Question

Given below is the diagrammatic representation of impulse conduction through an axon.

Select the option with the incorrect information about impulse conduction at points A and B.

• A. An electric current flows on the inner surface from site A to B and on outer surface, from site B to A to complete the circuit.
• B. Action potential generated at site A arrives at site B
• C. Permeability to $N{a}^{+}$ decreases and quickly followed by a rise in permeability to $K^{+}$ at site A. Polarity at the site B is reversed and hence called repolarised.
• D. $N{a}^{+}$ channels get closed and $N^{+}$ channels get opened at site-A
  
  Depolarisation and opening of $N^{+}$ channels at site-B

Answer 1

Answer: (C)

When a stimulus is applied at a site ( e.g., point A) on the polarised membrane, the membrane at the site A becomes freely permeable to Na⁺. This leads to a rapid influx of Na⁺ followed by the reversal of the polarity at that site, i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged. The polarity of the membrane at the site A is thus reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site A is called the action potential, which is in fact termed as a nerve impulse. At sites immediately ahead, the axon (e.g., 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 impulse (action potential) generated at site A arrives at site B. The sequence is repeated along the length of the axon and consequently the impulse is conducted. The rise in the stimulus-induced permeability to Na⁺ is extremely shortlived. 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 the site of excitation and the fiber becomes once more responsive to further stimulation.