Hello Prof Case,
Thank you so much for taking the time to reply.
Yes, of course, I am running sander TI with idecomp set (in my case idecomp =1).
I'm transforming a template ligand bound to a protein to a modeled ligand bound to the same protein with 9 ë windows (for starters). Say I look at the computed per-residue energies at ë = 0.1 and at ë = 0.9. I want to know what the reported energies mean at the specified ë values and how I can compare them, say for residue X at ë = 0.1 and at ë = 0.9.
As I said, I'm doing decomposition with idecomp = 1 and only the ligands have any significant internal energy contributions and I was wondering why that is.
A snippet of results:
...
resid internal vdw eel
TDC 129 -0.000006890 0.000233396 -0.024933102
TDC 130 -0.000004136 0.000222831 -0.072463416
TDC 131 0.000002817 0.000331271 0.038252823
TDC 132 -0.000004575 0.000238998 0.095537392
TDC 133 -0.000007337 0.000272501 0.121986048
TDC 134 -0.000006254 0.000278103 0.004401696
TDC 135 -0.000002067 0.000355450 0.159351534
TDC 136 -0.000000870 0.000120816 0.192155232
TDC 137 0.000002848 0.000073761 0.076219228
TDC 138 -29.988820451 3.635571771 40.819828714
Residue 138 is the ligand.
A sample input file:
TI simulation
&cntrl
imin = 0, nstlim = 1000000, irest = 1, ntx = 5, dt = 0.001, cut = 8.0,
ntt = 3, temp0 = 300.0, gamma_ln = 2.0, ig = -1, idecomp = 1,
ntc = 1, ntf = 1,
ntb = 2,
ntp = 1, pres0 = 1.0, taup = 2.0,
ioutfm = 1, iwrap = 1,
ntwe = 1000, ntwx = 1000, ntpr = 1000, ntwr = 1000,
icfe = 1, clambda = 0.1, scalpha = 0.5, scbeta = 12.0,
ifmbar = 1, bar_intervall = 1000, bar_l_min = 0.05, bar_l_max = 0.95, logdvdl=1,
bar_l_incr = 0.05,
ifsc=1,
crgmask='',
scmask=':N3C.C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C11,C12,C13,N14,Cl1,H2,H3,H5,H6,H8,H9,H10,H11,H14,H131,H132',
/
Residues considered for decomp
RRES 1 138
END
Residues to print
RES 1 138
END
END
/
The other TI input files are analagous for the rest of the lambda values and for the partner molecule.
Is there a paper that explains how decomposition is implemented in Amber? If not, how can I find out? Is there a flag that outputs decompositions for a specified set of atoms (per-atom), rather than for a set of residues?
Sincerely,
Stefan Ivanov
________________________________________
From: David A Case [david.case.rutgers.edu]
Sent: Friday, January 20, 2017 5:25 PM
To: AMBER Mailing List
Subject: Re: [AMBER] Interpreting energy decomposition in Amber thermodynamic integration
On Fri, Jan 20, 2017, Stefan Ivanov wrote:
>
> I would like to ask a few questions relating to the energy
> decomposition in TI with Amber.
Just to be clear: are you running sander/TI with the idecomp flag set?
A couple of other points:
1. Any single TI energy doesn't mean much: you always have to construct
thermodynamic cycles and compare two different legs of the cycle in order
to extract meaningful values; this is the case for both the total energy and
for its decomposition. But it wasn't clear to me exactly which comparisons
you are looking at (i.e. which cycle you are analyzing.)
2. The total free energy change obtained in the TI calculation is indepedent
of the path used through lambda-space (assuming perfect sampling); but the
decomposition is not independent of path. So, interpreting decomposition
results only makes good sense if you know you are using a "same path"
(whatever that means) in various circumstances.
> I am performing TI, transforming one strong protein binder to
> another. All protein internal energies are 0, whereas the internal
> energy for the template ligand and modeled ligand is strongly
> negative. Why are internal energies non-zero only for the ligands?
This results sounds wrong, but I'm not sure exactly what calculation you are
reporting the results of.
>
> In the .out files, are energy values relative or absolute? I.e., if the
> vdw value for a given residue at ë = 0.9 is negative and positive at
> ë = 0.1, are those values absolute or relative? Are they in kJ/mol or
> kcal/mol?
I'm unsure of your definition of the terms "relative" and "absolute" here.
All Amber energies are in kcal/mol.
Maybe others on the list can give better answers here: my experience with TI
decomposition is pretty limited, for reason 2 above.
...dac
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Received on Fri Jan 20 2017 - 10:00:02 PST
