Re: [AMBER] MMPBSA per-residue decomposition: how is the polar solvation contribution calculated?

From: Ray Luo, Ph.D. <ray.luo.uci.edu>
Date: Wed, 25 Jul 2012 10:30:36 -0700

As far as decomposition is concerned, a reasonable way is to view EPB in
two parts, self/desolvation energy and interaction/pairwise energy. This is
the same as GB.

Here, the self energy is the EPB when charging only the residue of interest
while all other residues remain as neutral; and the interaction energy is
the charging energy of the residue in the field of all other charged
residues. Apparently the procedure outlined here involves two numerical
solutions of the PB equation with differently charged state of the same
molecule for EACH residue of interest.

While this being said, the Amber PBSA program has been tailored/optimized
for general uses so it is very hard to perform these operations as an end
user.

All the best,
Ray

On Wed, Jul 25, 2012 at 3:07 AM, Jan-Philip Gehrcke <jgehrcke.googlemail.com
> wrote:

> Hello,
>
> I do MMPBSA per-residue decomposition for a receptor-ligand system (with
> MMPBSA.py from AT 1.5) and am wondering how the numbers in the Polar
> Solvation column in FINAL_DECOMP_MMPBSA.dat have actually been built.
>
> I understand the PB part of PBSA for a system comprised of a low
> dielectric volume containing some charge distribution placed in a high
> dielectric medium. Now, what is done to consider the contribution of
> only a single residue of the system?
>
> Considering the contribution of residue R: is the volume of the low
> dielectric medium still defined by all atoms (i.e. entire receptor or
> ligand or receptor+ligand) while the atomic charges of all residues
> except for R are considered to be zero?
>
> I would be happy to get some insights!
>
> Cheers,
>
> Jan-Philip
>
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Received on Wed Jul 25 2012 - 11:00:02 PDT
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