# Action Potential Through Ion Channel, Total Number of Ions Through

1. Oct 6, 2014

### Bluestribute

1. The problem statement, all variables and given/known data
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, [Broken] is opened for 1[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3A.png2 [Broken] ms, approximately how many singly-ionized potassium ions travel through the channel during this time?

2. Relevant equations
I = n*q*Vd*A

3. 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.

Last edited by a moderator: May 7, 2017
2. Oct 6, 2014

### BvU

Ever heard of Ohm's law ?

3. Oct 6, 2014

### Bluestribute

Yep. And I=V/R is not the amount of ions through a channel

4. Oct 6, 2014

### BvU

Nope, it isn't. But there is a very simple relation between charge and current.

5. Oct 6, 2014

### Bluestribute

Q = I*t (after googling).

I don't understand how this helps at all figuring out how many ions have gone through the channel . . .

6. Oct 6, 2014

### BvU

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!

7. Oct 6, 2014

### Bluestribute

I tried Q = I*t, (I*t)/1.602E-19 (fundamental charge). Nope.

8. Oct 6, 2014

### BvU

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.

 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: Oct 6, 2014
9. Oct 6, 2014

### Bluestribute

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. Oct 6, 2014

### BvU

A lot of digits for a single/two digit set of givens. I get 2.45e5 too. Previously I had 6.7e5.

11. Oct 6, 2014

### Bluestribute

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. Oct 6, 2014

### BvU

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. Oct 6, 2014

### Bluestribute

Thanks for the help. I just hate studying when they don't give us the proper tools to study with -__-

14. Oct 7, 2014

### BvU

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 ..