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*To*: ccp4bb@dl.ac.uk*Subject*: Re: cad in R32/ccp4 asymmetric units*From*: Ian Clifton <ian@biop.ox.ac.uk>*Date*: Wed, 29 Sep 1999 14:56:51 +0100*Sender*: owner-ccp4bb@dl.ac.uk

*** For details on how to be removed from this list visit the *** *** CCP4 home page http://www.dl.ac.uk/CCP/CCP4/main.html *** Erica Ollmann Saphire wrote: > I am running cad on a data set in the space group R32 and am > wondering why cad does what it does. >For example, input.mtz has these two reflections first: >4 -2 6 >4 -1 -7 with a certain F and sigF. > output.mtz (output from cad) has these two reflections first, with > the same F and sigF as the first dataset, so they ought to be > equivalent: >2 2 6 >3 1 -7 with the same F and sigF > So, the question is, how are these reflections equivalent? CAD is correct, and those pairs of reflexions are indeed equivalent. It's always easier to see what's going on in trigonal & hexagonal if you use four-index Miller-Bravais symbols (h k (-h-k) l): (4 -2 6) becomes (4 -2 -2 6) (4 -1 -7) becomes (4 -1 -3 -7) Next, study the general symmetry operations for 3bar m 1, the Laue class of R32. These show that: Rotating h, k, i in any reflexion gives an equivalent one. (h k i l) = (k i h l) = (i h k l) Swapping any two of h, k, i and then changing all three of their signs in any reflexion gives an equivalent one. e.g. (h k i l) = (-k -h -i l) Applying these two rules to your reflexions gives: (4 -2 -2 6) = (-2 -2 4 6) = (2 2 -4 6) (in the last step, the two -2's have been swapped!) (4 -1 -3 -7) = (-3 4 -1 -7) = (3 1 -4 -7) Finally, dropping the i index gives CAD's output. -- Ian Clifton Phone: +44 1865 275631 Dyson Perrins Laboratory Fax: +44 1865 275674 Oxford University OX1 3QY UK ian@biop.ox.ac.uk

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