Re: [ccp4bb] Rotation
Dear David, The following will be of interest:- *J. Chem. Soc., Faraday Trans.*, 1993, *89*, 2667-2675 (*DOI: *10.1039/FT9938902667 ) and Y P Nieh PhD Thesis 1997 available at :- http://ethos.bl.uk/Home.do [ie just search on his name and the pdf is available.] All best wishes, John On Fri, Dec 2, 2016 at 6:35 PM, David Schuller wrote: > Probably you were wondering what the diffraction patter would look like if > you rotated a lysozyme crystal 360 degrees rather than a fraction of a > degree. Wonder no more. > > > > -- > === > All Things Serve the Beam > === >David J. Schuller >modern man in a post-modern world >MacCHESS, Cornell University >schul...@cornell.edu > > -- Professor John R Helliwell DSc
Re: [ccp4bb] Rotation
Hey, that's cool! Now if you spin that image in the XY plane around the center, you get powder diffraction... JPK -Original Message- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of David Schuller Sent: Friday, December 02, 2016 1:35 PM To: CCP4BB@JISCMAIL.AC.UK Subject: [ccp4bb] Rotation Probably you were wondering what the diffraction patter would look like if you rotated a lysozyme crystal 360 degrees rather than a fraction of a degree. Wonder no more. -- === All Things Serve the Beam === David J. Schuller modern man in a post-modern world MacCHESS, Cornell University schul...@cornell.edu
Re: [ccp4bb] rotation peak number in new version phaser
Hi, I would only suggest using the brute rotation function if you want to use the "rotate around" option or do something other than a full rotation search, because the fast search followed by rescoring should give you essentially the same list of peaks (but faster). If you want to get a bigger list of orientations, you should be able to open the "Expert parameters" section of the ccp4i GUI for Phaser, then change the percent cutoff in the "Purge rotation peaks" line to something lower than the default of 75%. Alternatively, if you're running Phaser from a script, you can include commands like: PURGE ROT ENABLE ON PURGE ROT PERCENT 50.0 I've just tested and this works for me. Let me know if you're still having trouble with this. Best wishes, Randy Read On 5 Dec 2013, at 10:11, LISA wrote: > hi all, > > I try to solve the crystal structures of a mult-domain protein with phaser. I > cannot solve the structure by automatic search using domain structure as the > models. I tried to do the rotation search using the fast or brute rotation > function. I defined the rotation search peaks number is 1000. But I only get > no more than 100 peaks. I tried the phaser in ccp4-6.3.0 and ccp4-6.4.0. But > I can only get much less peaks than I required in the rotation. But the old > version phaser can give about 1000 peaks if I use brute rotation function ask > for 1000 peaks. How can I more rotation peaks in the new version phaser > ? > Thank you. > > Lisa -- Randy J. Read Department of Haematology, University of Cambridge Cambridge Institute for Medical Research Tel: + 44 1223 336500 Wellcome Trust/MRC Building Fax: + 44 1223 336827 Hills RoadE-mail: rj...@cam.ac.uk Cambridge CB2 0XY, U.K. www-structmed.cimr.cam.ac.uk
Re: [ccp4bb] Rotation angle and translation distance
LSQKAB gives something of what you want.. If you fit domain_open to domain_closed after the rest of the structure is aligned. you get the COM of the two domains. the rotation angles in euler and polar form I would say the rotation angle was the omega angle and the translation the difference between the COMs Eleanor peter hudson wrote: Hi all I am working with a multidomain protein whcih binds to a ligand. This protein exist in two different states, a) apo form of the protein is closed state and b) holo protein is an open state. One of the domain is mobile and have different conformation compared to its closed conformation. This domain have rotated by an angle of 15 degree and also translated with some Angstrom. Can any one explain me how to calculate the Translation distance as well as the rotation angle of the domain in the different states? I would appreciate the suggestions. Thanks in advance Peter
Re: [ccp4bb] Rotation Function of MOLREP
ALMN can plot the rotation function as a map - although I challenge anyone to make much sense of it! More or less the same methodology as MOLREP. More useful - it lists all peaks generated by the symmetry operators in Eulerian, and polar angles and gives the direct cosines of the rotation axes - you can then check which is related to which by trigonometry. Eleanor To repeat: Hmm - this is tricky. 1) Presumably the size of your cell and your molecule determine how many molecules you expect in an asymmetric unit. THE asym. unit for I4(1)22 is half that of I4(1) if the cell dimensions are the same so it is reasonable to have 2 mols in I41 which generate a tetramer using the crystallographic 2-fold -x,-y,z and 1 mol in I4122 which generates a tetramer using the 2 perpendicular crystallographic 2 folds of I4(1)22, -x,-y,z and y,x,-z. 2) A rotation function is not a Patterson - it is a function of intensities, and can be visualised as an integration of the product of overlapping patterson function within a spherical volume, but it cannot be simpler plotted as a map. ALMN ( an old MR program) did output a "map" of these values along the Eulerian angles alpha, beta, gamma, but it is not very useful for visualisation. 3) peak heights for the rotation function are not very reliable - and without having the values I cant comment. You often get shoulders of existing symmetry related peaks. Did you search with the monomer alone? The solution usually comes out of the translation search. Eleanor Rajan Pillai wrote: Hi All, I want to plot the rotation function that MOLREP uses. I cannot find any output of rotation function in the logfile or in moIrep.doc. I want to locate the peaks of the rotation function, that are shortlisted as solutions, which would help me in understanding the following problem. My protein is a tetramer with 222 point group symmetry. In I4(1) space group Molrep gives two molecules in the asymmetric unit with top two RF peak heights 10 sigma and 6 sigma. The final solution is obtained from these two peaks after TF search. Moreover in case of another protein with the same tetrameric assembly and quite the same unit cell parameters, but in space group I4(1)22, Molrep gives 1 molecule in the asymmetric unit; however, in this case the peaks from the RF are 14 and 12 sigmas and the solution is obtained from the first peak after TF search. In the input file I did not mention the number of monomers to be searched. It detects that based on Matthews coefficient. I am a bit confused as to why in space group I4(1) the RF values are so different. I would have expected them to be closer in values as they are dimer. And also, in case of I4(1)22, the RF values also should be closer. I was wondering if plotting the RF can help in understanding the relation of the peaks and their values based on their location. I believe MOLREP calculates RF over the whole unit cell, instead of the asymmetric unit. Thanks, Rajan
Re: [ccp4bb] Rotation axis
My original posting on this question generated quite a few replies, for which many thanks. I found the most useful pointer to the required algorithm was the link to Mark Gerstein's site given by Ian, and I have now programmed this in a C++ class using Clipper routines, which I can make available if anyone wants it. There are many programs which will give the direction of the rotation axis, but defining a point lying on the axis, such that the minimal translation is just along the axis (ie a screw) is a bit more elusive. Thanks Phil On 29 Jul 2008, at 12:30, Ian Tickle wrote: Phil What I suggested works only if the point x is transformed onto itself, i.e. there's no screw component. The general solution is here: http://bioinfo.mbb.yale.edu/geometry/screw-axis/ There may be a neater way of deriving this in the general case using a homogeneous matrix & co-ordinates but I haven't worked it out yet! Cheers -- IAn -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans Sent: 29 July 2008 09:11 To: CCP4BB@JISCMAIL.AC.UK Subject: Rotation axis If I've go a superposition transformation (x' = Rx + t), as it happens from a superposition in ccp4mg, how do I get the position & direction of the rotation axis (to draw in a picture)? I know that any (orthonormal) transformation can be represented as a rotation about an axis + a screw translation along that axis I'm sure I've done this before ... thanks Phil
Re: [ccp4bb] Rotation axis
Calc-ax (which I mentioned in my post a few days ago) calculates the position of the rotation axis such that translation perpendicular to that axis is zero; all remaining translation (if any) is parallel to the axis. Dave David Borhani, Ph.D. D. E. Shaw Research, LLC 120 West Forty-Fifth Street, 39th Floor New York, NY 10036 [EMAIL PROTECTED] 212-478-0698 http://www.deshawresearch.com > -Original Message- > From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On > Behalf Of Winn, MD (Martyn) > Sent: Sunday, August 03, 2008 10:39 AM > To: CCP4BB@JISCMAIL.AC.UK > Subject: Re: [ccp4bb] Rotation axis > > ROTMAT will also do this. But I am not sure that this is what > Phil wants? > > The superposition transformation includes a translation, so > there is no > locus of points that are left unchanged. You can generate an axis of > rotation for polar angles from the R which will be quite > different from t. > You should be able to translate the rotation axis (change the > origin of > the coordinate system) to get a new transformation x' = R'x + t' which > might be better for visualisation. If Phil's last sentence is > right, you > should be able to arrange it so that t' is parallel to the > rotation axis > of R' > > No, I don't know how to do this off the top of my head, but it sounds > very similar to the transformations done in the Schomaker and > Trueblood > TLS paper. > > Or maybe this is over-complicating things ;-) > > Martyn > > -Original Message----- > From: CCP4 bulletin board on behalf of > [EMAIL PROTECTED] > Sent: Tue 7/29/2008 1:32 PM > To: CCP4BB@JISCMAIL.AC.UK > Subject: Re: [ccp4bb] Rotation axis > > Dear Phil, > Because question keep popping up in the bullitin board about > conversion > from a rotation matrix into rotation angles, I decided to take the > relevant subroutines from an old program from Groningen and > make a jiffy > to do these conversions. It is a small fortran program and > does not need > any additional libraries or subroutines. The program will take a > rotation matrix and translation vector and print all kind of rotation > angles and also the component of the translation vector > parallel to the > rotation axis, which is the number you want. All other > components of the > translation vector can be made zero by choosing the right position of > the rotation axis. > > Best regards, > Herman Schreuder > > > > -Original Message- > From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On Behalf Of > Phil Evans > Sent: Tuesday, July 29, 2008 10:11 AM > To: CCP4BB@JISCMAIL.AC.UK > Subject: [ccp4bb] Rotation axis > > If I've go a superposition transformation (x' = Rx + t), as it happens > from a superposition in ccp4mg, how do I get the position & > direction of > the rotation axis (to draw in a picture)? > I know that any (orthonormal) transformation can be represented as a > rotation about an axis + a screw translation along that axis > > I'm sure I've done this before ... > > thanks > Phil >
Re: [ccp4bb] Rotation axis
ROTMAT will also do this. But I am not sure that this is what Phil wants? The superposition transformation includes a translation, so there is no locus of points that are left unchanged. You can generate an axis of rotation for polar angles from the R which will be quite different from t. You should be able to translate the rotation axis (change the origin of the coordinate system) to get a new transformation x' = R'x + t' which might be better for visualisation. If Phil's last sentence is right, you should be able to arrange it so that t' is parallel to the rotation axis of R' No, I don't know how to do this off the top of my head, but it sounds very similar to the transformations done in the Schomaker and Trueblood TLS paper. Or maybe this is over-complicating things ;-) Martyn -Original Message- From: CCP4 bulletin board on behalf of [EMAIL PROTECTED] Sent: Tue 7/29/2008 1:32 PM To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] Rotation axis Dear Phil, Because question keep popping up in the bullitin board about conversion from a rotation matrix into rotation angles, I decided to take the relevant subroutines from an old program from Groningen and make a jiffy to do these conversions. It is a small fortran program and does not need any additional libraries or subroutines. The program will take a rotation matrix and translation vector and print all kind of rotation angles and also the component of the translation vector parallel to the rotation axis, which is the number you want. All other components of the translation vector can be made zero by choosing the right position of the rotation axis. Best regards, Herman Schreuder -Original Message- From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans Sent: Tuesday, July 29, 2008 10:11 AM To: CCP4BB@JISCMAIL.AC.UK Subject: [ccp4bb] Rotation axis If I've go a superposition transformation (x' = Rx + t), as it happens from a superposition in ccp4mg, how do I get the position & direction of the rotation axis (to draw in a picture)? I know that any (orthonormal) transformation can be represented as a rotation about an axis + a screw translation along that axis I'm sure I've done this before ... thanks Phil
Re: [ccp4bb] Rotation axis
Hi Phil, If you have a reasonably recent version of the cctbx around (phenix, cci apps, maybe ccp4), try this: import scitbx.math rotation_matrix = (1,0,0,0,0,1,0,-1,0) fm = scitbx.math.r3_rotation_axis_and_angle_from_matrix(r=rotation_matrix) print fm.axis print fm.angle(deg=True) The implementation (with comments) is in scitbx/include/scitbx/math/r3_rotation.h Ralf - Original Message From: Phil Evans <[EMAIL PROTECTED]> To: CCP4BB@JISCMAIL.AC.UK Sent: Tuesday, July 29, 2008 1:10:50 AM Subject: [ccp4bb] Rotation axis If I've go a superposition transformation (x' = Rx + t), as it happens from a superposition in ccp4mg, how do I get the position & direction of the rotation axis (to draw in a picture)? I know that any (orthonormal) transformation can be represented as a rotation about an axis + a screw translation along that axis I'm sure I've done this before ... thanks Phil
Re: [ccp4bb] Rotation axis
And the little jiffy program you want Phil is calc-ax (Version 951120) by Joachim Meyer, University of Freiburg, Germany, which takes an RT mtx (in OMAT format), and gives back all kinds of useful info. A google search didn't turn this program up on the web, or Dr. Meyer; original & my slightly modified code attached, & binary for linux. Dave David Borhani, Ph.D. D. E. Shaw Research, LLC 120 West Forty-Fifth Street, 39th Floor New York, NY 10036 [EMAIL PROTECTED] 212-478-0698 http://www.deshawresearch.com > -Original Message- > From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On > Behalf Of Ian Tickle > Sent: Tuesday, July 29, 2008 5:18 AM > To: CCP4BB@JISCMAIL.AC.UK > Subject: Re: [ccp4bb] Rotation axis > > Hi Phil > > The rotation axis is the locus of points which the > transformation leaves > unmoved, i.e. the eigenvector of the transformation matrix which has a > unit eigenvalue. So writing the transformation in > homogeneous form for > convenience: x' = Sx you need to solve x' = x, or Sx = x, either > analytically or just plug the matrix S into a canned eigenvector > routine. > > Cheers > > -- Ian > > > -Original Message- > > From: [EMAIL PROTECTED] > > [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans > > Sent: 29 July 2008 09:11 > > To: CCP4BB@JISCMAIL.AC.UK > > Subject: Rotation axis > > > > If I've go a superposition transformation (x' = Rx + t), as > > it happens > > from a superposition in ccp4mg, how do I get the position & > > direction > > of the rotation axis (to draw in a picture)? > > I know that any (orthonormal) transformation can be > represented as a > > rotation about an axis + a screw translation along that axis > > > > I'm sure I've done this before ... > > > > thanks > > Phil > > > > > > > Disclaimer > This communication is confidential and may contain privileged > information intended solely for the named addressee(s). It > may not be used or disclosed except for the purpose for which > it has been sent. If you are not the intended recipient you > must not review, use, disclose, copy, distribute or take any > action in reliance upon it. If you have received this > communication in error, please notify Astex Therapeutics Ltd > by emailing [EMAIL PROTECTED] and destroy all > copies of the message and any attached documents. > Astex Therapeutics Ltd monitors, controls and protects all > its messaging traffic in compliance with its corporate email > policy. The Company accepts no liability or responsibility > for any onward transmission or use of emails and attachments > having left the Astex Therapeutics domain. Unless expressly > stated, opinions in this message are those of the individual > sender and not of Astex Therapeutics Ltd. The recipient > should check this email and any attachments for the presence > of computer viruses. Astex Therapeutics Ltd accepts no > liability for damage caused by any virus transmitted by this > email. E-mail is susceptible to data corruption, > interception, unauthorized amendment, and tampering, Astex > Therapeutics Ltd only send and receive e-mails on the basis > that the Company is not liable for any such alteration or any > consequences thereof. > Astex Therapeutics Ltd., Registered in England at 436 > Cambridge Science Park, Cambridge CB4 0QA under number 3751674 > calc-ax.tgz Description: calc-ax.tgz
Re: [ccp4bb] Rotation axis
Dear Phil, Because question keep popping up in the bullitin board about conversion from a rotation matrix into rotation angles, I decided to take the relevant subroutines from an old program from Groningen and make a jiffy to do these conversions. It is a small fortran program and does not need any additional libraries or subroutines. The program will take a rotation matrix and translation vector and print all kind of rotation angles and also the component of the translation vector parallel to the rotation axis, which is the number you want. All other components of the translation vector can be made zero by choosing the right position of the rotation axis. Best regards, Herman Schreuder -Original Message- From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans Sent: Tuesday, July 29, 2008 10:11 AM To: CCP4BB@JISCMAIL.AC.UK Subject: [ccp4bb] Rotation axis If I've go a superposition transformation (x' = Rx + t), as it happens from a superposition in ccp4mg, how do I get the position & direction of the rotation axis (to draw in a picture)? I know that any (orthonormal) transformation can be represented as a rotation about an axis + a screw translation along that axis I'm sure I've done this before ... thanks Phil rotax.f Description: rotax.f
Re: [ccp4bb] Rotation axis
Hi Phil The rotation axis is the locus of points which the transformation leaves unmoved, i.e. the eigenvector of the transformation matrix which has a unit eigenvalue. So writing the transformation in homogeneous form for convenience: x' = Sx you need to solve x' = x, or Sx = x, either analytically or just plug the matrix S into a canned eigenvector routine. Cheers -- Ian > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans > Sent: 29 July 2008 09:11 > To: CCP4BB@JISCMAIL.AC.UK > Subject: Rotation axis > > If I've go a superposition transformation (x' = Rx + t), as > it happens > from a superposition in ccp4mg, how do I get the position & > direction > of the rotation axis (to draw in a picture)? > I know that any (orthonormal) transformation can be represented as a > rotation about an axis + a screw translation along that axis > > I'm sure I've done this before ... > > thanks > Phil > > Disclaimer This communication is confidential and may contain privileged information intended solely for the named addressee(s). It may not be used or disclosed except for the purpose for which it has been sent. If you are not the intended recipient you must not review, use, disclose, copy, distribute or take any action in reliance upon it. If you have received this communication in error, please notify Astex Therapeutics Ltd by emailing [EMAIL PROTECTED] and destroy all copies of the message and any attached documents. Astex Therapeutics Ltd monitors, controls and protects all its messaging traffic in compliance with its corporate email policy. The Company accepts no liability or responsibility for any onward transmission or use of emails and attachments having left the Astex Therapeutics domain. Unless expressly stated, opinions in this message are those of the individual sender and not of Astex Therapeutics Ltd. The recipient should check this email and any attachments for the presence of computer viruses. Astex Therapeutics Ltd accepts no liability for damage caused by any virus transmitted by this email. E-mail is susceptible to data corruption, interception, unauthorized amendment, and tampering, Astex Therapeutics Ltd only send and receive e-mails on the basis that the Company is not liable for any such alteration or any consequences thereof. Astex Therapeutics Ltd., Registered in England at 436 Cambridge Science Park, Cambridge CB4 0QA under number 3751674
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
Ian Tickle schrieb: <... interesting information about AMORE deleted ...> Anyway this is all academic now, I always use Phaser which does all this properly (I have no experience using it for highly ellipsoidal models though). Cheers -- Ian My understanding is that Phaser by default does the full-blown maximum likelihood calculations (which are very slow) _only_ for the peaks identified in a "fast rotation function" (a scaled and variance-weighted version of the Patterson overlap function used in the traditional Crowther (1972) fast rotation function). Therefore the same "radius cutoff caveats" apply to this calculation; only if the correct peak is in the list that is scored with MLRF then it may be identified as such. It appears to me that you if you choose "brute rotation function" then the fast rotation function will not be calculated by Phaser. Kay -- Kay Diederichs http://strucbio.biologie.uni-konstanz.de email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz smime.p7s Description: S/MIME Cryptographic Signature
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
> > Ian, > > I agree with your description, but not quite with your > conclusion. In the > original poster's problem one of the cell axes is so short > that at most a < 20 > Angstrom Patterson outer radius can be used in a > Patterson-based rotation > function. This will inevitably lead to loss of many > interesting Patterson > vectors, and probably a low signal-to-noise ratio. > > So in this particular situation the "direct rotation function" > (http://dx.doi.org/10.1107/S0907444995001284) might have a > distinct advantage > compared to Patterson methods, which is why I suggested it. > Whether this is > enough to solve Pietro's problem remains to be seen. For me > it's a case of "know > your options, and try them". > > best, > > Kay > > -- > Kay Diederichs http://strucbio.biologie.uni-konstanz.de > email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 > Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz > Kay, But there's nothing in say AMoRe to stop you increasing the radius cutoff to whatever is required to optimise the signal/noise ratio, apart from this silly IF statement which I always used to delete (I just commented out the CALL CCPERR statement): ELSE IF (RMAX.GT. AOBS .OR. RMAX .GT. BOBS.OR. RMAX.GT.COBS) + THEN WRITE (6,*) 'Radius: ',RMAX, ' Cell: ', AOBS,BOBS,COBS CALL CCPERR(1,'Sphere radius > cell edge') There's also a program limit in AMoRe on the ratio radius-cutoff/high-res-cutoff due to max Bessel order but since this is currently 500 there will be few cases where it's a limit in practice (it's ~ 80, so max 200 Ang radius for 2.5 Ang data). There's obviously no point in having a radius cutoff bigger than the greatest molecular diameter (there are no intra-vectors longer than this). Some people object to including the transform of the origin peak vectors in adjacent cells, but the direct rotation function is effectively doing just that and no-one complains (as long as it solves their structure)! It seems to me it's particularly important to perform this radius optimisation for highly ellipsoidal molecules, because as you point out a limit of 20 Ang with a molecule 100 Ang long will only include a small fraction of the intra-vectors. There will always be a trade-off between increasing the number the intra-vectors thus improving the signal, and increasing the number of inter-vectors thus increasing the noise. It's just a question of finding where the optimal radius cutoff giving the maximum signal/noise ratio lies. The direct rotation function sometimes used to work better than the Crowther-type function not so much due to the lack of radius cutoff, but more because 1) it uses E's not F's; 2) it uses a correlation coefficient function rather than a product function (the latter is much easier to code using FFT's); 3) the radius cutoff for the Crowther function was often not optimised; and 4) care wasn't taken not to lose strong reflections at low res which several people have demonstrated has a disastrous effect on F-based functions, but is less critical with E-based functions. However use of a correlation coefficient function based on E's is now a feature of AMoRe (if not the CCP4 version then Jorge Navaza's version). Anyway this is all academic now, I always use Phaser which does all this properly (I have no experience using it for highly ellipsoidal models though). Cheers -- Ian Disclaimer This communication is confidential and may contain privileged information intended solely for the named addressee(s). It may not be used or disclosed except for the purpose for which it has been sent. If you are not the intended recipient you must not review, use, disclose, copy, distribute or take any action in reliance upon it. If you have received this communication in error, please notify Astex Therapeutics Ltd by emailing [EMAIL PROTECTED] and destroy all copies of the message and any attached documents. Astex Therapeutics Ltd monitors, controls and protects all its messaging traffic in compliance with its corporate email policy. The Company accepts no liability or responsibility for any onward transmission or use of emails and attachments having left the Astex Therapeutics domain. Unless expressly stated, opinions in this message are those of the individual sender and not of Astex Therapeutics Ltd. The recipient should check this email and any attachments for the presence of computer viruses. Astex Therapeutics Ltd accepts no liability for damage caused by any virus transmitted by this email. E-mail is susceptible to data corruption, interception, unauthorized amendment, and tampering, Astex Therapeutics Ltd only send and receive e-mails on the basis that the Company is not liable for any such alteration or any consequences thereof. Astex Therapeutics Ltd., Registered in England at 436 Cambridge Science Park, Cambridge CB4 0QA under number 3
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
The maximum likelihood method used in Phaser doesn't use any cut-off as it doesn't use the Patterson formulation. It might work better for such an elongated case Phil On 28 Nov 2007, at 17:04, Pietro Roversi wrote: Dear everyone, is any of the currently available molecular replacement programs capable of accepting a description of an ellipsoid (rather than the radius of a sphere) to define the portion of the Patterson around the origin to be used in a Molecular Replacement rotation search? Our search model is an elongated object and we are searching in a cell with a=205 b=100 c=21 Angstrom ... ;-) Ciao Pietro -- Pietro Roversi Sir William Dunn School of Pathology, Oxford University South Parks Road, Oxford OX1 3ER, England UK Tel. 0044-1865-275385
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
Ian Tickle schrieb: -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Kay Diederichs Sent: 02 December 2007 10:07 To: [EMAIL PROTECTED] Cc: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin Pietro Roversi schrieb: Dear everyone, is any of the currently available molecular replacement programs capable of accepting a description of an ellipsoid (rather than the radius of a sphere) to define the portion of the Patterson around the origin to be used in a Molecular Replacement rotation search? Our search model is an elongated object and we are searching in a cell with a=205 b=100 c=21 Angstrom ... ;-) Ciao Pietro Pietro, I believe that the "direct rotation function" in CNS should circumvent the problem entirely - there are no cutoffs AFAIK. sorry for answering late! HTH, Kay -- Kay Diederichs http://strucbio.biologie.uni-konstanz.de email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz Unlike the Rossman & Blow, Crowther and related rotation functions, the "direct rotation function" doesn't use any kind of radius cutoff. However the problem then is obviously that a spherical or ellipsoidal radius cutoff, which serves to reduce the noise level by removing intermolecular vectors in the Patterson, may be essential to improve the discrimination of the correct solution relative to wrong ones in order for the correct solution to appear reasonably close to the top of the list. This has certainly been my experience. In the "direct rotation function" no attempt is made to remove these intermolecular vectors, either those between instances of the search model related by lattice translations, or those between symmetry/lattice-related instances of the target molecule. The direct RF is easier to compute because you only need to compute normalised SF's for the model, and you don't have to worry about the complexities of spherical harmonics & Bessel functions, but otherwise it's not clear to me what its advantage is. -- Ian Disclaimer This communication is confidential and may contain privileged information intended solely for the named addressee(s). It may not be used or disclosed except for the purpose for which it has been sent. If you are not the intended recipient you must not review, use, disclose, copy, distribute or take any action in reliance upon it. If you have received this communication in error, please notify Astex Therapeutics Ltd by emailing [EMAIL PROTECTED] and destroy all copies of the message and any attached documents. Astex Therapeutics Ltd monitors, controls and protects all its messaging traffic in compliance with its corporate email policy. The Company accepts no liability or responsibility for any onward transmission or use of emails and attachments having left the Astex Therapeutics domain. Unless expressly stated, opinions in this message are those of the individual sender and not of Astex Therapeutics Ltd. The recipient should check this email and any attachments for the presence of computer viruses. Astex Therapeutics Ltd accepts no liability for damage caused by any virus transmitted by this email. E-mail is susceptible to data corruption, interception, unauthorized amendment, and tampering, Astex Therapeutics Ltd only send and receive e-mails on the basis that the Company is not liable for any such alteration or any consequences thereof. Astex Therapeutics Ltd., Registered in England at 436 Cambridge Science Park, Cambridge CB4 0QA under number 3751674 Ian, I agree with your description, but not quite with your conclusion. In the original poster's problem one of the cell axes is so short that at most a < 20 Angstrom Patterson outer radius can be used in a Patterson-based rotation function. This will inevitably lead to loss of many interesting Patterson vectors, and probably a low signal-to-noise ratio. So in this particular situation the "direct rotation function" (http://dx.doi.org/10.1107/S0907444995001284) might have a distinct advantage compared to Patterson methods, which is why I suggested it. Whether this is enough to solve Pietro's problem remains to be seen. For me it's a case of "know your options, and try them". best, Kay -- Kay Diederichs http://strucbio.biologie.uni-konstanz.de email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz smime.p7s Description: S/MIME Cryptographic Signature
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
> -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Kay Diederichs > Sent: 02 December 2007 10:07 > To: [EMAIL PROTECTED] > Cc: CCP4BB@JISCMAIL.AC.UK > Subject: Re: [ccp4bb] Rotation search using the Patterson in > a non-spherical neighbourhood of the origin > > Pietro Roversi schrieb: > > Dear everyone, > >is any of the currently available > > molecular replacement programs capable of accepting a > > description of an ellipsoid (rather than the radius > of a sphere) > > to define the portion of the Patterson around the > origin to be > > used in a Molecular > > Replacement rotation search? Our search model is an > elongated > > object and we are searching in a cell with a=205 b=100 c=21 > > Angstrom ... ;-) > > > > Ciao > > > > Pietro > > Pietro, > > I believe that the "direct rotation function" in CNS should > circumvent the > problem entirely - there are no cutoffs AFAIK. > > sorry for answering late! > > HTH, > > Kay > -- > Kay Diederichs http://strucbio.biologie.uni-konstanz.de > email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 > Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz > Unlike the Rossman & Blow, Crowther and related rotation functions, the "direct rotation function" doesn't use any kind of radius cutoff. However the problem then is obviously that a spherical or ellipsoidal radius cutoff, which serves to reduce the noise level by removing intermolecular vectors in the Patterson, may be essential to improve the discrimination of the correct solution relative to wrong ones in order for the correct solution to appear reasonably close to the top of the list. This has certainly been my experience. In the "direct rotation function" no attempt is made to remove these intermolecular vectors, either those between instances of the search model related by lattice translations, or those between symmetry/lattice-related instances of the target molecule. The direct RF is easier to compute because you only need to compute normalised SF's for the model, and you don't have to worry about the complexities of spherical harmonics & Bessel functions, but otherwise it's not clear to me what its advantage is. -- Ian Disclaimer This communication is confidential and may contain privileged information intended solely for the named addressee(s). It may not be used or disclosed except for the purpose for which it has been sent. If you are not the intended recipient you must not review, use, disclose, copy, distribute or take any action in reliance upon it. If you have received this communication in error, please notify Astex Therapeutics Ltd by emailing [EMAIL PROTECTED] and destroy all copies of the message and any attached documents. Astex Therapeutics Ltd monitors, controls and protects all its messaging traffic in compliance with its corporate email policy. The Company accepts no liability or responsibility for any onward transmission or use of emails and attachments having left the Astex Therapeutics domain. Unless expressly stated, opinions in this message are those of the individual sender and not of Astex Therapeutics Ltd. The recipient should check this email and any attachments for the presence of computer viruses. Astex Therapeutics Ltd accepts no liability for damage caused by any virus transmitted by this email. E-mail is susceptible to data corruption, interception, unauthorized amendment, and tampering, Astex Therapeutics Ltd only send and receive e-mails on the basis that the Company is not liable for any such alteration or any consequences thereof. Astex Therapeutics Ltd., Registered in England at 436 Cambridge Science Park, Cambridge CB4 0QA under number 3751674
Re: [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
Pietro Roversi schrieb: Dear everyone, is any of the currently available molecular replacement programs capable of accepting a description of an ellipsoid (rather than the radius of a sphere) to define the portion of the Patterson around the origin to be used in a Molecular Replacement rotation search? Our search model is an elongated object and we are searching in a cell with a=205 b=100 c=21 Angstrom ... ;-) Ciao Pietro Pietro, I believe that the "direct rotation function" in CNS should circumvent the problem entirely - there are no cutoffs AFAIK. sorry for answering late! HTH, Kay -- Kay Diederichs http://strucbio.biologie.uni-konstanz.de email: [EMAIL PROTECTED] Tel +49 7531 88 4049 Fax 3183 Fachbereich Biologie, Universität Konstanz, Box M647, D-78457 Konstanz smime.p7s Description: S/MIME Cryptographic Signature
[ccp4bb] RE : [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin
Hi, I think there is something like that described in this paper: F. M. D. Vellieux "An enveloped cross rotation function" J. Appl. Cryst. (1995). 28, 834-836 If I remember correctly, the calculations where done using x-plor scripts. Pierre Proxima-1 Team Synchrotron-SOLEIL Message d'origine De: CCP4 bulletin board de la part de Pietro Roversi Date: mer. 28/11/2007 18:04 À: CCP4BB@JISCMAIL.AC.UK Objet : [ccp4bb] Rotation search using the Patterson in a non-spherical neighbourhood of the origin Dear everyone, is any of the currently available molecular replacement programs capable of accepting a description of an ellipsoid (rather than the radius of a sphere) to define the portion of the Patterson around the origin to be used in a Molecular Replacement rotation search? Our search model is an elongated object and we are searching in a cell with a=205 b=100 c=21 Angstrom ... ;-) Ciao Pietro -- Pietro Roversi Sir William Dunn School of Pathology, Oxford University South Parks Road, Oxford OX1 3ER, England UK Tel. 0044-1865-275385
Re: [ccp4bb] Rotation Axis visualisation.
Hello David, In CCP4mg you can display any arbitrary vector in a variety of display styles - there is an example in the the online documentation: http://www.ysbl.york.ac.uk/~ccp4mg/ccp4mg_help/vector.html Liz On Tuesday 18 September 2007 10:00, David Briggs wrote: > Good morning ccp4bb-ers! > > I have two protein-protein complex structures, and the orientation of one > of the components shifts slightly with respect to the other component > between the two structures. > > If I run superimpose on the shifting component I get this: > > CROWTHER ALPHA BETA GAMMA 12.41607-9.09294-9.64895 > SPHERICAL POLARS OMEGA PHI CHI 73.10930 -78.96756 9.50379 > DIRECTION COSINES OF ROTATION AXIS 0.18311-0.93918 0.29055 > > Angle between rotation axis and Centroid vector93.14726 > > * Note: Since this angle between rotation axis and Centroid vector is > near to 90.0 this may represent a pure rotation *** > > This is fine - Its what I expected. However. I would like to know how I can > graphically represent this axis in Coot/Pymol/CCP4MG? > > This has probably been asked before, but a quick google of the archives > reveals nowt. > > Can anybody help? > > Cheers, > > Dave
Re: [ccp4bb] Rotation function calculation from 2 patterson maps
ALMN will do this .. It does not have a GUI - Gr #!/bin/csh -f # almn \ hklin /y/people/ccp4/projects/insmon/ins_p1_1,55A-I422cell.mtz \ hklin2 /y/people/jean/Youshang/monomeric/datproc/*mtz < Hi, I have two patterson maps created from two observed data sets of similar structures. I want to rotate one map against the other and hence to match them to get the corresponding rotation function. Does anyone have experience how I can do that? Raja
