A) \[{{K}^{+}}\]ions from extracellular fluid to intracellular fluid
B) \[N{{a}^{+}}\]ions from intracellular fluid to extracellular fluid
C) \[{{K}^{+}}\]ions from intracellular fluid to extracellular fluid
D) \[N{{a}^{+}}\] ions from extracellular fluid to intracellular fluid
Correct Answer: D
Solution :
During the propagation of nerve impulse when a stimulus of adequate strength is applied to a polarised membrane, the permeability of the membrane to \[N{{a}^{+}}\] is greatly increased at the point of stimulation. As a result the sodium ion channels permit the influx of Na+ by diffusion. Since, there are more \[N{{a}^{+}}\]ions entering than leaving, the electrical potential of the membrane changes from - 70 mV towards zero. At 0 mV the membrane is said to be depolarised. While the resting potential is determined largely by \[{{K}^{+}}\] ions, the action potential is determined largely by \[N{{a}^{+}}\] ions. Action potential is another name of nerve impulse. The stimulated negatively charged point on the outside of the membrane sends out an electrical current to the positive point adjacent to it. This local current causes the adjacent inner part of the membrane to reverse its potential from -70 mV to-+30 mV.
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