RE: AMBER: Negative dihedral coefficients in amber99 - Possible?

From: Ross Walker <ross.rosswalker.co.uk>
Date: Sun, 17 Aug 2008 22:15:45 -0700

Hi Ilyas,

> I am doing re-parametrization of some of the dihedral angles defined
> for the RNA molecules. In the parm99.dat file, I see that the coefficients
> of the dihedral terms are all positive. Is there a physical reason why
> they are positive? If not, can I define dihedral terms with negative
> coefficients? The fitting I am doing gives negative coefficients. As far
> as the calculations are concerned, with these choices, the QM and force
> field energies come pretty close to each other. In OPLS, there are
> negative coefficients defined for dihedrals, but seeing all the dihedral
> coefficients in parm99.dat positive makes me wonder if I am missing
> something. Any suggestions are greatly appreciated. Thanks in advance.

As far as I am aware there is nothing in the code that prevents you from
setting a negative barrier height for dihedrals, in fact I have done it
myself in the past when testing an automatic parameter fitting algorithm.
That said I would encourage you to check things against a system you can
calculate the energy of by hand just to be sure.

As for why parm99.dat is all +ve this is really, I believe, just an issue of
where you choose to place the origin for the total energy. If I am not
mistake a -ve barrier height would just be a +ve barrier + a phase angle
appropriate for the periodicity + a constant offset so really you just
ultimately have a shifted origin. Also the reason +ve values were chosen I
believe is just for simplicity. Someone will correct me if I am wrong here
but I believe Peter Kollman's original philosophy was that each dihedral
included in the force field should have some kind of physical reasoning
behind it - I.e. a 3 fold + a 2 fold term etc. Note, compare this to
automated parameter fitting where the dihedral terms effectively represent a
Fourier series so that you can just keep adding additional terms (of
different phases) until you get the exact energy surface you want (for
reasons you cannot easily justify). The net result of this though is that
you end up losing transferability.

Good luck,
Ross


/\
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|\oss Walker

| Assistant Research Professor |
| San Diego Supercomputer Center |
| Tel: +1 858 822 0854 | EMail:- ross.rosswalker.co.uk |
| http://www.rosswalker.co.uk | PGP Key available on request |

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Received on Wed Aug 20 2008 - 06:07:16 PDT
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