Action Potential Through Ion Channel, Total Number of Ions Through

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Homework Help Overview

The discussion revolves around calculating the number of singly-ionized potassium ions that travel through an ion channel during an action potential in a nerve cell. The problem involves concepts from electrophysiology and requires understanding of current, voltage, and resistance in the context of ion movement across a membrane.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between current and charge, referencing Ohm's law and the equation Q = I*t. There are attempts to apply various equations to determine the number of ions, with some questioning the assumptions made in their calculations.

Discussion Status

Several participants have shared their calculations and results, noting discrepancies in their answers. There is an ongoing exploration of the correct approach to use, with some guidance offered regarding the relationship between charge and the number of ions. Participants express uncertainty about their methods and results, indicating a lack of consensus on the correct solution.

Contextual Notes

Some participants mention constraints such as the need for specific values (e.g., area of the ion channel) and express frustration over the lack of instructional support in their studies.

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Homework Statement


Nerve signals in the body occur when a small voltage, called an action potential, is applied across the membrane of a cell. When this action potential is applied across a region of the cell membrane called an ion channel, current in the form of moving potassium ions will be established across the cell. If, during an action potential of 90 mV, a single ion channel with a resistance of 1 G[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, is opened for 1[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3A.png2 ms, approximately how many singly-ionized potassium ions travel through the channel during this time?

Homework Equations


I = n*q*Vd*A

The Attempt at a Solution


I first converted 90mV to 90E-3V.
Then I convert 1GOhm to 1E9 Ohm.
Then I converted 1.2 ms to .0012 s.
I'm assuming I'm solving for "n" in the equation, so did V/R to find I (90E-3/1E9)
I divided that by the fundamental charge, 1.602E-19 (*note, I also used simply a "1" here with no luck either) times time.

I got 4.68E4, and other multitudes of 10, as my answer with no luck. And it's a practice exam so I shouldn't have to google anything (i.e. the area of an ion channel?).

I also tried using J=n*q*Vd and J=0.5*Q/V, making the assumption my Q's could cancel. No cigar.
 
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Ever heard of Ohm's law ?
 
Yep. And I=V/R is not the amount of ions through a channel
 
Nope, it isn't. But there is a very simple relation between charge and current.
 
Q = I*t (after googling).

I don't understand how this helps at all figuring out how many ions have gone through the channel . . .
 
There is also a charge per singly-ionized potassium ion, call it e. total charge Q, one each has e, so their number is ...

Voila!
 
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I tried Q = I*t, (I*t)/1.602E-19 (fundamental charge). Nope.
 
Nope what ? Show your steps so I don't need telepathy.

Oh, sorry. The steps are there. I get something else, so checking might get you off the hook.

[edit] Yes, definitely. 9*1.2/1.6 should be closer to 7 than to 5. And you seem to prefer hosing the target over sharp-shooting, so the power of 10 is a matter of 1,2,3,4,5,6,7,8,...;)
 
Last edited:
I ran out of tries and switched my numbers: 85mV, 1.3GOhms,0.6ms

Q = I*t = (V/R)*t = 6.54E-11*0.0006 = 3.92E-14 => 3.92E-14/1.602E-19 = 244886.

Now when I did this with my previous numbers, I did NOT get the correct answer (despite the same exact method). Before my numbers have been convenient despite my formula being wrong. I can't tell if this is the same case. If my numbers were just convenient though my method wrong.
 
  • #10
A lot of digits for a single/two digit set of givens. I get 2.45e5 too. Previously I had 6.7e5.
 
  • #11
So that is the correct way to solve a problem like that than? Use Q=I*t and Q/1.602E-12? And it was just a mis type into my calculator for the first set of numbers?

I'm learning the equivalent of all the lectures they forgot to present right now . . . Hooray . . .
 
  • #12
I suppose so. Quite a lot of ions for such a small bit of nerve activity is what they try to teach us in this exercise.

Good luck! :-)
 
  • #13
Thanks for the help. I just hate studying when they don't give us the proper tools to study with -__-
 
  • #14
Most important tool is your brain. Common sense is equally important. And with experience in doing exercises you acquire some insight in the minds of those poor chaps who have to concoct them ..
 

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