AMBER: FEP and charges

From: Annette Höglund <hoeglund.informatik.uni-tuebingen.de>
Date: Wed, 03 Nov 2004 12:59:16 +0100

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


Ray Luo wrote:

> Annette and He,
>
> Could you let us know a bit more about your systems? How large are
> your systems? Are they solvated in water?
>
> Best,
> Ray
>
>> Dear David,
>>
>> Thanks for the response!
>> Yes, the perturbation file and everything is created ok (I have run
>> the same thing using gibbs FEP and then it works i.e. free energies
>> are calculated).
>> Now I just want to try it out with TI in sander and TI in gibbs.
>>
>> This is the gibbs FEP input that I use:
>> &cntrl
>> dt=0.002,
>> ntx=7,
>> ntpr=100,
>> temp0=300,
>> ntt=5,
>> tautp=2.0,
>> ntb=2,
>> ntp=1,
>> ntc=2,
>> ntf=2,
>> nsnb=5,
>> scee=1.2,
>> scnb=2.0,
>> nstlim=2000,
>> nstmeq=1000,
>> nstmul=1000,
>> intprt=1,
>> nrun = 21,
>> almda = 1.0,
>> almdel = 0.05,
>> isldyn = -3,
>> &end
>>
>> And I would want to have a corresponding TI input (for gibbs and one
>> for sander) for comparison.
>> Thanks in advance!
>> /Annette
>>
>>
>>
>>>>> Dear Amber users,
>>>>> I want to use thermodynamic free energy calculations,
>>>>>
>>>>> I read the manual and try a system like this
>>>>>
>>>>> nstlim=1000,ntpr=100
>>>>> icfe=1,
>>>>> clambda=0.5,
>>>>> klambda=4,
>>>>> ntb=0,
>>>>> temp0 = 300.0, tautp = 1.0,
>>>>> ntf = 2, ntc = 2,
>>>>> dt = 0.002,
>>>>> ntave = 25,
>>>>>
>>>>> But it seems it does not calculate the free energy.
>>>>
>>>>
>
>


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Received on Wed Nov 03 2004 - 12:53:00 PST
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