That was helpful. Thanks, Ray!
On 07/25/2012 07:30 PM, Ray Luo, Ph.D. wrote:
> 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 Thu Jul 26 2012 - 04:00:02 PDT
