On Thu, 2 Apr 2009 17:25:34 -0700 (PDT), Bill
<connelly.bill from gmail.com> wrote:+++
>>I've already satisfied myself that the function to explain the
>behaviour of a neuronal membrane (in current clamp) is
>dV/dt = I/C where I is the sum of all transmembrane currents
>>But what is the equation for a membrane under voltage clamp? That is
>how does one explain the behaviour of the holding current I, in
>response to membrane capacitance and changes in membrane conductances?
>>Thanks for any help.
I don't have my copy of Hodgkin/Huxley (1952) with me at the moment,
but that is exactly where you must begin. Here, to my recollection,
is how it goes:
Iclamp = Imembrane because any longitudinal current is eliminated by
Imembrane = Icap + Iionic because those are the only two ways current
can cross the membrane.
Icap = C dV/dt = 0 because during voltage clamp, the voltage is
ordinarily constant. Of course, there is an enormous transient
capacitative current at the time of the voltage step.
Iionic = INa + IK + Ileak because some ion has to carry the current
and Na and K are ordinarily the only significant players. Any
chloride current is hidden in the "leak".
So that reduces to Iclamp = INa + IK + Ileak.
Now throw in INa = gNa(Vm - ENa), IK = gK(Vm - EK) and Ileak =
gLeak(Vm = Eleak) where the ENa, EK and Eleak are the Nernst
equilibrium potentials dependent only on concentration, hence constant
during the clamp. You now have the clamp current as a function of
membrane potential and membrane conductances.
Once again, during the clamp the membrane potential is ordinarily
constant. Therefore, any changes in clamp current during the clamp
must necessarily be the result of changes in membrane conductance. The
trick now is to separate the total current into the separate ionic
components so you can study the separate ionic conductances; the
behavior of the ion channels.