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I have known for so long that Na+/ K+ pump restores the membrane potential. But as it pumps in 2 K+ for every 3+ Na+ moving out how can it make the membrane potential less negative when the net result is one positive charged ion moving out?

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    $\begingroup$ If you see a normal AP curve, you would be able to appreciate that the membrane potential is generally brought back - which is around -70mV. So the potential doesnt matter much since its restored by the ion fluxes. What matters most is the ion composition on either side. You might see that after depolarization, Na+ came in (it generally satys out) and after hyperpolarization K+ came in. The potential is safe but the distribution is altered. This is what the pump takes care of, not the potential (although slightly electrogenic) $\endgroup$ – Polisetty Mar 11 '16 at 20:35
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The Na,K-ATPase restores negative membrane potentials. When a cell is hyperpolarized, leaky K+ channels take care of that. Leaky K+ channels are always open. Generally, K+ has the tendency to diffuse back out of the cell along its chemical gradient through leaky K+ channels, after been pumped in, making the cell's inside more negative. However, in the reverse situation, where the cell is hyperpolarized, they may flow inwards, along the electrical gradient.

Note that while the Na,K-ATPase is indeed the driving force behind the membrane potential, it is not the direct consequence of the ratio of Na+ being pumped out and K+ being pumped into the cell. It is the leaky K+ channels that determine the membrane potential largely, together with the Na+ channels being permanently closed, except when an action potential is generated.

Suggested further readings
- What keeps the resting potential of neurons constant at -70 mV?
- If the average resting potential of a neuron is -70 mV, why is there such a high ratio of potassium ions inside relative to out?

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  • $\begingroup$ Could you cite some sources beyond SE posts? It would be helpful. $\endgroup$ – Tyto alba Mar 12 '16 at 8:16
  • $\begingroup$ @SanjuktaGhosh - There are in-text links here and the further readings contain useful references too, that's why they are added $\endgroup$ – AliceD Mar 12 '16 at 9:07

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