unit 7: nucleic acids and proteins
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Unit 7: Nucleic Acids and Proteins
Lesson 7.5 Proteins
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7.5.1 Explain the four levels of protein structure, including each level’s
significance. Primary- order of
individual amino acids
Secondary- helix or pleated sheet, due to hydrogen bonds.
Tertiary- asymmetrical, cluster-like shape, due to bonding occurrences between R groups.
Quaternary- combination of two or more individual polypeptide chains.
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7.5.2 Outline the difference between fibrous and globular proteins, with
reference to two examples of each protein type.
Fibrous protein- have consistant repeating sequences, which form long pieces of tissue, eg. muscle fiber, collagen.
Globular protein- asymmetrical, occur as individual units which may contain several polypeptide chains, eg. hormones, enzymes.
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7.5.3 Explain the significance of polar and non-polar amino acids.
Because the phospholipid bilayer of the plasma membrane has both hydrophilic and hydrophobic components, globular proteins will “line up”, with their hydrophilic and hydrophobic areas matching those of the plasma membrane. This helps position the proteins correctly.
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7.5.4 State four functions of proteins, giving a named example of
each. Enzymes- catalase
Structural- collagen
Transport- hemoglobin
Hormones- insulin
Hemoglobin molecule
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Unit 7: Nucleic Acids and Proteins
Lesson 7.6 Enzymes
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7.6.1 State that metabolic pathways consist of chains and cycles of
enzyme catalyzed reactions.
The products of the first reaction, become the reactants of the second reaction, and so on. Enzymes catalyze each step.
ABCDE
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7.6.2 Describe the induced fit model.
The induced fit model is an extension of the lock and key model. It is important in accounting for the broad specificity of some enzymes.
Courtesy of Jerry Crimson Mann
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7.6.3 Explain that enzymes lower the activation energy of the chemical
reactions that they catalyse.
Courtesy of Jerry Crimson Mann
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7.6.4a Explain the difference between competitive and non-competitive
inhibition, with reference to one example of each.
Competitive inhibition- an inhibiting molecule structurally similar to the substrate molecule binds to the active site, preventing substrate binding. Example- the antibiotic Prontosil in bacteria.
Courtesy of Jerry Crimson Mann
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7.6.4b Explain the difference between competitive and non-competitive
inhibition, with reference to one example of each.
Non-competitive inhibition- the inhibiting molecule binds to the enzyme, but not at the active site. This causes a conformational change in the overall enzyme, including its active site, which reduces activity. Example- cyanide binds to proteins in the cytochrome complex, inhibiting cell respiration.
Courtesy of Jerry Crimson Mann
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7.6.5 Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sites.
ABCDE An accumulation of product E goes back and
inhibits the conversion of A, slowing the rate of the whole sequence.
Allostery is a form of non-competitive inhibition. End products of a metabolic sequence can bind to allosteric sites earlier in the metabolic pathway, regulating the entire chain of events. Example- ATP can inhibit components of glycolysis.
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Unit 8: Cell Respiration and Photosynthesis
Lesson 8.1 Cell Respiration
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8.1.1 Explain oxidation and reduction.
Oxidation- involves the loss of electrons from an element. Also frequently involves gaining oxygen or losing hydrogen.
Reduction- involves a gain in electrons. Also frequently involves losing oxygen or gaining hydrogen.
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8.1.2 Outline the process of glycolysis including phosphorylation, lysis, oxidation and ATP formation.
Glycolysis- In the cytoplasm, one hexose sugar is converted (lysis) into two tree-carbon atom compounds (pyruvate) with a net gain of two ATP and two NADH + H+.
Phosphorylation- is a process in which ATP is produced from ADP. During glycolysis, this is a substrate level phosphorylation.
C6(molecule) 2C3 (molecules)
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8.1.3 Draw and label the structure of a mitochondrion as seen in electron
micrographs.
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8.1.4a Explain aerobic respiration.
Oxidative decarboxylation of pyruvate- one carbon is removed from the C3 molecule (link reaction).
Krebs Cycle- produces trios phosphate, precurser to glucose.
NADH + H+- carrier molecules created during the Krebs cycle.
Electron Transport Chain- chemiosmotic synthesis of ATP via oxidative phosphorylation
Role of oxygen- acts as a final electron acceptor for electrons which have gone through the ETC.
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8.1.4b Picture of link reaction and
Krebs Cycle.
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8.1.5a Explain oxidative phosphorylation in terms of
chemiosmosis. 1) NADH and FADH2 release high
energy electrons into the electron transport chain.
2) As the electrons move down the cytochrome chain toward oxygen, H+ ions are propelled against their concentration gradient from the matrix into the intermembrane space.
3)H+ ions flow back to the matrix via gated ATP synthase, which uses energy from the flow to make ATP. ATP synthase
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8.1.5b Picture of electron transport chain.
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8.1.6 Explain the relationship between the structure of the mitochondrion and its
function. 1) Cristae form a
large surface area fort he electron transport chain.
2) The space between the outer and inner membranes is small.
3) The fluid contains enzymes of the Krebs cycle.
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Unit 8: Cell Respiration and Photosynthesis
Lesson 8.2 Photosynthesis
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8.2.1 Draw the structure of a chloroplast as seen in electron
micrographs.
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8.2.2 State that photosynthesis consists of light-dependent and light-
independent reactions. Light dependent reaction (green) Light independent reaction (pink)
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8.2.3 Explain the light-dependent reaction.
1) photoactivation of photosystem II
2) photolysis of water
3) electron transport 4) cyclic and non-
cyclic phosphoryliation
5) photoactivation of photosystem I
6) reduction of NADP+
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8.2.4 Explain photophosphorylation in terms of chemiosmosis.
Electron transport causes the pumping of protons to the inside of the thylakoids. They accumulate (pH drops) and eventually move out to the stroma through ATP synthase. This flow provides energy for ATP synthesis.
Courtesy of The University of Salzburg
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8.2.5 Explain the light-independent reactions.
1) Carbon fixation- CO2 is fixed to RuBP to form glycerate 3- phosphate (GP).
2) Reduction- GP is reduced to trios phosphate (TP).
3) Regeneration- RuBP is regenerated, and able to begin another turn on the cycle (with the help of Rubisco). Courtesy of Mike Jones
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8.2.6 Explain the relationship between the structure of the chloroplast and its
function. 1) Thylakoids have a
large surface area for light absorption.
2) The area inside the thylakoid is small, which facilitates the buildup of protons used in chemiosmosis.
3) the fluid filled stroma surrounding the thylakoid contains enzymes which facilitate the calvin cycle.
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8.2.7 Explain the relationship between the action spectrum and the absorption
spectrum of photosynthetic pigments in green plants.
The absorption spectrum illustrates the efficiency with which certain wavelengths of color are absorbed by pigments.
The action spectrum is a measure of overall photochemical activity.
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8.2.8 Explain the concept of limiting factors in photosynthesis.
Light intensity- as light intensity increases, photosynthetic rate increases, until a maximum efficiency is reached.
Temperature- each plant species has an optimum temperature range at which photosynthesis operates. To deviate in either direction reduces photosynthetic rate.
Concentration of CO2- as concentration of CO2 increases, photosynthetic rate increases, until a maximum efficiency is reached.
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Unit 9: Plant Science
Lesson 9.1 Plant Structure and Growth
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9.1.1 Draw and label plant diagrams to show the distribution of tissues in the
stem and leaf of a dicotyledonous plant.
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9.1.2 Outline three differences between the structures of dicotyledonous and
monocotyledonous plants. Dicot Flowers in groups of four or five Seeds have two cotyledons Leaves have reticulate venation
Monocot Flowers in groups of three Seeds have one cotyledon Leaves have parallel venation
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9.1.3 Explain the relationship between the distribution of tissues in the leaf and the
functions of these tissues. Absorption of light- palisade
mesophyll at top of leaf. Gas exchange- spongy
mesophyll in lower portion of leaf near stomata.
Support- dense, structural tissue.
Water conservation- regulated by stomata.
Transport of water- through the xylem.
Products of photosynthesis- transported through the phloem.
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9.1.4 Identify modifications of roots, stems and leaves for different
functions. Bulb- modified leaf used
for food storage.
Stem tuber- thickened rhizome or stolon used to store nutrients.
Storage root- modified root used for food storage.
Tendril- modified stem, leaf or petiole used by climbing plants for support and attachment.
tendril
taproot
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9.1.5 State that dicotyledonous plants have apical and lateral
meristems.
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9.1.6 Compare the growth due to apical and lateral meristems in
dicotyledonous plants. Meristematic tissue
generates new cells for growth of the plant.
Apical (terminal) meristems are found in roots and shoots, and facilitate vertical growth.
Lateral meristems facilitate horizontal growth,
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9.1.7 Explain the role of auxin in phototropism as an example of the
control of plant growth. Auxin is a plant
hormone which elongates cells. When a plant is exposed to a light source, the auxin migrates away from the source. In this way, the side of the plant farther from the light elongates, bending the plant toward the light source.
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Unit 9: Plant Science
Lesson 9.2 Transport in Angiospermophytes
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9.2.1 Explain how the root system provides a large surface area for mineral ion and water
uptake. Branching- increases overall surface area Root hairs- increases surface area of individual roots Cortex cell walls- facilitates absorption.
Yucca plant roots.Root hairs.
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9.2.2 List ways in which mineral ions in the soil move to the root.
1) Diffusion of mineral ions.
2) Fungal hyphae (in a mutualistic relationship)
3) Mass flow of water in the soil carrying ions.
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9.2.3 Explain the process of mineral ion absorption from soil into roots by active
transport. Integral proteins
transport minerals from the soil into roots through active transport. One the minerals have crossed over into the plants, they attract water through a concentration gradient.
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9.2.4 State that terrestrial plants support themselves by means of thickened
cellulose, cell turgor and xylem.
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9.2.5 Define transpiration. Transpiration- the
loss of water vapor from leaves and stems of plants.
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9.2.6 Explain how water is carried by the transpiration stream.
Xylem vessel structure- dead, empty cells with no cytoplasm.
Transpiration pull- a vacuum is created by the evaporation of water from the stomata of the leaves. The water column moves up to fill the vacuum.
Cohesion- the hydrogen bonding in water causes it to ‘stick’ to itself.
Evaporation- works with transpiration as described above.
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9.2.7 State that guard cells can open and close stomata to regulate
transpiration.
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9.2.8 State that the plant hormone abscisic acid causes the closing of
stomata.
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9.2.9 Explain how the abiotic factors, light, temperature, wind and humidity affect the rate
of transpiration in a typical terrestrial mesophytic plant.
Direct relationship: light = rate temperature = rate wind = rateInverse relationship: humidity = ↓rate
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9.2.10 Outline four adaptations of xerophytes that help to reduce
transpiration. Reduced leaves and
spines Deep roots Thickened, waxy
cuticles Reduced number of
stomata
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9.2.11 Outline the role of phloem in active translocation of sugar and
amino acids. The phloem
transports the products of photosynthesis, primarily sugar. Movement is from source (leaves) to sink (fruits, seeds, roots).
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Unit 9: Plant Science
Lesson 9.3 Reproduction in Flowering Plants
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9.3.1 Draw and label a structure of a dicotyledonous animal-pollinated
flower. Identify: sepal, petal,
anther, filament, stigma, style, ovary.
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9.3.2 Distinguish between pollination, fertilization and seed
dispersal. Pollination- the transfer
of male gametes (pollen) from anther to stigma.
Fertilization- the fusion of pollen with a female gamete. Pollination does not always lead to fertilization.
Seed Dispersal- once fertilized, the fused ovule develops into a seed. This is then contained in a fruit, which facilitates seed dispersal.
Courtesy of Debivort
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9.3.3 Draw and label a diagram showing the external and internal structure of a
named dicotyledonous seed (non-endospermatic).
Identify: Testa Micropyle Embryo root Embryo shoot Cotyledon
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9.3.4 Explain the conditions needed for the germination of a typical seed. Hydration- seeds
need to absorb water to initiate the germination process.
Temperature/pH- optimum temperature and pH ranges contribute to the probability of germination.
Note: Light requirements (or the lack of light) vary among seeds, and are difficult to generalize.
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9.3.5 Outline the metabolic processes of germination in a typical
starchy seed. Absorption of water
precedes the formation of gibberellin in the cotyledon. This stimulates the production of amylase, which catalyses the breakdown of starch to maltose. This subsequently diffuses to the embryo for energy production and growth.
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9.3.6 Explain how flowering is controlled in long-day and short-day
plants, including the role of phytochrome.
Phytochrome- a plant protein which detects the length of daylight, and in turn, can trigger flowering based seasonal changes of light.
Long Day Plant- will not flower unless daylight hours extend past a certain number of hours.
Short Day Plant- will not flower unless daylight hours are capped below a certain minium.
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Unit 10: Genetics
Lesson 10.1 Meiosis
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10.1.1a Describe the behavior of chromosomes in the phases of
meiosis. Prophase I-
chromosomes start to supercoil. Homologous chromosomes pair up during synapsis.
Crossing over can occur at this stage at the chiasmata.
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10.1.1b Describe the behavior of chromosomes in the phases of
meiosis. Metaphase I-
homologous chromosomes line up along the equatorial plane.
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10.1.1c Describe the behavior of chromosomes in the phases of
meiosis. Anaphase I-
homologous chromosomes separate, and move toward opposite poles.
(Note: there is no uncoupling of centromeres, as chromatids are still attached to each other.)
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10.1.1d Describe the behavior of chromosomes in the phases of
meiosis. Telophase I-
chromosomes arrive at poles. Spindle microtubules disappear. Cytokinesis follows, resulting in two separate cells.
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10.1.1e Describe the behavior of chromosomes in the phases of
meiosis. Prophase II- new
spindle microtubules attach to the centromeres.
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10.1.1f Describe the behavior of chromosomes in the phases of
meiosis. Metaphase II-
chromosomes line up along the equatorial plane.
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10.1.1g Describe the behavior of chromosomes in the phases of
meiosis. Anaphase II-
chromosomes separate and move toward opposite poles.
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10.1.1h Describe the behavior of chromosomes in the phases of
meiosis. Telophase II- spindle
microtubules disappear. Nuclear membrane reforms. Chromosomes relax into chromatin.
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10.1.2 Outline the formation of chiasmata in the process of crossing
over. Crossing over
occurs when homologous chromosomes bend around each other. The crossing point is called the chiasmata. The result is that potions of each chromosome are interchanged.
Pictured: double crossing over.
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10.1.3 Explain how meiosis results in an effectively infinite genetic variety
in gametes. Crossing over in prophase I- Since crossing
over can occur at any point along the chromosome, there is unlimited potential for genetic variety when it occurs.
Random orientation in metaphase 1- Homologous chromosomes line up along the equatorial plane independently of each other, eg. If chromosome 1 from the mother is on the left, chromosome two on the left is not necessarily also from the mother.
Without crossing over, the number of different gametes able to be produced, is 2n, with n= haploid number.
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10.1.4 State Mendel’s law of independent assortment.
Law of independent assortment- homologous chromosomes separate independently of other homologous chromosomes, allowing for many combinations in gametes, and ultimately, in the zygote that if formed by egg and sperm.
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10.1.5 Explain the relationship between Mendel’s law of
independent assortment and meiosis.
Independent assortment occurs during metaphase I of meiosis, when homologous chromosomes line up along the equatorial plane.
As chromosomes sort randomly, they create opportunities for new recombinants during fertilization, in essence shuffling the genetic deck.
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Unit 10: Genetics
Lesson 10.2 Dihybrid Crosses and Gene Linkage
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10.2.1 Calculate and predict the genotypic and phenotypic ratios of
offspring of dihybrid crosses involving unlinked autosomal genes.
Pea seedlings: T = tall t = short Y = yellow y = green Predicted offspring ration is 9:3:3:1
TY Ty tY tyTY TTYY TTYy TtYY TtYy
Ty TTYy TTyy TtYy TtyytY TtYY TtYy ttYY ttYyty TtYy Ttyy ttYy ttyy
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10.2.2 Distinguish between autosomes and sex chromosomes.
Autosomes- chromosomes pairs #1-22.
Sex chromosomes- X and y chromosomes, found as pair #23 (either as XX or Xy).
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10.2.3 Explain how crossing over in prophase I (between non-sister chromatids of a
homologous pair) can result in an exchange of alleles.
Crossing over in prophase I- Since crossing over can occur at any point along the chromosome, there is unlimited potential for the exchange of alleles and genetic variety.
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10.2.4 Define linkage group.
Linkage group- a group of alleles located on the same strand of DNA.
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10.2.5 Explain an example of a cross between two linked genes.
Alleles are usually shown side-by-side in dihybrid crosses eg. TtBb. In representing crosses involving linkage it is more common to show them as vertical pairs:
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10.2.6 Identify which of the offspring in such dihybrid crosses are
recombinants. In a test cross of:
The recombinants will be:
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Unit 10: Genetics
Lesson 10.3 Polygenic Inheritance
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10.3.1 Define polygenic inheritance.
Polygenic inheritance- occurs when a phenotype is controlled by more than one gene, resulting in a mosaic of phenotypes.
Courtesy of Scientific American
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10.3.2 Explain that polygenic inheritance can contribute to continuous variation
using two examples.
1) Human skin color- is thought to be controlled by at least 3 independent genes.
AABBCC x aabbccF1 = AaBbCc , then perform a dihybird
cross (AaBbCc), and there are many possible outcomes, such as:
AABBCc, AABBcc, AABbcc, AAbbcc, etc.
2) Human hair color- is also thought to be controlled but multiple genes, accounting for the large variety in shade.
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Unit 11: Human Health and Physiology
Lesson 11.1 Defense Against Infectious Disease
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11.1.1 Describe the process of clotting.
1) Platelets and damaged cells release clotting factors.
2) Prothrombinthrombin 3) Fibrinogenfibrin, which captures red
blood cells.
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11.1.2 Outline the principle of challenge and response, clonal selection and
memory cells as the basis of immunity.
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11.1.3 Define active immunity and passive immunity.
Active immunity- immunity due to the production of antibodies by the organism itself after the body’s defense mechanisms have been stimulated by invasion of foreign microorganisms.
Passive immunity- immunity due to the acquisition of antibodies from another organism in which active immunity has been stimulated, including via placenta or in the colostrum.
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11.1.4 Explain antibody production.
1) Macrophage presents antigen to helper T cell
2) Helper T cell activates B cell
3) B cells divide to form clones of plasma cells and memory cells, which secrete antibodies.
Plasma cells- fight the pathogen immediately.
Memory cells- stay in body, armed and ready if the pathogen appears in again in the future.
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11.1.5 Describe the production of monoclonal antibodies, and include one
use in diagnosis and one use in treatment.
Monoclonal antibodies are produced by fusing cancerous tumor cells with B-cells. This hybrid cell then proliferates and produces antibodies in perpetuity.
Diagnosis- used to detect HIV in the blood stream, as well as HCG in pregnancy tests.
Treatment- emergency treatment of rabies, blood and tissue typing for transplants.
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11.1.6 Explain the principle of vaccination.
A vaccine introduces the disabled pathogen in some for to the body, stimulating an immune response. Memory cells are created and circulate in the body, in case the real pathogen ever shows up.
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11.1.7 Discuss the benefits and dangers of vaccination.
Benefits: total elimination of diseases, prevention of pandemics and epidemics, decreaded health-care costs and prevention of harmful side-effects of disease.
Dangers: possible toxic effects of mercury in vaccines, possible overload of immune system, possible links with autism.
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Unit 11: Human Health and Physiology
Lesson 11.2 Muscles and Movement
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11.2.1 State the role of bones, ligaments, muscles, tendons and nerves in human
movement.1) A nerve impulse reaches muscle.2) The impulse triggers muscle
contraction.3) Muscles are attached to bone by
tendon.4) Bone moves.5) Bones are attached to other bones
by ligaments.
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11.2.2 Draw a diagram of the human elbow joint.
Identify: cartilage, synovial fluid, tendons, ligaments, radius, ulna, bicep, tricep.
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11.2.3 Outline the function of each of the structures named in the elbow
joint. Cartilage and synovial
fluid- cushion against friction.
Tendons- connect bone to muscle.
Ligaments- connect bone to bone.
Humerous- connected to bicep and tricep muscle.
Radius/Ulna- help rotate forearm.
Bicep/Tricep- help lift and lower forearm.
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11.2.4 Compare the movements of the hip joint and the knee joint.
Hip joint- flexion, extension, abduction, adduction, medial and lateral rotation, circumduction.
Knee joint- flexion, extension.
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11.2.5 Describe the structure of striated muscle fibers.
Myofibrils- bundled muscle filaments
Light bands- primarily actin filaments
Dark bands- protein discs found between sarcomeres
Mitochondria- provide energy for contraction.
Sarcoplasmic reticulum- similar to smooth ER with large stores of calcium.
Nuclei- fibers are multinucleated.
Sarcolemma- membrane surrounding muscle fiber
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11.2.6 Draw the structure of skeletal muscle fibers as seen in electron
micrographs. Identify: sarcomere, light and dark bands,
actin (thin) filaments, myosin (thick) filaments, sarcoplasmic reticulum.
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11.2.7 Explain how skeletal muscle contracts by the sliding of filaments.
1) Calcum ions flood sarcoplasmic reticulum.
2) Myosin binds to ATPADP +PMyosin in high energy configuration (SET).
3) Actin/myosin cross-bridge forms.
4) Myosin releases ADP + Prelaxes to low energy state, cross bridge moves actin filament.
5) Myosin binds to new ATP releases cross-bridge.
6) ATPADP + PMyosin back in high energy configuration.
Courtesy of David Richfield
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11.2.8 Analyze electron micrographs to find the state of contraction of
muscle fibers.
Courtesy of Ronnie Burns
Courtesy of University of British Columbia
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Unit 11: Human Health and Physiology
Lesson 11.3 The Kidney
99
11.3.1 Define Excretion. Excretion- the removal
from the body of the waste products of metabolic pathways.
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11.3.2 Draw and label a diagram of the kidney.
Cortex Medulla Pelvis Ureter Renal blood vessels
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11.3.3 Annotate a diagram of a glomerulus and associated nephron to show the function of each part.
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11.3.4 Explain the process of ultrafiltration.
Ultrafiltration- Blood pressure from the pumping heart forces fluid and materials out of the glomerulus (across a semi-permeable membrane) into the nephron.
Fenestrated blood capillaries- are elastic in nature to help with ultrafiltration.
Basement membrane- thick, layer of negatively charges tissue which helps keep negatively charged particles from crossing into the nephron.
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11.3.5 Define osmoregulation.
Osmoregulation- the control of the water balance of the blood, tissue or cytoplasm of a living organism. An inability to osmoregulate may result in edema.
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11.3.6 Explain the reabsorption of glucose, water and salts in the proximal
convoluted tubule. Reabsorption- water and
solutes which have been removed from the blood from ultrafiltration are moved back into the blood. Reabsorption involves: Microvilli- increase surface
area to help facilitate reabsorption
Osmosis- water is diverted back into the blood due to a concentration gradient.
Active transport- some solutes are actively transported back into the blood.
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11.3.7 Explain the roles of the loop of Henle, medulla, collecting duct and ADH
in maintaining water balance of the blood.
ADH= antidiuretic hormone. ADH increase = more water reabsorbed.ADH decrease = more water released in urine.
Collecting duct- funnels water into the ureter for excretion.
The primary role of the Loop of Henle is to reabsorb water. Water leaves the descending loop due to a concentration gradient, sodium leaves the ascending side due to active transport.
106
11.3.8 Explain the differences in the concentration of proteins, glucose and urea between blood plasma, glomerular filtrate
and urine. The flow sequence is: blood plasma glomerular
filtrateurine
As fluid progresses through the renal system, nitrogenous waste (urea) moves into the filtrate and is eliminated through the urine. Glucose also moves into the filtrate but is reabsorbed back into the blood. Large proteins remain in the blood plasma, and are not moved into the glomerular filtrate.
107
11.3.9 Explain the presence of glucose in the urine of untreated
diabetic patients.A diabetic’s inability metabolize glucose can result in hyperglycemia. Elevated levels of glucose in the blood will move into the glomerular filtrate, but will not be reabsorbed back into the blood. Instead, excess glucose will be found in the urine.
Cross section of human ureter
108
Unit 11: Human Health and Physiology
Lesson 11.4 Reproduction
109
11.4.1 Annotate a light micrograph of testis tissue to show the location and function of interstitial
(Leydig) cells, germinal epithelium cells, developing spermatozoa and Sertoli cells.
110
11.4.2 Outline the processes involved in spermatogenesis within
the testes. 1) mitosis 2) cell growth 2) two cell
divisions 3) cell
differentiation
111
11.4.3 State the role of LH, testosterone and FSH in
spermatogenesis. FSH- secreted by the
pituitary gland, facilitates spermatogenesis
LH- secreted by the pituitary gland, facilitates development of interstitial cells. The interstitial cells then secrete testosterone.
Testosterone- secreted by the testes, facilitates spermatogenesis.
112
11.4.4 Annotate a diagram of the ovary to show the location and function of germinal epithelium,
primary follicles, mature follice and secondary oocyte.
Identify- developing oocytes, Graafian follicle, primary oocyte, zona pellucida.
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11.4.5 Outline the processes involved
in oogenesis within the ovary. 1) mitosis 2) cell growth 3) two
divisions of meiosis
4) unequal division of cytoplasm
5) degeneration of polar body
114
11.4.6 Draw and label a diagram of a mature sperm and egg.
115
11.4.7 Outline the role of the epididymis, seminal vesicle and prostate gland in the
production of semen.Epididymis- an area
above the testicle where sperm is stored until ejaculation.
Seminal vesicle- gland that contributes most of the fluid volume of semen (about 70%).
Prostate gland- also contributes to seminal fluid (about 10-30%).
116
11.4.8 Compare the processes of spermatogenesis and oogenesis.
Number of viable gametes formed from one stem cell:spermatogenesis 4oogenesis 1
Timing and formation of gametes:spermatogenesis- development of sperm is continuous from puberty onward.oogenesis- development occurs in a monthly cycle, beginning with puberty and ending with menopause.
117
11.4.9 Describe the process of fertilization.
1) acrosome reaction- acrosome releases enzymes which digest the surrounding layer of the egg.
2) penetration of egg membrane by sperm 3) cortical reaction- cortical granules are
secreted by the egg via exocytosis, rendering the egg impermeable to future sperm.
118
11.4.10 Outline the role of human chorionic gonadoprophin (HCG) in
early pregnancy. HCG is secreted by
the embryo during early pregnancy. HCG helps signals the corpus luteum to stay active by continuing to secrete progesterone, which maintains the pregnancy.
119
11.4.11 Outline early embryo development up to the implantation
of the blastocyst.
120
11.4.12 Explain how the structure and functions of the placenta, including it’s hormonal role in
secretion of estrogen and progesterone, maintain pregnancy.
The placenta’s primary purpose is to bridge the blood supply between mother and fetus.
Secretion of progesterone helps maintain the uterine lining and placenta.
Secretion of estrogen inhibits the development of new follicles.
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11.4.13 State that the fetus is supported and protected by the amniotic sac and
amniotic fluid.
122
11.14.14 State that materials are exchanged between maternal and
fetal blood in the placenta.
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11.4.15 Outline the process of birth and its hormonal control.
Reduction in the level of progesterone results in the release of oxytocin. Oxytocin causes uterine contractions that trigger further release of oxytocin. In this way, the contractions get stronger and more rapid. This is an example of positive feedback.
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