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A level biology help: hyperpolarisation

I was revising action potential and got a bit stuck.
How does the membrane return to resting potential after hyperpolarisation?
The membrane needs to become slightly more positive, and in the textbook it says that this is done by the Na/K pump. But surely the Na/K pump will make it more negative (as 3 positive ions are being pumped out for every 2)?
Thank you!
(edited 2 years ago)
Original post by Ylife
I was revising action potential and got a bit stuck.
How does the membrane return to resting potential after hyperpolarisation?
The membrane needs to become slightly more positive, and in the textbook it says that this is done by the Na/K pump. But surely the Na/K pump will make it more negative (as 3 positive ions are being pumped out for every 2)?
Thank you!

Theres K pumps as well
Reply 2
Original post by chris01928
Theres K pumps as well

Thank you so much!
Just to double check, does that mean that K+ is being pumped in by both the K+ pumps and the Na+/K+ pump?
Original post by Ylife
Thank you so much!
Just to double check, does that mean that K+ is being pumped in by both the K+ pumps and the Na+/K+ pump?

yes but one is active transport teh other is voltage gated
Reply 4
Original post by chris01928
yes but one is active transport teh other is voltage gated

Thank you!
Reply 5
Original post by Ylife
I was revising action potential and got a bit stuck.
How does the membrane return to resting potential after hyperpolarisation?
The membrane needs to become slightly more positive, and in the textbook it says that this is done by the Na/K pump. But surely the Na/K pump will make it more negative (as 3 positive ions are being pumped out for every 2)?
Thank you!

The Na/K pump has very little effect on the membrane potential. The resting membrane potential is due to the fact that the membrane is far more permeable to K+ ions than Na+ ions via leak ion channels. This means that more K+ ions can leak out of the cell than Na+ ions can leak in, this means you get a build of charge on the outside of the cell.

After the voltage-gated K+ channels have closed at the end of hyperpolarisation, there are lots of positively charged ions outside of the cell. This will cause more Na+ ions to move into the cell (down their electrochemical gradient) and reduce the movement of K+ ions from the inside to the outside of the cell. This will return the membrane potential to the resting state.

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