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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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