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9700 · 15.1

Control and coordination in mammals — practice questions

Practice and worked examples for 9700 Control and coordination in mammals. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Explain how a nerve impulse is transmitted along a myelinated neurone from the resting state to the end of repolarisation, including the role of the myelin sheath.

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  1. Resting Potential: The neurone membrane maintains a resting potential of approximately -70mV. This is established and maintained by the sodium-potassium pump, which actively transports three Na+ ions out for every two K+ ions pumped in, coupled with the greater permeability of the membrane to K+ ions (due to more K+ leak channels), leading to a net efflux of positive charge.
  2. Depolarisation: Upon receiving a sufficient stimulus (reaching threshold potential), voltage-gated Na+ channels at a node of Ranvier open. Na+ ions rapidly diffuse down their electrochemical gradient into the axon, causing the membrane potential to reverse and become positive (reaching around +30mV to +40mV).
  3. Repolarisation: Immediately after depolarisation, the voltage-gated Na+ channels close, and voltage-gated K+ channels open. K+ ions rapidly diffuse out of the axon down their electrochemical gradient, restoring the negative charge inside the axon.
  4. Role of Myelin Sheath and Saltatory Conduction: The myelin sheath, formed by Schwann cells, acts as an electrical insulator, preventing ion flow across the axon membrane between the nodes of Ranvier. This means action potentials can only be generated at the nodes. The local currents generated at one node quickly depolarise the next node, causing the impulse to 'jump' along the axon. This saltatory conduction dramatically increases the speed of impulse transmission compared to unmyelinated neurones, where continuous conduction occurs along the entire length of the axon.

Worked example 2

The sciatic nerve in a human can be approximately 0.9 m long. A large myelinated neurone (Type Aα) in this nerve can conduct an action potential at 120 m/s. A small unmyelinated neurone (Type C) conducts at 1.5 m/s. Calculate the time taken for a nerve impulse to travel the full length of the sciatic nerve for both neurone types and determine how much faster the myelinated neurone is.

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  1. Recall the formula: Speed = Distance / Time, therefore Time = Distance / Speed.
  2. Calculate time for the myelinated neurone:
    • Distance = 0.9 m
    • Speed = 120 m/s
    • Time = 0.9 m / 120 m/s = 0.0075 s (or 7.5 ms)
  3. Calculate time for the unmyelinated neurone:
    • Distance = 0.9 m
    • Speed = 1.5 m/s
    • Time = 0.9 m / 1.5 m/s = 0.6 s (or 600 ms)
  4. Calculate the difference in time:
    • Time difference = Time (unmyelinated) - Time (myelinated)
    • Time difference = 0.6 s - 0.0075 s = 0.5925 s
  5. Conclusion: The impulse in the myelinated neurone takes 7.5 ms to travel the length of the sciatic nerve, while the unmyelinated neurone takes 600 ms. The myelinated neurone transmits the signal approximately 0.59 seconds faster, highlighting the efficiency of saltatory conduction.
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