Why not just go to P1, then build up the symmetry? Is completeness low? JPK
On Mon, Oct 3, 2011 at 10:14 AM, Eleanor Dodson <c...@ysbl.york.ac.uk> wrote: > Further Qs. > > Do you have a noncryst translation parallel to the b axis (ctruncate will > list any such translation..) > > If the b shift is 0.5 then the 0k0 "absences" will be present whether the > spacegroup is P2 or P21. > > How many Xe sites do you expect? If there is only one then phasing is more > difficult in monoclinic SGs - you have to break the centrosymmetry of the > heavy atom distribution. > > Do you have native data without Xe? > > I always check for peaks which are consistent in the isomorphous and > anomalous difference pattersons. > > The NCS symmetry will doubtless make the exptl phasing more complicated, but > it should help you with averaging later! > Eleanor > > > > On 10/03/2011 03:33 PM, Marta Ferraroni wrote: >> >> Dear all, >> >> We collected some Xenon derivatives of a protein (an heterotetramer >> α2β2)that seems to crystallize in P21 since the 0k0 reflections with >> k=2n+1 are not present. However in the anomalous Patterson we found >> strong peaks in the section v=0 yet none in the Harker section v=1/2. >> Furthermore we weren't able to solve the structure both in P2 and P21 >> using these derivatives with the most commonly used programs. >> >> The cell is 89 125 90 90 102 90 so a is approximately equal to c that >> could permit pseudomerohedral twinning albeit the tests (Padilla-Yeates >> and Britton) estimate a fraction of twinning around 0.05. >> >> Data cannot be scaled as C orthorhombic even if the data reduction >> programs (XDS and imosflm) assign a higher score to the related >> orthorhombic cell with dimensions 112 139 124 90.00 90.00 90.00. In the >> native Patterson map there are not strong peaks. According to the >> Matthews coefficient the asymmetric unit contains 2 heterotetramers and >> the self rotation function may indicate a 222 non crystallographic >> symmetry with one two-fold axis perpendicular to the crystallographic >> one (see figure attached). >> >> Our questions are: >> >> 1) why the anomalous Patterson is not consistent with the space group? >> >> 2) Is there the possibility that the NCS could hamper the determination >> of the correct space group and eventually determine a lower estimate of >> the twinning fraction? >> >> thanks in advance for your help >> >> Regards, >> >> >> Marta Ferraroni >> Dept. of Chemistry >> University of Florence >> Italy >> > -- ******************************************* Jacob Pearson Keller Northwestern University Medical Scientist Training Program cel: 773.608.9185 email: j-kell...@northwestern.edu *******************************************
