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Q.

As soon as an action potential generated at site A arrives at site B which is immediately ahead of it, the resting potential of the membrane at site A is restored by the

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a

efflux of Na+

b

influx of K+

c

influx of Na+

d

efflux of K+

answer is A.

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Detailed Solution

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An action potential at site A is initiated by the influx of Na+ causing depolarisation of the axonal membrane (inner side becomes positively charged). At sites immediately ahead, the axon (site B) membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B. On the outer surface current flows from site B to site A to complete the circuit of current flow. Hence, the polarity at the site is reversed, and an action potential is generated at site B. Thus, the action potential generated at site A arrives at site B. The rise in the stimulus-induced permeability to Na+ is extremely short-lived. It is quickly followed by a rise in permeability to K+. Within a fraction of a second, K+ diffuses outside the membrane and restores the resting potential of the membrane at site A and the fibre becomes once more responsive to further stimulation.

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