Question

The intracellular and extracellular concentrations (in mM) of three important ions are given. The relative permeability of the cell membrane to each ion is provided. Universal gas constant \(R=8.31~\mathrm{J/(mol\cdot K)\) and Faraday’s constant \(F=96500~\mathrm{C/mol}\). What is the \emph{absolute value} of the resting membrane potential (in mV) across the cell membrane at \(27^\circ\mathrm{C}\)? (Round off the answer to one decimal place.)}

Hint:

- For GHK, anions (e.g., Cl\(^-\)) appear with inside and outside concentrations interchanged.
- At room temperature, \(RT/F\approx 25.8~\text{mV}\), useful for quick estimates.

Answer 1

Correct Answer: 82.6

Step 1: Use the Goldman–Hodgkin–Katz equation (for cations Na\(^+\), K\(^+\) and anion Cl\(^-\)):
\[ V_m=\frac{RT}{F}\ln\!\left(\frac{P_K[K^+]_o + P_{Na}[Na^+]_o + P_{Cl}[Cl^-]_i} {P_K[K^+]_i + P_{Na}[Na^+]_i + P_{Cl}[Cl^-]_o}\right). \] At \(27^\circ\mathrm{C}\Rightarrow T=300\) K, so \(\frac{RT}{F}=\frac{8.31\times300}{96500}\approx 0.02582~\text{V}=25.82~\text{mV}\).
Step 2: Substitute values (mM; permeabilities: \(P_{Na}=0.02\), \(P_K=1.0\), \(P_{Cl}=0.38\)): \[ \text{Numerator}=1\cdot3+0.02\cdot140+0.38\cdot3=6.94, \] \[ \text{Denominator}=1\cdot140+0.02\cdot10+0.38\cdot90=174.4. \] Hence \[ V_m=25.82\,\ln\!\left(\frac{6.94}{174.4}\right)\approx 25.82\times(-3.224)\approx -83.2~\text{mV}. \] The question asks for absolute value: \(|V_m|=83.2~\text{mV}\).