# Re: [AMBER] General question about MM-PB(GB)SA

From: Jason Swails <jason.swails.gmail.com>
Date: Sat, 7 Mar 2015 07:25:14 -0500

> On Mar 7, 2015, at 12:56 AM, Atila Petrosian <atila.petrosian.gmail.com> wrote:
>
> Dear amber users
>
> I am doing MM-PBSA calculation for my protein-ligand system based on TUTORIAL
> using AmberTools 14 (MMPBSA.py).
>
> I have some general question:
>
> 1) Should I use all of the frames in solvated complex trajectory? Binding
> delta G is different in various selections of frames. Is there a criterion
> for selection of frames for MM-PBSA and MM-GBSA calculation (startframe and
> endframe)? I have 2000 frames, what is your suggestion?

Depends. For MM/GBSA and MM/PBSA, I would use 1/10th of those frames (so startframe=10, endframe=2000, interval=10).

> 2) I calculated the binding free energy of my complex (protein-ligand) +
> Quasi-harmonic entropy for one frame of trajectory:

One frame for quasi-harmonic entropy does not make sense. The quasi-harmonic approximation is not like the normal mode analysis in that you can’t simply compute it for one frame. It builds a covariance matrix (mass-weighted) and treats the eigenvectors as normal modes with the eigenvalues as their frequencies. You need more frames than you have atoms if you even want to get the right number of eigenvectors from the covariance matrix.

> Energy Component Average
> --------------------------------------------
> BOND -0.0001
> ANGLE 0.0000
> DIHED -0.0000
> VDWAALS -22.8383
> EEL -6.8608
> 1-4 VDW 0.0001
> 1-4 EEL -0.0000
> EPB 18.8155
> ENPOLAR -15.1933
> EDISPER 23.7355
>
> DELTA G gas -29.6991
> DELTA G solv 27.3577
>
> DELTA TOTAL -2.3414
>
> -------------------------------------------------------------------------------
> Using Quasi-harmonic Entropy Approximation: DELTA G binding = 18.4960
> -------------------------------------------------------------------------------
>
> Since 1- Delta G = Delta H - T * Delta S and 2- Temperature is 298.15 K in
> this calculation, thus
>
> 18.4960 = -2.3414 - 298.15 * Delta S, ===> Delta S = 0.069 kcal/mol.K
>
> Is my consideration true?

No. The implicit solvent calculation includes solvent entropy (which is significant). What you are calling “Delta H” is really the entire free energy less some contributions to the solute entropy. And there’s no way to “back-out” the solvent entropy that’s included in the PB part of the calculation.

>
> 3) I calculated 1- the binding free energy of my complex (protein-ligand)
> for one frame of trajectory and 2- I calculated entropy of my complex
> (protein-ligand) using Normal Mode Analysis for the same frame:
>
> Energy Component Average
> ----------------------------------------------
> BOND -0.0001
> ANGLE 0.0000
> DIHED -0.0000
> VDWAALS -22.8383
> EEL -6.8608
> 1-4 VDW 0.0001
> 1-4 EEL -0.0000
> EPB 18.8155
> ENPOLAR -15.1933
> EDISPER 23.7355
>
> DELTA G gas -29.6991
> DELTA G solv 27.3577
>
> DELTA TOTAL -2.3414
>
>
> Entropy Term Average
> ----------------------------------------------
> Translational -12.1100
> Rotational -8.7200
> Vibrational 17.2921
>
> DELTA S total= -3.5378
>
>
> There are following in output file:
>
> All units are reported in kcal/mole.
> All entropy results have units kcal/mol (Temperature is 298.15 K).
>
> How to use and combine DELTA S total (-3.5378) with DELTA TOTAL (-2.3414)?
>
> Delta G = Delta H - T * Delta S
>
> Delta G = (-2.3414) - (-3.5378)
>
> Is my consideration true?

Again, no, for the same reason as above.

HTH,
Jason

```--
Jason M. Swails
BioMaPS,
Rutgers University
Postdoctoral Researcher
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Received on Sat Mar 07 2015 - 04:30:02 PST
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