中文（繁體）
中文（简体）
English
S5 Biology: Nervous System – Practice Paper 4 (Error Retry, 10 Questions)
S5 Biology: Nervous System – Practice Paper 4 (Error Retry, 10 Questions)
Paper four retrains the errors that show up most often in the first three papers: the size of a potential change, the order of the reflex arc, unit conversion in speed questions, and the sums for refractory periods and inhibitory synapses.
Quick answer Depolarisation is +30 minus -70, which is 100 mV; the arc runs receptor, sensory neurone, centre, motor neurone, effector; 1.5 m in 0.05 s is 30 m/s; four synapses cost 2.0 ms; 36 over 0.9 is forty times; a 2.5 ms refractory period caps the rate at 400; inhibition to -85 mV leaves 30 mV, fifteen more than before.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 20 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 Biology | Nervous System: Nerve Impulses, Synapses and Reflexes |
| Level | Error Retry |
02 Questions (10 items)
- FoundationThe resting potential is about -70 mV and the action potential peak about +30 mV: how large is the depolarisation?
- FoundationStarting from the receptor, what are the five parts of a reflex arc in order?
- FoundationWhat conduction speeds do myelinated and unmyelinated fibres typically reach?
- FoundationFrom which side of a synapse is neurotransmitter released, and by what process?
- IntermediateA 1.5 m pathway conducts in 0.05 s: what is the average velocity?
- IntermediateA reflex arc has four synapses at 0.5 ms each: what is the total synaptic delay?
- IntermediateA myelinated fibre runs at 36 m/s and an unmyelinated one at 0.9 m/s: how many times faster is the first?
- ChallengeA neurone has an absolute refractory period of 2.5 ms: what is the theoretical maximum impulse rate per second?
- ChallengeAn inhibitory transmitter pushes the membrane from -70 mV to -85 mV: how much depolarisation is then needed to reach the -55 mV threshold, and how much more than before?
- Error redoError retry: a student turned 0.9 m in 30 ms into 0.3 m/s. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 100 mV | Rising from -70 mV to +30 mV spans 100 mV, not 30 mV. |
| 2 | Receptor, sensory neurone, centre (interneurone), motor neurone, effector | The order runs from signal in to command out, and reversing it loses the whole item. |
| 3 | Myelinated about 15 to 120 m/s, unmyelinated about 0.5 to 2 m/s | Myelin lets the impulse jump between nodes of Ranvier, giving speeds tens of times higher. |
| 4 | From the presynaptic membrane, the axon terminal, by exocytosis | Vesicles sit only in the presynaptic terminal while the postsynaptic membrane carries receptors, which fixes the direction. |
| 5 | 30 m/s | Velocity is distance over time, and 1.5 divided by 0.05 gives 30. |
| 6 | 2.0 ms | Half a millisecond times four is 2.0, so the delay scales with the synapse count. |
| 7 | Forty times faster | 36 divided by 0.9 is 40, and myelin is the main reason for the gap. |
| 8 | 400 impulses per second | 2.5 ms is 0.0025 s, and one over 0.0025 gives 400, because the refractory period sets the ceiling. |
| 9 | 30 mV needed, which is 15 mV more than the original 15 mV | Rising from -85 mV to -55 mV takes 30 mV, against only 15 mV from -70 mV. |
| 10 | 30 m/s | Thirty milliseconds is 0.03 s, and 0.9 over 0.03 gives 30 rather than 0.3. |
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.
Find a tutor freeTutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.