Re: [ccp4bb] translational pseudo symmetry
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# chen0...@e.ntu.edu.sg 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 CCP4BB@JISCMAIL.AC.UK on behalf of Eleanor Dodson eleanor.dod...@york.ac.uk Sent: Monday, November 18, 2013 8:47 PM To: CCP4BB@JISCMAIL.AC.UK 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 rj...@cam.ac.uk 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# chen0...@e.ntu.edu.sg 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.:0.5000.5000.174 Distance to origin : 93.757 Height (origin=100) : 55.763 p_value(height) :3.018e-05 The reported
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 rj...@cam.ac.uk 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# chen0...@e.ntu.edu.sg 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.:0.5000.5000.174 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 zheight p-value(height) ( 0.500, 0.500, 0.174 ) : 55.763 (3.018e-05) ( 0.500, 0.500,
Re: [ccp4bb] translational pseudo symmetry
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 CCP4BB@JISCMAIL.AC.UK on behalf of Eleanor Dodson eleanor.dod...@york.ac.uk Sent: Monday, November 18, 2013 8:47 PM To: CCP4BB@JISCMAIL.AC.UK 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 rj...@cam.ac.uk 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# chen0...@e.ntu.edu.sg 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.:0.5000.5000.174 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
