AMBER Archive (2006)

Subject: AMBER: question about REDII

From: Raffaella D'Auria (rdauria_at_uci.edu)
Date: Thu May 11 2006 - 21:04:51 CDT

Hello All,

I am trying to use REDII to find the RESP charges of a non standard
residue (an alcohol which I have built using bits of aminoacids from the
Cornell force field). Ultimately I like to compare the difference between
this way of creating a new residue and the more standard way which goes
through antechamber.

I am quite puzzled by the name assignment in the pdb file that serves as
input. In particular suppose I have a series of C atoms which type in
amber ff94 is CT, now, since the REDII naming convention all of my C
become CT1 (which makes them all equivalent???)? Something about the
naming convention must be eluding me... here is the pdb I used as input:

ATOM 1 CT1 1BO 1 3.537 1.423 -0.000 1.00 0.00
ATOM 2 H1 1BO 1 3.556 1.394 -1.116 1.00 0.00
ATOM 3 H1 1BO 1 3.581 0.373 0.378 1.00 0.00
ATOM 4 H1 1BO 1 4.448 1.964 0.354 1.00 0.00
ATOM 5 CT1 1BO 1 2.293 2.115 0.494 1.00 0.00
ATOM 6 H1 1BO 1 2.281 3.176 0.124 1.00 0.00
ATOM 7 H1 1BO 1 2.305 2.156 1.615 1.00 0.00
ATOM 8 CT1 1BO 1 1.042 1.406 0.027 1.00 0.00
ATOM 9 H1 1BO 1 1.048 0.345 0.392 1.00 0.00
ATOM 10 H1 1BO 1 1.016 1.370 -1.093 1.00 0.00
ATOM 11 CT1 1BO 1 -0.204 2.109 0.534 1.00 0.00
ATOM 12 H1 1BO 1 -0.249 3.167 0.160 1.00 0.00
ATOM 13 H1 1BO 1 -0.220 2.115 1.658 1.00 0.00
ATOM 14 OT2 1BO 1 -1.387 1.510 0.039 1.00 0.00
ATOM 15 H2 1BO 1 -1.367 0.582 0.297 1.00 0.00

and this is what I get from REDII (punch2) note I am using the defaults
values for the different fields (such as optimization etc):

 Molecule

iqopt irstrnt ihfree qwt
  2 1 1 0.001000

 rel.rms dipole mom Qxx Qyy Qzz
    0.43037 2.01555 -7.62250 3.98593 3.63657

          Point charges before & after optimization
    NO At.No. q0 q(opt) IVARY d(rstr)/dq
     1 6 -0.092959 0.144680 0 0.005685
     2 1 0.020596 -0.047679 0 0.000000
     3 1 0.014715 -0.047679 2 0.000000
     4 1 0.028640 -0.047679 2 0.000000
     5 6 0.050488 0.144680 1 0.005685
     6 1 -0.015887 -0.047679 2 0.000000
     7 1 -0.009222 -0.047679 2 0.000000
     8 6 -0.020894 0.144680 1 0.005685
     9 1 0.010972 -0.047679 2 0.000000
    10 1 0.027654 -0.047679 2 0.000000
    11 6 0.167083 0.144680 1 0.005685
    12 1 0.072194 -0.047679 2 0.000000
    13 1 0.001573 -0.047679 2 0.000000
    14 8 -0.677305 -0.468635 0 0.002087
    15 1 0.422352 0.319021 0 0.000000

Off course something is very wrong because now the H's have become
negative and more so than the C's or O!!!

Note I also tried the following input pdb file:

HEADER
REMARK Spartan exported Wed Mar 4 00:35:13 1998
HETATM 1 H1 UNK 0001 5.302 -0.019 -3.477
HETATM 2 CT1 UNK 0001 5.283 0.010 -2.361
HETATM 3 H1 UNK 0001 5.325 -1.040 -1.983
HETATM 4 CT1 UNK 0001 4.040 0.705 -1.867
HETATM 5 H1 UNK 0001 6.195 0.549 -2.007
HETATM 6 CT1 UNK 0001 2.788 -0.002 -2.334
HETATM 7 H1 UNK 0001 4.030 1.765 -2.237
HETATM 8 H1 UNK 0001 4.052 0.746 -0.746
HETATM 9 H1 UNK 0001 2.792 -1.063 -1.969
HETATM 10 H1 UNK 0001 2.762 -0.038 -3.454
HETATM 11 CT1 UNK 0001 1.543 0.704 -1.827
HETATM 12 OT2 UNK 0001 0.359 0.107 -2.322
HETATM 13 H1 UNK 0001 1.500 1.762 -2.201
HETATM 14 H1 UNK 0001 1.527 0.710 -0.703
HETATM 15 H2 UNK 0001 0.377 -0.822 -2.064
CONECT 1 2
CONECT 2 1 3 4 5
CONECT 3 2
CONECT 4 2 6 7 8
CONECT 5 2
CONECT 6 4 9 10 11
CONECT 7 4
CONECT 8 4
CONECT 9 6
CONECT 10 6
CONECT 11 6 12 13 14
CONECT 12 11 15
CONECT 13 11
CONECT 14 11
CONECT 15 12
END

but I still get nonsense...

What am I doing wrong?

Thanks,

Raffaella.
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