If Na+/K+ ATP-ase were inhibited in a resting neuron, which voltage would the neuron ultimately reach over a long period of time?

Question

BlackTom AI · Accepted Answer

The Na+/K+ ATPase maintains the transmembrane ionic gradients (high K+ inside, high Na+ outside) by pumping 3 Na+ out and 2 K+ in; with the pump inhibited these gradients dissipate via passive leak channels and diffusion, driving the membrane potential toward 0 mV.

1) Relevant concepts
- Nernst equation: E_ion = (RT/zF) ln([ion]_out/[ion]_in). If [out] = [in], E_ion = 0 mV.
- Goldman (constant-field) equation for membrane potential: V_m is a weighted log ratio of outside to inside concentrations of permeant ions. If all ionic gradients collapse ([out] = [in] for all permeant ions), V_m = 0 mV.
- The Na+/K+ pump is the source of the steady-state concentration differences; without it, only passive fluxes remain.

2) Step-by-step reasoning
- At steady state with the pump working, [K+]_in >> [K+]_out and [Na+]_out >> [Na+]_in, producing resting V_m ≈ -70 mV (mainly due to high K+ permeability).
- Inhibition of the pump stops the active maintenance of these gradients. Passive K+ leaks out and Na+ leaks in until their concentrations equilibrate across the membrane.
- As [K+]_out rises toward [K+]_in and [Na+]_in rises toward [Na+]_out, the Nernst potentials for K+ and Na+ move toward 0 mV.

3) Intermediate calculation example
- Using the Nernst form at physiological temperature (approx): E_ion ≈ (61.5 mV/z) log([out]/[in]).
- If [out] = [in], log(1) = 0 → E_ion = 0 mV for each ion.
- Goldman then gives V_m = 0 because the numerator and denominator terms become equal when every [ion]_out = [ion]_in.

4) Why the correct answer is correct
- With the pump inhibited long enough, all ionic concentration differences collapse; therefore every ionic equilibrium potential becomes 0 mV, and the membrane potential (a combination of those) approaches 0 mV.

5) Why other choices are incorrect
- A value near the original resting potential (e.g., -70 mV) is incorrect because that resting potential depends on active maintenance of gradients by the pump.
- A K+ equilibrium value (e.g., -90 mV) is incorrect because K+ gradient dissipates without the pump.
- A Na+ equilibrium value (e.g., +60 mV) is incorrect because the Na+ gradient also collapses.
- Any nonzero steady potential assumes persistent gradients; with the pump inactive, gradients decay to equality and voltage → 0.

6) Conclusion
Because pump inhibition lets ionic gradients dissipate until [ion]_in = [ion]_out and each E_ion = 0, the membrane potential ultimately approaches 0 mV. This confirms the correct answer: 0 mV.

If Na+/K+ ATP-ase were inhibited in a resting neuron, which voltage would the neuron ultimately reach over a long period of time?Single choice

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