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Apologies for extending this somewhat but I think the question I have relates to the principles Ian and others are covering. My question is what are the correct coefficients to use if refining twinned data using twin related Icalc values. In this case, shouldn't coefficients for maps be 4mFo-3DFc? Colin > -----Original Message----- > From: CCP4 bulletin board [mailto:[log in to unmask]] On > Behalf Of Ian Tickle > Sent: 30 July 2010 11:51 > To: [log in to unmask] > Subject: Re: [ccp4bb] Will 3mFo-2DFc maps have less model > bias than 2mFo-DFc maps? > > Hi Fred > > You have to be careful here because believe it or not, not > all programs output the same coefficients for 'minimal bias' > maps, so depending on which program Hailiang is using for SF > calculation/refinement he may or may not get the right > answer! You are assuming the difference map coefficient is > (mFo-DFc) for both acentrics & centrics so you are expecting > to calculate: > > 3mFo-2DFc = (2mFo-DFc) + (mFo-DFc) for acentrics > 2mFo-DFc = Fo + (mFo-DFc) for centrics > > However as I have been at pains to point out on numerous > occasions the correct difference map coefficient is > 2(mFo-DFc) for acentrics (i.e. 2 times half the peak height > from an acentric mFo-DFc map), and > (mFo-DFc) for centrics (i.e. 1 times the full peak height > from a centric mFo-DFc map). This fact tends to be obscured > if you think of it as Fo+(Fo-Fc) instead of Fc+2(Fo-Fc), > which was my real objection to thinking of it in the way > Pavel suggested. > > In fact the last time I checked (recently) neither Refmac nor > Buster got it right (details on request!) - not only that but > they get it wrong in different ways: at least they are > inconsistent with what in my view are the correct > coefficients, which is based on my understanding of Randy > Read's 1986 paper, and no-one has yet provided me with a > rationale for the formulae used by Refmac & Buster. The > CCP4 version of Sigmaa now gets it right, but that's only > because I recently fixed it myself. I can't speak for > phenix.refine, I suspect it gets it completely correct, since > Pavel is on the case! So I think the safest CCP4 approach is > to use Sigmaa to recalculate the map coefficients, then use > FFT to combine them. This will require something like the > following input to FFT: > > LABIN F1=FWT F2=DELFWT PHI=PHIC > SCALE F1 1 0 F2 0.5 0 > > (check the FFT doc!) > > in other words: > 3mFo-2DFc = (2mFo-DFc) + 0.5*(2(mFo-DFc)) for acentrics > 1.5mFo-0.5*DFc = Fo + 0.5*(mFo-DFc) for centrics > > Note that this gives the coefficient 1.5mFo-0.5DFc for > centrics, not 2mFo-DFc as suggested in your paper (sorry I > couldn't see the rationale for that choice). Again this > becomes much clearer if you write 3mFo-2DFc as Fc + > 3(mFo-DFc) i.e. 3 times half height (= 1.5 times true > height), so to be consistent the centric coefficient should > also be 1.5 times true, or Fc + 1.5(mFo-DFc). I think it's > important to get the centric reflections right (particularly > in tetragonal and cubic space groups!) because obviously the > centric phases tend to be better determined than the acentric ones. > > Cheers > > -- Ian > > On Fri, Jul 30, 2010 at 9:24 AM, Vellieux Frederic > <[log in to unmask]> wrote: > > Hi, > > > > You take the output mtz from the refinement program (let's > assume it's > > called refine_1.mtz). > > > > Command line mode: > > sftools > > read refine_1.mtz col 1 2 3 4 # assuming the mtz contains H K L > > 2FOFCWT PHI2FOFCWT FOFCWT PHI2FOFCWT cal col 3FO2FCWT col 1 col 3 + > > set types F P F P R F write 3fo2fc.mtz col 5 2 3 4 quit (or stop, > > can't remember which) > > > > That's it... > > > > Fred > > > > Armando Albert de la Cruz wrote: > >> > >> Does anyone have got a script to compute 3fo2fc map with CCP4? > >> Armando > >> > >> > >> El 29/07/2010, a las 23:38, Ian Tickle escribió: > >> > >>> On Thu, Jul 29, 2010 at 8:25 PM, Pavel Afonine > <[log in to unmask]> wrote: > >>>> > >>>> Speaking of 3fo2fc or 5fo3fc, ... etc maps (see classic works on > >>>> this published 30+ years ago), I guess the main > rationale for using > >>>> them in those cases arises from the facts that > >>>> > >>>> 2Fo-Fc = Fo + (Fo-Fc), > >>>> 3Fo-2Fc = Fo +2(Fo-Fc) > >>>> > >>>> To be precise, it is actually > >>>> > >>>> 2mFo-DFc for acentric reflections > >>>> and mFo for centric reflections > >>> > >>> I prefer to think of it rather as > >>> > >>> 2mFo - DFc = DFc + 2(mFo-DFc) for acentrics and mFo = DFc + > >>> (mFo-DFc) for centrics. > >>> > >>> Then it also becomes clear that to be consistent the > corresponding > >>> difference map coefficients should be 2(mFo-DFc) for acentrics and > >>> (mFo-DFc) for centrics. > >>> > >>> Cheers > >>> > >>> -- Ian > >> > >> > > >