OCR A Level Biology A: Neuronal communication (H420) – 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 OCR A Level Biology A (H420) section 5.1.3, Neuronal communication, learning outcomes (a) to (d), from specification Version 4.1 (April 2026). Learn the content first from the study guide and the revision notes. Course pages: the course hub, the printable checklist and the free 10-minute diagnostics.

Unless told otherwise, take rest as −70 mV and threshold as −55 mV.

Questions

1. Explain how a Pacinian corpuscle acts as a transducer, from the application of pressure to the start of an action potential in the sensory neurone. [5]

2. A Pacinian corpuscle was pressed with increasing pressures. The membrane potential of the sensory neurone ending was recorded, before any action potential.

Pressure / kPa	0.5	1.0	1.5	2.0	2.5
Membrane potential / mV	−66	−61	−56	−51	−49

(a) Calculate the size of the generator potential at 1.5 kPa. [1] (b) State which pressures produced action potentials in the sensory neurone, and explain your answer. [2] (c) Explain why the neurone sent no impulses at 1.0 kPa although the membrane potential changed. [2]

3. (a) Describe the role of a relay neurone in a reflex pathway. [2] (b) State two structural differences between a sensory neurone and a motor neurone. [2]

4. Explain how the resting potential of a neurone is established and maintained. [4]

5. The table shows the membrane potential of one point on an axon after a stimulus at time 0.

Time / ms	0	0.4	0.8	1.2	1.6	2.0	2.4	2.8	3.2	4.0
Potential / mV	−70	−60	−48	+10	+38	+5	−40	−72	−78	−72

(a) Calculate the change in potential from threshold to the peak. [2] (b) Calculate the mean rate at which the potential rose from the 0.8 ms reading to the peak. [2] (c) Explain why the potential rises so steeply after threshold is passed. [3] (d) Explain the value of −78 mV at 3.2 ms. [2]

6. An impulse takes 7.0 ms to travel 0.84 m along a myelinated axon. An impulse takes 7.5 ms to travel 0.015 m along a non-myelinated axon.

(a) Calculate the speed of the impulse in the myelinated axon. [2] (b) Calculate how many times faster it is than the impulse in the non-myelinated axon. [2] (c) Explain why myelination increases the speed of conduction. [3]

7. A sensory neurone fires 24 impulses in 0.40 s with a light stimulus and 52 impulses in 0.40 s with a heavier stimulus.

(a) Calculate the frequency of impulses for each stimulus. [2] (b) Calculate the percentage increase in frequency, to the nearest whole number. [2] (c) The refractory period of this neurone is 2.5 ms. Calculate the maximum frequency of impulses. [2] (d) Explain why frequency, not the size of the action potential, shows the strength of a stimulus. [2]

8. Describe the sequence of events at a cholinergic synapse from the arrival of the impulse until the next neurone’s membrane depolarises. [6]

9. A postsynaptic neurone receives excitatory input. Each excitatory impulse depolarises the postsynaptic membrane by 2.0 mV. Assume all effects arrive together and none decays.

(a) Distinguish between spatial and temporal summation. [2] (b) Calculate the smallest number of simultaneous excitatory impulses needed to reach threshold. [2] (c) One inhibitory input that hyperpolarises the membrane by 4.0 mV is active at the same time. Calculate the new smallest number of excitatory impulses. [2]

10. (a) Explain how an inhibitory synapse makes an action potential in the postsynaptic neurone less likely. [2] (b) Give two reasons why synapses are important in the nervous system. [2]

11. A drug inhibits acetylcholinesterase. One impulse arrives at a cholinergic synapse. Without the drug, the postsynaptic neurone fires one impulse. With the drug, it fires six impulses in 10 ms.

(a) Explain the difference. [3]

At a different synapse onto the same neurone, an inhibitory neurotransmitter opens chloride channels and holds the membrane at −75 mV. Each excitatory impulse depolarises the membrane by 2.4 mV.

(b) Calculate how many simultaneous excitatory impulses are needed to reach threshold (i) from −70 mV and (ii) from −75 mV. [4]

12. In an experiment, a steady pressure was applied for 3.0 s to a Pacinian corpuscle. Impulses in the sensory neurone were counted in successive 0.5 s intervals. The experiment was repeated after the lamellae were removed.

Interval / s	0-0.5	0.5-1.0	1.0-1.5	1.5-2.0	2.0-2.5	2.5-3.0
Impulses, lamellae present	5	0	0	0	0	0
Impulses, lamellae removed	5	4	4	3	3	3

(a) Calculate the total number of impulses in each case. [2] (b) Calculate the frequency of impulses in the first interval. [1] (c) Use the data to explain what the lamellae do. [3] (d) Suggest why this is useful for a receptor in the skin. [2]

Answers

1. The stimulus is mechanical (pressure) and the output is electrical [1]. Pressure deforms the lamellae and stretches the membrane of the neurone ending [1]. Stretch-mediated sodium channels widen [1]. Sodium ions diffuse in, depolarising the membrane to give a generator potential [1]. If it reaches threshold, voltage-gated sodium channels open and an action potential passes along the sensory neurone [1]. Examiner insight: name the stretch-mediated channel and the ion; “sends a signal” gives no mechanism.

2. (a) −56 − (−70) = 14 mV [1]. (b) 2.0 and 2.5 kPa [1], because the membrane potential reached −51 and −49 mV, which is above the −55 mV threshold [1]. (c) At 1.0 kPa the potential was −61 mV, a change of 9 mV [1], which is below the 15 mV needed to reach threshold, so voltage-gated sodium channels did not open [1]. Examiner insight: quote the table values beside the threshold value.

3. (a) It connects the sensory neurone to the motor neurone [1], within the central nervous system [1]. (b) Any two from: the sensory neurone has a long dendron and the motor neurone has many short dendrites; the sensory cell body is off to the side of the fibre and the motor cell body is at one end; the sensory axon is short and the motor axon is long [2]. Examiner insight: a difference names both neurones.

4. ATP drives the sodium-potassium pump, which exchanges 3 sodium ions leaving for 2 potassium ions entering [1]. Potassium channels are open at rest, so potassium ions diffuse out down their concentration gradient [1]. Few sodium channels are open, so sodium ions diffuse back in slowly [1]. More positive charge leaves than enters, so the inside is negative, about −70 mV, and the pump maintains it [1]. Examiner insight: give the 3:2 ratio and the greater permeability to potassium.

5. (a) 38 − (−55) [1] = 93 mV [1]. (b) (38 − (−48)) / (1.6 − 0.8) [1] = 86 / 0.8 = 107.5 mV per ms [1]. (c) Threshold is passed, so many more voltage-gated sodium channels open [1]. Sodium ions diffuse in [1], and the depolarisation opens still more channels, which is positive feedback [1]. (d) Voltage-gated potassium channels stay open slightly too long [1], so extra potassium ions leave and the membrane is hyperpolarised [1]. Examiner insight: give units, and say what feeds back on what.

6. (a) 0.84 / 0.0070 [1] = 120 m per s [1]. (b) Non-myelinated speed = 0.015 / 0.0075 = 2.0 m per s [1]; 120 / 2.0 = 60 times [1]. (c) Myelin insulates, so depolarisation happens only at the nodes of Ranvier [1]. Local circuits carry the current from node to node [1]. The action potential appears to jump, so fewer points depolarise and the impulse travels faster (saltatory conduction) [1]. Examiner insight: convert milliseconds to seconds before dividing, and state the unit in the final line.

7. (a) 24 / 0.40 = 60 per s [1]; 52 / 0.40 = 130 per s [1]. (b) (130 − 60) / 60 × 100 [1] = 117% [1]. (c) 2.5 ms = 0.0025 s [1]; 1 / 0.0025 = 400 per s [1]. (d) All action potentials have the same size (all-or-nothing) [1], so a stronger stimulus is signalled by more impulses per second [1]. Examiner insight: in a percentage change, divide by the original value, here 60, not by the new one.

8. The action potential depolarises the presynaptic knob [1]. Voltage-gated calcium channels open and calcium ions diffuse in [1]. Vesicles of acetylcholine fuse with the presynaptic membrane [1], releasing the neurotransmitter by exocytosis [1]. Acetylcholine crosses the gap and attaches to receptors in the next membrane [1], which opens sodium channels so that sodium ions enter and depolarise it [1]. Examiner insight: write events in order and name the ion at each step.

9. (a) Spatial: several different presynaptic neurones release transmitter together [1]. Temporal: one presynaptic neurone releases transmitter again and again before the effect of the first release has gone [1]. (b) 15 / 2.0 = 7.5 [1], so 8 impulses [1]. (c) 15 + 4.0 = 19 mV needed [1]; 19 / 2.0 = 9.5, so 10 impulses [1]. Examiner insight: round up: seven impulses give 14 mV, short of threshold.

10. (a) The inhibitory neurotransmitter opens chloride (or potassium) channels in the postsynaptic membrane [1]. The membrane becomes more negative (hyperpolarised), so more depolarisation is needed to reach threshold [1]. (b) Any two from: transmission is one way; summation combines inputs from many neurones; weak background signals are filtered out; one neurone can connect to many others [2]. Examiner insight: “passes impulses on” restates the stem; give a property.

11. (a) Without the drug, acetylcholinesterase breaks acetylcholine down in the cleft [1]. With the drug, acetylcholine stays in the cleft and keeps binding to receptors [1], so sodium channels open repeatedly and the membrane repeatedly reaches threshold [1]. (b) (i) 15 / 2.4 = 6.25, so 7 impulses [1]; seven give 16.8 mV, which is above 15 mV [1]. (ii) 20 / 2.4 = 8.33, so 9 impulses [1]; nine give 21.6 mV, above 20 mV [1]. Examiner insight: showing that the rounded-down number fails makes the answer convincing.

12. (a) Present: 5 + 0 + 0 + 0 + 0 + 0 = 5 [1]. Removed: 5 + 4 + 4 + 3 + 3 + 3 = 22 [1]. (b) 5 / 0.5 = 10 per s [1]. (c) Both cases give impulses at the start of the pressure [1]. With the lamellae present, impulses stop even though the pressure is still applied [1], so the lamellae redistribute a steady pressure, and the neurone ending is deformed only briefly at the start [1]. (d) It responds to changes in pressure and to vibration, not to constant pressure [1], so the brain is not flooded with signals from steady pressure such as clothing [1]. Examiner insight: use both rows of the table, then link them to the cause.

Where marks are usually lost

• Not giving a unit, or not converting milliseconds to seconds, in speed and frequency questions.
• Leaving out positive feedback when explaining the steep rise at the start of an action potential.
• Saying a stronger stimulus gives a larger action potential.
• Saying the pump alone creates the resting potential, without membrane permeability.
• Rounding down in “how many inputs” questions.
• Mixing up depolarisation, repolarisation and hyperpolarisation.
• Describing only one neurone when asked for a difference.

Next steps

• Re-read the study guide for any question that cost you marks.
• Work through the revision notes.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Level 3 Advanced GCE in Biology A (H420), specification Version 4.1 (April 2026), first assessment 2025, published by OCR. These questions follow section 5.1.3, learning outcomes (a) to (d).