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

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