Action potentials

When an axon experiences hyperpolarisation the axon is more negative than when at resting potential. How is resting potential restored (axon becomes less negative), if only the sodium potassium pump is active? If there are 3 Na+ leaving and 2 K+ entering would the axon not become more negative if the difference in charge is increasing?

Reply 1

0gg

the axon experiences hyperpolarisation when the voltage-gated potassium ion channels are left open 'too long', so 'too many' potassium ions leave.

when they finally close, the non-voltage-gated potassium ion channels are still open, so when the sodium-potassium pump does its thing potassium ions are moved into the axon, but they have a shallower concentration gradient to leave the axon, since loads of potassium ions were just moved out.

hope this helps cus it's kind of the blind leading the blind lol

Reply 2

tk_3378

Original post by 0gg

the axon experiences hyperpolarisation when the voltage-gated potassium ion channels are left open 'too long', so 'too many' potassium ions leave.
when they finally close, the non-voltage-gated potassium ion channels are still open, so when the sodium-potassium pump does its thing potassium ions are moved into the axon, but they have a shallower concentration gradient to leave the axon, since loads of potassium ions were just moved out.
hope this helps cus it's kind of the blind leading the blind lol

Thank youu

Reply 3

Jpw1097

Original post by tk_3378

This is a very common misconception. The sodium-potassium pump has very little effect of the membrane potential. Yes, the pump is important in maintaining Na and K concentration gradients over a long period of time, but it has only a tiny effect on membrane potential. The membrane potential is determined by the permeability of the membrane to Na and K ions.

Depolarisation - voltage-gated Na channels open
Repolarisation - voltage-gated Na channels close/voltage-gated K channels open
Hyperpolarisation - delayed closure of voltage-gated K channels

Reply 4

tk_3378

Original post by Jpw1097

This is a very common misconception. The sodium-potassium pump has very little effect of the membrane potential. Yes, the pump is important in maintaining Na and K concentration gradients over a long period of time, but it has only a tiny effect on membrane potential. The membrane potential is determined by the permeability of the membrane to Na and K ions.
Depolarisation - voltage-gated Na channels open
Repolarisation - voltage-gated Na channels close/voltage-gated K channels open
Hyperpolarisation - delayed closure of voltage-gated K channels

Thank you for the reply. Could you explain what is happening when a resting potential is being restored after the axon is hyper polarised please?

Reply 5

Jpw1097

Original post by tk_3378

Thank you for the reply. Could you explain what is happening when a resting potential is being restored after the axon is hyper polarised please?

During hyperpolarisation, the inside of the cell is more positively charged than it is at the resting membrane potential. This excess of positive charge inside the cell increases the electrochemical gradient for K ions, meaning more K ions leave the cell through leak channels.

You have to remember that even at the resting membrane potential, Na ions are constantly flowing into the cell and K ions are constantly leaving via leak channels. However, the rate at which Na ions enter the cell and K ions leave the cell are the same, so the membrane potential remains the same - hence it is the resting membrane potential.