Re: [AMBER] Query about residue decomposition energy

From: Jason Swails <>
Date: Fri, 22 Nov 2013 10:17:32 -0500

On Fri, 2013-11-22 at 20:11 +0530, Kshatresh Dutta Dubey wrote:
> Dear Jason,
> Thanks for such detail explanation. I have still a query; in
> biomolecular systems, hydrogen bonds and salt bridges are assumed to
> be the strongest non covalent interactions whose value are still less
> than 5-6 kcal/mol.

This is an indirect measurement. There is no way to experimentally
measure a particular interaction and the way in which these strengths
were estimated experimentally is unclear.

> What will be the interpretation of this energy (~12 kcal/mol)?
> Another point; I think that your given example (-332 kcal/mol) is for
> two monovalent ions, not for the present case (two charged amino
> acids).

The example was just provided to give a (rough) estimate. At large
distances of course the monopole term dominates and charged amino acids
can be treated as ions with respect to each other.

> My issue is that, -12 kcal/mol looks fine if it is electrostatic
> interaction only, but for free energy between two residues it still
> seems very large (we have already compensated electrostatic energy
> with polar solvation term).

Here I have to disagree. Solvation effects are inherently non-local.
Residue A affects the 'solvated' interaction between residues B and C;
residue B affects the 'solvated' interaction between residues A and C;
residue C affects the 'solvated' interaction between residues A and B.

These pairwise energies, however, capture only part of the solvation
energy, and there's no way to rigorously quantify 'how much' is missed.
This all comes back to the assertion that GB energies are _not_ pairwise
decomposable. The only thing that these pairwise energies are useful
for is comparison to each other. For instance, if you make a mutation
will it form a stabilizing interaction with a bound ligand? Will it
destabilize it? Will another mutation be more or less stabilizing?
Even here you need to be careful though, since energy is an extensive
property and simply inserting a bigger residue will make the energy
contribution appear larger for no other reason than there are more


Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Fri Nov 22 2013 - 07:30:02 PST
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