[cctbxbb] Generating crystal copies
Pavel Afonine
pafonine at lbl.gov
Tue Apr 7 14:35:56 PDT 2015
Hi,
in addition, see implementation of
iotbx.show_distances your.pdb > all_distances
that will output potentially long file with distances "all with all"
(with some cut-off I think) including symmetry related.
Also, a method expand_to_p1() of xray.structure
(cctbx/xray/structure.py) may be of help: this will expand model into P1
(isn't it what you want!).
Actual code that does the calculations:
cctbx_project/cctbx/xray/scatterer_utils.h (expand_to_p1)
All in all, cctbx offers a variety of tools to do what you want. Perhaps
someone needs to add a command like
iotbx.expand_to_p1 your_model_in_any_space_group.pdb > model_p1.pdb
though it sounds like you want this at your script level, not as
end-user-ready application.
Let me know if you have more questions or need help with this.
Pavel
On 4/7/15 2:16 PM, Oleg Sobolev wrote:
> Hi Nicholas,
>
> Please find the script for searching contacts attached. It reads in
> .pdb file (file name is passed through command-line arg) and prints
> out symmetry contacts with some additional info. Hopefully this will
> help, or feel free to ask more questions. I believe that you have
> already read this description of the machinery you are trying to use:
>
> http://cci.lbl.gov/publications/download/iucrcompcomm_aug2004.pdf
>
> Best regards,
> Oleg Sobolev.
>
>
>
> On Tue, Apr 7, 2015 at 9:33 AM, <markus.gerstel at diamond.ac.uk
> <mailto:markus.gerstel at diamond.ac.uk>> wrote:
>
> Dear Nicholas,
>
> I found the CCP4 program ncont to be the easiest way to identify
> crystal contacts.
>
> If you’re interested in a completed unit cell I have a
> PDBCUR/PDBSET script somewhere, which might provide a useful
> starting point.
>
> I have tried using cctbx a couple of years back to compute either
> of these, but got inconsistent results. (Also at that time I
> didn’t really know what I’m doing. Which may still be true.)
>
> -Markus
>
> *From:*cctbxbb-bounces at phenix-online.org
> <mailto:cctbxbb-bounces at phenix-online.org>
> [mailto:cctbxbb-bounces at phenix-online.org
> <mailto:cctbxbb-bounces at phenix-online.org>] *On Behalf Of
> *Nicholas Pearce
> *Sent:* 07 April 2015 17:26
> *To:* cctbxbb at phenix-online.org <mailto:cctbxbb at phenix-online.org>
> *Subject:* [cctbxbb] Generating crystal copies
>
> Hi Everyone,
>
> I'm currently trying to generate crystal contacts/copies. What I
> want out at the end is a list of symmetry operations to generate
> the crystallographic copies that form contacts with my input model.
>
> I've been using the asu_mappings object, but I'm not sure I'm
> using the output correctly or if it's the right thing to use.
>
> Code snippet used to calculate the mappings:
>
> # Extract the xray structure from the reference hierarchy
> ref_struc =
> ref_hierarchy.extract_xray_structure(crystal_symmetry=crystal_symmetry)
>
> # Extract the mappings that will tell us the adjacent symmetry
> copies
> asu_mappings = ref_struc.asu_mappings(buffer_thickness=5)
>
> # Symmetry operations for each atom
> mappings = asu_mappings.mappings()
>
> I printed out all of the mappings, as I presumed the vast majority
> of the them would be the identity, meaning that any non-identity
> operations would be from atoms in the buffer area I have defined.
> These non-identity operations would presumably be the ones that
> would generate the crystal contacts.
>
> However, a large number of the atoms only had one mapping, which
> was not the identity - e.g. asu_mappings.get_rt_mx(x).as_xyz()
> might give '-x+1/2,-y+1/2,z-1/2' for an atom with only one mapping.
>
> The effect of this is that the operations given by this generate
> symmetry copies that do not contact my input model in real space.
>
> I presume this means that asu_mappings is mapping the atoms in my
> model to some other arbitrary asu, as well as the atoms in the
> buffer zone around it? Therefore the symmetry operations given are
> for these mapped atoms?
>
> Can anyone help, or suggest an alternative way to generate the
> crystal contacts?
>
> Thanks,
> Nick
>
> --
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