Re: AMBER: FEP and charges

From: Chunhu Tan <tanc.uci.edu>
Date: Thu, 4 Nov 2004 14:23:16 -0800

Hi, Annette,

      Here is some experience of mine in calculating the solvation free
energy and I hope it is something helpful.
      1. When there are atoms disappeared in the process, you should first
decouple the Coulombic contribution and then the van der Waals. When I did
this, I prepared two sets of parm files. In the first one, atoms' pert.type
were not changed, only the delta.charge were set to the minus atom's charge
(in Amber8), which means in the finial state, all the atoms have zero
charge. With this parm file, after the minimization and equilibration, you
can run the SanderTI with Dr. Case's script. After that, pick out the final
results of DV/DL in the out.o file of every lambda and do the integration by
hand. But keep in mind, this energy includes two parts: the Coulombic
interaction between the solute and solvent and the intra-molecular
electrostatic contribution. Usually, we need the first part, so, you need to
repeat all the process in vacuum, the result of which is the second part.
The difference between these two results, in water and vacuum, is the
solvation free energy (Coulomibc part) of the things disappeared. (This is
corresponds with the INTPRT=0 in Gibbs.) For example, the results of ASN
analog , 76.0759(vacuum)-87.2608(water)=-11.1849, which is comparable with
the Gibbs result, -11.2084.
      2. Van der Waals part. When you prepare the parm file, you should set
the atom's charge and pert.charge all to zero. The pert.type is set to dummy
atom which you defined in frcmod file, in which the foce constant and(or)
rmin of dummy shoud be zero. (I think this is the case of the test file of
SanderTI). But so far I only get the vdW results with Gibbs TI. Also, for
ASN analog, it is 1.7684.
      3. Following is the input files of Gibbs for these two parts.
Coulombic:
&cntrl
   irest = 0, ntx = 7, init = 4,
   ntt = 1,
   ntb = 2, ntp = 1, temp0 = 298.0,
   ntf = 2, ntc = 2, tol = 0.00001,
   scee = 1.2,
   nstlim = -1, dt = 0.001,
   ielper = 1,
   idifrg = 1, intprt = 0,
   almda = 1, isldyn = -3, almdel = 0.01,
   nstmeq = 40000, nstmul =40000,
   ntpr = 40000, isande = 1, ntwx = -1,
   cut = 9.
 &end

vdW:
 &cntrl
   irest = 0, ntx = 7, init = 4,
   ntt = 1,
   ntb = 2, ntp = 1, temp0 = 298.0,
   ntf = 2, ntc = 2, tol = 0.00001,
   scee = 1.2,
   nstlim = -1, dt = 0.001,
   ielper = -1, idsx0 = 200,
   idifrg = 1, intprt = 0,
   almda = 1, isldyn = -3, almdel = 0.01,
   nstmeq = 40000, nstmul =40000,
   ntpr = 40000, isande = 1, ntwx = -1,
   cut = 9.
 &end


Regards,
Chunhu



----- Original Message -----
From: "Annette Höglund" <hoeglund.informatik.uni-tuebingen.de>
To: <amber.scripps.edu>
Sent: Wednesday, November 03, 2004 3:59 AM
Subject: AMBER: FEP and charges


> Ray and other interested,
>
> My system contains a protein bound a to a ligand solvated in water.
> The protein is about 100 aa res. The watbox is 8 A layer containing about
> 20 000 WATs.
>
> FEP & Charges:
> One more question: when perturbing (using dielectric coupling) a charged
> side chain to an uncharged e.g. ASP to ALA and doing this for a complexed
> and uncomplexed state (in order to make the thermodynamic cycle complete)
> my ddG for the FEP is about 130 kcal/mol. (a difference of a few kcal/mol
> would be expected). The unpert system is neutral, the perturbed system is
> not. Can someone tell me how I can "account" for the charge in the
> conversion step?
>
> Thanks in advance!
> /Annette
>
>


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