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S5 Biology: Nervous System – Practice Paper 2 (Intermediate, 10 Questions)
S5 Biology: Nervous System – Practice Paper 2 (Intermediate, 10 Questions)
Paper two concentrates on synapses: the steps of transmission, how delay grows with the synapse count, excitation against inhibition, then total delay, speed ratios and the frequency ceiling set by the refractory period.
Quick answer Transmission runs from depolarisation through calcium entry and exocytosis to receptor binding; each synapse adds about 0.5 ms, so three synapses cost 1.5 ms; 24 m/s is forty times 0.6 m/s, and a 2 ms refractory period caps the rate at 500 impulses.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 25 minutes |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 2 · Error redo 2 |
| Chapter | Chapter: S5 Biology | Nervous System: Nerve Impulses, Synapses and Reflexes |
| Level | Intermediate |
02 Questions (10 items)
- FoundationIn order, what are the steps of transmission across a synapse?
- FoundationWhat is synaptic delay, and how does it relate to the number of synapses?
- FoundationHow do excitatory and inhibitory synapses differ in their effect on the postsynaptic membrane?
- IntermediateThe knee-jerk reflex uses only two neurones and one synapse: what is its synaptic delay?
- IntermediateA reflex arc has three synapses at 0.5 ms each: what is the total synaptic delay?
- IntermediateA myelinated fibre runs at 24 m/s and an unmyelinated one at 0.6 m/s: how many times faster is the first?
- ChallengeA neurone has an absolute refractory period of 2 ms: what is the highest impulse rate per second?
- ChallengeSeveral sub-threshold stimuli arrive at different points of the postsynaptic membrane at once and add up past the threshold: what is this called?
- Error redoError retry: a student wrote that transmitter is released by the postsynaptic membrane. What is correct?
- Error redoError retry: a student turned 0.8 m in 20 ms into 4 m/s. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Impulse reaches the presynaptic membrane, calcium enters, vesicles release transmitter by exocytosis, transmitter binds postsynaptic receptors, the postsynaptic membrane depolarises | Calcium entry is the trigger for exocytosis, so without calcium no transmitter is released. |
| 2 | The time for transmitter to diffuse and bind, about 0.5 ms per synapse, so more synapses mean a longer delay | The delay comes from the chemical step; conduction along the fibre itself takes almost no time. |
| 3 | Excitatory synapses depolarise it through sodium entry; inhibitory synapses hyperpolarise it through chloride entry or potassium exit, making the threshold harder to reach | Both release transmitter; the difference is which ion channel the receptor opens. |
| 4 | About 0.5 ms | A monosynaptic reflex has just one chemical step, which is why it is so fast. |
| 5 | 1.5 ms | Half a millisecond times three gives 1.5, so the delay grows with the synapse count. |
| 6 | Forty times faster | 24 divided by 0.6 is 40, and myelin is the main reason for the gap. |
| 7 | 500 impulses per second | One divided by 0.002 s gives 500, because the refractory period sets the ceiling. |
| 8 | Spatial summation | Rapid repeated stimuli at one point give temporal summation instead, and the two often act together. |
| 9 | It is released by the presynaptic membrane, the axon terminal, by exocytosis | The postsynaptic membrane carries receptors but no vesicles, so it cannot release transmitter. |
| 10 | 40 m/s | Twenty milliseconds is 0.02 s, and 0.8 over 0.02 gives 40 rather than 4. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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