[Chimera-users] RMSD for two ligands

Pavithra meetpavithramsc at gmail.com
Wed Aug 21 17:48:05 PDT 2013


Dear Ma'am,
Thank you so much. its working now.



On Thu, Aug 22, 2013 at 2:25 AM, Elaine Meng <meng at cgl.ucsf.edu> wrote:

> Dear Pavithra,
> Chimera cannot guess which atoms to use when there are different numbers
> of atoms in the two groups (#0 and #1). You would have to either
>
> (A) include all the names of the atoms, in the order you want them
> matched, in the rmsd command. It would be something like: rmsd #0:15 at c1,c2,c3
> #1:12 at c10,c5,c2  (if you wanted to match residue 15 atom C1 with residue
> 12 atom C10, residue 15 atom C2 with residue 12 atom C5, etc.). To see the
> atom names, show labels, for example command: label #0:15#1:12
>
> or
>
> (B) select all the atoms with the mouse (Shift-Ctrl-click) one by one,
> first the atoms you want to match in the first molecule, then the atoms you
> want to match in the second molecule in the corresponding order, and then
> use "sel" to specify the atoms in the rmsd command
>
> The "rmsd" manual page can be viewed here (or by using command: help rmsd):
> <http://www.rbvi.ucsf.edu/home/meng/docs/UsersGuide/midas/rmsd.html>
>
> ... and more about how to specify atoms in the command line:
> <
> http://www.rbvi.ucsf.edu/home/meng/docs/UsersGuide/midas/atom_spec.html#basic
> >
>
> I hope this helps,
> Elaine
> -----
> Elaine C. Meng, Ph.D.
> UCSF Computer Graphics Lab (Chimera team) and Babbitt Lab
> Department of Pharmaceutical Chemistry
> University of California, San Francisco
>
> On Aug 21, 2013, at 1:05 AM, Pavithra wrote:
>
> > Dear Users,
> > I want to calculate the RMSD of two ligand molecules of different atom
> counts but have a common scaffold.
> > I tried rmsd #0 #1
> > but I'm getting the error: unusual numbers of atoms chosen for
> evaluation.
> >
> > rmsd #0:15-17 #1:15-17
> > atleast one atom must be selected.
>
> > could somebody help me to find the RMSD between two ligands of different
> atom counts.
> > --
> > regards,
> > Pavithra.K.B.
>
>


-- 
regards,
Pavithra.K.B.
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