Question on 2 stage RESP fit

From: Ambarish Nag <ambarish_at_midway.uchicago.edu>
Date: Thu 12 Dec 2002 00:39:08 -0600 (CST)

I am trying to get the charges for a protonated non terminal Glutamic Acid
residue(the carboxylate group in the side chain is protonated)which is
capped at the N terminus by a acetyl group and at the C terminus by an
amine group.
How do I make the two amide hydrogens of the C-capping amine group
equivalent in the first stage?I do not have AMBER and I am
using the freely downloadable version of resp. Hence I do not have any
access to the example -i input files for the program resp.
 Also how do I impose the forced symmetry during the fit for the methyl
and methylene hydrogens during the second stage of the fit?
 I tried to solve the problem in the following way
I run the resp program two times.
During the first run I set iqopt = 1 (i.e. all initial charges are set to
zero) irstrnt = 1(hyperbolic restraint to charge of 0) and qwt =0.0005
I set the ivary values of all centers to be 0.
I did not impose any charge restraint on the 2 H's of the capping NH2
group
using the 7th and 7.1st section of
the -i input file because whatever I understood by reading the program
resp.f suggests that imposing the constraint in the above manner
corresponds to imposing the forced symmetry during the fit.What I
understood from reading papers on 2 stage resp fit is that the symmetry
for the charges of the 2 H's of the capping NH2 group should be obtained
by averaging
a posteriori to the initial fit.Also,even if I wanted to impose the forced
symmetry of the charges of the 2 NH2 hydrogen atoms I would also have no
idea about the charge to which I should constrain each of these two atoms.
 So in the -i file of the second resp run I set iqopt = 3(read in new
initial charges from -q (qin) and perform avergaing of the charges on the
NH2 hydrogen atoms for which ivary values are set to +n where qin is the
set of charges obtained from the previous resp run)and qwt =0.001.
The charges on all the atoms other than methyl, methylene and methine
groups are frozen by using -n ivary values.
For all the atoms in the methyl and methylene and methine groups I put Ivary =0
 also during the second run.I try to impose the forced symmetry on the
methyl and methylene hydrogenms by using sections 7 and 7.1 of the
 -i input file where I constrain each hydrogen atom of a particular methyl
group to the average charge of the hydrogens of that methyl group from the
previous run.But in the final set of charges obtained the different hydrogens of each
methyl group have different charges inspite of the charge constraint in
the sections 7 and 7.1 of the -i input file.
I shall be highly obliged if you please write to me about how to impose
the different symmetry restrictions in the 2 different runs.
Thanking you,
Ambarish Nag.
Graduate Student
Department of Chemistry
University of Chicago.
Received on Wed Dec 11 2002 - 22:39:08 PST
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