Dear Amber users
I am using antechamber to parametrize an inhibitor (PP1).
The inhibitor contains a condensate aromatic ring plus a benzene ring.
The mol2 file I obtained is the following:
.<TRIPOS>ATOM
1 N1 26.3700 31.6330 39.5860 n3 532 PP1 -0.447400
2 C2 26.5500 29.2320 36.9190 c2 532 PP1 0.598100
3 N3 26.9570 30.3440 37.5350 n2 532 PP1 -0.629900
4 C4 26.2520 30.5770 38.6810 c3 532 PP1 0.366400
5 C5 25.2080 29.7600 39.1940 c3 532 PP1 -0.281000
6 C6 24.8860 28.6240 38.4100 c3 532 PP1 0.710600
7 N7 25.5810 28.3850 37.2780 nh 532 PP1 -0.743600
8 N8 25.4840 31.5120 40.6190 n3 532 PP1 -0.350400
9 C9 24.8030 30.4120 40.4040 c3 532 PP1 0.277400
10 N10 23.8700 27.7270 38.7290 n3 532 PP1 -0.921800
11 C11 23.7730 29.9750 41.3480 c3 532 PP1 -0.026000
12 C12 22.7970 30.8850 41.8090 c3 532 PP1 -0.058700
13 C13 21.8040 30.4670 42.7160 c3 532 PP1 -0.096700
14 C14 21.7780 29.1340 43.1730 c3 532 PP1 -0.061000
15 C15 22.7550 28.2240 42.7220 c3 532 PP1 -0.098700
16 C16 23.7480 28.6430 41.8160 c3 532 PP1 -0.080700
17 C24 20.7120 28.6860 44.1510 c3 532 PP1 -0.064100
18 C28 27.2910 32.7860 39.5450 c3 532 PP1 0.277200
19 C29 27.0910 33.7450 40.7460 c3 532 PP1 -0.105433
20 C33 27.0480 33.5950 38.2550 c3 532 PP1 -0.105433
21 C37 28.7480 32.2800 39.5830 c3 532 PP1 -0.105433
22 H2 27.0700 28.9880 36.0040 h5 532 PP1 0.057800
23 H101 23.6470 26.9570 38.1150 hn 532 PP1 0.439800
24 H102 23.2700 27.8700 39.5280 hn 532 PP1 0.439800
25 H12 22.8060 31.9090 41.4640 hc 532 PP1 0.105700
26 H13 21.0630 31.1740 43.0590 hc 532 PP1 0.095700
27 H15 22.7480 27.2020 43.0720 hc 532 PP1 0.095700
28 H16 24.4970 27.9380 41.4860 hc 532 PP1 0.105700
29 H241 21.0410 28.8660 45.1740 hc 532 PP1 0.046367
30 H242 19.7820 29.2330 43.9880 hc 532 PP1 0.046367
31 H243 20.5020 27.6220 44.0390 hc 532 PP1 0.046367
32 H291 27.7800 34.5890 40.6970 hc 532 PP1 0.052033
33 H292 26.0820 34.1570 40.7700 hc 532 PP1 0.052033
34 H293 27.2710 33.2430 41.6980 hc 532 PP1 0.052033
35 H331 27.6850 34.4800 38.2170 hc 532 PP1 0.052033
36 H332 27.2600 33.0140 37.3580 hc 532 PP1 0.052033
37 H333 26.0130 33.9340 38.1910 hc 532 PP1 0.052033
38 H371 29.4540 33.1110 39.5930 hc 532 PP1 0.052033
39 H372 28.9330 31.6810 40.4750 hc 532 PP1 0.052033
40 H373 28.9920 31.6660 38.7170 hc 532 PP1 0.052033
.<TRIPOS>BOND
1 1 4 1
2 1 8 1
3 1 18 1
4 2 3 2
5 2 7 1
6 2 22 1
7 3 4 1
8 4 5 1
9 4 5 1
10 5 6 1
11 5 9 1
12 6 7 1
13 6 7 1
14 6 10 1
15 8 9 1
16 8 9 1
17 9 11 1
18 10 23 1
19 10 24 1
20 11 12 1
21 11 16 1
22 11 16 1
23 12 13 1
24 12 13 1
25 12 25 1
26 13 14 1
27 13 26 1
28 14 15 1
29 14 15 1
30 14 17 1
31 15 16 1
32 15 27 1
33 16 28 1
34 17 29 1
35 17 30 1
36 17 31 1
37 18 19 1
38 18 20 1
39 18 21 1
40 19 32 1
41 19 33 1
42 19 34 1
43 20 35 1
44 20 36 1
45 20 37 1
46 21 38 1
47 21 39 1
48 21 40 1
.<TRIPOS>SUBSTRUCTURE
1 PP1 1 TEMP 0 **** **** 0 ROOT
To me it looks like the aromatic parts are not recognized.
Any suggestion to solve that problem?
Thank you
Valentina
_______________________________
Valentina Romano | PhD Student |
Biozentrum, University of Basel & SIB Swiss Institute of Bioinformatics |
Klingelbergstrasse 61 | CH-4056 Basel |
Phone: +41 61 267 15 80
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Received on Fri Feb 20 2015 - 02:30:06 PST
