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*To*: "ccp4bb@dl.ac.uk" <ccp4bb@dl.ac.uk>*Subject*: Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix - advice retraction*From*: Daniel Lee <dlee@protein.biochem.queensu.ca>*Date*: Thu, 19 Oct 2000 14:22:44 -0400*References*: <39EE2ACF.11521DBB@protein.biochem.queensu.ca> <39EE47EA.426A839A@lbl.gov> <39EEBB11.6EB1ECAE@yorvic.york.ac.uk>*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 *** Dear all: Again, a million thanks to your warm suggestions. 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 ) And the result is: Translation vector = ( -88.6619 130.9580 0.0000) But I don't know where does (0.00000000 95.49215467 0.00000000 ) come from (or it is just an example). I tried to figure it out by myself from substracting the coordinates of my 2 heavy atom's pdb file (from CNS heavy_search) 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? Regarding the CNS get_ncs_matrices.inp program, I guess I would need it later on to refine the NCS data. Currently, I am unable to benefit from it because I am still solving the structure of the protein. Thanks all! Daniel

**Follow-Ups**:**Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix -advice retraction***From:*"David J. Schuller" <schuller@uci.edu>

**Re: [ccp4bb]: From (psi phi kappa) to 3 by 3 rotational matrix - advice retraction***From:*Edward Berry <eaberry@lbl.gov>

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