On Tuesday, October 24, Thomas Steinbrecher wrote:
>
> The free energy change is given by
>
> dG = dG^0 + RT ln K
>
> with K = ( [EI] / [E]*[I] )
>
> The dG^0 stays the same of course, but if you mix ligand and receptor at
> the indicated concentrations, an equilibrium would form in which dG is 0,
> thus
>
> exp(-dG^0/RT) = K ~= 125799
>
> meaning equilibrium lies on the far right side of E + I -> EI.
> This would lead to changing concentrations from the E0 and I0 you
> mentioned by x, forming EI at concentration x.
>
> simple math leads to an x of ca. 0.00999912, meaning that final
> concentrations of:
>
> [EI] = 0.009999 M
> [E] = 8.8*10^-7 M
> [I] = 0.09 M
>
> are obtained.
Thanks Thomas this is very helpful. I used R = 8.3144, cal2J = *4.184, and
T = 298, and got a Keq of 136114 instead of the 125799, but the rest makes
clear sense.
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Received on Wed Oct 25 2006 - 06:07:31 PDT
