Dirk Kostrewa wrote:
Dear Dean and others,
Peter Zwart gave me a similar reply. This is very interesting
discussion, and I would like to have a somewhat closer look to this to
maybe make things a little bit clearer (please, excuse the general
explanations - this might be interesting for beginners as well):
1). Ccrystallographic symmetry can be applied to the whole crystal and
results in symmetry-equivalent intensities in reciprocal space. If you
refine your model in a lower space group, there will be reflections in
the test-set that are symmetry-equivalent in the higher space group to
reflections in the working set. If you refine the (symmetry-equivalent)
copies in your crystal independently, they will diverge due to
resolution and data quality, and R-work and R-free will diverge to some
extend due to this. If you force the copies to be identical, the R-work
& R-free will still be different due to observational errors. In both
cases, however, the R-free will be very close to the R-work.
Ah- that's going way to fast for the beginners, at least one of them!
Can someone explain why the R-free will be very close to the R-work,
preferably in simple concrete terms like Fo, Fc, at sym-related
reflections, and the change in the Fc resulting from a step of refinement?
Ed
