[gmx-users] The negative state of a Phe molecule is not parametrized in gromos96 FF?

[Justin A. Lemkul]

What I would check is the ffG43a1-n.tdb file; maybe someone has been playing 
with it, because what pdb2gmx is doing is assigning the charges from the NH2 
terminus to the atoms of an NH3+ terminus, hence the excess charge.

-Justin

Hu Zhongqiao wrote:
> Dear all,
> 
>  
> 
> I wonder if a negative Phe molecule is parametrized in gromos96 FF 
> properly. I used
> 
>  
> 
> pdb2gmx –f phe.pdb –o phe.gro –p phe.top –ter
> 
>  
> 
> I selected “1: NH2” for N-terminus and “0: COO-” for C-terminus. But the 
> phe.top obtained is not correct
> 
>  
> 
>  
> 
> [ atoms ]
> 
> ;   nr       type  resnr residue  atom   cgnr     charge       mass  
> typeB    chargeB      massB
> 
>      1         NL      1    PHE      N      1      -0.83    14.0067   ; 
> qtot -0.83
> 
>      2          H      1    PHE     H1      1      0.415      1.008   ; 
> qtot -0.415
> 
>      3          H      1    PHE     H2      1      0.415      1.008   ; 
> qtot 0
> 
>      4          H      1    PHE     H3      1      0.415      1.008   ; 
> qtot 0.415
> 
>      5        CH1      1    PHE     CA      2          0     13.019   ; 
> qtot 0.415
> 
>      6        CH2      1    PHE     CB      2          0     14.027   ; 
> qtot 0.415
> 
>      7          C      1    PHE     CG      2          0     12.011   ; 
> qtot 0.415
> 
>      8          C      1    PHE    CD1      3       -0.1     12.011   ; 
> qtot 0.315
> 
>      9         HC      1    PHE    HD1      3        0.1      1.008   ; 
> qtot 0.415
> 
>     10          C      1    PHE    CD2      4       -0.1     12.011   ; 
> qtot 0.315
> 
>     11         HC      1    PHE    HD2      4        0.1      1.008   ; 
> qtot 0.415
> 
>     12          C      1    PHE    CE1      5       -0.1     12.011   ; 
> qtot 0.315
> 
>     13         HC      1    PHE    HE1      5        0.1      1.008   ; 
> qtot 0.415
> 
>     14          C      1    PHE    CE2      6       -0.1     12.011   ; 
> qtot 0.315
> 
>     15         HC      1    PHE    HE2      6        0.1      1.008   ; 
> qtot 0.415
> 
>     16          C      1    PHE     CZ      7       -0.1     12.011   ; 
> qtot 0.315
> 
>     17         HC      1    PHE     HZ      7        0.1      1.008   ; 
> qtot 0.415
> 
>     18          C      1    PHE      C      8       0.27     12.011   ; 
> qtot 0.685
> 
>     19         OM      1    PHE     O1      8     -0.635    15.9994   ; 
> qtot 0.05
> 
> 20         OM      1    PHE     O2      8     -0.635    15.9994   ; qtot 
> -0.585
> 
>  
> 
> The N-terminus is still NH3, not NH2, and the total charge of the system 
> is -0.585, not -1. I realized that if I removed one of first 3 H atoms 
> manually, then the total charge is -1. But I think it should be not so 
> simple. Anyone can give some clues on this? I just want to produce the 
> topology file of a negative Phe molecule in Gromos99 FF.
> 
>  
> 
> Best wishes,
> 
> Zhongqiao
> 
>  
> 
> 
> ------------------------------------------------------------------------
> 
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-- 
========================================

Justin A. Lemkul
Graduate Research Assistant
Department of Biochemistry
Virginia Tech
Blacksburg, VA
jalemkul[at]vt.edu | (540) 231-9080
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin

========================================