3. Restoring resting potential

• How is resting potential restored?

  • once +40mV is reached, voltage gated Na+ channels close
  • more voltage gated K+ channels open, and K+ leaves the axon by facilitated diffusion
  • this brings about REPOLARISATION of the axon membrane (+40mV returns to -70mV)
  • there is a K+ overshoot as K+ leaves the axon rapidly
  • this causes HYPERPOLARISATION, and the membrane potential goes below -70mV
  • while the axon membrane is hyperpolarised, this is called the refractory period
  • eventually K+ channels close
  • resting potential is restored using the sodium potassium pump
• Why is the refractory period important?

  • it ensures that an impulse only travels in one direction along an axon
  • this is because a new action potential cannot be generated when the axon membrane is hyperpolarised and not at resting potential
  • this prevents overstimulation of an axon (too many action potentials) which could damage the axon