Re: [AMBER] Issue with intramolecular pairwise-residue energy decomposition

From: Jason Swails <>
Date: Mon, 9 Sep 2013 09:32:45 -0400

On Mon, Sep 9, 2013 at 3:33 AM, <> wrote:

> Dear Amber users,
> I am trying to calculate the intramolecular pairwise-residue energy
> decomposition for a small protein. The protein is an apo monomer (no
> ligand). I have tried the method as explained in the manual.
> Since the system is not a complex, I used a single monomer for the complex
> file with the command:
> -O -i -cp -y receptor.crd
> In the output, the total energy for Res# n - Res# n is not null between
> identical residues as I would be expected.

Why would you expect this? This value would not even necessarily be 0 for
a binding calculation. [1] Two atoms within the same residue interact
with each other just like two atoms in different residues. (Bonds,
angles, dihedrals, and non-bonded interactions all exist between atoms in
the same residue). These energy contributions are assigned to that residue
interacting with itself.

I tried also to create a 'complex' pdb by putting the same protein as both
> receptor and ligand (superimposed) with exactly the same coordinates for
> all the atoms.

Don't do this. The complex now has an infinite energy, since it has many
atoms all on top of one another (with the same charge, no less).

> So I have 3 top files - (2xMonomer),
> (Monomer), (Monomer). In this case, the calculation exits with the
> following error:
> "Beginning GB calculations with /opt/amber12/bin/sander
> calculating complex contribution...
> CalcError: /opt/amber12/bin/sander failed with prmtop! "

Sander is dividing by zero many, many times here. I would not expect this
to succeed.

> So I would like to know if it is possible to calculate, using,
> the intramolecular pairwise-residue energy decomposition for each
> residue-pair for a protein which is not complexed with any ligand or
> another protein.

Yes, you already did it.


[1] The fact that a single residue interacting with itself does not cancel
completely in a binding free energy calculation comes from the fact that
the effective GB radius is calculated via a coulomb field integral over
nearby atoms. For a residue close to the active site of an enzyme, the
effective radii in the presence of the ligand will be larger than the radii
in the unbound state, thereby changing the polar solvation free energy in
that residue's interaction with itself. This is why I have repeatedly said
that GB is not pairwise decomposable

Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Mon Sep 09 2013 - 07:00:03 PDT
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