Question

Choose the correct order of changes in the potential on the inner side of the neuronal membrane during transmission of a nerve impulse.

• A. First negative, then positive and continues to be positive
• B. First positive, then negative and continues to be negative
• C. First positive, then negative and again back to positive
• D. First negative, then positive and again back to negative

Answer 1

The correct option is D First negative, then positive and again back to negative

In a neuron, the potential difference across the resting plasma membrane is termed as the resting membrane potential which is around $-60\mathrm{mV}$(ranges from $-60\mathrm{mV}$ to $-90\mathrm{mV}$) .

When the neuron is in the resting state, the inner side of the axonal membrane, which is similar to the plasma membrane of any cell, is electronegative relative to the layer of extracellular fluid (ECF) just outside the membrane. This is because, in the resting state, the axonal membrane is more permeable to $K^{+}$ions when compared to ${\mathrm{Na}}^{+}$ions. Therefore, the axoplasm (cytoplasm of the axon) contains a high concentration of negatively charged proteins along with $K^{+}$ ions and a low concentration of ${\mathrm{Na}}^{+}$ions, thereby, making the net charge inside the axoplasm negative. However, the fluid surrounding the axon has a higher concentration of ${\mathrm{Na}}^{+}$ions when compared to the axoplasm. Therefore, the presence of high concentration of ${\mathrm{Na}}^{+}$ions along with some $K^{+}$ions make the surrounding fluid more positive when compared to the axoplasm, thereby, creating a concentration gradient. This makes the neuron polarised in its resting state.

When the neuron is stimulated, the permeability of the axonal membane is reversed. Now, it is readily permeable to ${\mathrm{Na}}^{+}$ions. An action potential is generated due to the sudden influx of ${\mathrm{Na}}^{+}$ ions into the axoplasm.

This reverses the polarity inside the axonal membrane (since it becomes more positive inside due to the sudden influx of ${\mathrm{Na}}^{+}$ions when compared to the fluid outside which becomes relatively negative) thereby resulting in the process of depolarisation.
Later, due to opening of voltage-gated $K^{+}$channels, the $K^{+}$ ions from the inside rushes out of the cell. This causes the decrease in the membrane potential and gradually helps in the restoration of the resting potential at the site of excitation of the neuron. Hence, the axoplasm, again becomes negative (the polarised state).

Therefore, the order of changes in the potential on the inner side of the neuronal membrane during the transmission of nerve impulse is first negative, then positive and then negative again.