Thanks a lot. I found the manual you mentioned.
Yuhui
--- Lihua Wang <lhw.broyde.nyu.edu> wrote:
> 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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>
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Received on Thu Sep 02 2004 - 01:15:11 PDT
