From: lipengfei_mail <lipengfei_mail.126.com>
Date: Thu, 12 Nov 2015 22:45:41 +0800 (CST)

Dear all,
When I do TI calculations, I encountered some questions as below:

1.In the "Final Remarks"part of AMBER TI Tutorial A9, there is a word that"The author of this tutorial has found that the one¨Cstep protocol may fail in some circumstances".
What does it mean? Could you give me some examples in which one-step protocol may fail?
Are there some rules about how to choose one-step or three-step protocol?
2.After I accomplished this tutorial, I got much the same results as the tutorial gave.

But I think that for ligands, ¦¤Gligands (solution)= -38.99 kcal/mol, this seems unreasonable.
As a rough estimation, the solvation free energy for benzene and phenol are 0.59 and -4.68 kcal/mol,respectively.
The energy variation from benzene to phenol in gas phase is about -23.22 kcal/mol.
Then the total energy variation from benzene to phenol in the solvation can be estamitated as dG=-0.59-4.68 -23.22=-28.49 kcal/mol,
which has large difference with the result given out by the A9 tutorial (-38.99 kcal/mol ).

3.In the course of decharge, if only decharge the O1 and H6 of benzene where crgmask = ':1.O1,H6',
the rest atoms of benzene for MD will be charged which is obviously unreasonable.
One way is to set the crgmask=¡®£º1' . But if 1 refers to a large ligand, then there is less overlap between the initial and final state.
Could any advice be given out about the two choices as above or some better ways instead?

Best,
Thank you very much !

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Received on Thu Nov 12 2015 - 07:00:03 PST
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