Another cause of difficulty - nothing really to do with the spacegroup
selection - is when one copy of the model has much higher B factors than
others. Most MR searches assume that the copies contribute more or less
equally to scattering.
If you assign too high a symmetry this will make MR less likely to work.
This usually happens when the data is twinned.
Eleanor
On 02/16/2011 05:57 PM, Edward A. Berry 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.
>>
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