Neuron X is at its resting membrane potential of -65 mV, with standard resting membrane permeabilities, and standard intracellular and extracellular concentrations of sodium, potassium, chloride and calcium. In which direction will K+ move across the membrane when its selective channel opens?

Question

BlackTom AI · Accepted Answer

E_K (the potassium equilibrium potential) sets the voltage at which K+ would be at electrochemical equilibrium. Use the Nernst equation (37°C): E_K = 61.5 mV × log10([K+]out/[K+]in). With standard values [K+]out ≈ 4 mM and [K+]in ≈ 140 mM:

1) Compute the concentration ratio and its log:
[K+]out/[K+]in = 4/140 = 0.028571
log10(0.028571) = -1.5441

2) Multiply by 61.5 mV:
E_K = 61.5 × (−1.5441) ≈ −94.9 mV ≈ −95 mV

3) Determine the driving force for K+:
Driving force = Vm − E_K = (−65 mV) − (−95 mV) = +30 mV

For a positively charged ion, a positive value of Vm − E_ion corresponds to an outward driving force (positive current flowing out of the cell). Physically, intracellular K+ concentration is much higher than extracellular, so the concentration gradient pushes K+ out. The membrane electrical potential (−65 mV) attracts K+ inward, but it is not negative enough to balance the concentration gradient (equilibrium would be at ≈ −95 mV). Therefore the net electrochemical gradient drives K+ out of the cell when K+-selective channels open.

Why other answers would be wrong:
- “In” is incorrect because Vm (−65 mV) is less negative than E_K (≈ −95 mV); that condition produces an outward, not inward, net flux for a cation.
- “No net movement” is incorrect because Vm ≠ E_K, so the ion is not at equilibrium and there is a net driving force.

Thus K+ will move out of the cell. This matches the provided correct answer: Out.

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