robin082006 Forum Hero

Topics: 471 Posts: 5,125
| | 02/03/06 - 11:36 PM  
 
   
 
|   #4 |
let's me try according to kaplan note, I think Potassium conductance is decreased when extracellular K+ increases when extracellular K+ increases--->membrane potential becomes more positive (normally membrane potiental is negative)---> decreasing K+ efflux. Therefore it can be infered that K+ conductance is decreased. correct me if I'm wrong
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/04/06 - 04:33 AM  
 
   
 
|   #5 |
When extracellular K+ is increased, then difference between ECT and ICT K+ concentration is decreased so as concentration gradient. So, less K+ leaves the cell. K+ is a positive ion, so membrane potential becomes less negative (due to decreased efflux of positive ions i.e. K+) and thus closer to the threshold potential i.e. the cell is more excitable. So: A - not correct, because it is even opposite (less negative) B - incorrect, sodium conductance is increased only from th moment when threshold is reached and action potential is elicited (i.e. when voltage-gated channels open) C - not correct, it is even opposite D - correct, due to less negative membrane potential E - Na/K pump works always, maybe its activity can be slowed down but not inactivated completely
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| nadiabarati
| | 02/04/06 - 05:09 AM  
 
   
 
|   #6 |
Dear mildus, your explanation is quite reasonable and I like it. But this Q is posted by someone who says that the correct answer is C!!! that's why I was surprised and asked for your help. thanks a lot. can I ask you what source you used for physiology?
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| nadiabarati
| | 02/04/06 - 05:17 AM  
 
   
 
|   #7 |
The answer is c. The potassium conductance will increase. Increasing the extracellular concentration of K+ causes the cell to depolarize, that is, to become more positive. When the cell depolarizes, the activation gate on the K+ channel opens, causing K+ conductance to rise. Depolarization causes both the opening of the Na+ activation (m) gate and the closing of the Na+ inactivation (h) gate. As a result, Na+ conductance remains the same. The inactivation of the Na+ channels resulting from depolarization reduces membrane excitability. The activity of the Na-K pump will decrease if extracellular K+ concentration is decreased. this is the explanation that I got from the person who posted the Q! read it, plz mildus
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| skyhigh Forum Guru
Topics: 105 Posts: 561
| | 02/04/06 - 07:30 AM  
 
   
 
|   #8 |
doesn't the normal Na/K pump work by pumping 3 Na out and 2 K into the cell. so, if there is more K outside wouldn't it's conduction into the cell increase due to increased concentration.

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| Geroo Forum Guru
Topics: 114 Posts: 799
| | 02/04/06 - 08:23 AM  
 
   
 
|   #9 |
I agree with mildus,answer should be D.
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| skyhigh Forum Guru
Topics: 105 Posts: 561
| | 02/04/06 - 08:39 AM  
 
   
 
|   #10 |
i just read kaplan, pg 43. D makes more sense now
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| rm Forum Senior
Topics: 16 Posts: 77
| | 02/04/06 - 12:34 PM  
 
   
 
|   #11 |
it makes so much sense now! nice explanation
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| Stan Forum Senior
Topics: 16 Posts: 65
| | 02/04/06 - 01:39 PM  
 
   
 
|   #12 |
Increasing the extracellular concentration of K+ will make the cell less negative i.e. threshold will decrease, but it doesn't neccesarily mean that the degree of depolarization will be enough to open voltage-gated channels. So, correct is D
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/05/06 - 07:10 AM  
 
   
 
|   #13 |
As Stan said, increased extracellular K+ concentration will make the membrane potential less negative i.e. closer to the threshold, but it doesn't necessairly mean that the degree of depolarization will be enough to open voltage-gated channels and evoke action potential! (e.g. MMP is now -75mV instead of -90mV, but threshold is e.g. -55mV, so we are still 20mV far from threshold i.e. action potential). However, when threshold is reached, then action potential appears, so we have sodium influx, and voltage-gated K+ channels also open in order to allow K+ efflux. But, what happens then? K+ leaves the cell very slowly, because, remember, there is too much K+ outside i.e. concentration gradient is decreased; so repolarization is not so easy as it should be, and the cell can't be repolarized so easily and remains depolarized for a longer period of time, during which Na+ channels are closed because they close naturally very fast after the beginning of an action potential, and new action potential can't be elicited although in the beginning we had increased excitability! This is explanation for paralysis in case of familiar hyperkalemic periodic paralysis
Edited by mildus on 02/05/06 - 07:19 AM
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/05/06 - 07:10 AM  
 
   
 
|   #14 |
When we have hypokalemia, then MP is more negative, the cell is less excitable and we have paralysis (e.g. in hypokalemic periodic paralysis), but we have seen that even in hyperkalemia where MP is less negative and the cell more excitable, there is paralysis (e.g. hyperkaliemic periodic paralysis). Those two hereditary conditions are, at least it is my opinion, the best way to understand those concepts
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/05/06 - 07:13 AM  
 
   
 
|   #15 |
So, we have seen that in case of hyperkalemia, K+ conductance isn't increased not even during action potential repolarization phase
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/05/06 - 07:28 AM  
 
   
 
|   #16 |
Nadiabarati, after some more thinking, I begin to realize why that person didn't choose D. We initially have the cell with increased excitability (which means it is easier to reach the threshold and AP) but as soon as single action potential happens, the cell becomes inexcitable due to disturbed repolarization due to hyperkalemia. But, I can't accept C as a choice.
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| mildus Forum Guru
Topics: 19 Posts: 614
| | 02/05/06 - 07:30 AM  
 
   
 
|   #17 |
Sorry for writing so much, but I have really tried to explain it
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| Stan Forum Senior
Topics: 16 Posts: 65
| | 02/05/06 - 08:52 AM  
 
   
 
|   #18 |
well done mildus
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| doc179 Forum Guru
Topics: 67 Posts: 1,217
| | 04/12/06 - 11:38 AM  
 
   
 
|   #19 |
I agree with mildus. mildus you have strong good knowledge. great going
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| waqastariq
| | 04/25/06 - 11:47 AM  
 
   
 
|   #20 |
D What I think is that this question is a true representation of paralysis associated with familiar hyperkalemic periodic paralysis and as mildus has already told the explanation of the question...I cant do it better
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| doc179 Forum Guru
Topics: 67 Posts: 1,217
| | 04/28/06 - 01:16 PM  
 
   
 
|   #21 |
Hey everyone I found this Q in the pretest physio and as nadiabarati said the ans with the explanation is C. I think its a a mistake as I also agree that the ans should be D.
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