Re: [AMBER] decomposition energy

From: Jason Swails <>
Date: Thu, 19 Nov 2015 12:16:47 -0500

On Thu, Nov 19, 2015 at 11:32 AM, Fabian gmail <>

> Hi
> I am calculated the decomposition energy of a ligand, which gives for
> example the following LYS energies (LIGAND)
> LIGAND TOTAL Non-Polar Solv. Polar Solvation Electrostatic van der
> Waals
> L GLU 2 -0.83 -0.24 324.47 -324.93 -0.13
> L LYS 9 1.35 -0.05 -327.77 330.10 -0.94
> L LEU 7 -3.72 -0.46 2.29 -1.90 -3.65
> I have a couple of questions:
> 1) I don’t really understand why the electrostatics and polar solvation
> energies in teh case of LYS and GLU almost neutralize each other, so that
> the contribution of a charge residue is almost non-existent, why is so??
> They do form hydrogen bonds with water or the other receptor residues.

​That's the effect of solvation. Water will interact strongly with a
charged residue, since water is quite polar. This is the "polar solvation
free energy" term. You can also think about the solvent "screening"
electrostatic interactions (that's what a dielectric does, and PB/GB treat
the solvent like a dielectric medium). So the polar solvation term will
counteract the electrostatic interactions -- it will stabilize unfavorable
interactions (this is why highly negatively charged DNA strands form a
stable duplex rather than pushing each other away due to electrostatic
repulsion). It will also reduce the attraction between oppositely charged
groups (this is why zwitterions of amino acids are stable in solution while
the neutral form is preferred in the gas phase).

2) Looking at the total ligand decomposition the total dg on a different
> case comes almost exclusively from the vDWaals term, is that normal?



Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Thu Nov 19 2015 - 09:30:05 PST
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