Dear Ed,
although, I don't think that a comparison of refinement in a higher
and a lower symmetry space group is valid for general NCS cases, I
will try to answer your question. Here are my thoughts for two
different cases:
(1) You have data to atomic resolution with high I/sigma and low Rsym
(I assume high redundancy). The n copies of the asymmetric unit in
the unit cell are really identical and obey the higher symmetry (so,
not a protein crystal). When you process the data in lower symmetry
(say, P1), the non-averaged "higher-symmetry"-equivalent Fobs will
differ due to measurement errors, and thus reflections in the working-
set will differ to "higher-symmetry"-related reflections in the test-
set due to these measurement errors. If you then refine the n copies
against the working-set in the lower P1 symmetry, you minimize |Fobs
(work)-Fcalc|, resulting in Fcalcs that become closer to the working-
set Fobs. As a consequence, the Fcalcs will thus diverge somewhat
from the test-set Fobs. However, since this atomic model is assumed
to be very well defined obeying the higher symmetry, and,
furthermore, the working-set contains well measured "higher-symmetry"-
equivalent Fobs, the resulting atomic positions, and thus the Fcalcs,
will be very close to their equivalent values in the higher-symmetry
refinement. Therefore, the Fcalcs will also be still very similar to
the "higher-symmetry"-equivalent Fobs in the test-set, and I would
expect a difference between Rwork and Rfree ranging from "0" to the
value of Rsym. In other words, the Fobs in the test-set are not
really independent of the reflections in the working-set, and thus
Rfree is heavily biased towards Rwork.
In this case, I would not expect large differences in the outcome due
to the additional application of "NCS"-constraints/restraints.
(2) You have data to non-atomic lower resolution, weak I/sigma and
poor Rsym. It is impossible to say whether the n copies of the
asymmetric unit in the unit cell are really identical, but they are
treated so assuming the higher symmetry (so, a real protein crystal).
For data processing, the same holds true as for case (1). In
contrast, here I think that it makes a difference, whether you apply
"NCS"-constraints/restraints between the n copies in the lower
symmetry P1, or not. If you apply "NCS"-constraints or strong "NCS"-
restraints, the n copies are made equal and you get n times the
average structure. This is similar to the refinement in the higher
symmetry, except that again you minimize the discrepancy between
Fcalcs and working-set Fobs, which will increase the discrepancy to
the "higher-symmetry"-related Fobs in the test-set. But since the
Fobs in the test-set are still not really independent to the Fobs in
the working-set, I would again expect maximum differences between
Rwork and Rfree in the same order of magnitude as Rsym. So, Rfree is
still biased towards Rwork, but it might be more difficult to notice
this. But if you do not apply "NCS"-constraints/restraints, you give
the less well-defined atomic model more freedom to converge against
the working-set Fobs, resulting in a higher discrepancy between Rwork
and Rfree. But since the Fobs in the working set still contain
"higher-symmetry"-equivalent Fobs, you will end up with a model that
still shows some similarity to the refined structure in the higher
symmetry. As a result, the Rfree is even then not really independent
of Rwork, but it might be even more difficult to notice this,
depending on data resolution and quality. Here, I can't give a range
of differences between Rwork and Rfree.
So, this is still not quantitative, and I hope that I'm not
completely wrong with my argumentation.
These lower vs. higher symmetry examples given above are only
transferable to reality in special NCS-cases with pseudo-higher
symmetry (what Dale Tronrud discussed). Taking these special cases
aside, what do the NCS experts say to my original statement that
precautions against NCS bias in Rfree must only be taken if NCS-
constraints/restraints are really applied during refinement?
Best regards,
Dirk.
Am 08.02.2008 um 21:43 schrieb Edward A. Berry:
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
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Dirk Kostrewa
Gene Center, A 5.07
Ludwig-Maximilians-University
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Germany
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