Dear Amber Users,
I ran the antechamber tutorial on the
Sustiva molecule (using Amber ver.11) as mentioned in the URL
http://ambermd.org/tutorials/basic/tutorial4b/ using the coordinates
file given therein and attached below.
USER MOD reduce.3.13.080428 H: found=0, std=0, add=9, rem=0, adj=0
REMARK 800 SITE_DESCRIPTION: BINDING SITE FOR RESIDUE SUS A 999
HETATM 7759 CL SUS A 999 -4.685 -32.725 25.222 1.00 50.93 CL
HETATM 7760 F1 SUS A 999 -0.755 -36.632 25.697 1.00 62.80 F
HETATM 7761 F2 SUS A 999 1.078 -37.043 24.672 1.00 70.73 F
HETATM 7762 F3 SUS A 999 -0.784 -37.177 23.626 1.00 75.12 F
HETATM 7763 O1 SUS A 999 1.524 -34.934 20.910 1.00 73.73 O
HETATM 7764 O2 SUS A 999 0.989 -34.880 23.058 1.00 73.24 O
HETATM 7765 N SUS A 999 -0.681 -34.971 21.434 1.00 71.12 N
HETATM 7766 C1 SUS A 999 -1.662 -34.414 22.313 1.00 71.29 C
HETATM 7767 C2 SUS A 999 -2.915 -33.947 21.843 1.00 72.17 C
HETATM 7768 C3 SUS A 999 -3.838 -33.423 22.771 1.00 71.25 C
HETATM 7769 C4 SUS A 999 -3.533 -33.373 24.119 1.00 63.88 C
HETATM 7770 C5 SUS A 999 -2.310 -33.829 24.593 1.00 63.40 C
HETATM 7771 C6 SUS A 999 -1.386 -34.378 23.681 1.00 68.96 C
HETATM 7772 C7 SUS A 999 -0.002 -34.957 24.144 1.00 71.35 C
HETATM 7773 C8 SUS A 999 0.552 -34.236 25.315 1.00 75.97 C
HETATM 7774 C9 SUS A 999 0.985 -33.657 26.241 1.00 81.31 C
HETATM 7775 C10 SUS A 999 1.605 -32.802 27.586 1.00 80.51 C
HETATM 7776 C11 SUS A 999 2.808 -33.083 27.639 1.00 83.59 C
HETATM 7777 C12 SUS A 999 2.404 -31.980 27.123 1.00 79.58 C
HETATM 7778 C13 SUS A 999 -0.121 -36.472 24.539 1.00 72.76 C
HETATM 7779 C14 SUS A 999 0.665 -34.932 21.725 1.00 70.58 C
HETATM 0 H122 SUS A 999 2.472 -31.798 26.040 1.00 79.58
H new
HETATM 0 H121 SUS A 999 2.585 -31.017 27.623 1.00 79.58
H new
HETATM 0 H112 SUS A 999 3.348 -33.100 28.597 1.00 83.59
H new
HETATM 0 H111 SUS A 999 3.235 -33.882 27.015 1.00 83.59
H new
HETATM 0 H101 SUS A 999 0.784 -32.761 28.317 1.00 80.51
H new
HETATM 0 HN SUS A 999 -0.981 -35.402 20.583 1.00 71.12
H new
HETATM 0 H5 SUS A 999 -2.066 -33.763 25.664 1.00 63.40
H new
HETATM 0 H3 SUS A 999 -4.811 -33.050 22.419 1.00 71.25
H new
HETATM 0 H2 SUS A 999 -3.163 -33.993 20.772 1.00 72.17
H new
TER
END
I then used the command
antechamber -i sustiva_new.pdb -fi pdb -o sustiva.mol2 -fo mol2 -c bcc -s 2
The output mol2 file is here
.<TRIPOS>MOLECULE
SUS
30 32 1 0 0
SMALL
bcc
.<TRIPOS>ATOM
1 CL -4.6850 -32.7250 25.2220 cl 999 SUS -0.074400
2 F1 -0.7550 -36.6320 25.6970 f 999 SUS -0.231300
3 F2 1.0780 -37.0430 24.6720 f 999 SUS -0.214300
4 F3 -0.7840 -37.1770 23.6260 f 999 SUS -0.227300
5 O1 1.5240 -34.9340 20.9100 o 999 SUS -0.582500
6 O2 0.9890 -34.8800 23.0580 os 999 SUS -0.378900
7 N -0.6810 -34.9710 21.4340 n 999 SUS -0.474100
8 C1 -1.6620 -34.4140 22.3130 ca 999 SUS 0.107600
9 C2 -2.9150 -33.9470 21.8430 ca 999 SUS -0.172000
10 C3 -3.8380 -33.4230 22.7710 ca 999 SUS -0.068000
11 C4 -3.5330 -33.3730 24.1190 ca 999 SUS -0.017600
12 C5 -2.3100 -33.8290 24.5930 ca 999 SUS -0.053000
13 C6 -1.3860 -34.3780 23.6810 ca 999 SUS -0.175300
14 C7 -0.0020 -34.9570 24.1440 c3 999 SUS 0.314200
15 C8 0.5520 -34.2360 25.3150 c1 999 SUS -0.204100
16 C9 0.9850 -33.6570 26.2410 c1 999 SUS 0.013900
17 C10 1.6050 -32.8020 27.5860 cx 999 SUS -0.082600
18 C11 2.8080 -33.0830 27.6390 cx 999 SUS -0.108400
19 C12 2.4040 -31.9800 27.1230 cx 999 SUS -0.107400
20 C13 -0.1210 -36.4720 24.5390 c3 999 SUS 0.626900
21 C14 0.6650 -34.9320 21.7250 c 999 SUS 0.843600
22 H122 2.4720 -31.7980 26.0400 hc 999 SUS 0.086700
23 H121 2.5850 -31.0170 27.6230 hc 999 SUS 0.077700
24 H112 3.3480 -33.1000 28.5970 hc 999 SUS 0.077700
25 H111 3.2350 -33.8820 27.0150 hc 999 SUS 0.085700
26 H101 0.7840 -32.7610 28.3170 hc 999 SUS 0.103700
27 HN -0.9810 -35.4020 20.5830 hn 999 SUS 0.356500
28 H5 -2.0660 -33.7630 25.6640 ha 999 SUS 0.170000
29 H3 -4.8110 -33.0500 22.4190 ha 999 SUS 0.156000
30 H2 -3.1630 -33.9930 20.7720 ha 999 SUS 0.149000
If we calculate the net charge of the molecule it is -0.00199 which
is close to zero. My question is if I now generate the charges of a
new molecule and get the net charge to be close to the above value,
can I consider the net charge to be zero for all practical purposes?
Secondly, the charge distribution on the hydrogen atoms H121/H122 and
H111/H112 which is bonded to the carbon atoms C12 and C11 respectively
are not identical. I was of the understanding that the charge
distribution on these atoms have to be equal. Is this appropriate?
I will be glad if you can kindly guide met in these matter.
Thank you in advance,
Dilraj lama,
post doctoral researcher,
BII-Singapore.
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Received on Tue Feb 14 2012 - 00:30:01 PST
