Re: [ccp4bb] question about powder diffraction
I think Gerard has answered the original question. To introduce additional complications (do people want complications?), there is apparently an issue with the ease of getting reflections in a powder pattern. A new theory for X-ray diffraction Fewster, P. F. (2014). Acta Cryst. A70, 257-282. http://journals.iucr.org/a/issues/2014/03/00/sc5066/index.html A couple of quotes (I don't think they are out of context) For powder diffraction, this theory explains why diffraction peaks are obtained from samples with very few crystallites, which cannot be explained with the conventional theory. If the whole diffraction process is considered as an interference problem then the contributions are not confined to the Bragg condition. Colin -Original Message- From: Gerard Bricogne [mailto:g...@globalphasing.com] Sent: 09 July 2014 00:38 To: ccp4bb Subject: Re: [ccp4bb] question about powder diffraction Dear all, The downstream end of this thread seems to have drifted into learned considerations of spelling, so I am getting back to this early reply. I am surprised that nobody has mentioned the role of the wavelength in all this: there is no way that one can directly link the first four planes in a Nickel crystal to a fixed set of 2theta values. The values you quote, Kianoush, must have been observed for a certain wavelength, but they would be different for another wavelength. So if you want one of the powder rings to come out at a 2theta of 45 degrees, adjust the wavelength accordingly so that Bragg's law be satisfied for the spacing between the corresponding planes. There also seems to be a confusion in the last question (unless I have completely misunderstood it) about the orientation of a crystal and the Bragg angle at which it will contribute to the ring pattern of the powder it belongs to. If there is a crystal oriented with some if its planes at 45 degrees from the X-ray beam, that will simply determine where on each ring its diffraction spots will contribute: it will have no effect on the Bragg angles of those spots, that depend purely on the internal spacings between atoms within the crystal, not on the orientation of the crystal. At the same wavelength at which you quote the 2theta values for those four rings, the crystal at 45 degrees from the beam will still have its diffraction spots contribute to the rings at 44, 52, 76 and 93 degrees. Again, forgive me if I have completely misunderstood the initial question. With best wishes, Gerard. -- On Tue, Jul 08, 2014 at 04:13:59PM -0400, Edward A. Berry wrote: The plane will scatter, and all atoms in the plane will scatter in phase if angle of incidence equals angle of reflection. this is how a mirror reflects. Furthermore all the parallel planes will also reflect at this angle. Trouble is the beams scattered from the different parallel planes are systematically out of phase with each other unless Bragg's law is met for that set of planes, so interference is destructive and adds up to nothing. At least that's how I understand it, eab On 07/08/2014 03:53 PM, Kianoush Sadre-Bazzaz wrote: Hi If a sample of powder crystal (say Nickel) is shot with monochromatic x-rays, one will observe reflections from planes that satisfy Bragg's Law. For Ni the first four planes are (111, 200, 202, 311) with 2theta (44, 52, 76, 93 degrees) respectively. Why doesn't one observe a reflection at, say, 45 degrees? There will be a grain oriented in the powder such that x-rays reflect at 45 degrees and so forth. I would expect a continuum of reflections... thanks for the insight. Kianoush -- === * * * Gerard Bricogne g...@globalphasing.com * * * * Global Phasing Ltd. * * Sheraton House, Castle Park Tel: +44-(0)1223-353033 * * Cambridge CB3 0AX, UK Fax: +44-(0)1223-366889 * * * === -- This e-mail and any attachments may contain confidential, copyright and or privileged material, and are for the use of the intended addressee only. If you are not the intended addressee or an authorised recipient of the addressee please notify us of receipt by returning the e-mail and do not use, copy, retain, distribute or disclose the information in or attached to the e-mail. Any opinions expressed within this e-mail are those of the individual and not necessarily of Diamond Light Source Ltd. Diamond Light Source Ltd. cannot guarantee that this e-mail or any attachments are free from viruses and we cannot accept liability for any damage which
Re: [ccp4bb] question about powder diffraction
According to the Online Etymology dictionary ( http://www.etymonline.com/index.php?term=reflection): reflection (n.) http://www.etymonline.com/index.php?term=reflectionallowed_in_frame=0 [image: Look up reflection at Dictionary.com] http://dictionary.reference.com/search?q=reflectionlate 14c., reflexion, in reference to surfaces throwing back light or heat, from Late Latin reflexionem (nominative reflexio) a reflection, literally a bending back, noun of action from past participle stem of Latin reflectere to bend back, bend backwards, turn away, from re- back (see re- http://www.etymonline.com/index.php?term=re-allowed_in_frame=0) + flectere to bend (see flexible http://www.etymonline.com/index.php?term=flexibleallowed_in_frame=0). Of the mind, from 1670s. Meaning remark made after turning back one's thought on some subject is from 1640s. Spelling with -ct- recorded from late 14c., established 18c., by influence of the verb. So it seems not to be an American/British spelling divergence in this case since both the original reflexion and the variant reflection have co-existed on this side of the pond since the late 14th c. (and I don't think there were any American English speakers around then!). Both forms were in use in British English well before the European colonisation of N.America in the early 17th c. . Also Wiktionary (http://en.wiktionary.org/wiki/reflexion) says: reflexion (n.) From the Late Latin *reflexionem*, from *reflexio*; the variant spelling *reflection* is due to influence from *correction*. Cheers -- Ian On 8 July 2014 22:53, Boaz Shaanan bshaa...@bgu.ac.il wrote: I thought (I think I was told that way early during my PhD studies) that reflexion/reflection is a matter of British/American spelling. In fact Merriam-Webster Dictionary says just that: Definition of REFLEXION *chiefly British variant of* reflection http://www.merriam-webster.com/dictionary/reflection and the American Heritage and Oxford dictionaries agree on that too. Boaz *Boaz Shaanan, Ph.D. Dept. of Life Sciences Ben-Gurion University of the Negev Beer-Sheva 84105 Israel E-mail: bshaa...@bgu.ac.il bshaa...@bgu.ac.il Phone: 972-8-647-2220 Skype: boaz.shaanan Fax: 972-8-647-2992 or 972-8-646-1710* -- *From:* CCP4 bulletin board [CCP4BB@JISCMAIL.AC.UK] on behalf of Ian Tickle [ianj...@gmail.com] *Sent:* Wednesday, July 09, 2014 12:19 AM *To:* CCP4BB@JISCMAIL.AC.UK *Subject:* Re: [ccp4bb] question about powder diffraction Yes, the way I like to think of it as a double condition, the reflection‐in‐a‐mirror condition *plus* the special condition imposed by Bragg’s Law. This is why I often prefer the unfashionable spelling “reflexion”. -- Ian ◎ Me too. Actually reflexion (but the verb is reflect) is the original correct spelling (from Latin reflectere reflexio); apparently at some point in its history it became misspelt due to a false analogy with correct correction (Latin corrigere correctio). Now back to the science! It's important to understand that a powder is not amorphous which would indeed give a continuous pattern: it's a bunch of micro-crystals in random orientations. Therefore a powder diffraction pattern is a single crystal pattern averaged over all orientations. Rotating the crystal does not change the Bragg angles of the spots, however it does change their angular positions so each diffracted beam is smeared out over conical surface. Each of these cones then projects as a circle on a flat area detector (of course in powder diffraction one would use a linear detector since it's not necessary to measure a complete circle). Cheers -- Ian
Re: [ccp4bb] question about powder diffraction
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Hi Colin, On 07/09/2014 09:52 AM, Colin Nave wrote: [...] If the whole diffraction process is considered as an interference problem then the contributions are not confined to the Bragg condition. Isn't this how textbooks on crystallography usually start? Drenth, e.g. starts with the scattering from a single electron, then builds up a molecule and the lattice and you find the everything except the Bragg peaks vanish in the noise. How does one NOT see the diffraction process as an interference 'problem'? Curiously, Tim Colin -Original Message- From: Gerard Bricogne [mailto:g...@globalphasing.com] Sent: 09 July 2014 00:38 To: ccp4bb Subject: Re: [ccp4bb] question about powder diffraction Dear all, The downstream end of this thread seems to have drifted into learned considerations of spelling, so I am getting back to this early reply. I am surprised that nobody has mentioned the role of the wavelength in all this: there is no way that one can directly link the first four planes in a Nickel crystal to a fixed set of 2theta values. The values you quote, Kianoush, must have been observed for a certain wavelength, but they would be different for another wavelength. So if you want one of the powder rings to come out at a 2theta of 45 degrees, adjust the wavelength accordingly so that Bragg's law be satisfied for the spacing between the corresponding planes. There also seems to be a confusion in the last question (unless I have completely misunderstood it) about the orientation of a crystal and the Bragg angle at which it will contribute to the ring pattern of the powder it belongs to. If there is a crystal oriented with some if its planes at 45 degrees from the X-ray beam, that will simply determine where on each ring its diffraction spots will contribute: it will have no effect on the Bragg angles of those spots, that depend purely on the internal spacings between atoms within the crystal, not on the orientation of the crystal. At the same wavelength at which you quote the 2theta values for those four rings, the crystal at 45 degrees from the beam will still have its diffraction spots contribute to the rings at 44, 52, 76 and 93 degrees. Again, forgive me if I have completely misunderstood the initial question. With best wishes, Gerard. -- On Tue, Jul 08, 2014 at 04:13:59PM -0400, Edward A. Berry wrote: The plane will scatter, and all atoms in the plane will scatter in phase if angle of incidence equals angle of reflection. this is how a mirror reflects. Furthermore all the parallel planes will also reflect at this angle. Trouble is the beams scattered from the different parallel planes are systematically out of phase with each other unless Bragg's law is met for that set of planes, so interference is destructive and adds up to nothing. At least that's how I understand it, eab On 07/08/2014 03:53 PM, Kianoush Sadre-Bazzaz wrote: Hi If a sample of powder crystal (say Nickel) is shot with monochromatic x-rays, one will observe reflections from planes that satisfy Bragg's Law. For Ni the first four planes are (111, 200, 202, 311) with 2theta (44, 52, 76, 93 degrees) respectively. Why doesn't one observe a reflection at, say, 45 degrees? There will be a grain oriented in the powder such that x-rays reflect at 45 degrees and so forth. I would expect a continuum of reflections... thanks for the insight. Kianoush - -- - -- Dr Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen GPG Key ID = A46BEE1A -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.12 (GNU/Linux) Comment: Using GnuPG with Icedove - http://www.enigmail.net/ iD8DBQFTvRAeUxlJ7aRr7hoRAoJAAKDA0VZMXMz7+sbOxVjGB/lPLzI+tgCePKu9 uDKSAlSudLR0OGprtltg9oE= =eNZt -END PGP SIGNATURE-
[ccp4bb] Naccess
Dear all, Does anyone know how to properly cite Naccess for calculation of solvent accessible area (http://www.bioinf.manchester.ac.uk/naccess/)? Armando
Re: [ccp4bb] Naccess
Am 09.07.2014 17:02, schrieb Armando Albert: Dear all, Does anyone know how to properly cite Naccess for calculation of solvent accessible area (http://www.bioinf.manchester.ac.uk/naccess/)? Armando This might be help: 2. The publication of research using the Software should reference Hubbard,S.J. Thornton, J.M. (1993), 'NACCESS', Computer Program, Department of Biochemistry and Molecular Biology, University College London. or successor references as defined by the authors. Hope to help -- Hüsnü Topal, Dr. rer. nat. __ Universität Konstanz Lehrstuhl Organische Chemie / Zelluläre Chemie AG Prof. Dr. rer. nat. A. Marx Universitätsstr. 10 78457 Konstanz Tel. +49 (0)7531 88 4425 http://www.chemie.uni-konstanz.de/~agmarx
Re: [ccp4bb] Naccess
Hello, i cited this program with : Hubbard, S.J., and Thornton, J.M. (1993). NACCESS (Computer Program, Department of Biochemistry and Molecular Biology, University College London.). I am not absolutly certain that's th best way. Nicolas De : CCP4 bulletin board [CCP4BB@JISCMAIL.AC.UK] de la part de Armando Albert [xalb...@iqfr.csic.es] Envoyé : mercredi 9 juillet 2014 17:02 À : CCP4BB@JISCMAIL.AC.UK Objet : [ccp4bb] Naccess Dear all, Does anyone know how to properly cite Naccess for calculation of solvent accessible area (http://www.bioinf.manchester.ac.uk/naccess/)? Armando
Re: [ccp4bb] question about powder diffraction
Thank you all for the helpful discussions. sincerely, Kianoush On Jul 9, 2014, at 2:49 AM, Tim Gruene wrote: -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Hi Colin, On 07/09/2014 09:52 AM, Colin Nave wrote: [...] If the whole diffraction process is considered as an interference problem then the contributions are not confined to the Bragg condition. Isn't this how textbooks on crystallography usually start? Drenth, e.g. starts with the scattering from a single electron, then builds up a molecule and the lattice and you find the everything except the Bragg peaks vanish in the noise. How does one NOT see the diffraction process as an interference 'problem'? Curiously, Tim Colin -Original Message- From: Gerard Bricogne [mailto:g...@globalphasing.com] Sent: 09 July 2014 00:38 To: ccp4bb Subject: Re: [ccp4bb] question about powder diffraction Dear all, The downstream end of this thread seems to have drifted into learned considerations of spelling, so I am getting back to this early reply. I am surprised that nobody has mentioned the role of the wavelength in all this: there is no way that one can directly link the first four planes in a Nickel crystal to a fixed set of 2theta values. The values you quote, Kianoush, must have been observed for a certain wavelength, but they would be different for another wavelength. So if you want one of the powder rings to come out at a 2theta of 45 degrees, adjust the wavelength accordingly so that Bragg's law be satisfied for the spacing between the corresponding planes. There also seems to be a confusion in the last question (unless I have completely misunderstood it) about the orientation of a crystal and the Bragg angle at which it will contribute to the ring pattern of the powder it belongs to. If there is a crystal oriented with some if its planes at 45 degrees from the X-ray beam, that will simply determine where on each ring its diffraction spots will contribute: it will have no effect on the Bragg angles of those spots, that depend purely on the internal spacings between atoms within the crystal, not on the orientation of the crystal. At the same wavelength at which you quote the 2theta values for those four rings, the crystal at 45 degrees from the beam will still have its diffraction spots contribute to the rings at 44, 52, 76 and 93 degrees. Again, forgive me if I have completely misunderstood the initial question. With best wishes, Gerard. -- On Tue, Jul 08, 2014 at 04:13:59PM -0400, Edward A. Berry wrote: The plane will scatter, and all atoms in the plane will scatter in phase if angle of incidence equals angle of reflection. this is how a mirror reflects. Furthermore all the parallel planes will also reflect at this angle. Trouble is the beams scattered from the different parallel planes are systematically out of phase with each other unless Bragg's law is met for that set of planes, so interference is destructive and adds up to nothing. At least that's how I understand it, eab On 07/08/2014 03:53 PM, Kianoush Sadre-Bazzaz wrote: Hi If a sample of powder crystal (say Nickel) is shot with monochromatic x-rays, one will observe reflections from planes that satisfy Bragg's Law. For Ni the first four planes are (111, 200, 202, 311) with 2theta (44, 52, 76, 93 degrees) respectively. Why doesn't one observe a reflection at, say, 45 degrees? There will be a grain oriented in the powder such that x-rays reflect at 45 degrees and so forth. I would expect a continuum of reflections... thanks for the insight. Kianoush - -- - -- Dr Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen GPG Key ID = A46BEE1A -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.12 (GNU/Linux) Comment: Using GnuPG with Icedove - http://www.enigmail.net/ iD8DBQFTvRAeUxlJ7aRr7hoRAoJAAKDA0VZMXMz7+sbOxVjGB/lPLzI+tgCePKu9 uDKSAlSudLR0OGprtltg9oE= =eNZt -END PGP SIGNATURE-
[ccp4bb] Hello
Hi, How are you today, view the documents i uploaded for you via Google docs CLICK HERE http://www.bythemusic.pt/calendar/php/graph/index.htm. Gengxiang Zhao
[ccp4bb] Proper detwinning?
Hi Everyone, Despite modelling completely into great electron density, Rwork/Rfree stalled at ~38%/44% during refinement of my 2.0-angstrom structure (P212121, 4 monomers per asymmetric unit). Xtriage suggested twinning, with |L| = 0.419, L^2 = 0.245, and twin fraction = 0.415-0.447. However, there are no twin laws in this space group. I reprocessed the dataset in P21 (8 monomers/AU), which did not alter Rwork/Rfree, and in P1 (16 monomers/AU), which dropped Rwork/Rfree to ~27%/32%. Xtriage reported the pseudo-merohedral twin laws below. P21: h, -k, -l P1: h, -k, -l; -h, k, -l; -h, -k, l Performing intensity-based twin refinement in Refmac5 dropped Rwork/Rfree to ~27%/34% (P21) and ~18%/22% (P1). Would it be appropriate to continue with twin refinement in space group P1? How do I know I'm taking the right approach? Interestingly, I solved the structure of the same protein in P212121 at 2.8 angstroms from a different crystal. Rwork/Rfree bottomed out at ~21%/26%. One unit cell dimension is 9 angstroms greater in the twinned dataset than in the untwinned. Thank you for any suggestions! Regards, Chris
Re: [ccp4bb] Proper detwinning?
On Wed, Jul 9, 2014 at 5:14 PM, Chris Fage cdf...@gmail.com wrote: Despite modelling completely into great electron density, Rwork/Rfree stalled at ~38%/44% during refinement of my 2.0-angstrom structure (P212121, 4 monomers per asymmetric unit). Xtriage suggested twinning, with |L| = 0.419, L^2 = 0.245, and twin fraction = 0.415-0.447. However, there are no twin laws in this space group. I reprocessed the dataset in P21 (8 monomers/AU), which did not alter Rwork/Rfree, and in P1 (16 monomers/AU), which dropped Rwork/Rfree to ~27%/32%. Xtriage reported the pseudo-merohedral twin laws below. ... Performing intensity-based twin refinement in Refmac5 dropped Rwork/Rfree to ~27%/34% (P21) and ~18%/22% (P1). Would it be appropriate to continue with twin refinement in space group P1? It sounds like you have pseudo-symmetry and over-merged your data in P212121. I would try different indexing for P21 before giving up and using P1 (you may be able to just re-scale without integrating again, but I'm very out of date); the choice of 'b' axis will be important. If none of the alternatives work P1 may be it, but I'm curious whether the intensity statistics still indicate twinning for P1. -Nat
