AQA A-level Biology: Nerve Impulses and Synaptic Transmission (7402) – 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 3.6.2.1 (Nerve impulses) and 3.6.2.2 (Synaptic transmission) of the AQA A-level Biology (7402) specification, for first teaching 2015. Both sections are A-level only. Learn the content from the study guide and revision notes. See the AQA A-level Biology hub and the printable checklist.

Questions

1. Explain how an axon establishes a resting potential. [3]

2. A stimulus was applied to an axon and the potential difference across its membrane was recorded.

Time (ms)	0.0	0.6	0.9	1.2	1.6	2.0	2.5	3.5	5.0
Potential (mV)	−70	−55	+5	+30	0	−50	−80	−73	−70

(a) State the maximum potential and calculate the change from the resting potential to this value. [2] (b) Account for the rise in potential from 0.6 to 1.2 ms. [3] (c) Account for the fall in potential from 1.2 to 2.5 ms. [3]

3. A neurone has a refractory period of 1.6 ms. (a) Calculate the maximum number of impulses it can conduct per second. [2] (b) Explain two ways in which the refractory period affects how impulses travel along an axon. [2]

4. Stimuli of increasing strength were applied to a neurone. Strength is in arbitrary units.

Stimulus strength	2	4	6	8	10
Height of action potential (mV)	none	100	100	100	100
Impulses per second	0	20	45	70	90

(a) State the lowest stimulus strength in the table that produced an action potential. [1] (b) Explain what the table shows about the all-or-nothing principle, and how stimulus strength is signalled. [3]

5. Impulse A travels 0.45 m along a myelinated axon in 5.0 ms. Impulse B travels 0.04 m along a non-myelinated axon in 20 ms. (a) Calculate the speed of each impulse in m per s. [4] (b) Calculate how many times faster A is than B. [1]

6. (a) Explain how saltatory conduction increases the speed of an impulse. [3] (b) Explain why a wider axon conducts faster than a narrower one of the same type. [2] (c) In a frog nerve, the speed of conduction was 32 m per s at 20 °C and 48 m per s at 30 °C. Calculate the percentage increase in speed, and explain it. [4]

7. An impulse crosses a pathway made of a sensory neurone 0.90 m long (speed 75 m per s), a motor neurone 0.42 m long (speed 70 m per s) and three synapses, each with a delay of 0.5 ms. Ignore the length of any other neurone. (a) Calculate the time taken for an impulse to pass through the pathway. Give your answer to 3 significant figures in ms. [4] (b) Explain why each synapse causes a delay. [1]

8. Describe the events at a cholinergic synapse from the arrival of an action potential at the presynaptic knob to the start of an action potential in the postsynaptic neurone. [6]

9. (a) Explain why transmission across a synapse is unidirectional. [2] (b) Explain the importance of acetylcholinesterase in the synaptic cleft. [2]

10. At a synapse the resting potential is −70 mV and the threshold is −55 mV. Assume that each impulse from an excitatory presynaptic neurone depolarises the postsynaptic membrane by 4 mV, and that there is no loss of effect. (a) Calculate the smallest number of excitatory impulses that reach threshold. [2] (b) Name the type of summation if four different excitatory neurones fire at once. [1] (c) Explain why temporal summation needs impulses from one neurone to be close together in time. [2] (d) An inhibitory neurone adds −5 mV when four excitatory neurones fire at once. Calculate the potential and state whether threshold is reached. [2] (e) Explain how an inhibitory synapse produces a change in potential. [1]

11. Compare a neuromuscular junction with a cholinergic synapse between two neurones. [4]

12. Three drugs affect transmission at a neuromuscular junction. Predict and explain the effect on the muscle fibre of each. (a) Drug P prevents voltage-gated calcium channels from opening. [3] (b) Drug Q inhibits acetylcholinesterase. [3] (c) Drug R binds to the acetylcholine receptors but does not open their sodium channels. [2]

Answers

1. ATP drives the sodium-potassium pump, which exports 3 Na⁺ and imports 2 K⁺ [1]. K⁺ channels are open and most Na⁺ channels shut, so K⁺ leaks out faster than Na⁺ leaks in [1]. The inside becomes negative relative to the outside, giving a stable potential difference [1]. [3] Examiner insight: Name both ions and the direction of each movement. “Ions move” is too vague.

2. (a) Maximum is +30 mV [1]. Change = 30 − (−70) = 100 mV [1]. (b) Sodium channels open and Na⁺ diffuses in down its electrochemical gradient [1]. At −55 mV (threshold) more voltage-gated sodium channels open, which is positive feedback [1]. This rapid influx depolarises the membrane [1]. (c) The sodium channels shut [1] and voltage-gated K⁺ channels open, so K⁺ leaves down its gradient and the potential falls (repolarisation) [1]. Potassium channels are slow to close, so too much K⁺ leaves and the potential falls below −70 mV (hyperpolarisation) [1]. [8] Examiner insight: Use figures from the table. Linking “−80 mV” to slow-closing potassium channels shows you have read the data.

3. (a) 1.6 ms = 0.0016 s [1]; 1 / 0.0016 = 625 impulses per s [1]. (b) Sodium channels cannot open straight away, so impulses are separate from each other [1]. The region behind the impulse is refractory, so the impulse travels in one direction only [1]. [4] Examiner insight: Convert ms to s before dividing, and give the unit. A rate stated as “625” with no unit is easy to lose credit on.

4. (a) 4 units [1]. (b) The height of the action potential is 100 mV at every strength that produces one, so the response is all-or-nothing [1]. A larger stimulus does not give a bigger action potential [1]. Strength is signalled by frequency, which rises from 20 to 90 impulses per s [1]. [4] Examiner insight: Quote values from both rows of the table. “Frequency increases” with no data is weaker than the range.

5. (a) A: 5.0 ms = 0.0050 s [1]; 0.45 / 0.0050 = 90 m per s [1]. B: 20 ms = 0.020 s [1]; 0.04 / 0.020 = 2.0 m per s [1]. (b) 90 / 2.0 = 45 times [1]. [5] Examiner insight: Show the unit conversion as a separate step.

6. (a) Myelin is an insulating layer, so no ions pass through the membrane beneath it [1]. Depolarisation happens only at the nodes of Ranvier, and local currents flow from node to node [1]. The impulse jumps from node to node, so fewer depolarisations are needed and conduction is faster [1]. (b) A wider axon has less resistance to the flow of ions in the cytoplasm [1]. There is less leakage of ions out of the axon, so local currents spread faster [1]. (c) Increase = 48 − 32 = 16 m per s [1]; 16 / 32 × 100 = 50% [1]. Ions diffuse faster at the higher temperature [1], and enzymes make ATP faster for the sodium-potassium pump [1]. [9] Examiner insight: “Jumps” needs a reason. Link it to insulation and to local currents between nodes, or the explanation is only a description.

7. (a) Sensory neurone: 0.90 / 75 = 0.012 s = 12.0 ms [1]. Motor neurone: 0.42 / 70 = 0.0060 s = 6.0 ms [1]. Synapses: 3 × 0.5 = 1.5 ms [1]. Total = 12.0 + 6.0 + 1.5 = 19.5 ms [1]. (b) Time is needed for Ca²⁺ entry, vesicle fusion and diffusion of acetylcholine across the cleft [1]. [5] Examiner insight: Show each time before adding, and keep the unit with each figure.

8. The action potential depolarises the presynaptic knob [1]. Calcium channels in that membrane open and Ca²⁺ diffuses into the knob [1]. Vesicles of acetylcholine merge with the membrane and empty their contents into the cleft (exocytosis) [1]. The transmitter diffuses across the gap and attaches to receptors on the postsynaptic membrane [1]. This opens sodium channels, so Na⁺ enters and the membrane depolarises [1]. If the potential reaches threshold, a new action potential begins [1]. [6] Examiner insight: Follow the order of events and name the ion at each stage. Do not swap the calcium step and the sodium step.

9. (a) Vesicles of acetylcholine are only in the presynaptic knob [1]. Receptors are only on the postsynaptic membrane [1]. (b) It hydrolyses acetylcholine into choline and ethanoate [1]. This stops continuous stimulation of the postsynaptic membrane and allows the products to be recycled [1]. [4] Examiner insight: State the feature and its consequence. “It removes acetylcholine” gains less than saying what would happen without it.

10. (a) Change needed = −55 − (−70) = 15 mV [1]. 15 / 4 = 3.75, so 4 impulses [1]. (b) Spatial summation [1]. (c) The depolarisation from one impulse fades, and acetylcholine is broken down [1], so later impulses must arrive before the effect is lost for the effects to add [1]. (d) −70 + (4 × 4) − 5 = −59 mV [1]. This is below (more negative than) −55 mV, so threshold is not reached [1]. (e) Transmitter opens Cl⁻ and K⁺ channels, so Cl⁻ enters and K⁺ leaves, causing hyperpolarisation [1]. [8] Examiner insight: Round up in part (a), because 3 impulses cannot reach threshold. Writing 3.75 with no comment loses the final mark.

11. Both use acetylcholine, calcium channels, vesicles, receptors and acetylcholinesterase [1]. The postsynaptic cell of a neuromuscular junction is a muscle fibre, whereas at a synapse it is a neurone [1]. The membrane of the muscle fibre is folded and has more receptors [1]. A neuromuscular junction is always excitatory, whereas a synapse between neurones may be excitatory or inhibitory [1]. [4] Examiner insight: A “compare” needs similarities and differences stated side by side. Two separate descriptions leave the comparison undone.

12. (a) Calcium ions cannot enter the presynaptic knob [1], so vesicles do not fuse and less acetylcholine is released [1]. The muscle fibre is not depolarised, so it does not contract [1]. (b) Acetylcholine is not broken down [1], so it keeps binding to receptors and sodium channels stay open [1]. The muscle fibre stays depolarised, so contraction is prolonged [1]. (c) Acetylcholine cannot open sodium channels, so Na⁺ does not enter [1]. The muscle fibre is not depolarised, so there is no contraction even though acetylcholine is released [1]. [8] Examiner insight: Identify the step affected, then give the effect on the muscle fibre. A bare word such as “paralysis” shows no reasoning.

Where marks are usually lost

• Leaving out the permeability difference between K⁺ and Na⁺ when explaining the resting potential.
• Describing the pump as the cause of depolarisation.
• Quoting no data from tables or traces when asked to explain them.
• Not converting milliseconds to seconds.
• Writing “jumps” for saltatory conduction without mentioning insulation and local currents.
• Claiming that a stronger stimulus makes each action potential larger.
• Confusing the roles of Ca²⁺ in the presynaptic knob and Na⁺ in the postsynaptic membrane.
• Giving 3.75 as a number of impulses.

Next steps

• Review the revision notes and the study guide.
• Visit the AQA A-level Biology hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Biology (7401/7402) specification, for first teaching 2015, published by AQA. These questions follow sections 3.6.2.1 and 3.6.2.2. Check the current document on the AQA website for the latest version.