Re: [AMBER] Some common doubts about decomposition

From: Jason Swails <>
Date: Mon, 26 Sep 2011 20:45:24 -0400

On Mon, Sep 26, 2011 at 3:24 PM, Rajesh Raju <
> wrote:

> Dear AMBER users,
> I have some common doubt about the MMGB/PBSA decomposition methods.
> [1] Wen I did the free energy decomposition for pairs resid 1 and
> resid 2 similar values but slightly different values. for example: 1
> and 2 the total value is -6.173 and for 2 and 1 the value is -6.154.
> The small difference comes from the term non-polar solvation. All
> other values are exactly the same.

This is a known issue. There are existing issues with the gbsa=2 surface
area method used for decomposition. You will see other issues if you look
through the history of this mailing list carefully. There is no explanation
or fix as far as I know.

> [2] The previous peril scrpit will divide the total decomposition
> between the pair. Is that the same procedue follow here in

If any splitting is done, it's done in the sander code (at least for I don't know if the perl version does any additional processing
of the sander output (but I don't think it does). I recall doing some short
tests to verify that the perl and python versions give the same results for
the same input.

For example if we do the QM calculation on the residues 1 and 2, the
> calculated interaction energy is ~13 kcal mol-1. So i assume that the
> binding free energy splitted between the two residues..So in order to
> comapre with the QM values we have to double it. If I am wrong pls
> correct me..

It depends on how you calculate this interaction energy. Ideally, energy
differences between "good" methods should be the same or similar (a "good"
method is as close to possible just a translation of the potential energy
surface, but the shape and scale stays the same). If you're comparing
energy differences (as I think you are), then roughly speaking I'd say
you're probably right. Keep in mind, though, that there are a lot of ways
for these numbers to be wrong. (Is your QM method appropriate for the
problem at hand? Your force field? Are you actually calculating the exact
same thing with both hamiltonians? etc.) If I'm mistaken here, I'll
probably be corrected very quickly :)

> [3] How the solvation free energy terms decompose. ? Since solvation
> free energy does not do anything with the other residues, Wat is the
> significance of this decomposition terms?

I'm not quite sure what you're asking here. I do not actually know the
formula for how the non-polar part breaks down in a pairwise manner (nor is
it easy for me to get a picture of what a pairwise SA term would actually
mean), but the polar solvation term is pretty straightforward to decompose
in a pairwise fashion. If you look at the basic GB formula (excluding the
SA part), then the energy is calculated by a double-sum running over atom
pairs. The only reason the polar solvation term isn't strictly
pairwise-decomposable is because the GB radii are actually dependent on
every other atom in the system as well.

> [4] Is there any way to performe intra strand decomposition? I have
> performed decomposition between two modified DNA chains each
> containing 10 residues. When I did decompostion for all the 20
> residues, the decomposition energy values for the 1-2, 1-3, 1-4
> ...1-9, 1-10 are 0.000 for all terms..I thought it would print the
> intrastrad binding free energy terms too..

The code will print residue-residue interactions and that's all. If you
want strand-strand interactions, you have to be clever about what terms you
add up, making sure to count just the ones that contribute to the energy
you're looking for and taking care not to double-count any of the

> [5] The polar solvation decomposition value for the 2-2, 3-3, 4-4
> (dont know wat is the significance of the interaction of the residue
> with itself) prints some + values like +4 kcal mol-1..Wat does it
> mean?

A couple things. First, each residue can be composed of multiple atoms
(ions are a common exception), and all amino acids have at least 2 atoms
that are separated by more than 3 bonds, which means each residue actually
has full non-bonded interaction that's counted entirely in that residue's
interaction with itself. Also, GB has a self-energy term for each atom,
which is more or less the free energy of solvation of an equivalently
charged and sized ion. Thus, each atom's self-term contributes to that
residue's interaction with itself. Finally, when looking at the
differences, keep in mind what I said about GB energies not being pairwise
decomposable due to the radii-dependence on other atoms in the system. For
this reason, you can get a non-zero DELTA (for polar solvation) for a
residue paired with itself if the ligand is large enough to significantly
affect the calculated GB radii of atoms within that residue. (Same with the
non-polar solvation).


> I hope these questions can help others too as I could nt find any
> details from manual....
> Thanks
> Rajesh
> _______________________________________________
> AMBER mailing list

Jason M. Swails
Quantum Theory Project,
University of Florida
Ph.D. Candidate
AMBER mailing list
Received on Mon Sep 26 2011 - 18:00:04 PDT
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