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Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix - advice retraction
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Daniel Lee wrote:
> ATOM 1 1 SITE 1 20.536 19.662 -0.188 1.00 26.50 SITE
> ATOM 2 2 SITE 2 -10.541 64.429 7.276 1.00 27.46 SITE
> -----------------------
> (31.077 -44.767 -7.464)
>
> Since each molecule only has one heavy atom biding site, by connecting
> these two heavy atoms, I should be able to get a vector indicating the
> relative location of the two molecules in the asymmetric unit. Could
> somebody let me know whether my assumption is correct?
It's more complicated than that because even the pure rotation moves
an atom (if it is not at the origin). _IF_ you knew this rotation
related
these two atoms, you could operate on the first with the pure
rotation,
then subtract the result vectorwise from the second atom to get the
translational component.
However, unless your crystal is P1, I think there is an additional
complication due to arbitrariness of the asymmetric unit and the fact
that an NCS operator only applies to one asymmetric unit. (Another
name
for NCS symmetry is "Local Symmetry"). Suppose the ncs operator takes
atom A to its ncs-mate B. But the heavy atom sites you found may
correspond to A and B', where B' is related to B by crystallographic
symmetry. Or for that matter you may have A' and B' or A' and B''.
Ed