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
>>
>>
>
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