Differential Attenutation of EPSPs in VC vs IC

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Bill Connelly
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Differential Attenutation of EPSPs in VC vs IC

Post by Bill Connelly »

So I was playing around with neuron last night. I had a 3x500um long cable, with passive properties vaguely like a layer V dendrite. I applied alphaSynapses over the length of the cable, and had a voltage clamp or current clamp electrode at one end. I saw your typical exponential attenuation of the EPSP/C. However, against my assumption, the EPSP/C attenuated to a greater extent when the recorded was done in voltage clamp (i.e. soma.v(0)) vs voltage clamp (i.e. SEClamp[0].i)

Why is this?

I would have thought the (albiet partial) voltage control along the cable would mean that voltage clamp would reduce capacitive currents (i.e. less C*dv/dt), and therefore would retain more longitudial current and you would get less attenuation.

P.S. I measuring EPSP/C amplitude, as peak amplitude.
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ted
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Re: Differential Attenutation of EPSPs in VC vs IC

Post by ted »

measuring EPSP/C amplitude, as peak amplitude.
Peak amplitude of what?
Bill Connelly
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Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University

Re: Differential Attenutation of EPSPs in VC vs IC

Post by Bill Connelly »

Peak amplitude of the EPSC when in voltage clamp (basal holding current to maximum deflection in current), peak amplitude of the EPSP when in current clamp (resting membrane potential to maximum deflection in membrane potential).


Though I think I understand the cause of this now. The voltage clamp essentially acts like a huge conductance with E = Vcmd. And a leaky cable attenuates more.
ted
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Re: Differential Attenutation of EPSPs in VC vs IC

Post by ted »

Here's a thought experiment for you:

Imagine a sprinkler hose, with evenly spaced perforations along its length that allow little upward-pointing jets of water to escape. The "proximal" end is attached to a faucet, and the "distal" end is capped. Turn on the faucet ("clamp the proximal end to a high pressure") and observe how high the little jets are. Pressure at the proximal end being highest, the jet at that end is also highest, and both pressure and the height of the jets decrease progressively with increasing distance.

Now unscrew the cap at the distal end ("clamp the distal end to a pressure of 0"). What happens to the height of the jets?

This physical situation is mathematically equivalent to what you were doing with a model of a leaky cable. But unlike with the cable, you already had an intuitive insight into what would happen with the sprinkler hose, largely because of prior experience with similar physical situations. Familiarity breeds (hopefully) correct intuitions.
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