# Re: AMBER: equilibrium constants via MMPBSA

From: Julien Michel <j.michel.soton.ac.uk>
Date: Fri, 01 Jun 2007 17:48:55 +0100

Hi Sean,

The correct way to write Keq is
Keq = ( |AB|/|C0| ) / [ ( |A|/|C0| ) * ( |B|/|C0| ) ]
which you can also rewrite as
Keq = ( |AB| / (|A|*|B|) ) * ( ( |C0|*|C0| ) / |C0| )
and so Keq in 2 and 3 are unitless once the standard state is considered.

In an absolute binding free energy calculations using thermodynamic
integration or related methods you have to correct the computed free
energy of "decoupling" your ligand from a binding site by a term that is
proportional to the ratio of the volume your decoupled ligand could
occupy in the simulation to the standard state volume.

See for instance Boresch JPCB, 2003, 107, 9535-9551 or Gilson Biophys J
1997, 72, 1047-1069.

It is not clear to me at which stage a standard state correction would
occur in a MMPBSA calculation. That should be less of an issue if you
are interested in ranking ligands or differences in equilibrium
constants rather than their absolute values.

Julien

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
>

```--
Julien Michel
http://www.julienmichel.net
Dr J. W. Essex research group
University of Southampton
United Kingdom
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Received on Sun Jun 03 2007 - 06:07:38 PDT
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