Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
Thanks Randy, The a cell edge is 56.118, so not exactly half of 30.86. I am currently refining using NCS cartesian restraints as Phil suggested. Then I will visually inspect the model as well as compare b-factors. Thanks for your suggestions, I will look into them. -Yarrow > Hi, > > It's not uncommon for pseudosymmetry to be found together with twinning, > and the presence of pseudosymmetry perturbs the statistics used to test > for twinning. In that circumstance, as Phil suggests, a really good way > to see what is going on is to take the lower symmetry solution and see if > it really obeys higher symmetry, but you can do that either with > coordinates or calculated structure factors. > > Your NCS matrix specifies a 2-fold rotation around an axis that is about 1 > degree off the x axis. Whether that 1 degree matters or not depends on > how precisely the molecules are placed in the MR solution. If 30.8649 is > precisely half of the a-cell edge, then this corresponds to a 2(1) screw > axis, but whether or not that is crystallographic depends on whether the > origin of that axis is in the right place relative to the 2(1) you're > assuming is correct. Working all that out from coordinates can be a bit > of a challenge, which will really have you hitting the books! > > The other way we've approached this kind of problem is to take the Fcalcs > from an MR model (usually solved in P1 if possible to avoid making any > assumptions about which symmetry operators are correct) and then use > either pointless or xtriage to see if those Fcalcs obey higher symmetry. > Another good approach is to use the zanuda program in the CCP4 suite, > which is designed to answer questions about pseudosymmetry and other > related problems. > > Good luck! > > Randy Read > > - > Randy J. Read > Department of Haematology, University of Cambridge > Cambridge Institute for Medical ResearchTel: +44 1223 336500 > Wellcome Trust/MRC Building Fax: +44 1223 336827 > Hills Road > E-mail: rj...@cam.ac.uk > Cambridge CB2 0XY, U.K. > www-structmed.cimr.cam.ac.uk > > On 15 Oct 2013, at 22:31, Yarrow Madrona wrote: > >> 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_matrix1.0.0. rota_matrix0.1.0. rota_matrix0.0.1. tran_orth 0.0.0. center_orth 17.72011.4604 71.4860 RMSD = 0 (Is this the identity?) #2 new_operator rota_matrix0.9994 -0.02590.0250 rota_matrix -0.0260 -0.99970.0018 rota_matrix0.0249 -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 identic
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
Hi, It's not uncommon for pseudosymmetry to be found together with twinning, and the presence of pseudosymmetry perturbs the statistics used to test for twinning. In that circumstance, as Phil suggests, a really good way to see what is going on is to take the lower symmetry solution and see if it really obeys higher symmetry, but you can do that either with coordinates or calculated structure factors. Your NCS matrix specifies a 2-fold rotation around an axis that is about 1 degree off the x axis. Whether that 1 degree matters or not depends on how precisely the molecules are placed in the MR solution. If 30.8649 is precisely half of the a-cell edge, then this corresponds to a 2(1) screw axis, but whether or not that is crystallographic depends on whether the origin of that axis is in the right place relative to the 2(1) you're assuming is correct. Working all that out from coordinates can be a bit of a challenge, which will really have you hitting the books! The other way we've approached this kind of problem is to take the Fcalcs from an MR model (usually solved in P1 if possible to avoid making any assumptions about which symmetry operators are correct) and then use either pointless or xtriage to see if those Fcalcs obey higher symmetry. Another good approach is to use the zanuda program in the CCP4 suite, which is designed to answer questions about pseudosymmetry and other related problems. Good luck! Randy Read - Randy J. Read Department of Haematology, University of Cambridge Cambridge Institute for Medical ResearchTel: +44 1223 336500 Wellcome Trust/MRC Building Fax: +44 1223 336827 Hills RoadE-mail: rj...@cam.ac.uk Cambridge CB2 0XY, U.K. www-structmed.cimr.cam.ac.uk On 15 Oct 2013, at 22:31, Yarrow Madrona wrote: > 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_matrix1.0.0. >>> rota_matrix0.1.0. >>> rota_matrix0.0.1. >>> tran_orth 0.0.0. >>> >>> center_orth 17.72011.4604 71.4860 >>> RMSD = 0 >>> (Is this the identity?) >>> >>> #2 new_operator >>> >>> rota_matrix0.9994 -0.02590.0250 >>> rota_matrix -0.0260 -0.99970.0018 >>> rota_matrix0.0249 -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,
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
R factors cannot be used to detect twining. The traditional R is calculated using structure factors (roughly the square root of intensity) but you can't do that calculation in the presence of twining because each structure factor contributes to two intensities. The formula for the "R" in the presence of twining is very different than that of the formula used in its absence. It would have been better to have used a different name and prevent the confusion. If you are worried about your systematic absences you need to figure out which images they were recorded on and judge the spot for yourself. Everything you have said points to your crystal being P212121 (or very nearly P212121). Dale Tronrud On 10/15/2013 02:31 PM, Yarrow Madrona wrote: > 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_matrix1.0.0. >>> rota_matrix0.1.0. >>> rota_matrix0.0.1. >>> tran_orth 0.0.0. >>> >>> center_orth 17.72011.4604 71.4860 >>> RMSD = 0 >>> (Is this the identity?) >>> >>> #2 new_operator >>> >>> rota_matrix0.9994 -0.02590.0250 >>> rota_matrix -0.0260 -0.99970.0018 >>> rota_matrix0.0249 -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 >>> >>> >> >> > >
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
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_matrix1.0.0. >> rota_matrix0.1.0. >> rota_matrix0.0.1. >> tran_orth 0.0.0. >> >> center_orth 17.72011.4604 71.4860 >> RMSD = 0 >> (Is this the identity?) >> >> #2 new_operator >> >> rota_matrix0.9994 -0.02590.0250 >> rota_matrix -0.0260 -0.99970.0018 >> rota_matrix0.0249 -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
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
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_matrix1.0.0. > rota_matrix0.1.0. > rota_matrix0.0.1. > tran_orth 0.0.0. > > center_orth 17.72011.4604 71.4860 > RMSD = 0 > (Is this the identity?) > > #2 new_operator > > rota_matrix0.9994 -0.02590.0250 > rota_matrix -0.0260 -0.99970.0018 > rota_matrix0.0249 -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 >> >> > >
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
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_matrix1.0.0. rota_matrix0.1.0. rota_matrix0.0.1. tran_orth 0.0.0. center_orth 17.72011.4604 71.4860 RMSD = 0 (Is this the identity?) #2 new_operator rota_matrix0.9994 -0.02590.0250 rota_matrix -0.0260 -0.99970.0018 rota_matrix0.0249 -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
Re: [ccp4bb] A case of perfect pseudomerehedral twinning?
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
[ccp4bb] A case of perfect pseudomerehedral twinning?
Hello crystallography experts, I think I may have a case of perfect pseudomerehedral twinning where P222 Laue symmetry is actually P2 with twinning and aniostropic data masks intensity based twinning statistics. I would like some input to determine if my reasoning makes sense. I would like to perform a twin test using local intensity differences (J E Padilla and T O Yeates, 2003) but I am not sure how to do this. I am a little worried that using a twin law in a lower symmetry group than appropriate will give the drop in Rfree/Rwork that I am seeing and the crystal is not really twinned. 1. I have processed in P222 and P21. In each case I get good scaling statistics using XDS (seebelow) 2. Systematic absences clearly point to a single 2(1)-fold screw axis 3. There appears to "almost" be two more 2(1) screw axis except that there are some significant violations in these absences leading me to believe that they are not true srew axis. 4. Patterson map shows no peaks relating molecules by translation in P222 (1 molec/asu) 5. I have difficulty interpreting the patterson map in P2 (2molec/asu) (I am new to this so could be my own fault) 7. Intensity statistics show no twinning but the data is very aniosotropic suggesting that twinning can be masked. The cumulative distribuiton of z= (I/) deviates for centric reflections my a max of 0.104 and the curve falls well below for all values of (Phenix triage). 8. Molecular replacement (Phaser) in P222 gives a single solution in P212121. Every other choice of space group fails to give a solution. 9. Refinement in P212121 space group stalls around 34/31% Rfree/Rwork. Althrough there are geometry problems to correct and water has not been added. 10. Molecular replacement in P2 give single solution in P21 11. Refinement against the same model used above gives 36.5 33.8 Rfree/Rwork without twin law and 32/30 Rfree/Rwork with twin law = h, -k, -l I have shown scaling for P212121. Statistics are nearly the same for P222 scaling except XDS throws up a warning that I have outliers in my systematic absences. P212121 20-1.7A CELL 56.27967.257 106.264 Res OBS UNIQUE POS COMP R(obs) R(exp)I/Sig Rmeas CC1/2 4.96 13696 1911 1989 96.1% 2.6% 2.7% 13681 61.96 2.8% 99.9* 23* 1.044 1400 3.56 24387 3163 3167 99.9% 2.5% 2.8% 24386 64.65 2.7% 99.9*9 0.888 2672 2.93 22483 3969 3995 99.3% 3.2% 3.1% 22463 46.05 3.5% 99.9*1 0.880 3292 2.54 20108 4685 4712 99.4% 4.2% 4.1% 20076 30.45 4.8% 99.8*1 0.839 3710 2.28 20966 5221 5268 99.1% 5.5% 5.6% 20923 22.69 6.3% 99.7*-1 0.779 3932 2.08 21599 5702 5779 98.7% 7.3% 7.7% 21521 16.73 8.5% 99.4* 0 0.756 3932 1.93 21589 6216 6289 98.8% 11.1% 12.1% 21478 10.73 13.1% 98.5* 0 0.735 3755 1.80 19392 579 6709 98.1% 17.5% 19.7% 19089 6.32 21.3% 96.1* -3 0.695 2421 1.70 10889 5093 7126 71.5% 24.8% 27.6% 9561 3.63 31.7% 89.2*2 0.750 350 total 175109 42539 45034 94.5% 4.0% 4.2% 173178 23.25 4.6% 99.9*1 0.808 25464 0 210 3.203 0.6806E+03 0.3320E+0220.50 3* 0 220 3.057 0.9423E+04 0.3288E+0328.66 1 0 230 2.924 0.3205E+03 0.4586E+02 6.99 1* 0 240 2.802 0.1231E+03 0.4612E+02 2.67 1 0 250 2.690 0.2306E+02 0.4566E+02 0.51 1* 0 260 2.587 0.1540E+04 0.7494E+0220.56 1 0 270 2.491 0.3212E+03 0.5579E+02 5.76 1* 0 280 2.402 0.2706E+04 0.1156E+03 23.42 1 0 290 2.319 0.3761E+02 0.4931E+02 0.76 1* 0 300 2.242 0.4053E+03 0.6758E+02 6.00 1 0 310 2.170 0.4495E+03 0.6956E+02 6.46 1* 0 320 2.102 0.3573E+04 0.1442E+03 24.78 1 0 330 2.038 0.5625E+02 0.7052E+02 0.80 1* 0 340 1.978 0.2167E+04 0.1090E+0319.88 1 0 350 1.922 0.3802E+02 0.6082E+02 0.63 1* 0 360 1.868 0.1628E+04 0.9449E+0217.23 1 30018.760 0.6193E+01 0.5268E+01 1.18 2* 400 14.070 0.4882E+04 0.1672E+0329.19 1 500 11.256 0.3065E+02 0.8641E+01 3.55 2* 700 8.040 0.1511E+03 0.1280E+0211.81 2* 80 0 7.035 0.5131E+04 0.1255E+0340.87 2 90 0 6.253 -0.2260E-01 0.1419E+02 0.00 2* 1000 5.628 0.7113E+03 0.2502E+0228.43 2 1100 5.116 0.3705E+03 0.2093E+0217.70 2* 1300 4.329 0.4986E+03 0.2448E+0220.37 2* 1500 3.752 0.8513E+03 0.3547E+0224.00 2* 1600 3.517 0.7354E+03 0.3602E+0220.42 2 1700 3.311 0.2188E+03 0.3146E+02
[ccp4bb] A case of perfect pseudomerehedral twinning?
Hello crystallography experts, I think I may have a case of perfect pseudomerehedral twinning where P222 Laue symmetry is actually P2 with twinning and aniostropic data masks intensity based twinning statistics. I would like some input to determine if my reasoning makes sense. I would like to perform a twin test using local intensity differences (J E Padilla and T O Yeates, 2003) but I am not sure how to do this. I am a little worried that using a twin law in a lower symmetry group than appropriate will give the drop in Rfree/Rwork that I am seeing and the crystal is not really twinned. 1. I have processed in P222 and P21. In each case I get good scaling statistics using XDS (seebelow) 2. Systematic absences clearly point to a single 2(1)-fold screw axis 3. There appears to "almost" be two more 2(1) screw axis except that there are some significant violations in these absences leading me to believe that they are not true srew axis. 4. Patterson map shows no peaks relating molecules by translation in P222 (1 molec/asu) 5. I have difficulty interpreting the patterson map in P2 (2molec/asu) (I am new to this so could be my own fault) 7. Intensity statistics show no twinning but the data is very aniosotropic suggesting that twinning can be masked. The cumulative distribuiton of z= (I/) deviates for centric reflections my a max of 0.104 and the curve falls well below for all values of (Phenix triage). 8. Molecular replacement (Phaser) in P222 gives a single solution in P212121. Every other choice of space group fails to give a solution. 9. Refinement in P212121 space group stalls around 34/31% Rfree/Rwork. Althrough there are geometry problems to correct and water has not been added. 10. Molecular replacement in P2 give single solution in P21 11. Refinement against the same model used above gives 36.5 33.8 Rfree/Rwork without twin law and 32/30 Rfree/Rwork with twin law = h, -k, -l I have shown scaling for P212121. Statistics are nearly the same for P222 scaling except XDS throws up a warning that I have outliers in my systematic absences. P212121 20-1.7A CELL 56.27967.257 106.264 Res OBS UNIQUE POS COMP R(obs) R(exp)I/Sig Rmeas CC1/2 4.96 13696 1911 1989 96.1% 2.6% 2.7% 13681 61.96 2.8% 99.9* 23* 1.044 1400 3.56 24387 3163 3167 99.9% 2.5% 2.8% 24386 64.65 2.7% 99.9*9 0.888 2672 2.93 22483 3969 3995 99.3% 3.2% 3.1% 22463 46.05 3.5% 99.9*1 0.880 3292 2.54 20108 4685 4712 99.4% 4.2% 4.1% 20076 30.45 4.8% 99.8*1 0.839 3710 2.28 20966 5221 5268 99.1% 5.5% 5.6% 20923 22.69 6.3% 99.7*-1 0.779 3932 2.08 21599 5702 5779 98.7% 7.3% 7.7% 21521 16.73 8.5% 99.4* 0 0.756 3932 1.93 21589 6216 6289 98.8% 11.1% 12.1% 21478 10.73 13.1% 98.5* 0 0.735 3755 1.80 19392 579 6709 98.1% 17.5% 19.7% 19089 6.32 21.3% 96.1* -3 0.695 2421 1.70 10889 5093 7126 71.5% 24.8% 27.6% 9561 3.63 31.7% 89.2*2 0.750 350 total 175109 42539 45034 94.5% 4.0% 4.2% 173178 23.25 4.6% 99.9*1 0.808 25464 0 210 3.203 0.6806E+03 0.3320E+0220.50 3* 0 220 3.057 0.9423E+04 0.3288E+0328.66 1 0 230 2.924 0.3205E+03 0.4586E+02 6.99 1* 0 240 2.802 0.1231E+03 0.4612E+02 2.67 1 0 250 2.690 0.2306E+02 0.4566E+02 0.51 1* 0 260 2.587 0.1540E+04 0.7494E+0220.56 1 0 270 2.491 0.3212E+03 0.5579E+02 5.76 1* 0 280 2.402 0.2706E+04 0.1156E+03 23.42 1 0 290 2.319 0.3761E+02 0.4931E+02 0.76 1* 0 300 2.242 0.4053E+03 0.6758E+02 6.00 1 0 310 2.170 0.4495E+03 0.6956E+02 6.46 1* 0 320 2.102 0.3573E+04 0.1442E+03 24.78 1 0 330 2.038 0.5625E+02 0.7052E+02 0.80 1* 0 340 1.978 0.2167E+04 0.1090E+0319.88 1 0 350 1.922 0.3802E+02 0.6082E+02 0.63 1* 0 360 1.868 0.1628E+04 0.9449E+0217.23 1 30018.760 0.6193E+01 0.5268E+01 1.18 2* 400 14.070 0.4882E+04 0.1672E+0329.19 1 500 11.256 0.3065E+02 0.8641E+01 3.55 2* 700 8.040 0.1511E+03 0.1280E+0211.81 2* 80 0 7.035 0.5131E+04 0.1255E+0340.87 2 90 0 6.253 -0.2260E-01 0.1419E+02 0.00 2* 1000 5.628 0.7113E+03 0.2502E+0228.43 2 1100 5.116 0.3705E+03 0.2093E+0217.70 2* 1300 4.329 0.4986E+03 0.2448E+0220.37 2* 1500 3.752 0.8513E+03 0.3547E+0224.00 2* 1600 3.51
[ccp4bb] A case of perfect pseudomerehedral twinning?
Hello crystallography experts, I think I may have a case of perfect pseudomerehedral twinning where P222 Laue symmetry is actually P2 with twinning and aniostropic data masks intensity based twinning statistics. I would like some input to determine if my reasoning makes sense. I would like to perform a twin test using local intensity differences (J E Padilla and T O Yeates, 2003) but I am not sure how to do this. I am a little worried that using a twin law in a lower symmetry group than appropriate will give the drop in Rfree/Rwork that I am seeing and the crystal is not really twinned. 1. I have processed in P222 and P21. In each case I get good scaling statistics using XDS (seebelow) 2. Systematic absences clearly point to a single 2(1)-fold screw axis 3. There appears to "almost" be two more 2(1) screw axis except that there are some significant violations in these absences leading me to believe that they are not true srew axis. 4. Patterson map shows no peaks relating molecules by translation in P222 (1 molec/asu) 5. I have difficulty interpreting the patterson map in P2 (2molec/asu) (I am new to this so could be my own fault) 7. Intensity statistics show no twinning but the data is very aniosotropic suggesting that twinning can be masked. The cumulative distribuiton of z= (I/) deviates for centric reflections my a max of 0.104 and the curve falls well below for all values of (Phenix triage). 8. Molecular replacement (Phaser) in P222 gives a single solution in P212121. Every other choice of space group fails to give a solution. 9. Refinement in P212121 space group stalls around 34/31% Rfree/Rwork. Althrough there are geometry problems to correct and water has not been added. 10. Molecular replacement in P2 give single solution in P21 11. Refinement against the same model used above gives 36.5 33.8 Rfree/Rwork without twin law and 32/30 Rfree/Rwork with twin law = h, -k, -l I have shown scaling for P212121. Statistics are nearly the same for P222 scaling except XDS throws up a warning that I have outliers in my systematic absences. P212121 20-1.7A CELL 56.27967.257 106.264 Res OBS UNIQUE POS COMP R(obs) R(exp)I/Sig Rmeas CC1/2 4.96 13696 1911 1989 96.1% 2.6% 2.7% 13681 61.96 2.8% 99.9* 23* 1.044 1400 3.56 24387 3163 3167 99.9% 2.5% 2.8% 24386 64.65 2.7% 99.9*9 0.888 2672 2.93 22483 3969 3995 99.3% 3.2% 3.1% 22463 46.05 3.5% 99.9*1 0.880 3292 2.54 20108 4685 4712 99.4% 4.2% 4.1% 20076 30.45 4.8% 99.8*1 0.839 3710 2.28 20966 5221 5268 99.1% 5.5% 5.6% 20923 22.69 6.3% 99.7*-1 0.779 3932 2.08 21599 5702 5779 98.7% 7.3% 7.7% 21521 16.73 8.5% 99.4* 0 0.756 3932 1.93 21589 6216 6289 98.8% 11.1% 12.1% 21478 10.73 13.1% 98.5* 0 0.735 3755 1.80 19392 579 6709 98.1% 17.5% 19.7% 19089 6.32 21.3% 96.1* -3 0.695 2421 1.70 10889 5093 7126 71.5% 24.8% 27.6% 9561 3.63 31.7% 89.2*2 0.750 350 total 175109 42539 45034 94.5% 4.0% 4.2% 173178 23.25 4.6% 99.9*1 0.808 25464 0 210 3.203 0.6806E+03 0.3320E+0220.50 3* 0 220 3.057 0.9423E+04 0.3288E+0328.66 1 0 230 2.924 0.3205E+03 0.4586E+02 6.99 1* 0 240 2.802 0.1231E+03 0.4612E+02 2.67 1 0 250 2.690 0.2306E+02 0.4566E+02 0.51 1* 0 260 2.587 0.1540E+04 0.7494E+0220.56 1 0 270 2.491 0.3212E+03 0.5579E+02 5.76 1* 0 280 2.402 0.2706E+04 0.1156E+0323.42 1 0 290 2.319 0.3761E+02 0.4931E+02 0.76 1* 0 300 2.242 0.4053E+03 0.6758E+02 6.00 1 0 310 2.170 0.4495E+03 0.6956E+02 6.46 1* 0 320 2.102 0.3573E+04 0.1442E+0324.78 1 0 330 2.038 0.5625E+02 0.7052E+02 0.80 1* 0 340 1.978 0.2167E+04 0.1090E+0319.88 1 0 350 1.922 0.3802E+02 0.6082E+02 0.63 1* 0 360 1.868 0.1628E+04 0.9449E+0217.23 1 30018.760 0.6193E+01 0.5268E+01 1.18 2* 40014.070 0.4882E+04 0.1672E+0329.19 1 50011.256 0.3065E+02 0.8641E+01 3.55 2* 700 8.040 0.1511E+03 0.1280E+0211.81 2* 800 7.035 0.5131E+04 0.1255E+0340.87 2 900 6.253 -0.2260E-01 0.1419E+02 0.00 2* 1000 5.628 0.7113E+03 0.2502E+0228.43 2 1100 5.116 0.3705E+03 0.2093E+0217.70 2* 1300 4.329 0.4986E+03 0.2448E+0220.37
