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# <[log in to unmask]> 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 <[log in to unmask]> on behalf of Eleanor Dodson <[log in to unmask]>
> Sent: Monday, November 18, 2013 8:47 PM
> To: [log in to unmask]
> 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 <[log in to unmask]> 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# <[log in to unmask]> 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: [log in to unmask]
>> Cambridge CB2 0XY, U.K. www-structmed.cimr.cam.ac.uk
>>
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