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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 ******************************************************* Dirk Kostrewa Gene Center, A 5.07 Ludwig-Maximilians-University Feodor-Lynen-Str. 25 81377 Munich Germany Phone: +49-89-2180-76845 Fax: +49-89-2180-76999 E-mail: [log in to unmask] *******************************************************