Question

Match the following column I and column II:

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{A. Resting potential}& \text{i. Na ions are more on the inside of the membrane at the site}\\ \text{B. Depolarization}& \text{ii. Potassium channels gets opened}\\ \text{C. Repolarization}& \text{iii. More K ions inside of the cell and more Na ions outside of the cell}\end{array}$

• A. A-iii, B-i, C-ii
• B. A-ii, B-iii, C-i
• C. A-iii, B-ii, C-i
• D. A-i, B-iii, C-ii

Answer 1

The correct option is A A-iii, B-i, C-ii

The electrical potential difference across the resting plasma membrane is called the resting potential. At resting, membrane is more permeable to potassium ions $\left(K^{+}\right)$ and nearly impermeable to sodium ions ($N{a}^{+}$) and to negatively charged proteins present in the axoplasm. As a result the inside of the membrane contains high concentration of $K^{+}$ and negatively charged proteins while high concentration of $N{a}^{+}$ will be there outside the membrane.

When a stimulus is applied at a site on the membrane, the membrane becomes freely permeable to $N{a}^{+}$. This leads to a rapid movement of $N{a}^{+}$ to the inside of the membrane. The inward flow of sodium ions increases the concentration of positively charged in the cell leading to reversal of polarity i.e., the inside of the membrane becomes positively charged and outside becomes negatively charged. This is called depolarization of neurons.

The repolarization phase of the action potential is dependent on the opening of potassium channels. At the end of depolarization, the sodium channels close and potassium channels open. Potassium diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation.