Dear AMBER users,
For my protein-drug system: we are doing the free energy perturbations in a
transformation of ILE ---> VAL using the TI method. We used 14 windows of
clambda value as 0.05, 0.15, 0.25........0.85 and 0.90, 0.92.....0.98 with
klambda = 6, as per the AMBER tutorial for tolune-->nothing. But
strange thing in the result is that, in the last windows of clambda value
0.90 to 0.98 we are getting the DV/DL almost zero difference. The
different clambda values and there respective DV/DL values we found are as
follows:
clambda DV/DL
0.05 50.4736
0.15 25.9430
0.25 12.2901
0.35 5.2432
0.45 1.6762
0.55 0.5597
0.65 0.1154
0.75 0.0253
0.85 0.0020
0.90 0.0002
0.92 0.0001
0.94 0.0000
0.96 0.0000
0.98 0.0000
Here is the input file for the calculation.
Transformation of ILE 2 VAL Equilibration:
&cntrl
ntr=0,
nstlim =20000, nscm=2000, ntave=5000,
ntx=1, irest=0, ntb=2, ntpr=100, tempi=300.0, ig=974651,
ntp=1, taup=1.0,
dt=0.001, nrespa=1,
ntt=1, temp0=300., tautp=2.0,
ntc=2, ntf=2, tol=0.000001,
ntwr = 10000, ntwx=0,
icfe=1, clambda=0.96, klambda=6,
cut=9.0,
&end
Transformation of ILE 2 VAL production phase
&cntrl
ntr=0,
nstlim =100000, nscm=2000, ntave=5000,
ntx=7, irest=1, ntb=1, ntpr=100,
ntp=0, taup=2.0,
dt=0.001, nrespa=2,
ntt=0, temp0 = 300., tautp=2.0,
ntc=2, ntf=2, tol=0.000001,
ntwr = 10000, ntwx=0,
icfe=1, clambda=0.96, klambda=6,
cut=9.0,
&end
So can anybody please point out the mistake and suggest to overcome it. Thanks
for your suggestions.
Regards
Biswa Ranjan Meher
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Received on Sun Sep 24 2006 - 06:07:18 PDT
