Re: [AMBER] MMPBSA binding energy

From: Jason Swails <jason.swails.gmail.com>
Date: Wed, 29 Dec 2010 22:01:16 -0700

Hello,

My comments are below.

On Dec 29, 2010, at 6:11 PM, Eliac Brown <eliacbrown.yahoo.com> wrote:

> Dear AMBER
> MMPBSA binding energy is given by:
> dG = Gcomplex - Gligand - Greceptor
> We always use single trajectory approach to calculate G from each species where it is faster than the 2- or 3- trajectory approach.
> Here is my question:
> If we calculated the ddG for different inhibitors binding with a certain protein, so:
> ddG= (Gcomplex1 -Gcomplex2) - (Gligand1-Gligand2) - (Greceptor1-Greceptor2)
> because we study the inhibition of some ligands with the same protein, so we could consider:
> Greceptor1 = G receptor2 (as I believe it is the case in 3-trajectory approach)

This is true if you use the same trajectory for bot simulations (which makes sense).

> So,
> ddG= (Gcomplex1 -Gcomplex2) - (Gligand1-Gligand2)
> So, why when I used the later equation I got a horrible results using single trajectory? I would expect to get the same results.

You should get exactly the same results, since Greceptor1-Greceptor2 = 0. If you are not getting the same results, then you are making a mistake somewhere. If you are doing it by hand, make sure you use the parentheses on your calculator where need be, or be sure to carry the negative sign through the Gligand term.

Good luck,
Jason

> Any hint would be appreciated.
> Thanks
> Happy New Year to all
> Eliac
>
>
>
>
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--
Jason Swails
Quantum Theory Project,
University of Florida
Ph.D. Graduate Student
352-392-4032
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Received on Wed Dec 29 2010 - 21:30:02 PST
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