Re: [AMBER] TI problem

From: Fabrício Bracht <>
Date: Fri, 12 Oct 2012 10:12:53 -0300

But what about PME. Won't sander complain that the system charge is not zero?
Thank you

2012/10/11 David A Case <>:
> On Thu, Oct 11, 2012, Fabrício Bracht wrote:
>> I am trying to run a TI simulation, mutating a zinc bound water
>> molecule (H-O-H) into a hydroxyl group (O-H). I have set up the
>> simulation by following the instructions on the amber tutorial A9.
>> Once I set the two parameter files for the HOH and the OH groups and
>> run the mpirun -np $NP sander.MPI -O -ng 2 groupfile I get the
>> following error.
>> SANDER BOMB in subroutine setup_sc
>> All non-softcore atoms must be identical in both systems
>> Well, the number of atoms of both systems is exactly the same. But the
>> atoms themselves are not exactly equal, since the system with the HOH
>> group has "n" charge and the OH has "n-1". Being so, one system has
>> "n" sodium atoms and the other has "n-1" (so that the system charge
>> remains 0). The number of water molecules in both systems are the same
>> and the rest of the protein (non softcore part of the protein) is also
>> identical. Knowing this, how am I supposed to make both system's
>> non-softcore atoms identical?
> Either get rid of the extra sodium ion, or include it in the softcore region.
> There is no good reason to require that the charge be 0 in both states.
> Remember that you will be doing a parallel transformation for the other side
> of a thermodynamic cycle in order to obtain a number that has real physical
> reality. As long as you do things the same way in both calculations, things
> should cancel.
> ....dac
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Received on Fri Oct 12 2012 - 06:30:03 PDT
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