On Tue, Oct 20, 2015 at 2:33 AM, wliu <wliu.itcs.ecnu.edu.cn> wrote:
> Dear all,
>
> I have used the energy decomposition scheme in mm_pbsa to calculate
> the per-residue binding free energy contribution of a complex. But it
> seems that one PHE residue in the pocket has unreasonable polar
> solvation contribution. So I want to know the formulas of per-residue
> (idecomp=1) and pairwise-residue (idecomp=2) free energy decomposition
> used in mm_pbsa, then I can understand these two methods better. Thanks
> a lot.
>
For GB, it is the same formula used to calculate the entire energy. Only
for idecomp=1, the energies are accumulated in each residue only when one
atom in the interaction is part of that residue (each interaction gives
half of its value to the residue that contains atom 1 and the other half to
the residue that contains atom 2).
For idecomp=2, the pairwise interaction is assigned to a residue only when
one atom is part of one of the residues in that pair and the other atom is
part of the other residue.
For PB decomposition, I have no idea how it's done. The only way I can
*think* of would require multiple PB calculations for each printed residue
-- you would need to do a full PB calculation, then a separate PB
calculation *for each residue* (setting that residue's charges to zero) --
then the per-residue decomposed PB values would be, qualitatively, the
difference between the interactions between ALL residues minus the
interaction between all residues EXCEPT the one of interest. But this
would be massively expensive, as it would require NRES PB calculations.
The pairwise decomposition could be done similarly, although the number of
pairs grows faster than the number of residues, so this would be even
slower.
HTH,
Jason
--
Jason M. Swails
BioMaPS,
Rutgers University
Postdoctoral Researcher
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Received on Tue Oct 20 2015 - 07:00:04 PDT
