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Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix -advice retraction
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On Thu, 19 Oct 2000, Daniel Lee wrote:
> I got the same 3 by 3 rotational matrix as many of you mentioned. The
> only problem now is the translational vector. As Robert kindly pointed
> it out, the input is:
> translation=(0.00000000 95.49215467 0.00000000 )
> 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?
if i remember correctly, i believe you said that you got your rotation
from the self-rotation function. the self-rotation does not provide you
with a translation solution.
a) your rotation solution is correct
b) you know your heavy atom sites correctly (same origin?)
c) the heavy atom sites do correspond to the same position on each monomer
then yes, you can derive your translation solution from the given data.
choose one of your H.A. sites as a starting point,
apply your rotation matrix to it,
THEN translate the result to the second H.A. site.
i recommend testing any NCS solution you get by applying it to some actual
coordinates, to see if they go where you thought they should. and probably
you will want to refine your NCS solution using something like RAVE IMP.
"The Wrong Hat is like wearing an enormous sign, duct-taped to a plumber's
helper stuck to your head, that says, "I'm a moron." - Mimi Pond
David J. Schuller
modern man in a post-modern world
University of California-Irvine