AMBER Archive (2005)

Subject: Re: AMBER: D-enantiomers

From: justin litchfield (litch_at_stanford.edu)
Date: Wed Sep 21 2005 - 02:50:36 CDT

I agree in that this was my initial concern: that they are
diastereomers once you put the D-AA in with L-AA's. but as far as I
know, these force fields are basically built on a unit-by-unit basis
- with the units being the amino acids - and if that were not the
case, you would need one Ser force field that was for say xxx-Ser-Glu-
xxx and one that was for xxx-Ser-Lys-xxx. this is obviously not the
case so one must assume that the force fields are for each individual
unit in isolation. so though I cannot speak to this concern
specifically as it concerns the model, chemically I think that it is
sensible to use the same parameters for both D- or L- amino acids.
hope that makes sense,

justin

On Sep 20, 2005, at 11:53 PM, FyD wrote:

> Quoting "Thomas E. Cheatham, III" <cheatham_at_chpc.utah.edu>:
>
>
>>> I am relatively new to Amber, but have not found any information
>>> on the
>>> following topic. Is there an easy way to implement the D amino
>>> acids in
>>> Amber? I am looking to simulate a short peptide which contains
>>> one or more
>>> D-enantiomers along the backbone.
>>>
>>
>> As enantiomers have equivalent properties (outside of a chiral
>> environment
>> other than bending plane polarized light), one would not expect
>> that the
>> charges or intra-molecular parameters would be different for D vs.
>> L in an
>> empirical force field.
>>
>
> And what about introducing a D amino-acid (AA) in a sequence
> composed of L AA ?
> Each AA (D or L enantiomer), individually I mean, is optically
> equivalent, but a
> di-AA D-L compared to L-L are diastereoisomers. In this case,
> (compared to the
> "ALL" L sequence), I guess, you can expect charge differences
> (consequently FF
> param. differences).
> Since the charges are derived for each "UNIT" individually (L
> enantiomers), the
> problem could arise from mixing them which can lead to
> diateroeisomers. Am I
> wrong or is it simply "too much" for an empirical force field ?
>
> Regards, Francois
>
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