Quick question (membrane potential) KHAN Q

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br2pi5

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1) I'm having trouble understanding the "down the membrane potential" part. I'm sure it's really easy answer but for whatever reason I can't get quite get it.

2) I first thought it would be permeability because there should be a down the gradient movement but a good reason why it's not could be because no channels are opened, or there are no channels. Can anyone explain why A is correct?

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When K+ moves into the cell its moving into up its gradient because its moving from lower concentration outside the cell to higher concentration inside the cell. Up = Low -> high.. Down = High -> low. It is moving down its membrane potential because its going from a higher potential outside the cell to a lower potential inside the cell. High -> low = down. K+ is a cation so its moving from + to - which is also considered down. If we were to pump K+ out of the cell it would be moving up (or against) its electrochemical gradient. Also, they aren't talking about an action potential, they can be simply talking about the ATPase Na+/K+ channel which shuttles out 3 Na+ and 2K+ into the cell.

Answer this question. If Na+ were to move into the cell at resting membrane potential -70mV. Is it moving up or down its concentration gradient? Is it moving up or down its electrochemical gradient?
 
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Regarding the question for the cell injected with ion X, you know the cell must be permeable to ion X (allow it to cross the membrane) because you see that some ion X has diffused out into the extracellular solution. So B, C, D must be incorrect because you can see it is able to move out of the cell even though at a low level. This leaves you with A.

Now, you see that MOST of X still remains inside of the cell. Why is that? The chemical gradient would want X to move out so that there is an equal amount of X inside and outside the cell, but something is holding it back. The only other piece of information they give you is that the cell is -70mV and extracellular environment is 0mV so something about X must be very attracted to the negative potential and does not move out.
 
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