Explain why a neurone cannot be stimulated to produce a second action potential immediately after the first action potential has occurred.

OCR A-Level Biology A (H420) — 5.3 Neuronal communication · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

During a nerve impulse, ion movements across the axon membrane produce rapid changes in membrane potential. Immediately after an action potential, the neurone enters a refractory period during which it is unresponsive to further stimulation.

Model answer (4 marks)

1. Voltage‑gated Na⁺ channels are inactivated after depolarisation and cannot reopen immediately.
2. The membrane becomes hyper‑polarised, making the potential more negative than resting.
3. Voltage‑gated K⁺ channels stay open, allowing K⁺ to leave the cell.
4. Because of the above, a stimulus cannot reach threshold until the resting potential is restored.

Examiner tips

  • Use the exact terms: ‘inactivated Na⁺ channels’, ‘hyper‑polarisation’, ‘K⁺ channels remain open’, ‘threshold cannot be reached’.
  • Show the causal chain: depolarisation → Na⁺ inactivation + K⁺ opening → hyper‑polarisation → refractory period.
  • Keep each point brief and separate with numbers or bullet points.

Mark scheme (4 marks)

  1. Voltage-gated sodium ion channels are inactivated / closed and cannot be reopened immediately after depolarisation
  2. The membrane is hyperpolarised / the membrane potential is more negative than the resting potential during the refractory period
  3. Voltage-gated potassium ion channels remain open, causing continued outward movement of K⁺ ions
  4. A sufficiently large stimulus cannot generate a new depolarisation / threshold cannot be reached until the resting potential is restored

Key terms in this question

action potential

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