My question is, how does a neuron recover from AHP? It’s been puzzling me for a long time, and I really can’t find a single source that explains it in detail. Apparently the sodium potassium pump restores it following hyperpolarization, but I really can’t imagine how it would do that!

My only hypothesis as to how that would work would be that the sodium potassium pump increases the hyperpolarization beyond Ek, so as to reverse the direction of K current back into the cell, but that for me doesn’t make much sense, because that effect for one would be very temporary; once K flows in to the cell, it would depolarize the membrane, decreasing the potential back to something lower than Ek, and we will be back where we came from.

Also I read somewhere in a neurophysiology book that ENa is the ceiling of the action potential, and Ek is the floor of the action potential, so wouldn’t the hyperpolarization caused by the sodium potassium pump push it beyond Ek, contradicting the book?

Anyways I hope that wasn’t too long, and please someone help me out, I’m supposed to be studying anatomy but this whole thing is time consuming.

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Apr 4, 2022 at 17:50

1 Answer 1


The Goldman equation is your friend when understanding voltage changes in neurons. Forget about ion concentration changes unless you're focusing on them specifically, they are ordinarily too small to matter in the cycle of an action potential. What matters instead is permeability, and changes in permeability to different ions underlies all of the voltage changes you see in a neuron: responses to excitatory and inhibitory neurotransmitters, triggering and propagation of action potentials, repolarization, hyperpolarization, everything.

It's also helpful to think about the Nernst equation, which is very similar to the Goldman equation except it involves only one ion. The Nernst equation will give you the "reversal potential" (or "Nernst potential" or "equilibrium potential"; your "EK" is for "Equilibrium potential for K") for each ion. When you increase permeability of the membrane to some ion, the result will always move the cell's potential in the direction of that ion's reversal potential.

During an action potential, there is a positive feedback loop involving voltage-gated sodium channels; as more sodium permeability opens up in the membrane, the voltage moves towards the reversal potential for sodium, which is typically around +20 mV, far more positive than rest. In mammalian neurons, repolarization is aided by opening voltage-gated potassium channels, at the same that the sodium channels close. The reversal potential for potassium is typically around -90 mV, slightly more negative than rest.

Because there is more permeability to potassium while these channels are open compared to rest, the membrane potential will be closer to the reversal potential for potassium than at rest.

As the voltage-gated potassium channels close, that additional permeability to potassium is gone and the cell is back to having only resting "leak" channels open, so it returns to the resting voltage. Yes, these are also most permeable to potassium, but the resting cell also has some small permeabilities to other ions, hence why the resting potential (typically around -70 mV, -65 mV) is slightly more positive than just potassium's reversal.

Some textbooks and internet resources say stupid things about the role of the sodium/potassium pump, and say it's necessary for "returning the cell to rest" or something like that, which misleads students like you into thinking that the pumps are important for this recovery from afterhyperpolarization. You know how I mentioned that ion concentration changes are too small to matter? Well, if the Na/K pump wasn't working, those small changes would accumulate over time (think minutes to hours; not a fraction of a second) and would matter. Thanks to the pump, your life can be simpler as you understand neurophysiology, because the pump is always keeping the concentrations of sodium and potassium different inside the cell compared to outside. If you're just thinking about one action potential cycle, though, forget the pump!

  • $\begingroup$ Thank you for your answer! However, I have another small question, if you don’t mind. Let’s forget that I said the Na-K pump returns the membrane potential back to the resting one. Then, what actually does? I have an idea about that, which I tried to confirm by calculating the ratio between the Na and K currents as soon as the voltage gated K channels close, assuming the membrane returns to the resting conductances, and the Na current is larger due to the increased electrochemical driving force. So does that mean that Na influx is responsible for depolarizing the membrane back to the RMP? $\endgroup$ Commented Apr 4, 2022 at 16:55
  • $\begingroup$ @NeuroIsHard Ions of all types are moving all the time. The net flow for a given ion is always in the direction suggested by the Nernst equation for that ion, so sodium is always flowing in to the cell as long as the cell is more negative than E(Na), and potassium is always flowing out of the cell as long as the cell is more positive than E(K). You can calculate the potential at which the sum of all ions according to their charge is zero with the Goldman equation. $\endgroup$
    – Bryan Krause
    Commented Apr 4, 2022 at 16:58
  • $\begingroup$ If you think just in terms of sodium and potassium, at E(rest) you can say that the net flow of Na+ in to the cell equals the net flow of K+ out of the cell. If you are more negative than E(rest), then there is a little less K+ flowing out (because you're closer than rest to the reversal potential for K+) and a little more Na+ flowing in (because you're further from the reversal potential for Na+ than at rest). It's not fair to say it's "sodium flowing in" because "less potassium flowing out" is equally (or more) important; it's the overall balance that matters, given by the Goldman equation. $\endgroup$
    – Bryan Krause
    Commented Apr 4, 2022 at 17:01

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