AMBER Archive (2006)

Subject: Re: AMBER: charge constraints in resp

From: Ilyas Yildirim (yildirim_at_pas.rochester.edu)
Date: Thu Mar 30 2006 - 11:03:43 CST


Hi Kenley,

I think u should use the restraints on the first fit; not on the second
one. Most of the atom charges on the second fit are frozen; only the
H-atoms are recalculated. So, having constrainst on the first 20 atoms
to have a total zero charge does not have any meaning.

But still, I am not sure if it is correct what u have for the second fit.
What is the structure? U equivalence the H-atoms which are bound to the
same carbon, but u froze the charge of the C-atom in the second fit. U
should recalculate the whole group in the second fit; both the C-atoms
and H-atoms that are needed to be equilivalenced. Can u send a .pdb
file of the structure u are dealing?

Best,

On Thu, 30 Mar 2006, Kenley Barrett wrote:

> Dear AMBER community,
>
> I am having some difficulties with resp. I want to do a second-stage resp
> fitting in which atoms 1-20 are restrained to have a net charge of 0.
> However, once I've put the resp charges in a prep file and loaded it into
> xleap, the charge of these atoms comes out to be 0.1230, not 0.
>
> I created the constraints in my resp2 input file based on the peptoid
> example in /opt/local/amber8/examples/resp_charge_fit/. The &cntrl namelist
> is the same as what is automatically created by respgen (I've discovered
> that ioutopt needs to equal 1 in order for AMBER to go through all the steps
> to make a working prep file).
>
> Here are the first and second stage resp input files:
>
> First stage input file:
> First resp fitting for methylated histidine, following Duan et al.
> procedure.
>
> &cntrl
>
> nmol = 1,
> ihfree = 1,
> ioutopt = 1,
>
> &end
> 1.0
> Resp charges
> 0 26
> 7 0
> 1 0
> 6 0
> 6 0
> 8 0
> 1 0
> 6 0
> 1 0
> 1 0
> 6 0
> 7 0
> 6 0
> 1 0
> 7 0
> 6 0
> 1 0
> 1 0
> 1 0
> 6 0
> 1 0
> 6 0
> 8 0
> 1 0
> 7 0
> 1 0
> 1 0
>
>
> Second stage input file:
> Second fitting for methylated histidine, following Duan et al. procedure
> except
> restraining the non-blocking atoms to a total zero charge.
>
> &cntrl
>
> nmol = 1,
> ihfree = 1,
> ioutopt = 1,
> iqopt = 2,
> qwt = 0.001,
>
> &end
> 1.0
> resp charges
> 0 26
> 7 -99
> 1 0
> 6 -99
> 6 -99
> 8 -99
> 1 0
> 6 -99
> 1 0
> 1 8
> 6 -99
> 7 -99
> 6 -99
> 1 0
> 7 -99
> 6 -99
> 1 0
> 1 16
> 1 16
> 6 -99
> 1 0
> 6 -99
> 8 -99
> 1 -99
> 7 -99
> 1 -99
> 1 -99
> 20 0.0
> 1 1 1 2 1 3 1 4 1 5 1 6 1 7
> 1 8 1 9 1 10 1 11 1 12 1 13 1 14 1
> 15 1 16 1 17 1 18 1 19 1 20
>
>
> Here is the -t output of the second stage fitting:
>
> -0.030630 0.030630 0.002225 0.457386 -0.567234 0.115050 -0.136863
> 0.094573 0.094573 0.173194 -0.449827 0.022620 0.156302 0.161476 -
> 0.180110 0.099700 0.099700 0.099700 -0.302310 0.181925 0.335545 -
> 0.538195 0.053203 -0.593932 0.360673 0.260624
>
>
> Strangely, if you add up the charges of atoms 1-20, the net charge is
> actually 0.12208 rather than 0.1230--is this kind of difference normal after
> reading information into XLeap? But regardless of whether the charge is
> 0.12208 or .1230, I can't use this--I need to find a way to make the charges
> of atoms 1-20 equal 0.
>
> I would be very grateful for any suggestions. Thank you in advance for your
> help.
>
> Sincerely,
> Kenley
>

-- 
  Ilyas Yildirim
  ---------------------------------------------------------------
  - Department of Chemisty       -				-
  - University of Rochester      -				-
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