Thanks very much for your help!
Wei
On 2015-10-20 21:40, Jason Swails wrote:
> 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
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Received on Tue Oct 27 2015 - 23:30:04 PDT
