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From: <Yokota_Akihiro.takeda.co.jp>

Date: Thu, 10 Feb 2011 15:04:56 +0900

Dear Jason,

Thank you for your kind reply.

As you mentioned, PB energies can be decomposed to use geometrical

arguments to partition the grid points.

But I'd like to know how to do it and how accuracy it is,

because it might bring us ambiguous and inaccurate energies.

And you mean that I must read sander source code rather than MMPBSA.py

because sander calculates PB energies which are decomposed into each residue.

---------------------------------------------------------------

I checked temporary file of MMPBSA.py, and found 2 files below.

This shows me that PB energies are calculated by sander.

Input file: _MMPBSA_pb_decomp_rec.mdin

Output file: _MMPBSA_receptor_pb.mdout

---------------------------------------------------------------

So, I will read sander source code, and if I can't understand it,

I will ask someone who developed sander module.

Thank you for your kind reply.

And I am grateful to you for creating MMPBSA.py which can recognize

receptor and ligand automatically, and make sophisticated operations.

Best Regards,

Akihiro Yokota

-----Original Message-----

From: Jason Swails [mailto:jason.swails.gmail.com]

Sent: Thursday, February 10, 2011 11:46 AM

To: AMBER Mailing List

Subject: Re: [AMBER] AMBER11 MMPBSA.py tutorial section 3.6

Hi Yokota,

As the GB equations consist of pairwise, analytical terms, you're right that

they're much easier to decompose than PB energies.

Since I haven't looked at the PBSA/decomp code, I can't say any of this for

sure. If I'm mistaken here, I'm hoping that a PBSA developer will step in

and correct me.

PBSA works by placing your system in a grid and solving the PB equation via

a finite difference method. It assigns each grid point a specific charge,

dielectric constant, and *starting* electrostatic potential (and ionic

strength if you specified one), which it then solves iteratively to get the

converged electrostatic potential. Thus, at the end of the calculation,

each gridpoint has associated with it a charge, potential, and dielectric.

You can use geometrical arguments to partition the grid points into

different residues (i.e. each grid point belongs to its *closest* residue).

Once you've created this partition, you simply toss it into the equation

DELTA G = 1/2 integral(charge density * reaction field potential)

Since it's a discrete problem, the integral becomes a sum, and for each

residue you just sum over that residue's grid points. I agree that it's not

as straightforward as GB, but it seems reasonable to me.

Hope this helps (and that it's right :) ),

Jason

On Mon, Feb 7, 2011 at 2:54 AM, <Yokota_Akihiro.takeda.co.jp> wrote:

*> Dear AMBER users,
*

*>
*

*> I have a question about "Polar Solvation" energy for each residue in
*

*> MMPBSA.py tutorial section 3.6.
*

*> http://ambermd.org/tutorials/advanced/tutorial3/py_script/section6.htm
*

*>
*

*> In the middle of the file below, there is a comment "Energy Decomposition
*

*> Analysis (All units kcal/mol):
*

*> Poisson Boltzmann solvent", I think this means that "Polar Solvation(PB)"
*

*> of "each residue" can be calculated.
*

*>
*

*> http://ambermd.org/tutorials/advanced/tutorial3/py_script/files/FINAL_DECOMP_MMPBSA_perres.dat
*

*>
*

*> However PB energy cannot be decomposed straightly by the definition (On the
*

*> other hand, GB energies can be decomposed).
*

*> Does anyone teach me how the PB energies are decomposed into each residue?
*

*>
*

*> Thanks in advance,
*

*> Akihiro Yokota
*

*>
*

*> _______________________________________________
*

*> AMBER mailing list
*

*> AMBER.ambermd.org
*

*> http://lists.ambermd.org/mailman/listinfo/amber
*

*>
*

Date: Thu, 10 Feb 2011 15:04:56 +0900

Dear Jason,

Thank you for your kind reply.

As you mentioned, PB energies can be decomposed to use geometrical

arguments to partition the grid points.

But I'd like to know how to do it and how accuracy it is,

because it might bring us ambiguous and inaccurate energies.

And you mean that I must read sander source code rather than MMPBSA.py

because sander calculates PB energies which are decomposed into each residue.

---------------------------------------------------------------

I checked temporary file of MMPBSA.py, and found 2 files below.

This shows me that PB energies are calculated by sander.

Input file: _MMPBSA_pb_decomp_rec.mdin

Output file: _MMPBSA_receptor_pb.mdout

---------------------------------------------------------------

So, I will read sander source code, and if I can't understand it,

I will ask someone who developed sander module.

Thank you for your kind reply.

And I am grateful to you for creating MMPBSA.py which can recognize

receptor and ligand automatically, and make sophisticated operations.

Best Regards,

Akihiro Yokota

-----Original Message-----

From: Jason Swails [mailto:jason.swails.gmail.com]

Sent: Thursday, February 10, 2011 11:46 AM

To: AMBER Mailing List

Subject: Re: [AMBER] AMBER11 MMPBSA.py tutorial section 3.6

Hi Yokota,

As the GB equations consist of pairwise, analytical terms, you're right that

they're much easier to decompose than PB energies.

Since I haven't looked at the PBSA/decomp code, I can't say any of this for

sure. If I'm mistaken here, I'm hoping that a PBSA developer will step in

and correct me.

PBSA works by placing your system in a grid and solving the PB equation via

a finite difference method. It assigns each grid point a specific charge,

dielectric constant, and *starting* electrostatic potential (and ionic

strength if you specified one), which it then solves iteratively to get the

converged electrostatic potential. Thus, at the end of the calculation,

each gridpoint has associated with it a charge, potential, and dielectric.

You can use geometrical arguments to partition the grid points into

different residues (i.e. each grid point belongs to its *closest* residue).

Once you've created this partition, you simply toss it into the equation

DELTA G = 1/2 integral(charge density * reaction field potential)

Since it's a discrete problem, the integral becomes a sum, and for each

residue you just sum over that residue's grid points. I agree that it's not

as straightforward as GB, but it seems reasonable to me.

Hope this helps (and that it's right :) ),

Jason

On Mon, Feb 7, 2011 at 2:54 AM, <Yokota_Akihiro.takeda.co.jp> wrote:

-- Jason M. Swails Quantum Theory Project, University of Florida Ph.D. Graduate Student 352-392-4032 _______________________________________________ AMBER mailing list AMBER.ambermd.org http://lists.ambermd.org/mailman/listinfo/amber _______________________________________________ AMBER mailing list AMBER.ambermd.org http://lists.ambermd.org/mailman/listinfo/amberReceived on Wed Feb 09 2011 - 22:30:02 PST

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