I'm currently struggling with what I think is a variation on this theme,
would appreciate comments as to whether my thinking is reasonable.
The space group is basically rhombohedral, but along the lines of spots
in the (hexagonal)L direction, with some crystals there are two weak
spots between each pair of indexed spots, and denzo tends to index these
crystals with a hexagonal lattice whose primitive cell is the same as
the hexagonal setting of the rhombohedral space group.
There is a 2-fold operator (or perfect twin operator) perpendicular
to the 3-fold, so the hexagonal data can be indexed as P321 or H32
(R32 in denzo?). Molecular replacement with one small domain seems
to work in P3(1)21 or H32. but no density shows up outside the
search model so I'm not sure.
If indexed in P321, Xtriage finds native patterson peaks:
x y z height p-value(height)
( 0.332,-0.335, 0.333 ) : 73.464 (1.830e-06)
( 0.282,-0.440, 0.345 ) : 5.416 (8.402e-01)
The first of which vaguely resembles the translational operators in H32:
155 18 6 H32 PG321 TRIGONAL 'H 3 2'
X,Y,Z
2/3+X,1/3+Y,1/3+Z
1/3+X,2/3+Y,2/3+Z
So could this be a case of rhombohedral symmetry breaking down into
trigonal?
eab
herman.schreu...@sanofi-aventis.com wrote:
Dear Konstatin,
First I would reiterate what Fred has said: you only know the space
group once the structure has been solved and completely refined.
Especially bad maps and unsuccesful molecular replacements may point to
a wrong space group assignment.
Second, what causes strong and weak reflections? I once worked it out
for a case where only one axis was doubled, but I guess that in your
case it might be similar: for the small unit cell, your crystal may look
like AAAAAAAAA (all layers of A's, unit cell is A). In case of the
doubled unit cell, the crystal will look like ABABABABAB (alternating
layers of A and B's, unit cell AB). If A is identical to B, the
scattering of the A's will cancel the scattering of the B's for the odd
reflections and you have the small unit cell. If the A's are a little
different from the B's, your even reflections will have the sum of the
scattering of A and B, and the odd reflections will have the difference.
So if the odd reflections are weak, this means that the differences
between the A and B layers are small and you could consider to ignore
them for a preliminary structure, keeping in mind that the resulting
electron density will be the sum (superposition) of the densities of the
A and B layers. You might get clashes, since the differences in A and B
layers may be caused by the crystal packing so I would increase the
allowed number of clashes e.g. in Phaser. Once you have the small unit
cell, you could try to figure out how the big unit cell may look like.
Your situation might be different, but I would definitively try it.
Herman
-----Original Message-----
From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of
Marco Lolicato
Sent: Monday, June 13, 2011 5:16 PM
To: CCP4BB@JISCMAIL.AC.UK
Subject: Re: [ccp4bb] XDS question
Thank you to all!
@Frederic
I have a problem with the following sentence:
"if I collect all spots I get good map, but it is impossible to solve
the structure by molecular replacement" - if you have a good map (I
assume electron density map) then the structure is solved... for me a
good map is a map I can interpret.
You're right, I said "good map" instead of "good output values".
@Konstantin
It is possible to process diffraction spots from both crystals using
XDS. The procedure is described here (under 'Index and integrate
multiple-crystal diffraction'):
http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tr
icks
I tried but with no success! :(
@Kay and all the others
The following links are the images:
http://www.facebook.com/photo.php?fbid=2127189780277&set=o.1363238963856
79&type=1&theater
http://www.facebook.com/photo.php?fbid=2127189180262&set=o.1363238963856
79&type=1&theater
http://www.facebook.com/photo.php?fbid=2127188820253&set=o.1363238963856
79&type=1&theater
...then a little bit more details...
so, if I process only the strong spots I have those cell parameters:
a=b=96.66 c=112.26 alpha=beta=gamma= 90
f I process all the spots I have those cell parameters:
a=b=216.4 c=112.4 alpha=beta=gamma= 90
In both cases the space group is I422.
Thank you again to all, do you have any other suggestion?
Marco
