Re: [ccp4bb] Unit cell change - was Re: [ccp4bb] XDS question
It looks like a rhombehedral data set with small deviations from the exact H3 symmetry to give the weak spots. The standard H3 setting has origins at (0,0,0) (1/3, 2/3, 2/3) and (2/3,1/3,1/3) so to get your translation vector to match the conventional H3 one, you will have to reindex the P321 data as -h,-k,l (It isa standard default..) Then the translation vector will become (-1/3,1/3,1/3) Generate two equivalent data sets - one in P321 and one as H32 (2which will not have the weak spots) You should be able to decide whether it is twinned or not by checking the H32 data processing stats. Then I would solve the MR in the H32, and generate a P321 solution by adding 2 extra molecules at 1/3,-1/3,-1/3 and -1/3,1/3,1/3 Rigid body refinement should improve things and hopefully your structure will be OK Eleanor On 06/14/2011 08:07 PM, Edward A. Berry wrote: I'm currently struggling with what I think is a variation on this theme, would appreciate comments as to whether my thinking is reasonable. The space group is basically rhombohedral, but along the lines of spots in the (hexagonal)L direction, with some crystals there are two weak spots between each pair of indexed spots, and denzo tends to index these crystals with a hexagonal lattice whose primitive cell is the same as the hexagonal setting of the rhombohedral space group. There is a 2-fold operator (or perfect twin operator) perpendicular to the 3-fold, so the hexagonal data can be indexed as P321 or H32 (R32 in denzo?). Molecular replacement with one small domain seems to work in P3(1)21 or H32. but no density shows up outside the search model so I'm not sure. If indexed in P321, Xtriage finds native patterson peaks: x y z height p-value(height) ( 0.332,-0.335, 0.333 ) : 73.464 (1.830e-06) ( 0.282,-0.440, 0.345 ) : 5.416 (8.402e-01) The first of which vaguely resembles the translational operators in H32: 155 18 6 H32 PG321 TRIGONAL 'H 3 2' X,Y,Z 2/3+X,1/3+Y,1/3+Z 1/3+X,2/3+Y,2/3+Z So could this be a case of rhombohedral symmetry breaking down into trigonal? eab herman.schreu...@sanofi-aventis.com wrote: Dear Konstatin, First I would reiterate what Fred has said: you only know the space group once the structure has been solved and completely refined. Especially bad maps and unsuccesful molecular replacements may point to a wrong space group assignment. Second, what causes strong and weak reflections? I once worked it out for a case where only one axis was doubled, but I guess that in your case it might be similar: for the small unit cell, your crystal may look like A (all layers of A's, unit cell is A). In case of the doubled unit cell, the crystal will look like ABABABABAB (alternating layers of A and B's, unit cell AB). If A is identical to B, the scattering of the A's will cancel the scattering of the B's for the odd reflections and you have the small unit cell. If the A's are a little different from the B's, your even reflections will have the sum of the scattering of A and B, and the odd reflections will have the difference. So if the odd reflections are weak, this means that the differences between the A and B layers are small and you could consider to ignore them for a preliminary structure, keeping in mind that the resulting electron density will be the sum (superposition) of the densities of the A and B layers. You might get clashes, since the differences in A and B layers may be caused by the crystal packing so I would increase the allowed number of clashes e.g. in Phaser. Once you have the small unit cell, you could try to figure out how the big unit cell may look like. Your situation might be different, but I would definitively try it. Herman -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Marco Lolicato Sent: Monday, June 13, 2011 5:16 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] XDS question Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tr icks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.1363238963856 79type=1theater ...then a little bit more details... so, if I process only the strong spots
Re: [ccp4bb] XDS question
Marco, indeed, as Fred says, the space group I422 may not be the correct one for the large cell. Use the model from the small cell for molecular replacement in the large cell. You may try I4(1)22, I4(1), I4 first, then lower symmetries: look into the International Tables what subgroups of I422 are; I can think of I222, F222, P422, P4 A lazy way would be P1 and then let the symmetry be sorted out by Garib's Zanuda server, but that requires that you collected 180° of data (or at least expand to P1). HTH, Kay Original Message Subject: Re: XDS question Date: Mon, 13 Jun 2011 17:43:50 +0200 From: Vellieux Frederic frederic.velli...@ibs.fr the space group is I422 do you have any other suggestion? Yes, how certain are you of the space group? For myself, I'm never entirely certain of the space group until I have solved the structure... I always keep in mind the other possibilities for space group assignment, if need be. And sometimes the obvious space group is not the space group of the final structure. The computer programs we use only give hints of the solution, but these are only hints. Remember that crystals behave as they want, the fact for example that I(equiv.1) is approximately equal to I(equiv.2) is approximately equal to I(equiv.3) etc does not mean that the relationship between intensities is in fact an equality, it can be just an approximation... With crystals I have learned, everything is possible. It might be an idea to enclose parts of relevant XDS output files for our perusal. Using standard input parameters (for spot selection) as well as your spot selection input parameters. Fred. Marco Lolicato wrote: Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tricks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.136323896385679type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco smime.p7s Description: S/MIME Cryptographic Signature
Re: [ccp4bb] XDS question
Dear Konstatin, First I would reiterate what Fred has said: you only know the space group once the structure has been solved and completely refined. Especially bad maps and unsuccesful molecular replacements may point to a wrong space group assignment. Second, what causes strong and weak reflections? I once worked it out for a case where only one axis was doubled, but I guess that in your case it might be similar: for the small unit cell, your crystal may look like A (all layers of A's, unit cell is A). In case of the doubled unit cell, the crystal will look like ABABABABAB (alternating layers of A and B's, unit cell AB). If A is identical to B, the scattering of the A's will cancel the scattering of the B's for the odd reflections and you have the small unit cell. If the A's are a little different from the B's, your even reflections will have the sum of the scattering of A and B, and the odd reflections will have the difference. So if the odd reflections are weak, this means that the differences between the A and B layers are small and you could consider to ignore them for a preliminary structure, keeping in mind that the resulting electron density will be the sum (superposition) of the densities of the A and B layers. You might get clashes, since the differences in A and B layers may be caused by the crystal packing so I would increase the allowed number of clashes e.g. in Phaser. Once you have the small unit cell, you could try to figure out how the big unit cell may look like. Your situation might be different, but I would definitively try it. Herman -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Marco Lolicato Sent: Monday, June 13, 2011 5:16 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] XDS question Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tr icks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.1363238963856 79type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco
[ccp4bb] Unit cell change - was Re: [ccp4bb] XDS question
I'm currently struggling with what I think is a variation on this theme, would appreciate comments as to whether my thinking is reasonable. The space group is basically rhombohedral, but along the lines of spots in the (hexagonal)L direction, with some crystals there are two weak spots between each pair of indexed spots, and denzo tends to index these crystals with a hexagonal lattice whose primitive cell is the same as the hexagonal setting of the rhombohedral space group. There is a 2-fold operator (or perfect twin operator) perpendicular to the 3-fold, so the hexagonal data can be indexed as P321 or H32 (R32 in denzo?). Molecular replacement with one small domain seems to work in P3(1)21 or H32. but no density shows up outside the search model so I'm not sure. If indexed in P321, Xtriage finds native patterson peaks: x y zheight p-value(height) ( 0.332,-0.335, 0.333 ) : 73.464 (1.830e-06) ( 0.282,-0.440, 0.345 ) :5.416 (8.402e-01) The first of which vaguely resembles the translational operators in H32: 155 18 6 H32 PG321 TRIGONAL 'H 3 2' X,Y,Z 2/3+X,1/3+Y,1/3+Z 1/3+X,2/3+Y,2/3+Z So could this be a case of rhombohedral symmetry breaking down into trigonal? eab herman.schreu...@sanofi-aventis.com wrote: Dear Konstatin, First I would reiterate what Fred has said: you only know the space group once the structure has been solved and completely refined. Especially bad maps and unsuccesful molecular replacements may point to a wrong space group assignment. Second, what causes strong and weak reflections? I once worked it out for a case where only one axis was doubled, but I guess that in your case it might be similar: for the small unit cell, your crystal may look like A (all layers of A's, unit cell is A). In case of the doubled unit cell, the crystal will look like ABABABABAB (alternating layers of A and B's, unit cell AB). If A is identical to B, the scattering of the A's will cancel the scattering of the B's for the odd reflections and you have the small unit cell. If the A's are a little different from the B's, your even reflections will have the sum of the scattering of A and B, and the odd reflections will have the difference. So if the odd reflections are weak, this means that the differences between the A and B layers are small and you could consider to ignore them for a preliminary structure, keeping in mind that the resulting electron density will be the sum (superposition) of the densities of the A and B layers. You might get clashes, since the differences in A and B layers may be caused by the crystal packing so I would increase the allowed number of clashes e.g. in Phaser. Once you have the small unit cell, you could try to figure out how the big unit cell may look like. Your situation might be different, but I would definitively try it. Herman -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Marco Lolicato Sent: Monday, June 13, 2011 5:16 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] XDS question Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tr icks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.1363238963856 79type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco
Re: [ccp4bb] Unit cell change - was Re: [ccp4bb] XDS question
You may well have a rhombohedral obverse/reverse twin. See Acta Cryst. B58 (2002) 477-481. George On Tue, Jun 14, 2011 at 03:07:40PM -0400, Edward A. Berry wrote: I'm currently struggling with what I think is a variation on this theme, would appreciate comments as to whether my thinking is reasonable. The space group is basically rhombohedral, but along the lines of spots in the (hexagonal)L direction, with some crystals there are two weak spots between each pair of indexed spots, and denzo tends to index these crystals with a hexagonal lattice whose primitive cell is the same as the hexagonal setting of the rhombohedral space group. There is a 2-fold operator (or perfect twin operator) perpendicular to the 3-fold, so the hexagonal data can be indexed as P321 or H32 (R32 in denzo?). Molecular replacement with one small domain seems to work in P3(1)21 or H32. but no density shows up outside the search model so I'm not sure. If indexed in P321, Xtriage finds native patterson peaks: x y zheight p-value(height) ( 0.332,-0.335, 0.333 ) : 73.464 (1.830e-06) ( 0.282,-0.440, 0.345 ) :5.416 (8.402e-01) The first of which vaguely resembles the translational operators in H32: 155 18 6 H32 PG321 TRIGONAL 'H 3 2' X,Y,Z 2/3+X,1/3+Y,1/3+Z 1/3+X,2/3+Y,2/3+Z So could this be a case of rhombohedral symmetry breaking down into trigonal? eab herman.schreu...@sanofi-aventis.com wrote: Dear Konstatin, First I would reiterate what Fred has said: you only know the space group once the structure has been solved and completely refined. Especially bad maps and unsuccesful molecular replacements may point to a wrong space group assignment. Second, what causes strong and weak reflections? I once worked it out for a case where only one axis was doubled, but I guess that in your case it might be similar: for the small unit cell, your crystal may look like A (all layers of A's, unit cell is A). In case of the doubled unit cell, the crystal will look like ABABABABAB (alternating layers of A and B's, unit cell AB). If A is identical to B, the scattering of the A's will cancel the scattering of the B's for the odd reflections and you have the small unit cell. If the A's are a little different from the B's, your even reflections will have the sum of the scattering of A and B, and the odd reflections will have the difference. So if the odd reflections are weak, this means that the differences between the A and B layers are small and you could consider to ignore them for a preliminary structure, keeping in mind that the resulting electron density will be the sum (superposition) of the densities of the A and B layers. You might get clashes, since the differences in A and B layers may be caused by the crystal packing so I would increase the allowed number of clashes e.g. in Phaser. Once you have the small unit cell, you could try to figure out how the big unit cell may look like. Your situation might be different, but I would definitively try it. Herman -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Marco Lolicato Sent: Monday, June 13, 2011 5:16 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] XDS question Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tr icks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.1363238963856 79type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.1363238963856 79type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco -- Prof. George M. Sheldrick FRS Dept. Structural Chemistry, University of Goettingen, Tammannstr. 4, D37077 Goettingen, Germany Tel. +49-551-39-3021 or -3068 Fax. +49-551-39-22582
Re: [ccp4bb] XDS question
Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tricks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.136323896385679type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco
Re: [ccp4bb] XDS question
the space group is I422 do you have any other suggestion? Yes, how certain are you of the space group? For myself, I'm never entirely certain of the space group until I have solved the structure... I always keep in mind the other possibilities for space group assignment, if need be. And sometimes the obvious space group is not the space group of the final structure. The computer programs we use only give hints of the solution, but these are only hints. Remember that crystals behave as they want, the fact for example that I(equiv.1) is approximately equal to I(equiv.2) is approximately equal to I(equiv.3) etc does not mean that the relationship between intensities is in fact an equality, it can be just an approximation... With crystals I have learned, everything is possible. It might be an idea to enclose parts of relevant XDS output files for our perusal. Using standard input parameters (for spot selection) as well as your spot selection input parameters. Fred. Marco Lolicato wrote: Thank you to all! @Frederic I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. You're right, I said good map instead of good output values. @Konstantin It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tricks I tried but with no success! :( @Kay and all the others The following links are the images: http://www.facebook.com/photo.php?fbid=2127189780277set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127189180262set=o.136323896385679type=1theater http://www.facebook.com/photo.php?fbid=2127188820253set=o.136323896385679type=1theater ...then a little bit more details... so, if I process only the strong spots I have those cell parameters: a=b=96.66 c=112.26alpha=beta=gamma= 90 f I process all the spots I have those cell parameters: a=b=216.4 c=112.4 alpha=beta=gamma= 90 In both cases the space group is I422. Thank you again to all, do you have any other suggestion? Marco
Re: [ccp4bb] XDS question
Dear Marco, in terms of using XDS, STRONG_PIXEL=99 is _highly_ non-standard (I personally try to stick with the defaults ... although at synchrotrons I do use STRONG_PIXEL=6 instead of the default of 3). I realize this is just a way to pick the strong reflection for indexing (since STRONG_PIXEL is used by COLSPOT , but not by INTEGRATE), but in the long run it would be important to figure out the relation between the strong lattice and the weak lattice (you could post the cell parameters here, or even better upload a few frames to some Internet service so that people can take a look themselves). You seem to be able to solve the structure using the lattice corresponding to the strong spots, if I understand correctly. The lattice that covers both the weak and the strong spots certainly has longer cell axes, and I'd guess that there are maybe twice as many molecules in the ASU. Taking account of this, you may be able to solve the structure using all reflections, and you might get some important insight concerning the binding mode(s) of the substrate. best, Kay Am 20:59, schrieb Marco Lolicato: Dear all, I have a particular problem... so, I have a beautiful crystal with nice diffraction pattern at 2.7A. The diffraction images are composed by very strong spots and weak spots. With XDS, if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement. If I collect only the strongest spots (STRONG_PIXEL=9), I'm able to solve a very good structure... My problem is: I was trying to get the apo-structure of my protein. I obtained nice crystals of the apo-protein, but using the method above, in the structure I have found also the ligand!! (probably incorporated during the overexpression). My protein is a multimer and, biochemically, I found that the endogenus ligand bond to the protein is in the ratio 1:6. ...and I got a crystal in this way. So, is there a way to analyze all spots in the diffraction pattern to have a structure of the apo-protein? Is a good idea discard the strongest spots and try to analyze only the weak spots? If yes, how I can do it? All the best, Marco
[ccp4bb] XDS question
Dear all, I have a particular problem... so, I have a beautiful crystal with nice diffraction pattern at 2.7A. The diffraction images are composed by very strong spots and weak spots. With XDS, if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement. If I collect only the strongest spots (STRONG_PIXEL= 99), I'm able to solve a very good structure... My problem is: I was trying to get the apo-structure of my protein. I obtained nice crystals of the apo-protein, but using the method above, in the structure I have found also the ligand!! (probably incorporated during the overexpression). My protein is a multimer and, biochemically, I found that the endogenus ligand bond to the protein is in the ratio 1:6. ...and I got a crystal in this way. So, is there a way to analyze all spots in the diffraction pattern to have a structure of the apo-protein? Is a good idea discard the strongest spots and try to analyze only the weak spots? If yes, how I can do it? All the best, Marco
Re: [ccp4bb] XDS question
Hi, I have a problem with the following sentence: if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement - if you have a good map (I assume electron density map) then the structure is solved... for me a good map is a map I can interpret. There have been several reports in the literature of structures where you have layers of normal intensities interspersed with layers of very weak spots in the diffraction patterns. Sometimes with a change in space group assignment when you go from integrating the strong spots only to integrating the entire pattern including the very weak intensities. I don't have a reference off my head right now, some other people on the bb may have such references available. But differences in spot intensities (including reassignment of the space group) is described in Morales et al. (2000), Acta Cryst D56, 1408-1412. So the reply I can give is the following: if the diffraction pattern with these strong diffraction intensities interspersed with weak diffraction intensities only comes from one crystal, then processing the strong spots only does not explain the diffraction pattern and is wrong. Similarly with processing the weak diffraction spots only (which would be difficult to do in practice... see below). The model produced should explain the diffraction pattern seen (in terms of space group, layers of molecules arranged in the crystal...). If the diffraction pattern originates from 2 crystals (in different orientations, a case I've had with one large crystal plus a satellite crystal attached to it in the same loop), it is in principle possible to integrate only the diffraction spots from only one of the crystals. No problems for the larger crystal that diffracts more strongly (which is what I did with my data set - the second lattice was ignored). To process the diffraction from the smaller crystal would be tricky: you should have some version of the data frame processing software that processes first the spots coming from the large crystal and would produce a copy of the input frames where the optical density corresponding to the processed reflections is set (on the images, on the frames) to say 0, for the entire range of frames processed. Then you would repeat the autoindexing and frame processing (integration) to take care of the diffraction from the satellite crystal. I don't think such a modification of the data processing programs is available. But could you explain more clearly the problem? Fred. Marco Lolicato wrote: Dear all, I have a particular problem... so, I have a beautiful crystal with nice diffraction pattern at 2.7A. The diffraction images are composed by very strong spots and weak spots. With XDS, if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement. If I collect only the strongest spots (STRONG_PIXEL= 99), I'm able to solve a very good structure... My problem is: I was trying to get the apo-structure of my protein. I obtained nice crystals of the apo-protein, but using the method above, in the structure I have found also the ligand!! (probably incorporated during the overexpression). My protein is a multimer and, biochemically, I found that the endogenus ligand bond to the protein is in the ratio 1:6. ...and I got a crystal in this way. So, is there a way to analyze all spots in the diffraction pattern to have a structure of the apo-protein? Is a good idea discard the strongest spots and try to analyze only the weak spots? If yes, how I can do it? All the best, Marco
Re: [ccp4bb] XDS question
On Sat, 11 Jun 2011, Vellieux Frederic wrote: If the diffraction pattern originates from 2 crystals (in different orientations, a case I've had with one large crystal plus a satellite crystal attached to it in the same loop), it is in principle possible to snip crystal. I don't think such a modification of the data processing programs is available. It is possible to process diffraction spots from both crystals using XDS. The procedure is described here (under 'Index and integrate multiple-crystal diffraction'): http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/Tips_and_Tricks -Konstantin But could you explain more clearly the problem? Fred. Marco Lolicato wrote: Dear all, I have a particular problem... so, I have a beautiful crystal with nice diffraction pattern at 2.7A. The diffraction images are composed by very strong spots and weak spots. With XDS, if I collect all spots I get good map, but it is impossible to solve the structure by molecular replacement. If I collect only the strongest spots (STRONG_PIXEL= 99), I'm able to solve a very good structure... My problem is: I was trying to get the apo-structure of my protein. I obtained nice crystals of the apo-protein, but using the method above, in the structure I have found also the ligand!! (probably incorporated during the overexpression). My protein is a multimer and, biochemically, I found that the endogenus ligand bond to the protein is in the ratio 1:6. ...and I got a crystal in this way. So, is there a way to analyze all spots in the diffraction pattern to have a structure of the apo-protein? Is a good idea discard the strongest spots and try to analyze only the weak spots? If yes, how I can do it? All the best, Marco -- Konstantin Korotkov, Ph.D. Research Scientist University of Washington Department of Biochemistry Box 357742 Seattle, WA 98195-7742 (206)616-4512 k...@u.washington.edu --
Re: [ccp4bb] xds question: inverse beam, lots of wedges
Hi I'd just process it in iMosflm, and run the Quickscale task after integration. With almost no effort you should get a rapid visual indicator (in the graphs produced by Scala) of the discontinuities between the wedges. If the discontinuities are too big, then you might encounter some items of interest during the integration stage... On 31 Mar 2011, at 23:08, Patrick Loll wrote: We've just collected a number of inverse beam data sets. It turns out the crystals showed little radiation damage, so we have a lot of data: 2 x 360 deg for each crystal, broken up into 30 deg wedges. The collection order went like this: 0-30 deg, 180-210, 30-60, 210-240, etc. Now, assuming no slippage, I could simply integrate the first set of data (non-inverse?) in one run: 0-360 deg. However, since the 12 individual wedges making up this 360 deg sweep were not collected immediately one after the other, I don't expect the scale factors for individual images to vary smoothly (there should be discontinuities at the boundaries between wedges). If I do integrate the data in one fell swoop, am I in danger of introducing errors? For example, I seem to recall that denzo had built-in restraints to ensure that scale factors for adjacent images didn't vary by too much. Is there a similar restraint that in XDS that I might run afoul of? The alternative is to integrate each each wedge separately, but with 24 wedges per xtal, this is starting to look a little tedious. Cheers, Pat Harry -- Dr Harry Powell, MRC Laboratory of Molecular Biology, MRC Centre, Hills Road, Cambridge, CB2 0QH
Re: [ccp4bb] xds question: inverse beam, lots of wedges
On 03/31/11 18:08, Patrick Loll wrote: We've just collected a number of inverse beam data sets. It turns out the crystals showed little radiation damage, so we have a lot of data: 2 x 360 deg for each crystal, broken up into 30 deg wedges. The collection order went like this: 0-30 deg, 180-210, 30-60, 210-240, etc. Now, assuming no slippage, I could simply integrate the first set of data (non-inverse?) in one run: 0-360 deg. However, since the 12 individual wedges making up this 360 deg sweep were not collected immediately one after the other, I don't expect the scale factors for individual images to vary smoothly (there should be discontinuities at the boundaries between wedges). So? Isn't that the purpose of scale factors? -- === All Things Serve the Beam === David J. Schuller modern man in a post-modern world MacCHESS, Cornell University schul...@cornell.edu
[ccp4bb] xds question: inverse beam, lots of wedges
We've just collected a number of inverse beam data sets. It turns out the crystals showed little radiation damage, so we have a lot of data: 2 x 360 deg for each crystal, broken up into 30 deg wedges. The collection order went like this: 0-30 deg, 180-210, 30-60, 210-240, etc. Now, assuming no slippage, I could simply integrate the first set of data (non-inverse?) in one run: 0-360 deg. However, since the 12 individual wedges making up this 360 deg sweep were not collected immediately one after the other, I don't expect the scale factors for individual images to vary smoothly (there should be discontinuities at the boundaries between wedges). If I do integrate the data in one fell swoop, am I in danger of introducing errors? For example, I seem to recall that denzo had built-in restraints to ensure that scale factors for adjacent images didn't vary by too much. Is there a similar restraint that in XDS that I might run afoul of? The alternative is to integrate each each wedge separately, but with 24 wedges per xtal, this is starting to look a little tedious. Cheers, Pat
Re: [ccp4bb] xds question: inverse beam, lots of wedges
Pat, at least give it a try with the one sweep approach. We have collected plenty of 360deg data sets on a Rigaku system which requires two omega sweeps at phi 0 and 180 deg. These data sets are for in-house phasing. We haven't seen big issues with running XDS over these images as one continuous sweep. Monitoring scalefactors might be a good indicator. Good luck Jan On Mar 31, 2011, at 3:08 PM, Patrick Loll wrote: We've just collected a number of inverse beam data sets. It turns out the crystals showed little radiation damage, so we have a lot of data: 2 x 360 deg for each crystal, broken up into 30 deg wedges. The collection order went like this: 0-30 deg, 180-210, 30-60, 210-240, etc. Now, assuming no slippage, I could simply integrate the first set of data (non-inverse?) in one run: 0-360 deg. However, since the 12 individual wedges making up this 360 deg sweep were not collected immediately one after the other, I don't expect the scale factors for individual images to vary smoothly (there should be discontinuities at the boundaries between wedges). If I do integrate the data in one fell swoop, am I in danger of introducing errors? For example, I seem to recall that denzo had built-in restraints to ensure that scale factors for adjacent images didn't vary by too much. Is there a similar restraint that in XDS that I might run afoul of? The alternative is to integrate each each wedge separately, but with 24 wedges per xtal, this is starting to look a little tedious. Cheers, Pat -- Jan Abendroth Emerald BioStructures Seattle / Bainbridge Island WA, USA home: Jan.Abendroth_at_gmail.com work: JAbendroth_at_embios.com http://www.emeraldbiostructures.com
[ccp4bb] xds question
Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon !=== File Automaticaly generated by mxCuBE !=== X-Ray data collected at: ESRF_ID14-1 !=== Detector type: ADSC Quantum Q210 !=== Date: Fri Feb 04 03:39:09 2011 !=== User comments: JOB= ALL !XYCORR INIT COLSPOT IDXREF DEFPIX XPLAN INTEGRATE CORRECT !JOB= DEFPIX XPLAN INTEGRATE CORRECT DATA_RANGE= 1 190 SPOT_RANGE= 1 20 SPOT_RANGE= 1 4 !SPOT_RANGE= 187 190 BACKGROUND_RANGE= 1 4 SECONDS=60 MINIMUM_NUMBER_OF_PIXELS_IN_A_SPOT= 6 STRONG_PIXEL= 6.0 OSCILLATION_RANGE= 1.000 STARTING_ANGLE= 0.000 STARTING_FRAME= 1 X-RAY_WAVELENGTH= 0.93340 NAME_TEMPLATE_OF_DATA_FRAMES= ./data/sk1_1_???.img !STARTING_ANGLES_OF_SPINDLE_ROTATION= 0 180 10 !TOTAL_SPINDLE_ROTATION_RANGES= 60 180 10 DETECTOR_DISTANCE= 298.55 DETECTOR= ADSC MINIMUM_VALID_PIXEL_VALUE= 1 OVERLOAD= 65000 ORGX= 1014.79ORGY= 1029.10 NX= 2048 NY= 2048 QX= 0.10200 QY= 0.10200 VALUE_RANGE_FOR_TRUSTED_DETECTOR_PIXELS= 7000 3 DIRECTION_OF_DETECTOR_X-AXIS= 1.0 0.0 0.0 DIRECTION_OF_DETECTOR_Y-AXIS= 0.0 1.0 0.0 ROTATION_AXIS= 1.0 0.0 0.0 INCIDENT_BEAM_DIRECTION= 0.0 0.0 1.0 FRACTION_OF_POLARIZATION= 0.98 POLARIZATION_PLANE_NORMAL= 0.0 1.0 0.0 !== Default value recommended !AIR= 0.00026895 SPACE_GROUP_NUMBER= 0 UNIT_CELL_CONSTANTS= 0 0 0 0 0 0 INCLUDE_RESOLUTION_RANGE= 50.0 2.4 RESOLUTION_SHELLS= 15.0 8.0 4.0 3.0 2.8 2.6 2.5 2.4 FRIEDEL'S_LAW= FALSE !FRIEDEL'S_LAW= TRUE TRUSTED_REGION= 0 1.40 REFINE(INTEGRATE)= BEAM ORIENTATION CELL !== Default value recommended !DELPHI= 3.000 MAXIMUM_NUMBER_OF_PROCESSORS= 16 !MAXIMUM_NUMBER_OF_JOBS= 16
Re: [ccp4bb] xds question
Dear Simon, I don't know how to change the output written to the log-file (maybe with the TEST-card), but I wonder why you don't want to look at all of your data but only those with I/sigI 0? XDS reports all reflections with I/sigI -3 for a good reason. Cheers, Tim On Tue, Feb 08, 2011 at 12:17:35PM +, Simon Kolstoe wrote: Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon !=== File Automaticaly generated by mxCuBE !=== X-Ray data collected at: ESRF_ID14-1 !=== Detector type: ADSC Quantum Q210 !=== Date: Fri Feb 04 03:39:09 2011 !=== User comments: JOB= ALL !XYCORR INIT COLSPOT IDXREF DEFPIX XPLAN INTEGRATE CORRECT !JOB= DEFPIX XPLAN INTEGRATE CORRECT DATA_RANGE= 1 190 SPOT_RANGE= 1 20 SPOT_RANGE= 1 4 !SPOT_RANGE= 187 190 BACKGROUND_RANGE= 1 4 SECONDS=60 MINIMUM_NUMBER_OF_PIXELS_IN_A_SPOT= 6 STRONG_PIXEL= 6.0 OSCILLATION_RANGE= 1.000 STARTING_ANGLE= 0.000 STARTING_FRAME= 1 X-RAY_WAVELENGTH= 0.93340 NAME_TEMPLATE_OF_DATA_FRAMES= ./data/sk1_1_???.img !STARTING_ANGLES_OF_SPINDLE_ROTATION= 0 180 10 !TOTAL_SPINDLE_ROTATION_RANGES= 60 180 10 DETECTOR_DISTANCE= 298.55 DETECTOR= ADSC MINIMUM_VALID_PIXEL_VALUE= 1 OVERLOAD= 65000 ORGX= 1014.79ORGY= 1029.10 NX= 2048 NY= 2048 QX= 0.10200 QY= 0.10200 VALUE_RANGE_FOR_TRUSTED_DETECTOR_PIXELS= 7000 3 DIRECTION_OF_DETECTOR_X-AXIS= 1.0 0.0 0.0 DIRECTION_OF_DETECTOR_Y-AXIS= 0.0 1.0 0.0 ROTATION_AXIS= 1.0 0.0 0.0 INCIDENT_BEAM_DIRECTION= 0.0 0.0 1.0 FRACTION_OF_POLARIZATION= 0.98 POLARIZATION_PLANE_NORMAL= 0.0 1.0 0.0 !== Default value recommended !AIR= 0.00026895 SPACE_GROUP_NUMBER= 0 UNIT_CELL_CONSTANTS= 0 0 0 0 0 0 INCLUDE_RESOLUTION_RANGE= 50.0 2.4 RESOLUTION_SHELLS= 15.0 8.0 4.0 3.0 2.8 2.6 2.5 2.4 FRIEDEL'S_LAW= FALSE !FRIEDEL'S_LAW= TRUE TRUSTED_REGION= 0 1.40 REFINE(INTEGRATE)= BEAM ORIENTATION CELL !== Default value recommended !DELPHI= 3.000 MAXIMUM_NUMBER_OF_PROCESSORS= 16 !MAXIMUM_NUMBER_OF_JOBS= 16 -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A signature.asc Description: Digital signature
Re: [ccp4bb] xds question
Hi Tim, could you write a bit more about this cutoff. I've been using the 0 cutoff for the longest time as it seemed to be much more stringent to report. Jürgen - Jürgen Bosch Johns Hopkins Bloomberg School of Public Health Department of Biochemistry Molecular Biology Johns Hopkins Malaria Research Institute 615 North Wolfe Street, W8708 Baltimore, MD 21205 Phone: +1-410-614-4742 Lab: +1-410-614-4894 Fax: +1-410-955-2926 http://web.mac.com/bosch_lab/http://web.me.com/bosch_lab/ On Feb 8, 2011, at 7:45 AM, Tim Gruene wrote: Dear Simon, I don't know how to change the output written to the log-file (maybe with the TEST-card), but I wonder why you don't want to look at all of your data but only those with I/sigI 0? XDS reports all reflections with I/sigI -3 for a good reason. Cheers, Tim On Tue, Feb 08, 2011 at 12:17:35PM +, Simon Kolstoe wrote: Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon !=== File Automaticaly generated by mxCuBE !=== X-Ray data collected at: ESRF_ID14-1 !=== Detector type: ADSC Quantum Q210 !=== Date: Fri Feb 04 03:39:09 2011 !=== User comments: JOB= ALL !XYCORR INIT COLSPOT IDXREF DEFPIX XPLAN INTEGRATE CORRECT !JOB= DEFPIX XPLAN INTEGRATE CORRECT DATA_RANGE= 1 190 SPOT_RANGE= 1 20 SPOT_RANGE= 1 4 !SPOT_RANGE= 187 190 BACKGROUND_RANGE= 1 4 SECONDS=60 MINIMUM_NUMBER_OF_PIXELS_IN_A_SPOT= 6 STRONG_PIXEL= 6.0 OSCILLATION_RANGE= 1.000 STARTING_ANGLE= 0.000 STARTING_FRAME= 1 X-RAY_WAVELENGTH= 0.93340 NAME_TEMPLATE_OF_DATA_FRAMES= ./data/sk1_1_???.img !STARTING_ANGLES_OF_SPINDLE_ROTATION= 0 180 10 !TOTAL_SPINDLE_ROTATION_RANGES= 60 180 10 DETECTOR_DISTANCE= 298.55 DETECTOR= ADSC MINIMUM_VALID_PIXEL_VALUE= 1 OVERLOAD= 65000 ORGX= 1014.79ORGY= 1029.10 NX= 2048 NY= 2048 QX= 0.10200 QY= 0.10200 VALUE_RANGE_FOR_TRUSTED_DETECTOR_PIXELS= 7000 3 DIRECTION_OF_DETECTOR_X-AXIS= 1.0 0.0 0.0 DIRECTION_OF_DETECTOR_Y-AXIS= 0.0 1.0 0.0 ROTATION_AXIS= 1.0 0.0 0.0 INCIDENT_BEAM_DIRECTION= 0.0 0.0 1.0 FRACTION_OF_POLARIZATION= 0.98 POLARIZATION_PLANE_NORMAL= 0.0 1.0 0.0 !== Default value recommended !AIR= 0.00026895 SPACE_GROUP_NUMBER= 0 UNIT_CELL_CONSTANTS= 0 0 0 0 0 0 INCLUDE_RESOLUTION_RANGE= 50.0 2.4 RESOLUTION_SHELLS= 15.0 8.0 4.0 3.0 2.8 2.6 2.5 2.4 FRIEDEL'S_LAW= FALSE !FRIEDEL'S_LAW= TRUE TRUSTED_REGION= 0 1.40 REFINE(INTEGRATE)= BEAM ORIENTATION CELL !== Default value recommended !DELPHI= 3.000 MAXIMUM_NUMBER_OF_PROCESSORS= 16 !MAXIMUM_NUMBER_OF_JOBS= 16 -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A
Re: [ccp4bb] xds question
Hello Juergen, since sigma is always positive, a negative I/sigI means that the measured intensity is negative, hence I would refer you to the truncate reference On the treatment of negative intensity observations by S. French and K. Wilson, Acta Cryst (1978) A34, 517-525. Why exactly -3 is chosen as a cut-off I cannot say, though, but it's better than 0 as cut-off, if I am not mistaken. Cheers, Tim On Tue, Feb 08, 2011 at 08:33:13AM -0500, Bosch, Juergen wrote: Hi Tim, could you write a bit more about this cutoff. I've been using the 0 cutoff for the longest time as it seemed to be much more stringent to report. Jürgen - Jürgen Bosch Johns Hopkins Bloomberg School of Public Health Department of Biochemistry Molecular Biology Johns Hopkins Malaria Research Institute 615 North Wolfe Street, W8708 Baltimore, MD 21205 Phone: +1-410-614-4742 Lab: +1-410-614-4894 Fax: +1-410-955-2926 http://web.mac.com/bosch_lab/http://web.me.com/bosch_lab/ On Feb 8, 2011, at 7:45 AM, Tim Gruene wrote: Dear Simon, I don't know how to change the output written to the log-file (maybe with the TEST-card), but I wonder why you don't want to look at all of your data but only those with I/sigI 0? XDS reports all reflections with I/sigI -3 for a good reason. Cheers, Tim On Tue, Feb 08, 2011 at 12:17:35PM +, Simon Kolstoe wrote: Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon !=== File Automaticaly generated by mxCuBE !=== X-Ray data collected at: ESRF_ID14-1 !=== Detector type: ADSC Quantum Q210 !=== Date: Fri Feb 04 03:39:09 2011 !=== User comments: JOB= ALL !XYCORR INIT COLSPOT IDXREF DEFPIX XPLAN INTEGRATE CORRECT !JOB= DEFPIX XPLAN INTEGRATE CORRECT DATA_RANGE= 1 190 SPOT_RANGE= 1 20 SPOT_RANGE= 1 4 !SPOT_RANGE= 187 190 BACKGROUND_RANGE= 1 4 SECONDS=60 MINIMUM_NUMBER_OF_PIXELS_IN_A_SPOT= 6 STRONG_PIXEL= 6.0 OSCILLATION_RANGE= 1.000 STARTING_ANGLE= 0.000 STARTING_FRAME= 1 X-RAY_WAVELENGTH= 0.93340 NAME_TEMPLATE_OF_DATA_FRAMES= ./data/sk1_1_???.img !STARTING_ANGLES_OF_SPINDLE_ROTATION= 0 180 10 !TOTAL_SPINDLE_ROTATION_RANGES= 60 180 10 DETECTOR_DISTANCE= 298.55 DETECTOR= ADSC MINIMUM_VALID_PIXEL_VALUE= 1 OVERLOAD= 65000 ORGX= 1014.79ORGY= 1029.10 NX= 2048 NY= 2048 QX= 0.10200 QY= 0.10200 VALUE_RANGE_FOR_TRUSTED_DETECTOR_PIXELS= 7000 3 DIRECTION_OF_DETECTOR_X-AXIS= 1.0 0.0 0.0 DIRECTION_OF_DETECTOR_Y-AXIS= 0.0 1.0 0.0 ROTATION_AXIS= 1.0 0.0 0.0 INCIDENT_BEAM_DIRECTION= 0.0 0.0 1.0 FRACTION_OF_POLARIZATION= 0.98 POLARIZATION_PLANE_NORMAL= 0.0 1.0 0.0 !== Default value recommended !AIR= 0.00026895 SPACE_GROUP_NUMBER= 0 UNIT_CELL_CONSTANTS= 0 0 0 0 0 0 INCLUDE_RESOLUTION_RANGE= 50.0 2.4 RESOLUTION_SHELLS= 15.0 8.0 4.0 3.0 2.8 2.6 2.5 2.4 FRIEDEL'S_LAW= FALSE !FRIEDEL'S_LAW= TRUE TRUSTED_REGION= 0 1.40 REFINE(INTEGRATE)= BEAM ORIENTATION CELL !== Default value recommended !DELPHI= 3.000 MAXIMUM_NUMBER_OF_PROCESSORS= 16 !MAXIMUM_NUMBER_OF_JOBS= 16 -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A signature.asc Description: Digital signature
Re: [ccp4bb] xds question
Hi, I've pasted below the reasons from Dan Gewirth and the HKL2000 manual authors for having a -3 sigma cutoff... I'll add briefly that if you assume the weak data has a Gaussian distribution around zero a -3 sigma cutoff allows you to record ~99.8% of the data. -bob SIGMA CUTOFF Cutoff for rejecting measurements on input. Default = -3.0. Be very careful if you increase this. What is the rationale for using sigma cutoff -3.0 in SCALEPACK? Wouldn't you want to reject all negative intensities? Why shouldn't you use a sigma cutoff 1.0 or zero? The answer to these questions is as follows: The best estimate of I may be negative, due to background subtraction and background fluctuation. Negative measurements typically represent random fluctuations in the detector's response to an X-ray signal. If a measurement is highly negative (= -3[[sigma]]) than it may be more likely the result of a mistake, rather than just random fluctuation. If one eliminates negative fluctuations, but not the positive ones before averaging, the result will be highly biased. In SCALEPACK, sigma cutoff is applied before averaging. If one rejects all negative intensities before averaging a number of things would happen: 1. The averaged intensity would always be positive; 2. For totally random data with redundancy 8, in a shell where there was no signal, , there would be on average 4 positive measurements, with average intensity one sigma. This is because the negative measurements had been thrown out. So the average of the four remaining measurements would be about 2 sigma! This would look like a resolution shell with a meaningful signal; 3. R-merge would be always less than the R-merge with negative measurements included; 4. A SIGMA CUTOFF of 1 would improve R-merge even more, by excluding even more valid measurements. Why should this worry you? Exclusion of valid measurements will deteriorate the final data set. One may notice an inverse relationship between R-merge and data quality as a function of sigma cutoff. So much for using R-merge as any criterion of success. Even the best (averaged) estimate of intensity may be negative. How to use negative I estimates in subsequent phasing and refinement steps is a separate story. The author of SCALEPACK suggests the following: 1. You should never convert I into F. 2. You should square Fcalc and compare it to I. Most, but not all of the crystallography programs do not do this. That is life. In the absence of the proper treatment one can do approximations. One of them is provided by French and also by French and Wilson. An implementation of their ideas is in the CCP4 program TRUNCATE. A very simplified and somewhat imprecise implementation of TRUNCATE is this: if I [[sigma]](I), F=sqrt(I) if I [[sigma]](I), F=sqrt([[sigma]](I)) format SIGMA CUTOFF value default -3 example SIGMA CUTOFF -2.5 referenced from: http://www.hkl-xray.com/hkl_web1/hkl/Scalepack_Keywords.html
Re: [ccp4bb] xds question
Hi, oh, I'm also surprised people seem to use something else than '-3' as the cutoff, i.e. are throwing away data. This, obviously, brings into new light all the discussions (which I definitely don't wish to restart) on the 'cutoff values' in R(sym) and I/sI which you use to determine the 'resolution limit'...and gives one more thing for referees to think about/require when looking at Table 1. I am sure most of them, and the readers, take it for granted that no data were thrown out before calculating those numbers...and sure, the effects of actually using those data might occasionally be more severe than a drop of, say, 1% in the apparent overall R(sym) or an increase in I/sI. Petri On Feb 8, 2011, at 3:07 PM, Robert Immormino wrote: Hi, I've pasted below the reasons from Dan Gewirth and the HKL2000 manual authors for having a -3 sigma cutoff... I'll add briefly that if you assume the weak data has a Gaussian distribution around zero a -3 sigma cutoff allows you to record ~99.8% of the data. -bob SIGMA CUTOFF Cutoff for rejecting measurements on input. Default = -3.0. Be very careful if you increase this. What is the rationale for using sigma cutoff -3.0 in SCALEPACK? Wouldn't you want to reject all negative intensities? Why shouldn't you use a sigma cutoff 1.0 or zero? The answer to these questions is as follows: The best estimate of I may be negative, due to background subtraction and background fluctuation. Negative measurements typically represent random fluctuations in the detector's response to an X-ray signal. If a measurement is highly negative (= -3[[sigma]]) than it may be more likely the result of a mistake, rather than just random fluctuation. If one eliminates negative fluctuations, but not the positive ones before averaging, the result will be highly biased. In SCALEPACK, sigma cutoff is applied before averaging. If one rejects all negative intensities before averaging a number of things would happen: 1. The averaged intensity would always be positive; 2. For totally random data with redundancy 8, in a shell where there was no signal, , there would be on average 4 positive measurements, with average intensity one sigma. This is because the negative measurements had been thrown out. So the average of the four remaining measurements would be about 2 sigma! This would look like a resolution shell with a meaningful signal; 3. R-merge would be always less than the R-merge with negative measurements included; 4. A SIGMA CUTOFF of 1 would improve R-merge even more, by excluding even more valid measurements. Why should this worry you? Exclusion of valid measurements will deteriorate the final data set. One may notice an inverse relationship between R-merge and data quality as a function of sigma cutoff. So much for using R-merge as any criterion of success. Even the best (averaged) estimate of intensity may be negative. How to use negative I estimates in subsequent phasing and refinement steps is a separate story. The author of SCALEPACK suggests the following: 1. You should never convert I into F. 2. You should square Fcalc and compare it to I. Most, but not all of the crystallography programs do not do this. That is life. In the absence of the proper treatment one can do approximations. One of them is provided by French and also by French and Wilson. An implementation of their ideas is in the CCP4 program TRUNCATE. A very simplified and somewhat imprecise implementation of TRUNCATE is this: if I [[sigma]](I), F=sqrt(I) if I [[sigma]](I), F=sqrt([[sigma]](I)) formatSIGMA CUTOFF value default -3 example SIGMA CUTOFF -2.5 referenced from: http://www.hkl-xray.com/hkl_web1/hkl/Scalepack_Keywords.html --- Petri Kursula, PhD Group Leader and Docent of Neurobiochemistry (University of Oulu, Finland) Visiting Scientist (CSSB-HZI, DESY, Hamburg, Germany) www.biochem.oulu.fi/kursula www.desy.de/~petri petri.kurs...@oulu.fi petri.kurs...@desy.de ---
Re: [ccp4bb] xds question
Hi Simon, I've put my answer into the XDSwiki article FAQ http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/FAQ#why_do_the_latest_XDS.2FXSCALE_versions_only_give_a_single_table.2C_with_I.2Fsigma.3E.3D-3_cutoff.3F because I was asked this privately a couple of times, after this was changed for the May 2010 version. In short, the table describes the data that are written out by XDS/XSCALE, and giving more than one table obscures this fact and tends to confuse users (which table should I use? - the -3 sigma cutoff table - then why are there the others?). Others have already mentioned in this thread that SCALEPACK uses the same cutoff of -3 sigma. I'm not sure about the cutoff in MOSFLM/SCALA. I do realize now that some people have been using the tables for deciding on a suitable resolution cutoff. With a bit of scripting, this can be overcome - pls ask me by email. hope that helps, Kay Am 08.02.2011 20:58, schrieb Simon Kolstoe: Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon
Re: [ccp4bb] xds question
On Tue, Feb 8, 2011 at 7:17 AM, Simon Kolstoe s.kols...@ucl.ac.uk wrote: XDS [...] signal/noise = -3.0. i'd be interested to know if there is an equivalent in scala... perhaps 'REJECT 6 ALL -8' -Bryan
Re: [ccp4bb] xds question
No cutoffs in Scala Phil On 8 Feb 2011, at 20:13, Kay Diederichs wrote: Hi Simon, I've put my answer into the XDSwiki article FAQ http://strucbio.biologie.uni-konstanz.de/xdswiki/index.php/FAQ#why_do_the_latest_XDS.2FXSCALE_versions_only_give_a_single_table.2C_with_I.2Fsigma.3E.3D-3_cutoff.3F because I was asked this privately a couple of times, after this was changed for the May 2010 version. In short, the table describes the data that are written out by XDS/XSCALE, and giving more than one table obscures this fact and tends to confuse users (which table should I use? - the -3 sigma cutoff table - then why are there the others?). Others have already mentioned in this thread that SCALEPACK uses the same cutoff of -3 sigma. I'm not sure about the cutoff in MOSFLM/SCALA. I do realize now that some people have been using the tables for deciding on a suitable resolution cutoff. With a bit of scripting, this can be overcome - pls ask me by email. hope that helps, Kay Am 08.02.2011 20:58, schrieb Simon Kolstoe: Dear ccp4bb, I am quite a fan of XDS and have just upgraded to the latest version. Normally, to assess the quality of my data, I look at the tables in CORRECT.LP and especially the table SUBSET OF INTENSITY DATA WITH SIGNAL/NOISE = 0.0 AS FUNCTION OF RESOLUTION. However in my latest run I only get a single table for all the data i.e. for signal/noise = -3.0. Is there a command I can put in my XDS.INP that will give me all the other tables or has the CORRECT.LP logfile been altered in the most recent version of XDS? (FYI my xds.inp obtained from the ESRF last week is copied below) Thanks, Simon
