Hi all,
we noticed a strange behaviour of refmac5.7.0029 that comes with CCP4-6.3.
We run a most basic script: 5 rounds of restrained refinement with all default
parameters. The input mtz file contains besides H/K/L and FreeR_flag,
IMEAN/SIGIMEAN, and F/SIGF.
Refmac 5.6.0117 outputs F/SIGF,
Dear ccp4bb,
I am working on refinement of an enzyme structure. I notice an alternative
conformation (gauche+ and gauche-) on both residues Phe and Tyr.
Here is the image of electron density before refinement. Based on the
different map intensity, I am quite sure there is an alternative
Dear Ni Shi,
Did you test a crystal, or uncrystallized protein? It may be that due to
crystal packing effects, only protein without ADP crystallizes.
Best regards,
Herman
From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On
Behalf Of Ni shi
Dear Ngo,
What do you mean by no positive map? If you mean the green difference
map density, this should go away if the refinement program did its job
properly. The residual green density at the Tyr is probably due to the
fact that when the Tyr is not at this position, something else (e.g.
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Dear Jan,
could you attach or include the refmac script, please?
Cheers,
Tim
On 07/27/12 08:10, Jan Abendroth wrote:
Hi all, we noticed a strange behaviour of refmac5.7.0029 that comes
with CCP4-6.3. We run a most basic script: 5 rounds of
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Dear Ngo Duc Tri,
I would manually set the occupancy of chain A to about 30% and chain
B's to 70% and try again.There is density for chain A present in the
second picture, it is just weaker.
Cheers,
Tim
On 07/27/12 10:01, Tri Ngo wrote:
Dear
Or refine the occupancies with Refmac...
Roberto
From my iPhone
On 27 Jul 2012, at 11:04, Tim Gruene t...@shelx.uni-ac.gwdg.de wrote:
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Dear Ngo Duc Tri,
I would manually set the occupancy of chain A to about 30% and chain
B's to 70% and try
Not here. SIGF is in the mtz output from refmac 5.7.029 is right below F:
4 NONE0.032.0 0 100.0016.0316.03 117.37 1.90 I
FreeR_flag
5 NONE 20.0 11667.8 12429 96.42 805.15 805.15 117.37 1.90 F F
6 NONE8.6 311.4 12429 96.4288.74
Don't forget you will need to adjust the contour level to see something with
occupancy 0.3 - easily done with coot
eleanor
On 27 Jul 2012, at 11:25, Steiner, Roberto wrote:
Or refine the occupancies with Refmac...
Roberto
From my iPhone
On 27 Jul 2012, at 11:04, Tim Gruene
Hi,
you can't expect to see something with occupancy ~0.3 at the same cutoff
level as you use to see fully occupied sites. So most likely the solution
is to use a lower cutoff levels, as many suggested already. Two maps you
attached look totally expectable to me.
Pavel
On Fri, Jul 27, 2012 at
Or similarly, the conditions in the crystallization drop (pH, ions, etc.)
may kick out the ligand.
JPK
On Fri, Jul 27, 2012 at 3:34 AM, herman.schreu...@sanofi.com wrote:
**
Dear Ni Shi,
Did you test a crystal, or uncrystallized protein? It may be that due to
crystal packing effects, only
Dear all
I have a confusion on the space group R32 and H32. For a cell parameter of a =
b not equal to c, alpha=beta, not equal to gamma, is it considered as R32 or
H32?
I tried searching the mail list archives but it does not help a beginner
crystallographer like me.
I also have another
H32 indicates the hexagonal obverse setting (as you list) for a R centered
trigonal cell, which is 3x larger than the primitive R32 cell indexed a=b=c,
al=be=ga 90. Standard imho is the H32 setting, for which I will probably get
flamed.
The relation between H and R cells is depicted here:
Hi
Just to clarify - for H32, a=bc (though you could have a=b=c within
measurement error), alpha=beta=90º, gamma=120º. Just having alpha=beta
gamma is necessary but not sufficient.
BTW, there's nothing wrong with alpha=beta=gamma=90º for the R
setting. It's just not required.
Sorry if
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Recognition at NIH
A postdoctoral
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To add a little more clarity to the discussion.
The hexagonal obverse setting is 3X larger because it is triply primitive.
This means there are 3 lattice points per unit cell, similar to the face
or body centered lattices, which have two lattice points per cell.
To read BR's scripts:
For the
Thanks for the translation, and a correction of the translator:
similar to the face or body centered lattices, which have two lattice
points per cell.
Is this so? Ortho F and FCC have 4 Bravais vectors
All (in full translation) in chapter 5 of BMC.
BR
You are correct, sir. It is one lattice point for each centered face.
So, for the F centered lattices which have a face centered lattice point
on all faces you add 3 lattice points to the original primitive lattice
point for a total of 4 lattice points.
It's all in the Tables.
Cheers,
Chris
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