We have been solving a protein complex at 3.1 A by molecular replacement. One
copy of two chains in the asymmetric unit.
The two chains have fairly different average B factors. As would be expected isotropic B factors in refmac overfits the
data and leaves a large gap between R and Rfree. TLS refinement with overall B factor gave a more reasonable separation
of R and Rfree and more importantly more interpretable difference maps. Phenix. refine with TLS and group B factors
bought the Rfree down further. Refmac does not seem to have a group B factor option but I presume that by setting tight
limits on the B factor constraints (I tried 0.1 on all of them (at most 1/10 of the defaults)- HAS ANYONE DONE ANY
SYSTEMATIC WORK ON WHAT THIS SHOULD BE?) I think you effectively achieve two B factors per residue- main and side. Is
this correct?
Does anyone else have comments on Bfactor use at low resolution in refmac?
Best wishes
Nick
--
Prof Nicholas H. Keep
Executive Dean of School of Science
Professor of Biomolecular Science
Crystallography, Institute for Structural and Molecular Biology,
Department of Biological Sciences
Birkbeck, University of London,
Malet Street,
Bloomsbury
LONDON
WC1E 7HX
email n.k...@mail.cryst.bbk.ac.uk
Telephone 020-7631-6852 (Room G57 Office)
020-7631-6800 (Department Office)
Fax 020-7631-6803
If you want to access me in person you have to come to the crystallography
entrance
and ring me or the department office from the internal phone by the door
