equilibrium fraction of channels that are open

Dec 15, 2015
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I found the formula for the fraction of channels open at equilibrium for a three-state mechanism (Colquhoun's paper in the microelectrode book http://www.utdallas.edu/~tres/microelectrode/me.html).

A+R <---------> AR<------------>AR* .

k1 and beta are forward rate constants. k-1 and alpha are backward rate constants.

The formula is p1(equilibrium) = (Xa/Ka)(beta/alpha) / (1+ ((Xa/Ka)(1+beta/alpha)))

Can someone show me how to arrive at this by derivation. I tried and I got it wrong (obviously!!!).

Xa is the concentration of the agonist and Ka is the equilibrium constant k1/k-1