Re: AMBER: ptraj-acceptor mask format

From: Thomas E. Cheatham, III <cheatham.chpc.utah.edu>
Date: Tue, 26 Jul 2005 14:06:08 -0600 (Mountain Daylight Time)

> ptraj finds the correct number of donors but it can not find any
> acceptors (and there should be at least 1 acceptor based on the ptraj
> scripts that work).

> So, the problem seems to be with the mask in the acceptor line and it
> does not seem to like the wild card or range specifications. (But, a
> wild card/range works with the mask for the donor command so it would
> seem that I've got the format of the mask command correct.) So,
> assuming I do have the format correct, why, with the acceptor command,
> can't I specify/use a wild card or range with mask?

....

> acceptor mask :*.N :*.H

There has to be a 1-1 correspondence between the acceptors, i.e. the first
N matched is paired with the first H matched, the second N with the second
H and so on... as the program is not smart enough (while remaining
completely general) to auto-detect what N should be paired with what H
atom.

The problem is that in this case, with your matching

:1-159.N or @N matches residues 1-159.N

:1-159.H or @H matches residues 2-159.H
   (since residue 1, if it is a normal N-terminal replaces the H with H1, H2 and H3).

Therefore, your acceptor list would be

   :1.N :2.H
   :2.N :3.H
   :3.N :4.H
   ...

and these will never be satisfied; moreover, ptraj will likely chuck out
the acceptors since there are more "N" atoms chosen than "H" atoms and
therefore it doesn't have a complete 1-1 match and will complain about
this.


To get it to work, do

   acceptor mask :2-159.N :2-159.H
   acceptor mask :1.N :1.H1
   acceptor mask :1.N :1.H2
   acceptor mask :1.N :1.H3

To get more information, before specifying the donors/acceptors, add

   prnlev 4

and then set it back to zero after this (before doing the hbond command
unless you want a lot of debugging info)

   prnlev 0

--tom

p.s. also note that :* is redundant; you just need .N, however, as you saw
this will not work since the N-terminal does not have atom H but atoms H1,
H2 and H3 bound to the N and there is no 1-1 match.
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Received on Tue Jul 26 2005 - 21:53:01 PDT
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