On 6/26/12 11:36 AM, Shima Arasteh wrote:
I tried again as below:
1. defined FOR in .rtp file
2. defined FOR in aminoacids.dat file.
3. chose NH2 as N-Terminal
and then got this topology for FOR and VAL:
; residue 0 FOR rtp FOR q 0.0
1 CH1 0 FOR C 1 0.38 13.019 ; qtot
0.38
2 O 0 FOR O 1 -0.38 15.9994 ; qtot 0
; residue 1 VAL rtp VAL q 0.0
3 NL 1 VAL N 2 -0.83 14.0067 ; qtot
-0.83
4 H 1 VAL H1 2 0.415 1.008 ; qtot
-0.415
5 H 1 VAL H2 2 0.415 1.008 ; qtot 0
6 CH1 1 VAL CA 3 0 13.019 ; qtot 0
7 CH1 1 VAL CB 4 0 13.019 ; qtot 0
8 CH3 1 VAL CG1 5 0 15.035 ; qtot 0
9 CH3 1 VAL CG2 6 0 15.035 ; qtot 0
10 C 1 VAL C 7 0.38 12.011 ; qtot
0.38
11 O 1 VAL O 7 -0.38 15.9994 ; qtot 0
I guess the topology is correct. Isn't it?
No, it isn't. An NH2 terminal says the N-terminus of VAL is a neutral amine.
You should choose "None" for the terminus. The formyl group occupies the
N-terminus as would another amino acid in a peptide bond, but there is no
special protonation of FOR.
-Justin
--
========================================
Justin A. Lemkul, Ph.D.
Research Scientist
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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