The space group in which you do the RF makes no difference (assuming the model in each case is identical of course) because you're superposing the same amount of scattering matter: it has nothing to do with the fraction of the contents of the AU. In the high symmetry space group with say 1 instance of a differently oriented model per AU you are still only superposing 1 AU at a time, and each AU gives a separate peak in the RF (some of which may be related by the symmetry of the RF and some not: e.g. the 3-fold related peaks in cubic space groups). In the low symmetry space group you still get 1 peak per differently oriented instance of the model.
This assumes that the high symmetry space group is the correct one; if it turns out that the subgroup is correct then of course that will give the better solution. -- Ian On Wed, Feb 16, 2011 at 5:57 PM, Edward A. Berry <ber...@upstate.edu> wrote: > I think the question was not concerning right vs wrong space > group, but right vs a lower symmetry superset space group - > P4322 vs P43 for example. I would also be interested in > the answer. > At first glance it appears harder in the lower symmetry > because with each monomer you would be searching for a smaller > fraction of the contents of the AU. But if the rotation > function doesn't consider space group symmetry (?) > does that matter? > eab > > Vellieux Frederic wrote: >> >> Ting-Wei Jiang wrote: >>> >>> Dear experts, >>> >>> Sorry for a simple question but confusing me so much! >>> >>> Does it make bad effects on determining the number of identical >>> molecules in ASU by choosing low symmetry space group. >>> For example,If I choose lowest symmetry(p4) instead of higher >>> one(p43212). >>> Does it cause any trouble in determining structure while we try to >>> find the solution by molrep. >>> If molrep compare the patterson of observe and model in P1 space >>> group, it doesn't matter in choosing a lower s.g.? >>> >>> Any suggestion will be appreciated. Thanks in advance. >>> >>> Ting wei >> >> For molecular replacement: it does not matter if you carry out the >> rotation search in P422 or P43212. To understand this, ask youself the >> following question: what is the symmetry of the Patterson compared to >> that of the crystal? The symmetry of the Patterson is obtained from the >> symmetry of the crystal by converting all translation operators to the >> corresponding non-translation operators and adding a centre of symmetry. >> Hence for the rotation function, the symmetry of the Patterson will be >> P4/mmm for space group P422 and P4/mmm for P43212. Hence no difference >> for the rotation function. >> >> Different for P4 though: the symmetry of the Patterson there is P4/m. >> >> Hence the rotation function can be computed once for all space groups >> that have Patterson symmetry P4/mmm. No need to repeat the rotation >> function calculations several times. >> >> The situation is different for the translation function however. This is >> the stage at which you distinguish between all these space groups >> (including the pair of enantiomorphs P41212 and p43212). Failure to >> assign the proper space group cannot give you a satisfactory model. >> >> HTH, >> >> Fred. >> >
