On Tue, Mar 20, 2012 at 10:17 AM, bin wang <wang_p_z.yahoo.com.cn> wrote:
> Thank you very much.
>
> So can we say the "Electrostatic" term in the energy decomposition include
> the contribution from hydrogen bonds? I am just thinking maybe we can use
> the energy decomposition to see which residues or pairs have hydrogen bonds
> and contribute most to the total binding energy. If this is not the right
> way to do it, how should I explain the decomposition results?
>
There's no straightforward way to decouple the "hydrogen bond" part of the
nonbonded potentials from the "other" part of the nonbonded potentials.
Furthermore, what qualifies as a "hydrogen bond" depends on your definition
(i.e. the constraints that you put on what qualifies as a H-bond).
BTW, I found that the sum of total energy of each residue is not equal to
> the energy change of the "DELTA G binding" without decomposition. Could you
> let me know if the decomposition is a weighted summary or some special
> algorithm is used here?
>
It depends on what type of decomposition you do (idecomp=1/2?
idecomp=3/4?). For per-residue decomposition, I think that any interaction
that involves at least one atom from that residue is counted toward that
residue's decomposed energy contribution. Therefore, if 2 atoms are in
different residues, their pairwise nonbonded terms are added to both
residues. Therefore, a simple sum of decomposed energies will double-count
a large number of the pairwise terms (to be precise, it will double-count
all, and only, interactions between two atoms in different residues).
As a result, a simple sum of decomposed energies will not equal your total
energy. Nor will it equal double your energy, since the terms between
atoms in the same residue will only be counted once (I think -- I'm not
sure about this last part -- someone with more knowledge of that code would
have to comment).
HTH,
Jason
--
Jason M. Swails
Quantum Theory Project,
University of Florida
Ph.D. Candidate
352-392-4032
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Received on Tue Mar 20 2012 - 10:30:03 PDT
