Hi Yuhui,
I think the reason for your confusion is that instructions on how to make
a prep file has been abridged from the amber manual since version 6.
There is a special section dedicated to topology, atom type designation in
the amber5 manual (the PREP module). It is too long for me to present the
whole section here, but one principle of the parameters is that "loop
closing bonds are not counted as connections when assigning M, E, S, B, 3,
4,5,6 topological types".
I personally think that section is very important to users because as of
the current version (amber8), the antechamber doesn't seem to be able to
generate prep file for residues. So users still have to manually prepare
the prep file by hand (and I don't see an easy way to automate this). So
including the instructions on how to prepare a prep file (with examples,
as in amber5 manual) is extremely important. If you have access to the
amber5 manual, check out pages 76-85 of volume 2. It was very clearly
described.
Best,
Lihua Wang
On Wed, 25 Aug 2004, Yuhui Cheng wrote:
> Date: Wed, 25 Aug 2004 16:46:36 -0700 (PDT)
> From: Yuhui Cheng <amberuser3.yahoo.com>
> Reply-To: amber.scripps.edu
> To: amber.scripps.edu
> Subject: AMBER: About tree structure of TRP residue
>
> Hi, all,
> Did any one notice that tree structure of TRP
> amino acid in amber8?
> In $AMBERHOME/dat/leap/prep/all_amino02.in, it
> looks like:
> ........
> TRYPTOPHAN
>
> TRP INT 1
> CORR OMIT DU BEG
> 0.00000
> 1 DUMM DU M 0 -1 -2 0.000 0.000
> 0.000
> 2 DUMM DU M 1 0 -1 1.449 0.000
> 0.000
> 3 DUMM DU M 2 1 0 1.522 111.100
> 0.000
> 4 N N M 3 2 1 1.335 116.600
> 180.000
> 5 H H E 4 3 2 1.010 119.800
> 0.000
> 6 CA CT M 4 3 2 1.449 121.900
> 180.000
> 7 HA H1 E 6 4 3 1.090 109.500
> 300.000
> 8 CB CT 3 6 4 3 1.525 111.100
> 60.000
> 9 HB2 HC E 8 6 4 1.090 109.500
> 300.000
> 10 HB3 HC E 8 6 4 1.090 109.500
> 60.000
> 11 CG C* S 8 6 4 1.510 115.000
> 180.000
> 12 CD1 CW B 11 8 6 1.340 127.000
> 180.000
> 13 HD1 H4 E 12 11 8 1.090 120.000
> 0.000
> 14 NE1 NA B 12 11 8 1.430 107.000
> 180.000
> 15 HE1 H E 14 12 11 1.010 125.500
> 180.000
> 16 CE2 CN S 14 12 11 1.310 109.000
> 0.000
> 17 CZ2 CA B 16 14 12 1.400 128.000
> 180.000
> 18 HZ2 HA E 17 16 14 1.090 120.000
> 0.000
> 19 CH2 CA B 17 16 14 1.390 116.000
> 180.000
> 20 HH2 HA E 19 17 16 1.090 120.000
> 180.000
> 21 CZ3 CA B 19 17 16 1.350 121.000
> 0.000
> 22 HZ3 HA E 21 19 17 1.090 120.000
> 180.000
> 23 CE3 CA B 21 19 17 1.410 122.000
> 0.000
> 24 HE3 HA E 23 21 19 1.090 120.000
> 180.000
> 25 CD2 CB E 23 21 19 1.400 117.000
> 0.000
> 26 C C M 6 4 3 1.522 111.100
> 180.000
> 27 O O E 26 6 4 1.229 120.500
> 0.000
> ...............
> The tree label of atom CE2 is "S", and CD2 is
> "E". That means there is no bond between CE2 and CD2,
> Also no bond between CD2 and CG. But obviously, it's
> not correct. I'm totally confused.
>
> Thanks,
> Yuhui
>
>
>
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Received on Thu Sep 02 2004 - 01:15:11 PDT
