles.

C. The dogs undergoing this treatment wouldbecome sterile.

D. The dogs undergoing this treatment woulddevelop autoimmune disease.


Biology Fertilization Of The Ovum Passage m

Passage HI (Questions 16-21)

In fertile women, ovulation occurs approximately 12-16 days after the onset of the previous menses. The ovummust be fertilized within 24-48 hours if conception is toresult.

Fertilization of the ovum by the sperm usually occursin the mid portion (ampulla) of the uterine tube. Thefertilized ovum, now called a blastocyst, moves down theuterine tube into the uterus. Once in contact with theendometrium, the blastocyst becomes surrounded by anouter layer of syncytiotrophoblast and an inner layer ofcytotrophoblast. The syncytiotrophoblast erodes theendometrium, and the blastocyst burrows into it, a processcalled implantation. After this process, the developmentof the placenta then proceeds.

Human chorionic gonadotropin (hCG) is secreted bythe placenta, and during the early stages of pregnancy,acts to maintain the corpus luteum, an endocrine organthat was maintained by luteinizing hormone (LH) duringthe luteal phase of the menstrual cycle. The corpusluteum, now termed the corpus luteum of pregnancy,primarily secretes estrogens and progesterone. In additionto secreting hCG, the placenta, after about the 6th week ofpregnany, takes over the function of the corpus luteumand produces sufficient estrogen and progesterone tocontinue suppression of gonadotropin releasing hormone(GnRH), follicle stimulating hormone (FSH), and LH.

16. Removal of hCG during the first 6 weeks ofpregnancy would result in:

I. continuation of the pregnancy.n. regression of the corpus luteum.HI. termination of the pregnancy.

A.

B.

C.

D.

I onlyIlonlyI and II onlyII and HI only

17. The protein hormone hCG is very similar but NOTidentical in structure to:

A. LH

B. FSH

C. GnRH

D. Estrogen


18. During pregnancy the levels of progesterone andestrogen:

A. increase during the first trimester, but decreaseduring the remaining six months.

B. increase during the first six months, butdecrease during the last trimester.

C. increase steadily until parturition.D. mimic the levels found during the menstrual

cycle.

19. According to the passage, it can be inferred thatduring pregnancy leutenizing hormone and folliclestimulating hormone levels would:

A. remain relatively low, thereby eliminatingfurther follicle development and ovulation.

B. be undetectable in either fetal or maternal

blood.

C. increase in the maternal blood but decrease inthe fetal blood.

D. progressively increase until ovulation hastaken place.

20. Ovariectomy (removal of the ovaries) or damage tothe ovaries during the last six months of fetaldevelopment would:

A. have no effect at all on the pregnancy.B. lead to termination of the pregnancy.C. result in a decrease in the levels of hCG.

D. result in an increase in follicular development.

21.. If implantation of the blastocyst does not occur in theuterine lining, then the proper hormonal levelsneeded to maintain the lining are removed andmenses normally follows. Which of the followingstructures is sloughed off during menses?

A.

B.

C.

D.

Endometrium

MyometriumPrimary oocytePrimary follicle


Biology Pregnancy Fuel Utilization Passage IV


Human gestation is 40 weeks. Along with the changesapparent to the eye, fuel metabolism changes considerablyfrom early pregnancy (0-25 weeks) to late pregnancy (25-40 weeks). In early pregnancy, the main goal is to buildup maternal stores of adipose tissue. The fetus is quitesmall and has low energy needs during this time. Themother experiences increased insulin sensitivity whichincreases glucose and fatty acid uptake by adipose tissue,especially in the lower body depots.

However, in late pregnancy, the fuel needs of the fetusare increased. The mother becomes insulin resistant, sothat glucose is diverted to the fetus. The growing fetusfeeds continuously. Glucose, received through theplacenta via facilitated diffusion, is its preferred fuel. Themother mobilizes fatty acids from her adipose tissue tomeet her energy needs. Maternal gluconeogenic enzymesare increased, as well, during late pregnancy.

22. When triglycerides are mobilized from adiposetissue, which of the following compounds arereleased into the blood?

I.

II.

III.

A.

B.

C.

D.

H - C- O - C — (CH2),4— CHj

oII

H- C- O-C— (CH,)I4— CH,

oI II

H - C—O-C— (CH2)I4- CH3

H

HI

H-C-OH

IH-C-OH

IH-C-OH

H

©CH3-(CH2)14—COO

I onlyII onlyII and HI onlyI and IQ only


23.

24.

Insulin is important for energy storage during earlypregnancy. Which of the following statements isFALSE regarding the effects of insulin on adiposetissue?

A. Insulin increases synthesis of triglyceride.B. Insulin increases production of oc-glycerol

phosphate.C. Insulin increases uptake of lipid from

circulating lipoproteins.D. Insulin increases synthesis of glucose.

Some women experience gestational diabetes duringlate pregnancy. This is partially due to insulinresistance. Therefore, many obstetricians screen alltheir patients for abnormal glucose tolerance usingan oral glucose tolerance test. An oral load of 50grams (200 kcals) of glucose is given to a fastingperson. A blood sample is taken 1 hour later. Whatchanges in glucose and insulin compared to pretestvalues would be seen in a woman with gestationaldiabetes?

A. Increased glucose, increased insulin.B. Increased glucose, decreased insulin.C. Decreased glucose, decreased insulin.D. Decreased glucose, increased insulin.

25. Which organ produces insulin?

26.

A.

B.

C.

D.

Pancreas

Thyroid glandLiver

Spleen

Which of the following statements are TRUE forlate pregnancy?

I. Gluconeogenic enzymes provide glucose forthe fetus when the mother is not eating.

II. Fatty acids are a required fetal energy source.HI. Maternal fat stores are rapidly filled during

late pregnancy.

A. I onlyB. I and II onlyC. II and HI onlyD. I, n, and m


Biology Pregnancy Fuel Utilization Passage IV

27. How would untreated gestational diabetes mostlikely affect the size of the fetus?

A. Lower birthweight.B. Higher birthweight.C. No change in birthweight.D. During the first trimester fetal weight is

slightly above normal, but at parturition thebirthweight is much lower.

28. Which of the following is TRUE regarding glucosetransport across the placenta?

I. It is an energy requiring process.EL It occurs by facilitated difussion.

EOT. It is higher during early pregnancy.

A. I onlyB. II onlyC. I and ffl onlyD. I and IV only.

29. What type of cells contain gluconeogenic enzymes?

A. Placental cells

B. Neurons

C. HepatocytesD. Intestinal mucosal cells


Biology Female Reproductive Hormones Passage V


The female reproductive system undergoes a series ofregular cyclic changes termed the menstrual cycle. Themost obvious of these changes is periodic vaginalbleeding resulting from shedding of the endometrial liningof the uterus. It results primarily from the interaction ofhormones derived from the hypothalamus, pituitary gland,and ovaries. In most women in the middle reproductiveyears, menstrual bleeding recurs 25-35 days, with amedian cycle length of 28 days. The interval from theonset of menses to ovulation is termed the follicular orproliferative phase. The time proceeding ovulation to theonset of menstrual bleeding is termed the luteal orsecretory phase. Ovulation normally occurs at about the14th day of the cycle.

In normal menstrual cycle, serum concentrations ofboth leutenizing hormone (LH) and follicle stimulatinghormone (FSH) begin to increase prior to menses. FSHconcentrations attain maximum levels during the first halfof the follicular phase and, with the exception of a briefpeak at midcyle, continue to fall until the lowestconcentration in the cycle are reached during the secondhalf of the luteal phase. The preovulatory decline of FSHis due to the increasing concentration of estradiol. LHlevels increase gradually throughout the follicular phaseand at midcyle, there is a large peak in serumconcentration of LH. Subsequently, LH levels graduallydecline reaching their lowest concentration late in theluteal phase.

30. Women treated over a long period of time withrelatively large doses of progesterone and estrogendo not ovulate. This is most likely due to:

I. inhibition of FSH.II. direct inhibition of progesterone by estrogen.DI. over stimulation of FSH by estrogen.

A. I onlyB. II onlyC. I and II onlyD. II and HI only

31. Ovulation is marked by a maximal peak of LHconcentration. This is primarily due to:

A. positive feedback by progesterone.B. positive feedbackby estradiol.C. negative feedback by progesterone.D. negative feedback by estradiol.


32.

33.

34.

35.

According to the passage, the preovulatory declineof FSH is due to:

A. positive feedback of estradiol.B. negative feedback of FSH.C. negative feedback of estradiol.D. positive feedback of FSH secretion.

In the absence of pregnancy, menses normallyoccurs. This may be due to the decline of thehormonal requirement needed by the endometrium.The primary hormones required by theendometrium, so that menses does not occur, are:

I. progesterone.II. estradiol.

m. hCG.

IV. FSH and LH.

A.

B.

C.

D.

I and II onlyI and ffl onlyII and IV onlyffl and IV only

All of the following structures will either secreteestrogen alone, progesterone alone, or estrogen andprogesteronetogether EXCEPT the:

A. anterior pituitary.B. placenta.C. granulosa cells.D. adrenal cortex.

The stimulus for FSH and LH production andsecretion is governed by the pulsatile release ofgonadotropin releasing hormone (GnRH). Whichone of the following structuresproducesGnRH?

A. Anterior pituitaryB. Posterior pituitaryC. HypothalamusD. Pineal gland


Biology Endocrine Control Of Ovarian Cycle Passage VI


Endocrine control of the ovarian cycle is complex,with cases of both positive and negative feedback control.The following diagram illustrates interactions between thehypothalamus, the anterior pituitary, and the ovaries. Thefollowing abbreviations are used in Figure 1: GnRH(gonadotropin releasing hormone), FSH (follicle-stimulating hormone), and LH (luteinizing hormone).

Hypothalamus ^

GnRH

Anterior

Pituitary

r- FSH & LH

Ovaries

T Estradiol

Increased sensitivityof follicles to FSH

Growth of

Follicles

T Estradiol

Jl (+) Feedback

Anterior

Pituitary

LH Surge

Ovaries

Ovulation

^ (Day 14)

"\

f

17

I

Decreased

negative feedbackinhibition

(Day 1) i Estradiol j(Day 28) i Progesterone *\

Corpus luteumregresses

ftOvaries

only'jjiLH

I FSH

Anterior

pituitary

ft4 GnRH

ftHypothalamus

(-) Feedback4^T Estradiol

T Progesterone

ft> Empty follicle will

become the .

corpus luteum J

LH

Figure 1


2a3

>3r p

a-B3

I "a

242

36. What is meant by negative feedback?

A. The rate-limiting enzyme in a pathway isactivated by the product(s) of the pathway.

B. The rate-limiting enzyme in a pathway canonly be inhibited.

C. The rate-limiting enzyme in a pathway isinhibited by the product(s) of the pathway.

D. The rate-limiting enzyme in a pathway canonly be activated.

37. Where is the ovum released upon ovulation?

A. Uterus

B. Uterine tube

C. Cervix

D. Abdominal cavity

38. Thecorpus luteum produces which of the followinghormones?

A. GnRH

B. ProgesteroneC. LH

D. FSH

39. Oral contraceptives contain both estradiol andprogesterone. Usually, the pills are taken for 3weeks and stopped for 1 week to allow menstrualflow. What happens to the ovarian cycle when oralcontraceptives are taken?

I. Inhibition of ovulationn. Decreased FSH and LH

III. Increased growth of follicles

A. II onlyB. II and ffl onlyC. I and II onlyD. I, H, and ffl only

40. Menopause is the cessation of the menstrual cycle.Newfollicles cease to develop in the ovary, althoughthe anteriorpituitary continues to function normally.What levels of FSH and LH would be seen in theblood of a menopausal women who takes noexogenous hormones?

A. Increased FSH, decreased LH.B. Decreased FSH, increased LH.C. Decreased FSH, decreased LH.D. Increased FSH, increased LH.


Biology Endocrine Control Of Ovarian Cycle Passage VI

41. To what class of hormones do estradiol and

progesterone belong?

A. Steroid hormones.

B. Growth hormones.

C. Peptide hormones.D. Glucoregulatory hormones.

42. A hysterectomy is the surgical removal of the uterus.What effect would this have on the cycle depicted inFigure 1?

A. No changes in ovarian cycle, but nomenstruation.

B. Anovulation due to lack of negative feedbackfrom estradiol and progesterone.

C. Decreased development of corpus luteum.D. Prolonged menstrual flow.

43. What is the trigger for ovulation?

A. Regression of corpus luteum.B. Follicle sensitization to FSH.

C. LH surge.D. Increased estradiol.

Copyright © by TheBerkeley Review 243 The Berkeley ReviewSpecializing in MCAT Preparation

Biology Gestation Passage VII


By the 42nd day of gestation, the embryonic gonadsare distinguishable. Under the influence of the genes thatcode for male sex determination, the gonad will begintesticular differentiation by 43-50 days of gestation. In thegonad destined to be an ovary, the lack of differentiationpersists. At 77-84 days, a significant number of germ cellsenter meiotic prophase to characterize the transition ofoogonia into oocytes, which marks the onset of ovariandifferentiation from undifferentiated gonads.

In the 7th week of gestation, the embryo has both maleand female primordial genital ducts. In a normal femalefetus, the Miillerian duct system develops into the uterinetubes, uterus, cervix, and upper one-third of the vagina. Innormal male fetus, the Wolffian duct system on each sidedevelops into the epididymis, vas deferens, seminalvesicles, and ejaculatory ducts.

In the presence of a functional testes, the Miillerianducts involute under the influence of "Miillerian inhibitingfactor" secreted by Sertoli cells. The differentiation of theWolffian duct is stimulated by testosterone secretion fromthe testes. In the presence of an ovary or in the absence ofa functional fetal testis, Miillerian duct differentiationoccurs and the Wolffian ducts involute.

Experimental evidence has accumulated demonstratingthat a small portion of the distal short arm of the Ychromosome, termed H-Y antigen, is critical for testicularorganogenesis of the bipotential gonads. The followingexperiments are a summary that supports this hypothesis.

Experiment J

Using the "moscona" technique, dissociated cells derivedfrom either mouse or rat newborn testes in culturereorganized to form seminiferous tubules. Another grouptreated with anti-H-Y antibody resulted in thereorganization of cells into "follicular-like" structuresrather than seminiferous tubules.

Experiment 2

In a similar experiment, bovine or human fetal XXundifferentiated gonads formed testicular-like structures'when incubated with H-Y antigen.

44. The Sertoli cells, which produce "Miillerianinhibiting factor" are located in the:

A. seminiferous tubules.

B. epididymis.C. vas deferens.

D. corpus luteum.


45. One of the factors involved in the differentiation of

the Wolffian ducts is testosterone. The productionand secretion of testosterone, during this stage ofdevelopment, is under the influence of:

A. follicle stimulating hormone (FSH).B. human placental lactogen (hPL).C. human chorionic gonadotropin (hCG).D. gonadotropin releasing hormone (GnRH).

46.

47.

48.

Testosterone is produced by which of the followingstructures?

I. Corpus luteum.II. Interstitial cells of Leydig.m. Sertoli cells.

A.

B.

C.D.

I onlyIlonlyI and II onlyII and HI only

Treatment of an XY embryo at gestational day 77with anit-Mullerian inhibitingfactor antibody wouldresult in development of:

A. Wolffian duct structures

B. Miillerian duct structures and male externalgenitalia.

C. Wolffian duct structures and male externalgenitalia.

D. Miillerian duct structures and female externalgenitalia.

Administration of anti-H-Y antigen antibody atgestational day 80 to an XY embryo would result in:

A. undifferentiated testes.

B. formation of "follicular-like" structures.C. no change.D. undifferentiated testes and formation of

"follicular-like" structures.

49. Treatment of an XX embryo with H-Y antigenduring gestational day 90 would result indevelopment of:

I. Wolffian duct structures.

II. Miillerian duct'structures.in. female external genitalia.IV. male external genitalia.

A.

B.

C.

D.

I onlyIlonlyI and HI onlyII and III only


Biology Female Birth Control Vaccine Passage vm


A birth control vaccine for women has been

researched by the World Health Organization since 1974.This vaccine promotes antibodies against the hormonehuman chorionic gonadotropin (hCG) which is producedby the embryo. hCG maintains a functional corpusluteum, allowing it to continue to produce estrogen andprogesterone during pregnancy, and promotes normalimplantation of the embryo into the uterine lining.Normal, nonpregnant levels of hCG are 0.000 nmol/L inplasma, and levels increase during the first trimester ofpregnancy.

The vaccine incorporates a synthetic peptiderepresenting the amino acid sequence 109-145 of the C-terminal region of the B subunit of hCG. This particularregion was chosen to avoid homology between hCG andluteinizing hormone (LH). The hCG peptide is conjugatedto diphtheria toxin to form a hapten-protein carriercomplex. Antibodies to hCG are raised with this complexin female baboons and in women, following one injection.The vaccine was tested in 30 sterilized, human subjects,with the following results:

Anti-hCG antibody concentration

GroupVaccine

dose (^g)Dall*

(nmol/L)Day 7

(nmol/L)Week 5

(nmol/L)

1 50 0.000 0.033 0.226

2 100 0.000 0.025 0.268

3 200 0.000 0.022 0.229

4 500 0.000 0.044 0.950

5 1000 0.000 0.058 0.783

Injection

50. How could a researcher determine the concentrationof an antibody in a blood sample?

A. Affinity chromatography using LH.B. Ion exchange chromatography.C. Affinity chromatography using hCG.D. Bradford protein assay.

51. Researchers estimated that a concentration of 0.52nmol/L were required to provide a contraceptiveeffect. Which group achieved this effect by Week 5?

A. Group 1B. Group 2C. Group 3D. Group 4


52. If the study were continued by giving a repeatinjection of the same concentration of vaccine atweek 6, what would happen to the anti-hCGantibody titer?

A. Increase and stay constantB. Decrease and stay constantC. No change in titerD. Increase and then decrease

53. Why is diphtheria toxin used?

A.

B.

C.

D.

To make a larger complex for antibodies torecognize, allowing more effective antibodyformation

To protect LH against antibody formation viacross-reactivityThe toxin prevents implantation of the embryoTo provide simultaneous diphtheria immunityand birth control.

54. At-home pregnancy tests contain monoclonalantibodies in a test kit to which a small volume of

urine is added. Which of these compounds arepresent in urine only during pregnancy?

A. EstrogenB. Human chorionic villi

C. Human chorionic gonadotropinD. Progesterone

55. How would this anti-hCG vaccine affect a woman's

normal menstrual cycles?

A. The presence of hCG would mimic pregnancy,halting menstruation.

B. The anti-hCG antibody would also bind LH,preventing ovulation, and causing irregularmenstruation.

C. Although the vaccine would raise hCG levelsto pregnancy levels, it would allow normalmenstruation.

D. The vaccine would not change normalmenstrual cycles.

56. Which of these compounds can act as an antigen inthe vaccine?

I. The hCG synthetic peptideII. Diphtheria toxin

ra. LH

A.

B.

C.

D.

I onlyIlonlyI and II onlyII and III only


Biology Estrogen Passage IX


Estrogen, a steroid hormone, is not limited in its roleas being one of the major hormones present during awoman's reproductive cycle. Estrogen receptors can befound in liver cells, melanocytes, neurons, vascularendothelial and vascular smooth muscle cells. Aftermenopause, a prolonged estrogen deficiency results. Thisprolonged deficiency has been shown to promoteosteoporosis and atherosclerotic disease. Studies indicatethat women using estrogens after menopause have half therisk of dying of a myocardial infarction when compared tomatched women not using estrogens. Furthermore,women with established coronary artery disease show a70% decrease in mortality over a 10 year period whencompared to similar women who do not receive estrogen.

In the liver, estrogen stimulates enzyme productionthat affects cholesterol catabolism. The overall effect is to

impede the development of a lipid profile consistent withatherosclerosis. In addition, estrogens have direct effectson arterial walls. Recent cell culture studies suggest thatestrogen may inhibit platelet aggregation and adhesionseen in early atherosclerosis.

Endothelial derived relaxing factor (EDRF) is the mostimportant molecule produced by the endothelium. EDRFinhibits myofibril contraction in smooth muscle, leadingto vasodilation. Estrogen appears to stimulate EDRF.EDRF is opposite in function to endothelin, a potentvasoconstrictor released by the vascular endothelium.Estrogens have been shown to induce arteriolar relaxationin which the endothelium has been removed. Subsequentstudies have shown that estrogens block voltage-gatedcalcium channels.

57. An estrogen receptor complex within a cell willmost likely:

A. be carried through the blood in search of atarget organ.

B. result in the production of an intracellularsecondary messenger.

C. create mutations within estrogen regulatedelements of the genome.

D. bind to DNA, resulting in changes intranscriptional rates.

58. EDRF most probably inhibits myofibril contractionby:

A. permitting calcium entry into the muscle cell.B. inhibiting calcium entry into the muscle cell.C. permitting potassium entry into the muscle

cell.

D. inhibiting sodium entry into the muscle cell.


59.

60.

61.

62.

Throughout a women's life, estrogen is producedprimarily by the:

A. uterus.

B. ovaries.

C. anterior pituitary.D. posterior pituitary.

Increasing the diameter of a blood vessel by a factorof 2 results in a resistance which is:

A. /̂2 of the original resistance.B. ' I4 ofthe original resistance.C. J/8 of the original resistance.D. Vi6 of the original resistance.

Estrogen has been shown to induce arteriolarrelaxation in arteries lacking endothelial cells. TheBEST explanation for this is that:

A. EDRF is not a vasodilator.

B. EDRF is not produced by the endothelial cells.C. estrogen directly acts on vascular smooth

muscle to cause relaxation.

D. estrogen stimulates the production of EDRF.

Which of the following effects of estrogen is notconsistent with the idea that estrogens benefit thecirculatory system?

I. Introduction of estrogen leads to a rapiddecrease in vascular resistance.

II. Introduction of estrogen leads to decreasedcatabolism of LDL-cholesterol.

III. Estrogens inhibit endothelial cell expression ofadhesive molecules.

A. I onlyB. II onlyC. I and II onlyD. I and III only

63. Based on information in the passage, it could beconcluded that estrogen:

A. moderately stimulates release of endothelin.B. significantly stimulates release of endothelin.C. inhibits release of endothelin.

D. plays no role in regulating levels of endothelin.


Biology Isoflavone Experiment Passage X


The rate of breast cancer is about 75% lower in FarEastern countries compared to Western countries.Epidemiological studies from migrant populationssuggests that an environmental rather than a geneticexplanation is plausible. When people immigrate to othercountries, their rates of breast cancer come to resemblethe country where they are living rather than their homecountry.

In the Far East, a significant quantity of soybeanprotein is consumed in the diet in many different forms,including beans, miso, tofu, and soy milk. Soy provides arich supply of isoflavones, nonsteroidal compounds withestrogenic properties. Isoflavones are structurally similarto estrogens. They bind to the estrogen receptor and act topartially antagonize estrogen.

In a clinical trial, 6 women were studied during severalmenstrual cycles while living in a research center. Afterrecording usual changes during several menstrual cycles,they began the clinical trial. First they followed a controldiet for 1 menstrual cycle, then they followed another 1-month diet with similar composition of macronutrients,except the protein was largely from soy isolate (60 g of a98 g protein diet).

The effects of a soy diet on the concentrations of LH andFSH as compared to the control diet is shown in Figure 1.

~-4 -3 -2-1 0 1 2 3 4Day of Ovulation

201" o Control »Soy

-4-3-2-101234Day of Ovulation

Figure 1

Table 1 indicates the results of several blood parametersthat were followed during the study.

[Plasma Hormone] Control Diet Soy Diet

Testosterone (nmol/L)

Cholesterol (mmol/L)

1.25 ±0.38

4.27 ±1.08

1.46 ±0.52

3.86 ±1.01

Table 1

The soy protein for this dietary intervention provided45 mg of isoflavones. For comparison, a Japanese dietcontains about 150-200 mg of isoflavones/day.


64. Isoflavones are antagonists to estrogen. What is therole of an antagonist?

A. An antagonist mimics a hormone, interactswith the hormone's receptor, and leads to verysimilar intracellular effects.

B. An agonist mimics a hormone, interacts withthe hormone's receptor, and leads to verysimilar intracellular effects.

C. An antagonist mimics a hormone, interactswith the hormone's receptor, and leads to ablock of the intracellular effects of thehormone.

D. An agonist mimics a hormone, interacts withthe hormone's receptor, and leads to a block ofthe intracellular effects of the hormone.

65. What effect does the soy-protein diet have on theconcentration of FSH and LH at ovulation?

A. The soy diet group has suppressed levels ofFSH and LH compared to the controls.

B. The control diet group has suppressed levels ofFSH and LH compared to the soy group.

C. FSH is suppressed by the soy diet, but LH isunchanged.

D. LH is suppressed by the soy diet, but FSH isunchanged.

66. If the amount of soy protein could be increased tocompletely suppress LH and FSH, what would bethe theoretical effect on the subject?

A. The subject could become pregnant at anyphase of her menstrual cycle.

B. The normal menstrual cycle would continuewithout any change.

C. Menstruation would begin sooner thanpredicted by the past menstrual cycle.

D. The subject would not ovulate.

67. Which of the following statements could you inferfrom the passage?

A. The sole risk factor for breast cancer is geneticbackground.

B. Japanese women and European women havecomparable rates of breast cancer.

C. Exposureto estrogen is a probable determinantof developing breast cancer.

D. Breast cancer is not related to diet.


Biology Isoflavone Experiment

68. What percentage of the protein in the soy diet wasprovided by thesoy isolate?

A. 98%

B. 25%

C. 33%

D. 61%

69. What was the effect of the soy protein diet oncholesterol levels?

A. The soy diet was hypercholesterolemic.B. The soy diet was hypocholesterolemic.C. The soy diet was isocholesterolemic.D. The soy diet was microcholesterolemic.


Passage X


Biology Ovulation Passage XI


By studying physiological changes during themenstrual cycle, a woman may plan a pregnancybasedonthe time interval during which she is fertile. One changethat is frequently tracked by hopeful parents is the basalbody temperature, that is the temperature taken just afterawakening in the morning. The hormone progesteroneleads to a rise in basal body temperature of about 0.5°F.This rise signals ovulation has occurred. The temperatureremains high until menstruation. By developing atemperature log over several months, a woman may usethis information to predict times when she is most likelyto conceive.

Is.n E

•aco

II.5'2

i

"5b

15 7 14 21 28

Days of the Ovarian CycleFigure 1

Another predictor for ovulation is a changes in themucus produced by the cervix. Cervical mucus, under theinfluence of estrogen, becomes more clear and wateryimmediately prior to ovulation, to allow penetration bysperm. The mucus may be placed on a microscope slide,dried, and examined for a pattern the resembles a fernplant, a phenomenon called "ferning". This phenomenonis present at ovulation and signals fertility.


70. What structure is the source of the rise inprogesterone after ovulation?

A. Corpus luteumB. EggC. Corpus albicansD. Cervix

71. Suppose a woman is planning a pregnancy and isplotting her basal body temperature daily. Whatconditions should she observe to make an accuratemeasurement?

A. Eating or drinking nothing before arising.B. Lying still during the measurement.C. Taking measurement at same time daily.D. All of the above.

72. Which hormonal changes precede ovulation?

I. Rise in luteinizing hormone.II. Rise in progesterone.III. Rise in estrogen.

A. I onlyB. I and III onlyC. n and in onlyD. I, II, and m

73. Contraceptive methods either inhibit ovulation, actas barriers to prevent conception, or preventimplantation of a fertilized egg. Which of the followmethods inhibits ovulation?

A. Condoms

B. RU-486

C. Birth control pillsD. Diaphragm

74. If a clinician tells a woman planning a pregnancythat she is exhibiting ferning, what does this mean?

A. She is pregnant.B. She is anovulatory.C. She is infertile.

D. She is ovulating.


Biology Ovulation

75. Which of the following hormone(s) directly triggersovulation?

I. Luteinizing hormoneII. Follicle stimulating hormonem. Estrogen

A. I onlyB. I and II onlyC. H and m onlyD. I, H, and D3

76. What molecule is the precursor for the hormonesestrogen and progesterone?

A. SphingomyelinB. Cholesterol

C. Arachidonic acid

D. Phosphatidyl choline

Copyright © by The BerkeleyReview 250

Passage XI


Biology Vertebrate Gastrulation Passage XII


Gastrulation is one of the most dramatic and crucialstages of vertebrate embryogenesis. The embryo, at firstsimply a hollow ball of cells, undergoes a transformationinto a multilayered structure with a central gut tube andbilateral symmetry. The outer layer of cells formed duringgastrulation is considered ectoderm, while the middle andinner layers are considered mesoderm and endoderm,respectively. Since interactions between these three celllayers will determine the further developmental fate of theembryo, events occuring during gastrulation are vital forthe proper development of the organism.

Gastrulation begins when cells around the embryonicblastopore begin to invaginate, or move towards the insideof the embryo. The site where this invagination initiates isreferred to as the dorsal lip of the blastopore. In the firsthalf of this century, it was learned that the dorsal lip actsas an organizer, inducing the tissue directly around it tobegin invaginating while triggering tissue further away toadopt other specific cell fates.

It was later found that the cells of the dorsal lip secretea diffusible signalling substance that slowly degrades afterits secretion. The result is a concentration gradient of thesubstance, with higher concentrations existing closer tothe source of the secretion. Cells at different distances

from the dorsal lip are exposed to different concentrationsof the substance, leading to the signalling of differentbehaviors. Due to its effect on morphogenesis (the"shaping" of the embryo), this substance was referred toas a morphogen.

77. If the mesodermal cell layer is improperly formedduring gastrulation, which of the following laterstructures would most likely be affected?

A. Nervous systemB. Heart

C. Stomach liningD. Liver

78. Which of the following could NOT act as amorphogen?

A. A small, inorganic compoundB. A steroid hormone

C. A secreted proteinD. A transmembrane protein


79. During invagination, sheets of cells around theblastopore increase in length and extend themselvesinto the interior of the embryo. Which of thefollowing could be responsible for this process ofcell sheet extension?

I. Cell shape changes within the cell sheet.II. Mitotic division within the cell sheet.in. Unidirectional migration of every cell in the

sheet.

A. I onlyB. II onlyC. I and H onlyD. I, H, and HI

80. In an embryo that is about to undergo gastrulation,the outermost cells protect more interior cells fromthe external environment. This is partiallyaccomplished by tight junctions which form near theouter surface of these cells. This outer cell layer ismost similar in structure and function to:

A. epithelial cells.B. neuronal cells.

C. smooth muscle cells.

D. connective tissue cells.

81. The early stages of human embryogenesis look verysimilar to comparable stages in the formation ofother vertebrate embryos. For example, humanembryos have gill slits and a tail at a certain stage indevelopment. This observation does NOT supportwhich of the following statements?

A. Humans evolved from lower vertebrates.

B. Primate embryos should not have gill slits.C. Vertebrates may share a common ancestor.D. There is little selective advantage in altering

these stages of embryogenesis.

82. If the morphogen mentioned in the passage were notslowly degraded after being secreted, which of thefollowing would result?

A. A very steep concentration gradient ofmorphogen would form.

B. The concentration gradient of morphogenwould persist indefinitely.

C. A concentration gradient of morphogen couldnever be formed.

D. There would be no effect on gastrulation.


Biology Vertebrate Gastrulation

83. The dorsal lip from a frog embryo is removed andgrafted opposite the dorsal lip of a second frogembryo. Assuming this was done prior togastrulation, which of the following would belikelyto result?

A. Gastrulation would never occur and theembryo would die.

B. The grafted cells would be rejected by theimmune systemof the host embryo.

C. The embryo would develop into Siamesetwintadpoles, the result of two independentinvaginations during gastrulation.

D. The embryo would gastrulate and developnormally.

Copyright © by The BerkeleyReview 252

Passage XII


Biology Oxytocin and Labor Passage xm


Human pregnancy lasts on average 270 days. Duringthe last month of pregnancy, irregular uterine contractionsincrease in frequency. At the time of labor and delivery,the cervix dilates and softens, and the muscular body ofthe uterus contracts to push out the fetus.

Oxytocin plays both a direct and an indirect role inlabor. It acts directly on uterine smooth muscle cells tomake them contract. It alse^stimulates the production ofprostaglandins in the endometrium of the uterus, whichenhancethe contractionsproduced by oxytocin.

During early labor, the uterus contracts withoutsignificantly higher levels of oxytocin compared toprepregnancy. Once labor is initiated, a positive feedbackloop is set up between the cervix and secretion ofoxytocin. Signals of dilation from the cervix lead tosignals in afferent nerves that promote increased secretionof oxytocin. In late labor, the plasma oxytocin level rises,leading to higher concentrations of oxytocin presented tothe uterus.

Cys- Tyr— He— Gin- Asn- Cys- Pro — Leu- Gly- NH2

I s s 1

Oxytocin

84. Where is oxytocin produced?

A. Posterior pituitaryB. HypothalamusC. Anterior pituitaryD. Placenta

85. During early labor, plasma oxytocin levels are nothigher than prepregnancy levels of 25 pg/ml. Whatmechanism(s) could explain an increase in uterinecontractions without an increased plasma level ofoxytocin?

I. An increase in the number of uterine receptorsfor oxytocin during late pregnancy.

II. Paracrine release of oxytocin that does notcommunicate with the bloodstream.

IH. Increased conjugation of oxytocin in the liver.

A. I onlyB. I and II onlyC. II and HI onlyD. I, n, and ffl


86. Which of the following would NOT increasecontractions in a female in labor?

A. Intravenous injection of oxytocin.B. Intravaginal administration of prostaglandins.C. Intravenous injection of human chorionic

gonadotropin.D. Oraladministration of prostaglandins.

87. What strategy could be used to prevent a pre-termbirth?

A. Administration of an oxytocin secretagogue.B. Administration of prostaglandins such as RU-

486.

C. Administration of inhibitors of prostaglandinproduction.

D. Administration of an oxytocin agonist.

88. A prostaglandin is a twenty-carbon long fatty acidthat contains a five-membered carbon ring. Picturedbelow is the prostaglandin thromboxane B2.

89.

Which fatty acid is the precursor for prostaglandins?

A. Stearic acid

B. Myristic acidC. Linolenic acid

D. Arachidonic acid

Oxytocin is a polypeptide hormone. What does thisimply about its method of interaction with uterinecells?

I. It works through a second messenger system.II. It stimulates mRNA synthesis directly.III. It is binds to a cellular membrane receptor.

A. I onlyB. I and ffl onlyC. II and III onlyD. I, n, and ffl


Biology Oxytocin and Labor

90. Studies indicate that paraplegic women can go intolabor and deliver with few difficulties. What wouldthis finding suggest?

A. Delivery can occur without stretch signalsfrom cervix.

B. Delivery requires voluntary contractions of theabdominal muscles.

C. Prostaglandins play only a minor role in thedelivery of children to paraplegic mothers.

D. Smooth muscle contraction does not occur inthese women.

Copyright © by TheBerkeley Review 254

Passage xm


Biology Puberty


After an interval of childhood quiescence,hypothalamic-pituitary-gonadal activity intensifies in theperipubertal period leading to increased secretion ofgonadal sex steroids that cause secondary sexualdevelopment, the pubertal growth spurt, and fertility. Thisstage of development is often termed puberty oradolescence.

The first sign of puberty in the female is an increase ingrowth. This is accompanied by breast development, aprocess stimulated by increase in estrogen levels. Otherdevelopmental changes influence by estrogen include:enlargementof the labia minora and majora, dulling of thevaginal mucosa, production of a clear whitish vaginalsecretion prior to menarche, and changes in uterine sizeand shape. The development of pubic hair is primarilyinfluenced by adrenal and ovarian androgen secretion.

In males, the first sign of normal puberty is an increasein the size of the testes, primarily due to seminiferoustubular development. The stimulation of the interstitialcells of Leydig also plays a small component in theincrease in testicular size. Pubic hair development isunder the influence of the adrenal gland, as well as thetestes, through secretion androgens.

The dramatic increase in overall body growth duringthe growth spurt at the time of puberty is under theinfluence of complex endocrine control. In addition to sexsteroids, growth hormone (GH) plays an important role inthis phenomenon.

91. Breast development in females is stimulated by anincrease in estrogen secretion. Which of thefollowing hormones must increase in order for thisphenomenon to occur?

A. Thyroid stimulating hormone (TSH)B. Gonadotropin releasing hormone (GnRH)C. Prolactin

D. Progesterone


92.

93.

94.

95.

Passage XIV

Which of the following events does NOT occurwithin the seminiferous tubules of the malereproductive system?

A. Spermatogenesis.B. Testosterone synthesis.C. Secretion of androgen-binding protein by the

Sertoli cells.

D. Enhancement of sperm production by folliclestimulating hormone (FSH).

The development of the seminiferous tubules isunder the influence of:

I. leutenizing hormone (LH).II. follicle stimulating hormone (FSH).III. gonadotropin releasing hormone (GnRH).

A. I onlyB. II onlyC. I and II onlyD. I, II, and III

The stimulation of the interstitial cells of Leydig isunder the direct influence of:

A. follicle stimulating hormone (FSH).B. leutenizing hormone (LH).C. gonadotropin releasing hormone (GnRH).D. Miillerian inhibiting hormone (MIH).

In addition to growth hormone's direct effects onmuscle and bone tissue growth, it also acts indirectlyon these tissues by stimulating the production andsecretion of another hormone or factor which then

acts on these tissues. Which of the followinghormones mediate the indirect effects of growthhormone?

A. Insulin

B. ThyroxineC. Epidermal growth factorD. Insulin-like growth factor I (IGF-I)


Biology Testicular Cancer Passage XV


Cancer of the testes if found in only 1% of malignanttumors of male internal organs. This cancer, whichshowslittle basis in genetics, is ten times more likely to befoundin males afflicted with cryptorchidism, a conditionwherethe testes have failed to descend down into the scrotumduring development. Only 5% of testicular tumors do notoriginate from germ cells. Because germ cells undergospermatogenesis, it is difficult to determine the exactorigin of tumor cells. There is some evidence thatspermatogonia are the likely candidates giving rise totesticular cancer. However, researches have found thatearly carcinoma cells bear more of a similarity to fetalgerm cells than spermatogonia.

It has been discovered, through studying teste tissuesamples from infertile men, that certain morphologicallydistinct cells were precursors to invasive cancer. Thesemalignant cells were termed carcinoma in situ (CIS).Most CIS cells will evolve into invasive cancer. Lookingfor differences between invasive and CIS cells, researchesnoted that CIS cells have a tetraploid complement ofchromosomes.

The initial events transforming germ cells into tumorcells is not well understood, but has been likened toparthenogenesis in the female. Tumors arising fromparthenogenesis, known as teratomas, contain multipletissue types with little organization. It is postulated thattesticular cancers may have similar origins.

96. Which of the following statements supports thetheory that spermatogonia give rise to testicularcancer?

A.

B.

C.

D.

Sertoli cells give rise to tumors which differsignificantly from tumors originating fromgerm cells.Cryptorchidism is a risk factor for testicularcancer.

Testicular cancer is a rare cancer.

Tumor cells do not have a haploidchromosome number.


97.

98.

99.

100.

256

Which of the following statements is TRUEregarding testicular cancer?

A. At least 90% of testicular cancers are found inthe vas deferens.

B. At least 90% of testicular cancers are found inthe seminiferous tubules.

C. At least 90% of testicular cancers are found inthe epididymis.

D. At least 90% of testicular cancers are found inthe rete testes.

Based on information in the passage, one couldargue that germ cell tumors that come from fetalgerm cells have a retarded developmental fate.Which of the following DOES NOT support thishypothesis?

A. Cryptorchidism is a risk factor.B. Evidence shows that tumor development

occurs in gestation.C. In mouse studies, cancer genes act in fetal

germ cells.D. Infertile men run a greater risk for carcinoma

in situ.

How many chromosomes are found in a carcinomain situ cell?

A. 23

B. 46

C. 69

D. 92

Which statement is TRUE regarding the geneticbasis of testicular cancer?

A. Testicular cancer is an autosomal recessive

trait.

B. Testicular cancer is found commonly withinfamilies prone to testicular cancer.

C. Testicular cancer is rare among identical twins.D. Testicular cancer is largely solved by surgical

resection.


Biology Reproduction St Development Section IV Answers

Passage 1(1 - 8) Spermatogenesis, Oogenesis, & RU486

2.

3.

6.

Cis correct, spermatids. Cellular division does not occur in spermatids. During spermatogenesis male germ cellscalled spermatogonia are converted into mature spermatozoa. Asingle diploid (2N) spermatogonia is transformedinto a 1 spermatocyte which then undergo meiosis I to form two haploid (N) V spermatocytes The two 2°spermatocytes undergo meiosis II to form four haploid spermatids. Each spermatid next undergoes a transformationprocess called spermiogenesis to form the mature spermatozoa. The spermatids do not undergo a cellulardifferentiation to form a spermatozoa. They simply differentiate. The whole process of spermatogenesis takes about64 days.

During oogenesis female germ cells called oogonia are transformed into mature ova. This process involves thecellular divisions ofmeiosis I and meiosis II. Early in the development of the fetus oogonia undergo mitosis and aretransformed into 1° oocytes. The diploid (2N) 1° oocyte enters into meiosis I and is arrested at prophase untilpuberty. Ashort time before ovulation meiosis I is completed and a haploid (N) 2° oocyte is formed along with thefirst polar body (which may divide into two more polar bodies). At ovulation the 2° oocyte undergoes meiosis II butis arrested at metaphase. At fertilization the 2° oocyte completes meiosis II and forms the mature ovum and a secondpolar body. Union ofspermatozoa and ovum gives rise to the zygote which is diploid. The correct choice is C.

C is correct, metaphase. See the explanation for oogenesis given above. The correct choice is C.

Biscorrect, II and III only. Fertilization leads to the formation ofa zygote. The zygote undergoes numerous mitoticdivisions to increase the number of cells but not the cytoplasmic mass. When the zygote divides into two cells(about 30 hours after fertilization) those two cells are referred to as blastomeres. About 3 days after fertilization thedividing zygote has formed about 16 blastomeres (a specialized cell-type). At this stage the cell mass is referred toas a morula and it begins to enter into the uterus.

Roughly four days after fertilization a fluid-filled cavity, called the blastocyst cavity or a blastocoel, begins to formin the central portion of the morula. As this cavity forms the cells within the morula arrange themselves into twoparts. One group of cells forms the outer layer of cells called the tropoblast. The other group of cells forms an innercell mass called the embryoblast. The tropoblast gives rise to the placenta while the embryoblast gives rise to theembryo. By this stage in development the morula is referred to as a blastocyst. Around the sixth day afterfertilization the blastocyst (normally) attaches to the endometrium of the posterior wall of the uterus. The correctchoice is B.

D iscorrect, I, II, III, and IV. As the primary follicle begins to develop, an antrum is formed and the primary oocyteis moved to one side. The formation of the antrum signals the development of the secondary follicle. Towards theend of the follicular phase the granulosa cells (i.e., follicle cells) begin to secrete large amounts of progesterone(17oc-hydroxyprogesterone) into the antrum. As the fluid in the antrum increases the pressure builds up and places astress on the stigma. Prostaglandins, which are synthesized and released from almost all cells in the body, can act tostimulate the smooth muscle of the uterus and the smooth muscle surrounding the outer layer of thecacells (calledthe theca externa). Contraction of this muscle can cause an increase in pressure in the antrum and rupture the stigma.The contents of the antrum, including the secondary oocyte and the progesterone, are released into the peritonealcavity (the space outside the ovary). The correct choice is D.

B is correct, be triploid and therefore exhibit aneuploidy. If two sperm fertilize an ovum, then each sperm willcontribute a chromosomal complement of 23. Since the ovum also has a chromosomal complement of 23, thisresults in an embryo that is triploid and contains 69 chromosomes. If there is a deviation from the normalchromosomal complement of 46, then that deviation is referred to as aneuploidy.

Trisomy 21 involves the presence of three chromosomes instead of the usual pair of chromosomes. This type ofchromosomal abnormality is referred to as Down's syndrome and usually results in a germ cell with 24 instead of 23chromosomes. Monosomy involves embryos that are missing a chromosome. Cells which are polyploid containmultiples of the haploid number of chromosomes. The haploid number of chromosomes is 46. If a cell contained 69chromosomes, it would be triploid. A triploid cell is a cell which exhibits polyploidy. The correct choice is B.

D is correct. Oxytocin is considered to be apeptide hormone, as are the other hormones of the hypothalamus andpituitary gland. In general, peptide hormones contain from 3 to over 200 amino acid residues.

Choice A represents adrenaline (epinephrine), a water-soluble amine hormone. It stems from a class of hormonescalled catecholamines. In the brain catecholamines act as neurotransmitters. Choice B is the methyl ester of a



7.

dipeptide called aspartame (trade name is NutraSweet®), an artificial sweetener which is commercially synthesizedin huge quantities. This is not ahormone. Choice C is estradiol, a lipid soluble steroid hormone that is synthesized inthe ovaries. The correct choice is D.

C iscorrect, progesterone and allows administered prosta-glandins to cause uterine contractions. RU 486 resemblesboth progesterone (a progestin) and Cortisol (a glucocorticoid) and will bind to both of their receptors with highaffinity. RU 486 does not resemble testosterone or estradiol as well as it resembles progesterone and Cortisol.Therefore, binding to the receptors of these two steroids is minimal (if at all). The reasoning and structuresassociated with question #7 wouldhelp you with this answer, but it is not necessary.

RU 486 is a progesterone antagonist and binds to progesterone receptors. Recall that progesterone aids in theconversion of the growing endometrial lining of the uterus into a tissue that canaccept a developing embryo (i.e., theblastocyst). Progesterone allows the cells of the endometrium to synthesize and store glycogen, increases thesecretory activity of the endometrial lining, and causes an increase in the vascularization of the tissue. Progesteronealso acts to inhibit expulsionof the implanted embryo by reducing uterinecontractionsand constricting the openingof the cervix. If RU 486 binds to progesterone receptors, then the developing embryo will be facing a hostileenvironment. Furthermore, 36 to 48 hours after RU 486 has been taken by the pregnant woman, prostaglandins areadministered. Recall that prostaglandins (released by most cells of the body) cause contraction of smooth muscle,especially the smooth muscle of the uterus.

Glucocorticoids, such as Cortisol, have a variety of effects on metabolism. In the liver Cortisol increasesgluconeogenesis and glycogen synthesis. In skeletal muscle it decreases glucose uptake, increases proteindegradation, and decreases protein synthesis. In adipose tissue it decreases glucose utilization and increases lipidmetabolism. Glucocorticoids tend to impair wound healing. An antagonist to glucocorticoid receptors might actuallyincrease wound healing. Glucocorticoids do not cause uterine contractions. The correct choice is C.

D is correct, Cortisol and progesterone receptors. Glucocorticoid and progestin receptors bind RU 486 with thegreatest affinity. Look for similarities in the molecules. First, note the three substituents at the C-3, C-11, and C-17carbons of RU 486 (see below). Estradiol has a hydroxyl function at the C-3 position; RU 486 does not. Estradioldoes not have a substituent at the C-l 1 position; RU 486 has a bulky dimethylaminophenyl group. Estradiol only hasa hydroxyl group at the C-17 position. RU 486 has both a hydroxyl and a bulky alkyne residue. A similar analysiscan be done for testosterone. We will find that the substituents on RU 486 do not match well with those on both

estradiol and testosterone. If we now compare RU 486 to both progesterone and Cortisol, we will find moresimilarities. For example, both progesterone and Cortisol have a carbonyl function at the C-3 carbon. They also havea (somewhat) bulky substituent at the C-17 position. The correct choice is D.

">C= C-CH3

Passage II (9 - 15) Male Contraception

9. D is correct, sperm production is unaffected, and sperm secretion stops. This would be a pretty poor form of birthcontrol if sperm secretion was unchanged. Immediately eliminate choices A and C as incorrect answers. Since thevas deferens is cut and closed, no sperm will be present in the semen several months following a vasectomy. Usuallythe patient is checked for aspermiabefore the procedure is considered a success.Sperm production is not affected bya vasectomy. Choice B is incorrect. The correct choice is D.

10. A is correct, antibodies to sperm may reduce fertility. Antibodies to sperm are produced when macrophages enterthe testis to catabolize the extra sperm. These would hamper fertility if they immobilized sperm. Choice B isincorrect. There is no effect of vasectomy on birth defects, since the sperm turnover cycle is about 10 weeks. Theman simply makes new sperm. Choices C and D are incorrect. The correct choice is A.



11.

12.

13.

14.

15.

B is correct, he has reduced fertility status compared to the population average. The prefix "oligo" means few.Oligospermia is the condition of having few sperm. (This is compared to aspermia, the condition of having nosperm). Although this man could impregnate a woman, his chances are less than that of the population average.However, his chances are not zero. Choice A is incorrect. Choices C and D are incorrect. The correct choice is B.

Dis correct, both FSH and LH are unchanged. Since sperm production continues along its normal course, really theonly interruption is in the transport of sperm outside of the body. This has no effect on the hormones controllingsperm production. Choices A, B, and C are incorrect. The correct choice is D.

B is correct, seminiferous tubules. The testis is the site of sperm production. It is compartmentalized into theseminiferous tubules and the interstitium. The interstitium contains androgen-secreting Leydig cells.Spermatogenesis occurs in the seminiferous tubules, not the interstitium. Choice D is incorrect. The vas deferensdrains sperm from the testis. Choice A is incorrect. The prostate gland is not involved in spermatogenesis. Choice Cis incorrect. The correct choice is B.

A is correct, I only. Following a vasectomy, the sperm count will eventually drop to zero, since the sperm cannotexit the testis. Choice I iscorrect. The female counterpart ofvasectomy is the ligation of the fallopian tubes (a tuballigation). Choice II is incorrect. Since sperm production continues normally, there is no reason tosuspect decreasedtestosterone levels. Choice III is incorrect. The correct choice is A.

C is correct, the dogs undergoing this treatment would become sterile. This treatment achieves the same effect as avasectomy. The vas deferens is blocked, so that sperm cannot exit the testis. The effects would be the same asexpected for vasectomy. The dogs would eventually have a sperm count of zero and be sterile. Choice A is incorrect.Normal sperm production continues, but the cells are destroyed by phagocytes in the testis. There would be noenlargement of the testicles. Choice B is incorrect. There is noevidence that an autoimmune disease would develop,since the phagocytic cells are localized in their attack on the sperm cells, not the testis itself. Choice D is incorrect.The correct choice is C.

Passage III (16 - 21) Fertilization of the Ovum

16. D is correct, II and III only (regression of the corpus luteum and termination of pregnancy). The corpus luteum isquite dependent on hCG during the first 8 weeks of pregnancy. During the luteal phase of the uterine cycle thecorpus luteum was maintained by LH, even though the LH levels were relatively low. The corpus luteum isresponsible for the secretion of large amounts of estrogen and progesterone. These two steroids, in concert, inhibitrelease of GnRH. The end result is that FSH and LH levels are kept low and follicular development and ovulationare suppressed. Increased levels of estrogen and progesterone lead to further endometrial development.

After fertilization the outer layer of cells (trophoblast) of the blastocyst begin to secrete hCG. This peptidestimulates and the corpus luteum to continue its synthesis of estrogen and progesterone. Why? Because thedeveloping placenta will not be able to secrete adequate amounts of estrogen or progesterone until about the 3rdmonth of pregnancy. High levels of these two steroids allow the endometrial lining to proliferate and implantation ofthe blsatocyst is maintained. In other words, pregnancy will continue. What happens if hCG is removed?

Removal of hCG during the first 6 weeks of pregnancy means that the corpus luteum is not being stimulated. Thecorpus luteum begins to regress and the output of estrogen and progesterone decreases. Uterine blood vessels beginto constrict, depriving cells of nutrients and oxygen. The uterine lining begins to slough. Pregnancy is terminated. Inturn, GnRH, LH, and FSH levels begin to rise and the cycle begins again. The correct choice is D.

17. A is correct, LH. We can reason out this answer based on information in the first sentence of the second paragraphof the passage. We are told that during the luteal phase of the uterine cycle LH maintains the corpus luteum. Afterfertilization we are told that hCG maintains the corpus luteum. We could assume that if both hormones maintain thecorpus luteum, then they must have similar structures if not similar receptors. It turns out that hCG is a glycoproteinthat consists of an alpha and a beta chain. The alpha chain is almost identical in sequence to the alpha chains ofTSH, FSH, and LH. The beta chain, which determines biologic effect, has a 67% homology with the beta chain ofLH. Consequently, hCG and LH have very similar biologic actions. The correct choice is A.

18. C is correct, increase steadily until parturition. Although FSH and LH levels are low during pregnancy, humanchorionic gonadotropin takes over the functions of LH to stimulate estrogen and progesterone secretion. In additionto the placenta, the fetal and maternal adrenal cortices may also produce estrogen and progesterone. The correctchoice is C.



19. A is correct, remain relatively low, thereby eliminating further follicle development and ovulation. LH and FSH areinhibited by high levels of estrogen and progesterone during pregnancy. Therefore, they remain relatively lowthroughout the duration of pregnancy. The correct choice is A.

20. A is correct, have no effect at all on the pregnancy. The ovaries produce estrogen and progesterone during the first 6weeks of pregnancy, after which, other tissues such as the placenta take over this role. Therefore, pregnancy canoccur in the second term without the ovaries. The correct choice is A.

21. A is correct, endometrium. Unless implantation of the blastocyst occurs, the continuing secretion of progesteroneand estrogen by the corpus luteum is stops. The result is that the glandular epithelial cells that make up theendometrium begin to slough. Immediately below the endometrium is the myometrium, which is the smooth muscletissue of the uterus. A primary oocyte will become surrounded by a layer of granulosa cells to become a primaryfollicle. The primary follicle will be converted to a secondary oocyte, and it is the secondary oocyte that will becomethe ovum. If an ovum is not fertilized, it will be removed in the monthly menstrual flow. Thus, the only structurementioned that is sloughed during menses is the endometrium. The correct choice is A.

Passage IV (22 - 29) Pregnancy Fuel Utilization

22. C is correct, II and III only. Triglycerides in the adipose tissue are hydrolyzed by the enzyme triglyceride lipase.This yields 3 fatty acids and 1 glycerol molecule per triglyceride. All the glycerol escapes to the blood. Some fattyacids may remain in the adipocyte for re-esterification, and the remainder will escape to the blood. The correctchoice is C.

23. D is correct, insulin increases synthesis of glucose. We are looking for the FALSE choice. We have learned fromthe passage that the goal of early pregnancy is storage of energy in adipose tissue. If a person eats, insulin levelsincrease, and some of the energy from the food will be storedas fat. Insulin does increasesynthesis of triglycerideinadipose tissue. It activates triglyceride synthase. Choice A is correct. Glucose that is taken up by the adipocyte isused to make a glycerol phosphate, an activated glycerol backbone for triglyceride synthesis. Choice B is correct.Food lipids are packed into chylomicrons and are taken up by cells, including adipocytes. Insulin increases theactivity of lipoprotein lipase. Choice C is correct. Finally, the false choice is D. The synthesis of glucose isgluconeogenesis, and that occurs in the hepatocytes. Insulin does not increase gluconeogenesis after a meal. Thecorrect choice is D.

24. A is correct, increased glucose, decreased insulin. Insulin resistance means that the insulin in less effective inproducing its effects compared to normal insulin sensitivity. The effects of insulin are to lower blood glucose byactivating glucose transporters and moving glucose into cells for processing. A feed-back loop is set up so thatinsulin will decrease as glucose decreases. Since the woman with gestational diabetes is quite insulin resistant,glucose will remain higherlonger after the test load. Insulin will also remain high because the feed-back loop to turnit down is not completed. The correct choice is A.

25. A is correct, pancreas. Insulin is produced by the betacells in the pancreas. The correct choice is A.

26. A is correct, I only. Glucose is the required fetal fuel source. The. mother builds fat stores during early pregnancy.This makes choices B, C, and D incorrect. The correct choice is A.

27. B is correct, higher birthweight. Since the fetal preferred source of energy is glucose and glucose is higher inwomen with gestational diabetes, the fetus will be able to "overeat". This condition is called macrosomy. Choice Bis correct. This makes choicesA and C incorrect. As for choiceD, we have no way of inferring from the passage (orfrom our own knowledge) that during the first trimester fetal weight will be slightly above normal. Not only that, butthe second part of choice D states that the birthweight is lower when we know it should be higher. The correctchoice is B.

28. B is correct, II only. Glucose transport is by facilitated diffusion, which is passive transport. It occurs throughoutpregnancy, not just following a meal. It is lower in early pregnancy, when the fetal needs are lower, and mother isinsulin sensitive. The correct choice is B.

29. C is correct, hepatocytes. The liver's cells are called hepatocytes. The hepatocyte is the only location ofgluconeogenic enzymes listed here. There is some evidence that the kidney can also do gluconeogenesis, but that isnot an issue in this question.The correct choice is C.



Passage V (30 - 35) Female Reproductive Hormones

30.

31.

A iscorrect, I only (inhibition of FSH). Together, estrogen and progesterone act at the level of the brain to inhibitthe secretion GnRH from the hypothalamus and FSH and LH from the anterior pituitary. Suppression of FSHprevents follicular growth and suppression of LH prevents ovulation. There is no mention inthe passage thatestrogen inhibits progesterone or that estrogen stimulates the synthesis of FSH. The correct choice is A.

Bis correct, positive feedback by estradiol. Estradiol is normally an inhibitor of LH secretion, but at increasinglevels it becomes a stimulus for LH secretion. The LH peak is due to the rise in estrogen levels during thepreovulatory period. The correct choice is B.

32. C is correct, negative feedback of estradiol. Estradiol inhibits FSH secretion. Thecorrect choice is C.

33.

34.

35.

A is correct, I and II only (progesterone andestradiol). Progesterone andestradiol are the main hormones that act onthe endometrium. FSH, LH, and hCG primarily act on the ovaries. The correct choice is A.

A is correct, anterior pituitary. The anterior pituitary is responsible for the release of FSH, LH, growth hormone(GH), thyroid stimulating hormone (TSH), prolactin, and adrenocorticotropic hormone (ACTH). All of thesehormones are protein hormones. This is information that is important to remember. Notice that the steroids estrogenand progesterone are not produced at the level of the brain. Right away this allows us to select choice A as thecorrect answer. The placenta produces both estrogen and progesterone. Granulosa cells (as well as thecal cells) canproduce estrogen. Recall that the granulosa cells are those cells which immediately surround the oocyte while thethecal cells are those cells that surround the granulosa cells. Together, the granulosa and thecal cells are called thefollicle cells. The adrenal cortex helps to define the outer region of the adrenal glands. This tissue secetes a varietyof steroid hormones, including androgens (male sex hormones) and estrogens (female sex hormones). The correctchoice is A.

C is correct, hypothalamus. We have learned that both FSH and LH are synthesized and released from the anteriorpituitary. The release of these two hormones (and the other hormones of the anterior pituitary) are regulated byhypothalamic hypophysiotropic hormones, suchas GnRH. GnRH is released from the hypothalamus onceevery twoto three hours in a pulsatile fashion. In turn, the release of both FSH and LH from the anterior pituitary is seen in apulsatile fashion as well. The poterior pituitary releases vasopressin (ADH) and oxytocin. These two peptidehormones are synthesized within the hypothalamus, packaged into secretory granules, and transported to theposterior pituitary for release. The pineal gland, derived from the roof of the diencephalon and posterior to thethalamus, secretes melatonin, an amino acid derivative. The function of this hormone has not been entirely workedout yet. The correct choice is C.

Passage VI (36 - 43) Endocrine Control of the Ovarian Cycle

36. C is correct, the rate-limiting enzyme in a pathway is inhibited by the product(s) of the pathway. First of all,eliminate choices B and D. Enzymes may be both activated or inhibited, turned up or down. Negative control meansthat the product of a pathway acts to decrease its own synthesis. This means the pathway is inhibited, not activated.Choice A is wrong. The correct choice is C.

37. D is correct, abdominal cavity. The ovum is released from the mature follicle directly into the abdominal cavity. Itis picked up and directed into the uterine tube by thefimbriated ends of the tube. Finally it passes into the uterus andleaves by the cervix if it is not fertilized. The correct choice is D.

38. B is correct, progesterone. From the diagram, we can see that the hypothalamus produces GnRH, so choice A isincorrect. We can also see that the anterior pituitary produces LH and FSH, so choices C and D are incorrect. Thisleaves B, progesterone, which is indeed made by the corpus luteum. The correct choice is B.

39. C is correct, I and II only. Look at the diagram. At around Day 1, the levels of estradiol and progesterone'arefalling. This decreases the negative inhibition on the hypothalamus. Then the normal cycle of follicle developmentand ovulation follows. Therefore, if estradiol and progesterone are kept artificially high with oral contraceptives, thenegative feedback control is not removed. The brief time between cycles of pills does not allow the normal time forall the ovulatory steps to get in sync. So there is no ovulation. Choice I is correct. There is also decreased FSH andLH because the fall in progesterone and estradiol does not occur in pill users. Choice II is correct. Choice III isincorrect, because we have already established that normal follicle development and ovulation does not occur. Thecorrect choice is C.



40. D is correct, increased FSH, increased LH. This is another question regarding feedback control. During thesecretory phase, high levels of estradiol and progesterone feed back on the hypothalamus and decrease GnRH. This,in turn, decreases FSH and LH. After menopause, levels of estradiol and progesterone are low because normalfollicular development has ceased. This means FSH and LH would remain high due to lack of feedback inhibition.This means choices A, B, and C are incorrect. The correct choice is D.

41. A is correct, steroid hormones. The precursor of estradiol and progesterone is cholesterol, which is a steroid. Thismeans choice A is correct. Growth hormone is a specific peptide hormone. So choice B is incorrect. Peptidehormones are made of chains of amino acids, so choice C is incorrect. Finally, glucorcgulatory hormones controlplasma glucose levels. Estradiol and progesterone do not do this. The correct choice is A.

42. A is correct, no changes in ovarian cycle, but no menstruation. Removal of the uterus alone will not affect the cycleinvolving the ovaries, hypothalamus, and anterior pituitary. Therefore, choices B and C are incorrect. There shouldbe no changes in the ovarian cycle. Choice D is incorrect because there can be no menstrual flow without the uterus,which is the source. The correct choice is A.

43. C is correct, LH surge. Although the follicle is prepared by FSH and LH, the final stimulus for ovulation is a surgeof LH about 16 hours before ovulation. Choices B and D precede ovulation, but are preparatory steps, not thetrigger. Choice A is about 10days following ovulation, so it could not be a trigger. The correct choice is C.

Passage VII (44 - 49) Gestation

44. A is correct, seminiferous tubules. Sertoli cells are located within theseminiferous tubules. Theircell body extendsfrom the base of the tubules to thecytoplasm. In addition to producing Miillerian inhibiting factor, theyalso providenutrients to the developing sperm. The correct choice is A.

45. C is correct, human chorionic gonadotropin (hCG). The production of steroid during the early part of gestation isdependent on hCG derived from the placenta. The correct choiceis C.

46.

47.

48.

B is correct, II only (interstitial cells of Leydig). The interstitial cells of Leydig are the testosterone producing cellslocated in the testes, outside of the seminiferous tubules. The correct choice is B.

C is correct, Wolffian duct structures, and male external genitalia. According to the passage, testiculardifferentiation occurs at43-50 days ofgestation and MIF has already taken effect. Therefore, the fetus will developmale genitalia. The correct choice is C.

C is correct, no change. According toexperiment 1, treatment of anti-H-Y antigen to XY, newborn testes would notdifferentiate into seminiferous tubules. In the passage, testicular differentiation occurs at 43-50 days of gestation.Theoretically, anti-H-Yantigen should not havean effectat gestational day 80. The correct choice is C.

49. D is correct, II and III only (Miillerian duct structures and female external genitalia). Similarly, there should not bean effect since treatment occurs beyond the critical period of ovarian differentiation (day 77-84 of gestation). Thecorrect choice is D.

Passage VIII (50 - 56) Female Birth Control Vaccine

50. C is correct, affinity chromatography using hCG. Affinity chromatography consists of an inert column whichsupports compounds that bind your molecule of interest. Since you want anti-hCG, you attach hCG to the support.This will allow binding of your antibody. LH is not the hormone of interest, so choice A is incorrect. Ion exchangechromatography is for separatingchargedmolecules, so choice B is incorrect. The Bradfordprotein assay only givestotal protein, so choice D is incorrect. The correct choice is C.

51. D is correct. Group 4. We can read this answer off the table. Only group 4 had antibody concentrations above the0.52 level. Only group 4 could be considered protected against pregnancy by the anti-hCG antibody. The correctchoice is D.

52. D is correct, increase and then decrease. Compare this to what we know about being immunized. We get a tetanusseries when we are young, and then we get booster shots to elevate our antibody level as it decreases over time. Witha booster, as in this question, wc seen an increase and then a decrease over time. The correct choice is D.


BlOlOgy Reproduction St Development Section IV Answers

53.

54.

55.

Ais correct, to make a larger complex for antibodies to recognize, allowing more effective antibody formation. Thisis called a hapten-carrier complex. Small molecules that an immunologically inert alone are bound to larger,immunologically robust carriers. This elicits the strongest antibody response and allows your small molecule ofinterest to have antibodies produced against it. Choice Bis incorrect, because the toxin does not protect LH. ChoiceC is incorrect, the toxin has no effect on reproductive status. Choice D is incorrect, the goal is not to providesimultaneous immunity, but contraception. The correct choice is A.

Cis correct, human chorionic gonadotropin. hCG is present only during pregnancy. Estrogen and progesterone arepresent throughout pregnancy and also during the normal menstrual cycle. Thiseliminates choices A and D. Humanchorionic villi arepartof the developing tissues accompanying the fetus. The correct choice is C.

D is correct, the vaccine would not change normal menstrual cycles. The vaccine causes the body to produceantibodies to hCG, so that any hCG produced by an embryo is trapped by the antibodies. hCG is critical for theprogression of pregnancy, so the pregnancy is terminated at this stage. Choice A is incorrect, the vaccine does notcontain the hCG molecule, merely a piece of it, made synthetically. Choice B is incorrect, the peptide is chosen toavoid cross-reactivity with LH. Choice C is incorrect, the hCG levels will be reduced to nonpregnant levels, 0.000nmol/L. The correct choice is D. •

56. C is correct, I and II only. The intended antigen is the hCG synthetic peptide. Choice I is correct. However, sincethe diphtheria toxin is present, a small antibody response occurs to it as well, so choice II is correct. LH is ahormone and is not an antigen. Choice III is incorrect. The correct choice is C.

Passage IX (57 - 63) Estrogen

57. D is correct, bind to DNA, resulting in changes in transcriptional rates. As stated in the passage, estrogen is asteroid hormone. Therefore, we must think about how steroid hormones bring about changes on a cellular level.Since steroids are very hydrophobic in nature, they can cross the cell membrane and bind to receptors within thecell'scytoplasm. Thesesteroid/receptor complexes have theability to bind to DNAand alter transcriptional levels ofcertain proteins. This is very different from protein hormones which bind to receptors found on the cell surface.These types of interactions lead to the formation of secondary messengers that bring about change within the cell.Eliminate choice B as a possible answer. Answer choices A and C are a bit non-sensible and do not fit the generalscheme of how hormones bring about change. The correct choice is D.

58. B is correct, inhibiting calcium entry into the muscle cell. Entrance of calcium into a muscle cell is what triggerscontraction. We have seen this in skeletal muscle cell. The question asks what inhibits myofibril (there are manymyofibrils in a muscle cell, and each myofibril contains many sarcomeres) contraction. Well, the only realpossibility should involved the calcium ion. As for sodium and potassium, we have no evidence they are involvedwith the contraction of a myofibril. They will be involved in the nerve transmission to stimulate contraction, but wecannot assume they take part in the actual contraction of the muscle. If we want to inhibit the myofibril contractionwhich will lead to vasodilation, we will want to inhibit calcium entry into the cell. The correct choice is B.

59. B is correct, ovaries. This is based entirely on our knowledge of hormones. Estrogen is produced primarily by theovaries, and its target tissue is the general female reproductive structures. Its principal actions stimulate developmentof secondary sex characteristics in females and growth of sex organs at puberty. It also primes the uterus forpregnancy on a monthly basis. Hormone questions are not going to be tough in you know your hormones, wherethey come from, and what they accomplish. The correct choice is B.

60. D is correct, '/^ of the original resistance. If we recall from physics or discussions of the cardiovascular system,the resistance of a tube is inversely proportional to the radius of the tube to the fourth power. If we increase thediameter by a factor of two, we are increasing the radius by a factor of two. We are decreasing the resistance by 2 tothe fourth power. Thus, the resistance is one sixteenth of what it was previously. The correct choice is D.

61. C is correct, estrogen directly acts on vascular smooth muscle to cause relaxation. We know from the passage thatthe endothelial cells produce this EDRF which causes relaxation of the smooth muscle, and ultimately leads tovessel dilation. Since this is stated in the passage, wc have to take it as the truth and thus we can eliminate choices Aand B. If one removes the endothelial layer, one is removing the ability to produce EDRF. If estrogen still acts as adilator in this condition (no EDRF because no endothelial layer), then estrogen must have the ability to directly acton the smooth muscle to cause its relaxation and ultimately vessel dilation. Considering choice D, we run into afamiliar problem. Estrogen cannot stimulate the production of something that cannot be produced without thepresence of endothelial cells. Since these have been removed, estrogen must act directly on smooth muscle. Thecorrect choice is C.



62. B is correct, II only (introduction of estrogen leads to decreased catabolism of LDL-cholesterol.). Statement I isconsistent with the idea the estrogen benefits cardiac tissue because less resistance to blood flow allows blood toflow easier, allowing your heart to work easier (assuming our heart was working harder than it should). In otherwords, more resistance to blood flow would force our heart to work harder, placing stress on the vital organ.Statement III is consistent with this idea because inhibiting the endothelial cell expression of adhesion moleculesinhibits platelet aggregation and other adhesion that is normally seen in the early stages of atherosclerosis. This ideais that if less "stuff' can stick to the walls of the vessel, the less chance exists of blocking blood flow

Statement II is inconsistent because the statement claims a decreased breakdown, or catabolism of LDL cholesterolis beneficial to the vascular system. An LDL particle is a low density lipoprotein which transports cholesterol fromthe liver to the body cells via the blood. The cells take up these particles through receptor mediated endocytosis.However, high levels of cholesterol will inhibit LDL receptor synthesis, and the particles will remain in thecirculatory system. The build up of cholesterol in the blood is thought to be a prime cause of atherosclerosis, acondition where the arteries are no longer compliant and blood moves though with difficulty. This condition isclearly not beneficial to the vascular system, making statement II inconsistent. The correct choice is B.

63. C is correct, inhibits release of endothelin. Throughout the entire passage, we have seen how estrogen benefits thecardiovascular system. It is stated in the passage that endothelin is a potent vasoconstrictor. In keeping with thetheme of the passage, it is most likely that estrogen will act to block the activity of endothelin. Answer choices Aand B call for estrogen to stimulate this potent vasoconstrictor. This is no evidence for such an interaction, andchoices A and B can be eliminated. Choice D wants us to believe thatestrogen will not affect endothelin. Again,this is not keeping with the theme that estrogen benefits cardiac tissue. Therefore, estrogen will most likely inhibitthe release of endothelin. This has been demonstrated in rabbit coronary arteries. The correct choice is C.

Passage X (64 - 69) Isoflavone Experiment

64.

65.

66.

67.

68.

69.

C is correct, an antagonist mimics a hormone, interacts with the hormone's receptor, and leads to a block of theintracellular effects of the hormone. Choices B and D are incorrect, because they refer to agonists, not antagonists.Choice B is the correctdefinition of an agonist. An antagonist acts against the action of a hormone and blocks itseffects. Choice A is incorrect. The correct choice is C.

A is correct, the soy diet group has suppressed levels of FSH and LH compared to the controls. Examination ofFigure 1 indicates that both FSH and LH levels are decreased (suppressed) in the soy diet trial compared to thecontrol trial. The correct choice is A.

D is correct, the subject would notovulate. Ovulation is triggered by a precise sequence of hormonal events. LH isthe direct stimulus for ovulation. The subject would not ovulate if LH were somehow suppressed (similar to thesuppression ofLH by the birth control pill). This isa theoretical question only. The amount ofsoy protein may nevercompletely hall LH production. The focus of this question is more on ovulation than on soy protein. If the subjectdid not ovulate, she could not become pregnant. Choice Ais incorrect. Since she did not ovulate, there isa change inthe normal menstrual cycle. Choice B is incorrect. One cannot predict when menstruation would occur due to thelack of influence of the corpus luteum. Choice C is incorrect. The correct choice is D.

C is correct, exposure to estrogen is a probable determinant of developing breast cancer. From the passage, welearned that women who moved tocountries other than their native countries gotbreast cancer at rates comparable tonatives of the adopted country. This implies an environmental cause rather than a genetic cause. Choice A isincorrect. The fact of importance in this passage is women in different countries do not contract breast cancer at thesame rate. Choice B is incorrect. Diet can modify estrogen levels, as seen in the experiment. Choice D is incorrect.The correct choice is C.

D is correct, 61%. This is taken directly from a tiny facet in the passage. The diet provided 98 grams of protein, ofwhich 60 grams was the soy protein. 60/98 = 61%. The correct choice is D.

B is correct, the soy diet was hypocholesterolemic. This question tests our knowledge of terminology. Hyper meansincreased. Hypo means decreased. Iso means the same or unchanged. Choiesterolemic means level of cholesterol inthe blood. From the data table, we can see that cholesterol levels in the blood were lowered by the soy-protein diet.The correct choice is hypocholcstcrcmic. Choices A and C are incorrect. Choice D is a nonsense answer and isincorrect. The correct choice is B.



Passage XI (70 - 76) Ovulation

70.

71.

72.

Ais correct, corpus luteum. The corpus luteum secretes estrogen and progesterone after the egg is released from thefollicle. The egg itself does not secrete hormones. Choice Bis incorrect. The corpus is adying corpus luteum whichis no longer secreting hormones. ChoiceC is incorrect. The cervix secretes mucus, but not hormones. Choice D isincorrect. The correct choice is A.

D is correct, all of the above. This answer goes back to our understanding of basal metabolic rate. A basalmeasurement is made at rest, before getting out of bed after a night's sleep, after all your food is digested, whilelying quietly. The correct choice is D.

B is correct, I and III only. Look at Figure 1. We can see that LH and estrogen rise dramatically right beforeovulation. Choice I and III are correct. However, progesterone is low before ovulation and rises dramaticallyafterwards. Choice II is incorrect. The correct choice is B.

73. C is correct, birth control pills. Both condoms and diaphragms are barrier methods of contraception. RU-486prevents continued implantation of a fertilized egg. The levels of estrogen and progesterone in birth control pillsinhibit ovulation, so no egg is produced. The correct choice is C.

74. D is correct, she is ovulating. Ferning is an indicator of ovulation. The woman is therefore not pregnant. Choice Ais incorrect. She is not anovulatory (not ovulating). Choice B is incorrect. She is ovulating and therefore fertile.Choice C is incorrect. The correct choice is D.

75. A is correct, I only.Although FSHandestrogen play critical roles in preparing the follicle and positive feedback onthe hypothalamus, respectively, LH is the actual trigger for release of the egg from the follicle. The correct choiceis A.

76. B is correct, cholesterol. The ring structure of cholesterol is the base molecule for the synthesis of sex hormones inmen and women. Arachidonic acid is a long-chain fatty acid, a precursor of the class of molecules calledprostaglandins. Phosphatidyl choline is a phospholipid, used in membranes. Sphingomyelin is a complex lipid foundin the brain. The correct choice is B.

Passage XII (77 - 83) Vertebrate Gastrulation

77. B is correct, the heart. The question is asking what structure would be affected if the embryonic mesoderm weredefective. This is really a matter of memorizing which structures are derived from each of the germ layers(ectoderm, mesoderm, endoderm). Choice, A, the nervous system, is derived from ectodermal precursors, as is theskin and the lens of the eye. Choice C, the stomach lining, is derived from the endodermal germ layer, which alsogives rise to other linings of the digestive and respiratory tracts, as well as differentiating into the major glands of thebody such as the pancreas and the liver (choice D). This leaves choice B, the heart, as the structure which ismesodermal in origin (along with the notochord, skeleton, muscle, outer coverings of internal organs, andreproductive organs). The correct choice is B.

78. D is correct, a transmembrane protein. From the passage, we learn that a morphogen is defined as a diffusible"substance." Answer choices A, B, and C could all fit this broad definition. Answer choice D, however, is far from adiffusible substance. Rather, a transmembrane protein is anchored into the plasma membrane and cannot either besecreted or diffuse anywhere except within the confines of the membrane. Such a confined protein would not be veryeffective at signalling distant cells. The correct choice is D.

79. C is correct, I and II only. The question inquires about the processes which could bring about the extension, or theincrease in length, of a sheet of cells. Statement I, cell shape changes, is a correct possibility. If individual cellswithin the cell sheet were to change their shape such that they were narrower and longer, the entire sheet would getnarrower and longer, resulting in extension. Statement II is also a valid possibility. Mitotic division of cells withinthe sheet would cause an increase in the total number of cells in the sheet. If these cell divisions were in strategiclocations, the entire cell sheet would become longer and extension would occur. Statement III, on the other hand, isnot a valid explanation for sheet elongation. The unidirectional migration of every cell in the sheet would simplyresult in the entire sheet of cells moving somewhere synchronously. This would not result in the elongation andextension described in the question. The correct choice is C.

80. A is correct, epithelial cells. The question gives us some clues about the type of cells on the surface of the embryo.We learn that they have tight junctions, which are cell-cell adhesions which prevent molecules from slipping in



81.

82.

83.

between these cells, as well helping to establish an apical/basolateral polarity. We also learn that they serve a role ofprotecting inner embryonic cells from the external environment. From these clues alone, the most likely candidatewould be answer choice A, epithelial cells. Epithelial cells can have tight junctions, often to aid them in their majorrole of protecting underlying tissues from arbitrary inward diffusion of lumenal, or gut, contents. We can alsoapproach this problem via a process of elimination. Neuronal, smooth muscle, and connective tissue cells are allhighly differentiated cell types. This question inquires about an embryo which has yet to undergo gastrulation. It isvery unlikely that these advanced cell types would have differentiated at this stage. The correct choice is A.

B is correct, primate embryos should not have gill slits. The question asks which statement is NOT supported by theobservation that human embryos have gill slits and a tail at some time during their development. This observationdoes suggest that humans and the other vertebrates share a common ancestry (choice C) as well as supporting thetheory that humans evolved from lower vertebrates (choice A). The fact that human embryos pass through stageswhich resemble the embryonic stages of lower vertebrates (fish and tailed mammals in particular) suggests thathumans might have evolved from these species. Since thesestages of developmenthave not been drastically alteredfrom fish to human, it could be said that there is little selective advantage in altering these stages of embryogenesis.This supports choice D. Since we are looking for the unsupported statement, by a process of elimination we are leftwith choice B. Since primates are evolutionarily close to humans, we would expect that they would display similarembryonic stages. Therefore, they should havegill slits at some embryonic stage. Incidentally, the conservation ofembryonic stages is often referred to as "ontogeny recapitulating phylogeny," that is, early developmental stagesreiterate the developmental stages of other species in the same phylogenctic tree. The correct choice is B.

C is correct, a concentration gradient ofmorphogen could never be formed. First, let's think about what normallyoccurs. If the morphogen isproduced at a point source (i.e., the dorsal lip cells), it is degraded as it diffuses away.The result isa concentration gradient of morphogen with a maximum near the source, because morphogen moleculesthat get further away are more likely to be degraded because they have been present longer. If the morphogen werenot slowly degraded, as the question postulates, what would happen? The gradient would never form (making choiceB incorrect). This is because the morphogen would diffuse away from the dorsal lip cells and equalize inconcentration throughout the embryo. This would affect gastrulation because proper signalling depends on theestablishment ofa gradient ofmorphogen to which cells at different distances could respond to differently; thereforechoice A is incorrect. Choice Dis likewise invalid because there would be no gradient, especially not a very steepone. The correct choice is C.

C is correct, the embryo would develop into two Siamese twin tadpoles, the result oftwo independent invaginationsduring gastrulation. This question is referring to Hilde Mangold's famous dorsal lip transplantation experiments.First of all, we must keep in mind that the function of the dorsal lip cells is to induce the invagination event thatbegins gastrulation. This is accomplished via the secretion of a morphogen gradient. From this, it follows thatintroducing an additional dorsal lip would induce a separate invagination event, in essence causing the embryo todevelop into two attached organisms (Siamese twins). Answer choice A is incorrect because not only wouldgastrulation occur, it would occur in two separate places. Answer choice Dis incorrect because two dorsal lips inone embryo would lead to two invaginations. Answer choice B is a tempting one, but remember that an embryo atthis early stage doesn't have a differentiated immune system yet. The correct choice is C.

Passage XIII (84 - 90) Oxytocin St Labor

84. B is correct, hypothalamus. Oxytocin is produced within the hypothalamus by neurons and transported to theposterior pituitary, where it is secreted. Choice A is incorrect. Choice C is incorrect. The placenta itselfdoes notproduce oxytocin, but the fetus docs. Choice D is incorrect. The correct choice is B.

85. B iscorrect, I and II only. If the uterus isresponding more strongly to normal, nonpregnant levels ofoxytocin, it isreasonable to expect that the number of oxytocin receptors has increased. This would mean that a small dose ofoxytocin would be very effective at stimulating uterine contractions. ChoiceI is correct. If oxytocin levels did notincrease in the blood, there is the possibility that some sort of paracrine transfer is occurring. This means thathormone is secreted from one cell to anotherwithout actually moving through the bloodstream. Choice II is correct.Usually when things are conjugated by the liver, they are being processed for excretion. If oxytocin wereincreasingly conjugated, then there would be less present, leading to fewer contractions. Choice III is incorrect.Choice B is the correct choice. The correct choice is B.

86. C is correct, intravenous injection of human chorionic gonadotropin. You are looking for the FALSE answer.Injection of oxytocin (intravenously) would increase the circulating levels of oxytocin. This would increase


BlOlOgy Reproduction St Development Section IV Answers

contractions. Choice Ais true. Intravaginal administration ofprostaglandins would increase contractions by actingon the uterus, just as if the prostaglandins were produced by the endometrium. (This is the idea behind oral orvaginal prostaglandin administration to allow abortion.) Choice B is true. Prostaglandins also work orally (seeabove). Choice Dis true. Human chorionic gonadotropin (hCG) is produced during early pregnancy to make sureestrogen and progesterone levels remain high. Although it tapers off after about 10 weeks, it alone would cause nouterine contractions. Choice C is FALSE. The correct choice is C.

87. C is correct, administration of inhibitors of prostaglandin production. Both oxytocin and prostaglandins must bepresent for labor to occur. By eliminating one or the other from the equation, one can avoid an early labor in whichthe fetus is premature. A secretagogue encourages the secretion of something. This would not be favorable in aperson in whom labor threatens. Choice Ais incorrect. An oxytocin agonist would act on the oxytocin receptors justas oxytocin would. Another unfavorable choice. Choice D is incorrect. RU-486 is used to promote abortion. ChoiceB is incorrect. Finally, inhibitors of prostaglandin production, such as indomethacin, can be used to prevent thepositive interactions between prostaglandins and oxytocin. Choice C is correct. The correct choice is C.

88. D is correct, arachidonic acid. This is a bit of trivia, and you will either know it or not. Anyway, stearic is 18:0,myristic is 14:0, and linolenic is 18:2. Only arachidonic has the required 20 carbons (20:4). It is the precursorprostaglandins. Choices A, B, and C are incorrect. Choice D is correct. The correct choice is D.

89. B is correct, I and III only. Peptide hormones do not cross the lipid bilayer directly. They usually interact withcellular membrane receptors and signals are transferred to the intracellular medium by second messengers. Choices Iand III are correct. If the hormone can't enter the cell, then it cannot stimulate mRNA synthesis directly. Choice 11 isincorrect. The correct choice is B.

90. A is correct, delivery can occur without stretch signals from cervix. A paraplegic woman may not be able tovoluntarily contract her abdominal muscles to assist in labor. This is not a requirement. Choice B is incorrect. Thewhole passage discusses the importance of the interactions between oxytocin and prostaglandins. They are required.Choice C is incorrect. The smooth muscle of the uterus contracts without voluntary stimulation. Choice D isincorrect. What is missing in these women is the positivefeedback (through afferent nerves) from cervical stretchingthat increases oxytocin secretion. Choice A is correct. The correct choice is A.

Passage XIV (91 - 95) Puberty

91. B is correct. Gonadotropin releasing hormone (GnRH). Estrogen is under primarily under the control of leutenizinghormone (LH) derived from the anterior pituitary. The production and release of LH, in turn, is under the influenceof gonadotropin releasing hormone (GnRH). Therefore, estrogen production and secretion is indirectly stimulatedthrough the indirect effect of GnRH. The correct choice is B.

92. B is correct, testosterone synthesis. The seminiferous tubules are located in the male testes. These tubules are thesite of sperm production, or spermatogenesis. The important anatomical features of each seminiferous tubule are theLeydig cells, the basement membrane, and the Sertoli cells. The Leydig cells are found between the seminiferoustubules, in the interstitial space. These cells respond to luteinizing hormone (LH) and it helps regulate theconversion of cholesterol to testosterone. Therefore, testosterone synthesis does NOT occur within the seminiferoustubules. Once synthesized, testosterone can cross the basement membrane and influence the Sertoli cells. The Sertolicells secrete androgen-binding protein which binds testosterone (an androgen). This helps to concentratetestosterone within the seminiferous tubules. The Sertoli cells also respond to FSH, which helps to controlspermatogenesis. The correct choice is B.

93. D is correct, I, II, and III. The seminiferous tubules are under the influence of LH, FSH, and GnRH. GnRH, derivedfrom the hypothalamus, stimulates the secretion of LH and FSH from the anterior pituitary. LH stimulates theinterstitial cells of Leydig to produce testosterone. FSH and testosterone, in turn, stimulate the development of theseminiferous tubules and spermatogenesis. The correct choice is I).

94. B is correct, LH. As mentioned in the previous answers, LH binds to receptors on the interstitial cells of Leydig andstimulates the synthesis of testosterone from cholesterol. The correct choice is B.

95. D is correct, Insulin-like growth factor I (IGF-I). The clue to answering this question comes from the question itself.We are looking for the secretion of a hormone ovfactor thai can indirectly act on muscle and bone tissue. The wordfactor might clue us into what we are looking for. Insulin is secreted from the (3-cells of the islets of Langerhans inthe pancreas. Insulin has a direct effect on the cellular uptake of glucose, fatty acids, and amino acids, and their



subsequent conversion into glycogen, triglycerides, and protein. Thyroxine (or T4, tetraiodothyronine) is a hormonethat has a global effect on basal metabolic rate and is critical for normal growth. Thyroxine directly influences thesecretion of growth hormone and helps it to promote protein synthesis and bone growth. Epidermal growth factoracts at the level of skin cells and promotes their growth. Insulin-like growth factor I (IGF-I), also called asomatomedian, is similar to insulin in its structure. IGF-I is synthesized in the liver and acts as a mediator for growthhormone. In other words, GH stimulates the production of IGF-I which in turn promotes things like proteinsynthesis, cellular division, and overall growth. The correct choice is D.

Passage XV (96- 100) Testicular Cancer

96. D is correct, tumor cells do not have a haploid chromosome number. We are looking for a statement that supportsthe idea that germ cell tumors arise from spermatogonia. In the process of spermatogenesis, spermatogonia will giverise to primary spermatocytes. In this process, the genetic complement will be reduced. In other words,spermatogonia have 46 chromosomes while primary spermatocytes have 23. Therefore, if tumor cells did have theirorigin in spermatogonia, they should be diploid cells. The fact that tumor cells do not have a haploid chromosomenumber supports this theory. The correct choice is D.

97. B is correct, at least 90% of testicular cancers are found in the seminiferous tubules. The passage tells us that only5% of testicular cancers do not originate from germ cells. That means that about 95% do originate from germ cells.Where are germ cells located? They are located in the seminiferous tubules of the teste. The correct choice is B.

98.

99.

100.

D is correct, infertile men run a greater risk for CIS. This answer could be arrived at by a process of elimination.Cryptorchidism is a risk factor. From the passage, we know that this condition arises from a developmental error.This supports our theory. The fact that tumors have been shown to develop in gestation supports the theory thatgerm cells tumors have their origins in fetal germ cells and have a misguided development. Furthermore, theevidence that cancer genes act in fetal germ cells supports the theory that tumors come from fetal germ cells andhave a disrupted developmental fate. The only statement which does not support the theory is that infertile men run agreater risk for CIS. This really says nothing about misguided development and its connection to testicular cancer.The correct choice is D.

Dis correct, 92. This question is very straightforward. The number ofchromosomes in a human haploid cell is 23.We are looking for a human cell that has a tetraploid number of chromosomes. Therefore, 23 X 4 = 92chromosomes. The correct choice is D.

C is correct, testicular cancer is rare among twins. The passage informs us that testicular cancer has little basis ingenetics. We are looking for a statement that provides evidencefor this notion. The fact that testicularcancer is rareamong identical twins is such a statement. If one twin developed testicular cancer, and the cancer had its basis ingenetics, it is likely the other twin develops the cancer. Since testicular cancer is rare among twins, this statementssupports the idea that testicular cancer has little basis in genetics. The correct choice is C.


BiologySection V

Endocrinologyand

Immunology

A. The Endocrine System

1. Catecholamine Hormones

2. Peptide Hormones3. Steroid and Thyroid Hormones4. Regulatory Mechanisms5. The Pituitary Gland

B. The Immune System

1. Cell Types2. Antibody Structure3. Antibody Action4. T Cells

5. Humoral and Cellular Immunity


BerkeleyXJ R-E-V-KE-W®


Endocrinology and ImmunologyTop 10 Section Goals

Jjr Cate<amiliar with catecholamine hormones and their actions.

Catecholamines likedopamine,norepinephrine, and epinephrineall containa catecholring (abenzenering with two adjacentnydroxyl groups).They act at thelevel of the CNSand PNS.

Ql^ Be familiar with peptide hormones and their actions.j5r Peptide hormones contain two or more amino acids linked together by a peptide bond. These* hormones are secretedby cells throughout the body and have a variety or actions.

©>im Be familiar the actions of the G protein and adenylate cyclase. __jBr Understand what happens when ahormone like epinephrine binds to a cell surface receptor. Be* familiar with the action of the G protein and conversion of ATP to cAMP by adenylate cyclase.

Q'fm Understand in a general way how phosphorylation events activate proteins.jSr Aclassic example involves the phosphorylation of glycogen phosphorylase by protein kinase A.* Phosphorylated glycogen phosphorylase isactivatedandcanconvert glycogen intoglucose.

§&* Be familiar with steroid hormones and their actions.jjjr The common precursor of all steroid hormones is cholesterol. Steroid hormones are not stored afterI their synthesis, butinstead are used immediately. Know the major groups ofsteroid hormones.

@#H Be familiar with the mechanisms bywhich homeostasis is maintained.jgr Understand the concept ofpositive and negative feedback. Understand what ismeant byendocrine,

•» neuroendocrine, paracrine, and autocrine regulation.

•*Know the peptide hormones of the pituitary.

It is important to know where the8 major hormones of the anterior and posterior pituitary aresynthesized and stored. Know why and how they are released into the circulatory system.

@ijk Understand how a plnripotent stem cell gives rise to the cells of the immune system.jar White blood cells, orleukocytes, are involved inthe body's immune response. The 5different types* ofleukocytes areneutrophils, eosinophils, basophils, monocytes, and lymphocytes.

(§) <&* Understand the basic structure and function of an antibody.Jr Know the5 classes ofantibodies (IgA, IgD, IgE, IgG, andIgM) andwhat they do. Have a general* grasp of their anatomy.

(g)<^ Be familiar withthe differences between humoral and cellular immunity. •jar Understand the workings ofmacrophages, MHC receptors, the different interleukins, theTcells and* Bcells, the different interferons, and howtheyeach interact withforeign antigens.

Biology Endocrinology & Immunology

EndocrinologyCatecholamine Hormones iis&y-

One way that cells can communicate with one another over long distances isthrough extracellular products called hormones. Hormones can generally bedivided into peptide hormones (e.g., insulin), amine hormones (e.g., epinephrineor adrenaline which are classified as catecholamines), and steroids (e.g., themale and female sex hormones). Hormones are released by endocrine organs intothe blood and travel by way of the circulatory system to various target tissues.When a particular endocrine cell is stimulated to release a hormone into theblood the concentration of that hormone increases. Once the stimulation is

terminated the hormone is no longer released the concentration falls back tosome normal resting level. Hormones generally have a very short lifetime in theblood. Even though the time spent in the blood may be relatively short,hormones can act within seconds or they can take hours and quite possibly daysto act on their target tissue. By comparison the action of the nervous system ismuch faster.

When a hormone acts on a target cell it can either do so at the level of the cell'smembrane by binding to a specific receptor in that membrane or by passingthroughthe membrane itselfand bindingwith a specific targetproteinwithin thecell's cytoplasm. After the hormone has bound to its specific receptor thatcomplex undergoes a conformational change that allows for the synthesis of anintracellular messenger. This second messenger passesthe information mediatedby the hormone (i.e., the first messenger) to some specific reaction within thecell. For example, the second messenger might conveyinstructions that allow aparticular set of reactions to release glucose into the bloodor it mighteven act atthe level of the gene to turn off or turn on gene expression.

Let's consider the action of the catecholamine epinephrine (adrenaline) on atypical hepatic (liver) cell. When the body is under some type of stress likephysical exercise or evenfright, an increased needfor glucose arises. Glucose isstored in the body in the form of glycogen and through a series of reactionscanbe mobilized and used as a source of energy during these times of stress.

Once a stress has been perceived the nervous system responds by signaling theadrenal medulla (part of the adrenal gland which sits on top of the kidneys) torelease epinephrine into the extracellular fluid. Epinephrine diffuses into thebloodand is carried to the hepatic cellswhere it binds to a specific cellmembranereceptor called a ^-adrenergic receptor. This action causes the activation of theenzyme adenylate cyclase (bound on the cytoplasmic membrane surface) whichincreases the concentration of the second messenger cyclic adenosinemonophosphate (cAMP)within the cell.

Thebinding of epinephrine, a water solublehormone, to the cellsurfacereceptorand the synthesis of cAMPwithin the cytosol is coupled through the action of aG protein. The name of this proteinstems from the fact that it binds guanosinetriphosphate (GTP) and GDP. GTP and GDP arebothnucleotides justlike ATPand ADP.

,Initially, before epinephrinebinds to its receptor, GDP is bound to a particularsubunit of the G protein. However, once epinephrine binds to its target receptor

Catecholamine Hormones


Biology Endocrinology & Immunology Catecholamine Hormones

this hormone-receptor complex interacts with the G protein and stimulates theexchange of GDP for GTP. The G protein-GTP complex diffuses through themembrane and activates the membrane bound adenylate cyclase enzyme. Inturn, adenylate cyclase catalyzes the conversion of ATP to cAMP. These eventsare shown in the sequences (a) through (d) in Figure 5-1.

Hormone

(Epinephrine)

Active

(d)

ATP cAMP + PPj

Extracellular Space

Membrane of

Liver Cell

Cytosol

Extracellular Space

Membrane of

Liver Cell .

Cytosol

Extracellular Space

Membrane of

Liver Cell

Cytosol

Extracellular Space

Membrane of

Liver Cell

Cytosol

Figure 5-1Action of adenylate cyclase.

After a brief period of time the G protein hydrolyzes the bound GTP to GDP andPi (not shown in Figure 5-1) thus returning the G protein back to its inactivestate (Figure 5-la). However, during this period of activation (Figure 5-ld) the Gprotein-GTP complex was able to activate many adenylate cyclase enzymes thusforming many molecules of cAMP. In other words, by the time G protein is



inactivated the hormonal signal induced by epinephrine has been amplifiedmany times.

Since cAMP is synthesized within the cytosol of the liver cell it can diffuse withinthat cytoplasm and interacts with a variety of molecules. One of the moleculesthat cAMP interacts with is a particular type of protein kinase called proteinkinase A (abbreviated as PKA). A protein kinase is simply an enzyme thattransfers the end phosphate (the gamma phosphate) of ATP to a specific aminoacid residue of a substrate protein. cAMP will bind to PKA and stimulate it tophosphorylate (add a phosphate group) an enzyme called glycogenphosphorylase. Once glycogen phosphorylase has become phosphorylated it isnow an active enzyme and will catalyze the conversion of glycogen into glucose.This is shown in Figure 5-2. Glucose is released into the blood and travelsthroughout the body by way of the circulatory system. [Recall that glycogen issimply the storage form of glucose in mammals. Also, there are other reactionsinvolved in this amplification process. Only the important ones have been shownhere.]

Inactive

(cAMP)PKA

f Glycogen |I phosphorylase J

H

Glycogen c

©

u

P( hGITger 1 Active~ I phosphorylasey

^ Glucose

Figure 5-2Activationby phosphorylation.

Each enzyme in the cascade of reactions shown in Figure 5-2 promotes theactivation of many more molecules in the next step in the sequence. This isexactly what we saw when epinephrine bound to its receptor. This rapidamplification process allows thebinding ofjusta few epinephrine molecules torelease grams ofglucose into the blood. [In Figure 5-1 andFigure 5-2 thesites ofamplification are indicated by asterisks (*).]

Cholera is an intestinal disorder caused by a bacterium (Vibrio cholerae). Themajor symptom of this disorder is diarrhea and if left untreated will result insevere dehydration andeventual death. This toxin binds to the active state of theG protein and prevents the GTP from being hydrolyzed to GDP and Pf. Thismeans that the adenylate cyclase enzyme is continually active and massiveamounts of cAMP are synthesized. cAMP causes the intestinal cells to secrete(digestive) fluids.

Catecholamine Hormones


Biology Endocrinology & Immunology Peptide Hormones

Peptide Hormones -;r •<•:..

The water soluble peptide hormone gastrin stimulates the secretion of HCl andpepsinogen from the stomach in response to stimulation from the vagus nerveand partially digested protein.

Gastrin(thefirstmessenger) binds to a specific cellsurfacereceptor in the plasmamembraneand activates a particular G protein (differentthan the one in Figure5-3). Theactivated Gproteininteracts with themembrane enzymephospholipaseC (abbreviated as PLC) and induces that enzyme to hydrolyze phosphatidyl-inositol-4,5-bisphosphate (PIP2) to inositol-l,4,5-triphosphate (IP3) and 1,2-diacylglycerol (DAG).

Hormone

(Gastrin) N

Extracellular Space

(a) :

(b)

(c)

Protein Protein-P

(inactive) (active)

HCl secretion

into the lumen

of the stomach

Figure 5-3Activation of a receptor by a peptidehormone.

Next, the second messenger, IP3, which was released into the cytoplasm,interacts with the endoplasmic reticulum and stimulates the release of Ca2® into


Biology Endocrinology St Immunology

the cytoplasm by some unknown mechanism. Meanwhile DAG, diffusingthrough the plasma membrane, interacts with a molecule called protein kinase Cand stimulates that kinase, with the help of Ca2® just released from theendoplasmic reticulum, to phosphorylate an unknown protein which in turncauses HCl secretion into the lumen of the stomach. These mechanistic events are

outlined in (a) through (c) in Figure 5-3.

Insulin is a water soluble peptide hormone that binds to a specific transmembrane receptor in the cell membranes of liver, fat, and muscle cells. Onceinsulin binds to the receptor on the cell surface the cytoplasmic portion of thereceptor is converted into a tyrosine kinase that autophosphorylates the aminoacid tyrosine found within the cytoplasmic portion of the receptor. This acts tofurther enhance the activity of the tyrosine kinase. Presumably the insulinreceptor can also internalize and (somehow) act as a second messenger. Thisaction, along with enhanced tyrosine kinase activity, leads to the internalizationof glucose into these cells (see Figure 5-4). The actual events in the insulinsignaling mechanism that leads to the uptake of glucoseare somewhat obscure atthe present time. [Recall that if glucose is in abundance (after a meal), it can bestored in the form of glycogen.]

Hormone

(Insulin) \

Receptor

Tyrosine kinaseactivity

Phosphorylationactivity

Figure 5-4Insulin signaling mechanism.

Extracellular Space

Plasma Membrane

Cytosol

Glucose uptake

Peptide Hormones


Biology Endocrinology St Immunology Steroid & Thyroid Hormones

Steroid & Thyroid Hormones

These are also lipid soluble hormones which can pass through the cell's plasmamembrane and interact with a receptor either in the cytosol or in the nucleus.Thyroid hormones, secreted from the thyroid gland, can diffuse across theplasma membrane and into the nucleus where they bind with specific receptors.The hormone receptor complex then activates transcription essential for certainmetabolic processes. Thyroid hormones help to regulate growth anddifferentiation and they can stimulate the breakdown of proteins, fats, andglucose.

Steroid hormones all originate from a single precursor molecule, cholesterol. Twoimportant classes of steroid hormones are the androgens (male sex hormones )and estrogens (female sex hormones). Estrogen, which is lipid soluble, candiffuse across the membrane of a target cell and bind with a specific receptorwith the cytoplasm. Thissteroid-receptor complex can then enter into the nucleuswhere it influences the transcription of mRNA forcertain protein products. Thegeneral scheme for this actionis shown in Figure 5-5.

Hormone

-C

So

Hormone-ReceptorComplex

Nucleus

Figure 5-5General action of a steroid hormone.

mRNA

Proteins



f,'- ihiii^' f'A

General MechanismsHow is homeostasis maintained in the body? Homeostasis is simply themaintenance of stability within the body. One way that homeostasis ismaintained is by negative feedback. In this case there is some controlled variablewhich is sensed by a particular sensor in the body. Through a variety ofmechanisms these sensors send input to control centers which in turn regulatethe controlled variable (see Figure 5-6). For example, if a certain product wasbeing produced in a large quantity before negative feedback was initiated, thenafter initiation of negative feedback a smaller quantity would be produced.Negative feedback reverses the output from effectors such as glands andmuscles.

Controlled

Variable

(-)fI j/ Sensor I j/

Negative Feedback

Mechanism

Figure 5-6Negative feedback.

Endocrine RegulationIn the endocrine system we might have a gland secreting a hormone into theblood. This hormone, as we have seen, can influence other cells in the body. Forexample, let's consider the regulation of blood glucose levels. Why are the bloodlevels of glucose so important? If glucose supplies to the brain are interrupted foran extended period of time, brain damage can result. Glucose turns out to be theonly fuel utilized by the brain. [That is, except during extended fasting in whichcase the brain can use compounds called ketone bodies.] Thus, the controlledvariable in this regulatory mechanism is the concentration of blood glucose.

Regulatory Mechanisms

B Cells c=> Insulin cBl00d% Liver, Fat, &Muscle cells

-S Uptake of:Glucose

Blood glucoseincreases

[Glucose]

Blood glucosedecreases

A cells c

(-)

(•)

Bloodw^ Glucagon • ^ Liver &

Fat cells

Release of:

^ GlucoseFatty Acids

Figure 5-7

Effectsof glucose on the p-cells and a-cells of the pancreas.

The pancreas secretes two hormones that are important in maintaining theproper levels of blood glucose. Both of these hormones are secreted from clusters


Biology Endocrinology & Immunology Regulatory Mechanisms

of specialized cells called the islets of Langerhans. Insulin is secreted by the (5cells (or B cells) while glucagon is secreted by the a cells (or A cells). After ameal the levels of blood glucose begin to increase. This increase stimulates the Pcells to release insulin into the blood. Insulin binds to specific receptors on liver,fat, and muscle cells and through a complicated mechanism promotes the uptakeof glucose into these cells. This action tends to lower (by negative feedback) theblood glucose levels. See Figure 5-7.

Conversely, if the levels of blood glucose levels begin to decrease below somenormal value, then the a cells are stimulated to secrete glucagon. This hormonecirculates in the blood and binds to specific receptors on liver and fat cells.Glucagon not only stimulates the liver cells to degrade glycogen to glucose but italso stimulates the fat cells to release fatty acids. Fatty acids can be metabolizedand used in the Krebs cyclewhere they supply energy. This helps to alleviate theneed for glucose as an energy source. Through these negative feedback actionsthe levels of glucose will begin to increase. See Figure 5-7.

Neuroendocrine RegulationIn this case the hormone is not release for an endocrine cell but rather from anerve cell which releases its neurotransmitter in the form of a hormone into theblood. For example, the adrenal medulla can receive sensory input from asympathetic nerve, which tells it to release epinephrine into the blood. Otherexamples of neuroendocrine regulation involve the hypothalamus and thepituitary gland. The pituitary gland can be divided into the anterior andposterior pituitary.

Paracrine RegulationIn paracrine regulation the chemical that acts as a signal is released from one celland influences a cell immediately adjacent to it (Figure 5-8). An exampleof sucha paracrine cell would be mast cells, which contain large amounts of histamine.The substances released by the paracrine cell are generally dumped into theextracellular space and not into the bloodstream. Other examples of a paracrinesignal would be neurohormones and neurotransmitters.

Paracrine

Signaling -compound

SecretoryCell

Autocrine

Figure 5-8Paracrine and autocrine regulation.

Autocrine RegulationIn autocrine regulation cells can release certain chemicals which they can thenrespond to themselves (Figure 5-8). For example, certain cells (like tumor cells)can release growth factors which can then bind to specific receptors on themembrane of that same cell.Thus, the cells that released the growth hormone arestimulated to grow. Prostaglandins, which are lipid-soluble chemicals, alsoappear to show autocrine regulation.



Pituitary Gland iipn

Posterior PituitaryThe posterior pituitary releases oxytocin and antidiuretic hormone (Figure 5-9).These two hormones are synthesized in specific cells of the hypothalamus. Themajor effect of oxytocin is to stimulate female uterine contraction whileantidiuretic hormone (ADH) stimulates water reabsorption in the kidneys andalso helps to increase the blood volume (pressure).

Capillaries CZ^

Anterior

Pituitary

TSH, ACTH, FSH, LH

GH.PRL

Hypothalmic SecretoryNeurons

PituitaryStalk

Artery

ftm \3 Capillaries

U/f^y Posterior^̂ ^ Pituitary

OxytocinADH

Figure 5-9The hormones released from the pituitary gland.

The ADH that is synthesized in the nerve cell bodies in the hypothalamus arepackaged into vesicles and transported down the axon to the terminal bouton inthe posterior pituitary. A nerve impulse (which propagated down the same axon)causes the release of these hormones into a system of nearby capillaries. ADHcirculates in the blood and eventually reaches the kidneys where it stimulateswater and Na® reabsorption.

The nerve cell bodies in the hypothalamus where ADH was synthesized cansense a change in blood volume (and a change in the concentration of Na®). Ifthese cells sense a low blood volume, they release ADH into the blood. At thelevel of the kidneys ADH will stimulate the retention of as much water aspossible. In other words, ADH prevents diuresis which is the excessive loss

The Pituitary Gland


Biology Endocrinology &Immunology The Pituitary Gland

urine. By preventing diuresis water can be reabsorbed and the blood volumeincreased.

Anterior PituitaryTheanterior pituitary can secrete six hormones (Figure 5-39). They are thyroidstimulating hormone (TSH), adrenocorticotropic hormone (ACTH), folliclestimulating hormone (FSH), luteinizing hormone (LH), growth hormone (GH),and prolactin (PRL). We will discuss the function of these hormones as weproceed in the course.

These hormones are regulated by a second set of hormones which are stored inhypothalamic nerves. Forexample, let's considerTSH. Within the hypothalamusare nerve cells that contain thyrotropin releasing hormone (TRH). When thisnerve is stimulated it secretes TRH into a set of capillaries which extend into theanterior pituitary. TRH stimulates the synthesis and release of TSH from theanterior pituitary into the blood. TSH binds to specialized receptors in thethyroid gland and causes the release of thyroxine. This hormone, when releasedfrom the thyroid gland, influence metabolismand growth.

If the metabolic rate of the body is too low, TSH is released from the anteriorpituitary and stimulates the release of thyroxine from the thyroid gland. Oncethyroxine increases metabolism to just the right level (homeostasis) it acts backon the nerve cells in the hypothalamus and the cells in the anterior pituitary andinhibits the release of TRH and TSH, respectively. Again, this is another exampleof negative feedback.



Immunology

Every day your body is subject to countless attacks by microorganisms andviruseswhich can cause a variety of diseases. Bacteria can enter into your systemand disrupt cellular function. One example that we have mentioned is massivediarrhea caused by the bacterium Vibrio cholerae. The bacterial toxin from thisorganism can cause over a liter of fluid to be lost every hour. Parasitic viruses canalso reap havoc on the body. Viruses are much smaller than bacteria and containa nucleic acid core (either RNA or DNA) surrounded by a protein coat. Viruseslack their own metabolic machinery. Once they invade an organism they can takeover the host's metabolic machinery for their own use and generate moreprogeny viruses which, after the infected cell lyses, can spread and infect othercells. However, there are certain conditions where the viral genome integratesinto the chromosome of the host cell. In this case the viral genome can bereplicated when the host chromosome is replicated. If this viral genome removesitself from the host genome, it can take over the metabolicmachinery of the hostcell and produce more progeny viruses. Cancer causing viruses can insert theirgenetic material in to host chromosomes as well. Viruses can cause a variety ofdiseases ranging from small pox and influenza to measles and the common cold.

If we were to examine some blood under the microscope, we would find twodifferent cells types—the erythrocytes (red blood cells)and the leukocytes (whiteblood cells). Erythrocytes in adults are produced in the marrow of the sternum,ribs,and vertebrae. Leukocytesare produced partially in the tissues of the lymphand partly in bone marrow. We have already considered the erythrocytes inprevious sections. Let's now briefly consider the leukocytes.

There are 6 types of leukocytes found in the blood. The three types of leukocytesthat we are interested in are the monocytes, neutrophils, and lymphocytes. Themonocytes and neutrophils are considered to be phagocytes. These cells, alongwith mast cells and a variety of other cell types, participate in the immuneresponse.

Mast Cells: These cells are derived from leukocytes and then migrate out into thetissues where they reside. When mast cells are stimulated they release histaminewhich acts on endothelial cells and causes an increased permeability to cells likethe neutrophils. This increased permeability allows the neutrophils easy access tothe surrounding tissue in order to defend against foreign bacteria or viruses.

Phagocytes: As we have mentioned, monocytes and neutrophils are bothphagocytes. When monocytes leave the blood through pores in the blood vesselsand enter into the tissues, they can be transformed into macrophages.Neutrophils circulate in the blood. However, they too can leave the bloodthrough pores in the blood vessels and enter into the tissues. The macrophagesand neutrophils are the primary cell types that attack and destroy foreignbacteria and viruses. They do this by the process of phagocytosis, engulfing theforeign invader by endocytosis. Once inside the phagocyte lysosomal enzymesor hydrogen peroxide is released which can then degrade the foreign objects.

Cell Types


Biology Endocrinology &Immunology Cell Types

Lymphocytes: There are2types oflymphocytes. Lymphocytes whicharederivedfrom the thymus glandare called T lymphocytes(orT cells). The other type oflymphocytes are called B lymphocytes (or B cells). Let's briefly consider eachtype.

T lymphocytes: These cells are responsible for cell-mediated immunity. Thesecells are responsible for the destruction of foreign microorganisms and othersuch agents harmful to the body. There are 3 types of T lymphocytes which wewill be considering: (1) cytotoxic T cells (also called killer T cells), (2) helper Tcells, and (3)suppressor T cells.

B lymphocytes: These cells circulate in the blood and to the lymph organs likethe spleen and lymph nodes and are responsible for humoral mediatedimmunity. [Humor is another way of saying fluid.] Upon infection the Blymphocytes can differentiate into plasma cells which have the ability tosynthesize and secrete antibodies. Antibodies are proteins that are synthesizedin response to an antigen that has been introduced into the body. An antigen issimply a foreign substance (protein or polysaccharide) which has a highmolecular weight that has entered the body and induces a particular immuneresponse. Antigens interact with specific antibodies.

General Immune ResponseLet's consider a general response to the invasion of a bacterium or even a virus.These invaders have on their surface an antigen. The cellular immune responsebegins with a macrophage engulfing a foreign particle and phagocytizing it. Theantigenic fragments released into the cytosol of the macrophage are transportedto the macrophage's membrane where they bind to a specific surface protein.This specific surface protein is called a major histocompatibility complex(MHC) protein (of the Class I type). The antigen is being presented or"displayed" on the surface of the macrophage.

Certain receptors on the class of T lymphocytes called cytotoxic T cells canrecognize this antigen-surface protein complex on the macrophage. When thecytotoxic T cell binds to the antigen-Class I MHC protein complex of themacrophage, a growth factor called interleukin-1 is released by the macrophage.The cytotoxic T cell itself releases a growth factor called interleukin-2.Interleukin-1, interleukin-2, and interleukins released by helper T cells(discussed below) stimulate the synthesis of more cytotoxic T cells. These killer Tcells proliferate and bind to the invading foreign cells bearing the antigen andinduce them to lyse. This type of response is referred to as cell mediatedimmunity.

On the surface of B lymphocytes are Class II MHC proteins and antibodies.When a B lymphocyte finds an antigen that has specifically bound to itsantibody, it engulfs that antigen-antibody complex. After degrading the antigen-antibody complex the B lymphocyte transports a portion of that antigen to theClass II MHC protein and "displays" the complex on the surface of itsmembrane.

Helper T cells with the right receptors are able to bind to the antigen-Class IIMHC protein complex. This binding stimulates the helper T cells to releaseinterleukins (a lymphokine). This stimulates the B lymphocytes to proliferateand form plasma cells. The plasma cells in turn produce a vast amount of


Biology Endocrinology fir: Immunology Cell Types

antibodieswhich are specific for the antigen. When these circulating antibodiesbind to the antigen they actasa tagthatsignals circulating phagocytes toengulftheantigen-antibody complex and destroy it. This type ofresponse is referred toas humoral mediated immunity.

The human immunodeficiency virus (HIV), thought to be responsible for theacquired immune deficiency syndrome (AIDS), acts at the level of the helperTcells by infecting them.


Biology Endocrinology St Immunology Antibody Structure

fc*#^Antibody Structure

Most human antibodies, or immunoglogulins (abbreviated as Ig), are composedof 4 subunits arranged in a "Y" configuration as shown in Figure 5-10. There are 2light chains and 2 heavy chains. These subunits are joined to one another bydisulfide bonds. Within each heavy and light chain are variable domains andconstant domains.

At the terminal ends of the heavy and light chains are variable (V) regions thatcan differ in amino acid sequence from immunoglobulin to immunoglobulin. Theconstant (C) regions of the heavy and light chains are found in the lower portionsof the immunoglobulin. The antigen binding site for a particular antibody islocated at the end of the variable regions of the heavy and light chains.

Two other regions of the immunoglobulin are important. They are the diversity(D) region and the joining (J) region. The genes that code for the J, D, and Vregions of the variable domain greatly increase the diversity one finds among theimmunoglobulins.

Antigenbinding

site

Light Chain

V = variable

D = diversityJ = joining

4^

i„[^>Heavy Chain

B Cell Membrane

^

Antigenbinding

site

Constant

Domains

Disulfide bond

Figure 5-10Generalized antibody structure.

Classes of AntibodiesWe have 5 classes of immunoglobulins which differ in the composition of theirheavy chains.They are (inalphabeticalorder) IgA, IgD, IgE,IgG, and IgM. Theirfunctions are:


BlOlOgy Endocrinology fif Immunology Antibody Structure

IgA: Found in milk and helps to protectnursing infants.IgD: Has an unknown function.IgE: Binds to mast cellsand is involved in the allergic reaction.IgG: Theonly antibody able to cross the placenta. It is also the most abundant

and is produced withindays after the IgMantibody is secreted.IgM: Produced a few days after detection of an antigen and it is the first

antibody produced in response to an antigen.

Combinatorial DiversificationThe light chain contains at least 300 different genes that can determine thevariable region and about 4 different genes that can determine that joiningregion. Therefore, there can be about (300) x (4) or 1,200 different combinationsfor the lightchain. The heavychain contains about 1,000 different genes for thevariable region, about 12 different genes for the diversity region, and about 4different genes for the joining region. Thus, we find that there are about (1,000) x(12) x (4) or 48,000 differentcombinations for the heavychain. If we associate allthe different possiblelight chainswith all the different possible heavychains, wewill geta combinatorial diversification ofabout(1,200) x (48,000) or roughly 5.8 x107 different possible antibody combinations. In other words, the immunesystem can generate an antibody for practically any antigen that invades thebody. [The point behind this is the enormous antibody diversity that isgenerated. Depending on how the calculation is done and what values are usedthediversity canbe as high as 1.1x 1011.]


Biology

Figure 5-13Cell surface coating.

Endocrinology St Immunology Antibody Action

Antibody Action

Antibodies do not destroy foreign antigens. They simply recognize and identifythem. Antibodies can do this is a number of ways. One method is by directlyblocking the foreign invader from gaining access to host cells. This isaccomplished by the antibodies binding to the antigens of, say, a virus as shownin Figure 5-11.

Figure 5-11Direct block.

Complement is a rather complicated system for disposing of invaders. Theessence of this process is as follows: An antibody has already recognized andbound to a specific antigen on a bacterial cell that is considered an intruder. Acomplement protein (a plasma protein) recognizes this antigen-antibody complexand binds to the Fc domain of the antibody. After a series of reactions thecomplement protein is activated and triggers an immune response. Furtherreactions form a membrane attack complex (MAC) that inserts into the bacterialcell's membrane and forms a channel that lets water into the cell. The bacterial

cell swells with water and eventually lyses (bursts). See Figure 5-12.

ComplementProtein

FCdomain

Activation

ComplementProtein

MAC

Water

Figure 5-12Complement.

Antibodies can bind to specific antigens on the surface of a bacterial cell and coatthe cell surface. Once the antibodies have attached to the bacterial cell,phagocytes and/or killer T cells can bind to the terminal portion of the Fcdomain of the antibodies and begin to engulf the foreign invader as shown inFigure 5-13.



Wi$$

The receptors for T cells are composed of two polypeptide chains, each with aconstantand a variabledomain. Within the variable domainin eachpolypeptidechain we find a variable (V) region and a joining (J) region. On one of thepolypeptide chainis a diversity (D) region. See Figure 5-14. This is analogous tothe immunoglobulin structure in that a great amount of (antigen binding)diversity canbe generatedfromdifferent combinations of the genes that producethese polypeptide chains.

V =

D =

J =

Variable

Domain

variable

diversityjoining

T Cell Membrane

Antigen bindingsite

Cytoplasm

Figure 5-14T Cell receptor.

Cytotoxic T cells have cell receptors like the one shown in Figure 5-15. If a virusinfects a host cell, then that virus will begin to take over the host's metabolicmachinery. As this happens some of the viral antigens are transported to thesurface of the host cell where they can complex with a Class I MHC proteinreceptor. Class I MHC receptors are found on almost every one of our cells.

MHC I receptor

"Sc3

MHC I receptorantigen complex

- ViralJk proteins

Figure 5-15Cytotoxic T Cell interaction with MHC I receptor.

The antigen-Class I MHC protein complex is recognized by the cytotoxic T cell.Binding occurs between the cytotoxic T cell and the antigen-bearing host cell asshown in Figure 5-15. The cytotoxic T cells induce lysis in the host cell contain

T Cells


Biology Endocrinology St Immunology T Cells

the viral particles in order to prevent more progeny viral particles fromspreading.

Immature helper T cells recognize macrophages which have presented anantigen on their Class II MHC protein receptors as shown in Figure 5-16.Binding induces themacrophage to synthesize and release interleukin-1 whichacts on the immature helper T cell and causes it to synthesize and releaseinterleukin-2. Interleukin-2 further stimulates the immature helper T cell toproliferate into a mature helper T cell. The mature form of the helperT cellsecretes interleukin-2 which can activate cytotoxic T cells, B cells, and morehelper T cells.

MHC II receptor CZ^

MHC II receptorantigen complex

j= 260 •*ca «a

•5 &

0 60c5 C

<K^> v/ Helper \^•"V T cel1 J

cvxrX v-«*. cytotox

Viral

proteins ^ A

Interleukin-1

Figure 5-16Cytotoxic T Cell interactionwith MHC I receptor.

Activates:

cytotoxic T cells, B cells,helper T cells


BlOlOgy Endocrinology St Immunology Humoral fir Cellular Immunity

Humoral & Cellular Immunity>rO

Generalized ReviewLet's summarize the basic events in humoral and cellular immunity. Thenumbers in these steps willcorrespond to thenumbers in Figure 5-17.

1. Avirus enters thebodyby theblood andisengulfed bya macrophage.

2. On the surface of the macrophage areClass I MHC receptors and ClassII MHC receptors.

3. A Class II MHC receptor presents the viral antigen to the receptor of ahelper T cell.

4. This causes the macrophage to release interleukin-1 (IL-1).

5. IL-1 stimulates the helper T cells to proliferate.

6. The helper T cell is stimulated to release IL-2 which enhancesproliferation of helper T cells.

7. A Bcellwith a Class II MHC proteinpresents viralantigen to helper Tcell.

8. IL-2 releasedfrom helper T cellstimulatesB cells to proliferate.

9. B cells produce plasma cells and memory B cells. Memory B cells"remember" antigen and proliferate faster during a future invasion ofthe same virus.

10. Plasma cells secrete antibodies specific for the viral antigen.

11. The antibodies respond by direct block, complement, and cell surfacecoating.

12. IL-2 from the helper T cells stimulates cytotoxic T cells which havebound to the Class I MHC protein-antigen complex of an infected cellto lyse that infected cell.

13. Interferon is secreted by the infected host cell and acts on the cytotoxicT cell to help enhance the immune response.

14. Cytotoxic T cell also make memory T cells (not shown) which willproliferate faster during a future invasion of the same virus.


Biology Endocrinology & Immunology Humoral fie Cellular Immunity

Class I (2)MHC Receptor CZ^ "

Cell Surface Coating

Figure 5-17Review of humoral and cellular immunity.

Antibodies

(10)

Complement

Mast

Cells

Histamine

LeakyEndothelial

Cells


Endocrinologyand Immunology

15 Passages

100 Questions

Passage TitlesI. Antibody Structure

II. Acquired ImmunityIII. Major Histocompatibility Complex (MHC)IV. IgAAntibody ExperimentV. Complement System

VI. Myasthenia Gravis/Autoimmune DiseaseVII. Glucose, Glucagon, & Insulin

VIII. Calcium, PTH, Calcitonin, and CaleitrioIIX. Erythroblastosis FetalisX. Septic Shock

XI. Calcitonin and OsteoporosisXII. Vertebrate Immune System

XIII. Insulin ReceptorXIV. Vitamin D3, PTH, and CalcitoninXV. Sepsis Syndrome

BerkeleySpecializing in MCAT Preparation

Questions

1 -7

8- 14

15-21

22-27

28-33

34-4041 -46

47-54

55-61

62-67

68-7576-83

84-90

91 -95

97 - 100

Suggestions

The passages thatfollow are designed toget you tothink ina conceptual manner about theprocessesof physiology at the organismal level. If you have a solid foundation in physiology, many of theseanswers willbe straightforward. Ifyouhavenot had a pleasant experience with the topic, someof theseanswers mightappeartocome from thevoidpasttheOort field ofthesolarsystem.

Picka fewpassage topics at random. Forthese initial few passages, do not worry about the time. Justfocus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, orgraphs. Third, read each question and the accompanying answers carefully. Fourth, answer thequestions thebestyoucan. Check thesolutions andsee howyoudid. Whether yougot theanswers rightor wrong, it is important to read the explanations and see if you understand (and agree with) what isbeing explained. Keepa record of your results.

After you feel comfortable with the format of those initial few passages, pick another block ofpassages and try them. Be awarethat timeis going tobecome important. Generally, you willhave about1 minute and 15 seconds to complete a question. Bea little more creative in how you approach this nextgroup. If you feel comfortable with the outline presented above, fine. If not, then try differentapproaches to a passage. Forexample, youmightfeel wellversed enoughto read the questions first andthen try to answersomeof them, withouteverhaving read the passage. Maybe you can answer someofthe questionsby just lookingat the diagrams, charts,or graphs that are presented in a particular passage.Remember, we are not clones of one another. You need to begin to develop a format that works best foryou. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where thelevel of distraction is at a minimum. Get out your watch and time yourself on these passages, eitherindividually or as a group. It is important to have a feel for time, and how much is passing as you try toanswer each question. Never let a question get you flustered. If you cannot figure out what the answer isfrom information given to you in the passage, or from your own knowledge-base, dump it and move onto the next question. As you do this, make a note of that pesky question and come back to it at the end,when you have more time. When you are finished, check your answers and make sure you understandthe solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tendsto stay with you longer. (For example, what is the Oort field?)


Section V



>12 86-100

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8-9 65-78

7 59-64

6 54-58

5 48-53

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Biology Antibody Structure Passage I


In order to protect themselves from pathogens, animalshave developed an array of protective mechanisms withintheir immune system. Immunity can be divided into twotypes: innate (nonspecific) and acquired (specific).

Innate immunity affords protection against a multitudeof different pathogens and is not dependent on specificrecognition of foreign macromolecules. The skin, mucousmembranes, and enzymes in secretions are all examples ofnonspecific immunity.

Acquired immunity enhances the protection providedby innate immunity and is the result of an initial encounterwith a foreign macromolecule (immunization). If animmune response is activated, the agent inducing thatresponse (an immunogen) will initiate a series of reactionsthat lead to the production of antibodies (proteins) that arespecific against the inducing agent. The ability of theinducing agent to combine with an antibody is calledantigenicity, and an agent with this characteristic is calledan antigen. Vaccinations and a recovery from a diseaselike the measles are examples of specific immunity.

Antibodies are globular glycoproteins that participatein the immune response. These immunoglobulins (Ig) arepresent in the y-globulin component of blood serum andare produced by plasma cells. All antibodies are related toone another through a common motif (Figure 1).

Fab

Figure 1

Key for IgG

C = Constant regionV = Variable regionL = Light chainH = Heavy chainFab = antigen binding fragment

Fc = crystallizable fragment•• = Disulfide bonds

= Papain cleaves

Pepsin cleaves


Fab

293

Immunoglobulins fall into five classes (isotypes). Theapproximate percentage of total antibody in the serumimmunoglobulin pool is indicated in Table 1.

Table 1: Antibody Classes

Isotype % of plasma Ig

IgG 70-75

IgA 15-20

IgM 10

IgD <1

IgE = 0.005

In humans IgG is the only antibody that can cross theplacental membrane. IgA is primarily found in externalsecretions such as tears, saliva, colostrum, and milk. IgMis the first antibody produced by the plasma cells duringan immune response and is the predominant antibodyproduced by the fetus during development. IgD has anunknown function, but since it is found as a surfaceprotein in many B cells, it may play a role in B celldifferentiation into a more mature form like plasma cellsor memory B cells. IgE is thought to confer immunity tocertain parasites, such as worms. IgE can also bind tobasophils and mast cells where, in mast cells, it promotesthe release of histamine, a chemical that causes peripheralvasodilation and increased capillary permeability, leadingto anaphylactic reactions.

The fragment of the antibody that binds antigen is theantigen binding domain. The binding occurs between theparatope of the antibody and the epitope (antigenicdeterminant) of the antigen. An antigen may possess just asingle epitope on its surface, making it a unideterminant,univalent antigen. Antigens may also be unideterminantand multivalent, multideterminant and univalent, or multi-determinant and multivalent. Antibody-antigen interactioncan result in a variety of outcomes, including precipitationand agglutination. These complexes can be dissociated byhigh salt concentration or either high or low pH.

1. Which of the following is NOT a component of thenonspecific immune system?

A. Keratinized and epidermal cells.B. Beating of epithelial cell cilia.C. Placental transfer of antibody.D. Normal body temperature.

2. Which of the following types of immunity wouldresult from a vaccination?

A. Active immunity that is naturally acquired.B. Passive immunity that is artificially acquired.C. Active immunity that is artificially acquired.D. Passive immunity that is naturally acquired.


Biology Antibody Structure Passage I

3. The binding of antibody to antigen occurs throughall of the following EXCEPT:

A. Covalent bonding.B. Hydrogen bonding.C. Hydrophobic forces.D. Van der Waals forces.

4. A paratope canbest bedescribed as that area of the:

A. antigenwhich reactswith the antibody.B. antibody designated by the Vh regions of the

Fab.C. antigen which exhibits a multideterminant and

univalent expression.D. antibody which has similar dimensions as the

antigenic determinant.

5.

6.

Papain, a protease extracted from papaya plants,cleaves primarily on the carboxyl side of lysine andarginine amino acid residues. Treatmentof IgG withpapain before IgG is exposed to a unideterminant,multivalent antigen will result in:

A. cross-linking, because the antigen possessesjust one type of determinant but many of them.

B. an Fc fragment and a bivalent Fab fragment,which can participate in cross-linking.

C. antigens that are not cross-linked, becauseeach Fab fragment is monovalent.

D. two monovalent Fab fragments, which canparticipate in cross-linking.

Treponema pallidium, an organism that causessyphilis, can cross the placental membrane and enterinto the fetal circulation. Which of the followingimmunoglobulins would be expected to increase inconcentration in the blood serum of the fetus?

A. IgGB. IgAC. IgMD. IgE


7.

294

Lysozyme inhibits peptidoglycan synthesis in thecell wall of gram negative bacteria. Which of thefollowing immunoglobulins would act agonisticallywith lysozyme?

A. IgGB. IgAC. IgMD. IgE


Biology Acquired Immunity Passage n


Acquired immunity is the resistance to disease that anorganism develops during its lifetime. Immunity may beacquired actively or passively. Active immunity occurswhen an organism is attacked by a foreign substance or amicroorganism (an antigen). Antibodies and specializedlymphocytes are produced by this exposure that confer a"memory" when the foreign agent attacks again at a latertime. Passive immunity occurs when an organism receivespre-formed antibodies from another organism.

Acquired immunity may be achieved through naturalor artificial means. For instance, a person may be exposedto the chicken pox virus, develop a rash, and becomeimmune to chicken pox for the rest of his life. Thisimmunity is actively acquired through a natural means.Conversely, a new chicken pox vaccine would provideartificiallyacquired active immunity. Vaccines do containmicroorganisms that may be dead or weakened. Thesubstances in vaccines stimulate an active immune

response. An example ofpassive natural immunity is thematernal provision of antibodies to the fetus via the bloodand to the neonate via breastmilk. The baby has animmature immune system, and these antibodies functionfor several months as the baby's system matures. Anexample of passive artificial immunity is receivingantibodies from a rabbit that was immune to a spidervenom, as a treatment for a spider bite.

8. When a person is passively immune to a foreignagent, is this a permanent immunity?

A. Yes, receiving antibodies confers permanentimmunity.

B. Yes, making antibodies confers permanentimmunity.

C. No, receiving antibodies confers transientimmunity.

D. No, making antibodies confers transientimmunity.

9. A newborn child is accidentally kissed by hercousin, who is contagious with measles. Why doesthe newborn stay healthy?

A. Maternal antibodies provided immunity tomeasles in this situation.

B. The newborn child quickly made antibodies tothe measles virus.

C. The cousin provided antibodies by kissing thenewborn child.

D. Paternal antibodies provided immunity tomeasles in this situation.


10.

11.

12.

295

The following diagram indicates the primary andsecondary response to an antigen. IgG, a monomer,crosses the blood vessels easily, while IgM, apentamer, does not cross as well.

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Which of the following statements is FALSE?

A. A "booster shot" for an adult exposure totetanus promotes formation of mainly IgG.

B. Before the primary exposure, there are noantibodies to a particular antigen.

C. Both IgG and IgM remain at high levelsfollowing a "booster shot".

D. IgM provides the primary response to anantigen.

A susceptible adult individual received pre-formedantibodies isolated from the serum of an immune

individual. What type of acquired immunity doesthis confer on the first individual?

A. Artificially acquired active immunity.B. Artificially acquired passive immunity.C. Naturally acquired active immunity.D. Naturally acquired passive immunity.

The first vaccine by Edward Jenner in 1798protected against smallpox. He used a related strainof microorganisms that caused cowpox as avaccination against the microorganisms of smallpox.What type of immunity was conferred in this case?

A. Artificially acquired active immunity.B. Artificially acquired passive immunity.C. Naturally acquired active immunity.D. Naturally acquired passive immunity.


Biology Acquired Immunity

13.

14.

Blood serum is subjected to electrophoresis in orderto separatethe proteins.Antibodies are presentin thegamma globulin fraction. The following diagramshowsserumproteins following gel electrophoresis:

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Direction of migration

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In this diagram, which protein is the largest, basedon its migration pattern?

A. albumin

B. a-globulinC. p-globulinD. y-globulin

If a newborn is orphaned at birth, which processwould provide more antibodies for the child?

I. Feed the child breastmilk from another nursingmother.

II. Seclude the child in a sterile hospital unit.m. Vaccinate the child immediately to all

childhood diseases.

A. I onlyB. I and II onlyC. II and III onlyD. I, II, and ffl


Passage n


Biology Major Histocompatibility Complex (MHC) Passage in


Transplantation antigen proteins expressed on thesurface of a cell and recognized by the immune systemare encoded by the major histocompatibility complex(MHC) genes. The MHC genes are of two fundamentaltypes, class I and class II. While mouse and humanpopulations carry more than one hundred forms of themolecules, only between three and six of each class areexpressed.

Both classes of MHC molecules are involved with

antigen processing, which includes the ingestion ofantigens, the fragmentation of antigens into peptides, andthe binding of these peptides to MHC molecules. Theformation of the MHC-peptide complex is a critical eventin the effective elimination of intracellular parasites. Thepeptides associated with MHC class I have invariablybeen found to originate from a cell's own proteins, whilethe peptides found bound to MHC class II are normallylocated on the outer membrane.

The MHC class I protein usually binds peptides thatare eight to nine amino acids long. The two ends of thepeptide chain bind to discrete binding sites located in thecleft of the MHC class I molecule (Figure 1). The bindingcleft of an MHC class II molecule is similar in shape tothat of an MHC class I molecule, but the MHC class IImolecule usually binds peptides in the middle of the cleft(Figure 2).

Peptide chains

Figure 1 Figure 2

MHC class II molecules have been shown to assemblein the endoplasmic reticulum. However, immediately aftertheir synthesis, the MHC class II subunits associate with athird molecule known as the invariant chain. The

invariant chain inhibits peptides from binding to the MHCclass II molecules and causes the MHC class II molecules

to leave the Golgi complex and fuse with endosomes.


Endosomes are membrane vesicles that often containsurface proteins and their associated ligands. Endosomesare known to contain proteases, which are believed todegrade the invariant chain and allow the molecule tobind peptide. However, a small part of the invariant chain(known as CLIP) remains at the binding cleft until a DMmolecule (an MHC class H-like molecule) enters theendosome and binds CLIP, actively removing it from theMHC class II molecule. At this time, the MHC class IImolecule, bound with peptide, is placed on the cellsurface.

15.

16.

17.

18.

The amino acid sequence of both the MHC class Iand MHC class II molecules should show:

A. hydrophilic regions.B. hydrophobic regions.C. both hydrophobic and hydroapathetic regions.D. both hydrophilic and hydrophobic regions.

On the surface of a cell infected by a virus, themajority of MHC class I molecules have boundpeptides that originate from:

A. the virus.

B. a killer T cell.

C. a B cell.

D. a host cell.

The two protein subunits that constitute the MHCclass II molecule are MOST likely to contain a:

A. Shine-Dalgarno sequence.B. signal peptide sequence.C. pyrimidine-rich sequence.D. purine-rich sequence.

Endosomes are formed by which of the followingprocesses?

A. MegacytosisB. ExocytosisC. EndocytosisD. Transcytosis


Biology Major Histocompatibility Complex (MHC) Passage III

19. The protein DM is structurallysimilar to:

A. the invariant chain.

B. MHC class II.

C. CLIP.

D. MHC class I.

20.

21.

Which of the following statements is true regardingpeptides bound to MHC class I and MHC class IIproteins?

I. Peptides bound to MHC class I should exhibita greater size variation.

II. Peptides bound to MHC class II should exhibita greater size variation.

III. The amino acids binding to a cleft should beconserved despite the variety of peptides thecleft can bind.

A. I onlyB. II onlyC. I and III onlyD. II and in only

Which of the following is LEAST likely to be adefense mechanism used by a pathogen to deter theantigen processing system?

A. Suppression of MHC molecules early in aninfection

B. Production of molecules that bind to MHC

class I molecules in the ER and prevent cellsurface expression

C. Production of a transcription factor thatincreases the transcription rate of the MHCgene early in an infection

D. Suppression of the DM molecule


Biology IgA Antibody Experiment Passage IV


In the following experiment, researchers studied therole of the cytokine, interleukin-6, as a factor in theresponse of immunoglobulin A (IgA) to foreign moleculesin mice. IgA is the antibody group that is released fromepithelial surfaces in secretions such as saliva or breastmilk. IgA often represents a first line of defense againstinvading pathogens.

Experiment 1

The IgA responses of mice unable to make IL-6 (IL-6")and control mice (IL-6+) were studied. Ovalbumin, an eggprotein, was added to the mouse intestinal mucosa.Immunoglobulin response is depicted in Figure 1.

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Experiment2

IL-6" and IL-6+ mice were immunized with a virusconstruct carrying the mouse IL-6 gene. The virus wastheorized to insert the IL-6 gene into the DNA of the hostcells it infected.

IL-6+

With virus carrier With virus carrieralone plus IL-6 gene

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Weeks following innoculation of mice

IL-6"

With virus carrierplus IL-6 gene

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Weeks following innoculation of mice

Figure 2


The graphs in Figure 2 represent the number of IgAsecreting cells in mouse lung tissue 1 and 2 weeksfollowing exposure to a virus carrier alone or a viruscarrier plus the BL-6 gene.

22. Which of the following tissues are lined by epithelialtissue?

I. Mouth

II. Small intestine

m. Urethra

A. I onlyB. I and II onlyC. n and m onlyD. I, II, and ffl

23. Which of the following statements is TRUE ofFigure 1?

A. IL-6" mice produce higher levels of IgA thanIL-6+ mice.

B. IL-6" mice produce lower levels of IgA thanIL-6+ mice.

C. The presence of IL-6 did not affect theproduction of IgA.

D. IL-6" mice respond strongly to ovalbumin.

24. IgA is composed of what type of molecules?

A. Amino acids

B. Fatty acidsC. SphingomyelinsD. Phospholipids

25. Which statement is TRUE of Figure 2?

A. The virus carrier alone transformed both

strains of mice.

B. Restoring IL-6 in the IL-6" mice improved IgAproduction.

C. IL-6" mice were hypersensitive to the viruscarrier.

D. IL-6+ mice did not respond to the virus carrier.


Biology IgA Antibody Experiment Passage IV

26.

27.

The researchers did not report BL-6 concentrationsinthe blood. Why is this the case?

A. Cytokines are hormones and act whilepassingthrough the entire circulatory system.

B. Hormone concentrations cannot be measuredin blood.

C. EL-6never leaves the cell that produces it.D. Cytokines are local hormones and often act

without passing through the entire circulatorysystem.

What term refers to the medical alteration of genesto correct an inherited or acquired disease?

A. Gene therapyB. Vaccination

C. Genetic immunization

D. Pleiotropy


Biology Complement System Passage V


Complement was given its name because itcomplements the action of antibodies and is the principalmeans by which antibodies defend vertebrates againstmost bacterial infections. A system of serum proteins areactivated to form a membrane attack complex (MAC)which forms holes in microorganisms. Complement alsoamplifies the defense system by dilating the blood vesselsand attracting phagocytic cells to the site of infection.Individuals who are complement deficient also sufferfrom immune complex diseases, in which antibody-antigen complexes precipitate in small blood vessels inskin, joints, and brain causing the destruction of tissue.

Complement consists of about 20 interacting proteins.The components involved in reactions are known as Cl-C9, factor B, and factor D. The rest of the proteins areinvolved in the regulation of this system. These proteinscirculate in the blood in an inactive form unless activateddirectly by an invading microorganism or indirectly by animmune response. The final result of activation isassemblage of the late complement components (C5-C9)into a membrane attack complex.

The early complement components are activated byeither antibodies bound to a microorganism or bypolysaccharides on a microbial envelope. There are twodistinct pathways of early component activation. CI, C2,and C4 belong to the classical pathway and is triggered byantibody binding. Factor B and D belong to the alternativepathway and are triggered by microbial polysaccharides.Both pathways will act on C3, a central component in thecomplement system.

All early components and C3 are proenzymes that areactivated by each other through proteolytic cleavage. Aseach proenzyme is cleaved, it is activated to generate aserine protease which cleaves the next proenzyme in thesequence. Each activated enzyme cleaves many moleculesof the next proenzyme in the chain. The cleavagenormally exposes a membrane binding site on the largerfragment and liberates a small peptide fragment into theblood stream. The C3 molecule is eventually cleaved,with its larger fragment binding both the cell membraneand C5. Activation of C5 will initiate the spontaneousassembly of C5 through C9, forming the membrane attackcomplex.

28. According to the passage, complement maynormally help:

A. produce antigen-antibody complexes.B. destroy antigen-antibody complexes.C. solubilize antigen-antibody complexes.D. precipitate antigen-antibody complexes.


29. The proteolytic cascade described in the passage:

A. focuses the complement system away from cellmembranes.

B. does not take place in the alternate pathway.C. provides a means of amplification, ultimately

leading to many MACs.D. uses serine proteases at all serine residues in a

protein.

30.

31.

32.

33.

The production of antibodies used in thecomplement process would be greatly affected by adisease of the:

A. lymph nodes.B. bone marrow.

C. thymus.D. spleen.

The classical pathway is usually activated by IgG orIgM antibodies bound to antigens on the surface of amicroorganism. The CI complex most likely bindsto the:

A. antigenic determinant.B. variable region of the antibody.C. cell membrane.

D. constant region of the antibody.

As stated in the passage, protease cleavage acts toexpose a membrane binding site on the largerfragment. The most likely reason for this is to:

A. avoid precipitation of complement proteins.B. have the larger fragment act as a diffusible

signal in the bloodstream.C. confine complement activation to the cell

surface where it began.D. inhibit the next reaction in the cascade

sequence.

During the proteolytic complement cascade, severalsmall biologically active fragments are generated.One of the end results of these molecules' activity isan increase the permeability of local blood vessels.Which is the most likely explanation for such anincrease?

A. These molecules stimulate the secretion ofhistamine from T lymphocytes.

B. These molecules stimulate the secretion ofhistamine from basophils.

C. These molecules stimulate the secretion ofhistamine from macrophage.

D. These molecules stimulate the secretion ofhistamine from erythrocytes.


Biology Myasthenia Gravis/Autoimmune Diseases Passage VI


Myasthenia gravis is a rare, chronic, neuromusculardisease. It is characterized by skeletal muscle weaknessand fatigability in response to repeated contraction.Resting partially restores muscle strength. The muscles ofthe eyes, face, jaw, and neck are usually affected first. Asthe disease progresses, weakness spreads to theextremities and the diaphragm. In severe cases, all themuscles are weakened.

Research indicates myasthenia gravis is anautoimmune disorder in which antibodies are produced tothe acetylcholine receptor that is present at theneuromuscular junction. Antibodies to the acetylcholinereceptor have been found in 85% of patients withgeneralized myasthenia. Although the mechanism forantibody production is unclear, one hypothesis is thatcertain thymus cells that resemble muscle (myoid cells)are damaged by a virus. The virus may have a molecularregion that mimics part of the acetylcholine receptor, suchas the herpes simplex virus. A virus may damage myoidcells so that antibodies are produced against them directly.By whatever mechanism, the viral infection inducesantibody production.

The actual interaction between the antibody and thereceptor is not fully understood. The antibody may blockthe receptor, it may cause faster receptor breakdown, or itmay promote complement-mediated damage.

Autoimmune diseases as a whole are relativelycommon. In the following table is a list of someautoimmune diseases and the antibodies produced in thedisease state:

Disease Antibody Against Effects

Type Idiabetes

B cells of pancreas Destroys B cells

Graves'

disease

Thyroid stimulatinghormone receptor

Stimulates TSH

receptor on thyroid

Multiplesclerosis

Myelin basic protein(hypothesized)

Disruptsmyelination

Glomerulo

nephritisBasement membrane

of glomular capillariesDestroys a variablenumber of glomeruli

34. Which drug could be given to counteract the effectsof the antibody produced in myasthenia gravis?

A. An inhibitor of acetylcholinesterase.B. An immunostimulant.

C. A paralytic agent, like curare.D. An inhibitor of acetylcholine synthase.


35. Shown below is a diagram of a neuromuscularjunction. Which number indicates an acetylcholinereceptor?

A. I

B. II

C. HI

D. TV

••••••

# \v m

36. Which of the following is NOT an example of anautoimmune disease?

A. Type II diabetes.B. Addison's disease.

C. Myasthenia gravis.D. Graves' disease.

37. Which of the following is NOT a symptom ofGraves' disease?

A. Increased metabolic rate.

B. Weight loss.C. Lethargy.D. Hyperactivity.

38. Secretion of which of the following from thepancreas is halted by antibodies to the B cells?

A. Glucagon.B. Insulin.

C. Bicarbonate.

D. Digestive enzymes.


Biology Myasthenia Gravis/Autoimmune Diseases Passage VI

39. How can a viral infection lead to an autoimmunedisease?

I. The virus resembles a "self molecule, leadingto antibodies that cross-react with other bodymolecules

n. The virus damages a cell so that unrecognizedcell proteins are released, causing antibodyproduction

HI. The virus resembles a "non-self molecule,leading to antibodies that cross-react withother body molecules

A. I onlyB. I and II onlyC. II onlyD. II and in only

40. Often patients with autoimmune diseases are treatedwith corticosteroids to reduce immune responses.Which organ in the body produces corticosteroids?

A. Pancreas.

B. Adrenal gland.C. Pituitary gland.D. Thyroid gland.


Biology Glucose, Glucagon, St Insulin Passage VH

Passage VTI (Questions 41-46)

The regulation and metabolism of carbohydrates in thebody are controlled by the pancreas. The endocrinecomponent of the pancreas, the islets of Langerhans, arespecifically responsible for carbohydrate control. Thesesmall clusters of cells imbedded within the exocrine

portion of the pancreas contain peptides with specifichormonal activity.

Glucagon, secreted by the a cells (or A cells), liberatesglucose from storage areas in the body, stimulates glucoseproduction, increases lipid concentration in the bloodstream by releasing free fatty acids, and increases theproduction of ketones, a cells are stimulated to secreteglucagon during increases in plasma amino acid levels,Cortisol secretion, exercise, and sympathetic nervoussystem stimulation. Inhibition of glucagon is promoted byincreases in plasma glucose, ketone, free fatty acids,insulin, and somatostatin levels.

Insulin is secreted by the P cells (or B cells) andfunctionally is important in increasing the storage ofglucose, fatty acids, and amino acids in the cells of targettissues. Furthermore, insulin decreases the release ofglucose, mannose, amino acid, and glucagon plasmalevels. Many intestinal hormones also stimulate insulinsecretion. Parasympathetic stimulation of the (3 cells willincrease insulin secretion, while sympathetic stimulationwill inhibit secretion.

Somatostatin is a peptide secreted by 8 cells (or Dcells) in the islets. Somatostatin inhibits the production ofboth insulin and glucagon, and it acts as a regulator ofislet secretion.

The effects of insulin and glucagontarget veryspecificregions of the body where their cellular actions occur.Insulin's effects are generally associated with muscle andadipose tissue, leukocytes, fibroblasts, and mammaryglands. Insulin does not directly affect brain and kidneytissue, intestinal mucosa, and red blood cells. Glucagon'seffects are targeted mainly on the liver.


41. During times of stress the importance of havingadequate levels of glucose, for utilization as energy,is dependent on hormonal secretions from thepancreas. Which choice below will BEST ready thebody for stressful situations?

A. Increased levels of glucagon and insulin.B. Increased levels of glucagon and decreased

levels of insulin.

C. Decreased levels of glucagon and decreasedlevels of insulin.

D. Decreased levels of glucagon and increasedlevels of insulin.

42. Ketosis is developed from an increase in theconversion of free fatty acids to ketone bodies.These ketone bodies are an important source ofenergy in times of fasting. However, prolongedketosis will lead to a plasma acidosis due to liberatedhydrogen ions from ketone bodies. A patient withacidosis will develop shortness of breath,dehydration, hypervolemia, and hypotension. Insevere cases the acidosis and dehydration willdepress consciousness to the point of coma. Whichchoice below will lead to the development of ketosisand acidosis?

A. Increased levels of glucagon and insulin.B. Increased levels of glucagon and decreased

levels of insulin.

C. Decreasedlevels of glucagon.D. None of the above.

43. Symptoms reported by a patient include weakness,dizziness, confusion, and hunger. Furthermore, sometremors, palpitations, and nervousness are alsoreported. These last symptoms are characteristic ofhypoglycemia. The development of these symptomsis due to:

A. abnormal increases in insulin secretion.

B. abnormal decreases in insulin secretion.

C. abnormally high levels of plasma glucose.D. abnormally low levels of free fatty acids.


Biology Glucose, Glucagon, St Insulin Passage VD

44. What pancreatic hormonal response is expected aftera heavy protein intake?

A. Increased levels of insulin and glucagon.B. Increased levels of insulin and decreased levels

of glucagon.C. Decreased levels of insulin and glucagon.D. Decreased levels of insulin and increased

levels of glucagon.

45.

46.

From the data reported on glucose utilization oftissue and responsiveness of tissue to insulin, it maybe determined that the main function of insulin

secretion is:

A. increasing glucose uptake in the brain.B. increasing glucose release from the liver.C. access and storage of glucose in cells of the

peripheral tissues.D. increasing glucose loss in the kidney.

Patients diagnosed as having diabetes mellitus aresaid to be in "a state of starvation in the midst of

plenty." This analogy refers to:

A. extracellular glucose excess.B. extracellular glucose deficiency.C. decreased effects of insulin on intestinal

mucosa uptake of glucose.D. deficiency of fatty acids in neural tissue.


Biology Calcium, PTH, Calcitonin, Calcitriol Passage VII


Calcium and phosphate both play an important role inthe mineralization of bone in vertebrates. Calcium is

obtained from the diet and is largely absorbed at the levelof the intestine. The intracellular concentration of calcium

is about 10*7 mol/L while the extracellular concentrationof calcium is about 10'^ mol/L. Blood calcium levels areprimarily determined by bone metabolism and urinaryexcretion.

Calcium metabolism is regulated through the actionsof parathyroid hormone (PTH), calcitonin, and vitamin D3(cholecalciferol). PTH is synthesized and secreted by theparathyroid glands, usually located in the central region ofthe thyroid gland near the trachea. Calcitonin issynthesized and secreted from parafollicular (or C) cellslocated in the thyroid gland. Cholcalciferol is synthesizedin an inactive form in the skin of animals from a

photolytic reaction involving UV light and a sterolderivative. In order for this inactive hormone to beactivated, it must first be hydroxylated in the liver andthen hydroxylated in the kidney. The activity of thehydroxylase enzyme in the kidney is regulated andincreased by PTH.

Bone is composed of an organic matrix consisting ofcollagen fibers and a ground substance composed ofextracellular fluid and proteoglycans. The collagen fibershelp to give bone its great tensile strength, while theground substance helps to control the deposition ofcalcium salts, like hydroxyapatite (CastPO^OH). Thecalcium salts help to provide for the great compressionalstrength found in bone.

The collagen matrix and ground substance is laid downby bone cells called osteoblasts. The tissue which isformed, called an osteoid, can entrap some of theosteoblasts. Entrapped osteoblasts are called osteocytes.As the bone grows, calcium salts precipitate on thecollagen fibers. Bone is also undergoing resorption bycells called osteoclasts.

PTH causes calcium absorption from bone bystimulatingosteoclastic activity and transiently inhibitingosteoblastic activity. At the level of the kidney, PTHincreases calcium absorption in the distal tubules andcollecting ducts and greatly decreases the reabsorption ofphosphate at the proximal tubules.

The activated form of cholecalciferol (l,25-(OH)2-D3)has target receptors in the intestine, bone, and kidney (toname but a few tissues). In the intestine this hormonepromotes absorption of calcium and phosphate (followingas the counterion), while in bone it promotes resorption ofboth calcium and phosphate. In the kidneys this hormonepromotes the reabsorption of both calcium and phosphateso that little is excreted in the urine.


Calcitonin acts to inhibit osteoclast activity and reducebone resorption. This hormone also inhibits calcium andphosphate reabsorption in the kidney and increases theexcretion of these ions in the urine.

47. Administration of PTH leads to changes in plasmacalcium and phosphate concentrations. Which of thefollowing graphs BEST represents these changes?

A.

B.

C.

D.

Phosphate

0 \fr Time (hours)PTH added

0 \/ Time (hours)PTH added

0 \J Time (hours)PTH added

0 \/ Time (hours)

PTH added


Biology Calcium, PTH, Calcitonin, Calcitriol Passage VD

48. Which of the following structures BEST representsvitamin D3?

A.

B.

C.

D.


49. Calcium and phosphate absorption in the intestinesis stimulated by an increase in:

I. l,25(OH)2D3n. Calcitonin

IE. PTH

A. I onlyB. I and II onlyC. Ill onlyD. I and III only

50. In order for the secretion of calcitonin to have a

greater effect on the concentration of calcium ions inthe plasma, which of the following statements mustbe true regarding osteoclast activity and plasmacalcium levels?

A. Increased osteoclast activity coupled with ahypercalcemic plasma.

B. Decreased osteoclast activity coupled with ahypercalcemic plasma.

C. Increased osteoclast activity coupled with ahypocalcemic plasma.

D. Decreased osteoclast activity coupled with ahypocalcemic plasma.

51. Hypophosphatemic rickets is an X-linked dominanttrait that leads to decreased levels of phosphatereabsorption in the kidneys. Which pedigree shownbelow BEST represents this disease?

A. B.

II O-r-1 B-r-O II 6-r-a D

rm

c.

I O-r-B

n #-r-a • •

in

rv

6-r-a i i-r-o

D.

n

m

rv

oTTV


Biology Calcium, PTH, Calcitonin, Calcitnol Passage VD

52. Familial hypophosphatemia is BEST treated by dietmodification and supplying adequate amounts of:

A. calcium.

B. phosphate.C. calcium and phosphate.D. phosphate and l,25(OH)2D3.

53. Hypoparathyroidism is BEST characterized by:

I. increased osteoblast activity.II. increased osteoclast activity.III. increased neural excitability.IV. increased plasma calcium concentrations.

A. I onlyB. H and m onlyC. m onlyD. I and IV only

54. In the graph shown below, all of the followingstatements concerning the relationship betweenPTH, calcitonin, and calcium are true EXCEPT:

A.

B.

C.

D.

Total Ca2+ concentration inplasma

PTH is a hypercalcemic hormone.calcitonin is a hypocalcemic hormone.a positive linear relationship exists betweencalcitonin secretion and the concentration ofplasma calcium.a positive linear relationship exists betweenPTH secretion and the concentration of plasmacalcium.


Biology Erythroblastosis Fetalis Passage IX


Erythroblastosis fetalis (EF) is also known ashemolytic disease of the newborn. Hemolysis is rupture ofred blood cells, so in hemolytic disease, anemia iscommon, due to the lysis of red blood cells. In the EFcondition, maternal antigens cross the placental barrier,attack proteins on the surface of the red blood cells of thefetus, and lyse the cells..A very specific set of conditionsmust exist for this disease to occur.

The mother must be Rhesus (Rh) factor negative, thefetus must be Rh factor positive, and the mother's immunesystem must be sensitized to the Rh positive antigenthrough previous full-term pregnancy or abortion. The Rhfactor antigen is transmitted as a dominant trait, so thatonly people who are homozygous recessive are Rh factornegative.

Roughly 90% of the cases of EF result fromsensitivities to the D antigen on the Rh factor. When Rhpositive blood enters the circulation of an Rh negativemother, antibody formation against D may be induced.This exposure may be during an accidental infusion,during pregnancy, delivery, or during a miscarriage orabortion. During a first pregnancy, there is usually littleexchange of fetal and maternal blood, except near the endof the pregnancy or during delivery.

This time frame does not allow for antibody formationfollowed by an attack on the fetal blood cells. Theproblem lies in subsequent pregnancies. Small amounts ofantigen, even the amount in 1 mL of fetal blood enteringthe mother's circulation, promote rapid increases in heranti-D antibody titer. IgG is produced, and it can easilycross the placental barrier into the fetal blood supply.Even in the ideal conditions for EF, sometimes the disease.does not manifest, due to variable physiologicalconditions.

55. A Rh-negative woman and a heterozygous Rh-positive man have one child together. The womanhas never been pregnant before. What is thelikelihood that the child will be born with EF?

A. 100%

B. 75%

C. 50%

D. 0%


56. Which of the following statements is TRUE?

A. The placenta allows passage of all bloodproducts from the mother to the fetus.

B. The Rh factor is not a component of the ABOblood group system.

C. The mother makes red blood cells for the fetusin the placenta.

D. Terminated pregnancies have no effect onfuture development of EF.

57. What preventive measure could protect subsequentfetuses if an Rh-negative mother gave birth to anvvnai picvcuuvc measuie uuuiu pruieui

fetuses if an Rh-negative mother gaveRh-positive fetus?

58.

A. Give the mother a blood transfusion with Rh-

positive blood.B. Give the fetus a blood transfusion with Rh-

negative blood.C. Treat mother with a set of antibodies directed

against the anti-Rh antibodies.D. Treat fetus with antibodies against the anti-Rh

antibodies.

An infant with severe EF has jaundice, a yellowcoloring due to excess bilirubin, a breakdownproduct of heme. In what tissue or organ is hemedegraded into bilirubin?

A. SpleenB. Bone marrow

C. Liver

D. All of the above

59. An Rh-positive mother is pregnant with an Rh-negative fetus. Will the fetus develop EF?

A. No, there are no maternal antigens to the Rhfactor antigens.

B. Yes, the mother can still make antigens to theRh factor of the fetus.

C. No, the fetal antibodies protect its red bloodcells.

D. No, but fetal antibodies attack the maternal redblood cells.


Biology Erythroblastosis Fetalis Passage IX

60.

61.

What variables could affect the severity ofhemolysis in an Rh-positive fetus whose mother isRh-negative?

I. Amount of blood transferred.II. Sensitivity of mother to D antigen.m. Number of pregnancies.

A. I onlyB. I and II onlyC. II and m onlyD. I, H, and HI

Which of the following clinical signs could beconsistent with a diagnosis of EF in a newborn?

I. High levels of bilirubin in the blood.II. Low levels of hemoglobin in the blood.HI. Increased levels of erythrocytes.

A. I onlyB. I and II onlyC. H and ffl onlyD. I, II, and HI


Biology Septic Shock Passage X


Septic shock, a disease characterized by hemodynamicderangements and multi-organ malfunction, has generallybeen associated with a gram-negative infection. However,it is becomming increasingly clear that gram positiveorganisms are equally responsible for sepsis. Many ofthese gram positive ogranisms release molecules knownas superantigens. These superantigens can induce T cellproliferation without regard to the antigenic specificity ofthe cell.

A common sign of septic shock is widespreadactivation of coagulation leading to widespreadintravascular clotting. Microbial products activate FactorXn, a molecule involved in blood clotting. Activation ofthis factor initiates the intrinsic coagulation pathway andalso the bradykinin pathway. Bradykinin is a potentvasodilator and also increases the permeability of vascularendothelial cells.

Cytokines, such as interleukin-1 and tumor necrosisfactor, activate tissue factor HI. This factor is found on theouter membrane of macrophage and endothelial cells, andstimulates the extrinsic coagulatory pathway.

62. A Gram-positive cell differs from a Gram-negativecell in that a Gram-positive cell:

A. does not have an outer membrane on its cellwall.

B. does have an outer membrane on its cell wall.

C. contains a thin peptidoglycan layer adjacent tothe plasma membrane.

D. contains no peptidoglycan layer adjacent to theplasma membrane.

63. Macrophagesdestroy microorganisms through:

A. exocytosis, secreting toxins which eventuallyform membrane attack complexes.

B. xxocytosis, surrounding the foreign particlewith a lipid bilayer which protects the hostorganism.

C. endocytosis, engulfing foreign particles whicheventually will fuse with a lysosome.

D. endocytosis, engulfing foreign particles whicheventually will fuse with a peroxisome.


64. A defect in which of the following cells wouldinhibit the production of soluble antibodies?

A. Mast cellsB. Cytotoxic T cellsC. Plasma cells

D. Erythrocytes

65. In an experiment, a sepsis patient is treated withanti-factor XII antibodies. After treatment, onewould expect to see a:

A. total lack of intravascular clotting.B. rise in the level of intravascular clotting.C. rise in the patient's blood pressure.D. decrease in the patient's blood pressure.

66. Bradykinin acts to increase the radius of a givenblood vessel by a factor of 2. The flow of bloodthrough the vessel should increase by a factor of:

A. 2.

B. 4.

C. 8.

D. 16.

67. The activation of bradykinin most likely results in:

A. hypotension in the patient.B. cytokine release.C. macrophage activation.D. stimulation of the extrinsic coagulation

pathway.


Biology Calcitonin and Osteoporosis Passage XI


Calcitonin (CT) is a polypeptide hormone secreted bythe parafollicular cells of the thyroid gland in mammals.32 amino acids make up the hormone, and a disulfidebridge links residues 1 and 7. The entire CT molecule andthe disulfide bridge are required for full biologicalactivity.

CT works as an antagonist of parathyroid hormone(PTH). In response to small increases in plasma calcium,CT is released and acts on the kidney and bone todecrease the calcium level. In the bone matrix, osteoblastssynthesize bone, and osteoclasts catabolize bone. Themain effects of CT are (1) inhibition of osteoclasts and (2)a transient increase in urinary calcium and phosphate.

Calcitonin is used pharmacologically as a treatmentfor osteoporosis. Salmon CT is commonly used. Althoughsalmon CT differs markedly from human CT, it is about30 times more potent when used in non-allergic humansto treat osteoporosis. Treatment with CT is not without itsown side effects. CT treatment for osteoporosis increasesplasma PTH and requires simultaneous calciumsupplementation to avoid hyperparathyroidism.

68. Osteoporosis is a disorder of bone characterizedby adecrease in bone quantity, most common in womenfollowing menopause and in elderly men andwomen. Which of the following conditions wouldlead to the GREATEST decrease in bone quantity?

A. Decreased osteoblast activity, decreasedosteoclast activity.

B. Increased osteoblast activity, decreasedosteoclast activity.

C. Decreased osteoblast activity, increasedosteoclast activity.

D. Increased osteoblast activity, increasedosteoclast activity.

69. What is the most abundant mineral in the humanbody?

A. B12B. Water

C. Calcium

D. Zinc

70. Based on the passage, what is the role of PTH?

A. PTH increases plasma calcium.B. PTH decreases plasma calcium.C. PTH increases urinary calcium.D. PTH increases urinary phosphate.


71. Which two amino acids form a disulfide bridge?

A. Methionine-threonine

B. Methionine-methionine

C. Cysteine-cysteineD. Cystine-cystine

72. Which of the following choices is a probableexplanation why salmon CT is 30 times more activein humans than human CT?

I. Salmon CT is more resistant to degradation byhuman enzymes.

U. Salmon CT attaches more strongly to the DNAof the osteoclasts.

m. Salmon CT attaches more tightly to the CTreceptor.

A. I onlyB. I and m onlyC. H and ffl onlyD. I, H, and ffl

73. If a person begins a calcium supplement regimenand doubles calcium intake, what would be theresponse in CT secretion?

A. CT secretion increases.B. CT secretion decreases.

C. CT secretionremains unchanged.D. CT secretion halts completely.

74. How is salmon CT, a polypeptide,administered?

A. Ingestion to avoid allergic reaction.B. Injection to avoid hydrolysis.C. Ingestion to avoid hydrolysis.D. Injection to avoid allergic reaction.

75. What are the symptoms of an allergic reaction to aforeign protein in the bloodstream?

I. Flushing or reddening of the skin.H. Skin welts (hives).m. Difficulty breathing.

A. I onlyB. I and II onlyC. ffl onlyD. I, H, and ffl


Biology Vertebrate Immune System Passage XII


In order to protect vertebrates from infection, theimmune system hasevolved bothan antibody-based andacell-mediated response to foreign antigens. B cells, whichoriginate and develop in hemopoietic tissues (bonemarrow and fetal liver),are responsible for producing andsecreting antibodies which bind to antigen particles. Tcells also originate in hemopoietic tissues but latermigrate to and mature in the thymus during earlydevelopment. CytotoxicT cells are mainly responsible formounting a cell-mediated defense by directly causing thedeath of infected cells. While B cells can be activated bythe binding of extracellular antigen to special receptors onthe plasma membrane, T cells must come in direct contactwith infected cells in order to become activated.

How the immune system differentiates between selfand foreign antigens has been the topic of much study.Immunologists in the first half of this century proposedtwo main theories:

Theory1

Vertebrates inherit genes that encode receptors(present on the surface of B and T cells) that are capableof binding only foreign antigens. The immune systemdoesn't react against host tissues because it geneticallylacks the receptors which bind self antigens.

Theory 2

The immune system is inherently capable ofresponding to both self and foreign antigens, but itbecomes "tolerant" to self antigens during earlydevelopment. Since foreign antigens aren't present duringembryonic stages, the immune system does not develop atolerance to them.

76. According to the passage, removal of the thymusfrom an adult human would most likely result in:

A. drastically decreased antibody-mediatedimmune response.

B. drastically decreased cell-mediated immuneresponse.

C. drastically decreased antibody and cell-mediated immune response.

D. little or no change in the effectiveness of eithertype of immune response.


77.

78.

79.

When certain types of antibodies bind to their targetcells, they attract a series of proteins collectivelyknown as complement. These proteinscan then formpores which allow small molecules to freely diffuseacross the plasma membrane. What effect would thishave on a target cell?

A. The cell would die due to its inability toinitiate action potentials.

B. The cell would become hyperpolarized.C. The cell would lyse due to an upset water

balance.

D. The cell would shrink due to an upset waterbalance.

Which of the following statements would beconsistent with BOTH Theory 1 and Theory 2?

I. A foreign cell line is injected into a mouseembryo. Further injections of the cell line intothe adult mouse do not elicit an immune

response.

n. Transplantation of organs betweenmonozygotic twins does not result in organrejection by the immune system.

in. Cells transplanted between mice which aregenetically identical (i.e., from the same inbredstrain) are tolerated by the new host's immunesystem.

A. I onlyB. II onlyC. II and III onlyD. I, II, and ffl

Tolerance to self antigens breaks down in the humanautoimmune disease myasthenia gravis, resulting inthe production of antibodies against the patient'sskeletal muscle acetylcholine receptors. Which ofthe following is a likely symptom of this disease?

A. Irregularities in heart contraction.B. Weakness and difficulty breathing.C. Paralysis of the gastrointestinal tract.D. Dementia.


Biology Vertebrate Immune System Passage xn

80.

81.

82.

Recently activated B and T cells are examined viaelectron microscopy. Which of the following wouldbe the most likely observations?

A. The B cells have a greater number ofmitochondria than the T cells.

B. The B cells have considerably more roughendoplasmic reticulum than the T cells.

C. Both the T and B cells lack nuclei.D. No differences between the two types of cells

are revealed at the level of electronmicroscopy.

A sample of B cells is removed from an adultmouse. A highly radioactive antigen X is added tothe B cells, killing the few that bind strongly(<0.01%). The remaining B cells are injected intomice whose own B cells were destroyed byirradiation. These mice can now make no antibodies

to antigen X but do respond to other antigens. Whichof the following can most likely be concluded?

A. Each B cell is predetermined to bind a specificantigen.

B. B cells can recognize new antigens and "learn"to bind to them.

C. T cells must interfere with B cell binding.D. Antigen X is normally non-immunogenic.

Secreted antibodies often act collectively to bind tolarge target antigens and cross-link them intoinsoluble masses which are easily phagocytosed.Which of the following would NOT be consistentwith this phenomenon?

A. Antibodies each have two identical bindingsites.

B. Antigens can be bound by only one antibody ata time.

C. Antibody binding sites are connected by aflexible "hinge" region.

D. Large antigens contain multiple antibody-binding sites.


83. Virally infectedcells are usually killed by cytotoxicT cells. The T cells can most likely target:

A. most types of cells in the body.B. only cells lining the blood vessels.C. epithelial cells only.D. other blood cells only.


Biology Insulin Receptor Passage XHI

Passage XUI (Questions 84-90)

The human insulin receptor is an integraltransmembrane protein located in the plasma membraneof many cells throughout the body. Studies have shownthat a functional receptor consists of two alpha and twobeta subunits which are connected using disulfidebonding. The receptor consists of several domains,including one which binds insulin, a transmembranebinding region, and an intracellular tyrosine kinasephosphokinase (TPK) segment.

How does ligand binding create intracellular changeswhich bring about the effects of the insulin hormone?There is experimental evidence which suggests that thehuman insulin receptor is a insulin activated tyrosinekinase. While many details remain unclear, the binding ofthe hormone leads to both an autophosphorylation of twotyrosine residues on the TPK and a phosphorylation ofcytoplasmic constituents. In addition, the binding ofinsulin to its receptor is believed to activate phospholipaseC, which converts glycan phosphatidyl inositol phosphateto glycan inositol phosphate (GIP) and 2-diacylglycerol.

GIP is known to mimic certain insulin responses and2-diacylglycerol, in the presence of elevated calciumlevels, is known to activate protein kinase C. Proteinkinase C is known to influence the activity of otherenzymes and metabolic pathways.

84. Which of the following procedures is used on cellsto isolate the entire insulin receptor for experimentalstudies?

A. Acidic wash.

B. Oxidizing wash.C. Detergent wash.D. Reducing wash.

85. Through genetic manipulations, the insulin bindingregion of the insulin receptor (IR) was fused with thekinase region of the epidermal growth factor (EGF)receptor (also a ligand activated kinase). Theresulting hybrid receptor was transfected into a celland the addition of insulin resulted in a functionalEGF kinase. From this experiment, it can beconcluded that IR and EGF have similar:

A. ligand binding regions.B. signal transduction mechanisms.C. secondary messenger cascades.D. effects on carbohydrate, fat, and protein

metabolism.


86.

87.

315

The response to glucagon is believed to involve theproduction of cAMP as a secondary messenger.Which of the following graphs BEST describes thesequence of events associated with the binding ofglucagon to its receptor? [Note: CA = cAMP; PP =proteinphosphorylation; CR = cellular response.]

A.

B.

C.

D.

•a lCO

Q

> t

< «

% 52

> X

< ca

Time —

Time —•

Time —

Time —

In a type I, insulin dependent diabetes, the individualafflicted with the disease most likely has:

A. a mutated insulin binding site.B. a mutated membrane anchoring site.C. a normal insulin receptor.D. an inactive phospholipase C.


Biology Insulin Receptor

Whichof the following is consistentwith the actionsof GIP?

A. Inhibition of hexokinase.B. Inhibition of phosphofructokinase.C. Stimulation of glycogen breakdown.D. Stimulation of pyruvate dehydrogenase.

89. In cell #1, the addition of bivalent anti-insulinreceptor antibodies induces a response without theaddition of insulin. In cell #2, addition of a fragmentanti-receptor antibody (monovalent) induces noresponse without insulin. To cell #2, antibodiesagainst the monovalent anti-receptor antibodies areadded. This will result in:

A. no response.B. decreased glucose uptake .C. increased CO2 production.D. increased insulin binding.

90. To disrupt the dimer in order to study itscomponents, a researcher will most likely use:

A. oxidizing conditions.B. reducing conditions.C. high speed centrifugation.D. high pH.


Passage xm


Biology Vitamin D3, PHI, &t Calcitonin Passage XIV


In all vertebrates the maintenance of calcium (Ca2®),phosphate (P043e), and magnesium (Mg2®) homeostasisis governed primarily by the vitamin D3 derivative 1,25-dihydroxycholecalciferol (l,25-(OH)2-D3), calcitonin(CT), and parathyroid hormone (PTH).

Humans obtain vitamin D3 either by ingestion orthrough synthetic mechanisms that are initiatedby cells inthe epidermis of the skin. Vitamin D3 is hydroxylated firstin the liver and then is transported to the kidney where theenzyme 1-hydroxylase converts it to the active form ofthe hormone, l,25-(OH)2-D3. The activity of la-hydroxylase is enhanced during hypocalcemia andhypophosphatemia. l,25-(OH)2-D3 passes into a targetcell and then into the nucleus where it complexes with areceptor protein that has a DNA binding site. Genetranscription can either be enhanced or suppressed.Enhancement leads to the synthesis of calcium bindingproteins (calbindins) in intestinal villi and crypt cells.Calbindin promotes the uptake of Ca2® from theintestinal lumen. A similar mechanism allows for theabsorption of phosphate and magnesium from theintestinal lumen.

PTH is released from the chief cells of the parathyroidglands while CT is released from the parafollicular cells(or C cells) of the thyroid gland. Both hormones havegenes encoded in the short arm of human chromosome 11and are synthesized as a preprohormone from differentprimary RNA transcripts.

PTH acts on bone and on the distal tubule of the

kidney to promote Ca2® reabsorption and inhibits thereabsorption of P043e in the proximal tubule of thekidney. PTH also stimulates the synthesis of l,25-(OH)2-D3 in the kidney. CT acts to lower both plasma calciumand plasma phosphate levels.

91. Ultraviolet light stimulates vitamin D3 production inwhich of the following cell types?

A. HepatocytesB. KeratinocytesC. LeukocytesD. Lymphocytes


92. The primary precursor for l,25-(OH)2-D3 is locatedin which oneof the following organs?

A. Liver

B. Intestine

C. IntegumentD. Kidney

93. Whichof the following categories represent the threehormones described in the passage?

I. Amine

II. PolypeptideIII. Steroid

A.

B.

C.

D.

II onlyI and III onlyII and III onlyIII only

94. Which of the following conditions best describes apatient with a deficiency in PTH secretion?

I. HypocalcemiaII. HypercalcemiaIII. Hyperphosphatemia

A.

B.

C.

D.

I onlyIII onlyII and III onlyI and III only

95. Which of the following curves BEST represents therelationship between blood levels of parathyroidhormone (PTH) and calcitonin (CT) in terms of total

plasma Ca2® concentration levels?

A.

Q

low high

Total Plasma [Ca2+]

c.

low high

Total Plasma [Ca2+]

B.

D.

6

low high

Total Plasma [Ca2+]

low high

Total Plasma [Ca2+]


Biology Sepsis Syndrome Passage XV


The normal response to an infection is a complexsystem of activations and inactivations of the members ofthe immune system. Neutrophils are the first to respond toan infection by squeezing through blood vessel walls anddestroying pathogens directly by the release of toxicoxygen products, such as superoxide, hydrogen peroxide,and nitric oxide, and by the release of proteolyticenzymes. These white blood cells follow chemicalattractants given off by the pathogen. Monocytes, anothertype of white blood cell, arrive next. They may betransformed into macrophages in the tissue and engulfpathogens and cellular debris. Monocytes andmacrophages release cytokines, a powerful class ofchemicals that modulate responses of different membersof the immune system. T cells and B cells are attractedand activated by cytokines at the infection site. Anotherset of cytokines are released to signal "stop" when thepathogens are neutralized. This is the normal progressionof events when the body attacks an invading bacteria.

However, in the U.S. each year, about 175,000 peopledie from sepsis syndrome, in which the sepsis process isactually amplified rather than terminated. The cytokinesfor signal termination are not released correctly, leadingto an imbalance with the amplifying cytokines. Thisattracts all the white blood cell participants to the area,which release more activating cytokines. Some of thesubstances that are released lead to increased permeabilityof the blood vessels. In normal situations, this allowsWBCs to more easily enter the infected tissue. In excess,this leads to blood cell leakage and damage, andpositively feeds back to attract more WBCs. Ultimately, ifthis process is unchecked, blood vessels will deteriorate,leading to tissue death, organ death, or even patient death.


96. Which is the term for the process of white bloodcells squeezing through blood vessels?

A. Mitosis

B. DiapedesisC. PhagocytosisD. Endocytosis

97. What purpose do proteolytic enzymes serve incellular defense?

A. Inactivate bacterial lipid signals.B. Digest bacterial cell surface carbohydrates.C. Destroy bacterial DNA.D. Hydrolyze bacterial proteins.

98. Which cell products are used in free radicalreactions to destroy pathogens?

A. Chemical attractants.

B. Proteolytic enzymes.C. Cytokines.D. Toxic oxygen products.

99. Researchers are attacking sepsis syndrome frommany perspectives. In one approach, syntheticreceptors for tumor necrosis factor (TNF), acytokine, are injected to bind to TNF and halt thecytokine cascade. What assumption does this modelmake?

A. TNF halts the abnormal cytokine cascade ofsepsis syndrome.

B. TNF receptors enter cells by endocytosis andbind intracellular TNF.

C. TNF is mainly present in the generalcirculation.

D. TNF moves from cell to cell only through gapjunctions.

100. What is the most abundant type of white blood cell?

A. NeutrophilsB. ErythrocytesC. LeukocytesD. Macrophages


Biology Immunology St Endocrinology Section V Answers

Passage 1(1 - 7) Antibody Structure

l.

3.

5.

C is correct, placental transfer of antibody. Innate (nonspecific) immunity is what the individual has developedfrom birth. In the second paragraph of the passage three examples ofnonspecific immunity were given. The skinconsists ofan outer layer of dead keratinized cells and an internal layer of live epidermal cells. All of the bodycavities and organs that open up to the skin surface are lined with mucous membranes. The epithelial lining ofthesestructures contain goblet cells which secrete mucous and the epithelial cells themselves have hair-like projectionscalled cilia that move particulate matter over the epithelial surface. In the respiratory tract the mucous membranes ofthe trachea and bronchi are lined with ciliated epithelium that beat towards the opening of the mouth. This allows forthe upward movement (away from the lungs) ofmucous and particulate matter so itcan be expelled form the system.Normal body temperature is about 37 °C. Many organisms (pathogens) simply do not grow well at this temperature.The only component which is not part of the nonspecific immune system is the placental transfer ofantibody. This isapassive process which occurs from mother to fetus via the placenta. The fetus acquires antibodies synthesized bythe mother. The correct choice is C.

C is correct, active immunity which is artificially acquired. In the first paragraph of the passage we see thatacquired immunity is specific immunity. In the third paragraph we see that vaccinations and a recovery from adisease like the measles are examples ofspecific immunity. A vaccination is simply an immunization against apathogen. It is something that is artificially acquired. The individual who receives the vaccination then begins toactively produce his or her own immunological agents against the attenuated (weakened) pathogen. Even though thepathogen has been made innocuous, itstill retains its antigenicity (i.e., its epitopes). Recovery from a disease like themeasles is an example of active immunity which is naturally acquired. Antitoxin administration would be anexample ofpassive immunity which is artificially acquired. Transfer ofIgG from the mother to the fetus through theplacental membrane would be an example ofpassive immunity which is naturally acquired. The correct choice is C.

Ais correct, covalent bonding. In the 7th paragraph ofthe passage itsays that antibody-antigen complexes can bedissociated by high salt concentrations or either high orlow pH. This must mean that the binding forces between anantibody and an antigen are relatively weak. All of the bonds and/or bonding forces listed in the answer choices arcweak except for the covalent bond. We would not expect a covalent bond to be broken by a change in(physiological) pH or by treatment witha high salt concentration. The correct choice is A.

Dis correct, antibody which has similar dimensions as the antigenic determinant. This question is simply asking foran understanding ofa paratope. In the 7th paragraph of the passage we find that the paratope is associated with theantibody. This allows us to immediately eliminate choices Aand C. Look at the structure of the IgG molecule inFigure 1ofthe passage. This antibody can bind two epitopes ofthe same specificity (unideterminant). The portion ofthe antibody that binds the epitope resides near the N-terminus of both the Vh and Vl regions of the Fab. Theepitope does not just bind the Vh regions alone. Wc can eliminate choice B. The average size of antigenicdeterminants is roughly equivalent to about 6 amino acids. Therefore, the size of the paratope must be similar indimensions. If this were not the case, then binding between the epitope and paratope would be unfavorable.Remember, hydrophobic forces, Van der Waal forces, electrostatic forces, and hydrogen bonding all requirerelatively close proximities between participating parties before they can be ofany use. The correctchoice isD.

C is correct, antigens that are not cross-linked, because each Fab fragment is monovalent. In order to answer thisquestion, we must consider the structure of the IgG molecule in Figure 1of the passage. Note that there are two Fabdomains. It is a bivalent structure. As it stands, this antibody could participate in cross-linking with aunideterminant, multivalentantigen (see the diagram below).

However, after addition of the protease papain, the antibody is cleaved into an Fc segment and two Fab segments.The important point here is that the Fab segments are now independent ofeach other. They can now independentlybind to the antigen's epitope (see above). Cross-linking will not result.

Inchoice A the first part of the answer (cross-linking) is wrong, but the second part is correct. This makes the wholeanswer choice incorrect. In choice B the answer is correct, but for treatment of IgG with pepsin, not papain. Inchoice Dtwo monovalent fragments are produced, but they cannot participate in cross-linking. The correct choiceisC.



Fab

epitope

-£I

Before IgG is treated with papain. After IgG is treated with papain.

C is correct, IgM. In the 6th paragraph of the passage we find that IgG is the only antibody to cross the placentalmembrane. Even though this antibody enters into fetal circulation and can confer fetal immunity to certain diseases,it is IgM which is the first antibody produced during an immune response. Also, it is the predominant antibodyproduced by the fetus during fetal development. The correct choice is C.

B is correct, IgA. In the 6th paragraph of the passage we see that IgA is primarily found in external secretions suchas tears, saliva, colostrum, and milk. This question is asking us to recall a little information about lysozyme. Invertebrates, this enzyme is widely distributed in a variety of cells and secretions. The connection that needed to bemade was that both lysozyme and IgA are found in bodily secretions. The actions of the other immunoglobulins areoutlined in the passage. The correct choice is B.

Passage II (8- 14) Acquired Immunity

8. C is correct, no, receiving antibodies confers transient immunity. Passive immunity means receiving antibodiesfrom another organism. The recipient does not make antibodies in this situation. Thus, choices B and D areincorrect. Injected foreign antibodies eventually degrade and are not replaced, since an immune response was nottriggered in the recipient. The effect is not permanent. It is transient. Eliminatechoice A. The correct choice is C.

9. A is correct, maternal antibodies provided immunity to measles in this situation. The newborn has an immatureimmune system. She is protected by passive immunity via maternal antibodies she received in the uterus and inbreastmilk, if she is fed that way. She cannot respond quickly to primary exposures of viruses until after about 1month of age. Choice B is incorrect. Although antibodies may be present in saliva in an extremely small amount,this is not how passive immunity is acquired. Choice C is incorrect. Paternalantibodies stay with the father. ChoiceD is incorrect. The correct choice is A.

10. C is correct, both IgG and IgM remain at high levels following a "booster shot". We are looking for the FALSEstatement. All of the questions can be answered by consulting the graph. A "booster shot" means receiving avaccination to boost the antibody effects of an earlier vaccination. On the second (and further) exposures, we can seein the figure that IgG provides the primary response. Choice A is TRUE. Before a person is exposed to an antigen,there are no antibodies for that particular antigen. ChoiceB is TRUE. The primary response to an antigen is a rise inIgM. This, as you can see in the figure, precedes a rise inIgG. Choice D is TRUE. Both IgG and IgM fall off in theirlevels following each expose. They do not remain at high levels. Choice C is FALSE. Since weare looking for theFALSE answer, choose C. The correct choice is C.

11. B is correct,artificially acquired passive immunity. Since the person receives antibodies from someone else who isimmune to the particular antigen, then it must be passive. Choices A and C are incorrect. Passive natural immunitycomes from a mother-baby relationship. This person is an adult, so it must be artificial. Choice D is incorrect. Thecorrect choice is B.

12. A is correct, artificially acquired active immunity. A vaccine provides artificially acquired immunity. Choices Cand D are incorrect. Active immunity is created by the immune response of the body to the antigens in the vaccine.



When a second exposure occurs, the body has antibodies and lymphocytes ready to attack. Choice B is incorrect.The correct choice is A.

13. D is correct, y-globulin. A gel electrophoresis retards larger molecules, while allowing smallerones to migrate morequickly. The larger molecules are the slowest. The y-globulin moved the least in a given time, so they must be thelargest. For your information, the y-globulin fraction contains antibodies which can be used to confer passiveimmunity. The correct choice is D.

14. A is correct, I only. Breastmilk, especially colostrum, the breastmilk of several days immediately following birth, isimportant for all babies. This child could benefit from the antibody protection conferred by another mother'sbreastmilk. Choice I is correct. Seclusion would protect the baby from exposure to antigens, but it still would notmake more antibodies. When the baby's immune system matures, it needs exposure to antigens in order to makeantibodies. Eliminate choice II. Immediate vaccination would notproduce considerably greaterantibodies, since thebaby's immune system cannot function really effectively at birth. Choice III is incorrect. The correct choice is A.

Passage HI (15- 21) Major Histocompatibility Complex (MHC)

15. D is correct, both hydrophilic and hydrophobic regions. The first sentence of the passage states that the proteinswhich are encoded from the class I and class II MHC genes are expressed on the surface of a cell. This should strikea nerve in your brain. In particular, it should tell you that proteins expressed on the surface of the membrane musthave both a hydrophilic region and a hydrophobic region. The hydrophilic region must exist because, in this case,most of the protein is sitting in the extracellular medium, which will most likely bepolar due to extracellular fluid.The hydrophobic region is necessary because the protein is anchored to thecell by the plasma membrane of thecell,which isofcourse lipid in nature. Therefore, the protein should contain both a hydrophobic and hydrophilic region.Knowing this information, choices A and Bare merely incomplete. Choice C is redundant because hydroapathetic issimilar to hydrophobicity. The correct choice is D.

D is correct, host cellular in origin. This question is not really a trick question, it simply requires a little thinking.There are about one halfof a million to one million class I MHC proteins. There is no way you should know thatnumber, but you should realize that there are a lot of these proteins on the surface. Furthermore, the passage statesthat the peptides found with class I MHC are invariably the cell's own proteins. When the cell is infected with virus,the viral proteins/peptides will be presented to stimulate an immune response. But this does not change the fact themostclass I MHC molecules will still have theirownproteins being presented on the cell surface. Choices B and Cdo not make any sensebecause thosecells are involved with the recognition/destruction of cells infected with virus.There is no evidence to suspect thosecells will have their peptidespresented. The correct choice is D.

17. B is correct, signal peptide sequence. This question requires a solid knowledge of cellular action. It is stated in thepassage thatclass II MHC molecules have been shown toassemble in the endoplasmic reticulum. This requires thatthe proteins get to the endoplasmic reticulum (ER). Proteins that are destined for the ER have a special sequencecalled the signal peptide sequence which causes the ribosome they are being translated on to bind to the ER. Thesubsequent translation of the protein occurs, and the entire protein then finds itself within the ER. Consider the otheranswers. Choice A is incorrect because a Shine-Dalgarno sequence is a segment of RNA found in prokaryotes that isrecognized by ribosomes as a place of binding. Therefore, choice A is incorrect. Choices C and D are incorrectbecause the question asks about proteins, not DNA sequences. The correct choice is B.

18. C is correct, endocytosis. Recall that the ingestion of material by the invagination of the plasma membrane is calledendocytosis. This is exactly the process that occurs for the formation of these endosomes. This is a verystraightforward question. Consider the other answers. Choice A is a non-sensical answer and can automatically beeliminated. Choice B is the oppositeof endocytosis, and would call for the expulsion of cellular material through theplasma membrane. Transcytosis is the transport of material by a vesicle from one side of the cell to another. Thisusually occurs in epithelial cells. The correct choice is C.

19. B is correct, structurally similar to that of the class II MHC. The passage states that the CLIP part of the invariantchain is actively removed by binding to the DM molecule. The clip used to be bound to the class II MHC molecule.Based on this information, one could logically assume that the DM and class II MHC molecule are structurallysimilar, if they bind the same molecule. Considering the other answers, there is no evidence for any of thestatements. The DM molecule is binding CLIP, which does not mean they are structurally similar. It just means they

16.



form a bond. Choice D is incorrect because class I MHC is not involved with DM. Finally, DM is not likely similarto the invariant chain. Remember, CLIP is merely a part of the invariant chain, and if DM is not like CLIP, there isno reason to believe it will be like the invariant chain. The correct choice is B.

20. D is correct, II and III only. A given class I MHC molecule will bind to a variety of peptides. However, the cleftwhere the peptides binddoes notchange for a given class I molecule. In other words, the amino acids that make upthe cleft are invariant. The peptides that bind to the cleft can only do so if certain amino acids in their chain are ableto bind to the amino acids which make up the cleft. That is, amino acids at certain positions in the peptides must behighly conserved if a large variety of peptides arcable to bind toa class I molecule which doesnotchange its aminoacid composition. Therefore, choice III is correct. We find that choice II is correct based on the following logic:looking at Figure 1 and Figure 2, we can see that in class I molecules, the peptides are bound in the cleft by theirends. Therefore, the peptides canonly be so long, and as stated in thepassage, abouteight or nineamino acids long.However, with class II, the binding occurs in the middle, and therefore we will see more of a varietyof lengths in thepeptides which bind. The peptides which bind to class II molecules are not constricted to binding to the ends, so theydo not have to be of a certain length. The correct choice is D.

21. C is correct, production of a transcription factor which increases the transcription rate of the MHC gene early in aninfection. If there was a transcription factor that promoted the transcription of the MHC gene early in infection, thiswould promote an immune response against the pathogen. This isclearly not a defense mechanism for the pathogen.All of theother answers result ina depression of the immune response, and therefore can becategorized as a defensemechanism used by a pathogen to disrupt the antigen processing system. Thecorrect choice is C.

Passage IV (22 - 27) IgA Antibody Experiment

22. D is correct, I, II, and III.Any body surface that can contact the outside environment is lined with epithelial tissue.The mouth, is, of course, lined with epithelium. Choice I is correct. The entire GI tract is essentially "outside thebody." It iscompletely lined with epithelial tissue. Choice IIiscorrect. The urethra also has contact with the externalenvironment. It, too is linedwith epithelium. Choice III is correct. The correct choice is D.

23. B is correct, IL-6" mice produce lower levels of IgA than IL-6+ mice. Read the bar graph. The IL-6" mice havealmost no IgA compared to the IL-6+ group. Choice A is incorrect. Choice C is incorrect. The response toovalbumin was almost nil in the IL-6" mice. Choice D is incorrect. The correct choice is B.

24. A is correct, amino acids. Immunoglobulins areproteins, made of amino acids. Fatty acids arepartof many types oflipid molecules. Choice B is incorrect. Sphingomyelins are special fats that are found in nervous system tissue.Choice C is incorrect. Phospholipids arepartof cell membranes. Choice D is incorrect. The correct choice is A.

25. B is correct, restoring IL-6 in the IL-6" mice improved IgA production. The virus carrier alone did not change thegenetic makeup ofeither strain ofmice. Choice Ais incorrect. The IL-6+ mice simply responded to a viral infectionby increasing IgA slightly. Choice D is incorrect. The IL-6" mice did not respond at all to the virus carrier. Theywerehyposensitive. Choice C is incorrect. The correct choice is B.

26. D is correct, cytokines are local hormones and often act without passing through the entire circulatory system.Cytokines are special "local hormones" that do not always pass through the entire blood supply. Measurements madeon blood samples may not reflect what is happening at the tissue level. Choice A is incorrect. Hormoneconcentrations can be measured in blood samples by radioimmunoassay. Choice B is incorrect. IL-6 can and doesleave the cell that produces it, like other cytokines. How could they interact with IgA if they never left the cell andIgA is present in secretions? Choice C is incorrect. Thecorrect choice is D.

27. A is correct, gene therapy. You may have read about the first gene therapy trials in humans during the last year orso. This was to alter genes in people with inherited, disease-causing genetic defects. Specific viruses carried the newgenetic material. Hopefully, these viruses will infect the host's DNA with a good copy of their defective gene.Vaccination is just plain, old immunization with a modified form of a pathogen. Choice B is incorrect. Geneticimmunization is a fake answer, and Choice C is incorrect. Pleiotropy having a gene that affects many differentcharacteristics in an individual. Choice D is incorrect. The correct choice is A.



Passage V (28 - 33) Complement System

28.

29.

30.

31.

32.

33.

C is correct, solubilizc antigen-antibody complexes. This question just requires one to read the passage carefully.The passage states that persons who are complement deficient suffer from immune diseases in which antigen-antibody complexes form precipitates in the blood vessels of skin, joints and brain. These precipitates can go on tocause tissue damage. One can assume from this statement that complement plays a role in solubilizing antigen-antibody complexes because when the complement is not present, the precipitatesform. The correct choice is C.

C is correct, provides a means of amplification, ultimately leading to many MACs. The key word in this question isthe word cascade. One should be familiar with the idea of biological cascades. Each activated enzyme in thepathway cleaves many molecules of the next proenzyme in the chain. The activation of early complementcomponents consists of an amplifying proteolytic cascade where each molecule activated at the beginning of thesequence leads to the production of many membrane attackcomplexes. The correct choice is C.

B iscorrect, bone marrow. This answer isarrived at through one's knowledge of immune system development. Thequestion is asking about what disease would affect the production of antibodies. This would clearly be the bonemarrow. The bone marrow is the site of B-cell birth and maturation. B-cells are responsible for the production ofantibodies. Therefore, a disease of the bone marrow would greatly affect B-cell maturation and thereby affectantibody production. The thymus is the site ofT-cell maturation, while the spleen and the lymph nodes are what arecalled secondary lymphoid organs, where mature B-cells and T-cells "hang out" to react with foreign antigens. Thecorrect choice is B.

D is correct, constant region of the antibody. We know that there exists interaction/activation between the CIcomplement component and either IgG or IgM. The question is asking what is the most likely means by which CI isactivated. CI does not change depending on the microorganism. In other words, we do not have a different CI fordifferent microorganisms. However, we do have different IgGs and IgMs for different microorganisms. Thedifference lies in the variable region of the antibody. That variable region binds to the antigenic determinant of themicroorganism. The antibodies also have a constant region, which does not change (except for class switching) ordepend on the particular microorganism. It most likely that the CI, which remains constant, will recognize and beactivated by the constant region of the antibody. If this were not the case, CI would itself have to change torecognize the variable region of the antibody. The correct choice is D.

C is correct, confine complement activation to the cell surface where it began. The early components and C3 areproenzymes that are activated sequentially by limited proteolytic cleavage. When each proenzyme in the sequence iscleaved, it is activated to generate aserine protease which cleaves the next protein in the sequence. Many cleavagesliberate a small peptide fragment and expose a membrane-binding site on the larger fragment. The larger fragmentcan now bind tightly to the target cell membrane and carry out the next reaction in the sequence. The reactioncascade is constrained to the cell surface where it began, and where we want it to be. The correct choice is C.

B is correct, these molecules stimulate the secretion of histamine from basophils. The question tells us that anincrease is seen in the permeability of blood vessels. From the answers given, we know this is due to histaminesecretion. This histamine that is secreted increases the permeability of local blood vessels which will allow whiteblood cells, antibodies, and more complement to enter the site of infection. Thequestion becomes what cell secreteshistamine. The answer is that mast cells and basophils secrete histamine. Most are familiar with mast cells, butbasophils are a type of white blood cell. They fall under the category of granulocytes and they secrete histamine tohelp mediate inflammatory responses. The other types of cells in the answers are simply not involved in thesecretion of histamine. The correct choice is B.

Passage VI (34 - 40) Myasthenia Gravis/Autoimmune Diseases

34. A is correct, an inhibitor of acetylcholinesterase. Since thereceptors are blocked or not as numerous, an increase inacetylcholine (ACh) should remove some of the effects. One way to accomplish this is to give an inhibitor ofacetylcholinesterase so the ACh is not degraded as quickly. This would give higher concentrations of ACh at thereceptors. An immunostimulant would further complicate the problem by encouraging more antibody production.Choice B is incorrect. A paralytic agent would add to the muscle weakness. Choice C is incorrect. Decreasing theamount of ACh would be a bad idea, so choice D is incorrect. The correct choice is A.


Biology Immunology &? Endocrinology Section V Answers

35. C is correct, III. The nerve is the top or presynaptic part, while the muscle is the bottom, or postsynaptic part.Choice I indicates a vesicle containing ACh. Choice II indicated vesicle fused with the membrane and releasingACh into the synaptic cleft. Choice III is the acetylcholine receptor on the muscle cell. Choice IV is a receptor onthe nerve cell. The correct choice is C.

36. A is correct, Type IT diabetes. Type II diabetes is not caused by an autoimmune agent. It is mostly related to insulinresistance. Choices B, C, and D are clearly autoimmune diseases from the passage. The correct choice is A.

37. C is correct, lethargy. The effect of stimulating the TSH receptor causes the thyroid to secrete more thyroidhormone. This increases metabolic rate and can lead to hyperactivity in severe cases. Weight loss is probablyconnected due to increased metabolic rate. The person is not lethargic, which means tired and sluggish. The correctchoice is C.

38. B is correct, insulin. The beta cells of the pancreas produce insulin. Lack of insulin is the defect in Type I diabetes.You could figure this out from the table if you forgot. The other hormones are produced by the pancreas, but not inthe beta cells. The correct choice is B.

39. C is correct, I and II only. A virus can resemble the structure of a body molecule and lead to autoimmune attack.Choice I is correct. A virus can damage a cell so that cell components that do not usually travel in the blood areplaced in the blood.This could also lead to an autoimmune response. Choice II is correct. Choice III is the oppositeof choice I and is incorrect. The correct choice is C.

40. B is correct, adrenal gland. The adrenal cortex secretes steroid hormones. The correct choice is B.

Passage VII (41 -46) Glucose, Glucagon, St Insulin

41. B is correct, increased levels of glucagon and decreased levels of insulin. It is important to remember that glucose isneeded by the body as an energy source. Thus, in times of stress, when the body needs energy to deal with thestressor, glucoseplasma levels must increase. Sympathetic stimulation of glucagon secretion will increase glycogenbreakdown and increase glucose plasma levels. This will make glucose available for tissues to utilize as an energysource. The sympathetic system will inhibit insulin secretion. The correct choice is B.

42. B is correct, increased levels of glucagon and decreased levels of insulin. The major function of glucagon secretionis to increase the amount of products in the plasma that can be utilized for energy production. During fasting, bloodglucose levels are low, stimulating glucagon secretion. Glucagon secretion will increase the release of free fattyacids, which can be converted to ketone bodies and utilized for energy. However, if insulin levels are low, as indiabetes, ketone utilization is low and ketone products increase in the plasma beyond normal levels leading toketosis and acidosis. The correct choice is B.

43. A is correct, abnormal increases in insulinsecretion. Hypoglycemia, by its name, means "low blood glucose."Thus,abnormal increases in insulin secretion will decrease levels of blood glucose. Decreases in insulin secretion and highlevels of plasma glucose are signs of hyperglycemia. Low levels of free fatty acids may indicate increases in insulinsecretion or decreases in glucagon secretion. The correct choice is A.

44. A is correct, increased levels of insulin and glucagon. Increases in plasma amino acid concentrations will stimulateinsulin secretion to promote protein production in cells. Furthermore, glucagon secretion from the pancreas is alsoincreased. Glucagon secretion is needed to prevent hypoglycemia after a heavy protein meal since amino acids alsostimulate insulin secretion. The correct choice is A.

45. C is correct, access and storage of glucose in cells of the peripheral tissue. Insulin stimulates the storage of glucoseand eases the entry of glucose into tissue. The tissue affected most readily appears to be muscle and connectivetissue of the periphery. The brain does not need insulin for glucose uptake, as brain tissue is not affected by insulin.Insulin stimulates glucose uptake and storage by the liver. Insulin will decrease glucose loss in the kidney bydecreasing plasma glucose levels and increasing uptake of glucose by the tubular cells of the kidney. The correctchoice is C.



46. A is correct, extracellular glucose excess. The lack of insulin secretion will decrease the uptake of glucose by cells.Even with extracellular concentrations of glucose rising, cells cannot utilize the glucose because it cannot enter thecells. Therefore, tissues die of starvation, with high concentration of glucose located just outside the cell. Thecorrect choice is A.

Passage VIII (47 - 54) Calcium, PTH, Calcitonin, Si Calcitriol

47. B is correct, (see the graph below). In the 5th paragraph we read the following: "PTH causes calcium absorptionfrom bone by stimulating osteoclastic activity and transiently inhibiting osteoblastic activity. At the level of thekidney, PTH increases calcium absorption in the distal tubules and collecting ducts and greatly decreases thereabsorption of phosphate at the proximal tubules." This means that at the time PTH is administered we wouldexpect to see an increase in plasma calcium levels and a decrease in plasma phosphate levels. Since PTH is ahormone and goes everywhere the blood goes, we would also expect to see these effects happen at about the sametime, which is exactly what we observe in the graph for choice B.

0 'LT Time (hours) 6PTH added

We would not expect to see the levels of phosphate or calcium lag behind one another. This immediately allows usto eliminate choices C and D. Wecan eliminate choice A simply because we know (from thepassage) that the levelsof plasma calcium will increase. The correct choice is B.

48. B is correct, (see the reaction below). The answer to this question is found in the first paragraph of the passage. Weare told that vitamin D3 is also called cholecalciferol and that it was synthesized from a sterol derivative in aphotolytic reaction involving UV light. A sterol derivative is based on the ring structure of cholesterol. A photolyticreaction involves the breaking of bonds. Think of "lytic" as something being "lysed" or broken. If you compare thefour structures which are given as answers, you will note that choices B, C, and D are rather similar in that theycontain just three complete rings. Choice A contains four complete rings. Oneof the rings (called the "B" ring) inchoice A has been open up by the breaking of a bond. In this case that bond was broken by the action of UV light.Since vitamin D3 is formed from a photolytic reaction, we can eliminate choiceA as a possible answer. [It turns outthat choice A is called 7-dehydrocholesterol and is the precursorto vitamin D3.]

H,C

Vitamin D3 (cholcalciferol)

We are left with choices B, C, and D. It was stated in the passage that vitamin D3 is the inactive form of thehormone. It was activated by first hydroxylating it in the liver and then by hydroxylating it in the kidney. In otherwords, the active form of vitamin D3 (called lcc,25-dihydroxycholecalciferol (or l,25-(OH)2-D3)) has two more


Biology Immunology & Endocrinology Section V Answers

49.

50.

51.

52.

53.

hydroxyl groups on it than the inactive form. Therefore, we look for that structure which has not had twohydroxylation events. Based on this analysis we can eliminate choices C and D. The correct choice is B.

D is correct, I and HI only. In the second paragraph it was mentioned that in order to synthesize the activated formof vitamin D3 (l,25-(OH)2-D3), the hydroxylase enzyme in the kidneys must be activated (stimulated) by PTH.Once the activated hormone is synthesized, calcium and phosphate can be absorbed across the mucosa membrane ofthe intestines. The phosphate ion follows the calcium ion across the membrane so electrical neutrality is maintained.We can eliminate choice C because PTH by itself has no direct effect on either calcium or phosphate absorption inthe intestines. We can also eliminate choice B because calcitonin does not allow for intestinal absorption of calciumor phosphate either. Even though choice A is a correct answer, it is not the BEST answer. In order to get theactivated form of vitamin D, PTH is needed. Therefore, an increase in both PTH and subsequently an increase inl,25-(OH)2-D3 is needed for absorption in the intestine. The correct choice is D.

A is correct, increased osteoclast activity coupled with a hypercalcemic plasma. Be careful with what this questionis asking you to answer.The bottomline of whatis beingasked is: "Whatdoes it take in order to secrete calcitonin?"Calcitonin acts to loser plasma calcium levels. In order to lower those levels they must first be high. What leads tohigh calcium levels? Increasedosteoclast activity (i.e., bone resorption and the return of calcium and phosphate tothe blood) and a hypercalcemic plasma (i.e., a plasma in which the level of calcium is above normal). These twoconditions demand a decrease in the levels of plasma calcium. The secretion of calcitonin by the parafollicular cellsof the thyroid gland reduces the concentration of calcium in the blood. Calcitonin does this by inhibiting osteoclastactivity (i.e., inhibits bone resorption) and inhibiting calcium reabsorption at the level of the kidney. The correctchoice is A.

C is correct, (see the pedigree below). The characteristic to look for in an X-linked dominant trait is that all of thedaughters and none of the sons of males who carry the trait are affected. This is exactly what we see in thispedigree. Interestingly, the phosphate levels are not as low and the rickets are not as severe in an afflictedheterozygous female as in the afflicted male.

II D-r-# B-i-O n O-r-D D

fmAutosomal Dominant Autosomal Recessive

I #J-D

n i-r-a D

DDEO • 1 O-r-n

rv ODD

Autosomal Dominant

If any son is affected or if any daughter is unaffected, then the inheritance is autosomal and not X-linked.Remember, autosomal recessive is only expressed in homozygotes. The correct choice is C.

D is correct, phosphate and l,25-(OH)2-D3. Familial hypophosphatemia refers to the fact that there is inadequatephosphate in the plasma. As mentioned, this is an X-linked recessive disease and stems from the fact thatreabsorption ofphosphate at the level of the kidneys isimpaired. The addition ofcalcium will not affect the levels ofphosphate and so wecan eliminate choice A. Addition of phosphate alone will not make that much of a differencebecause it will not have an adequate means by which to be absorbed. We can eliminate choice B. We cam alsoeliminate choice C for the same reasoning. The addition of l,25-(OH)2-D3 along with the phosphate insuresadequate absorption by the intestines and (presumably) an increased reabsorption at the level of the kidney. Thecorrect choice is D.

C is correct, increased neural excitability. Let's consider the listof possible answers. In orderto do thiswe need toknow the meaning of hypoparathyroidism. The parathyroid glands secrete PTH and PTH allows for calciumabsorption from bone by stimulating the osteoclasts and transiently inhibiting osteoblasts. PTH stimulates calciumreabsorbtion at the kidneys and indirectly allows for calcium absorption at the intestine through the action of theactive form of vitamin D. A hypoparathyroid gland is one thatdoes not secrete sufficient amounts of PTH. As a



54.

result the levels of calcium in the body decrease below normal values. Not only is osteoclast activity decreased, butbecause calcium levels are low, osteoblast activity is decreased as well. With this information we can eliminatechoices A, B, and D. We are left with choice C as the correct answer. Why would hypoparathyroidism lead toincreased neural activity? A decrease in calcium in the extracellular fluid allows more sodium to flow through themembranes of nerve cells. Some nerves, like those in the peripheral nervous system, begin to spontaneouslydischarge. This will eventually give rise to tetany and if not correct immediately can lead to death. The correctchoice is C.

D is correct, a positive linear relationship exists between PTH secretion and the concentration of plasmacalcium.This statement is false. In order to make it true we would say that: "an inverse relationship exists between PTHsecretion and the concentration of plasma calcium. We can see this from the graph shown below. The first threechoices are all true as discussed in the passage and in the previous answers. The correct choice is D.

O

Calcitonin

-

Total Ca2+ concentration in plasma

Passage IX (55 - 61) Erythroblastosis Fetalis

55. D is correct, 0%. If this is the woman's first child, she is very unlikely to have antigens to the Rh factor. Exposureoccurs during late pregnancy, delivery, abortion, or miscarriage. Even though a Punnett square would indicate a 50%likelihood of an Rh-positive child, the mother must also have the anti-Rh antibodies from a priorsensitization event.In this case, she is probably not sensitized. The correct choice is D.

56. B is correct, the Rh factor is not a component of the ABO blood group system. The placenta acts as a barrier forsome things and allows passage to others. There is not indiscriminate mixing of maternal and fetal blood. Nor doesthe mother make the red blood cells for the fetus. Choices Aand C are incorrect. As we learned in the passage, anypregnancy, whether it is taken to completion or not, can trigger antibody formation in Rh- women who have Rh+fetuses. Choice D is incorrect. The correct choice is B.

57. C is correct, treat mother with antibodies against the anti-Rh antibodies. If the antibodies can be trapped andremoved by other antibodies, then the mother is not sensitized against Rh antigen for subsequent pregnancies.Treating the fetus would not help future fetuses. Choice D is incorrect. In extreme circumstances, the fetus, not themother, receives a blood transfusion. This would not help future pregnancies, either. The correct choice is C.

58. D is correct, all of the above. All three of these organs produce bilirubin, a break-down product of heme. The liverconjugates the bilirubin with glucuronic acid, so it is soluble, and secretes it into the gall bladder. The correctchoice is D.

59. A is correct, no, there are no maternal antigens to the Rh factor antigens. The term positive means the antigen ispresent on the RBC. The mother does not have antigens to herself. The fetus is negative, so there are no Rh antigenson the RBCs to make antibodies against. Choice B is incorrect. The newborn is not immunologically competent,able to make antibodies, until about 1 month after birth. It relies on the antibodies donated by the mother. Choices Cand D are incorrect. The correct choice is A.

60. D is correct, I, II, and III. Hemolysis in the Rh-positive fetus depends on the amount of antibody produced by itsRh-negative mother. If a large amount of blood has been transferred between the fetus and the mother, then she ismore likely to produce anti-Rh factor antibodies. Choice I is correct. How her antibody production responds to agiven stimulus is also a factor. Choice II is correct. As the number of pregnancies increases, her exposure to Rh


Biology Immunology fie Endocrinology Section V Answers

positive blood increases, and the likelihood of developing antibodies increases. Choice III is correct. The correctchoice is D.

61. B is correct, I and II only. The primary problem in EF is lysis of red blood cells. This leads to anemia and lowerhemoglobin level. Choice II is correct. If RBCs are destroyed, this does not lead to an increased number.Erythrocytes are RBCs. Choice III is incorrect. Bilirubin is a breakdown product of heme. It is elevated in neonatesand fetuses with EF. Choice I is correct. The correct choice is B.

Passage X (62 - 67) Septic Shock

62. A is correct, does not have an outer membrane on its cell wall. This question requires one to know a little about thedifferences between Gram-positive and Gram-negative cells. A Gram-positive cell does not have an outer membraneon its cell wall. This information eliminates choice B. Furthermore, a Gram-positive bacterium has a very thickpeptidoglycan layer adjacent to the plasma membrane. The correct choice is A.

63. C is corrrect, endocytosis, engulfing foreign particles which eventually will fuse with a lysosome. The questionrequires us to use our knowledge about how macrophages are functioning. They are phagocytes in the body alongwith the neutrophils. The cells engulf microorganisms (this eliminates choices A and B) which then becomephagocytic vesicles. These vesicles fuse with the cell's lysosome (eliminating choice D), an organelle with containshighly reactive molecules like superoxides. The fustion of thephagocytic vesicles with the lysosome exposes theforeign particles to these caustic molecules, acting to destroy the microorganism. The correct choice is C.

64. C is correct, plasma cells. This question requires us to draw on our knowledge of the immune system and the cellswhich comprise that system. Antibodies are associated with B cells, which are lymphocytes. Antibodies can befound in two forms. One is the membrane bound form, attached to the B cell at its plasma membrane. When a B cellbecomes activated, it will proliferate. Some of the new B cells will produce the antibody (it will be the sameantibody) in a soluble form, enabling the antibody to circulate in the lymph and blood. B cells which make solubleantibody are called plasma cells. Therefore, a defect in plasma cells would affect production of soluble antibodies.The correct choice is C.

65. C is correct, rise in the patient's blood pressure. Anti-Factor XII antibodies will take out (render non-functional)some of the available Factor XII. This will affect the productionof bradykinin and the intrinsic coagulatory pathway.Since not all Factor XII is non-functional, some intrinsic pathway leading to coagulation may occur. We still havethe entire extrinsic pathway. Therefore, we should not see either a total loss (choice A) or a rise (choice B) in theintravascular clotting of the sepsis patient. As stated above, we are attenuating the role of bradykinin and one of itsrole is to increase the permeability of the vessel. Increasing the permeability leads to a loss of fluid and hypotension.Since we are attenuating the role of bradykinin, we should therefore see a rise in the patient's blood pressure. Thecorrect choice is C.

66. D is correct, 16. The flow of blood through a vessel is proportional to the radius to the fourth power. If we see anincreasein the radius of a factorof two, the increase in flow is 24 =16. The correct choice is D.

67. A is correct, hypotension in the patient. One should realize from the passage that bradykinin acts in two ways. Itincreases the amount of blood flowing as a vasodilator, and it increases the permeability of the endothelial cellsmaking up the blood vessels. The increased permeability causes a lot of the plasma in the blood to leaves the vessel(through osmotic forces ) and enter into the tissue. The result of this is that the patient becomes hypovolumeic whichwill lead to a drop in blood pressure. Therefore, bradykinin will lead to hypotension in the patient. The correctchoice is A.

Passage XI (68 - 75) Calcitonin St Osteoporosis

68. C is correct, decreased osteoblast activity, increased osteoclast activity. When bone synthesis is low (decreasedosteoblast activity) and bone breakdown is high (increased osteoclast activity), the amount of bone decrease is thegreatest. The correct choice is C.

69. C is correct, calcium. The question asks for a mineral. Eliminate choice B, water, because water is a compound, nota single atom. B \2 is a vitamin, not a mineral, so choice A is incorrect. Since calcium was mentioned in the passage,



consider it. Calcium is the major mineral making up bones. Our body contains about 1 kg of calcium. Although zincis part of many enzymes, our body contains only a lew grams of zinc. The correct choice is C.

70. A is correct, PTH increases plasma calcium. The passage states the CT is an antagonist of PTH. This means theyhave opposite actions. CT works to decrease plasma calcium, and PTH works to increase plasma calcium. Choice Bis incorrect. To increase plasma calcium, the kidney decreases urinary calcium and phosphate. Choices C and D areincorrect. The correct choice is A.

71. C is correct, cysteine-cysteine. This is a biochemistry trivia question. Two cysteine residues link together via theirsulfur atoms to form a cystine molecule that contains a disulfide bridge. Cysteine is the single amino acid, whilecystine refers to a dimer of cysteine. The correct choice is C.

72. B is correct, I and III. Choice II is incorrect. Since CT is a polypeptide hormone, it does not enter the cell, but ratheracts on a receptor on the plasma membrane. Choice I is correct. The salmon CT is less recognizable to the humanenzymes, so it resists degradation more and has a longer-lasting action. Choice III is correct. Salmon CT has astronger interaction with the CT receptor and promotes a longer signal on the receptor. The correct choice is B.

73. A is correct, CT secretion increases. In the beginning of a supplement program, a person would have increasedcalcium levels. In response to these increased levels, CT secretion would increase to bring the levels back down tonormal. The correct choice is A.

74. B is correct, injection to avoid hydrolysis. Since CT is a polypeptide, it is subject to digestion (hydrolysis) in thestomach. Ingestion means to take orally, so choices A and C are incorrect. There is a possibility of an allergicreaction when proteins are injected, so choice D is incorrect. The correct choice is B.

75. D is correct, I, II, and III. In a systemic allergic reaction, histamine, leukotrienes, and prostaglandins are releasedand lead to skin reddening, hives, and constriction of the bronchioles (leads to breathing difficulty), among othersymptoms. The correct choice is D.

Passage XII (76 -83) Vertebrate Immune System

76. D is correct, little or no change in the effectiveness of either type of immune response. The question asks about theconsequences of removing the thymus from an adult human. From the passage we learn that T cells are produced inthe bone marrow and later mature in the thymus during early development. The thymus is therefore crucial duringearly development for the production of T cells, which are responsible for the cell-mediated response. If the thymusis removed from a mature adult, however, there will be little or no impact on the immune system since most T cellformation has already occurred. The tempting answer is choice B, but remember to read all of the answer choicesbefore making a decision. A careful reading of the question in this case is also a must. The correct choice is D.

77. C is correct, the cell would lyse due to an upset water balance. When pores are formed in the membrane of thetarget cell, small molecules (i.e. charged ions) are suddenly able to flow freely along their electrochemical gradients.The effect of this would be to disrupt the osmotic balance maintained by the cell. This balance is necessary becausethe large number of negatively charged macromolecules in the cell (proteins, nucleic acids, etc.) would normallyattract a large number of cations in order to balance the charge. The entry of such a large number of cations woulddraw water into the cell with them, a catastrophe which would cause lysis, or bursting, of the plasma membrane. Innormal cells, this is prevented because the membrane is mostly impermeable to small charged ions, and the action ofmembrane transporters serves to maintain an appropriate osmotic balance. The action of the complement proteins isto poke holes in the membrane, allowing ions and water to rush into the cell, causing lysis; this is an efficient way ofkilling invading microorganisms. Choice D is the opposite of what was just described and is therefore incorrect.Choice A is also wrong; an inability to initiate action potentials is not lethal to the cell. Additionally, most cellswhich are attacked by the antibody/complement system are invading microorganisms which do not have anything todo with the neuronal function of conducting action potentials. Choice B is wrong because if the cell is normallynegatively charged inside, holes in the membrane would allow cations to rush in, leading to depolarization ratherthan hyperpolarization. The correct choice is C.

78. C is correct, II and III only. Statement II states that organ transplants between monozygotic twins do not result inimmune rejection of the transplanted organ. This implies that the recipient's immune system regards the new organ


BlOlOgy Immunology &Endocrinology Section vAnswers

as being "self." Since monozygotic twins aregenetically identical (as opposed to dizygotic twins), this statement isconsistent with Theory 1, which states that individuals lacks the genes toproduce a response against selfantigens.Since both twins have the same genetic structure, they both lack the genes necessary tomount an immune responseagainst theirowntissues; therefore, according to Theory 1 thetransplanted organis not rejected. Statement II is alsoconsistent with Theory 2, which inessence states that the immune system "learns" nottoattack itsown tissues earlyin development. Since both twins had theexactly the same tissues (the result of their identical genes), both of theirimmune systems "learned" not to attack the same tissues. This also explains why the transplanted organ was notrejected. Statement HI is, for all intents andpurposes, an identical situation to thatpresented in Statement n. In thiscase, genetically identical mice from the same inbred strain are able to tolerate transplants from each other. Aninbred strain, incidentally, is a strain of mice that has been inbred formany generations, resulting in siblings whicharegenetically identical to one another. Statement I states thatcellsfrom a foreign cell lineare injected intoa mouseat theembryonic stage. When themouse matures, it does not mount an immune response to further injections of thesame cells. Thissupports Theory 2 in that it provides evidence that the immune system "learns" during embryonicdevelopment not to attack cells that are present at the time (i.e., self cells). The immune system is fooled intothinking theforeign cells areselfcells. This contradicts Theory 1.According to this theory, theforeign cells wouldstill be attacked even if they were injectedat the embryostage.The correct choice is C.

79. B is correct, weakness and difficulty breathing. The question states that antibodies are produced against skeletalmuscle acetylcholine receptors. This would most likely result in muscle weakness (due to an inability of musclefibers to receive signals from effector nerves). Difficulty breathing would also be likely because many skeletalmuscles (i.e., the intercostals) are involved in the respiratory process. This problem could also be solved via aprocessof elimination. AnswerchoiceA can be eliminated becausethe heart is made up of cardiac muscle. Since theantibodies only attack skeletal muscle receptors, the heart should not be affected. Likewise, the smooth muscle ofthe gastrointestinal tract should not be affected, ruling out answer choiceC. ChoiceD is unlikely to be true as well.Neurons in the brain should not be affected by skeletal muscle receptor antibodies, and therefore no dementia shouldoccur. The correct choice is B.

80. B is correct, the B cells have considerably moreroughendoplasmic reticulumthan the T cells. Since the function ofactivated B cells is to produce and secrete large amounts of antibody, it would be likely that under electronmicroscopy they would appear to have more rough endoplasmic reticulum than activated T cells, which do not haveas great of a secretory role. Rough ER is where proteins thatare to be secreted are mainly produced. The "rough"appearanceof this organelle is due to the presenceof ribosomes dotting the convolutedmembrane of the ER. Theseribosome produce the antibody proteins. Choice A is not necessarily a likely observation, as neither cell has anespecially great need for mitochondria relative to the other. Answer choice C is also incorrect as T and B cells bothhave nuclei. Do notconfusethemwithanothertypeof blood cell which does lacka nucleus: the erythrocyte. Answerchoice D is incorrect because electron microscopy is the most powerful means of visualizing cells and as such isvery likely to detect differences between B and T cells. The correct choice is B.

81. A is correct, each B cell is predetermined to bind a specific antigen. First, let's go over the experiment in thequestion. A population of B cells is removed from an adult mouse and mixed with an effectively lethal antigen X.The B cells that have the appropriate surface receptors bind the antigen are killed due to the radioactivity. Very fewB cells are killed, therefore very few out of the total population (<0.01%) recognize antigen X specifically. When thesurviving B cells are injected into an irradiated mouse (lacking B cells), the mouse is now capable of responding toother antigens, but not antigen X. It can.be implied that the reason for this was the death of the B cells whichspecifically recognized antigen X. The remaining population of B cells had no members which recognized antigenX, though they could recognize other antigens. Therefore we can conclude that each B cell is predetermined to binda specific antigen. B cells can not recognize new antigens and "learn" to bind to them, making answer choice Bincorrect. Answer choice C is wrong because the above experiment implies nothing about T cell interference with Bcell binding. Answer choice D is wrong because an antigen, by its very definition (antibody generator), isimmunogenic. Therefore antigen X does elicit an immune response normally. The correct choice is A.

82. B is correct, antigens can only be bound by a single antibody at a time. The question is asking which conditionwould not hold true if antibodies were capable of "teaming up" and cross-linking antigens together. We canapproach this problem by first theorizinghow antibodies could accomplish this feat. First, if antigens could be boundby more than one antibody,and each antibody had two antigenbindingsites connected by a flexible "hinge" region,cross-linking could easily occur. Each antibody would bind two antigen particles, which would then bind anadditional two antibodies, which would then bind another two antigen particles, etc. The flexible hinge region wouldallow greater flexibility during the cross-linking process. The above description would make answer choices A, C,



83.

and D all consistent with the cross-linking phenomenon. This leaves choice B as the answer. If antigens could onlybe bound by a single antibody at a time, the above scheme would not be possible and there could be no cross-linking. The correct choice is B.

A is correct, most types of cells in the body. The key concept to remember here is thatviruses can infect most typesofcells in thebody. Since the question states that cells infected with viruses are usually killed by cytotoxic T cells, itlogically follows that the T cells should be able to target most cells in the body. The other answer choices mentionspecific cell types and are therefore incorrect. The correct choice is A.

Passage XIII (84 - 90) Insulin Receptor

84. C is correct, detergent wash. The passage informs us that the insulin receptor is an integral transmembrane protein.The receptor is thus embedded in the lipid bilayer and actually spans the entire layer from the extracellular to theintracellular face. These proteins arc tightly bound to the lipid bilayer by hydrophobic forces and harsh conditionsare needed to remove them. In particular, detergents must be used. The reason is that these molecules areamphipathic, containing both a polar and non-polar region. The non-polar region binds to the hydrophobic regionsofthe membrane proteins, disrupting the lipid bilayer. The polar region creates a water soluble micelle with themembrane protein (hydrophobic portion) in the middle. The receptor is thus isolated and ready to study. Thecorrect choice is C.

85. B is correct, transduction mechanisms. We knowfrom thequestion that we have a hybrid receptor which is made upof a insulin binding region and a EGF tyrosine kinase segment. From the passage, we know that the insulin receptoris believed to be a insulin activated tyrosine kinase. Recall that the binding of insulin leads to tyrosine kinaseactivity. The mechanism which brings about this activity is termed the signal transduction pathway. With our hybridreceptor, the binding of insulin leads to kinase activity in the EGF segment. This information indicates a similarsignal transduction mechanism. The correct choice is B.

86. C is correct. The question tells us that the binding of glucagon will result in the production of the secondarymessenger cAMP. The values of cAMP should thus rise with time. Using this piece of information, we can eliminatechoices A and B. The question then becomes one of knowing the sequence of events. To answer this, one has to useprevious knowledge and/or a little common sense. The secondary messengers produced lead to changes in the cell.Recall that to truly bring about change for the cell, there must be some change in proteins, as proteins are the key tofunction. One very common way to regulate proteins is through their phosphorylation and their dephosphorylation. Itis the secondary messenger that promotes the phosphorylation of certain peptides, converting these proteins intotheir active forms. The active forms of the proteins bring about the change. From this information, the correctsequence of graphs can be seen. The correct choice is C.

87. C is correct, a normal insulin receptor. The question refers to a type I, insulin dependent diabetes. These individualsdo not produce insulin. That is why they are dependent. This should clue us into the fact that their receptors morethan likely have nothing wrong with them. The problem is that these individuals have no beta cells in the pancreaswhich release insulin. These individuals take insulin and their body responds appropriately. From this information,we can conclude that the receptors are more than likely normal. The correct choice is C.

88. D is correct, stimulation of pyruvate dehydrogenase. We know from the passage that GIP mimics certain insulinresponses. Therefore, we should look for an insulin response. Choice D indicates a stimulation of pyruvatedehydrogenase, the enzyme converting pyruvate into acetyl CoA. This would indicate an acceleration of glycolysisand the Krebs cycle. This is certainly one of the effects of insulin, as the hormone promotes the use of glucose as asource of fuel. The correct choice is D.

89. C is correct, increased CO2 production. The question tells us that the addition of bivalent anti-insulin receptorantibodies induces a response without insulin. What will antibodies cause? The key word here is bivalent. Thebivalent structure will induce cross-linking, leading to a clustering of insulin receptors. This conclusion is bolsteredby the next sentence of the question which states that monovalent antibodies, which cannot produce cross-linking,do not induce a response in the absence of insulin. However, if we add antibodies against these monovalentantibodies/fragments, we should be able to induce cross-linking and thus clustering. Therefore, we should be able toinduce an insulin-like response, which involves the promotion of glucose as a fuel. When glucose is oxidized, thecarbons are released as carbon dioxide. The correct choice is C.



90. B is correct, reducing conditions. This question is answered from our understanding of the conditions which arenecessary to form or break disulfide bonds. When two cysteine residues are oxidized, the bond is formed and theyform cystine. In order to break the bond, cystine must be placed in reducing conditions. From the passage, we knowthat the dimer insulin receptor is held with disulfide bonds. To isolate a component of the receptor for study,reducing conditions wouldbe used to breakthesedisulfide connections. The correct choice is B.

Passage XIV (91 -95) Vitamin D3, PTH, & Calcitonin

91. B is correct, keratinocytes. Hepatocytes are cells found in the liver. The liver has numerous functions, such as theregulation of metabolism of carbohydrates, lipids, and proteins. The liver stores glycogen and is the primary site ofgluconeogenesis. The liver is a major source of cholesterol in the body and is a major storage site for iron. The liveris also the site where many hormones are degraded and where many toxins are inactivated. Leukocytes (white bloodcells) are a cellular constituent of blood, along with erythrocytes and platelets, and they play an important role in thebody's defensive system. Leukocytes come in five classes: neutrophils, eosinophils, basophils, monocytes, andlymphocytes. Lymphocytes can be categorized as being either B-cells, which mature in the bone marrow, or T-cells,which mature in the thymus. Lymphocytes participate in the immune response.

Keratinocytes are skin cells of the epidermis. This cell type (of all the cell types listed) is closest to the exposure ofultraviolet light and allows for the conversion of 7-dehydrocholesterol to previtamin D3. In the skin previtamin D3

can be converted to vitamin D3, which is the precursor to the active form, l,25-(OH2)-D3. The correct choice is B.

92. C is correct integument. In the passage it states that humans obtain vitamin D3 either by ingestion or throughsynthetic mechanisms that are initiated by cells in the epidermis of the skin. If we ingest vitamin D3, it will end upbeing absorbed by the cells of the intestine. However, vitamin D3 is not the primary precursor for l,25-(OH2)-D3. Inthe skin (integument) the primary precursor (7-dehydrocholesterol) reacts with UV light to form previtamin D3,which in turn will form vitamin D3. The correct choice is C.

93. C is correct, II and III only (polypeptide and steroid). In the second paragraph of the passage we are told that 1,25-(OH)2-D3 passes into a target cell and then into the nucleus where it complexes with a receptor protein that has aDNA binding site. This is the general mechanism for action of steroid hormones. We can eliminate choice A as apossible answer.

In the third paragraph we find that both PTH and CT have genes encoded in the short arm of human chromosome 11and are synthesized as a preprohormone from different primary RNA transcripts. If they are both synthesized as apreprohormone from an RNA transcript, it must mean that amino acids are being linked together in peptide linkagesat the ribosome. In other words, a polypeptide is being synthesized. The mature PTH polypeptide contains 84 aminoacids while the mature CT polypeptide contains 32 amino acids. These three hormones fall under the steroid andpolypeptide classes of hormones. The correct choice is C.

94. D is correct, I and III only (hypocalcemia and hyperphosphatemia). What does PTH do under normalcircumstances? Based on information in the passage we find that PTH acts on bone and on the distal tubule of thekidney to promote Ca2® reabsorption and inhibits the reabsorption of PO4-*® in the proximal tubule of the kidney.PTH also stimulates the synthesis of l,25-(OH)2-D3 in the kidney. We saw that l,25-(OH)2-D3 promotes the uptake

of Ca-® and PO^® from the intestinal lumen. If we did not have adequate levels of PTH secretion, we wouldexpect low levels of Ca2® reabsorption (hypocalcemia) and a decrease in urinary PO^® excretion. A decrease inurinary PO^® excretion means that PO^® is being reabsorbed by the proximal tubules of the kidney. This willlead to high (hyperphosphatemia) plasma levels ofPO^®. The correctchoice is D.

95. B is correct, (see the graph below). Be careful when reading the graph. Along the x-axis we see the total plasma[Ca2®], ranging from low to high values. What is going to happen when the plasma levels of Ca2® are low? We aregoing to want to place more Ca2® in the plasma. How can we do this? We increase the levels of PTH. In the fourthparagraph of the passage it states that PTH acts on bone and on the distal tubule of the kidney to promote Ca2®



reabsorption. As the levels of Ca2® begin to increase the levels of PTH will begin to decrease. This isexactly whatwe see in the graph for parathyroid hormone.

low high

Total Plasma [Ca2+]

What happens when plasma Ca2® levels are high? The parafollicular cells begin to secrete calcitonin which acts onosteoclasts of bone tissue. As the calcium levels are lowered (by a mechanism that is not completely understood) thecalcitonin levels begin to drop. This is exactly what is shown in the graph for calcitonin. The correct choice is B.

Passage XV (96 - 100) Sepsis Syndrome

96. B is correct, diapedesis. Although the answer is not given in the passage, let's use our test-taking skills. Choice A isincorrect because mitosis is the process of duplication of genetic material and make 2 cells in somatic cells.Phagocytosis and endocytosis involve a cell engulfing some outside particle. Choices C and D are incorrect. Finally,the correct term for WBCs squeezing out is diapedesis. The correct choice is B.

97. D is correct hydrolyze bacterial proteins. Think of what enzymes do and how they are named. A lipase hydrolyzeslipids, so choice A is incorrect. An amylase or glucosidase hydrolyzes carbohydrates, so choice B is incorrect. Anuclease would hydrolyze DNA, so choice C is wrong. A protease or proteolytic enzyme hydrolyzes proteins. Thecorrect choice is D.

98. D is correct, toxic oxygen products. Toxic oxygen products are toxic because they are free radicals or decompose toform free radicals. The passage explains the communication role of cytokines, so eliminate choice C as incorrect.The proteases directly attack bacteria protein and hydrolyze bonds, but do not use free radicals. Eliminate choice B.Chemical attractants signal neutrophils to move to infection area, so eliminate choice A. The correct choice is D.

99. C is correct, TNF is mainly present in the general circulation. In the question, we are told that binding TNF halts theabnormal cytokine cascade. Therefore choice A is incorrect. Usually proteins that enter cells by endocytosis aredegraded in lysosomes. If the receptors were taken up in this manner, then they would be degraded and not able tointeract with TNF. Choice B is incorrect. If the TNF receptors are injected in the general circulation and work, thenTNF must be present in the general circulation. This means choice C is correct and must contradict choice D. Thecorrect choice is C.

100. A is correct, neutrophils. By correct reading, we can eliminate choices B and C. Erythrocytes are red blood cells, sochoice B is incorrect. Leukocyte is simply another name for white blood cell, so choice C is incorrect. Neutrophilsare 60-70% of the WBC population, making them the most abundant. The correct choice is A.


Physiology

Diagnostic Passagesand Questions

Set I

BerkeleyJJr-e-v-le-w®


DIRECTIONS: A descriptive passage precedes many of the questions found in theBiological Sciences section of the exam. The questions are arranged into groups.Examine a passage before selecting the one best answer choice to each question inthe group. Therewill be some questions which areindependent of a descriptive passageand independent of each other. The one best answerchoice must be selected for thesequestions as well. If you are not sure of an answer, eliminate the alternative choicesthat youknowto be incorrect andthen choose an answer from the remaining alternatives.Indicate your choice by darkening the corresponding bubble on your answer sheet.A periodic table of the elements is provided for you at the end of this book, and you

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Physiology

Diagnostic Passagesand Questions

Set I

7 Passages with 39 Questions

13 Independent Questions

52 Questions Total

70 Minutes

BerkeleyUr-e-v-i^e-w®



The thyroid, one of the largest endocrine glands in thehuman body, is responsible for secreting thyroid hormone(thyroxine). This hormone is primarily responsible forregulating basal metabolic rate. Excess secretion of thyroidhormone leads to an abnormally high rate of metabolism,resulting in a range of symptoms including weight loss,increased appetite, and profuse sweating.

The secretion of thyroxine is stimulated by thyroid-stimulating hormone (TSH), a peptide hormone. TSHsecretion, in turn, is stimulated by thyrotropin-releasinghormone (TRH), a peptide containing 3 amino acids. Theinteraction of TRH, TSH, and thyroxine is shown below inFigure 1.

Hypothalamus

TRH ICIAnterior Pituitary

TSHIThyroid Gland

Thyroxine

Figure 1

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H H 0

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TSH stimulates the cells of the thyroid gland to take upiodine from the blood for use in the synthesis of thyroxine.Due to this iodine uptake, patients with cancer of the thyroidusually have a good prognosis even if the cancer has spread.This is because after surgical removal of the cancerousthyroid, radioactive l3,I can be administered, effectivelykilling (via radiation) the remaining cancerous thyroid cellsthat absorb it.


Radioactive iodine treatment is extremely effectivewhile having relatively few side effects. This is mainlybecause:

A. most cells of the body can take up iodine.B. once inside a cell, ,31I is non-toxic.C. iodine is only absorbed by thyroid cells.D. cancerous thyroid cells lose their ability to take

up iodine from the blood.

Surgical removal of the .thyroid without hormonalsupplementation might lead to:

A. increased appetite and food intake.B. weight gain.C. profuse sweating.D. increased oxygen usage by the tissues.

Thyroid hormone is most likely a derivative of theamino acid:

A. threonine.

B. tyrosine.C. histidine.

D. methionine.

Radioactive iodine treatment is most effective several

weeks after the surgical removal of the cancerousthyroid gland. This is most likely because by this time:

A. cancerous thyroid cells have had a chance todivide and spread further.

B. thyrotropin-releasing hormone levels havedropped.

C. thyroxine levels have risen, allowing increasediodine uptake.

D. thyroid-stimulating hormone levels have risen,allowing increased iodine uptake.

Can radioactively labeled TSH be used as a thyroidcancer therapy?

A. Yes; TSH selectively enters cells of the thyroid,allowing them to be destroyed by the radiation.

B. Yes; TSH stimulates the production ofradioactive thyroxine.

C. No; TSH binds to surface receptors on thyroidcells, thereby never actually entering these cells.

D. No; TSH is taken up by many tissues other thanthe thyroid.

GO ON TO THE NEXT PAGE


Over-the-counter pregnancy tests which are simple touse and fairly accurate have been available for several years.Although there are a large variety of such tests, all of themshare a set of common principles.

These home pregnancy tests are basicallyimmunoassays which detect the presence of the peptidehormone human chorionic gonadotropin (hCG). hCG isproduced by the fertilized ovum and released into themother's bloodstream, preventing the degradation of thecorpus luteum and thereby preventing menstruation. By thetime of the mother's first missed period, hCG levels in theurine are high enough to detect.

The home pregnancy test generally involves dipping astick containing immobilized hCG monoclonal antibodiesinto a urine sample. If hCG is present in the urine, it willbind to the monoclonal antibodies on the dipstick. Thedipstick is then placed in a solution containing a secondmonoclonal antibody which recognizes the bound hCG-antibody complex on the surface of the dipstick. This secondantibody is conjugated to colloidal gold particles whichchange color when they are immobilized, indicating apositive test result (i.e., pregnancy).

6.

7.

Why isn't the presence of progesterone used as anindicator of pregnancy in the home pregnancy test?

A. Antibodies to steroid hormones can't be made.B. Progesterone is secreted too late in the pregnancy

to be of any diagnostic value.C. Progesterone is not made exclusively by the

fertilized zygote.D. Progesterone is not present in the bloodstream.

The mechanism by which hCG acts on its target cells ismost similar to that of:

A. testosterone.

B. glucagon.C. progesterone.D. estradiol.


8. As described in the passage, the home pregnancy test:

A. lacks positive and negative controls.B. is only effective after at least one month of

pregnancy.

C. is inaccurate because hCG can be producednormally in the non-pregnant mother.

D. is inaccurate because hCG is sometimes notproduced by the fertilized zygote.

9.

10.

Which of the following can be concluded frominformation given in the passage?

A. hCG is also secreted by the mother.B. hCG does not enter the mother's bloodstream.

C. The tubules of the kidney do not reabsorb all thehCG filtered in the glomerulus.

D. The actions of progesterone are counteracted byhCG.

The same techniques used in the home pregnancy testcan be implemented in other clinical tests. Which ofthe following could NOT be detected using suchtechniques?

A. Strep throat caused by a particular strain ofbacteria

B. Autoimmune diseases such as rheumatoid

arthritis

C. HIV infection

D. Tryptophan blood levels

11. The monoclonal antibodies used in the home

pregnancy test:

A. come from a single T-cell clone and bind to asingle antigen site on the hCG molecule.

B. come from several different T-cell clones andbind to multiple antigen sites on the hCGmolecule.

C. come from a single B-cell clone and bind to asingle antigen site on the hCG molecule.

D. come from a single B-cell clone and bind tomultiple antigen sites on the hCG molecule.


Questions 12 through 15 are NOT based on adescriptive passage.

12.

13.

Whales are capable of diving to great depths andremaining underwater for extremely long periods oftime. Which of the following would LEAST contributeto these abilities in whales?

A. Large muscle myoglobin concentrationsB. High basal metabolic rateC. High cellular tolerance for CO2D. Large lung capacity

Which of the following digestive enzymes are NOTproduced by the pancreas?

A. PepsinB. ChymotrypsinC. LipaseD. Carboxypeptidase


14.

15.

Colchicine is a drug which prevents the polymerizationof microtubules. Which of the following processeswould be most affected by colchicine?

A. DNA synthesisB. Protein synthesisC. Mitosis

D. Muscle contraction

Which of the following represents a physiologicaleffect of breathing air with a slightly increased partialpressure of CO2?

A. Increased breathing rateB. Decreased breathing rateC. Increased blood pressureD. Decreased blood pressure



The bacterium Vibrio cholerae is responsible forcausing the intestinal disease cholera, a major cause ofmortality in developing nations. The enterotoxin secreted bythe bacterium causes an increase in chloride ion secretion byintestinal crypt cells. Water and sodium follow chloride intothe intestinal lumen, dehydrating the victim's blood andcausing diarrhea. The resulting water and electrolyte loss canlead to death within hours.

Studies of the intestinal sodium/glucose symport led tothe development of oral rehydration therapy (ORT). Thistherapy involves administering liquids with the componentsshown in Table 1.

Ingredients Grams /Liter of H20

Sodium Chloride

Trisodium Citrate"*"Potassium Chloride

Glucose

3.5

2.9

1.5

20.0

"'"Note: trisodium citrate is used in modern ORT solutionsinstead of the more unstable sodium bicarbonate.

Table 1: Standard ORT solution

The standard ORT solution, when administered early inthe disease, prevents dehydration of cholera victims by usingthe sodium/glucose symport to co-transport Na+ and glucosefrom the lumen of the intestine into the blood. Water follows

by osmosis, exactly replacing the fluid that is lost due to thetoxin-induced diarrhea. In order to avoid an increase of

water flow into the intestinal lumen, the ORT solution has anosmolarity (i.e., solute concentration) that equals that ofnormal blood. This makes it hypotonic relative to the choleravictim's partially dehydrated blood.


16.

17.

18.

19.

After administration of the standard ORT solution, thevolume of fluid lost through toxin-induced diarrhea:

A. increases due to a net flow of water out of theblood and into the intestinal lumen.

B. increases due to the increased osmolarity of thefluid in the intestinal lumen.

C. does not change, because ORT exactly replaceslost fluid withoutaffecting the diarrheadirectly.

D. decreases, because more fluid is reabsorbed intothe blood than is lost through diarrhea.

In addition to dehydration and diarrhea, cholera victimsare most likely to experience which of the followingsymptoms?

A. Increased glomerular filtration and urineproduction

B. Increased blood pressureC. Increased rate of breathingD. Metabolic alkalosis

The most important function of trisodium citrate in thestandard ORT solution is to:

A. provide Na+ ions for co-transport with glucose.B. help raise the pH of the cholera victim's blood.C. increase the osmolarity of the ORT solution.D. decrease the osmolarity of the ORT solution.

Would the addition of more glucose and sodium to theORT solution result in increased net fluid uptake by theblood?

B.

C.

D.

Yes; increased Na+ and glucose would increaseco-transport and draw more water into the blood.No; increased Na+ and glucose would have noeffect on fluid uptake.Yes; the increased glucose would provide ATPfor the active transport of water.No; an increase in Na+ and glucose wouldincrease the osmolarity of the ORT solution.

20. Potassium chloride is added to the ORT solution

primarily to:

A. help maintain muscle function and membranepotentials.

B. increase intestinal sodium absorption.C. boost the low chloride concentration of the

intestinal lumen.

D. activate K+/C1* antiporters in the intestinalepithelium.



When the bacterium Clostridium botulinum is grownanaerobically, it produces botulinum toxin (BT), one of themost potent natural poisons. For adults, BT is lethal in dosesas small as 2-3 |ig. The active portion of the toxin is aprotein with a molecular weight of 150,000 that isconjugated to a variety of accessory proteins. This proteincomplex functions by disrupting calcium-mediatedacetylcholine release, causing an irreversible blockage ofsynaptic transmission.

Despite being a dangerous poison, botulinum toxin hasrecently been used as a therapeutic agent in the treatment ofa series of conditions collectively known as focal dystonias.Focal dystonias are neuromuscular disorders caused by theinvoluntary, sustained contraction of localized (focal)regionsof skeletal muscle.The resulting twisting, turning, orsqueezing movements can result in physical impairment (asin severe "writer's cramp", a debilitating focal dystonia).

As a therapeutic agent, highly diluted botulinum toxin isinjected intramuscularly at the site of the focal dystonia.Within a short period, abnormal muscle contraction ceasesand the patient experiences considerable relief. Histologicalstudies show that BT causes the atrophy of nerves fibersinnervating the affected muscles, thereby preventingcontraction and relieving the focal dystonia.

21. Injection of botulinum toxin for the treatment of focaldystonia often provides relief which only lasts severalweeks. Reinjection after this time provides renewedrelief for most patients, but some no longer respond tothe therapy. Which of the following best explains whythese particular patients lose responsiveness tobotulinum toxin injections?

A. Neurotransmitters other than acetylcholine areused by the effector nerves in these patients.

B. These patients produce antibodies against theactive portion of the botulinum toxin.

C. Nerves regenerate over the course of severalweeks in these patients.

D. Affected muscle fibers become permanentlydisabled in these patients.


22.

23.

24.

25.

The physiological effects of botulinum toxin onmuscles can best be mimicked by:

A.

B.

C.

D.

addition ofCa2+ directly to the space between theendplate membrane and the presynapticterminal.addition of acetylcholinesterase directly to thespace between the endplate membrane and thepresynaptic terminal.injection ofCa2+ into muscle fibers.injection of acetylcholine into muscle fibers.

Botulinum toxin most likely exerts its effects directlyon the:

A. presynaptic nerve terminal.B. postsynaptic nerve terminal.C. calcium-sequestering organelles in muscle cells.D. nodes of Ranvier.

According to information given in the passage, focaldystonias are most likely caused by:

A. a defect in the sarcoplasmic reticulum of affectedmuscle fibers.

B. constant overstimulation by afferent nerve fibers.C. a malfunction in the central nervous system.D. an oversecretion of acetylcholinesterase.

Food poisoning can often be caused by Clostridiumbotulinum. Which of the following would carry thegreatest risk of botulinum toxin contamination?

A. Dairy products.B. Fruit.

C. Canned food.

D. Food left uncovered

overnight.and unrefrigerated


Passage V (Questions 26 - 32)

Circulatory shock occurs when blood flow through thecirculatory system is inadequate to meet the oxygendemands of the tissues. Without physiological compensationor emergency medical intervention, shock can rapidly lead todeath.

Hypovolemic shock, a type of circulatory shock, occurswhen the volume of the blood is too low, resulting in lowblood pressure and decreased cardiac output. Since arterialblood pressure is directly proportional to blood flow, theimmediate result of hypovolemic shock is inadequate bloodflow to the organs and tissues.

Sensing the drop in blood pressure, baroreceptors in thecirculatory system trigger a series of sympathetic responsesto hypovolemic shock which help to compensate for the lowblood flow. One such response involves the vasoconstrictionof arterioles leading to most tissues other than the brain andheart. This affects the total peripheral resistance (i.e., theresistance to blood flow encountered in all of the vessels inthe circulatory system). Blood pressure (BP) is related tocardiac output (CO) and total peripheral resistance (TPR) asshown below:

BP oc (CO) (TPR)

Resistance to blood flow is inversely proportional to theradius of the blood vessel raised to the fourth power. Cardiacoutput is equal to the heart rate times the stroke volume ofthe ventricles.

26. Which of the following would most likely NOT resultin hypovolemic shock?

A. Severe hemorrhagingB. DehydrationC. Heart failure

D. Blood loss due to injury

27. In order to prevent any further decrease in bloodvolume during hypovolemic shock, which of thefollowing should most likely occur?

A. Sympathetic stimulation of the renal afferentarterioles.

B. Parasympathetic stimulation of the renal afferentarterioles.

C. Sympathetic stimulation of the renal efferentarterioles.

D. Parasympathetic stimulation of the renal efferentarterioles.


28.

29.

Which of the following would represent an effectivemedical treatment for hypovolemic shock?

A. Administration of diuretic drugs.B. Transfusion of blood plasma.C. Injection of epinephrineantagonists.D. Injection of acetylcholine antagonists.

Another physiological compensation for hypovolemicshock involves the production of ADH andaldosterone. These hormones:

A. increase urine volume.

B. decrease salt reabsorption.C. increase blood pressure.D. inhibit the sympathetic nervous system.

30. Anaphylactic shock can occur during severe allergicreactions, resulting in the widespread release of thevasodilating hormone histamine. Histamine causes thispotentially deadly type of circulatory shock by:

A. increasing total peripheral resistance.B. decreasing total peripheral resistance.C. increasing cardiac output.D. increasing blood pressure.

31.

32.

Which of the following represents a likelyphysiological compensation for hypovolemic shock?

A. Extensive sweating.B. Dilation of arterioles leading to the skin.C. Decreased stroke volume.

D. Increased heart rate.

Despite the majority of blood volume being in thevenous system at any one time, the mean venouspressure is considerably lower than the mean arterialpressure due, in part, to the:

A. greater ability of the venous walls to expand.B. greater ability of the arterial walls to expand.C. presence of one-way valves in the veins.D. larger water permeability of the veins.



33.

34.

In order to maintain a constant body temperature,humans must employ regulatory mechanisms to copewith an increase in temperature. Which of thefollowing would NOT represent such a mechanism?

A. Dilation of arteriole sphincters leading to skincapillary beds

B. Activation of sweat gland secretionC. Decrease in skeletal muscle activityD. Piloerection of hairs on the skin surface

In severe asthma, a patient's bronchioles can becomechronically inflamed and obstructed, increasing theeffort necessary to inflate the lungs with air. Which ofthe following might be a consequence of severeasthma?

A. Enlargement of chest accessory muscles.B. Reduction in the size of the diaphragm.C. Low CO2 levels in the blood.D. High O2 levels in the blood.


35. Progesterone is primarily secreted by which of thefollowing structures?

A. Primary follicleB. Secondary follicleC. Corpus luteumD. Ovum

36. Which of the following are NOT components of cell-mediated immunity?

A. MacrophagesB. Killer T-cells

C. Helper T-cellsD. B-cells



In order for physiological processes to functioncorrectly, the pH of blood must be precisely maintained. If astress on the body works to change blood pH, certainmechanisms act to compensate for the stress. One suchmechanism is known as the CO2/HCO3' buffer system:

® 0H + HCO3 = H2C03 H20 + C02

Although effective when changes are small, thisbicarbonate buffering systemcan be overwhelmed by a largeacid or base change in the blood. The result can be eitheracidosis (lower-than-normal blood pH) or alkalosis (higher-than-normal blood pH). Respiratory acidosis or alkalosisresults from impaired pulmonary gas exchange or anabnormal rate of breathing. Metabolic acidosis or alkalosisrefers to changes in pH caused by non-respiratory factors.Acid/base disorders trigger renal and/or respiratorycompensation mechanisms that tend to return blood pHtowards normal levels (Figure 1).

45

40

,J 35

0C?UDC

30

S, 25 -

20 -

15 -

10 J I I I I I I I

7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

pH

Figure 1: Effects of renal and respiratory compensation on blood

pH. N = normal pH and HCO3® levels. A = respiratory acidosis.B = respiratory alkalosis. C = metabolic acidosis. D = metabolicalkalosis. Letters with subscripts depict compensatory effects.


37. Which of the following might lead to metabolicalkalosis?

A. VomitingB. Diarrhea

C. Rapid breathingD. Slow breathing

38. The decrease in blood pH associated with acidosiswould most likely NOT alter the:

A. tertiary structure of blood proteins.B. secondary structure of blood proteins.C. oxygen affinity of hemoglobin.D. isoelectric point of endocrine peptide hormones.

39. In response to respiratoryalkalosis, renal compensationcan help restore normal pH through increasedbicarbonate excretion in urine. Which point on thegraph in Figure 1 shows complete renal compensationfor respiratory alkalosis?

A. Aj

B. B2C. C\D. Dj

40. A patient's blood bicarbonate levels are 45 mmol/Lwhile his blood pH is 7.4. According to the informationgiven in the passage, this patient most likely:

A. is perfectly healthy.B. suffers from metabolic acidosis.

C. has elevated pC02 levels.

D. has blood alkalosis.

41. Which of the following statements is NOT true withregards to the graph shown in Figure 1?

A. The metabolic acidosis depicted by point C maybe caused by an abnormal excretion ofbicarbonate.

B. The underlying cause of the acid/base disorderhas been cured at points A2 and B2.

C. Point A2 represents respiratory acidosis withcomplete renal compensation.

D. Point Ci represents metabolic acidosis withpartial renal compensation.


Passage VII (Questions 42 - 47)

Multiple sclerosis is a debilitating disease which affectsover 250,000 Americans. During the course of the disease,neurons in the white matter of the brain are progressivelydemyelinated, resulting in axons which are unsheathed, ornaked.

Recent evidence suggests that multiple sclerosis may bean autoimmune disease. Autoimmune diseases are caused bya malfunction of the patient's own immune system, resultingin an immune response to self (i.e., non-foreign) tissue. Theprogressive steps leading to nerve demyelination duringmultiple sclerosis are outlinedbelow:

1. The patient is infected by a virus which contains anantigen that closely resembles a component of the myelinsheath. Through exposure to this virus, the patient's immunesystem is somehow provoked into attacking the myelincomponent which is mimicked by theviral antigen.

2. T cells which recognize the viral antigen are activatedand proliferate. They then travel through the blood stream tothe brain, where they cross the blood-brain barrier.

3. Once in the brain, the T cells recognize antigens presenton the surface of the myelin sheaths. These T cells thensecretedamaging peptidecytotoxins which gradually destroythe myelin,exposing the naked axon.

4. T cells also stimulate B cells to proliferate and secreteantibodies against oligodendroglial cells,which arenormallyresponsible for repairing the myelin covering.

Because infection by viruses which mimic neuralantigens doesn't always trigger the development of multiplesclerosis, it has been suggested that there is a geneticpredisposition to the disease. In other words, those who arepredisposed may be more likely to develop multiple sclerosisafter exposure to viral antigens.


42.

43.

44.

348

Based on information given in the passage, which ofthe following would NOT be a likely symptom ofmultiple sclerosis?

A. Muscle spasmsB. ParalysisC. FatigueD. Numbness

According to the passage, which of the followingwould NOT induce multiple sclerosis-like symptoms inwild-type, normal mice?

A. Injection of antibodies against components of themyelin sheath.

B. Infection with a virus whose antigen mimics acomponent of the myelin sheath.

C. Autoimmunization against components of themyelin sheath.

D. Injection of cytotoxins in the vicinity of neuralcells.

An individual with multiple sclerosis receives atransplantof oligodendroglial cells from her unaffectedidentical twin. Would such a transplant cure thedisease?

A. Yes; the transplanted oligodendroglial cellswould prevent further autoimmune attack.

B. Yes; the transplanted oligodendroglial cellswould repair the damaged myelin sheaths.

C. No; the transplanted cells would be rejected bythe patient's immune system because they comefrom a foreign source.

D. No; the transplanted cells would eventually bedestroyed by the same autoimmune attack thatoriginally caused the disease.


45. Based on the passage, which of the following wouldrepresent potential treatment(s) for a patient withmultiple sclerosis?

I. Injection of antibodies against secreted T cellcytotoxins.

II. Injection of antibodies against viruses whichmimic myelin-related antigens.

III. Injection of antibodies against T cells whichrecognize myelin-related antigens.

A. I onlyB. II onlyC. I and III onlyD. I, n, and III

46. Rheumatic fever is a condition in which the immune

system attacks and scars the patient's heart valves.Based on information given in the passage, what mighttrigger this autoimmune disease?

A. Increased blood pressure due to bacterialinfection.

B. Demyelination of nerve fibers innervating thecardiac tissue.

C. Infection by an agent which displays antigenssimilar to those present on the heart valves.

D. Direct infection of the heart valves by viruses.


47. For a virus, which of the following would be anevolutionary advantage of mimicking a host antigen?

A.

B.

C.

D.

Such mimicry would cause the death of the hostdue to multiple sclerosis.The mimicked antigen would act as a decoy,drawing attack by the immune system but leavingthe virus unharmed.

The virus could evade attack by the host'simmune system because the antigen would beconsidered as "self."

There would be no evolutionary advantage; it ispurely coincidental that these viral antigensresemble host antigens.



48. Individual cardiac muscle cells are electrically linkedwith one another through:

A. synaptic junctions.B. gap junctions.C. axonal extensions.D. acetylcholine-mediated neurotransmitters.

49. The renal afferent arteriole is of a larger radius than theefferent arteriole. What effect does this difference inradius have on the functions of the kidney?

A. Reabsorption in the tubules is decreased.B. Glomerular filtration rate is increased.C. ADH secretion is decreased.

D. Blood flow to the kidney is decreased.


50.

51.

During embryogenesis, the separation of cells into thethree individual germ layers (endoderm, mesoderm,and ectoderm) first occurs during:

A. cleavage.B. gastrulation.C. organogenesis.D. neurulation.

Vertebrates are evolutionarily well adapted to life onland. Which of the following adaptations is LEASTlikely to contribute to the vertebrates' land dominance?

A. Internal lungsB. Impermeable outer skinC. Short loops of HenleD. Internal fertilization


52. During the digestive process, the stomach andintestines undergo peristaltic contractions which helpto push food along the digestive tract. Thesecontractions are:

A. caused by the stimulation of skeletal musclefibers.

B. inhibited by the stimulation of the vagus nerve.C. increased during times of fear or stress.D. enhanced by the actions of the parasympathetic

nervous system.

Copyright© by The Berkeley Review 351 STOP, YOU ARE FINISHED

Biology Physiology Diagnostic Set I Answers

Passage I (Questions 1 - 5) Thyroid Hormone

C is the best answer. When treating thyroid cancer, the goal is always to destroy the cancerous cells while leaving normalbody cells unharmed. Radioactive iodine is such an effective treatment because only thyroid cells allow it to enter them.When it gets inside these cells, the radiation given off by the iodine isotope destroys the tumorous cells. Other body cellsdon't take up iodine, so they are left unharmed by the radioactivity. Radioactive iodine represents a "magic bullet,"selectively entering and destroying only thyroid cells, no matter where they have spread throughout the body. Let's considereach of the answer choices in light of this knowledge. Choice A can be eliminated because it would result in extremelysevere side effects. If most cells of the body could take up radioactive iodine, then treatment with 13II would result in thedestruction of most normal body cells, in addition to the tumorous thyroid cells. Choice Bcan be ruled out because if l3IIwere non-toxic once inside cells, it would be an extremely ineffective treatment for cancer. It would be unable to kill thecancerous thyroid cells at all. Choice D can likewise be eliminated. If cancerous thyroid cells can't absorb iodine from theblood, treatment of thyroid tumors with radioactive iodine (which chemically behaves exactly the same as regular iodine)would be useless. The 131I wouldn't enter the tumor cells and therefore wouldn't kill them. The best answer is C.

B is the best answer. Surgical removal of the thyroid gland would result in a decline in thyroid hormone (thyroxine)production. From the first paragraph ofthe passage, we learn that thyroxine regulates basal metabolic rate. We are also toldthat excess thyroxine secretion causes weight loss even though food intake is increased. This implies that thyroxine tends toincrease metabolic rate, leading to a higher rate ofcellular respiration. The excess heat produced in this process may lead tothe profuse sweating which is the other symptom of excess thyroid hormone secretion. Knowing what an excess ofthyroxine may cause, we can infer what a decline in thyroxine production would do. Lower thyroxine levels would result ina lower metabolic rate. This lower rate of cellular respiration would use less ATP. Therefore, the excess calories that aperson consumes would not be "burned" by this lower metabolism, resulting in weight gain. Normally, when the thyroid issurgically removed, thyroxine pills (i.e., hormonal supplements) are taken to maintain the basal metabolic rate at appropriatelevels. Let's consider the otheranswer choices. Choice A represents theconsequences of increased thyroxine production. Ahigher metabolic rate would lead to an increased appetite and food intake (necessary to fuel the increased metabolism).Since we are looking for the effect of decreased thyroid hormone production, eliminate A. Choices C and D can beeliminated for the same reasons. Profuse sweating is the result of excess thyroxine secretion (as mentioned above).Likewise, excess oxygen usage by the tissues would be the result of increased cellular respiration caused by excessthyroxine levels. The best answer is B.B is the best answer. Figure 1 in the passage gives us the structure of thyroxine (a.k.a. thyroid hormone). Notice thehydroxylated benzyl rings, reminiscent of the structure of tyrosine, shown below:

HO

II II ()

I I II-C-C-C-OH

NH,

Tyrosine

It can be seen how thyroxine might bederived from tyrosine. The other answer choices, shown below, do not resemble thestructure of thyroid hormone:

ii

i

ii ii o

I I IIC-C-C-OH

I IH NH,

Histidine

H NH2

Thyroxine

II H II 0I I I II

H,C —S—C - C - C - C - OHI I I

H H NH,

Methionine

The best answer is B.

D is the best answer. From the passage, we learn that TSH stimulates thyroid cells to increase their uptake of iodine inorder to produce thyroxine. Therefore, high TSH levels would be beneficial to l3lI treatment ofthyroid cancer because TSHwould increase the uptake of radioactive iodine by cancerous cells. After the surgical removal of the thyroid, thyroxinelevels drop offover the course ofa few weeks. According.to Figure 1. if thyroxine levels drop, there would be less of aninhibitory effect on TSH secretion. This means more TSH would be secreted, resulting in more effective radioactive iodinetherapy. This is why the therapy is most effective several weeks after the removal ofthe thyroid. By this time, TSH levelshave risen, promoting the uptake of 13II by any remaining cancerous cells. Choice Ais invalid because ifcancerous thyroidcells spread further, the iodine treatment would be less effective at killing them all. Choice B can be eliminated because

,@Copyright © byThe Berkeley Review11 352 The Berkeley ReviewSpecializing in MCAT Preparation


5.

decreased thyroxine levels (as the result ofthyroid removal) would not lead to adecrease in TRH (see Figure 1). Choice Cis wrong because thyroxine levels do not rise when the thyroid is removed. In any case, rising levels ofthyroxine would notstimulate iodine uptake. The best answer is D.

C is the best answer. In order for a radioactively labeled compound to destroy a cell, it must first actually enter the cell. Inthe passage, we learn that TSH is a peptide hormone. Recall that peptide hormones can't cross the plasma membrane oftarget cells. Generally, they act by binding to receptors on the surface of target cells. This binding then triggers secondmessenger production, which leads to changes in the cell (e.g., gene transcription). In the case of radiolabeled TSH, theradioactive molecule would never be able to enter its target (i.e., cancerous thyroid cells); rather, it would bind to a receptoron the surface of these cells. This is not sufficient to destroy the tumorous cell. Let's consider the other answer choices.Choice A can beeliminated because TSH does not actually enter the cells of the thyroid, as mentioned above. Choice B isinvalid because radiolabeled TSH can't make radiolabeled thyroxine, and even if it could, thyroxine's targets are notcancerous thyroid cells. Choice D can also be ruled out because the only target of thyroid-stimulating hormone is thethyroid. The best answer is C.

Passage II (Questions 6-11) Home Pregnancy Test

6. C is the best answer. In order to be indicative of pregnancy, a hormone must only be produced either by a mother who ispregnant or the fertilized embryo itself. Otherwise, false test results might be possible. Progesterone is normally producedby females whoare not pregnant during the course of their monthly menstrual cycles. It is not purely a hormone associatedwith pregnancy. The other answer choices do not offer valid explanations why progesterone can't be used. Choice A isincorrect because antibodies can be made to almost any molecule, as long as that molecule will fit into the antibody bindingsite. Choice B is incorrect because progesterone is secreted by the corpus luteum throughout pregnancy. Choice D can beeliminated because progesterone is definitely present in the bloodstream; this is how endocrine hormones reach their targetcells. The best answer is C.

7. B is the best answer. From the passage we learn that hCG is a peptide hormone. Peptide hormones act by binding toreceptors on the plasma membranes of their target cells. This binding then triggers a signaling cascade which results in thetranscription of specific genes. Peptide hormones can't cross the plasma membrane directly because they are not lipidsoluble. Glucagon is a peptide hormone secreted by the alpha cells of the pancreas; it therefore uses a similar mechanism tothat of hCG. The other answer choices can be eliminated because they are steroid hormones. These act by directly crossingthe plasma membrane and binding to receptors either in the cytoplasm or in the nucleus. The best answer is B.

8. A is the best answer. The test, as described in the passage, simply assays for the presence of hCG in the urine. The personusing the test would either obtain a positive or negative result. Whether or not this result is valid remains in questionbecause there are no experimental controls involved. In other words, there is nothing to test whether the test reagents arefaulty or the procedure was done incorrectly. A positive control would involve using a sample liquid containing hCG andtesting to see whether the pregnancy test could actually detect it. If it can't, then something is wrong. A negative controlmight involve using a urine sample with no hCG in it and seeing if the test gives a false positive result. Such controls arenot included in the test as described in the passage. The other answer choices can be eliminated. Choice B is wrong becausethe passage states that hCG levels are high enough to be detected by the time of the first missed period. This could beanywhere from 7-14 days from the time of fertilization. Choice C can be eliminated because the mother does not producehCG. If she did, the test would be entirely useless. Only the embryo produces the hormone. Choice D is also incorrectbecause the embryo must produce hCG in order to prevent menstruation and maintain the corpus luteum. If it doesn't,menstruation will occur and the embryo will be sloughed off with the uterine lining. The best answer is A.

9. C is the best answer. In the passage it is stated that hCG is detected in the urine. In order to get there, the hormone must befiltered from the blood into the kidney tubules (via the glomerulus). The tubules must then not reabsorb all of it back intothe blood; instead, the hCG must be allowed to leave in the urine. The other answer choices can be eliminated afterconsideration. Choice A is wrong because hCG is only secreted by the fertilized embryo; if it was also secreted by themother, the pregnancy test would not be effective because hCG could be detected in non-pregnant mothers. In either case,the passage does not imply that the mother makes hCG. Choice B can be eliminated because the passage directly states thathCG is released into the mother's bloodstream. Choice D is wrong because hCG works to maintain the corpus luteum,which secretes progesterone. If anything, hCG works to maintain the production and function of progesterone, notcounteract it. The best answer is C.

10. D is the best answer. Answering this question simply requires picking the answer choice that doesn't have anything incommon with the others. Choices A, B and C all represent conditions in which specific antibodies are produced whichmight be tested for using the same technique described in the passage. After all, antibodies are proteins, and monoclonalantibodies can be made to detect them. HIV infection can be detected by assaying for the antibodies against HIV. Arthritiscan be detected by looking for antibodies against "self" proteins. Strep throat can be detected by directly assaying for the

®Copyright © by The Berkeley Review 353 The Berkeley ReviewSpecializing in MCAT Preparation


presence of bacterial proteins. The same technique which was used to detect hCG can be used in these instances.Tryptophan blood levels, however, can not be determined using the same methodology. For one thing, tryptophan (an aminoacid) does not elicit antibody production. Secondly, the actual levels of substances in the blood can't be detected by the typeof test described in the passage. This test yields either a"present" or "not-present" result. The best answer is D.

11 C is the best answer. Antibodies are made by B-cells, not T-cells (eliminate choices Aand B). Monoclonal antibodies, astheir name implies, are produced by a single clone of B-cells. In other words, a single B-cell is provoked to divide,producing a clone of identical cells. These cells then crank out one single type of antibody. These are monoclonalantibodies. Monoclonal antibodies only strongly recognize and bind to one specific antigenic site. The hCG molecule, beinga polypeptide, may have multiple antigenic sites that can be recognized by antibodies, but a given set of monoclonalantibodies only recognizes one ofthese sites. Eliminate choice D. The best answer is C.

(Questions 12 -15) Not Based on a Descriptive Passage

12. Bis the best answer. The question asks what would LEAST contribute to the whale's ability to remain submerged for longperiods of time. Ahigh basal metabolic rate represents a large demand for oxygen by the rapidly respiring tissues. Thiswould be detrimental to long stays underwater where no new oxygen can be obtained. Stores ofoxygen obtained on thesurface would more rapidly be consumed by a high metabolism, decreasing the ability to remain underwater. The otheranswer choices all represent advantages for prolonged submersion. Choice A can be ruled out because large musclemyoglobin concentrations would help to store oxygen and deliver it most efficiently to the muscles which need it during thedive. Choice C can be eliminated because a high cellular tolerance for CO2 would be beneficial to a diving animal, as CO2would build up during the dive since itcould not be exhaled. Choice Dcan also be thrown out because a large lung capacitywould facilitate a better initial oxygen intake before the dive, allowing more oxygen to be available during submersion. Thebest answer is B.

13. A is the bestanswer. Pepsin isa protease secreted by the chief cells lining the stomach (initially in the form of the zymogenpepsinogen). All of the other answer choices represent enzymes secreted by the pancreas. Lipase aids in the digestion offats. Chymotrypsin and carboxypeptidase are both proteases (i.e., they break down proteins). The pancreas secretes itsenzymes intothe small intestine. The best answer is A.

14. C is the best answer. Colchicine affects microtubules, the strands of polymerized protein which make up part of thecytoskeleton. We must therefore pick the answer choice that is associated with some sort of microtubule-driven process.DNA and protein synthesis are not such processes. These do not depend on microtubules and therefore would not beaffected by colchicine (eliminate choices A and B). Muscle contraction depends on actin and myosin, not microtubules(eliminate choice D). This leaves us with mitosis as the answer. During mitosis, chromosomes are attached to microtubuleswhich serve to line them up and pull them apart. In fact, the spindle apparatus is predominantly made up of microtubules.The polymerization and depolymerization of microtubules is vital to mitotic division. Colchicine thereby prevents mitosis.The best answer is C.

15. A is the best answer. Slightly increased levels ofCO2 in the air we breathe stimulates an increase in breathing rate. This isbecause breathing such air causes arterial levels ofCO2 to rise, telling the breathing centers of the brain that CO2 buildup isoccurring. The brain then stimulates an increase in ventilation (hyperventilation) to try to exhale as much of the blood COtas possible. Choice B is therefore incorrect. Choices C and Dcan also be eliminated because small changes in CO2 levelswould not significantly affect blood pressure in any way. The best answer is A.

Passage III (Questions 16 - 20) Oral Rehydration 1 herapy

16. C is the best answer. The question asks whether administration of an oral rehydration therapy solution would reduce theextent of fluid loss through diarrhea. From the last paragraph of the passage, we learn that the standard ORT solutionoutlined in Table 1exactly replaces the fluid that is lost due to toxin-induced diarrhea. This means that the diarrhea is stilloccurring and fluid is being lost, but whatever is lost from the body is replaced by intestinal absorption of the ORT solution.Since the volume of liquid leaving through diarrhea is unchanged, we can conclude the ORT does not reduce the extent ofdiarrhea; it simply prevents the patient from experiencing a net loss of fluid while the diarrhea is continuing. A cholerapatient who is given an ORT solution still experiences the same amount ofdiarrhea; the ORT solution simply prevents totaldehydration by keeping blood fluid and electrolyte levels fairly constant in the face of this massive loss. We can eliminatechoice A because ORT does not promote a net flow of water out of the blood and into the intestinal lumen. Rather, thepassage tells us that it promotes the bloodward flow ofwater and electrolytes by taking advantage of sodium/glucose co-transport mechanisms. Additionally, the ORT solution is hypotonic (i.e., it has a lower osmolarity. orsolute concentration)relative to the dehydrated (hypertonic) blood; water flows by osmosis from a hypotonic solution to a hypertonic one (or

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BlOlOgy Physiology Diagnostic Set I Answers

from a solution oflow osmolarity to a solution ofhigh osmolarity). Since the ORT solution does not increase the osmolarityof the intestinal lumen, we can eliminate choice B. Choice D can be ruled out because in the passage we learn that ORTexactly replaces fluid that is lost through diarrhea. The best answer is C.

17. C is the bestanswer. Diarrhea causes massive fluid loss. This lost fluid contains intestinal secretions (i.e., electrolytes andbicarbonate, HCO3"). Recall that the pancreas produces bicarbonate ions and secretes them into the lumen of the intestine.In the case of cholera, however, this bicarbonate is excreted at a high rate in the diarrhea. Removal of bicarbonate alters theacid/base composition of the blood; the blood becomes more acidic (less basic) andacidosis results. This type of acidosis isconsidered metabolic because it is not caused by respiratory problems (as in respiratory acidosis). In order to compensatefor the increased acidity of the blood, the breathing rate becomes faster. This allows CO2 to be exhaled at a greater rate,lowering the acidity of the blood. The other answer choices can be eliminated after consideration. A cholera victim wouldbe dehydrated. This means that the victim's blood volume would be low, lowering blood pressure. This directly eliminatesanswer choice B and indirectly eliminates answer choice A. Recall that glomerular filtration rate (GFR) would decrease ifthe hydrostatic pressure of the blood were lower, as it is in cholera. This means that a cholera victim would not experiencean increased GFR, and therefore there would be no increase in urine production. Choice D can be ruled out because, as wediscussed above, loss of bicarbonate (a basic anion) lead to metabolic acidosis, not alkalosis. The best answer is C.

18. B is the best answer. The question asks for the mostimportant function of trisodium citrate. Our major clue comes from thefootnote to Table 1, which states that trisodium citrate is used instead of the more unstable sodium bicarbonate. As weknow, bicarbonate is a basic anion which is an important part of the blood buffering system. In cholera, vast quantities ofbicarbonate are lost due to diarrhea, and the patient's blood becomes acidotic. The ORT solution must address this acid/baseimbalance. It does so by including trisodium citrate, which dissociates to form basic citrate anions, removing some of theexcess H+ in the blood by forming citric acid (the conjugate acid of citrate). We can infer that this is the function oftrisodium citrate by thinking about what bicarbonate would do if included in the ORT solution. Bicarbonate would raise thepH of the blood by combining with H+. It is not included in the ORT solution because it is unstable; since trisodiumcitrateis its replacement, we can infer that trisodium citrate has a similar pH-raising function. Answer choice A is extremelytempting; after all, trisodium citrate has plenty of sodium to put into solution. This is not its most important function,however. Sodium chloride provides most of the sodium needed in the solution. Trisodium citrate's most important functionis to fix the blood's abnormally low pH. Choice C can be eliminated because although trisodium citrate does increase theosmolarity of the ORT solution (by increasing the concentration of solute), this is not its most important function (nor is it afunction exclusive to trisodium citrate). Every component of the solution raises the osmolarity of the solution, but onlytrisodium citrate alters acid/base balance. Choice D can be thrown out because adding more solute causes an increase, not adecrease, in osmolarity. The best answer is B.

19. D is the best answer. Recall from the passage that the osmolarity, or solute concentration, of the standard ORT solution isequal to that of normal blood. This makes the ORT solution hypotonic relative to the dehydrated blood of the patient. Fromour knowledge of osmosis, we know that water flows from a hypotonic solution to a hypertonic solution. The oralrehydration therapy is successful because water is not drawn into the lumen of the intestine by osmosis. The lower soluteconcentration of the ORT solution relative to the higher solute concentration of the blood prevents this flow (known as anosmotic penalty). Since we are trying to increase the bloodward flow of water and electrolytes, such an osmotic penaltywould be counterproductive. This is why the concentrations of glucose and sodium in the ORT solution can't be muchhigher than what is given in Table 1. If they are, the osmolarity of the solution will exceed that of the blood and water willflow into the lumen of the intestine from the blood. Even though the sodium/glucose co-transporter would function at ahigher rate (due to the higher concentrations of its substrates), it would not entirely make up for this osmotic penalty. If wedon't consider this, we might have picked choice A; knowing that increased sodium and glucose concentrations would notresult in net increased fluid uptake (due to osmotic penalty), we can eliminate choice A. Choice B can also be ruled outbecause the passage states that sodium and glucose do affect fluid uptake. Choice C is incorrect because water itself cannotdirectly be actively transported. The best answer is D.

20. A is the best answer. Potassium is one of the major electrolytes lost through cholera-induced diarrhea. Potassiumreplacement is important because K+ is vital in maintaining membrane potentials and muscle function (recall that actionpotentials involve potassium and potassium channels). The other answer choices can be eliminated after carefulconsideration. Choice B is incorrect because neither K+ nor CI"can increase Na+ absorption in the small intestine. Choice Ccan be thrown out because the lumenal chloride concentration in cholera victims is already too high. Recall that choleratoxin induces crypt cells to secrete chloride ions into the intestinal lumen. There is no therapeutic need to further boost theconcentration of CI"in the lumen. Choice D can likewise be eliminated because there is no K+/C1* antiporter. Additionally,why would we add KC1 just to activate these hypothetical antiporters? This would accomplish nothing of therapeutic use.The best answer is A.

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Passage IV (Questions 21-25) Botulinum loxin

21.

22.

23.

24.

B is thebest answer. Recall from the passage that the active portion of the botulinum toxin isa large protein conjugated toa series ofaccessory proteins. Certain patients, upon initial injection of BT, respond well but become tolerant to furtherinjections, making the therapy ineffective in the long run. What could be happening in these patients? One explanation isthat upon initial injection, the immune system recognizes the protein toxin as foreign and begins synthesis ofantibodiesagainst it. By the time the antibodies are made, the botulinum toxin has already served its function of irreversibly preventingacetylcholine secretion from the presynaptic terminals of the neuromuscular junction (causing these terminals to atrophyfrom lack ofuse). This initially relieves the focal dystonia by preventing the muscle contractions which cause it. Over time,however, new terminals sprout to replace the old damaged ones, and the muscle eventually becomes renervated. This is whyBT therapy only lasts a few weeks. For the patients who produced antibodies to the toxin, further injections ofBT would beuseless becausethe antibodies wouldbind to and inactive the active portion of the toxin before it could perform its function.These individuals have become tolerant to the therapy. Let's eliminate the other answer choices. Choice A is incorrectbecause acetylcholine is the only major neurotransmitter used in the neuromuscular junctions to trigger muscle contraction.Even if another transmitter were used, this would not account for the initial effectiveness of BT therapy in these patients.Choice C is a true statement; atrophied nerves do regenerate over the course of several weeks (this causes the return of thefocal dystonia by restoring innervation of the affected muscles). This would not, however, explain the tolerance to BT thatcauses it to become ineffectual in certain patients. Afterall, several weeks is a long time for nerves to regenerate; we wouldexpect BT to cause alleviation of symptoms for at least that long. Eliminate choice C. Choice D can easily be thrown outbecause if affected muscle fibers became permanently disabled in these patients, there would be no need for further therapybecause the dystonia would be cured. The best answer is B.

B is the best answer. From the passage, we learn that botulinum toxin acts by causing an irreversible blockage ofacetylcholine release. Recall that acetylcholine is normally released from the presynaptic terminal; after crossing thesynaptic cleft, it binds to receptors on the endplate membrane (on the muscle side of the neuromuscular junction). Thisbinding triggers a series of depolarizations that ultimately lead to muscle fiber contraction. If acetylcholine doesn't bind toits receptors on the endplate membrane, muscle contraction can't occur; this is the normal physiological effect of botulinumtoxin. In order to mimic this effect, we must prevent acetylcholine from binding to its receptors. One way to do this is tointroduce a large amount ofacetylcholinestera.se directly into the space between the endplate membrane and the presynapticterminal. Recall that acetylcholinesterase breaks down acetylcholine. Adding extraneous acetylcholinesterase would preventacetylcholine from binding to its endplate membrane receptors, effectively mimicking the effects of botulinum toxin. Wecan also approach this problem through a process of elimination. Choice A can be ruled out because calcium, whenintroduced into the neuromuscular junction, would not cause cessation of acetylcholine release, nor would it preventacetylcholine from binding to its receptors. Choice C is invalid because injection of Ca-+ into muscle fibers would causecontraction of the muscle fibers; remember that calcium binding allows interaction between actin and myosin. Musclecontraction is normally prevented by botulinum toxin, so injection of calcium would not mimic BT's effects. Choice D canbe eliminated because acetylcholine, when injected into muscle fibers, would have no effect because there would be nobinding to outer membrane receptors. The best answer is B.

A is the best answer. Recall from the passage that botulinum toxin exerts its effects by blocking the calcium-mediatedrelease of acetylcholine. Acetylcholine is normally released from the presynaptic terminal underthe influence of a changingintracellular calcium concentration; therefore, BT must exert its effects here. Choice B is incorrect because acetylcholine isnot released from the postsynaptic terminal (or endplate membrane, in this case). Choice C can be ruled out becausecalcium-sequestering organelles in muscle cells (i.e., sarcoplasmic reticulum) have no influence on acetylcholine secretion.They are instead responsible for altering intracellular calciumconcentrations in order to regulate muscle contraction. ChoiceD can be ruled out because the nodes of Ranvier do not secrete acetylcholine. They simply facilitate faster transmission ofnerve action potentials. The best answer is A.

C is the best answer. Focal dystonias are localized, sustained abnormal contractions of muscle fibers. Whatever causesthem must continuously stimulate the muscle fibers to contract. From the passage, we learn that botulinum toxin, whichoperates on the level of the presynaptic terminal of the neuromuscular junction, can relieve the abnormal contractionsencountered in focal dystonias. The cause of the dystonias must therefore lie somewhere at or before the level of thepresynaptic terminal. Since signals for muscle contraction originate in the central nervous system and then travel to theneuromuscular junction, a defect at the level of the CNS could cause the focal dystonia. A defective contraction signal sentfrom the CNS would be blocked at the level of the presynaptic terminal (via acetylcholine blockage) by botulinum toxin.We can eliminate the other answer choices. Choice A is incorrect because a defect in the sarcoplasmic reticulum would lieat the level of the muscle fibers themselves. This can't be the case, because we learn from the passage that botulinum toxincounteracts focal dystonias by preventing acetylcholine secretion. BT would have no effect if the dystonia were caused by adefect in the muscle fibers themselves. Choice B is a tempting answer, but we must make sure we understand the termsused. Afferent nerves lead away from the muscles and towards the central nervous system (as opposed to efferent nerves

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which lead from the CNS to the effector organs or muscles). Overstimulation by afferent fibers would therefore affect theCNS, not the muscle fibers; this is most likely not a cause of the sustained muscle contraction seen in focal dystonias.Choice D can be eliminated because an oversecretion of acetylcholinesterase would cause increased degradation ofacetylcholine and therefore would induce paralysis, not contraction, of muscle fibers. This can't be the cause of focaldystonias. The best answer is C.

25. C is the best answer. According to the first paragraph of the passage, botulinum toxin is produced when the bacterium C.botulinum grows anaerobically. In order to be contaminated with BT, a food product must be present in an environment thatdoes not contain oxygen, forcing any C. botulinum that are present to grow anaerobically (without O2). This would causethe production of botulinum toxin. Canned foods are packaged such that there is very little oxygen inside the can. This is anideal anaerobic environment. Incidentally, the potentially fatal disease resulting from ingestion of BT is called botulism.Choices A and B can be ruled out because these foods are generally exposed to some oxygen. Choice D can be eliminatedbecause food left uncovered is also exposed to oxygen. None of these answer choices represent foods that would provide agood anaerobic environment. The best answer is C.

Passage V (Questions 26 - 32) Circulatory Shock

26. C is the best answer. Recall from the passage that hypovolemic shock results from a low blood volume which tends tolower blood pressure and cardiac output. The less blood volume present in the system, the less blood the heart is able topump at any given time, thereby lowering cardiac output: since blood pressure is directly proportional to cardiac output,blood pressure is lowered as well. The ultimate result is decreased blood flow to the tissues, resulting in circulatory shock.Since hypovolemic shock is the result of lowered blood volume, we can eliminate responses that tend to lower the volumeof the blood. Choice A, hemorrhaging, refers to excessive bleeding as the result of some sort of damage to blood vessels.Since hemorrhaging results in blood loss, it decreases blood volume. Eliminate choice A. Choice B, dehydration, involvesthe loss of water from the blood plasma. This also would tend to lower the volume of the blood, so eliminate choice B. Forthe same reasons, choice D can be ruled out because blood loss due to injury (similar to hemorrhaging) could causehypovolemic shock due to decreased blood volume. This leaves us with choice C, heart failure, as the correct answer.Although heart failure may lead to some type of circulatory shock (i.e., inadequate blood flow to the tissues), it would notreduce the volume of the blood present in circulation. Therefore, it would not directly cause hypovolemic shock. The bestanswer is C.

27. A is the best answer. The question asks about preventing any further decrease in blood volume during hypovolemic shock.Recall that the kidneys are normally very active in regulating the volume of the blood through both the reabsorption andexcretion of water (as urine). The kidneys can lower blood volume by excreting more urine or they can maintain bloodvolume by excreting less urine (and reabsorbing more water in the tubules). In this case, we want to prevent blood volumeloss and therefore we want to prevent excessive urination. This can be accomplished by constricting the renal afferentarterioles. Recall that these arterioles bring blood into the glomeruli for filtration into the glomerular capsule. If thesearterioles are constricted, less blood can get to the glomeruli and produces a lower glomerular filtration pressure, therebyfiltering less fluid and resulting in a decrease in urine production. From the passage, we learn that the physiologicalresponses that compensate for hypovolemic shock are sympathetic in nature. We can therefore conclude that sympatheticstimulation of the renal afferent arteriole would cause its constriction, thereby reducing the amount of urine produced in thekidneys. Remember that the parasympathetic nervous system does NOT innervate the kidney at all, and therefore can't be ananswer choice. If we couldn't recall this fact, we might also remember that parasympathetic stimulation generally has arelaxing effect anyway, and probably wouldn't cause constriction of the renal arterioles. Eliminate choices B and D. Withregard to choice C, constriction of the efferent arteriole would not have as great an effect on urine production. This isbecause the efferent arteriole takes blood away from the glomerulus. Constricting it wouldn't affect urination as greatly asconstriction of the afferent arteriole (which takes blood to the glomerulus). The best answer is A.

28. B is the best answer. Recall that hypovolemic shock is caused by a decrease in blood volume. In order to treat it, we mighttry to bring the volume of the blood back up. One way to do this is to transfuse (inject) more blood plasma into the personsuffering from the shock. This would add more to the volume of the blood, raising blood pressure and increasing cardiacoutput. Choice A is incorrect because administration of diuretic drugs would cause diuresis, or water loss due to urination. Ifmore water were lost from the blood, the blood volume would decrease further. This would not be an effective treatment; itwould in fact be very harmful. Choice C is also invalid because epinephrine (a.k.a. adrenaline) is a major effector moleculeof the sympathetic nervous system; adding an epinephrine antagonist would decrease the activity of epinephrine, therebydecreasing sympathetic function. Recall from the passage that the mechanisms by which the body tries to compensate forhypovolemic shock are mainly sympathetic in nature. By blocking these beneficial responses, the patient suffering fromshock would actually get worse. Choice D, injection of acetylcholine antagonists, does not represent a valid treatment forhypovolemic shock either. By blocking acetylcholine (a neurotransmitter), the patient might become paralyzed, a situationof no benefit in this case. The best answer is B.

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29. C is the best answer. Recall that ADH (anti-diuretic hormone) and aldosterone are both hormones which tend to decreasewater excretion in the form of urine. These hormones thereby increase blood pressure by promoting the reabsorption ofwater in the kidney (and preventing urine production). We could answer this question even if we had forgotten the functionsof ADH and aldosterone. Recall that the question states that secretion of these hormones represents a physiologicalcompensation for hypovolemic shock (i.e., low blood volume). Raising blood pressure is the only answer choice that wouldcompensate for the loss ofblood volume and blood pressure encountered in shock. Choice Ais incorrect because if the twohormones increased urine volume, they would increase water secretion and therefore lower blood pressure even further.Choice B can be ruled out because if salt reabsorption were decreased, the excreted salt would cause water to follow,lowering blood volume and pressure further. Aldosterone actually promotes an increase in salt reabsorption. Choice D canbe eliminated because inhibiting the sympathetic nervous system would not represent a physiological compensation forhypovolemic shock. Recall from the passage that the sympathetic system actually helps to compensate for shock; inhibitingit wouldcause moreharm than good for the patient suffering from hypovolemic shock. The best answer is C.

30. B is the best answer. In the question, we are told that histamine is a vasodilating hormone. This means that it promotes thedilation (expansion) of blood vessels. Anaphylactic shock is a severe allergic response (e.g., to a bee sting) which results inbody-wide histamine release. This release promotes the dilation of many blood vessels in the body. According to the lastparagraph of the passage, an increase in the radius of a blood vessel (as occurs during dilation) results in a decrease in theresistance of that vessel. From the equation relating blood pressure to cardiac output and peripheral resistance, we can seethat this decrease in resistance would cause a drop in the blood pressure. Since histamine causes a lot of blood vessels todilate, the total peripheral resistance is greatly reduced, causing blood pressure to plummet, resulting in inadequate bloodflow to the tissues and thus shock. With this knowledge, we can eliminate choice A, because histamine does not cause anincrease in total peripheral resistance. Choice C can be eliminated, because an increase in cardiac output would not causeshock. Likewise, we can rule out choice D, because an increase in blood pressure would not necessarily cause shock.Dilation of blood vessels by histamine neither raises cardiac output nor increases blood pressure. The best answer is B.

31. D is the best answer. In order to compensate for hypovolemic shock, we must somehow try to increase blood pressure sothat the tissues may be adequately provided with blood. One such way to do this is to raise the heart rate. From the lastparagraph of the passage, we learn that cardiac output is equal to the heart rate times the stroke volume. If heart rateincreases, cardiac output goes up. Since blood pressure is directly proportional to cardiac output, blood pressure would riseas well. Therefore, increasing the heart rate might be an effective physiological compensation for hypovolemic shock.Choice A is actually counterproductive because extensive sweating would rob the blood of yet more water, decreasing itsvolume even further. Choice B would result in decreased total peripheral resistance. As we learn from the equation given inthe passage, decreasing such resistance leads to a further lowering of blood pressure. Choice C can be ruled out because adecrease in stroke volume leads to a reduction in cardiac output which leads to a decrease in blood pressure. The bestanswer is D.

32. A is the best answer. Even though there is more blood volume in the veins and venules at any given time, they have a lesserpressure than the arteries and arterioles, which have less blood volume in them at any given time. One reason (amongseveral) for this is that the venous blood vessels have a greater compliance; that is, they can expand more easily. When theyexpand, their radius increases and the resistance to blood flow decreases, thereby lowering the pressure according to thevarious relationships detailed in the passage. The arterial vessels are less compliant; they don't expand as easily (due to theirmore extensive smooth muscle component) and their radii are thereby small, increasing the pressure of the blood insidethem. Knowing this, we can eliminate choice B. Choice C has little to do with the decreased venous blood pressure; the oneway valves simply keep blood flowing back to the heart, preventing backflow (due to the low pressure of venous blood).Choice D is a smoke screen; veins aren't especially permeable to water. The best answer is A.

(Questions 33 - 36) Not Based on a Descriptive Passage

33. D is the best answer. This question asks which mechanism of thermoregulation does NOT defend the body against elevatedtemperatures. Let's consider the answer choices. Choice A represents an effective defense against increased temperature. Ifthe sphincters regulating blood flow to the skin capillary beds were to dilate, or open, blood would rush to the surface of theskin, carrying heat with it. This heat could then be dissipated to the environment, cooling the body. Choice B is also aneffective cooling mechanism. Sweat glands secrete liquid onto the surface of the skin. When the liquid evaporates, it carriesheat away with it. Choice C would also help to cool the overheating body. If activity in the skeletal muscles were todecrease, this would lead to less metabolic heat production. This leaves us with choice D as the correct answer. Piloerectionof hairs on the skin surface (i.e., goosebumps) result in hairs which stand perpendicular to the skin. This interferes with theflow of air over the skin, resulting in a conservation of body heat. Goosebumps form when the body is trying to keep warm,not when it is overheating. The best answer is D.

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34. A is the best answer. In severe asthma, it takes more effort to inflate the lungs. Recall that such inflation is an activeprocess, carried out by contractions of the diaphragm and chest accessory muscles (such as the intercostals). If it is moredifficult to carry out inflation (or inspiration), these muscles have to work harder, and as a result they eventually becomelarger (hypertrophy). This gives certain asthmatics a "barrel chested" look. Choice B therefore can be eliminated, becausethe size of the diaphragm will not be reduced if it must work harder all the time. Choice C can be ruled out becauseasthmatics have a difficult time breathing; this would lead to higher levels of CO2 in their blood because their lungs can'tefficiently excrete it. Choice D is likewise incorrect because asthmatics can't breathe efficiently enough to raise their bloodO2 levels above normal. The best answer is A.

35. C is the best answer. The corpus luteum is formed from the secondary follicle after the egg is released during ovulation.After its formation, it begins to secrete progesterone. The other answer choices do not represent structures that secreteappreciable amounts of progesterone. The primary follicle will later differentiate into the secondary follicle, which willeventually release the ovum, or egg. As a note, when the secondary follicle bursts and releases the ovum, a small amount ofprogesterone is released. This amount is far less than what is later secreted by the corpus luteum, however. The bestanswer is C.

36. D is the best answer. Cell-mediated immunity is the branch of the immune response that involves direct cellular attack oninvading organisms or infected body cells. This is opposed to humoral, or antibody-mediated immunity. B-cells secreteantibodies; they are the main source of humoral immunity. B-cells do directly participate in cell-mediated immuneresponses. Macrophages are lymphocytes which phagocytose foreign material. Killer T-cells are activated by helper T-cellsto destroy cells of the body which are infected. Both participate in the cell-mediated response. Incidentally, helper T-cellsalso serve a role in activating B-cells, and are therefore part of the humoral immune response. The best answer is D.

Passage VI (Questions 37-41) Acid/Base Disorders

37. A is the best answer. Vomiting involves ejecting the contents of the stomach. These stomach contents contain a largeamount of H+ due to the acidic secretion of HCl that occurs in the stomach. When these acidic contents are lost, more HCl

must be secreted to offset the loss, causing the blood H+ concentrations to drop; this raises the pH of the blood, causingalkalosis. We refer to this condition as metabolic alkalosis because it results from a problem that does not have to do with

the lungs (as in respiratory alkalosis). Incidentally, notice that by lowering the concentration of H+ in the blood, theconcentration of HCO3" increases according to the equation given in the passage.This occurs because the removal of H+(a

reactant) shifts the entire equation to the left, producing more HCO3" (which is also a reactant). To confirm this, take a look

at point D on the graph. This point depicts uncompensated metabolic alkalosis. Notice that pH levels are raised, as areHCO3" levels. The other answer choices can be ruled out after consideration. Choice B can be eliminated because diarrhea

would result in the excretion of large amounts of bicarbonate (HCO3") , thereby lowering the pH (by increasing the |H+])and causing metabolic acidosis. Choice C, rapid breathing, would lead to respiratoryalkalosis. This is because CO2 would

be lost in large amounts during rapid breathing, shifting the equation shown in the passage to the right, lowering both the H+and bicarbonate concentrations (see point B on the graph in Figure 1). Since the [H+] is lower, the pH is raised. Choice D,slow breathing, is also incorrect because this would lead to respiratory acidosis. Slow breathing allows CO2 to build up in

the blood, shifting the bicarbonate equation to the left, producing more H+and bicarbonate (see point A on the graph inFigure I). The best answer is A.

38. D is the best answer. Endocrine hormones, by definition, circulate in the blood in order to reach their target tissues. Adecrease in pH associated with acidosis would not affect the isoelectric point of these peptides, though. Recall thatisoelectric point is an inherent property of a peptide, determined by the pKa's of the various side chains. The isoelectricpoint is the point at which a protein has no net charge. By lowering the pH of the blood, we do not directly change theisoelectric point of individual peptides; the only way to do that is to change the amino acid sequence of these peptides. Let'sconsider the other answer choices. Both the secondary and tertiary structures of proteins can be altered by changes in bloodpH. This is because secondary structure is largely determined by hydrogen bonding and ionic interactions. If the mediumbecomes "more acidic, more side chains may become protonated and hydrogen bonding and ionic interactions can be altered.This alteration of secondary structure (i.e., alpha helices and beta-pleated sheets, etc.) can lead to a change in the tertiary, orfolded, structure of the protein. Therefore, eliminate choices A and B since the question asks for the answer choice thatwould NOT be affected by a lower pH. Choice C can also be eliminated because hemoglobin's affinity for oxygen isdefinitely altered by a drop in pH. Recall that hemoglobin binds oxygen less tightly at lower pH's. This is called the BohrEffect. Th The best answer is D.

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39. B is the bestanswer. Recall that respiratory alkalosis involves an increase in blood pH that is due to an increase in the rateof breathing. Such an increased breathing rate would excrete an excess amount of blood CO2, shifting the bicarbonateequation given in the passage to the right (because CO2 levels are low). This lowers both the H+ and HCO3" concentrationsin accordance with Le Chatelier's principle. In other words, since we have taken away products (CO2) we must shift theequation to the right to compensate for this stress to the equilibrium system; reactants (H+ and bicarbonate) are consumed.The drop in H+ levels raises the pH. The question states that the kidney can help to compensate for this higher pH byexcreting HCO3" (renal compensation). Such excretion would cause the equation given in the passage to shift to the left,raising H+ concentrations. According to Figure 1, point Bdepicts uncompensated respiratory alkalosis (notice the higher pHand lower bicarbonate levels). Letters with subscripts represent compensatory effects. Point B2 represents complete renalcompensation for respiratory alkalosis. We can tell this because at point B2, the blood pH has returned to a normal 7.4,while the HCO3" levels have dropped even further (due to excretion by the kidney). We can rule out the other answerchoices. Point A2 depicts compensation for respiratory acidosis. Eliminate choice A. Point Ci represents partialcompensation for metabolic acidosis. Eliminate choice C. Point D\ represents partial compensation for metabolic alkalosis.Eliminate choice D. Remember to carefully read the legend to Figure 1 in order to stay on the right track. The best answerisB.

40. C is the best answer. The best way to approach this problem is to use the values given in the question and locate thecorresponding point on the graph given in Figure 1. Bicarbonate levels of 45 mmol/L and a pH of 7.4 correspond to pointA2 on the graph. Notice that at this point, blood pH has normalized but HCO3" levels are above normal. What could begoing on with this patient? From the legend to Figure 1we learn that point A represents uncompensated respiratory acidosis,while the letters with the subscripts represent compensatory effects. Point A2 therefore depicts an underlying respiratoryacidosis that has been compensated for by an increase in bicarbonate levels (shifting the reaction shown in the passage to theright and lowering the abnormally high H+ levels; see answer explanations to previous questions). Recall that respiratoryacidosis involves the failure of the lungs to excrete enough CO2. This leads to a buildup of CO2 in the blood, shifting the

bicarbonate reaction (shown in the passage) to the left, increasing the H+ and bicarbonate concentrations, thereby loweringthe pH and causing acidosis. In this patient, although the pH levels have normalized, the underlying high blood CO2 levelsare still present; their effect is simply minimized by the compensation of higher bicarbonate levels. The patient is thereforenot perfectly healthy; eliminate choice A. He also does not suffer from metabolic acidosis. His problem is respiratory innature; eliminate choice B. Likewise, he does not suffer from alkalosis; rule out choice D. The best answer is C.

41. B is the best answer. It might be helpful to refer to the answer explanations for the previous questions in order to get abetter background for approaching this one. Points A2 and B2 depict areas where the pH has been normalized bycompensatory mechanisms that either increase or decrease bicarbonate levels in response to the underlying acidosis oralkalosis. While these bicarbonate compensation mechanisms do return the pH to its normal value, the underlying diseasethat disrupted the pH in the first place has not been cured. Its effects are simply being masked by compensation mechanisms(such as renal compensation mentioned in an earlier question). An example of this is given in the previous question aboutthe patient who has underlying respiratory acidosis with complete compensation. Let's eliminate the other answer choices,keeping in mind that the question is looking for the FALSE statement. Choice A is a true statement and can therefore beeliminated. Metabolic acidosis (point C) may be caused by an abnormal HCO3" excretion (as in chronic diarrhea). This

would raise H+ levels, thereby lowering the pH and causing acidosis that is not related to respiratory problems (hencemaking it metabolic acidosis). Choice C is a true statement because point A2 depicts a normal pH with elevated bicarbonatelevels, representing respiratory acidosis with complete renal compensation (the kidneys reabsorb more bicarbonate, raisingHCO3" levels and lowering H+ levels). Eliminate choice C. Choice D is also a true statement. Point C\ represents a partialrenal compensation for the metabolic acidosis depicted in point C. Eliminate choice D. The best answer is B.

Passage VII (Questions 42 - 47) Multiple Sclerosis

42. A is the best answer. The question asks for the answer choice which does NOT represent a potential symptom of multiplesclerosis. Recall from the passage that the disease causes the destruction of the myelin sheaths which surround the axons ofneurons in the white matter of the brain. If the axons are no longer wrapped in myelin, the efficiency of the conduction ofaction potentials decreases. Therefore, muscles are stimulated less, resulting in fatigue and eventual paralysis (choices B andC). Likewise, incoming nervous impulses from skin sensory organs would be disrupted once they arrived at the "naked"axons. This might lead to choice D, numbness. This leaves us with choice A as the correct answer. Muscle spasms would

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not be a likely symptom of multiple sclerosis because they would require excessive nervous stimuli to the muscle fibers,something that would be impossible if action potentials were not conducted efficiently. In the case of multiple sclerosis,such excessive nervous stimuli could not occur due to the demyelination of nerve axons. The best answer is A.

43. B is the best answer. Notice that the question asks about inducing multiple sclerosis symptoms (i.e., demyelination) inwild-type, normal mice. From the last paragraph of the passage, we learn that individuals with multiple sclerosis may havebeen genetically predisposed to acquire the disease. Such individuals would get symptoms after being infected by a viruswhichmimicsa myelinantigen. Normal individuals would not get the disease after infection by the same virus.Therefore, ifwe can extrapolatethis information to our animal model, the mouse, we can safely assume that wild-type animals would notbe genetically predisposed to acquiring an autoimmune attack on myelin. Therefore, infection by an antigen-mimickingvirus such as the one described in the passage would have no effect (on the myelin sheaths, at least). We can eliminate theother answer choices because they all represent methods that would induce multiple-sclerosis like symptoms in mice.Choice A can be ruled out because injection of antibodies against components of the myelin sheath would causedemyelination. Choice C and choice D both also mimic the effects of multiple sclerosis by causing the destruction ofmyelin. The best answer is B.

44. D is the best answer. It is essential that we understand what the question asks: would transplantation of healthyoligodendroglial cells from an identical twin result in a cure for multiple sclerosis? Although such a transplant mayrepresent a treatment, it does not represent a cure. The underlying autoimmune disease is still present, and the newlytransplanted oligodendroglial cells would eventually be attacked by the host's immune system in much the same way thatthe original cells were attacked. In order to actually cure the disease, we must address the autoimmunity issue; this is not asimple task, and thus multiple sclerosis remains incurable at present. Let's eliminate the other answer choices. Choice A canbe ruled out because nowhere in the passage is it implied that oligodendroglial cells prevent immune system attack. They donot cure the underlying autoimmune disease. Choice B can be eliminated because although the new oligodendroglial cellsmight assume their normal role of repairing myelin sheaths, they would eventually be destroyed by the same autoimmuneattack that killed the original cells. Thus, such a transplant represents a treatment (which may temporarily eliminate somesymptoms) but not a cure for the disease. Choice C is incorrect because it states that the patient's immune system woulddestroy the newly transplanted cells because they came from a foreign source. These transplanted cells would not beconsidered foreign because they are genetically identical to the host's original cells; recall from the question that the cellswere transplanted from an identical twin. The best answer is D.

45. C is the best answer. In this case we are asked about potential treatments, that is, methods that would temporarily relievethe symptoms of multiple sclerosis. Let's consider each of the statements in the context of the information given in thepassage. Statement I suggests that the injection of antibodies against secreted T cell cytotoxins might pose a potentialtreatment. From the passage, we learn that these cytotoxins are responsible for myelin sheath degradation. We also learn thatthey are peptides; therefore antibodies can be manufactured which bind and inactivate the peptide cytotoxins. This mightindeed constitute and effective treatment, as it could prevent degradation of the myelin sheath. Statement III is likewise avalid treatment option. T cells which recognize myelin-relatedantigens are the ones which, according to the passage, travelto the brain and attack myelin-covered neurons. Injecting antibodies which bind to and inactivate these T cells might slow orprevent the progression of the disease by preventing demyelination. Statement II, on the other hand, is not a valid treatmentoption. Although these antigen-mimicking viruses may be responsible for triggering the disease (according to the passage),theyare notdirectlyresponsible for the demyelination that causes the symptoms of multiple sclerosis. Therefore, if someonealready has the disease, injecting antibodies which inactivate the virus that initially triggered autoimmune attack would notaccomplish anything; the damage has already been done by the virus at this stage. Such antibody injection might constitute apossibleprevention of the disease, but it would not constitute a treatment. The best answer is C.

46. C is the best answer. The key to answering this question correctly is picking up on the word autoimmune. Rheumatic feveris said to be an autoimmune disease. Therefore, we can eliminate answer choices which have nothing to do with the attackof the immune system on "self tissues. Choice A is therefore incorrect. Choice B is invalid because demyelination of nervefibers innervating the heart would not cause scarring of the heart valves. Choice D, direct infection of the heart valves byviruses, is invalid because this is does not represent an autoimmune attack on the valves. Choice C is therefore the correctanswer. Infection by an agent which mimics antigens present on the heart valves might provoke the immune system intomistakenly attacking the heart valves. We know this is a possibility because from the passage we learn that a similar modeof action may trigger multiple sclerosis. Recall that the question states, "Based on information given in the passage, whatmight trigger this autoimmune disease?" Choice C best answers this question. The best answer is C.

47. C is the best answer. Recall that an evolutionary advantage is anything that allows an organism to better survive, adapt, andreproduce. Viruses that mimic host antigens probably evolved because they had some selective advantage. In normalindividuals who are not affected by any autoimmune disease, self antigens are ignored by the immune system. Therefore, ifa virus could mimic these self antigens, it might also be ignored by the immune system of its host. This would be anenormous evolutionary advantage because the virus could then reproduce and survive unchecked. We can eliminate the

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otheranswer choices. Choice A is invalid because antigen mimicry, from the last paragraph of the passage, does notalwayslead to multiple sclerosis. Additionally, killing the host would not be the most evolutionarily advantageous event for a virus;if the host dies, it is most likely that the virus will die too. Choice Bis incorrect because the mimicked antigen appears to bea self antigen to the immune system; therefore, it couldn't act as a decoy because it would be normally ignored by theimmune system. Choice Dcan also be eliminated because there is an evolutionary advantage to mimicking host antigens, aswe have alreadydiscussed. The best answer is C.

(Questions 48 - 52) Not Based on a Descriptive Passage

48. B is the best answer. Cardiac muscle in considered syncytial, that is, the cytoplasms of each of the individual cardiac cellscan "communicate" electrically and chemically with one another almost instantly. This ability is due to gap junctions, whichare transmembrane protein channels connecting each of the cardiac muscle cells to each other. We can eliminate the otheranswer choices on the basis that each would require that the cells be connected to each other by nerves; instead, the cellsdirectly contact oneanother andareelectrically-coupled by gap junctions. The best answer is B.

49. B is the best answer. The renal afferent arteriole leads to the glomerulus, while the renal efferent arteriole leads away fromit. If the afferent arteriole has a larger radius than the efferent arteriole, blood flowing into the glomerulus must leave itthrough a smaller vessel. This can be thought of as trying to "squeeze" a lot of liquid through a small pipe; a back-pressureis created. This pressure increases the force with which the blood is filtered through the glomerulus. The more force, thegreater the rate of filtration. The other answer choices can be eliminated after consideration. Choice A is incorrect becausethe diameter of the renal afferent and efferent vessels should have little effect on reabsorption in the tubules; if anything,reabsorption would increase because more material is being filtered by the glomerulus. Choice C can be ruled out becauseADH secretion has little to do with renal arteriole diameters. Choice D is wrong because blood flow to the kidneys isincreased because the afferent arteriole bringing blood in has a larger radius. The best answer is B.

50. B is the best answer. Answering this question requires background knowledge of the stages of embryonic development.Recall that after fertilization occurs, the zygote undergoes a series of cell divisions (via mitosis), dividing the embryo into aseries of subcompartments. This division stage is referred to as cleavage. The next stage of development involves theformation of the blastula, a hollow ball of cells with a fluid-filled interior (the blastocoel). After this stage, gastrulationoccurs. During gastrulation, the embryo is subdivided into the three germ layers: endoderm on the inside, mesoderm in themiddle, and ectoderm on the outside. Hence, it is during gastrulation that the separation of cells into the three individualgerm layers occurs (making choice B the correct one). After gastrulation, the formation of the notochord and the dorsalnerve chord begins. This stage is known as neurulation. Organogenesis occurs afterwards, during which the three germlayers differentiate into the various organs of the mature organism. The best answer is B.

51. C is the best answer. Vertebrates are extremely well adapted to life on land mainly because they have solved the problemof maintaining water inside their bodies, eliminating the need to be surrounded by it. Recall from renal physiology that theloops of Henle are responsible for concentrating urine; the longer the loops, the more the urine can be concentrated and theless water is needlessly excreted. Hence, longer loops of Henle are an evolutionary advantage for life on land, asconservation of water is extremely important when one does not live in a body of water. Shorter loops of Henle would notconserve water and concentrate urine as effectively, making them least likely to contribute to the success of the vertebrates'land existence. Let's consider the other answer choices. Choice A, internal lungs, would definitely represent an adaptation toliving on land versus living in the water. Internal lungs are protected from the elements and do not lose as much moisture.Choice B also represents an advantageous land adaptation, protecting the organisms internal tissues from desiccation(drying out). Choice D, internal fertilization, does indeed represent an evolutionary adaptation to life on land. In the sea,external fertilization is easy, but on land, gametes must be protected from the harsh external environment. The best answerisC.

52. D is the best answer. The parasympathetic is responsible for stimulating digestive functions such as the peristalticcontractions of the smooth muscle of the stomach and small intestine. Let's consider the other answer choices. Since the

muscles responsible for the contractions of the digestive tract are predominantly smooth muscle, we can eliminate choice A.Choice B can be ruled out because the vagus nerve is parasympathetic in nature, and stimulation of the gastrointestinal tractby the vagus actually increases digestive activity. Choice C can be eliminated because during times of fear and stress, thesympathetic nervous system takes over and inhibits digestive function (so blood can be diverted to the skeletal muscles andbrain: a "fight or flight" response). This means that digestive function is decreased during times of fear and stress. The bestanswer is D.

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Biology

Copyright ©by The Berkeley Review®

Physiology Diagnostic Set I

Passage TopicsPhysiology

Diagnostic Set I

Passage I Thyroid HormonePassage II Home Pregnancy TestPassage III Oral Rehydration TherapyPassage IV Botulinum ToxinPassage V Circulatory ShockPassage VI Acid/Base DisordersPassage VII Multiple Sclerosis

Score Physiology

>13 45-52

11-12 40-44

10 37-39

9 34-36

8 31-33

7 28-30

6 24-27

5 22-23

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