AMBER Archive (2005)Subject: Re: AMBER: ptraj-acceptor mask format
From: Thomas E. Cheatham, III (cheatham_at_chpc.utah.edu)
Date: Tue Jul 26 2005 - 15:06:08 CDT
> 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_at_N or @N matches residues 1-159_at_N
:1-159_at_H or @H matches residues 2-159_at_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_at_N :2_at_H
:2_at_N :3_at_H
:3_at_N :4_at_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_at_N :2-159_at_H
acceptor mask :1_at_N :1_at_H1
acceptor mask :1_at_N :1_at_H2
acceptor mask :1_at_N :1_at_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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