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Neurobiology homework help (Na and K conductance question)

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psychkub
Posts: 1788 - Karma: 16
Alright guys, I'm desperate here, and NS has never steered me wrong (HAHAHA!). This is due in like 7 hours and I can't seem to figure out Part D.

Click for larger image

Can anyone figure out what they're looking for?

Isn't Na the main excitatory ion in most if not all neurons? So if you equalize the concentration on both sides of the membrane and provide a pulse there won't be any Na conductance, but how will that affect K conductance? Will the reversal potential be more towards the equilibrium potential for Na or K?

Gah, I'm so fucked.

Halp!


Feb 9 2010 9:28AM
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user098123
Posts: 6072 - Karma: 18
Real channels are not perfectly selective. The selectivity of ion channels is not perfect and, for example in K+ channels, for every 20 K+ ions that flow through the channel, one Na+ ion can get through. This means that we cannot apply the Nernst equation to compute the potential that produces zero flow across the channel because more than one ion is involved. Instead, we could use the Goldman-Hodgkin-Katz equation (with PK/PNa=20, in the case of the K channel) and the potential predicted by the equation would be called reversal potential, instead of equilibrium potential, at which the net flow of charge through the channel is zero.

Sounds like equalizing the concentration of the ions would not cause a net charge flow of 0 so you would still get charge flow and you would not be at a reverse potential.
Feb 9 2010 9:38AM
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psychkub
Posts: 1788 - Karma: 16
First of all, thanks for even taking a look at this major +K

Second, I don't think you can use the Goldman eq b/c permeability of each ion is not given.

Third, I think you misunderstood the question, it's asking to only equalize the concentration of 1 ion.

I'm think the professor just wants like a qualitative answer where, say we hold Na concentration inside and outside the cell constant, then the only flow of ions at that time will be K+out of the cell. This net flow will try to bring the cell more towards equilibrium potential for K thereby lowering the reversal potential of the cell to something lower, say 0mv....



Feb 9 2010 10:34AM
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user098123
Posts: 6072 - Karma: 18
Ah ok, I see what you are saying.  I did misinterpret the question.
Feb 9 2010 5:18PM
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ksnow
Posts: 313 - Karma: 125
You asked: if you equalized the concentration gradient on both sides of the membrane and provide a pulse, what happens to K+ conductance?


The net diffusion of K+ will stop rather quickly due to the developing electrical potential difference. Ions, because they are charged, are influenced not only by their concentration gradient but also by electrical potential differences. As K+ diffuses due to its concentration gradient, it causes the development of an electrical potential difference that acts in the opposite direction on positive ions. Indeed, the diffusion of K+ down its concentration gradient quickly creates an electrical potential difference that completely counterbalances the tendency of K+ to diffuse due to its concentration gradient. The membrane potential becomes equal to the K+ equilibrium potential. This is because only K+ can cross the membrane and influence the electrical potential difference. Therefore, K+ should, once reached equilibrium, won't diffuse. Yet, change the concentration of NA and the K+ will diffuse to the lower levels of concentration.

Took AP and college bio-hope this helps answer that question.
Feb 9 2010 5:29PM
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