# Re: [AMBER] How to calculate the energy contributed by hydrogen bonds?

From: bin wang <wang_p_z.yahoo.com.cn>
Date: Thu, 22 Mar 2012 11:58:22 +0800 (CST)

Thank you so much for your help.  I really appreciate it.

________________________________
From: Jason Swails <jason.swails.gmail.com>
To: bin wang <wang_p_z.yahoo.com.cn>; AMBER Mailing List <amber.ambermd.org>
Sent: Tuesday, March 20, 2012 1:01 PM
Subject: Re: [AMBER] How to calculate the energy contributed by hydrogen bonds?

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 Wed Mar 21 2012 - 21:00:03 PDT
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