[gmx-users] Total potential energy: 1/2 protein-solvent interactions??

Xavier Periole X.Periole at rug.nl
Thu Mar 6 10:38:51 CET 2008


On Thu, 06 Mar 2008 09:42:09 +0100
  David van der Spoel <spoel at xray.bmc.uu.se> wrote:
> Xavier Periole wrote:
>> On Wed, 05 Mar 2008 20:38:59 +0100 (MET)
>>  pascal.baillod at epfl.ch wrote:
>>>
>>> Dear developers,
>>>
>>> I would like to know the exact definition of the total potential 
>>> energy with
>>> respect to protein-solvent interactions, in an explicit solvent 
>>> protein simulation.
>> The definition of protein-solvent interaction is the sum of the
>> pair-wise interaction (non-bonded) involving on one side the protein
>> atoms and on the other side the solvent atoms.
>> 
>>>
>>> For protein atom Pm and solvent atom Wn, there are two interactions:
>>>
>>> Pm to Wn
>>> Wn to Pm
>> Those two terms are identical! You can not separate them! Here you just
>> express them in two different ways which are totally identical.
>> It is like A+B=B+A, can you differentiate the sum of B on A and the
>> sum of A on B?
>>>
>>> Are both interaction energies counted in the total potential energy 
>>> given in the log file or by g_energy?
>> As they are the same they are both counted but only one time!
>>>
>>> If I want to compute the total potential enery of the protein, plus
>>> protein-solvent interactions, should I then only add half of the 
>>> protein-solvent terms given by g_energy? I am only interested in the 
>>> effect of solvent "felt" by the protein, and not in the effect of the 
>>> protein felt by the solvent.
>> Again, how would you differentiate those two terms? Counting
>> the interactions from protein to solvent or solvent to protein is
>> exactly the same.
>>
> 
> Still, if you want to partition the energy over molecules you have to make 
>some kind of division. For instance, if you calculate the potential energy 
>for 216 water molecules you will find that is is roughly -9000 kJ/mol at room 
>T, and hence you can derive the potential energy per molecule to be -42 
>kJ/mol, which agrees with heat of vaporization. 
Agreed.
>If you however would do as 
>you suggest, and take one water molecule and compute all its intermolecular 
>interactions you would end up with an energy of -84 kJ/mol, because all terms 
>are counted double! 
There no terms counted 2 times in this calculation!? They become doubled if
you use this number x 216 H2O to obtain the total potential energy. The
average interaction energy of one molecule with 215 others is still -84 
kJ/mol!
Isn't it?
>Therefore it is entirely reasonable (though this is not a 
>rigorous derivation!) to partition the Protein-Solvent energy equal between 
>protein and solvent, in order to get an estimate of the Protein energy. 
I totally miss the concept of partitioning the interaction energy between
two parts into each one!
>As an 
>extra indication that this is reasonable, the linear interaction energy 
>method by Aqvist (Prot. Eng. 7 (1994) p. 385-391) derives that the 
>contribution to the Gibbs energy of solvation involves 0.5 times the 
>protein-solvent Coulomb interaction.
>
> A proper derivation would probably involving computing the heat of solvation 
>for the protein, and compare that to potential energies that come directly 
>from the simulation, (and obviously to experimental data).
> 
> Cheers,
> -- 
> David van der Spoel, Ph.D.
> Molec. Biophys. group, Dept. of Cell & Molec. Biol., Uppsala University.
> Box 596, 75124 Uppsala, Sweden. Phone:	+46184714205. Fax: +4618511755.
> spoel at xray.bmc.uu.se	spoel at gromacs.org   http://folding.bmc.uu.se
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-----------------------------------------------------
XAvier Periole - PhD

NMR & Molecular Dynamics Group
University of Groningen
The Netherlands
http://md.chem.rug.nl/~periole
-----------------------------------------------------



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