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(B) $1$. In the resting state,the axolemma is polarized,meaning the inside is negative relative to the outside (resting membrane potential is approxim

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Initially negative,then positive,and then negative again.

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(B) $1$. In the resting state,the axolemma is polarized,meaning the inside is negative relative to the outside (resting membrane potential is approximately $-70 \ mV$).$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing depolarization,where the inside becomes positive relative to the outside (action potential).$3$. Following this,the membrane undergoes repolarization,where $K^+$ ions move out,restoring the original negative charge inside the axolemma.$4$. Therefore,the sequence of potential change is negative $\rightarrow$ positive $\rightarrow$ negative.

For the propagation of a nerve impulse along a nerve fiber,what is the sequence of change in the electrical potential difference across the axolemma?

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An action potential in a nerve fiber is generated when the $+ve$ and $-ve$ charges on the outside and inside of the axon membrane are reversed because ...........

Which of the following statements is/are correct?
$(a)$ The electrical potential difference across the resting plasma membrane is called the resting potential.
$(b)$ $Na^+$ ions are responsible for the generation of an action potential.
$(c)$ $A$ nerve impulse is an action potential.

During the resting state,the axonal membrane allows the movement of which ions due to the concentration gradient?

$A$ nerve impulse is generated in a nerve fiber when the membrane becomes more permeable to:

The rate of rise of action potential and its amplitude are determined by the concentration of

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