Clarification-
Someone wrote:
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
Hi Ed,
Here's what I think they're saying:
If the NCS is almost crystallographic, then one wedge of spots will be
almost identical to another wedge. If spot "a" is in the test set, but
the almost-crystallographically identical spot "a' " in the 2nd wedge
isn't, then because you're refining directly against a', spot a doesn't
really count as "free".
Was that the question?
Thanks, but,
Here we are talking about refining a structure in an artificially low
space group, to get away from the complexities of the G-function and
degree of overlap. The "NCS" brings a reflection in the test set exactly
onto a reflection in the work set. I'm asking "so what?"
Think about what you mean when you say "spot a and spot a' are
crystallographically identical".
Do you mean the Fo are identical?
They are not, because if we consider it a lower space group then
we will not average these spots, but have separate experimentally
determined values for them. However as pointed out by Jon Wright
and Dean Madden yesterday, the difference between sym-related Fobs
is usually much smaller than the difference between Fo and Fc, so
the sym-related Fobs can be considered almost the same in comparison
to Fc. Specifically,they are likely to be both on the same side of Fc,
so changing two Fc in the same direction will have the same effect on
|Fo-Fc| at the two reflections.
Do you mean the Fc are identical? If we start with the symmetrical
structure refined in the higher space group, their initial values will
be the same. However if we do not enforce NCS, then the changes induced
by refinement will be asymmetric, and the two "NCS-related" Fc will
start to diverge. A change which is made because it improves the fit for
some reflections in the working set may well make the fit worse for
the related reflections in the test set. The only way they are coupled
is through the fact that if a change makes the model more like the real
structure, then the expected value of the resulting change in |Fo-Fc|
is negative for all reflections.
Remember R and Rfree will be statistically the same before refinement,
and start to diverge once refinement begins. Dirk's lesson seems
to imply they will diverge less if there is (perfect) NCS, even if
the NCS is not applied.
(I'm probably wrong, but I want someone to show me,and not with hand-waving
arguments or invocation of crystallographic intuition or such)
To convince me, someone needs to show that the expected value of the change
in |Fo-Fc| at a test reflection upon a change in the model (a step of
refinement)
is negative, even in the absence of any real improvement in the model,
simply because the change reduces |Fo-Fc| at a sym-related working
reflection.
Ed
