PR Biochemistry textbook question on ATPase

[DingDongD]

Hi guys,
I was reading through chapter 7 of the PR biochem book and came across with free standing question:

If an inhibitor of Na+/K+ ATPase is added to cells, which of the following may occur?

A. The cell will shrink and lose water
B. The interior of the cell will become less negatively charged
C. Secondary active transport processes will compensate for the loss of primary active transport.
D. The cell will begin to proliferate.

I just don't understand why it cannot be A. I mean, the reason we are using a ATPase pump is it is against concentration gradient. So, the outside of the cell should have more Na+. Inside the cell should have more K+ than outside. But, it just seems like the outside should be more concentrated than the inside.

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[nkafeel]

The function of Na+/K+ ATPase is to pump 3 Na+ out and 2 K+ into the cell, so if there is an inhibitor present, Na+ K+ flow is prevented to its respective areas outside and inside the cell.

Although water follows Na+, the net flow of Na+ will be into the cell because Na+ will start to flow down its concentration gradient (into the cell) due to the inactive Na+/K+ ATPase causing the cell to be less negative/more positive.

 

[StarFall]

First thing that popped into my head when I read Na/K pump was 3Na out, 2K in. Has a negative potential inside the cell.
Inhibit the pump = inactive pump -> means Na goes into cell -> means inside cell is now less negative.
I think A is meant to trick you into thinking it's an osmosis-related question, but really it's focusing on the Na/K pump.

 

[azor ahai]

Yeah, A isn't directly related to the pump in question. A is talking about osmosis, which occurs through simple diffusion and you don't know if it will shrink and lose water or gain water and burst. That's dependent on the environment the cell is in.

The pump in question's sole duty is to actively pump 3 Na+ out while pumping 2 K+ in simultaneously using energy. This is to keep the resting membrane potential which would otherwise equilibrate. The resting membrane potential is maintained at a net negative charge because there is a net loss of one cation. Inhibiting this pump will stop that net loss of a cation, leading to a less negative cell interior.

The rest of the choices are things that may happen depending on various factors other than the pump, but B is the only choice that addresses the pump directly.

 

[DingDongD]

I don't get how the interior becomes less negative. Wouldn't K leave the cells through leak channels?

 

[StarFall]

I don't get how the interior becomes less negative. Wouldn't K leave the cells through leak channels?

The purpose of the pump is to pump 3 Na out and 2 K in, giving the inside of the cell a negative potential which is active transport so it's going AGAINST it's gradient and using energy/ATP.

When you turn the pump OFF, it can't pump 3 Na out and 2 K in anymore. Think of the opposite now. Na will go in and K will go out now since that's WITH the concentration gradient now. (Na leaks into the cell and K leaks out of the cell naturally even when the pump is active, but the pump is more efficient or faster in pumping Na out and K in so the total net is still a negative potential inside the cell.)
Na naturally wants to enter the cell and K naturally wants to leave the cell. This nets in a total of, let's say, 3 Na IN and 2 K OUT for a net of +1 inside of the cell, making the inside of the cell less negative (positive).

Correct me if I'm wrong please.
Hope this helps! Thanks.