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Studying biochemistry I noticed that equilibrium is reached for K and Na separately (not taking into account electric potential), as would be the case I assume for any molecule or atom. This is of course entropy.

Eg: My understanding is, you have two types of atoms, and a membrane separating them, with different concentrations on each side. Each atom will reach equilibrium with itself across the membrane, as if the other type of atom doesn't exist (barring the influence of charge).

Why is it that, for instance, a K atom is so keen to reach equilibrium with another K atom in an area of lower concentration, and yet seemingly ignores the Na, even if the Na is in high concentration where the K is low? Both Na and K still have the same charge, and would also collide with each other I imagine? Why do identical atoms behave like this but non-identical ones with very similar properties don't?

Update #2:

My understanding was incorrect regarding K and Na having separate forces. The two atoms seem to be treated the same, except for selective permeability.

I'm still not understanding why the two atoms have separate gradient potentials, if they can be treated as the same. I made the following images to illustrate my point a little clearer.

There is a compartment with a left and right side, separated by a wall containing two selectively permeable channels that may open or close. Only Na can cross the green channel when it is open, only K can cross the orange channel when it is open.

First, the assumption: The main assumption behind all this is that, in answering this question, we can treat Na and K as being the same. They are effectively two marbles with identical physical and chemical properties, just with different colours. This is what I have come to understand based upon the answer by orthocrestol.

The cell on the left has a -70mv potential energy compared to the right compartment, caused by invisible negative ions which will always remain on the left side.

enter image description here

All that has been summarized by the discussed assumption is seen in figure 1.

Now in figure 2, we open the selective channel for Na. Since Na has a gradient potential, and there is also a potential difference, the positive Na flows into the left hand compartment.

In figure 3, an equilibrium is reached at an arbitrary value of -20mv. On average, no more Na will move to the left or the right, it's at equilibrium.

Figure 4, channels closed. Now this is where my understanding ends.

enter image description here

Apparently now, from what I gather, in figure 5 if we open the selective channel for orange (K+), since K+ has a gradient potential, it moves to the right. In the process, it takes its positive charge with it and restores more or less the original potential difference of -70mV on the left hand side.

But, if this is what happens, this must surely invalidate the marble analogy of the two atoms being the same. Because if they're the same, then whether we open the Na channel (which is already at equilibrium) or the K channel should make no difference, because once gain, they're the same atom with a different colour (except for being separately permeable via their selective channels).

The only difference that should be apparent is that if we leave both channels open, then the colours will eventually reach an equilibrium on both sides. But if an equilibrium is already reached for the charge (which it has in figures 3 and 4), I'm failing to grasp how opening then closing a different channel suddenly causes that equilibrium to change (as in figure 5 and 6). The explanation is the K potential gradient, but this is still the same marble as Na is it not? How come the potential difference changes?

Please direct me where my thinking has gone wrong, or if this is what even happens. And if it is, how is this possible? Many thanks.

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    $\begingroup$ There are a lot of different sub-questions embedded here. It might be prudent to try to split it into multiple posts. For example, the mechanism of diffusion should really be a completely separate question from the requirement for both $\ce{Na+}$ and $\ce{K+}$ in action potentials. $\endgroup$ – hBy2Py Jul 8 '15 at 17:50
  • $\begingroup$ I agree @Brian. Problem is though I thought I understood diffusion in terms of two atoms, as orthocresol correctly described, and yet in the neuron which I thought was just a biological application of this same system it seems actual charge transfer takes place (in depolarization/repolarization, absent the Na/K pump, which I imagine would play a trivial role in this discussion). My misunderstanding is lying either in the neuron or the diffusion, but I'm not sure which. $\endgroup$ – user4779 Jul 9 '15 at 1:01
  • $\begingroup$ I'm insufficiently knowledgeable to say for sure, but I think the key aspect lies in the relative concentrations of $\ce{Na+}$ and $\ce{K+}$ in the system. I'm pretty sure there's a lot more sodium than potassium, meaning that the overall potential is dominated by the instantaneous $\Delta C_{\ce{Na+}}$, but I haven't been able to turn up a good reference to confirm it. $\endgroup$ – hBy2Py Jul 9 '15 at 1:45
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    $\begingroup$ On a high level, the differences in their movement and equilibria boil down to differences in the number and type of selective channels present for each ion, and how those channels respond to the dynamics of the ion movements and charge distribution. It's highly non-trivial. Will write it up as soon as I can, but it may be a day or three. $\endgroup$ – hBy2Py Jul 9 '15 at 3:05
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    $\begingroup$ It's much easier if you edit out the action potential portion and reask it on Biology, though I think you'd find a few similar questions there already. It's always best to tailor your questions for the target site. $\endgroup$ – jonsca Jul 10 '15 at 1:11
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This answer has been edited quite a bit. New additions are put in blockquotes.

To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: Any mention of "potential" in this answer refers to an electric potential, not a chemical potential. The definition of electric potential can be found here on Wikipedia.


Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to $\ce{Na+}$ ions. In the left compartment, we put 1 M of $\ce{K+}$ ions and in the right compartment, we put 1 M of $\ce{Na+}$ ions (with appropriate counterions). If diffusion of $\ce{Na+}$ entirely ignored the presence of $\ce{K+}$ as you say, then we would expect that after a while, we would have 1 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ on the LHS, and on the RHS we would have 0.5 M $\ce{Na+}$.

This does not happen. In actual fact, there would be a concentration gradient that favours net movement of $\ce{Na+}$ from RHS->LHS, and there would be a electric potential difference that favours net movement of $\ce{Na+}$ from LHS->RHS. We could model the concentration at any point in time with:

  • LHS - 1 M $\ce{K+}$ and $x$ M $\ce{Na+}$
  • RHS - $0.5 - x$ M $\ce{Na+}$

where $x$ is between 0 and 0.5.

When $x = 0$, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so $\ce{Na+}$ starts diffusing from RHS-> LHS, causing $x$ to increase. Now as $x$ increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of $x$, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of $\ce{Na+}$ across the membrane. This is the point where we could say that diffusion has stopped.

Now, let me change the scenario and say that the membrane is permeable to BOTH $\ce{K+}$ and $\ce{Na+}$ ions. In that case, clearly the equilibrium concentrations will be 0.5 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.

It is important to note that the diffusion of $\ce{Na+}$ is NOT directly affected by the pre-existing concentration gradient of $\ce{K+}$. It is indirectly affected only because the concentration gradient of $\ce{K+}$ leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of $\ce{Na+}$. If we were to replace $\ce{K+}$ with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of $\ce{Na+}$.


Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.

The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as $\ce{Na+}$, $\ce{K+}$, $\ce{Ca^2+}$, and $\ce{Cl-}$. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.

The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of $\ce{Na+}$ is much higher than the intracellular concentration of $\ce{Na+}$; the converse is true for $\ce{K+}$. Due to the membrane having different permeabilities to $\ce{Na+}$ and $\ce{K+}$ (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.

Neurons react to certain stimuli by having $\ce{Na+}$ enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated $\ce{Na+}$ channels open, making the membrane permeable to $\ce{Na+}$.

At this stage, we have to consider the two factors that influence the diffusion of $\ce{Na+}$:

  1. Since the potential difference is still negative, it favours influx of $\ce{Na+}$.
  2. Since the extracellular concentration of $\ce{Na+}$ is larger than the intracellular concentration, the concentration gradient also favours influx of $\ce{Na+}$.

Hence more $\ce{Na+}$ enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated $\ce{Na+}$ channels close and the voltage-gated $\ce{K+}$ channels open. Again, we need to consider the two factors that influence the diffusion of $\ce{K+}$:

  1. The potential difference, which is now positive, favours efflux of $\ce{K+}$.
  2. Since the intracellular concentration of $\ce{K+}$ is higher than the extracellular concentration, the concentration gradient also favours efflux of $\ce{K+}$.

Therefore $\ce{K+}$ leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both $\ce{Na+}$ and $\ce{K+}$ are not the same as they were at the very start of this discussion. They are restored by the $\ce{Na+}$/$\ce{K+}$ pump.

The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.

Prior to my edit you might have seen that I wrote that "diffusion does not ignore similarly charged species". This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when both X and Y are charged species, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.

I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.

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    $\begingroup$ After I finished typing out everything I reread your question and I think this might not really answer your question, so I apologise if this is the case. I think maybe my comment on the other answer might be more useful... $\endgroup$ – orthocresol Jul 3 '15 at 8:05
  • $\begingroup$ That's really helpful orthocresol, thank you. But just two points I'm not getting. First, if side A has 1M K, side b 1M Na, and a semi-permeable membrane for Na, wouldn't both these sides be in equilibrium in terms of their number of particles (atoms) ? If any Na did move to the left, wouldn't this break this equilibrium of 1M? I think there's something simple I'm not seeing in what makes the concentration gradient have potential like that. I assume this would also explain why in the neuron, if the Na channel is open we have an influx of Na, but the moment we open the K channel an efflux of K. $\endgroup$ – user4779 Jul 3 '15 at 15:41
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    $\begingroup$ @user4779 The equilibrium that is established is Na (left) <=> Na (right), not for total particles. I could dump 1 M glucose and 1 M urea in the LHS and since both species are uncharged it wouldn't affect the diffusion of Na at all. $\endgroup$ – orthocresol Jul 3 '15 at 17:26
  • $\begingroup$ Really helpful explanation in the edit, thanks. So I see now that each atom does in fact have its own unique concentration gradient, provided it has no charge. This brings me to my original question. How could this be possible? Since there are so many more atoms, as you say, if we dump moles upon moles of some similar neutral atoms on one side, wouldn't those collisions near the membrane prevent the atoms on the other side from crossing? And even if not completely, surely moreso than if they weren't there in the first place? $\endgroup$ – user4779 Jul 10 '15 at 11:38
  • $\begingroup$ If we have 10 marbles on side A, 10 marbles on side B different colours, shake them and they will be at equilibrium with enough trials, I'm fine with that. But put 1000 marbles on side B and 10 marbles on side A, how could they even get into side B? Is there some other analogy that could better describe this? $\endgroup$ – user4779 Jul 10 '15 at 11:41
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Each atom does its own diffusion across the membrane. There is a probability that a $\ce{K+}$ or $\ce{Na+}$ ion will move in the "forward" or the "reverse" direction. The probability of getting through the membrane may differ between the types of ions (e.g. if the membrane pores allow one species through more easily than another because of ionic size), but, barring any external force, eventually each type of ion should come into equilibrium, with as many going forward as backward.

Imagine a box of marbles of two colors, with a partial separator between, and that the box is shaken. It is simple chance that a marble crosses the separator, and if many marbles of one color are on one side, there are fewer to migrate backward from the other.

See this online demo of diffusion across a semipermeable membrane.

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    $\begingroup$ So, this is what I"m failing to grasp. In a biological membrane of a neuron for example, if a sodium channel is open, the sodium ions will rush in. Then if it shuts and the potassium ion channel opens, the potassium ions will rush out. I can appreciate the channels being selective, but why is this particular behaviour observed? My understanding is because there is a lower concentration of Na inside the cell, lower concentration of K outside the cell. But why would the Na go in if there's already so much K in there? How can it differentiate between K and Na? $\endgroup$ – user4779 Jul 3 '15 at 5:02
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    $\begingroup$ And if it's just a different colour of the same marble to use the analogy, wouldn't the marbles already be in equilibrium, and the colour that moves is then just random? If this is the case, how come a gradient potential exists for that specific atom then? $\endgroup$ – user4779 Jul 3 '15 at 5:07
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    $\begingroup$ Oh, just to directly answer your question "But why would the Na go in if there's already so much K in there? How can it differentiate between K and Na?" - it's actually just all about statistics. Let's say you and I sit across a table and you have 20 red marbles and I have 20 blue marbles. We blindfold ourselves and spend the next 1 hour exchanging random marbles. At the end of the 1 hour we take off our blindfolds and guess what the most likely outcome is? Of course - the most likely outcome is that each of us have 10 blue and 10 red marbles. $\endgroup$ – orthocresol Jul 3 '15 at 8:00
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    $\begingroup$ Of course, there is a chance that maybe you have 15 blue and 5 red, and by implication I have 15 red and 5 blue. That's always possible and if we kept the blindfolds on, there would be no way for us to know that that happened. It's just that statistically speaking it's more likely that we have nearly equal numbers. The same is true for the neuron - where the numbers are much, much larger than 20. The entire concept of entropy is actually just statistics and probability. @user4779 I should have tagged you in the earlier comment to alert you, sorry. $\endgroup$ – orthocresol Jul 3 '15 at 8:08
  • $\begingroup$ Just regarding the marble analogy, so the net number of particles remain the same. But in the neuron for example, the depolarization will release a lot of Na into the cell. And then release K in the repolarization. If K and Na are just different colours, then is this to say that if we hypothetically left the Na channel open, depolarization would also occur with Na items then moving out the membrane just like K would through its channel? Since they'd be interchangeable? $\endgroup$ – user4779 Jul 3 '15 at 15:51

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