On Tuesday 16 February 2010 10:54:49 Dale Tronrud wrote:
> I'm puzzled about this. If the B's listed by Refmac in the PDB
> file are the residuals that are to be added to the values derived
> from the TLS model, shouldn't they have a distribution with a mean
> value of zero? Why would one what a minimal residual of 2A^2?
> Aren't negative residuals not only possible but inevitable?
The residuals are not normally distributed.
But you are correct that refmac should allow the residuals to go negative.
This is a recurring problem. It can sometimes be overcome by
resetting the TLS parameters to 0 and starting a new refinement
from scratch. But this isn't guaranteed either.
Ethan
>
> Dale Tronrud
>
> Eleanor Dodson wrote:
> > Two possible points. Could there be a problem with twinning, or
> > spacegroup? The Rfactors seem rather high..
> >
> > You dont say whether you have a non-crystallographic translation, but it
> > is faintly possible with 2 molecules that the SG is actually C222 ..
> >
> > Or that twinning could be present - look at the ctruncate plots - espec
> > the L test - for infdications..
> >
> > But re the TLS. You need to set the overall B to the wilson value more
> > or less. All B factors after REFMAC are relative to that initial starting B
> >
> > If you TLSANL you should get back sensible B values in the PDB slot, but
> > of course any which were restricted to the default value of " will be
> > corruppted..
> >
> > Eleanor
> >
> >
> >
> >
> > Christophe Wirth wrote:
> >> Dear CCP Community,
> >>
> >>
> >> We have processed protein scattering data in space group C2221 to a
> >> resolution of 2.4 Å. The structure shows two protein molecules (chain
> >> A and B) in the asymmetric unit, related by a local two-fold symmetry
> >> axis. Initial rigid body refinement and subsequent full refinement of
> >> isotropic atomic B-factors (including all water and ligand molecules;
> >> using also non-crystallographic symmetry restraints) with the program
> >> REFMAC5 yielded R and Rfree values of 32.1and 33.9, respectively.
> >>
> >> In the next step, after setting the B-factor to 30 Å^2, we carried out
> >> 6 cycles of TLS refinement followed by 10 cycles of isotropic atomic
> >> B-factor refinement (two TLS-bodies were defined: the two
> >> symmetry-related streptavidin chains A and B in the asymmetric unit).
> >> As expected, the R and Rfree values droped down by 1-2% to 29.6 and
> >> 32.5%, respectively. Checking the residual atomic B-factors their
> >> global average had changed from 13 Å^2 (without TLS refinement) to 2
> >> Å^2, the lower B-factor boundary. In fact, all atoms in the .pdb file
> >> showed B-values of 2 Å^2.
> >> Interestingly, when we previously processed the same molecule in C2 we
> >> did not have this TLS problem.
> >>
> >> Does anybody has an explanation why after TLS refinement, individual
> >> B-factor refinement is failing? Is this due to wrongly refined (i.e.
> >> too large) TLS parameters? Has this been observed before? Any
> >> suggestions to solve the problem?
> >>
> >> Many Thanks,
> >>
> >>
> >> Christophe
> >>
> >>
>
--
Ethan A Merritt
Biomolecular Structure Center
University of Washington, Seattle 98195-7742
