a biologist investigated the stimulation of a pacinian

A biologist investigated the stimulation of a Pacinian corpuscle in the skin of a fingertip.She used microelectrodes to measure the maximum membrane potential of a Pacinian corpuscleand its sensory neurone when different pressures were applied to the fingertip.

The figure below shows the Pacinian corpuscle, its sensory neurone and the position of themicroelectrodes.

The table below shows some of the biologist’s results.

Pressure applied to thefingertip

Membrane potential at P /millivolts

Membrane potential at Q /millivolts

None –70 –70

Light –50 –70

Medium +30 +40

Heavy +40 +40

(a) Explain how the resting potential of –70 mV is maintained in the sensory neurone when nopressure is applied.

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(2)

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of 39Catalyst Tutors

(b) Explain how applying pressure to the Pacinian corpuscle produces the changes inmembrane potential recorded by microelectrode P.

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(3)

(c) The membrane potential at Q was the same whether medium or heavy pressure wasapplied to the finger tip. Explain why.

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(2)

(d) Multiple sclerosis is a disease in which parts of the myelin sheaths surrounding neuronesare destroyed. Explain how this results in slower responses to stimuli.

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(2)

(Total 9 marks)


(a) Describe how a Pacinian corpuscle produces a generator potential when stimulated.

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(3)

2

Doctors investigated the relationship between heart rate and arterial blood pressure. Theyrecruited healthy volunteers. For each volunteer, they recorded their normal arterial bloodpressure at rest. With each volunteer, they then carried out the following experiments.

Experiment 1Experiment 2

Experiment 3

They recorded heart rate at different blood pressures.They repeated experiment 1 after injecting a drug that inhibited theparasympathetic nervous system.They repeated experiment 1 after injecting a drug that inhibited thesympathetic nervous system.

The graph shows the results for one volunteer.


(b) Calculate the ratio of heart rate in experiment 2 to heart rate in experiment 3 at an arterialblood pressure of 10 kPa.Show your working.

Answer = ____________________

(2)

(c) What do these data suggest about the control of heart rate by the parasympathetic andsympathetic nervous systems in response to changes in arterial blood pressure?

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(3)

(Total 8 marks)

(a) The blink reflex is caused by stimulation of receptors in the eye or eyelid.Suggest two types of stimuli to which these receptors might respond.

1. _________________________________________________________________

2. _________________________________________________________________

(1)

3


(b) In humans, resting blink rate varies widely from 8 to 24 blinks per minute.This variation could result in the investigations into effect of stimulation on blink rateproducing means that are not significantly different. Explain why.

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(2)

(c) Some diseases cause changes in blink rate. Doctors do not often use blink rate todiagnose these diseases. Suggest two reasons why.

1 __________________________________________________________________

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2 __________________________________________________________________

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(2)


(d) A student completed an investigation to determine if the length of time eyes are closedbefore opening them affected blinking rate. His results are shown below.

The student did not draw a line of best fit.Suggest two reasons why.

1 __________________________________________________________________

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2 __________________________________________________________________

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(2)

(e) The student did not carry out repeats. He was still able to carry out a statistical test.Explain why.

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(1)


(f) The blink reflex can be stopped by drugs which prevent the opening of sodium ion channelproteins in the axons of motor neurones.Suggest how these drugs affect the passage of nerve impulses along the axons.

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(2)

(g) The blink reflex involves synapses. Channel proteins on presynaptic neurones are involvedin reflex responses.Explain how.

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(3)


(h) A student wanted to investigate the resting blink rate in people 60 years of age and people15 years of age.Describe how the student could find out whether there was a significant difference in blinkrates between the two age groups.

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(3)

(Total 16 marks)

CREB is a transcription factor in the mitochondria of neurones.

(a) What is a transcription factor?

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(2)

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(b) CREB leads to the formation of a protein that removes electrons and protons from reducedNAD in the mitochondrion.

Huntington’s disease (HD) causes the death of neurones. People with HD produce asubstance called huntingtin. Some scientists have suggested that binding of huntingtin toCREB may lead to the death of neurones.

Suggest how binding of huntingtin to CREB may lead to the death of neurones.

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(3)

(c) CREB is a protein synthesised in the cytoplasm of neurones. Transport of CREB from thecytoplasm into the matrix of a mitochondrion requires two carrier proteins.

Use your knowledge of the structure of a mitochondrion to explain why transport of CREBrequires two carrier proteins.

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(2)

(Total 7 marks)


A gardener accidentally pricks a finger on a thorn. She quickly pulls the finger away. This reactionresults from a simple reflex arc involving three neurones.

The diagram shows part of the pathway involved in this reaction.

5

(i) Complete the diagram to show the rest of the simple reflex arc.

(1)

On your diagram

(ii) name and label the three neurones;

(iii) label the effector.

(2)

(Total 3 marks)


Alzheimer’s disease (AD) is a non-reversible brain disorder that develops over anumber of years. At the start of 2014 the number of Americans with AD wasestimated to be 5.4 million. Every 30 seconds another person in Americadevelops AD.

5 In the brain of a person with AD there is a lower concentration of acetylcholine.This affects communication between nerve cells and initially results in memoryloss and confusion. Some of the symptoms of AD that are associated withcommunication between nerve cells are reduced by taking the drug donepezil.Donepezil inhibits the enzyme acetylcholinesterase.

10 A gene mutation called E280A found on chromosome 14 causes early-onset ADat a mean age of 49 years. The age at which the E280A mutation is expressedto cause AD varies.Yaramul is a town in a historically isolated region of the Andes Mountains. Thepopulation of this town has the highest frequency of the E280A mutation in the

15 world. The origin of the E280A mutation in this population has been traced backto a common ancestor in the 17th century. Natural selection has not reducedthe frequency of the E280A mutation in the population.

This autosomal dominant mutation involves a change in triplet 280 from GAA toGCA. Scientists analysed chromosome 14 from 102 individuals from Yaramul.

20 They recorded a sample size of 204 and detected 75 E280A mutations but only74 potential AD cases. The scientists identified individuals with the mutation bywhole genome sequencing. They had decided that a DNA probe would not be asuitable method to detect the E280A mutation.

6

(a) Assuming no one with AD died in 2014, calculate the annual percentage increase in ADcases in America for 2014 (lines 2–4).

Answer = ____________________ %

(2)


(b) Explain how donepezil could improve communication between nerve cells (lines 7–9).

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(3)

(c) Suggest and explain two reasons why there is a high frequency of the E280A mutation inYaramul (lines 13–15).

1. _________________________________________________________________

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2. _________________________________________________________________

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(2)

(d) Explain why natural selection has not reduced the frequency of the E280A mutation in thepopulation (lines 16–17).

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(2)


(e) The age at which the E280A mutation is expressed to cause AD can vary (lines 11–12).

Suggest and explain one reason for this.

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(2)

(f) One scientific study which analysed chromosome 14 involved 102 individuals. Thescientists recorded a sample size of 204. In this sample they detected 75 E280A mutationsbut only 74 potential AD cases (lines 19–21).

Suggest explanations for the figures the scientists recorded.

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(2)

(g) Suggest why a DNA probe for the mutated triplet was not considered a suitable method fordetection of the E280A mutation (lines 22–23).

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(2)

(Total 15 marks)


(a) When a nerve impulse arrives at a synapse, it causes the release of neurotransmitter fromvesicles in the presynaptic knob.

Describe how.

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(3)

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(b) The presynaptic knob contains actin filaments and myosin molecules.

The myosin molecules can attach to mitochondria and move them towards the presynapticmembrane, as shown in the diagram.


Use your knowledge of how myosin and actin interact to suggest how the myosin moleculemoves the mitochondrion towards the presynaptic membrane.

Do not include the roles of calcium ions and tropomyosin in your answer.

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(2)

(c) This movement of mitochondria happens when nerve impulses arrive at the synapse.

Suggest and explain one advantage of the movement of mitochondria towards thepresynaptic membrane when nerve impulses arrive at the synapse.

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(2)

(Total 7 marks)


The diagram below shows a nerve pathway in an animal.

8

(a) The nerve pathway shown in the diagram may be regarded as a simple reflex arc.

Use the diagram to explain why.

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(1)

(b) Suggest two advantages of simple reflexes.

1. _________________________________________________________________

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2. _________________________________________________________________

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(c) In the nerve pathway in the diagram, synapses ensure that nerve impulses only traveltowards the muscle fibre.

Explain how.

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(2)

(d) Axon P was found to conduct impulses much faster than other axons in the nerve pathwayshown in the diagram.

Describe and explain one feature of axon P that might cause this difference.

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(2)

(Total 7 marks)


Malaria is a disease that is spread by insects called mosquitoes. In Africa, DDT is a pesticideused to kill mosquitoes, to try to control the spread of malaria.

Mosquitoes have a gene called KDR. Today, some mosquitoes have an allele of this gene, KDRminus, that gives them resistance to DDT. The other allele, KDR plus, does not give resistance.

Scientists investigated the frequency of the KDR minus allele in a population of mosquitoes in anAfrican country over a period of 10 years.

The figure below shows the scientists’ results.

Year

9

(a) Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygousfor the KDR gene in this population in 2003.

Show your working.

Frequency of heterozygotes in population in 2003 ____________________

(2)


(b) Suggest an explanation for the results in the figure above.

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(4)

The KDR plus allele codes for the sodium ion channels found in neurones.

(c) When DDT binds to a sodium ion channel, the channel remains open all the time.Use this information to suggest how DDT kills insects.

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(2)

(d) Suggest how the KDR minus allele gives resistance to DDT.

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(2)

(Total 10 marks)


The blink reflex involves synapses. Below is a diagram of a synapse.

Identify A, B and C.

A ____________________________

B ____________________________

C ____________________________

(Total 3 marks)

10


The blink reflex can be affected by anaesthetics. Local anaesthetics are used to stop peoplefeeling pain but do not make them unconscious. General anaesthetics make people unconsciousand stop them feeling pain.

Doctors investigated two ways of measuring the effect of general anaesthetics.

They gave:

• anaesthetic S to 18 people• anaesthetic Q to 29 people

They recorded how long it took for the people to stop blinking.

The doctors then repeated the investigation. This time, they used a machine that measures brainactivity to decide when a person was unconscious, rather than when blinking stopped. For eachperson, they recorded how long it took for the machine’s readings to show that the person wasunconscious.

Their results are shown in the table. A value of ±2 × SD from the mean includes over 95% of thedata.

AnaestheticMean time taken to stop

blinking / minutes (±2 × SD)

Mean time taken formachine to show that

person was unconscious /minutes (±2 × SD)

S 0.24 (±0.01) 0.48 (±0.11)

Q 0.28 (±0.02) 0.44 (±0.07)

11

(a) Blinking involves cholinergic synapses. Anaesthetic S is a similar shape to acetylcholine.Suggest how anaesthetic S stops the transmission across the synapse.

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(b) Should time taken to stop blinking be used as an indicator of when to start surgery?Explain your answer.

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(c) Each person was given the same volume of anaesthetic per kg of body mass.Suggest why.

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(1)

(Total 6 marks)

The graph shows the distribution of rod cells and cone cells across the retina of a human eye.

Use the diagram to explain why

(i) no image is perceived when light is focused on the retina at Y;

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(1)

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(ii) an image formed at X is perceived in more detail than an image formed at Z.

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(Total 3 marks)

The diagram shows part of the retina in a human eye.

(a) Explain each of the following observations.

(i) When light falls on cells 1 and 2, only one spot of light is seen. But, when light falls oncells 2 and 3, two spots of light are seen.

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(1)

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(ii) When one unit of light energy falls on cell 3, no light is seen. But, when one unit oflight energy falls on cell 3, one unit falls on cell 4 and one unit falls on cell 5, light isseen.

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(3)

(b) Cells of the same type as cells 6 and 7 are found in large numbers at the fovea. Thisresults in colour vision with high visual acuity.

Explain what causes vision using the fovea.

(i) to be in colour;

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(1)

(ii) to have high visual acuity.

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(1)

(Total 6 marks)


The diagram shows the distribution of cone cells across the retina of a human eye.

(a) On the diagram draw a line to show the distribution of rod cells across the retina.

(2)

14

(b) Nocturnal mammals are active at night. Describe how the number and distribution of rodsand cones across the retina would differ in a nocturnal mammal from the number anddistribution in a human. Explain your answer.

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(3)

(Total 5 marks)


Mark schemes

(a) 1. Membrane more permeable to potassium ions and less permeable to sodium ions;2. Sodium ions actively transported / pumped out and potassium ions in.

2

1

(b) 1. (Pressure causes) membrane / lamellae to become deformed / stretched;2. Sodium ion channels in membrane open and sodium ions move in;3. Greater pressure more channels open / sodium ions enter.

3

(c) 1. Threshold has been reached;2. (Threshold or above) causes maximal response / all or nothing principle.

2

(d) 1. Less / no saltatory conduction / action potential / impulse unable to ‘jump’ fromnode to node;

2. More depolarisation over length / area of membranes.2

[9]

(a) 1. (Increased pressure) deforms / changes stretch-mediated sodium (ion) channel;2. (Sodium channels open and) sodium ions flow in;

Accept Na+

3. Depolarisation (leading to generator potential).

Accept correct description of depolarisation3

2

(b) Value between 2.17:1 and 2.29:1;

Accept rounding up to 2.2 or 2.3

Accept: number without : 1

Correct working showing answer but incorrect rounding in answerline = 1

Values between 117 to 119 and between 52 to 54 found but ratio wrong way round =1 mark.

Wrong way round gives answer between 0.35:1 and 0.46:12


(c) 1. Parasympathetic greater effect than sympathetic;

Ignore: descriptions of graph

2. Parasympathetic keeps heart rate down / lower / decreases heart rate (as bloodpressure increases);

3. Sympathetic keeps heart rate up / higher / increases heart rate (as bloodpressure increases);

2. and 3. Accept converse for blood pressure decreases

4. Parasympathetic greatest / greater effect at high blood pressure / sympatheticgreatest effect at low blood pressure.

3 max

[8]


(a) Any two from:

• light

• pressure

• touch

• temperature

• chemicals

• (loud) noise

• smell;

Two required for 1 mark

Do not accept unqualified reference to dust / particles / objects

Accept (rapid) movement (of particles / air) towards the eye

Accept humidity / moisture / tears1

(b) 1. Standard deviations / standard errors;

2. (So) likely to overlap;2

(c) 1. Would not know the patient’s / human’s normal blink rate sounable to make a comparison;

2. Blink rate could be affected by stress of seeing a doctor;

3. Many factors could affect blink rate so it would be difficult totell if blink rate was due to illness

2 max

(d) 1. Not possible to predict intermediate values;

2. Only one result for each time period / not mean values;2

(e) Collected paired data;1

(f) 1. No / low influx of sodium ions;

2. So no depolarisation / action potential;

2. ‘so no impulses’ insufficient2

(g) 1. Allows calcium ions in;

2. At end of presynaptic neurone;

3. Causing release of neurotransmitter;

1. Accept Ca2+/Ca ions but not Ca/Ca+

3


2. The idea of the end of the presynaptic neurone must be givene.g. presynaptic knob

3

(h) 1. Reference to large group size;

2. Reference to matching a specific, named variable;

3. Applying a statistical test to the data;

1. Accept ‘≥ 20 / many / lots’ but not ‘several / less than 20’2. Accept any named variable other than age.

3. Accept ‘use SE / 95% confidence limits’3

[16]

(a) 1. (Protein / molecule) that moves from cytoplasm to DNA;

Accept ‘it’ as TF.

Accept moves into nucleus

2. (TF) binds to specific gene / genes / to specific part of / site on DNA / binds topromoter / RNA polymerase;

Accept regulator / enhancer region

3. Leads to / blocks (pre)mRNA production / allows / blocks binding of RNApolymerase (to DNA) / allows RNA polymerase to work;

Ignore translation unless context wrong

Max 1 if refer to oestrogen as a transcription factor2 max

4

(b) 1. (Binding to CREB) prevents transcription / mRNA formation;

Accept that lack of protein leaves NAD reduced

2. (Binding of huntingtin) prevents production / translation of protein (that removeselectrons / protons from NAD);

3. Fewer electrons to electron transport chain / electron transport chain slows /stops / stops / slower oxidative phosphorylation;

4. Fewer protons for proton gradient;

5. Not enough ATP produced / energy supplied to keep cells alive / anaerobicrespiration not enough to keep cell alive;

Accept neurones require ATP for active transport of ions

Ignore references to resting potential3 max


(c) 1. Mitochondrion has two membranes / inner and outer membranes;

Accept cristae for inner membrane

2. For each (different) membrane a (different) carrier required;

Ignore reference to channel proteins2

[7]

(i) arc shows 3 neurones;

(3 distinct neurones, one of which is in the grey matter, with correctroute through dorsal and ventral roots and indication of synapses.Ignore position of cell bodies.)

1

(ii) neurones labelled sensory, relay / intermediate, motor;1

(iii) muscle labelled as effector;1

5

[3]

(a) 1. Correct answer of 19.4 / 19.41%OR19.47 / 19.5% = 2 marks;

2. Incorrect answer but shows increase of1,048,320 OR 1,051,200 = one mark;

Accept: 19.46% for one mark.2

6

(b) 1. Less / no acetylcholine broken down;2. Acetylcholine attaches to receptors;

3. (More) Na+ enter to reach threshold / for depolarisation / action potential /impulse;

1. Accept: more acetylcholine present / remains.

1 and 2. Accept: remains attached for longer = 2 marks.

3. Must be sodium ions.3

(c) 1. Isolated so inbreeding / low genetic diversity / small gene pool;2. Allele inherited (through generations) from (common) ancestor;

1. Ignore: Founder effect.

1. Accept: no interbreeding with other populations.

1. Reject: interbreeding within the population.2

(d) 1. AD / symptoms develops late / at 49;2. Have already reproduced;

Note: ‘It’ is not equivalent to AD / symptom as the question stemrelates to the mutation.

2


(e) 1. Epigenetics / environment / named factor e.g. stress, alcohol, toxins, diet,exercise, smoking;

2. methylation (of genes)ORacetylation (of histones);

1. Ignore: gender and lifestyle.

2. If further details are provided the context must be correct e.g.increased methylation or decreased acetylation inhibit geneexpression / transcription.

2

(f) 1. One person was homozygous dominant / has two dominant alleles = 2 marks;2. For one mark has two alleles / chromosomes;

1. Accept; homozygous dominant genotype e.g. ‘one person hasAA’ for 2 marks.

2. Accept: is diploid or has two copies of the gene.2

(g) 1. (GCA / triplet) is common / found in other places;2. Would not know if it was the mutation / allele / gene

ORProduces ‘false positives’

1. Accept: Probe will bind elsewhere.2

[15]

(a) 1. (Nerve impulse / depolarisation of membrane) causes Ca 2+ channel (proteins) toopen;

2. Ca 2+ enter by (facilitated) diffusion;3. Causes (synaptic) vesicles to fuse with (presynaptic) membrane;

Accept single reference to ions to cover 1 and 2

Penalise once for no reference to ions

1. Reject carrier proteins

3. Reject ref to release of vesicles

3. Ignore vesicles bind to membrane (but accept merge with)3

7

(b) 1. Myosin head attaches to actin and bends / performs powerstroke;2. (This) pulls mitochondria past / along the actin;3. Other / next myosin head attaches to actin (and bends / performs powerstroke);

1. Accept change shape / change angle

2. Ignore pulls actin along

2. Ignore refs to cytoskeleon

Accept plural or singular statements2 max


(c) 1. (Mitochondria) supply (additional) ATP / energy;2. To move vesicles / for active transport of ions / for myosin to move past actin

ORRe-synthesis / reabsorption of neurotransmitter / named neurotransmitter;

1. Reject produces energy

2. Ignore ref. to ATP for opening calcium ion channels/makingvesicles fuse with membrane

2

[7]

(a) Only 3 neurones / nerve cells (in reflex arc)1

(b) 1. Rapid;

2. Protect against damage to body tissues;

3. Do not have to be learnt;

4. Help escape from predators;

5. Enable homeostatic control.2 max

8

(c) 1. Neurotransmitter only made in / stored in / released from pre-synaptic neurone;

2. (Neuro)receptors only on the post-synaptic membrane;2

(d) 1. Axon P is myelinated;

2. So shows saltatory conduction / impulses jump between nodes of Ranvier

OR

3. Axon P has a larger diameter;

4. So less resistance to flow of ions.

Mark as 1 & 2 OR 3 & 42

[7]

(a) 0.32.

Correct answer = 2 marks

Accept 32% for 1 mark max

Incorrect answer but identifying 2pq as heterozygous = 1 mark2

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(b) 1. Mutation produced KDR minus / resistance allele;2. DDT use provides selection pressure;3. Mosquitoes with KDR minus allele more likely (to survive) to reproduce;4. Leading to increase in KDR minus allele in population.

4


(c) 1. Neurones remain depolarised;2. So no action potentials / no impulse transmission.

2

(d) 1. (Mutation) changes shape of sodium ion channel (protein) / of receptor(protein);

2. DDT no longer complementary / no longer able to bind.2

[10]

A Vesicle;

B Neurotransmitter;

C Synaptic cleft;

B Accept named neurotransmitter[3]

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(a) 1. Complementary to receptor for acetylcholine;

2. Binds to receptor;

3. On postsynaptic (membrane);

4. Prevents acetylcholine from binding;

5. No action potential in postsynaptic neurone;

2. Accept description of ‘binds’

3. Must be in context of membrane

5. Accept ‘depolarisation’ but not ‘impulse’3 max

(b) 1. Takes longer to become unconscious than it does to stopblinking;

2. No overlap of standard error;

1. Accept reference to 0.24/0.28 and 0.48/0.44 in place of longer2

(c) Different body masses but need to have comparable effects;

Do not accept ‘same’ effects or unqualified references to ‘bias /comparison / fair test’.

1[6]

11


(i) no (photo)receptor cells at Y / no rods and cones;1

(ii) X has many / only cones / more cones than Z;which each synapse to a single neurone / bipolar cell / noretinal convergence;ORZ has mainly rods / more rods than cones;which share / converge on neurones / bipolar cells;

2

[3]

12

(a) (i) 1 and 2 share neurone but 2 and 3 have separate neurones (to brain);

Ignore wrong names of neurones1

(ii) 1 unit is sub-threshold / 3 units are above threshold / give sufficientdepolarisation;(1 unit) No impulses / no action potential / in (sensory) neurone /does not stimulate (sensory) neurone / 3 units → impulses;(Spatial) summation / sufficient neurotransmitter released / from3 receptors / insufficient N-T from one;Reject ‘temporal’

3

13

(b) (i) (Three) different types of (cone) cells / types 6 and 7 sensitiveto different wavelengths / different frequencies / different colours;

(ii) Impulses along separate neurone from each receptor cell / eachreceptor cell connects to separate neurone;

2

[6]

(a) no rods at blind spot or fovea;greater distribution of rods at edge;

2

(b) more rods and no / fewer cones present;rods at the fovea / rods not mainly at periphery;

rods have high sensitivity / show retinal convergence /converse for cones;

rhodopsin ‘bleached’ at low light intensities / iodopsin ‘bleached’;at high light intensities;

3 max

[5]

14


Examiner reports

(a) Just over a third of students obtained 3 marks. The examiners were looking for a referenceto stretch-mediated sodium channels. This is the term used in the specification and isappropriate for this pressure receptor. Some students wrote about movement of sodium,rather than sodium ions, or did not state that they moved / diffused / flowed in. Others failedto note that this produces depolarization.

(b) About 30 percent of students obtained 2 marks in this part. As has been noted in otherexaminations, ratios seem to pose problems for many students. Many do not realize that aratio should be given as something : 1. Other students did not realize that it mattered inwhich order the numbers were given when they were asked to give the ratio of heart rate inexperiment 2 to that in experiment 3. These problems resulted in about 30 percentobtaining one mark. Many students had no idea how to calculate a ratio.

(c) Nearly 50 percent obtained 2 marks for stating that the parasympathetic slows heart rateand the sympathetic speeds it up. Very few noted from the graph that the parasympathetichas a much greater effect / influence on heart rate than the sympathetic; resulting in only10 percent obtaining all 3 marks.

2

Most students gave correct answers to this question. The most common incorrect answer was“dust” without an explanation of what stimulus this would be. Some students named a relevantstimulus and then gave an incorrect answer, which meant that they did not gain a mark since itwas a one mark question.

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(a) In this part, many students expressed themselves unsatisfactorily and there were a numberof misconceptions; almost half failed to score. Some simply said that a transcription factoraffects transcription. Others attempted to use oestrogen as an example of a transcriptionfactor, rather than a hormone that binds to a receptor to form a transcription factor. Yetothers failed to state that a transcription factor binds to a specific site (or sites) on DNA(however expressed). Only about 20% of students obtained both marks.

(b) There were some very good, clear and concise answers to this part that obtained all threemarks. All the points on the mark scheme were seen but perhaps the commonestobservations were that binding of huntingtin to CREB stops production of the protein (thatremoves protons and electrons from reduced NAD). This stops / slows the electron transferchain. This leads to not enough ATP being produced and the nerve cells die. A lot ofstudents did not read the stem carefully and thought that CREB removes electrons andprotons. Others made general references to respiration slowing but did not mention theelectron transfer chain, or proton gradients, or ATP production.

(c) It was pleasing in this part to find many students connecting the requirement for twocarriers for CREB to the two membranes of a mitochondrion; just over 50% obtained bothmarks. Some students had problems expressing themselves and others seemed genuinelyconfused about the membranes of the mitochondrion; for example, some wrote about amembrane round the cytoplasm and then the cristae.

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Considering that this topic would probably have been covered at GCSE, answers weresurprisingly poor. Many candidates were unable to complete the diagram of the reflex arc, andeven the better candidates frequently failed to show the relay neurone as being in the greymatter. A number tried to show the relay neurone apparently aiming out of the spinal cordtowards the brain. The names of the neurones were generally better known, and these werecredited wherever a reasonable attempt at a diagram had been made. Most correctly identifiedthe effector, but some mistakenly labelled the receptor. It would help if candidates could betrained to label structures with clear (preferably straight) guidelines which actually touch therelevant structures.

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(a) Over a third of students obtained both marks for this question, for answers of 19.41 / 19.4%or 19.47 / 19.5%, depending on whether the student used 7 x 52 (weeks) or 365 as thenumber of days in a year. Almost a third of students gained one mark for correctlycalculating the increase in AD cases per year as being 1 048 320 or 1 051 200, dependingon the number of days used. Incorrect rounding to give 19.46% was quite common, to gainone mark.

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(b) The majority of students gained at least one mark for stating that less acetylcholine wouldbe broken down, or that more acetylcholine would be present. Almost half of these studentsobtained a second mark for stating that the acetylcholine binds to receptors. However, only10% of students obtained maximum marks by describing how an impulse would beproduced in the postsynaptic neurone. Many students did appreciate that sodium ionchannels would open, but then failed to mention that sodium ions would then enter tocause depolarisation.

(c) Over 50% of students scored zero for this question, usually due to answers lackingcomplete explanations for the valid suggestions they outlined. Less than 2% of studentsgained both marks. The most frequently credited response was that isolation had resultedin a small gene pool or low genetic diversity. Poor use of terminology also prevented manystudents gaining both marks. Invariably, students referred to the gene or mutation, ratherthan the allele being inherited from a (common) ancestor. There was also considerableconfusion in the use of the terms inbreeding and interbreeding. There were also manyresponses which referred to genetic bottlenecks, the Founder effect, and an increase in therate of mutation in isolated areas. A significant number of students suggested that the lateonset of AD enabled individuals still to reproduce and pass on the mutation. This wouldexplain why the frequency of the mutation had not been reduced (part d), rather than whythere is a high frequency of this mutation to begin with.

(d) Almost a third of students obtained both marks, clearly expressing the idea that, due to thesymptoms of AD developing late on, affected individuals would have already reproduced.Over 50% of students scored zero, often providing responses that suggested that themutation was not harmful, or indeed that it was beneficial. An improvement in health carewas also provided as an explanation for the frequency of the mutation not being reduced.Students obtaining one mark often did not refer to the late onset of AD, but did understandthat individuals with the mutation could still reproduce and pass on the allele.

(e) Over 50% of students scored zero on this question. Many of these responses suggestedthat differences in the ‘level’ of acetylcholine, or exposure to mutagenic agents, caused thevariation in the age at which the mutation is expressed. Over a third of students obtainedone mark, usually by naming an environmental factor such as diet, smoking or stress.Answers specifically referring to epigenetics for at least one mark were infrequent. Theseresponses often gained a second mark, 10% of students, for mentioning methylation oracetylation. Students who described these processes generally provided correct details.


(f) Almost three out of ten students obtained one mark for explaining the sample size of 204 interms of two copies of chromosome 14 or two copies of an allele. A common error was torefer to two chromatids. Explaining why there were only 74 potential AD cases when 75mutations had been detected proved very challenging, with only 2% of students gaining thissecond mark. Many students suggested that the allele causing AD is recessive, despite line18 of the comprehension passage stating that it is dominant. A common misconceptionwas that one individual was heterozygous for the condition. Other incorrect responsesfocused on AD not having yet developed, or attempted to explain the data in relation to thedegeneracy of the genetic code.

(g) Over 10% of students did not attempt this question, and over 5% omitted the parts (e) and(f). It seems likely that some students had difficulty completing the paper, but it was alsoevident that these last three questions were demanding. Only 25% of students obtained amark on this question. Almost all of these students gained one mark for realising that theGCA triplet would occur in a number of different places. Half of these students thenexplained that you could not then determine if the mutation was present or not. As in part(f), a number of incorrect ideas were linked to the degenerate nature of the genetic code.The misconception that probes were being used to sequence the whole genome arose, solots of different primers would be needed and it would be very time consuming and costly.The probe was sometimes thought only to be able to identify the mutation if the gene hadbeen expressed to cause the disease, or that the probe would not bind because themutation had not occurred yet. A surprising number of students said the sequence of themutation was not known, so therefore a probe could not be made, or that the mutation wasdifferent in different people.

(a) This was a factual recall question that discriminated well. Examiners expected students touse appropriate A-level terminology, including references to calcium ions, channel proteinsand the (facilitated) diffusion of calcium ions in through the (presynaptic) membrane. Weakanswers included statements about calcium crossing the membrane.

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(b) This question required application of knowledge of how myosin and actin interact. Manystudents treated it as a factual recall question, not infrequently referring to shortening of thesarcomere. The best answers (about 12% of students) scored both marks by focusing onthe information given and noted that, in this circumstance, myosin moves past actin andpulls the mitochondrion along.

(c) Around half of the students obtained one mark for noting that mitochondria produce ATP.Just over 40% got a second mark, usually for relating the use of ATP to some aspect ofrecycling of neurotransmitter, or the active transport of an appropriate ion. Some studentsfailed to score the point relating to ATP because of references to ‘the production of energy’,rather than ‘the supply of energy’.

This question was exceptionally well answered, with the majority of students gaining at least twomarks and a substantial number scoring full marks. Very few students failed to score any marks.10

This question was generally well answered. Some students failed to gain marks because theyconfused enzyme terminology, e.g. active site, with specialist language relevant to a moleculebinding to a receptor. Some students were unclear about the location of the receptor. Otherscame close to stating that there would be no action potential in the postsynaptic membrane butdid not write an answer that was sufficiently clear to gain credit for that marking point.

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Unit 6

The question in part (i) was well answered. Only those referring to area Y as the ‘blind spot’failed to gain a mark. In (ii), generally candidates realised the importance of there being cones atX and rods at Z. Marks were not given when candidates failed to specify which area they werewriting about, X or Z, when explaining that there were more cones. About a third of thecandidates failed to gain the second mark. Candidates appeared to be aware of the greateracuity of cones but did not relate this to the neural connections.

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Unit 7

In (i), although most were able to explain that region Y lacked receptors, some fell short by justdescribing it as the ‘blind spot’ or as the point where the optic nerve was joined to the retina.

Descriptions of the two types of photoreceptors were sometimes less than precise, with somecandidates hinting that there were some rods present at X (the fovea) or that there were nocones at point Z. In (ii), while most understood about retinal convergence of rods, or lack of it forcones, some had a fundamental lack of understanding and thought that the reason cones gavemore precise vision was due to the way they differentiated between different colours of light.

Some candidates included the term ‘acuity’ in their answers but did not explain the basis of this.

Others used inaccurate terminology and referred to cones each being connected to a single‘nerve’.

Unit 6

(a) In (i), some told only half the story and others were clearly muddled, particularly concerningthe distinction between a neurone and a nerve (inappropriate terminology was notrewarded). Most, however, were able to deduce from the diagram which pair of rod cellsshared a single sensory neurone and which pair was connected via separate neurones.In (ii), the phenomenon of summation appeared to be well understood, although essentialdetails, such as the attainment of the necessary threshold potential, were sometimesomitted.

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(b) Very few candidates were able to express their ideas unambiguously regarding thesensitivity of different cone cells to different ranges of wavelength: many gave theimpression that each cone cell was able to respond to a full range of wavelengths of light.Greater success was enjoyed in explaining that visual acuity was achieved via thestimulation of separate neurones leading from the individual cone cells, as illustrated in thediagram.

Unit 7

(a) Part (i) asked for an explanation of two events. Many candidates gave a good descriptionof stimulation of only one or of two bipolar cells and obtained the mark. Some responsesonly gave an account of the one event and therefore could not obtain the mark. Part (ii)was answered well and many candidates obtained the three marks. In this instance theterm bipolar neurone was not demanded though it is worth noting that these cell werevariously described as ‘neurones’, ‘nerve cells’, ‘sensory nerves’, ‘interneurones’, ‘relaycells’ and ‘motor neurones’. Candidates should have knowledge that the bipolar cells aresensory neurones.


(b) The most common response to part (i) was the inadequate statement that ‘the fovea hascones’. Better answers referred to the absorption of different wavelengths by differentcones. Part (ii) was answered much better and most candidates obtained the availablemark.

This question produced a wide range of marks and proved to be an effective discriminator.

(a) Although many candidates indicated a lack of rods at the blind spot, far fewer appreciatedthat there would be either a reduction or an absence of rods at the fovea.

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(b) Many candidates gained up to two marks for suggesting that there would be a greaternumber of rods and fewer cones across the retina of a nocturnal mammal. Very fewcandidates gained a third mark by referring to retinal convergence or the breakdown ofrhodopsin at low light intensities.


a biologist investigated the stimulation of a pacinian

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