I'm trying to understand how all of the potentials during an action potential are created. My question specifically is about the sodium potassium pumps, however I would also be grateful if someone could verify if I have grasped the concepts correctly.

From what I understand, the resting membrane potential is at around $-60\rm\,mV$, close to the equilibrium potential for potassium (from the Nernst equation) because the permeability of potassium is much greater than that of sodium, even though the permeability is low.

However, the resting potential is not as negative as the equilibrium potential for potassium because of some permeability of sodium and the permeability of potassium is very low so the tiny permeability of sodium has some impact in preventing the potential reaching the equilibrium potential of potassium.

When the threshold potential of $-50\rm\,mV$ is reached, the voltage gated $\ce{Na+}$ ion channels open, and sodium ions enter the cell, causing depolarisation. The potential rises to about $+40\rm\,mV$, which is close to the equilibrium potential for sodium (because the permeability for sodium is much greater than for potassium). However, it doesn't quite reach the equilibrium potential for sodium because of some potassium ions leaving the cell.

When the $+40\rm\,mV$ threshold is reached, sodium channels close and potassium channels open. Because of the now much higher permeability of potassium ions compared with when the membrane is at its resting potential, the potential becomes much closer to the equilibrium potential for potassium, which I think is about $-90\rm\,mV$, and therefore the cell hyperpolarises. Then the potassium channels close.

Now I'm unsure as to how the resting membrane potential is reached from the hyper polarised state, and I also do not see where the action of the sodium potassium pumps comes into this.


2 Answers 2


The Sodium-Potassium Pumps are always at work. One can think of them as a continuous process that maintains the equilibrium potential for the individual ions. They always are grabbing internal sodium and exchanging it with external potassium at the cost of ATP.

However a neuron's rest state (in your example -60 mV) is a combination of the equilibrium of the Sodium, Potassium, Chlorine, and other ions. Thus when the membrane hyperpolarizes beyond the rest potential, it is actually the leak potential that brings the membrane potential back up, not the Sodium-Potassium pump.

Leak potentials arise from ions (usually chorine) that pass through the membrane via channels that are always open. Furthermore, sodium channels reactivate and a small amount open to sodium to enter. (Recall as a population there is usually a small amount of sodium channels open at rest. Another contributing factor is as the potassium channels close the other to factors dominate and slowly bring the membrane back to rest state. Remember that rest is defined as the balance of currents, there will always be small amounts of currents flowing. So even though the neuron is rest small amounts of sodium leak in and less potassium out. The sodium potassium pump maintains the equilbrium potential that allows these currents to flow.

The Sodium-Potassium pump is a slower process, so it usually can be ignored over a single spike. But If there is a high frequency spike train then the small amount of sodium that enters the cell and potassium that exits the cell can add up and effect the equilibrium potentials of the individual Ions. This of course changes the firing properties of the neuron.

  • $\begingroup$ I am having trouble finding non academic sources for my last claim, I shall come back and add more sources later once I find them. $\endgroup$
    – xelo747
    Nov 28, 2015 at 17:12
  • 1
    $\begingroup$ thank you for your reply! I'm just slightly confused as you said that it is the membranes leakiness that restores the equilibrium potential, however from what I understand there is only significant movement of potassium ions when the ion channels are 'closed', in which case the positive potassium ions would be leaving the cell and it would become more hyper polarised? Please correct me where I have misunderstood! $\endgroup$
    – Meep
    Nov 28, 2015 at 19:18
  • 2
    $\begingroup$ @21joanna12 The wrong idea that you have that is blocking you is that at rest there are only potassium channels open. That isn't the case. There are a small amount of sodium channels and a decent number of chloride channels, too, and putting them all together gives rise to a resting membrane potential of around -65mV. It would be great if there were some kind of equation that took all this into account. Hey, there is! paynesnotebook.net/Research/Neurotools/GHKEquation $\endgroup$
    – Chelonian
    Dec 2, 2015 at 5:53
  • $\begingroup$ "Thus when the membrane hyperpolarizes beyond the rest potential, it is actually the leak potential that brings the membrane potential back up, not the Sodium-Potassium pump." - hey xelo747 do you have a reference to this statement? $\endgroup$
    – bonCodigo
    May 20, 2019 at 22:41
  • $\begingroup$ @bonCodigo Here is a link that describe everything in great detail. khanacademy.org/science/biology/human-biology/… $\endgroup$
    – xelo747
    May 21, 2019 at 5:14

Very good question. Most of your arguments, to the best of my knowledge are accurate. As to answer your questions, I'll provide a basic model of understanding. (Disclaimer:- I'm sorry if the explanation seems overly-messed up and confusing)

At any moment, the potential difference across the cell membrane has to be such that it makes all fluxes balanced. Let us assume there is a cell with nothing but potassium and non-diffusible negative porteins. Since potassium is the only ion that can have a flux, the only balanced position will be the one with zero flux, because any net ionic movement will cause a change of potential and hence will not be the balanced state. Zero flux will be reached when the concentration ratio of potassium outside and inside is such that the potential difference is equal to potassium's Nernst potential.

To complicate it, let's add sodium. Now the balanced state should have zero net flux. But this does not mean that the fluxes of both the ions is zero. They can be equal and opposite. Let us assume that the permeability of the membrane is equal for both. Then, the balanced potential would be equidistant from the Nernst potential of both the ions. This is because, permeability, in an abstract way, is a measure of the resistance. Flux is the potential difference (between the Nernst potential, where the flux would be zero, and the membrane potential) divided by the resistance (like a simple current), which would be equal only if this difference is same, which is, for a membrane potential midway between their Nernst potentials. Hence, the ion which has a higher permeability, will have a lesser resistance, and would hence require the potential difference to be low, so that the ratio of $PD$ and $R$ be equal to the other ion with a large $PD$ and a high $R$ (low permeability). This $PD$, is not the membrane potential but the difference between this potential and the Nernst potential.

Now, for your first question

...resting membrane potential is reached from the hyper polarised state....

This is simple to understand because the opening of the potassium channels which caused the potential drop to $-90$ has now been reset and the permeabilities have been reset to the original resting values. Since the balanced state depends on the permebilities only, the balanced state of this is the RMP ($-60$), which the cell will achieve. Since $-90$ is unbalanced, the slightly unbalanced flux will cause a drift of the potential towards $-60$ and when it reaches it, the fluxes will match and hence won't deviate any further. The drift is because more of one ion is moving than the other, causing a net movement of charges across the membrane.

.... action of the sodium potassium pumps comes into this....

This is easy to understand but tough to compute. Since the sodium potassium pump is unbalanced, (it puts out 3 sodiums and takes in 2 potassiums), it contributes a net flux always. Hence, the remaining channels, instead of having exactly equal sodium potassium fluxes, will have to have a slight potassium excess to offset the minor flux contributed by the pump. As a result, the effect can be computed in two ways.

  1. We find the net flux contribution of the pump. Now we know by what amount should the potassium flux be greater than the sodium in the remainign channels which follow the PD/R formula and hence we can caluclate the PD such that the fluxes have a certain difference (equal to the flux contributed by the pump)

  2. We perform experiments and directly find a net amount of potential "correction" which has to be added to the potential calculated by matching fluxes because of the pump, which will remarkably stay constant. This method is easier and the value of the correction can be obtained from literature online.

Hope this helped. Feel free to ask for clarifications. If you want net links or references, ask. :)


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