OxfordAQA International AS and A-level Biology: Nerve impulses and synapses (9610) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.4.3.1 and 3.4.3.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International A-level exams May/June 2018 onwards. Unit 4, Control, is listed as International A-level only. All figures are data supplied. Learn the ideas in the study guide and use the revision notes for recall. Course pages: hub, checklist and free diagnostics.

Questions

1. The table gives ion concentrations in mmol dm⁻³ for a resting axon.

Ion	Inside	Outside
Na⁺	15	145
K⁺	140	4

(a) Calculate the ratio inside : outside for K⁺ and outside : inside for Na⁺. [2] (b) Explain how these concentrations and the resting potential are established. [3]

2. An axon was stimulated and the potential difference across its membrane recorded. Threshold is −55 mV.

Time / ms	0	0.4	0.8	1.2	1.6	2.0	2.4	2.8	3.2	4.0
Potential / mV	−70	−66	−55	+20	+34	−8	−66	−79	−75	−70

(a) Calculate the amplitude of the action potential. [1] (b) Calculate the mean rate of repolarisation between 1.6 and 2.4 ms. [2] (c) Explain the change in potential from 0.8 ms to 1.6 ms. [3] (d) Explain the value at 2.8 ms. [2]

3. A neurone has a refractory period of 4.0 ms.

(a) Calculate its maximum frequency of impulses. [2] (b) Explain two ways in which the refractory period affects how impulses travel along an axon. [2]

4. Axon X is myelinated: an impulse crosses 0.72 m in 9.0 ms. Axon Y is not: an impulse crosses 0.018 m in 12 ms.

(a) Calculate the speed in each axon. [2] (b) Explain why conduction in X is faster. [3] (c) Axon Y has a larger diameter than a second non-myelinated axon Z. Explain why Y is expected to conduct faster than Z. [2]

5. Conduction speed in a non-myelinated axon from an ectothermic animal was measured.

Temperature / °C	5	15	25	45
Speed / m per s	8	14	22	3

(a) Calculate the percentage increase in speed from 15 °C to 25 °C. [2] (b) Explain the increase. [2] (c) Explain the result at 45 °C. [2]

6. During a burst of impulses 2.4 × 10⁶ Na⁺ enter an axon.

(a) Calculate the ATP needed for the sodium-potassium pump to remove them. [2] (b) Calculate the K⁺ the pump imports in this process. [1] (c) A poison stops ATP production in the axon. Explain the long-term effect on impulses. [2]

7. Explain the role of each in transmission at a cholinergic synapse.

(a) Calcium ions [2] (b) Acetylcholinesterase [2] (c) Mitochondria in the presynaptic knob [2]

8. Resting potential is −65 mV and threshold is −52 mV. A presynaptic neurone fires at regular intervals. Each impulse adds 5.0 mV to the postsynaptic membrane, and 40 % of the depolarisation present has faded by the next impulse.

(a) Calculate the depolarisation after each of the first three impulses. [3] (b) Explain why threshold is never reached by this neurone alone. [2] (c) A second presynaptic neurone fires at the same times with identical effect. Calculate how many impulses from each are needed and name the types of summation involved. [3]

9. Describe two ways the postsynaptic membrane at a neuromuscular junction differs from that at a synapse between neurones and explain the advantage of each. [4]

10. An inhibitory synapse acts on a neurone with resting potential −65 mV and threshold −52 mV. Each excitatory impulse adds 2.4 mV and all arrive together.

(a) Explain how an inhibitory synapse makes an action potential less likely. [3] (b) Calculate the number of excitatory impulses needed without inhibition. [2] (c) Calculate the number needed while inhibition holds the membrane 3 mV lower. [2]

11. Three substances act at a neuromuscular junction. Predict and explain the effect of each on the muscle fibre.

(a) A blocks voltage-gated calcium ion channels in the presynaptic membrane. [2] (b) B binds to acetylcholine receptors and keeps their sodium channels shut. [2] (c) C inhibits acetylcholinesterase. [3]

Answers

1. (a) K⁺: 140 / 4 = 35 : 1 [1]. Na⁺: 145 / 15 = 9.7 : 1 [1]. (b) Powered by ATP, the pump exports 3 Na⁺ and imports 2 K⁺ per cycle [1]. Many K⁺ channels are open and few Na⁺ channels, so K⁺ diffuses out faster than Na⁺ diffuses in [1]. Net loss of positive charge leaves the inside negative [1]. [5] Examiner insight: Give the direction and the mechanism for each ion. “Ions move” earns little.

2. (a) 34 − (−70) = 104 mV [1]. (b) Change = 34 − (−66) = 100 mV in 0.8 ms [1]. 100 / 0.8 = 125 mV per ms [1]. (c) At threshold, voltage-gated Na⁺ channels open [1]. Na⁺ diffuses in down its electrochemical gradient [1]. The depolarisation opens more channels, which is positive feedback [1]. (d) Voltage-gated K⁺ channels are slow to close [1]. Extra K⁺ leaves, so the potential falls below resting (hyperpolarisation) [1]. [8] Examiner insight: Quote times and values from the table, and give the unit with a rate.

3. (a) 4.0 ms = 0.0040 s [1]. 1 / 0.0040 = 250 impulses per s [1]. (b) Na⁺ channels cannot open straight away, so impulses stay separate and frequency is limited [1]. The region just passed is refractory, so the impulse travels in one direction only [1]. [4] Examiner insight: “Two ways” needs two distinct effects. Repeating one in different words does not count twice.

4. (a) X: 0.72 / 0.0090 = 80 m per s [1]. Y: 0.018 / 0.012 = 1.5 m per s [1]. (b) Myelin is an insulator, so ions cross only at nodes of Ranvier [1]. Local circuits carry depolarisation between nodes [1]. The action potential jumps between nodes (saltatory conduction), so fewer depolarisations per metre [1]. (c) A wider axon has less resistance to ion flow in the cytoplasm [1]. There is less leakage of ions, so local circuits spread further and depolarise ahead faster [1]. [7] Examiner insight: Convert ms to s before dividing. A speed that is 1000 times too large is a common slip.

5. (a) (22 − 14) / 14 [1] × 100 = 57.1 % [1]. (b) Ions have more kinetic energy and diffuse faster [1]. Respiration and ATP supply for the pump are faster [1]. (c) The heat denatures proteins such as channels and pumps [1]. Action potentials are no longer conducted properly, so the speed falls [1]. [6] Examiner insight: Percentage change divides by the starting value, here the 15 °C value, not the final value.

6. (a) 3 Na⁺ per ATP, so 2.4 × 10⁶ / 3 [1] = 8.0 × 10⁵ ATP [1]. (b) 8.0 × 10⁵ × 2 = 1.6 × 10⁶ K⁺ [1]. (c) The pump stops [1]. The concentration gradients gradually run down, so the resting potential cannot be maintained and the axon can no longer fire impulses [1]. [5] Examiner insight: Say “gradually”. The axon does not fail on the first impulse, because one impulse moves few ions.

7. (a) Ca²⁺ diffuses in through voltage-gated channels when the knob depolarises [1]. It makes vesicles fuse with the presynaptic membrane, releasing acetylcholine by exocytosis [1]. (b) It hydrolyses acetylcholine to choline and ethanoate [1]. This clears the receptors so the response stops and the next impulse gives a separate response [1]. (c) They make ATP by aerobic respiration [1]. ATP is used to resynthesise acetylcholine from choline and ethanoate and to package it in vesicles [1]. [6] Examiner insight: “Release” is too vague for calcium. State that vesicles fuse and acetylcholine leaves by exocytosis.

8. (a) After impulse 1: 5.0 mV [1]. After 2: 5.0 + 0.6 × 5.0 = 8.0 mV [1]. After 3: 5.0 + 0.6 × 8.0 = 9.8 mV [1]. (b) The total rises, but so does the amount that fades [1]. It levels off at 12.5 mV, where the 5.0 mV faded equals the 5.0 mV added, which is below the 13 mV needed [1]. (c) Two neurones give 10.0 mV per volley, which is below 13 mV [1]. After the second volley: 10.0 + 0.6 × 10.0 = 16.0 mV, above 13 mV, so 2 impulses from each neurone [1]. Spatial (two neurones) and temporal (repeated impulses) summation [1]. [8] Examiner insight: Work out the threshold gap first (13 mV). Showing each step of the sequence makes your reasoning visible.

9. Folding of the postsynaptic membrane increases its surface area [1], so there are more acetylcholine receptors and a bigger depolarisation [1]. More acetylcholinesterase is present in the folds [1], so acetylcholine is cleared quickly and each impulse gives a separate response [1]. [4] Examiner insight: Pair each difference with its advantage. A list of differences alone leaves the “explain” unanswered.

10. (a) The inhibitory transmitter opens Cl⁻ channels and K⁺ channels [1]. Cl⁻ diffuses in and K⁺ diffuses out, so the inside becomes more negative (hyperpolarisation) [1]. More depolarisation is then needed to reach threshold [1]. (b) Needed rise = 13 mV; 13 / 2.4 = 5.4 [1], so 6 impulses [1]. (c) Needed rise = 16 mV; 16 / 2.4 = 6.7 [1], so 7 impulses [1]. [7] Examiner insight: Round up. Six impulses give 14.4 mV in (c), which is short of the 16 mV required.

11. (a) Ca²⁺ cannot enter the knob, so fewer vesicles fuse and less acetylcholine is released [1]. The depolarisation is smaller and the muscle fibre is less likely to reach threshold, so contraction is reduced [1]. (b) Acetylcholine cannot open the sodium channels [1]. There is no depolarisation, no action potential in the muscle fibre and no contraction, even though acetylcholine is released normally [1]. (c) Acetylcholine is not hydrolysed and stays in the cleft [1]. It keeps binding to the receptors, so the sodium channels stay open [1]. The fibre is stimulated repeatedly, giving prolonged contraction [1]. [7] Examiner insight: Follow the chain from the step affected to the muscle. Do not stop at “less acetylcholine”.

Where marks are usually lost

• Writing that ions “move” or “flow” instead of diffusing through channels, or being pumped.
• Crediting the sodium-potassium pump with the rapid rise of the action potential.
• Omitting positive feedback in explaining the rise after threshold.
• Forgetting to convert ms to s in speed and frequency calculations.
• Dividing by the wrong value in a percentage change.
• Rounding down when asked how many impulses are needed.
• Saying that a larger stimulus gives a larger action potential.
• Describing a drug effect without naming the step it changes.
• Leaving out Ca²⁺ and exocytosis when describing release of acetylcholine.

Next steps

• Revision notes
• Study guide
• OxfordAQA A-level Biology hub and checklist
• Exam preparation
• Free diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International A-level exams May/June 2018 onwards, published by OxfordAQA. These questions follow sections 3.4.3.1 and 3.4.3.2.