The propagation of an action potential during | vedclass

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Question

(D) During the conduction of a nerve impulse,the resting membrane potential is disturbed by a stimulus. This stimulus causes the voltage-gated $Na^+$

Vedclass · Accepted Answer

$Na^+$ ions from extracellular fluid to intracellular fluid

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(D) During the conduction of a nerve impulse,the resting membrane potential is disturbed by a stimulus. This stimulus causes the voltage-gated $Na^+$ channels to open,leading to a rapid influx of $Na^+$ ions from the extracellular fluid $(ECF)$ into the intracellular fluid $(ICF)$. This process is known as depolarization,which generates the action potential. Therefore,the propagation of the action potential is primarily driven by the movement of $Na^+$ ions from the $ECF$ to the $ICF$.

The propagation of an action potential during nerve impulse conduction is made possible by the movement of:

Similar Questions

Which of the following processes are responsible for the transport of $Na^+$ from outside to the inner side of the axon membrane during the generation of a nerve impulse?

$A$: Presence of myelin sheath increases the rate of conduction of nerve impulse.
$R$: Ionic channels are absent in the area covered by myelin sheath; therefore,depolarization occurs only at the nodes of Ranvier,resulting in saltatory or jumping conduction.

What is the conduction of an action potential at the Nodes of Ranvier called?

$A$ : The imbalance in the concentration of $Na^+, K^+$ and proteins generates the resting potential.
$R$ : To maintain the unequal distribution of $Na^+$ and $K^+$,the neurons use electrical energy.

Which of the following acts is responsible for the depolarisation of the neuron?

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