Hi Sean,
let me try this, it's been a while since I practised my freshman
chemistry:
> Just to clarify, say we model the reaction E + A --> EA as single molecular
> species in the standard state (everything is at 1M concentration).
>
> 2. We obtain a dG^0 of -7kcal/mol from 1M standard state terms.
>
> So, as David says in the above exerpt, if we wanted to compute the free
> energy change for any other concentrations, how would we do this? Would
> sombody be willing to provide a simple (hypothetical) example? Say our
> ligand is 0.1M and our receptor is 0.01M ...
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.
Kind Regards,
Thomas
Dr. Thomas Steinbrecher
The Scripps Research Institute
10550 N. Torrey Pines Rd.
San Diego CA 92037, USA
-----------------------------------------------------------------------
The AMBER Mail Reflector
To post, send mail to amber.scripps.edu
To unsubscribe, send "unsubscribe amber" to majordomo.scripps.edu
Received on Wed Oct 25 2006 - 06:07:30 PDT
