[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix - advice retraction

***  For details on how to be removed from this list visit the  ***
***    CCP4 home page http://www.dl.ac.uk/CCP/CCP4/main.html    ***

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
these two atoms, you could operate on the first with the pure
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
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''.