From owner-biophysics@net.bio.net Tue Mar 16 22:00:00 1999 Path: biosci!news.stanford.edu!newsfeed.stanford.edu!logbridge.uoregon.edu!netnews1.nw.verio.net!netnews.nwnet.net!news.ohsu.edu!not-for-mail From: Matt Jones Newsgroups: bionet.biophysics Subject: Re: conductance change Date: 17 Mar 1999 02:41:53 GMT Organization: Vollum Lines: 94 Distribution: world Message-ID: <7cn4ph$cte$1@fremont.ohsu.edu> References: <19990316150516.20725.00000270@ng-fi1.aol.com> NNTP-Posting-Host: 137.53.99.50 Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit X-Newsreader: Nuntius 2.0.4_PPC X-XXDate: Tue, 16 Mar 1999 02:49:36 GMT In article <19990316150516.20725.00000270@ng-fi1.aol.com> Dirac1pt6, dirac1pt6@aol.com writes: >Subject: conductance change >From: Dirac1pt6, dirac1pt6@aol.com >Date: 16 Mar 1999 20:05:16 GMT >>Hi All, > >I'm having trouble with a real basic calculation. I have not done any of this >type of work in a long time, and all my old textbooks seem a bit hazy on the >point, so I was hoping someone in netland could help. > >If a population of ligand gated channel open, and produce a depolarization of >say 10 mV, and the input resistance is 100 MegaOhms, then it stands to reason >by Ohms law, that a 100 pA current must have crossed the membrane. If I then >want to calculate the conductance (g) change necessary to produce this current: > >delta I = delta gx (Vm-Ex) where x indicates the permeant ion, and Ex is the >Nerst equillibrium potential for that ion. > >2 problems: First, this seems pretty circular (calculate current to calculate >conductance). Second it gives some fairly unrealistic numbers (assume >Ek=-90mv, Vm=-70mV , that gives a 5 nS change in conductance which would mean a >200 MegaOhm change in input resistance!) > >I'm sure I am forgetting something obvious, but I've been banging my head >against this one for a while now. > >Thanks in advance for any help. Hmm. You're right. It's circular, and gives bogus numbers. Here's what I think the problem is: The input resistance is a _passive_ property. It's defined around the resting potential, under conditions of teeny little current/voltage excursions that don't activate any additional conductances. It also doesn't include any ligand-gated or otherwise externally stimulated conductances. So it's misleading to base the whole calculation on Rin, because the problem is defined as a voltage change in response to the opening of additional channels (other than the resting conductances). So instead of calculating the current that flows as 100pA, that current is really an unknown (100 pA is what would flow through Rin in the absence of any new conductances, like if you were using an electrode to inject current, which wouldn't change the membrane conductance). When the channels open, the total conductance is really more than the original input conductance. Here's some math (assuming the depolarization has reached steady-state and that there's no change in ionic concentrations) (by the way, I'm not good at math, so I apologize in advance if I screw it all up): Define: dV = 10mV. Gin (input conductance) = 1/Rin = 1/100e6 = 10 nS Gx = the unknown ionic conductance Gtot = the unknown total membrane conductance during the depolarization = Gin + Gx Im = the unknown total membrane current Ohm's Law: dV = Im / Gtot = Im / (Gin + Gx) Rearranging and collecting etc.: Im = dVGin + dVGx; dV and Gin are known, so substituting gives Im = 10 mV * Gx + 100 pA. This is the equation of a straight line (i.e., y = mx + b) that plots current versus the ionic conductance. The Y-intercept is 100 pA, which is the current that would flow for zero ionic conductance (no channels open, and you inject the current through an electrode). The slope is the voltage excursion that you observed (10 mV). So ANY choice of total membrane current and ionic conductance that falls on this line will give you a 10 mV deltaV. In other words, there isn't any unique solution to this problem. On the other hand, if the experiment were done in voltage-clamp, your amplifier would (in an ideal world) determine the voltage change, and you would know exactly what the total membrane current was, because you'd see it directly on your oscilloscope. You'd get the ionic current by performing a "leak subtraction". Then you could read the value of the ionic conductance right off the graph, or by plugging all the known numbers into the equation above. Hope that helps, but if I screwed it all up, let me know what my mistakes were. Cheers, Matt Jones .