You would get a different MR solution in P41212 than in P43212 so you shouldnt test the SAME pdb in both SGS? Not sure I am understanding this though. Eleanor
On 19 November 2013 05:02, #CHEN DAN# <[email protected]> wrote: > Hi Eleanor, > > I checked P43212 and P41212 by changing the header of mtz file and running > refmac for the same PDB input. P43212 is a better match than P41212. > > Sincerely, > Dan > > ________________________________________ > From: CCP4 bulletin board <[email protected]> on behalf of Eleanor Dodson > <[email protected]> > Sent: Monday, November 18, 2013 8:47 PM > To: [email protected] > Subject: Re: [ccp4bb] translational pseudo symmetry > > I guess you have checked that P43212 is a better match than P41212? > (And that you are running REFMAC against an mtz file with the same > symmetry as the input PDB - you may need to change the SG in the mtz > header by hand. > mtzutils hklin P41212.mtz hklout P43212.mtz > symm P43212 > end > > Or vice versa.. > > Sorry - THIS IS CRAZY but there you are.. > > Re the pseudo translation -Randy summs up the situation very clearly. > I would build my model by hand actually but I am sure PHASER does itwell too! > > > Something I dont understand but maybe it is to do with your patterson > sampling. > > > Peak 3 is a consequence of Pk 1 and Pk2 - > Pk 5 is the consequence of Pk 1 and Pk4 > but the peak heights dont exactly fit.. > > Eleanor > > On 18 November 2013 10:19, Randy Read <[email protected]> wrote: >> Dear Dan, >> >> First, you don't want to reprocess in the smaller cell. What xtriage is >> saying is that, if *and only if* the translation detected in the Patterson >> map were an exact crystallographic translation, then you would get the >> smaller cell. However, in order for that to be a plausible hypothesis, the >> Patterson peaks would have to be near to 100% of the origin peak. >> >> You actually seem to have a very interesting case, where the Patterson peaks >> are related by multiples of approximately the same translation. If you take >> a translation of 1/2,1/2,1/6 and multiply it by 1, 2 and 3, you get >> something close to the three biggest peaks in your Patterson (taking account >> of lattice translations), and these are related by the Patterson inversion >> centre to what you get if you multiply by 4 and 5. So the six molecules >> should be related to each other by something close to a repeated translation >> of 1/2,1/2,1/6. (You should check this in the solution that you already >> have.) If this were exact, you would have a smaller cell, but it's not >> exact, and one way in which it is not exact is that the translations along z >> are not exactly multiples of 1/6. >> >> This is reminiscent of a structure that we recently collaborated with >> Mariusz Jaskolski and Zbyszek Dauter to solve (paper accepted for >> publication in Acta D). In that case, there are seven translations of >> approximately 0,0,1/7. The difficulty with cases like this is figuring out >> how to break the exact symmetry. Any solution that has approximately the >> right translations will basically fit the data, but you need to find the >> right combination of deviations from the exact symmetry to get an optimal >> answer. If you get the wrong deviations from exact symmetry, the refinement >> will stall, and this may be the problem that you're facing. >> >> You can deal with problems like this in Phaser by using the TNCS NMOL 6 >> command (to say that there are 6 copies related by repeated applications of >> the same translation). You should tell Phaser to use the 1/2,1/2,0.174 >> vector (TNCS TRA VECTOR 0.5 0.5 0.174), and hopefully this will break the >> symmetry in a way that subsequent rigid-body refinement can deal with. I'm >> happy to give you more advice on this, off-line, because this kind of case >> isn't something that we've figured out how to deal with automatically yet. >> The optimal approach probably involves getting a deeper understanding of >> commensurate modulation, which is another way of thinking about >> pseudo-translations. >> >> Best wishes, >> >> Randy Read >> >> On 18 Nov 2013, at 09:19, #CHEN DAN# <[email protected]> wrote: >> >> Dear experts, >> >> I am working on one dataset (2.5A) which was processed using space group >> P43212 ( 107.9, 107.9, 313.7; 90, 90, 90). >> After running MR with 6 molecules in ASU and one round of refmac, the R >> factors are high (38%/45%). >> I ran phenix.xtriage and found that translational pseudo symmetry is likely >> present. It suggested that the space group is I4122 with the unit cell about >> 1/3 smaller (I paste the patterson analyses below). >> I tried to reprocess the data to get the suggested space group and unit cell >> using HKL2000. But the index always gives a long c axis about 313A. >> Could you provide any suggestions on how to proceed? >> >> Patterson analyses >> ------------------ >> >> Largest Patterson peak with length larger than 15 Angstrom >> >> Frac. coord. : 0.500 0.500 0.174 >> Distance to origin : 93.757 >> Height (origin=100) : 55.763 >> p_value(height) : 3.018e-05 >> >> >> The reported p_value has the following meaning: >> The probability that a peak of the specified height >> or larger is found in a Patterson function of a >> macro molecule that does not have any translational >> pseudo symmetry is equal to 3.018e-05. >> p_values smaller than 0.05 might indicate >> weak translational pseudo symmetry, or the self vector of >> a large anomalous scatterer such as Hg, whereas values >> smaller than 1e-3 are a very strong indication for >> the presence of translational pseudo symmetry. >> >> The full list of Patterson peaks is: >> >> x y z height p-value(height) >> ( 0.500, 0.500, 0.174 ) : 55.763 (3.018e-05) >> ( 0.500, 0.500, 0.500 ) : 51.209 (5.796e-05) >> ( 0.000, 0.000, 0.326 ) : 32.915 (8.699e-04) >> ( 0.000, 0.000, 0.348 ) : 18.765 (1.266e-02) >> ( 0.500, 0.500, 0.151 ) : 11.396 (9.756e-02) >> >> If the observed pseudo translationals are crystallographic >> the following spacegroups and unit cells are possible: >> >> space group operator unit cell of reference setting >> I 41 2 2 (a+1/4,b+1/4,3*c) x+1/2, y+1/2, z+1/6 (107.94, 107.94, 104.58, >> 90.00, 90.00, 90.00) >> >> >> Thanks, >> Dan >> >> >> ------ >> Randy J. Read >> Department of Haematology, University of Cambridge >> Cambridge Institute for Medical Research Tel: + 44 1223 336500 >> Wellcome Trust/MRC Building Fax: + 44 1223 336827 >> Hills Road E-mail: [email protected] >> Cambridge CB2 0XY, U.K. www-structmed.cimr.cam.ac.uk >>
