During the transmission of a nerve impulse th | vedclass

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(D) The correct answer is $D$.$1$. At the resting state,the inner side of the axonal membrane is negatively charged relative to the outer side due to

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First negative,then positive and again back to negative.

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(D) The correct answer is $D$.$1$. At the resting state,the inner side of the axonal membrane is negatively charged relative to the outer side due to the differential permeability of the membrane to $K^+$ and $Na^+$ ions.$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing an influx of $Na^+$. This leads to depolarization,where the inner side becomes positively charged.$3$. Subsequently,the membrane becomes permeable to $K^+$ ions,which move out,leading to repolarization,where the inner side returns to its original negative state.$4$. Thus,the potential changes from negative to positive and back to negative during the transmission of an action potential.

During the transmission of a nerve impulse through a nerve fibre,the potential on the inner side of the plasma membrane has which type of electric change?

Similar Questions

When a nerve fiber is stimulated,what does the inner surface of the membrane become?

In which condition,the axonal membrane is comparatively more permeable to potassium ions and nearly impermeable to sodium ions?

Repolarization of a nerve fiber is caused by which of the following?

Assertion $(A)$: The axonal membrane of the neuron is more permeable to sodium ion $(Na^+)$ and nearly impermeable to potassium $(K^+)$.
Reason $(R)$: In a resting state,a neuron does not conduct any impulse.

The potential difference in the membrane,which is responsible for the conduction of an impulse,is brought about by a change in the membrane:

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