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[ccp4bb] Does NCS bias a randomly-chosen test set (even if not enforced)? |
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CCP4bb navigationCCP4bb <-- 2008 <-- February 2008 <-- 08 February 2008Subject: Does NCS bias a randomly-chosen test set (even if not enforced)? From: "Edward A {- dot -} Berry" EABerry {- at -} LBL {- dot -} GOV Date: 2008-02-08 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 CCP4bb navigationCCP4bb <-- 2008 <-- February 2008 <-- 08 February 2008 |
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