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Calculate the number of sodium ions entering an non-myelinated axon

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate the number of sodium ions entering a non-myelinated axon during the action potential, per metre of axon length. The change in potential is 100 mV, the axon membrane capacitance per unit length is 3 × 10−7 F/m, and the charge on an ion is 1.6 × 10−19 C.
    What concentration change does this produce and how does it compare with the concentration of sodium ions in the cell in its resting state.
    During the resting state of the axon, typical concentrations of sodium and potassium ions are 15 and 150 millimole/litre, respectively. Avogadro’s number is 6.02 × 1023 per mole, and the radius of the axon is 5 microns. (1 micron = 1 × 10−6 metres.)

    2. Relevant equations

    ∆Q = C ∆V


    3. The attempt at a solution

    First I found the change in charge which is 3 × 10−7 F/m X 0.1V = 3x10^-8 C/m
    I then divided this by the charge of 1 ion to find the amount of entering ions per axon meter: (3x10^-8 C/m)/(1.6 × 10−19 C) = 1.875x10^11 Na+ ions per m.

    Now to find the number of ions in a resting axon with radius 5microns. 0.015mole/L Na+ x 6.02x10^23 divided by 1000cm3 = 9.03x10^18 Na ions per cm3.
    Axon Volume: pi X (5x10^-4cm)^2 x 100cm = 7.853981634x10^-5cm3.
    7.85x10^-5cm3 x 9.03x10^18 Na ions/cm3 = 7.09x10^14 Na ions in a resting axon
    7.09x10^15 K ions in a resting axon (150mM is 10 fold higher than 15)

    So the amount of ions entering is almost insignificant (11 power vs 14) - I guess this question is meant to show us how easily charge is changed while concentration remains the same.

    I'm happy with my working so far (please tell me if I've done something wrong!) but can someone help me figure out the concentration change produced by the entering sodium ions. Does the concentration change significantly at all? I'm a bit confused on how to account for the exiting K+ ions.

    Thanks
     
  2. jcsd
  3. Oct 18, 2014 #2

    BvU

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    Concentration = moles/volume, right ? You have 2 1011 ions/meter, you have the volume for one meter and you have the number of ions per mole (valence is +1, so this is just NA).

    And yes, this shows that small changes in concentration give hefty voltages. Having to do with single electron charge times NA being BIG!
     
  4. Oct 18, 2014 #3
    So is it:
    1.875x10^11 Na+ ions per m/6.02x10^23 divided by avagadros number to go back to moles = 3.12x10^-13 moles per m. As 1 meter = volume we figured out earlier divide this number by the axon volume (divide it by 1000l first as we want to compare it to the 15milimole/litre). So 3.12x10^-13 / (7.853981634x10^-5cm3 /1000L = 7.8540x10^-8L) = 3.97x10^-6 Moles per L.
    = 3.97x10^-3 Milimoles per L.
    This number seems to be in line with the insignificant concentration change I pointed out earlier (3.97x10^-3 Milimoles per L vs 15milimoles/L is tiny) . So would I say that an action potential increases the intracellular sodium concentration by 3.97x10^-6 Moles per L.
    Does the fact that you didn't correct any of my working mean it was all right? Thanks a lot for the help by the way.
     
  5. Oct 18, 2014 #4

    BvU

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    I think you're doing excellent work.
    Can't think of any improvement, but have two tips:
    1. Work in Si units as much as possible and (if necessary at all) convert only at the last moment.
    2. Since you're given values are 1 or two digits, round off (again, at the end) to two digits.
     
  6. Oct 18, 2014 #5
    Thanks so much for all of the help! Yeah I was just being lazy with the rounding because typing everything out on here takes ages. It's crazy how a 1000 fold smaller change in concentration can cause such a large shift in potential difference. Just to make sure, I didn't have to do anything with the K+ ion concentration did I? Like comment on how much leaves the cell (I assume it's the same amount as the sodium ions that enters the cell due to potential difference (opposites attract) rather than concentration gradient?) or am I over thinking it haha.
     
  7. Oct 18, 2014 #6

    BvU

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    I don't see anything about the K+ in the problem statement (except the mentioning of the concentration), so I don't feel qualified to help here. I'd have to read up an awful lot to find out how these things work. From what I found in a short time, it's a sequential process (Na in, then K out), otherwise there would be no potential pulse.
     
  8. Oct 18, 2014 #7
    You've definitely done more than enough, no need to do any more reading! I'll just have a look, I'm pretty sure the K out matches the Na in but I'll have another read to be sure. You're a life saver, cheers :)
     
  9. Oct 18, 2014 #8

    BvU

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    You're welcome. Reading up (actually: getting to know about) is no punishment, it's quite an interesting subject ! So thanks for posting :-)
     
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