Question

In the axon of a giant squid neuron,
(i) The resting membrane potential is -70 mV.
(ii) At the equilibrium potential of $\text{Na}^+$ ($\text{E}_{\text{Na}^+} = +55$ mV) there is no net movement of $\text{Na}^+$ ions across the membrane.
In an experiment, when the axon is stimulated, the voltage-gated $\text{Na}^+$ channels open. The membrane potential peaks at +30 mV, where the resistance of the axonal membrane for $\text{Na}^+$ flow is $1 \times 10^6\ \Omega$. The net $\text{Na}^+$ current ($\text{I}_{\text{Na}^+}$) across the membrane and the direction of ionic movement at the peak is:

• A. $\text{I}_{\text{Na}^+} = 25\ \text{nA}$; direction = into the cell
• B. $\text{I}_{\text{Na}^+} = 100\ \text{nA}$; direction = out of the cell
• C. $\text{I}_{\text{Na}^+} = 25\ \text{nA}$; direction = out of the cell
• D. $\text{I}_{\text{Na}^+} = 100\ \text{nA}$; direction = into the cell

Hint:

Whenever the membrane potential ($V_m$) is more negative than the equilibrium potential ($E_{\text{ion}}$) of a cation (like $\text{Na}^+$), the driving force is negative, meaning the positive ions will flow inward (into the cell) to try to depolarize the cell toward $E_{\text{ion}}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the net sodium current ($I_{\text{Na}^+}$) and its direction at the peak of the action potential ($+30\ \text{mV}$) in a giant squid axon, given the resting potential, the sodium equilibrium potential ($E_{\text{Na}^+} = +55\ \text{mV}$), and the membrane resistance.

Step 2: Key Formula or Approach:
The ionic current ($I$) flowing through a membrane is given by Ohm's Law for ionic flow: \[ I = \frac{V_m - E_{\text{ion}}}{R} \] where $V_m$ is the membrane potential, $E_{\text{ion}}$ is the equilibrium potential of the ion, and $R$ is the membrane resistance.

Step 3: Detailed Explanation:

• We are given: itemize

• Membrane potential at the peak, $V_m = +30\ \text{mV} = +30 \times 10^{-3}\ \text{V}$

• Equilibrium potential of sodium, $E_{\text{Na}^+} = +55\ \text{mV} = +55 \times 10^{-3}\ \text{V}$

• Resistance of the membrane, $R = 1 \times 10^6\ \Omega$

The driving force acting on the sodium ions is: \[ V_m - E_{\text{Na}^+} = 30\ \text{mV} - 55\ \text{mV} = -25\ \text{mV} = -25 \times 10^{-3}\ \text{V} \]
Now, calculate the net sodium current ($I_{\text{Na}^+}$): \[ I_{\text{Na}^+} = \frac{V_m - E_{\text{Na}^+}}{R} = \frac{-25 \times 10^{-3}\ \text{V}}{1 \times 10^6\ \Omega} = -25 \times 10^{-9}\ \text{A} = -25\ \text{nA} \]
The magnitude of the current is $25\ \text{nA}$.
The negative sign indicates that the current is inward (positive sodium ions flowing into the cell). This occurs because the membrane potential of $+30\ \text{mV}$ is less positive than the sodium equilibrium potential of $+55\ \text{mV}$, creating an inward electrochemical gradient for $\text{Na}^+$.
itemize

Step 4: Final Answer:
Therefore, $I_{\text{Na}^+} = 25\ \text{nA}$; direction = into the cell.