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S5 Biology: Nervous System – Practice Paper 3 (Challenge, 10 Questions)
S5 Biology: Nervous System – Practice Paper 3 (Challenge, 10 Questions)
Paper three puts numbers into real situations: transmitter breakdown, myelin damage, working back from total time to velocity, the share of time taken by synapses, and how far inhibition pushes the threshold away.
Quick answer Acetylcholinesterase ends the acetylcholine signal; 0.9 m in 30 ms is 30 m/s and two synapses take only 3.3 per cent of it; unmyelinated at 0.6 m/s needs 1.5 s; a 1.5 ms refractory period caps the rate near 667; inhibition to -80 mV leaves 25 mV to the threshold.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 2 · Error redo 2 |
| Chapter | Chapter: S5 Biology | Nervous System: Nerve Impulses, Synapses and Reflexes |
| Level | Challenge |
02 Questions (10 items)
- FoundationWhich enzyme breaks down acetylcholine, and why must it be broken down?
- FoundationWhy is saltatory conduction faster than continuous conduction?
- FoundationWhy does myelin damage, as in multiple sclerosis, slow responses down?
- IntermediateA reflex pathway is 0.9 m long and conduction takes 30 ms: what is the average velocity?
- IntermediateThe same pathway has two synapses at 0.5 ms each: what percentage of the 30 ms total is synaptic delay?
- IntermediateIf the same 0.9 m pathway used unmyelinated fibres at 0.6 m/s, how long would conduction take?
- ChallengeA neurone has a refractory period of 1.5 ms: what is the theoretical maximum impulse rate per second?
- ChallengeAn inhibitory transmitter lowers the membrane from -70 mV to -80 mV: how much excitatory depolarisation is then needed to reach the -55 mV threshold?
- Error redoError retry: a student divided 0.9 by 0.03 and wrote 0.03 m/s. Which step is wrong?
- Error redoError retry: a student claimed a longer refractory period allows more impulses per second. What is the real relation?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Acetylcholinesterase; the breakdown ends the stimulus so the membrane does not stay depolarised | Without breakdown the membrane would stay depolarised and could not pass on the next impulse. |
| 2 | Myelin insulates the fibre so depolarisation happens only at the nodes, and the impulse jumps node to node, saving the time needed to depolarise the whole membrane | Myelin does not conduct at all; it confines the current to the gaps between nodes. |
| 3 | Without insulation the impulse can no longer jump, so it must depolarise the whole membrane point by point and speed falls sharply | Damaged myelin can also stop the impulse midway, so the response may disappear altogether. |
| 4 | 30 m/s | Thirty milliseconds is 0.03 s, and 0.9 over 0.03 gives 30. |
| 5 | About 3.3 per cent | One millisecond over thirty is 0.033, so conduction takes almost all the remaining time. |
| 6 | 1.5 s | 0.9 over 0.6 is 1.5 s, fifty times slower than the original 0.03 s. |
| 7 | About 667 impulses per second | One over 0.0015 s gives 666.7, so about 667 impulses a second at most. |
| 8 | 25 mV | Rising from -80 mV to -55 mV needs 25 mV, ten more than starting from -70 mV. |
| 9 | The unit conversion: 30 ms is 0.03 s, so 0.9 divided by 0.03 is 30 m/s | Convert everything to seconds first, then divide once, so the result is not divided a second time. |
| 10 | The opposite: the maximum rate is one over the refractory period, so a longer period gives a lower rate | The refractory period is recovery time, so the longer it lasts the longer the wait before the next impulse. |
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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