Free energy calculation using MM-GBSA with NAMD

From: Sneha Menon (sneha.menon15_at_gmail.com)
Date: Wed Jun 15 2016 - 05:50:41 CDT

Dear NAMD community,

We are trying to calculate the solvation free energy as well as binding
free energy, of a protein-carbon nanotube (CNT) system using MM-GBSA
(single trajectory) method.

We followed a post in NAMD mailing list regarding a similar system, link
given below:
http://www.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2014-2015/2726.html

It is mentioned in the post that the potential energy obtained from the
MM-GBSA log file is the GBSA free energy. As we know, the free energy of a
system is given by:

*∆G = ∆E(internal) + ∆E(vdw) + ∆E(ele) + ∆G(solv)*

So if we consider the potential energy as the GBSA free energy; *∆G *will
be equal to* [ ∆E(internal) + ∆E(vdw) + ∆E(ele)]*. Hence, *∆G(solv)* will
become zero. Is our understanding wrong here?

On a different note, for our protein-CNT system, MMGBSA calculation for
only the CNT does not yield zero for the electrostatic contribution
*∆E(ele)* (CNT does not carry any charge). What could be the reason?

Our input file for only CNT system is as follows:

coordinates nt.pdb
structure nt.psf
paraTypeCharmm on
parameters par_all22_prot_cmap.inp
outputname gb_nt
numsteps 0

GBIS on
solventDielectric 74.69
ionConcentration 0.3
alphaCutoff 14
switching on
switchdist 15
cutoff 16
pairlistdist 18
sasa on
surfaceTension 0.0072
timestep 1
nonbondedFreq 2
fullElectFrequency 4
exclude scaled 1-4
temperature 310

set ts 0

coorfile open ____.dcd
while { ![coorfile read] } {
firstTimestep \$ts
run 0
incr ts 1
}
coorfile close

Any help would be appreciated.

Regards,
Sneha

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