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Equal membrane permeability Na+, K+


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Hi guys

 

My professor in physiology told us that if the permeability for Na+ and K+ in a neuron would be equal, the cell would depolarize to a resting membrane potential difference of about -10 mV.

 

However, applying Goldman's equation and concentrations intracellular K+ of 139 mM, intracellular Na+ of 12 mM, extracellular K+ of 4 mM and extracellular Na+ of 145 mM gives us:

 

[math]-60\cdot\log{\frac{a\cdot 139 + a\cdot 12}{a\cdot 4 + a\cdot 145}}\approx -0.347[/math] mV.

(The a is to indicate that the permeability for both K+ and Na+ is the same)

 

Quite a big difference, over a factor of 30 in difference.

 

My professor told me he used the average of the K+-permeability-potential and the Na+-permeability-potential.

However, I'm convinced the Goldman equation MUST be applicable. Somehow.

 

Can someone tell me where I'm wrong?

Or how the hell he got -10 mV?

 

Thanks.

 

F.

Edited by Function
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I'm not sure what your prof means but the resting potential of neurons is not 10mV.

 

http://en.wikipedia.org/wiki/Resting_potential#Generation_of_the_resting_potential

I may have messed up his words, don't really recall exactly what he said; but the key problem is how he gets to -10 mV with an equal permeability for Na and K, and he's not the type to make mistakes so...

 

Anyone?

Edited by Function
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Hi guys

 

My professor in physiology told us that if the permeability for Na+ and K+ in a neuron would be equal, the cell would depolarize to a resting membrane potential difference of about -10 mV....

 

I'm not sure what your prof means but the resting potential of neurons is not 10mV.

 

http://en.wikipedia.org/wiki/Resting_potential#Generation_of_the_resting_potential

 

To discuss Studiot's objection - the Prof specifies a cell with equal permeability for Sodium and Potassium; as far as I recall the permeability in a neuron in vivo for these two ions is not equal - thus the difference between the calculated -10mV and the wiki entry stating in vivo -70mV. Why the Prof has stipulated this I do not quite grasp yet.

Again from depths of memory - it was last millennium when I was in your position as a first year medic - I think you are calculating using incorrect method. I would be using three (not two) separate E=61.5log(M_out/M_in) . To get total E_m each is multiplied by the relative permeability (in this case equal - but not calc'd as you have done) and the three are added; BTW I would want Cl- included

 

I get about -9.5mV using that method

Function - I have just reread mine above; if it is not followable let me know and I will engross my notes so that it becomes more obvious.

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To discuss Studiot's objection - the Prof specifies a cell with equal permeability for Sodium and Potassium; as far as I recall the permeability in a neuron in vivo for these two ions is not equal - thus the difference between the calculated -10mV and the wiki entry stating in vivo -70mV. Why the Prof has stipulated this I do not quite grasp yet.

Again from depths of memory - it was last millennium when I was in your position as a first year medic - I think you are calculating using incorrect method. I would be using three (not two) separate E=61.5log(M_out/M_in) . To get total E_m each is multiplied by the relative permeability (in this case equal - but not calc'd as you have done) and the three are added; BTW I would want Cl- included

 

I get about -9.5mV using that method

Function - I have just reread mine above; if it is not followable let me know and I will engross my notes so that it becomes more obvious.

 

But Goldman's equation states that, leaving Cl out of the question,

 

[math]E=-60\cdot\log{\frac{p_{K^+}\cdot \left[K^+\right]_i + p_{Na^+}\cdot\left[Na^+\right]_i}{p_{K^+}\cdot\left[K^+\right]_e+p_{Na^+}\cdot\left[Na^+\right]_e}}[/math]

 

At least, that's the equation in the syllabus. Again, my prof is not one to make mistakes, let alone writing down a whole wrong equation.

So I dropped the p's, filled in the values I gave in my original post, and got -0,8 mV. Why not?

Edited by Function
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[latex]E=61.5\cdot\log{\frac{p_{K^+}\cdot \left[K^+\right]_{out} + p_{Na^+}\cdot\left[Na^+\right]_{out}+p_{Cl^-}\cdot\left[Cl^-\right]_{in}}

{p_{K^+}\cdot\left[K^+\right]_{in}+p_{Na^+}\cdot\left[Na^+\right]_{in}+p_{Cl^-}\cdot\left[Cl^-\right]_{out}}}[/latex]

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[latex]E=61.5\cdot\log{\frac{p_{K^+}\cdot \left[K^+\right]_out + p_{Na^+}\cdot\left[Na^+\right]_out+p_{Cl^-}\cdot\left[Cl^-\right]_in}{p_{K^+}\cdot\left[K^+\right]_out+p_{Na^+}\cdot\left[Na^+\right]_out}+p_{Cl^-}\cdot\left[Cl^-\right]_out}[/latex]

[latex]E=61.5\cdot\log{\frac{p_{K^+}\cdot \left[K^+\right]_out + p_{Na^+}\cdot\left[Na^+\right]_out+p_{Cl^-}\cdot\left[Cl^-\right]_in}{p_{K^+}\cdot\left[K^+\right]_out+p_{Na^+}\cdot\left[Na^+\right]_out}+p_{Cl^-}\cdot\left[Cl^-\right]_out}[/latex]

 

Could you give me values that would give a potential of about -10 mV?

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Sorry Function my latex was askew. Revised equation

Well yes okay, but my Prof didn't mention anything about the permeability of chlorine... only said that "if the permeability for potassium would be equal to that of sodium..."

 

(Actually, if I'd talk about kalium and natrium, rather than potassium and sodium, would you understand me?)

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I would do it as follows

 

 

Kalium and Natrium - took a second or so but I get it :)


 

 

[latex]

 

E=\frac {p_{Na^+}}{p_{tot}} \cdot 61.5 \cdot log \frac{Na^+_{out}}{Na^+_{in}}+\frac {p_{K^+}}{p_{tot}} \cdot 61.5 \cdot log \frac{K^+_{out}}{K^+_{in}} +\frac {p_{Cl^-}}{p_{tot}} \cdot 61.5 \cdot log \frac{Cl^+_{out}}{Cl^-_{in}}

[/latex]


NB ptot is the total permeability of all ions, in this case pK+ + pNa+ +pCl


But I might be leading you up garden path - it really is 25 years since I studied any physiology. I am pretty sure on those equations but I would not want you spending precious study time figuring out where I have gone wrong.

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I would do it as follows

 

 

Kalium and Natrium - took a second or so but I get it :)

 

 

[latex]

 

E=\frac {p_{Na^+}}{p_{tot}} \cdot 61.5 \cdot log \frac{Na^+_{out}}{Na^+_{in}}+\frac {p_{K^+}}{p_{tot}} \cdot 61.5 \cdot log \frac{K^+_{out}}{K^+_{in}} +\frac {p_{Cl^-}}{p_{tot}} \cdot 61.5 \cdot log \frac{Cl^+_{out}}{Cl^-_{in}}

[/latex]

NB ptot is the total permeability of all ions, in this case pK+ + pNa+ +pCl

But I might be leading you up garden path - it really is 25 years since I studied any physiology. I am pretty sure on those equations but I would not want you spending precious study time figuring out where I have gone wrong.

 

Thank you :) So px/ptot would be 1, since they're all the same, aye?

Might I ask you; did you complete your medicine studies?

Edited by Function
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No they are not all 1. Think of it in these terms - pick a number at random and decide that is P_Na and so is P_K etc. P_Na/P_tot must be a fraction of one. NB ptot is the total permeability of all ions, in this case pK+ + pNa+ +pCl It will be a dimensionless ratio and clearly will be same for both Na+ and K+

 

And no - I left Med School after my 2nd MBs ie before I started clinical. Half my family are medics so I was able to create a pretty good idea of what lay ahead and decided to do something with a little less stress. My brother runs one of the largest neonatal care units in the country so I see in him both the incredible highs and the bad sad days. If you can finish and cope it is one of the best careers in the world - but I knew that I wouldn't

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No they are not all 1. Think of it in these terms - pick a number at random and decide that is P_Na and so is P_K etc. P_Na/P_tot must be a fraction of one. NB ptot is the total permeability of all ions, in this case pK+ + pNa+ +pCl It will be a dimensionless ratio and clearly will be same for both Na+ and K+

 

And no - I left Med School after my 2nd MBs ie before I started clinical. Half my family are medics so I was able to create a pretty good idea of what lay ahead and decided to do something with a little less stress. My brother runs one of the largest neonatal care units in the country so I see in him both the incredible highs and the bad sad days. If you can finish and cope it is one of the best careers in the world - but I knew that I wouldn't

 

So how did you get -9.5 mV then?

 

If someone quits medicine, that's a wise decision. As is the decision to push on. One should do what one wants to do.

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thats odd, google gets the same answer as I do


Ah - I know the confusion. I use ln to stand for natural logarithm and log to stand for log base 10. If you are multiplying by 61.5 you must be using log base 10


And the one third - as I see it there must be three ions Na+ K+ and Cl- and I have set their permeability equal. It was this step I was unsure of; this was the reason I was asking you to seek other help as I felt I might be leading you astray

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thats odd, google gets the same answer as I do

Ah - I know the confusion. I use ln to stand for natural logarithm and log to stand for log base 10. If you are multiplying by 61.5 you must be using log base 10

And the one third - as I see it there must be three ions Na+ K+ and Cl- and I have set their permeability equal. It was this step I was unsure of; this was the reason I was asking you to seek other help as I felt I might be leading you astray

 

http://www.wolframalpha.com/input/?i=61.5*%5B1%2F3*log%28145%2F12%29%2B1%2F3*log%284%2F139%29%2B1%2F3*log%28149%2F151%29%5D

 

Unless Wolfram|Alpha uses log for natural logarithm? This seems unusual...

 

Edit: Wolfram|Alpha is incorrect. Recalculated with my calculator and got indeed -9,5 mV

Edited by Function
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http://www.wolframalpha.com/input/?i=61.5*%5B1%2F3*log%28145%2F12%29%2B1%2F3*log%284%2F139%29%2B1%2F3*log%28149%2F151%29%5D

 

Unless Wolfram|Alpha uses log for natural logarithm? This seems unusual...

 

Edit: Wolfram|Alpha is incorrect. Recalculated with my calculator and got indeed -9,5 mV

 

Wolfram Alpha DOES interpret log() as the natural log of () - which is unusual and a bit disconcerting

 

post-32514-0-67539300-1418413120_thumb.jpg

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