Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equilibrium potentials and different permeabilities.

  1. Nov 26, 2011 #1
    This has been driving me crazy for aaaaaaaaaaaaaaaaaaaaaaaaaaaages. It's about equilibirum potential.
    Okay, say I have a cell with High K+ concentration inside and low K+ concentration outside, and the opposite for Na+. If for each cation there is an anion, then the voltage across the cell membrane will be zero. If K+ is then allowed to move, an accumulation of negative charge takes place in the cell. This opposes the move of K+ out of the cell; and an equilibirum is reached (i.e. there is no net movement, the movement of ions out of the cell due to random movement is opposed by the negative charge acting on the K+ ions to move them back into the cell, and these two opposing 'factors' are equal; although I guess there is a charge repulsion between K+ driving them out to some extent and random movement moving them into to some extent).
    The Na+ ions have no role, since they do no cross the membrane. I think all this is correct; I get confused when we increase the permeability of the membrane to Na+. If Na+ had the same permeability as K+, then the cell would not be able to distonguidh between the two cations, and so they can be treated as a single type of ion. However, if the membrane is less permeable to Na+, apparently they make unequal contributions to the voltage across the membrane at equilibirum. I don't undertsnd this. If Na+ moves more slowly, it just means that say, a K+ ion moves across, back and forth, more frequently; so it will approach the voltage that would be obtained for K+ since it moves muuch more quickly than Na+. However, if the Na+ concentration is unequal in the opposite set-up to K+, then the negatve charge in the cell will act to move the Na+ into the cell, as will its concentration gradient. Overtime, the Na+ will move in, this neutralises the negative charge and allows more K+ to move out.
    See what I mean? We would keep going until we get to a point where the net movement is zero and the rate of movement of Na+ in and out is equal, and the same is true for K+, except K+ has a much higher rate of movement. The euilibrium then, is when the proportion of total cations that are Na+ is the same for both inside and outside (also true for K+); the total cation conentrations do not have to be equal, but the net movement is zero, as determined by random movement and the voltage across the membrane.
    This is annoying me no end. Any help appreciated.
    Last edited: Nov 26, 2011
  2. jcsd
  3. Nov 26, 2011 #2


    User Avatar
    Gold Member

    that was a long stream of consciousness. Not completely sure about the question, but it seems like you may be neglecting the forces caused by the concentration gradient. Just because Na and K are in electric equilibrium does not mean they are in equilibrium.

    In a real cell, there's also ion pumps, Cl-, and other antions belonging to protein constituents.

    If I didn't answer your question it may help to restate it a single sentence.
  4. Nov 28, 2011 #3
    Thanks for the response Py.
    Sorry about that. I'm not sure if it's possible to communicate what I mean in a single sentence.
    I'll try to make it succinct.
    I'm confused about the Equilibrium potential of a cell. The Goldman equation, from my understanding, identifies the potential difference across the membrane for given ion concentrations and permeabilities at which the net flux of ions is zero: so there's no change in the system.
    This doesn't make sense to me. Let's say we have a cell with only two ions, K+ and Na+ and that their charges at exactly countered by the presence of some negative charge (i.e. to begin with, the potential difference across the membrane is zero); with the K+ concentration higher inside than outside and the opposite is true for Na+. At the equilibrium potential given by the Goldman equation, the net flux of cations (the only ions in the scenario that can move) is equal in both directions. However, the concentration of K+ is different on the two sides of the membrane, and so K+ is more likely to move out, and the opposite is true for Na+. K+ might move 'back and forth' more frequently than Na+ if it has a higher permeability, but over time Na+ will nonetheless creep across, and K+ will also move out. The only point I can see an equilibrium being a 'true equilibrium' is when the proportion of K+ and Na+ is the same on both sides, and the net flux of cations is equal in both directions.
    I found this:
    "It is important to keep in mind that neither sodium ions nor potassium ions are at equilibrium at that steady value of potential: sodium ions are continually leaking into the cell and potassium ions are continually leaking out. If this were allowed to continue, the concentration gradients for sodium and potassium would eventually run down and the membrane potential would decline to zero as the ion gradients collapsed."

    http://books.google.co.uk/books?id=...oQ6wEwAg#v=onepage&q=goldman equation&f=false

    I'm not sure if this is referring to the same thing as me, it's not that clear. I think it is saying the same thing as me concerning the concentrations changing because they would move with greater probability in one direction compared to the other. I don't know why the potential would go to zero, though.
    Hopefully this is a little clearer.
    Thanks again,
  5. Nov 28, 2011 #4


    User Avatar
    Science Advisor
    Gold Member

    I'm not sure what you are asking. In the reference you provide they state the potential across the membrane would go to zero if the ion gradients collapse. Fair enough. Is that what you are having trouble understanding? Cells work very hard and require energy to keep the ion gradients they have (sodium-potassium ATP pumps). If cells didn't keep those ions in disequilibrium then an equilibrium would be reached reducing the potential across the membrane and reducing the cells ability to utilize that potential to do work.
  6. Nov 28, 2011 #5
    Partly, there's not a single question as such, it's more how I imagine the process to work, but for some reason it may not be consistent with other sources. I was under the impression that the Goldman equation identifies the potential difference at which the net flux of ions is zero.

    Say there was only K+ that could move across the membrane, and the concentration was higher inside than outside. The positive charges all have a corresponding negative charge that is impermeable, so despite the difference in concentration, the potential difference across the membrane is zero.

    Since the concentration of K+ is higher inside the cell, the K+ ions would move down their concentration gradient due to random motion; but as this happens, a potential difference across the membrane would form. This would oppose their movement out of the cell; and when the negative charge inside the cell relative to outside the cell has a certain value, the charge will draw the ions into the cell with a strength such that the net flux is zero. Yet the concentrations will not be equal inside and outside; and it is an equilibrium. As far as I am aware, a stable one.

    The difficult part is extending this to more than one ion. If the Goldman equation represents the potential difference at which the net flow of ions is zero, I don't see how it represents a stable equilibrium. The Goldman equation takes into account permeabilities, and, from my understanding at least, puts the potential difference for equilibrium at a position between the equilibrium potentials for the individual ions under 'conditions' in which they were only permeable. Its weighted by the permeabilities; but why? I don't see why the permeabilities are important. Overtime, the ions will continue to move down their concentration gradients. As I said, the only 'conditions' I can see that give rise to a stable equilibrium is when the net fluc of ions is zero (i.e. they head in both directions at the same rate), and the concentrations have equal proportions on both sides (i.e. the probability that a given ion moving across the membrane is K+ is the same in both directions) - if they didn't the concentrations would continue to change.

    Something in my reasoning may be incorrect. So, if there is a single question, its where is the reasoning wrong? If it is wrong. Which it must be somewhere, since I don't see why the potential difference across the membrane would have to be zero. You could have a potential difference across the membrane if the proportions of each specific ions where equal both sides but the concentrations were different, by my understanding.

    Thanks for the quick response, by the way. I'll need to do some more reading on this, but the books I find skip an explanation.
  7. Nov 28, 2011 #6


    User Avatar
    Gold Member

    Well, no, the concentration of Na outside the membrane can be considered an infinite supply, so it doesn't go down at all (in terms of significance) even thought it causes the concentration inside the cell to go up significantly.

    Does this help?
  8. Nov 28, 2011 #7


    User Avatar
    Gold Member

    My teacher actually has us do the problem that shows the concentration change for each ion for inside and outside the cell. It's very tiny amount of ion flow that results in huge potential changes.
  9. Nov 28, 2011 #8


    User Avatar
    Science Advisor

    It's the same as what you are asking. If potassium and sodium keep flowing down their concentration gradients, the concentration gradient will eventually be zero. If there is no concentration gradient, then there will of course be no potential difference.
  10. Nov 28, 2011 #9

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    Maybe it would help to consider a complete circuit: consider epithelia. On the apical side, there are ENaC proteins that let only Na+ into the cell. On the basolateral side, there is the NA-K-ATPase molecule that pumps 2 Na+ out and 3 K+ in, and as a result generates the ~60mV membrane potential. To complete the circuit, basolateral RomK channels let the excess K+ leak back out. To ensure electroneutrality due to the vectoral transport of Na+, Cl- ions passively diffuse through the paracellular region.
  11. Dec 2, 2011 #10
    Thanks for all the responses! They are very helpful.
    I can make the pictures smaller if necessary, but doing so makes the text to small to read (on the screen I'm using, at least).

    This is useful to know; however, I have not come across this. Can we set this aside for the moment? I'm not sure what the consequences of this are. If this were always taken to be the case, how can the equilibrium potential collapse?

    I realise the change in voltage induced is major for a negligible change in the concentrations, although this depends on the scale you look at I would have thought: a small change at, say, a millimolar scale is not at a nanomolar scale. Anyway, the concentrations moving in opposite directions down their concentration gradients neutralise the charge as they move: K+ moving out generates a negative charge inside the cell, and Na+ moves down it's concentration gradient - additionally attracted by the negative charge - into the cell. If K+ has a higher permeability, it might reach a temporary 'stable state', but as Na+ leaks in, it alters the system, and this continues. So eventually there is a substantial change in concentration.

    This is where I am confused: why would the concentration gradient have to be zero?
    I drew some pretty pictures to illustrate my point. The blocks represent 'concentrations of ions', the larger the block, the higher the concentrations:


    This is for a single ion. The concentration might be greater on the left, but as the ions move to the right, a negative charge accumulates in the cell, which increases cation movement to the left. Equilibirum is reached when the random movement to the left (which is lower due to a lower concentration) coupled with the increased movement to the left due to being attracted to the negative charge is equal to the random movement to the right. By my understanding, this is a stable equilibirum: it goes unchanged.


    Here, there are two cations. The set-up is essentially the same. The system is in equilibirum for the reasons given above.
    The system is in a stable state: the rate of cation movement is the same both ways.
    In the above system, for any given cation moving across the membrane, the probability of it being the pink cation is equal in both directions, since it constitutes the same proportion of the total number of cations on both sides (remember, flux is equal both ways). The same is true for the green cations.
    Even if they have different permeabilities, the probability of the pink cation being the cation moving in a given direction at any time is equal for both sides: it has the same permeabilites both sides of the membrane (again, the same applies to the green cation).
    So the fluc in bboth directions is equal, and the probability of a specific cation crossing the membrane is equal both sides, this means that, for every pink cation that crosses to the left, one will cross to the right. As such, this seems to me to be a stable equilibirum, not just a steady state, since it doesn;t change over time. Additionally, it has a non-zero voltage.

    Hopefully a visual illustration is better.
    What's incorrect about the above reasoning? If it's true, then an equilibirum is possible without zero voltage across the cell.

    Thanks in advance,
    Last edited: Dec 2, 2011
  12. Dec 2, 2011 #11


    User Avatar
    Science Advisor

    Are you proposing that there is no concentration gradient for either ion, yet there is an electric potential difference?

    Ooops, I see. You are proposing a concentration gradient. However, in your initial setup potassium was high in the cell and low outside, and sodium high outside and low inside. In your new diagram, it looks like both sodium and potassium are high on the same side of the membrane?
    Last edited: Dec 2, 2011
  13. Dec 2, 2011 #12
    Thanks atyy.
    Yes, sorry; through all my time spent pondering this I have played with different ideas and forget sometimes that I have not communicated this!
    I have indeed moved both concentrations to be higher on one side. I was interested in knowing if both scenarios above provide a stable equilibrium.
    If the concentrations of the two ions could arrange themselves into this concentration is something I am having difficulty with, and I was wondering whether the objection to their being an equilibrium with a voltage across the membrane was precisely because the above scenario could not be established, but the books I have found make no mention of any of this.
    As mentioned, I am interested in knowing if both of the above are stable equilibria. The next step is then to determine if something analogous is possible under artificial, but reasonably approximate, cellular conditions.

    Thanks for the response!
  14. Dec 2, 2011 #13


    User Avatar
    Gold Member

  15. Dec 2, 2011 #14


    User Avatar
    Science Advisor

    Yes, looks correct to me. Basically it seems we are asking whether the zero net flux potential for two ions is the same as the zero net flux potential for each ion separately. This seems true under the assumption of the GHK equation and for the case where the ratio of concentrations on each side of the membrane is the same for each ion. However, I've never worked out this particular answer before, so would especially love commentary and corrections.
    Last edited: Dec 2, 2011
  16. Dec 3, 2011 #15


    User Avatar
    Science Advisor

    Under GHK assumptions such as a linear electric potential across the membrane, the current for one ion has the form

    I=PV(Coe-V-Ci)/(e-V-1), where P is the permeability.

    For I=0, we need (Coe-V-Ci)=0, ie. V=ln(Co/Ci), which depends only on the ratio of external to internal concentrations.

    For two ions of the same valence, because the GHK assumptions include independence, I=I1+I2, where I1 and I2 have the same form as the one ion current. If we need I=0, as well as I1=0 and I2=0, this can be achieved if the ratio of external to internal concentrations of each of the two ions is the same, even though V is not zero.

    In more general conditions I=0 is achieved by I1 and I2 being opposite in sign with equal non-zero magnitude, so the external and internal concentrations will eventually change considerably, unless they are replenished.
    Last edited: Dec 3, 2011
  17. Dec 5, 2011 #16
    Nobahar - you seeing things clearer now or are you still having issues?
    Most equilibrium conditions are indeed not at zero millivolts.
  18. Dec 6, 2011 #17
    Thanks for the responses. I will return to this thread, I didn't just want to post a "Thanks!" and leave it at that as I was hoping to explore it a little more, it's just that I have been preoccupied with huge amounts of other work, and haven't been able to look into it yet. I will return to this when some time frees up and hopefully others will want to continue with this.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook