[Neuroscience] Re: Equation that explains the behaviour of a
circuit in voltage clamp
(by r_s_norman from comcast.net)
Sun Apr 5 01:14:11 EST 2009
On Sat, 4 Apr 2009 18:30:36 -0700 (PDT), "Bill.Connelly"
<connelly.bill from gmail.com> wrote:
>On Apr 5, 10:51 am, r norman <r_s_nor... from comcast.net> wrote:
>> On Sat, 4 Apr 2009 14:57:34 -0700 (PDT), "Bill.Connelly"
>> Whether or not there is Rs, the membrane current is still Im = V/Rm +
>> C dV/dt where V is the true transmembrane voltage, not a measured
>> voltage contaminated by the I * Rs term.
>Ohhhhh... okay then.... So
>Im = Vm/Rm + C dVm/dt
>And Vm... well
>No, that still isn't right, because Vm tends absolutely towards Vcmd.
>Somehow I need to take into account Im*Rs
>I'm sorry I'm being stupid here. I'm trying to grasp what you're
>saying, I'm just not very good at the math and the physics.
We may be talking about different things here.
I am assuming you have a series resistance, Rs, in series with a
parallel combination of membrane resistance, Rm, and membrane
capacitance, C. The current through the series resistance is
necessarily exactly the same current that flows through the parallel
circuit of the membrane and is both membrane current and clamp
current. The membrane current is, as already described, Vm/Rm +
C dVm/dt. the voltage across the entire thing is IRs + Vm.
You should be able to measure Rs. You also measure I. Therefore you
can calculate I Rs. You know the clamp voltage. Therefore you can
calculate Vm. And you can calculate dVm/dt. You should be able to
measure C. Therefore you can calculate C dVm/dt. Therefore you
should be able to calculate Iionic = Vm/Rm. You already calculated Vm
so you should be able to calculate Rm = 1/gm. That is, from the
voltage clamp data you should be able to see the membrane conductances
change. Also knowing Iionic, you should be able to separate that into
its components, INa and IK using the same tricks Hodgkin and Huxley
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