Thank you Dale,

I will "hit-the-books" to better the rotation matrices. I am concluding
from all of this that the space group is indeed P212121. So I still wonder
why I have some outliers in the intensity stats for the two additional
screw axis and why R and Rfree both drop by 5% when I apply a twin law to
refinement in P21.

Thanks for your help.

-Yarrow


>    Since Phil is no doubt in bed, I'll answer the easier part.  Your
> second matrix is nearly the equivalent position (x,-y,-z).  This
> is a two-fold rotation about the x axis.  You also have a translation
> of about 31 A along x so if your A cell edge is about 62 A you have
> a 2_1 screw.
>
> Dale Tronrud
>
> On 10/15/2013 12:29 PM, Yarrow Madrona wrote:
>> Hi Phil,
>>
>> Thanks for your help.
>>
>> I ran a "Find-NCS" routine in the phenix package. It came up with what I
>> pasted below:
>> I am assuming the the first rotation matrix is just the identity. I need
>> to read more to understand rotation matrices but I think the second one
>> should have only a single -1 to account for a possible perfect 2(1)
>> screw
>> axis between the two subunits in the P21 asymetric unit. I am not sure
>> why
>> there are two -1 values. I may be way off in my interpretation in which
>> case I will go read some more. I will also try what you suggested.
>> Thanks.
>>
>> -Yarrow
>>
>> NCS operator using PDB
>>
>> #1 new_operator
>> rota_matrix    1.0000    0.0000    0.0000
>> rota_matrix    0.0000    1.0000    0.0000
>> rota_matrix    0.0000    0.0000    1.0000
>> tran_orth     0.0000    0.0000    0.0000
>>
>> center_orth   17.7201    1.4604   71.4860
>> RMSD = 0
>> (Is this the identity?)
>>
>> #2 new_operator
>>
>> rota_matrix    0.9994   -0.0259    0.0250
>> rota_matrix   -0.0260   -0.9997    0.0018
>> rota_matrix    0.0249   -0.0025   -0.9997
>> tran_orth   -30.8649  -11.9694  166.9271
>>> Hello Yarrow,
>>>
>>> Since you have a refined molecular replacement solution I recommend
>>> using that rather than global intensity statistics.
>>>
>>> Obviously if you solve in P21 and it's really P212121 you should have
>>> twice the number of molecules in the asymmetric unit and one half of
>>> the
>>> P21 asymmetric unit should be identical to the other half.
>>>
>>> Since you've got decent resolution I think you can determine the real
>>> situation for yourself: one approach would be to test to see if you can
>>> symmetrize the P21 asymmetric unit so that the two halves are
>>> identical.
>>>   You could do this via stiff NCS restraints (cartesian would be better
>>> than dihedral).  After all the relative XYZs and even B-factors would
>>> be
>>> more or less identical if you've rescaled a P212121 crystal form in
>>> P21.
>>>   If something violates the NCS than it can't really be P212121.
>>>
>>> Alternatively you can look for clear/obvious symmetry breaking between
>>> the two halves: different side-chain rotamers for surface side-chains
>>> for example.  If you've got an ordered, systematic, difference in
>>> electron density between the two halves of the asymmetric unit in P21
>>> then that's a basis for describing it as P21 rather than P212121.
>>> However if the two halves look nearly identical, down to equivalent
>>> water molecule densities, then you've got no experimental evidence that
>>> P21 with 2x molecules generates a better model than P212121 than 1x
>>> molecules.  An averaging program would show very high correlation
>>> between the two halves of the P21 asymmetric unit if it was really
>>> P212121 and you could overlap the maps corresponding to the different
>>> monomers using those programs.
>>>
>>> Phil Jeffrey
>>> Princeton
>>>
>>>
>>
>>
>
>


-- 
Yarrow Madrona

Graduate Student
Molecular Biology and Biochemistry Dept.
University of California, Irvine
Natural Sciences I, Rm 2403
Irvine, CA 92697

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