Re: AMBER: equilibrium constants via MMPBSA

From: David Mobley <>
Date: Mon, 4 Jun 2007 06:11:25 -0700


Julien's comment is right on track. People usually try and use MM/PBSA
for relative affinities only (so the standard state will not be an
issue, since it will be the same for all compounds). Usually MM/PBSA
'binding free energies' have, at best, some large offset relative to
the actual binding free energies, so the best you can hope for is that
the slope (the relative free energies) is roughly correct. If you want
absolute binding free energies you need to go a different route.


On 6/1/07, Sean Rathlef <> wrote:
> Dear Amber,
> A quick question regarding Keq computations via MMPBSA. I have an
> association reaction (A + B --> AB). I am modeling the free energies for
> AB, A, and B, and generating my deltaG from:
> 1 delG(AB) - [delG(A) + delgG(B)].
> Keq is then computed via:
> 2 Keq = e^(-delG/RT).
> In this instance, Keq will have units of M-1, because of the relationship to
> the concetrations of the species, i.e.,
> 3 Keq = [AB] / [A]*[B]
> The Keq leveraged from equation 2 however is unitless, so would it be
> appropriate to simply divide this value by standard state concentraion, C_0,
> of 1M?
> Thanks for any responses,
> Sean
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Received on Wed Jun 06 2007 - 06:07:17 PDT
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