# Re: cad in R32/ccp4 asymmetric units

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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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