6
$\begingroup$

I'm reading about Goldman Equation and I've seen in some resources that it calculates the resting potential (not membrane potential in general) and that this equation cannot be used when membrane potential is changing. One example is in Kandel textbook:

This equation applies only when Vm is not changing. Principles of Neural Science. Eric Kandel, Fifth edition. p. 135.

I don't really understand why this equation cannot apply when Vm is changing. Precisely, I understand that this equation defines how membrane potential changes taking into account permeability and concentration of ions. If you "give" to it the concentration-permeability changes for each moment, it'll show the Vm for each of these moments.

Could someone explain that to me? Thank you very much!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

What's the purpose of the Goldman equation?

I think that statement you posted from Kandel is a bit misleading, though not incorrect - I think he's just warning against a pitfall he's seen students make.

The Goldman Equation is an equation for steady-state voltage or in other words the reversal potential, not for Vm. It will only give you Vm in a (mostly) steady-state condition, like in the context of a resting membrane potential. However, that doesn't mean it isn't useful to calculate the Goldman equation voltage when Vm is changing or that it doesn't apply, but rather that it is not sufficient to use. It will still accurately tell you what the reversal potential is at any moment in time, but it can't tell you what Vm is at that moment.

How the Goldman equation can be helpful in understanding dynamic voltage changes:

Importantly, the rate of change of the voltage of the cell is going to be proportional to the difference between the reversal potential (from the Goldman equation) and the current voltage Vm, and after a long time the voltage will eventually asymptote to the Goldman voltage.

Therefore, if you know the current Vm, and you calculate the reversal potential with the Goldman equation, you can make some inference about what will happen to the voltage at the next moment in time: it will change in the direction of the reversal potential by a magnitude proportional to

Vm - Ereversal

Why the Goldman equation is not sufficient while Vm is changing:

I suspect what Kandel and others are warning about is that this isn't quite enough information to know what happens in the next moment of time:

It takes time to reach a steady voltage, and you need to know the membrane capacitance, which isn't represented in the Goldman equation, to know what the actual change in voltage is after $\delta$t, or how long it will take to reach a particular voltage.

Also, because of voltage-dependent conductances in a neuron, the permeabilities to different ions are also changing as a function of Vm, so the reversal potential is also changing constantly.

Furthermore, simulations of Vm sometimes need to consider the changing concentrations of individual ions, because the reversal potential will change as ion concentrations change. Therefore, it typically makes more sense to think about the driving force of individual ions given by the Nernst equation rather than the Goldman equation which gives you the net reversal potential across all ions.

Summary

In summary, authors that warn against using the Goldman equation when Vm is changing are reminding you that voltage changes take time and depend on membrane capacitance as well as changes in conductance from voltage-gated channels. However, the Goldman equation is still a correct way to calculate the instantaneous reversal potential at any moment in time (but not Vm), even if Vm is dynamic.

You could edit Kandel to say:

The Goldman equation only outputs the present Vm when Vm is not changing; also, when Vm is changing, the output of the Goldman equation is also changing due to changing conductances and ion concentrations.

..but you can see this is a bit verbose.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thank you Bryan for your very complete response. Do we have an equation that permits us calculate Vm when the cell is not at resting levels? $\endgroup$
    – Alexei
    Dec 27, 2017 at 23:15
  • $\begingroup$ @Alexei You need to use differential equations, because the Vm will always be a function of itself as well as other parameters. There are lots of different ways to express these equations, you can find some of them here: cns.nyu.edu/~david/handouts/membrane.pdf In practice, though, changing voltage-dependent conductances mean that a lot of the quantities that seem to be constants in these equations are also functions of time and the other parameters, so in that context you need a model like the en.wikipedia.org/wiki/Hodgkin%E2%80%93Huxley_model $\endgroup$
    – Bryan Krause ♦
    Dec 27, 2017 at 23:23
  • $\begingroup$ I've found this resource physiologyweb.com/lecture_notes/resting_membrane_potential/… Focusing in the text that refers to Figure 1, would you say that this part is wrong or, at least, not totally correct? Reading this resource I thought I've understood the lesson, but your responses have made me see that it's no so easy :( $\endgroup$
    – Alexei
    Dec 28, 2017 at 0:04
  • $\begingroup$ @Alexei Without going into it line by line, that figure seems fine, but what is important to realize is that this is a toy example showing you how relative permeabilities influence the membrane potential if one sets them magically in a model. It's really important to understand how this works, because it's an important building block to understanding the rest. Then, when you move on to real cells, it's important to recognize that the permeabilities are all changing dynamically because they are driven by physical channels rather than experimenter-set variables. $\endgroup$
    – Bryan Krause ♦
    Dec 28, 2017 at 16:49
  • $\begingroup$ Then I understand that de GHK equation can serve as an introduction to the "picture" but isn't the whole picture at all. Thank you Bryan for your kind responses :) You've been very helpful. $\endgroup$
    – Alexei
    Dec 28, 2017 at 17:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .