Dear AMBER,
I have been trying to test a new RISM based approach for solvation free energies by using
thermodynamic integration. Being allowed to use AMBER10 on my problem has helped a lot because
I now no longer have the problem with enormous errors at the lambda endpoints. Dr. Steinbrecher
has explained to me how to divide the perturbation into three sections, removing partial charges,
changing the other partial charges and disappearing/appearing atoms, amd finally adding new
partial charges. However I am still doubtful about the values that I am calculating. These differ a lot
from the MM-PBSA values and the ones I'm testing. For example I am comparing substituting a
methyl group on an oxygen for a hydrogen, and the charging energy (third step) is much more
negative (by 24.8kcal/mol) in the gas phase than in the solution. At first this seemed
counterintuitive to me as I expected that turning on charges would favor solvation. However since in
turning back on the charges on part of the molecule it goes from an overall charged state to a
neutral state, the charging part of the solvation energy is positive. However this should be
compensated for somehow in the second step by the difference in charging energies for the other
atoms and this step only gives about -16 in favor of the methyl to hydrogen change. I should
mention that the first step of taking the charge off of the methyl group gives about -6.5kcal/mol, so
that the total change is about positive 2.8kcal/mol. MM-PBSA on the other hand gives a solvation
energy of about -7kcal/mol. I would expect the molecule with the OH to be more solvated than the
one with OCH3, as predicted by MMPBSA but not by TI.
I am pasting the charging input files below; I used a three step integration on the first and third
steps mentioned above, and a nine step integration on the second step. I would appreciate any
suggestions on this. Prof. Case, Dr. Steinbrecher, or Dr. Mobley, I would especially like to get your
advice, as you have been very helpful to me in the past with thermodynamic integration.
Thanks in advance,
Holly
--
Department of Physics, University of Alberta
Edmonton CANADA
Without charge in water:
&cntrl
imin = 0,
ntx = 5, irest = 1, ntr = 0, iwrap=1,
ntpr = 100, ntwr = 100, ntwx = 5000, ntwe = 5000,
ntf = 2, ntb = 2, cut = 10,
nstlim = 500000, dt = .002, ntc = 2,
tempi = 298.0, temp0 = 298.0, ntt = 3, gamma_ln = 2,
tautp = 1.0, ntp = 1,
icfe = 1, clambda = 0.11270, klambda = 1,
ifsc = 0,
crgmask = ':1.H10', scmask = '', noshakemask = ':1.H10',
&end
With charge in water:
&cntrl
imin = 0,
ntx = 5, irest = 1, ntr = 0, iwrap=1,
ntpr = 100, ntwr = 100, ntwx = 5000, ntwe = 5000,
ntf = 2, ntb = 2, cut = 10,
nstlim = 500000, dt = .002, ntc = 2,
tempi = 298.0, temp0 = 298.0, ntt = 3, gamma_ln = 2,
tautp = 1.0, ntp = 1,
icfe = 1, clambda = 0.11270, klambda = 1,
ifsc = 0,
crgmask = '',
scmask = '',
noshakemask = ':1.H10',
&end
Without charge in vacuum:
&cntrl
imin = 0,
ntx = 5, irest = 1, ntr = 0, iwrap=0,
ntpr = 100, ntwr = 100, ntwx = 5000, ntwe = 5000,
ntf = 2, ntb = 0, cut = 10,
nstlim = 100000, dt = .002, ntc = 2,
tempi = 298.0, temp0 = 298.0, ntt = 3, gamma_ln = 2,
tautp = 1.0, ntp = 0,
icfe = 1, clambda = 0.11270, klambda = 1,
ifsc = 0,
crgmask = ':1.H10', scmask = '', noshakemask = ':1.H10',
&end
With charge in vacuum:
&cntrl
imin = 0,
ntx = 5, irest = 1, ntr = 0, iwrap=0,
ntpr = 100, ntwr = 100, ntwx = 5000, ntwe = 5000,
ntf = 2, ntb = 0, cut = 10,
nstlim = 100000, dt = .002, ntc = 2,
tempi = 298.0, temp0 = 298.0, ntt = 3, gamma_ln = 2,
tautp = 1.0, ntp = 0,
icfe = 1, clambda = 0.11270, klambda = 1,
ifsc = 0,
crgmask = '',
scmask = '',
noshakemask = ':1.H10',
&end
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Received on Wed Oct 24 2007 - 06:07:34 PDT
