Input resistance

Matt Jones jonesmat at physiology.wisc.edu
Tue Aug 17 14:36:26 EST 2004


BilZ0r <BilZ0r at TAKETHISOUThotmail.com> wrote in message news:<Xns9548CCC6B2C2BBilZ0rhotmailcom at 202.20.93.13>...
> I can't find the answer to this anywhere, I've been googling for an hour or 
> so...
> 
> In the sentance "5-HT bath applied on the hippocampal neurons caused a 
> dose-dependent hyperpolarisation and a drop in input resistance"; what do 
> they mean by input resistance? And what are the physiological correlates of 
> it?
> 
> I get the feeling its basically means a drop in membrane resistance, and 
> hence less current is needed to alter the cells potential differnce?


I think my last post got cancelled, so here it is again:

Yes, they basically mean membrane resistance. The input resistance is
however much resistance the headstage encounters as it tries to
"input" current into the system. If the system is considered as the
pipette plus the cell, the input resistance will be the sum of the
series resistance (pipette to cell) and the membrane resistance (cell
to ground). Assuming the series resistance is small, this will be
dominated by the membrane resistance. If series resistance is big
compared to the membrane resistance, there are a bunch of measurement
problems that arise (e.g., series resistance errors).

The physiological correlate of a drop in input resistance is the
opening of ion channels. The fact that they get a hyperpolarization,
coupled with a drop in input resistance, means that the ion channels
that are opening have a reversal (or equilibrium) potential negative
to the resting potential. Therefore either chloride or potassium
channels. Probably potassium channels for 5HT3 receptors.

PS: 
> 
> I get the feeling its basically means a drop in membrane resistance, and 
> hence less current is needed to alter the cells potential differnce?

Right idea, but wrong direction. A drop in resistance means that
*more* current is needed to alter the potential. Ohm's law: V = IR. If
R goes down, then V goes down for the same amount of I. Need to make I
bigger to get the same V.



Matt



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