Hello,
My comments are below.
On Tue, Jan 4, 2011 at 3:41 AM, George Tzotzos <gtzotzos.me.com> wrote:
> Happy New Year to all,
>
> I've been running MMPBSA.py.MPI to determine Delta G for a protein-ligand.
> I've also run the program to determine per residue decomposition of
> entropy.
>
> The same input files have been used in both cases. The the Delta G results
> obtained from Generalized Born differ by ~ 2kcal/mol
>
Is this the difference between parallel and serial or the difference between
decomp turned on and decomp turned off? Also, what individual terms differ
between the two calculations? It could be that the algorithm used to
compute the surface area between the two methods is slightly different.
>
> For example,
>
> Differences (Complex - Receptor - Ligand):
>
> DELTA G binding = -46.6213 +/- 3.1112
> 0.1663 (given by per-residue entropy
> decomposition)
> DELTA G binding = -44.2279 +/- 2.7619
> 0.1476 (without per-residue entropy
> decomposition)
>
> The same discrepancy of ~2kcal/mol has been observed using the same ligand
> with two (2) other receptors.
>
> The Poisson Boltzmann calculations with and without per residue
> decomposition gave identical values
>
As far as I know, the surface area is non-decomposable for PBSA, so this is
not really a factor. Hence, you get similar/identical results.
>
> Differences (Complex - Receptor - Ligand):
>
> DELTA G binding = -34.0898 +/- 3.1112
> 0.1663
>
> My specific questions are the following:
>
> 1. Is there an explanation for this discrepancy in the case of Generalized
> Born while this discrepancy is not observed in the Poisson Boltzmann
> calculations?
>
> 2. The Delta Gs given by the two methods are different by ~10 kcal/mol.
> That strikes me as being too much of a difference.
>
This is not unusual, and reflects the method's shortcomings when calculating
absolute binding free energies. A better comparison to make would be the
DELTA Delta G between different receptors with the same ligand or different
ligands with the same receptor.
>
> 3. Is temperature (say 300K) factored in the ENTROPY calculations?
>
Yes. There should be a comment in the output file saying exactly that. On
the tutorial website, the last line of text in the output file says
NOTE: All entropy results have units kcal/mol. (Temperature has already been
multiplied in as 300. K)
> 4. Can one assume that the enthalpy for the six translational and
> rotational degrees of freedom is 6*(1/2)*RT=1.8 kcal/mol at 300K?
>
No. The entropy is calculated from statistical mechanical equations using
the partition function assuming that the translational, rotational, and
vibrational parts of the Hamiltonian are separable. This introduces a
mass-dependence of the translational entropy.
Note that this email was begun in response to your first message, so my
later email will address future questions.
Good luck!
Jason
> Thanks in advance and best regards
>
> George
>
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--
Jason M. Swails
Quantum Theory Project,
University of Florida
Ph.D. Graduate Student
352-392-4032
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Received on Tue Jan 04 2011 - 11:30:02 PST
