Biology Question

[tommyngu]

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The answer says that if you inhibit the Na/K ATPase, the cell will lyse because of an increase in the Na concentration inside the cell.

My understanding is that the Na/K ATPase pumps out 3 Na and in 2 K against their concentration gradients. If the Na/K ATPase is inhibited, then no movement of Na nor K will occur. The question doesn't say there are other channels for ions to diffuse so I'm confused with how there is an increase of Na in the cell if ATPase action is inhibited (looking at only the Na/K ATPase).

 

[UndergradGuy7]

The answer says that if you inhibit the Na/K ATPase, the cell will lyse because of an increase in the Na concentration inside the cell.

My understanding is that the Na/K ATPase pumps out 3 Na and in 2 K against their concentration gradients. If the Na/K ATPase is inhibited, then no movement of Na nor K will occur. The question doesn't say there are other channels for ions to diffuse so I'm confused with how there is an increase of Na in the cell if ATPase action is inhibited (looking at only the Na/K ATPase).

I think what they mean is...
Well normally you pump 3 Na out of the cell and 2 K into the cell. If the pump is inhibitted you do not pump Na out and it all stays inside the cell. So water flows in and it lyses.

 

[Avery07]

The Na/K pump is there to restore the resting membrane potential.

At rest, there is a high concentration of K inside the cell (negative) and a high concentration of Na outside of the cell (positive).

During an action potential, Na channels open, causing an influx of Na ions into the cell. The cell then becomes positive on the inside and negative on the outside. Next, K channels open thus allowing K ions to move out of the cell. This causes the cell to return to being negative on the inside and positive on the outside.

At the end of all this, we have the Na/K pump functioning to restore the resting potential. To do this, it pumps 3 Na out for every 2 K into the cell.

If this pump were not working, you would have a severe excess of Na inside the cell, causing water to flow into the cell via osmosis. The influx of water would then cause the cell to lyse.

I believe you should assume that the other two channels are working and that in needing the Na/K pump, you have a high concentration of Na and a need to restore membrane polarity.

Hope that helps.

 

[RCT PC CRN]

The Na/K pump is there to restore the resting membrane potential.

At rest, there is a high concentration of Na inside the cell (negative) and a high concentration of K outside of the cell (positive).

Are you sure? You might be thinking other way?

 

[UndergradGuy7]

Are you sure? You might be thinking other way?

I think he mixed it up. Not sure what you mean by the (negative) and (positive) either. It is Na+ and K+.

Everything else sounds good.

 

[LuckMC11]

I think what they mean is...
Well normally you pump 3 Na out of the cell and 2 K into the cell. If the pump is inhibitted you do not pump Na out and it all stays inside the cell. So water flows in and it lyses.

that is correct, i remember reading the answer solution to this same question only yesterday

 

[Avery07]

Yeah.. I got it switched around. Sorry about that.

 

[Avery07]

I think he mixed it up. Not sure what you mean by the (negative) and (positive) either. It is Na+ and K+.

Everything else sounds good.

When I talk about negative and positive I'm referring to the charge on one membrane relative to the other.

At rest, the inside of the nerve cell is less positive than the outside of the nerve cell which is the same as saying more negative.

In textbooks, they usually teach this as just being "more negative" on the inside. It took me a while to get my head wrapped around this as both molecules are cations so I couldn't figure out how they are getting negative but it's just a reference to the other side of the membrane.

 

[tommyngu]

I think the negative/positive stuff is referring to the voltage across the membrane. Anyway, there is a high concentration of K inside the cell and a high concentration of Na outside the cell. The pump moves Na and K into areas of higher concentration (across their conc. gradient). If the pump were to stop working, there would be no movement of these ions, therefore there will still be a high concentration of K inside the cell and a high conc. of Na outside the cell. According to the answer, water enters the cell because of an increase of Na ions inside the cell. What increase? Na concentration was lower inside the cell before and after the pump was inhibited.
If water enters the cell via osmosis, then that implies that there is a greater concentration of total solutes inside the cell as it stands "inactive" with respect to the pump. The only explanation I could come up with besides this assumption, that incorporates the excess Na inside is that there are channel proteins or co-transporters that allow the Na to enter the cell along its concentration gradient bringing in whatever else that causes the cell to lyse.

I think I have another idea. When the pump is working, there is a net movement of Ions out of the cell (3 out vs 2 in). This constant movement of ions out the cell keeps the total concentration of ions outside of the cell higher (even though K ions are in higher conc. inside the cell), which in turn, keeps the water from entering the cell. If the pump were to stop working, this net movement of ions out the cell is lost...but the hiccup here for me is that it's still assuming that the total concentration of solutes is higher inside the cell (the cell is hypertonic to the exterior). This is true naturally isn't it? I have a thing for keeping all of my reasoning limited to the info that is provided in the question.

 

[Avery07]

I think the negative/positive stuff is referring to the voltage across the membrane. Anyway, there is a high concentration of K inside the cell and a high concentration of Na outside the cell. The pump moves Na and K into areas of higher concentration (across their conc. gradient). If the pump were to stop working, there would be no movement of these ions, therefore there will still be a high concentration of K inside the cell and a high conc. of Na outside the cell. According to the answer, water enters the cell because of an increase of Na ions inside the cell. What increase? Na concentration was lower inside the cell before and after the pump was inhibited.
If water enters the cell via osmosis, then that implies that there is a greater concentration of total solutes inside the cell as it stands "inactive" with respect to the pump. The only explanation I could come up with besides this assumption, that incorporates the excess Na inside is that there are channel proteins or co-transporters that allow the Na to enter the cell along its concentration gradient bringing in whatever else that causes the cell to lyse.

I think I have another idea. When the pump is working, there is a net movement of Ions out of the cell (3 out vs 2 in). This constant movement of ions out the cell keeps the total concentration of ions outside of the cell higher (even though K ions are in higher conc. inside the cell), which in turn, keeps the water from entering the cell. If the pump were to stop working, this net movement of ions out the cell is lost...but the hiccup here for me is that it's still assuming that the total concentration of solutes is higher inside the cell (the cell is hypertonic to the exterior). This is true naturally isn't it? I have a thing for keeping all of my reasoning limited to the info that is provided in the question.

Read my first post again. I explained the answer and reason for having a high Na content inside the cell.

 

[tommyngu]

Thanks, I understand it now. It just seems awkward that the answer requires the use of "other" information, especially since it was not covered in the chapter.