Frankly when faced with these problems of generating symmetry
equivalents i revert to almn, where a) I can guarantee the
orthogonalisation is as I expect, and b) it generates an exhaustive set
of symmetry equivalent peaks.
But that is old technology..
If you have two dimers in the asymmetric unit, and only one self
rotation peak, it is reasonable to assume the 2 fold NCS axes are in
the same orientation. That doesnt necessarily mean the dimers have to
be in exactly the same orientation about that 2 fold, s0 I guess you
might expect a split peak for a pseudo translation.
another trick that sometimes helps - is there a dimer model with exptl
data available? again back to almn but sometimes you can fit the two
sets of hklin together. Although theoretically you should get the same
information from search with a dimer model matching Fcalc to Fobs.
Eleanor
Derek Logan wrote:
Thanks to everyone who helped with the self RF problem: Eleanor, Ian,
Claudine, Pietro & Alexei.
Eleanor wrote:
1) It is a bit hard to find out how MOLREP defines its orthogonal
axes - many programs use X0 || a, Yo || b* and in P21 hence Zortho is
|| to c*
If that is what Molrep does then your 2 fold is in the a c* plane, 21
degrees or 111 degrees from c*.
The 2 peaks you see are symmetry equivalents.
This was my interpretation. Glad we agree ;-) The documentation says
"A parallel to X , Cstar parallel to Z"
As for the Patterson - what height are those peaks relative to the
origin?
The peaks are u = 0.129, v = 0.473, w = 0.220 (20% of origin peak
height) and u = 0.180, v = 0.500, w = 0.248 (19%). What I don't get is
why there are two and only one strong 2-fold. 2 dimers in the AU gives
50% solvent, 1 dimer 75%. The crystals diffract to 2.3Å, which would
tip the balance in favour of 50% solvent in my opinion.
With 2 dimers in the asymm unit and with the non-cryst 2-fold
perpendicular to b* you could have such translations between one
monomer and another.
Would the 2-folds of both dimers have to be very similarly oriented?
Maybe one peak masks the other at this resolution?
is there a model - easiest to solve it then analyse this sort of
stuff later!
Believe me, we've been trying for a very long time! The problem is
that it's a leucine rich repeat protein with under 30% sequence
identity to any of the other LRR models out there. I think the failure
of MR is down to a combination of a) the low homology, b) the
pseudosymmetry, c) the nature of the LRR, which means you can get MR
solutions that are out by one or more repeats. Maybe even the internal
symmetry of the whole LRR structure can add to this pathology? We've
had some solutions that looked almost right, but we can never see much
more than what's already in the MR solution.
Ian wrote:
The symmetry of the self-RF is explained in detail in the
documentation for POLARRFN, in fact I would advise you to use this
because you can then plot monoclinic space groups with the unique b
axis along the orthogonal Z axis (NCODE = 3) and then the symmetry is
*much* easier to interpret.
The reason I started using Molrep was that POLARRFN always used to
choke on these data. However that problem seems to have disappeared.
Using ORTH 3 indeed gives a more interpretable plot, as you say.
According to polarrfn.doc the symmetry generated by a 2-fold along b
parallel to Z is (180-theta, 180-phi, kappa) so the peak in the list
(159,180,180) is the same as (21,0,180) which is a NCS 2-fold that
you can see just below centre. The peak (111,0,180) is thus the same
as (69,180,180) near the top which is another NCS 2-fold perp to the
first generated by the crystallographic 2-fold.
Indeed, I see the peak (69, 180, 180) but I don't find it in the list
in the log file from Molrep. I thought that list was supposed to be
exhaustive. Also the plot is not well documented for Molrep. I wrote
to the BB a while ago to ask what the contour levels were but no-one
answered. By Googling I found a crystallisation paper where it was
described as "from 0.5 sigma in steps of 0.5 sigma" but that
information appears to have come by word of mouth. Also, is it just
the "north hemisphere", as Claudine put it, that is plotted?
Anyway, I feel somewhat wiser now...
Derek
