[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
[ccp4bb]: SUMMARY for Patterson P3121
*** For details on how to be removed from this list visit the ***
*** CCP4 home page http://www.dl.ac.uk/CCP/CCP4/main.html ***
Dear CCP4 users,
Thank you very much for all the suggestions. I got really a
good replies from everybody. Thanks a lot for thier efforts.
I could solve the problem now. It was a hand problem so I have
to change the space group to P3221. I am giving the summary
of the responses which could be helpful to others.
Eleanor wrote :
As far as the space group goes, there is no special tricks, all
Pattersons obey the same rules; peaks at vectors between atoms.
The Harkers follow from the symmetry positions;
-Y,X-Y,Z+1/3 - X,Y,Z > -Y-X, X-2Y, 1/3
Y-X,-X,Z+2/3 - X,Y,Z > Y-2X,-X-Y,2/3 and because of the centre of
symmetry there will also be a vector at 2X-Y,X+Y,1/3
And Y,X,-Z - X,Y,Z > Y-X,X-Y,-2Z
So you can look at the Harker section at Z/W=1/3 and see if you can
find peaks which satisfy the required algebra, and scan for peaks u,-u,W
and then solve for Y-X.
If the algebra gets too heavy ( and I think it does in P3121!) you can
use SHELX patterson search ( it is set up as part of the GUI) but ALWAYS
first look at the peak list and Harker sections of your patterson; if
there are no clear peaks then you wont be able to find a solution..
But remember that you can get a set of solutions (x,y,z) in P3121 or
the related set (-x,-y,-sz) in P3221 so when you test the other hand you
need to also chang ethe symmetry..
this is alll set up ve5y nicely in the GUI..
Andrea wrote :
I guess the easiest thing is to throw everything into Solve, and most
likely the program will come up with the right solution. Unfortunately,
this will take away all the fun with solving the Patterson by hand.
Whatever you choose to do, the correct choice of spacegroup is critical,
and I'm not sure what Solve does to handle this problem. If you find your
sites by hand, and then refine them with MLPHARE, you will need to pay
attention to the signs of the anomalous accupancies. There are four
possible solutions, corresponding to a left- and a right-handed
configuration in each of the two spacegroups. You will need to select for
strong and positive anomalous occupancies (there should also be strong and
negative, weak positive, and weak negative solutions in the two
spacegroups). Throughout, you can initially choose one of the two
spacegroups, because even if you picked the wrong one, you will at least
get the high anomalous occupancy right (whether negative or positive), if
your heavy atom positions are correct.
Ralf wrote :
You have at least three choices for automatically finding your
Go to the tutorial section and look for
Phasing by Multiple Anomalous Diffraction (MAD)
Heavy atom search
Phraenquex wrote :
If the Patterson has such clear peaks, just run it through Shelx - it will
probably do it for you in a minute (literally). Or else SnB, which is
very powerful and also very easy to use, although it doesn't use the
Patterson as such.
Web pages are (in case they're not installed already):
Or you could try the ccp4 program VECSUM, but it's unsupported,
apparently. Haven't used it myself, so wouldn't know how effective it is.
RANTAN in ccp4 could also do the job.
Arjan wrote :
As with any patterson map you can start with ths symmetry opperators. These should be pair wise
substracted from each other. Yielding 6 times 6 combinations of (u,v,w)'s (which are combinations
of x,y,z). It will become obvious that there are certain harker sections. The two obvious ones in
this space group are z=1/3 and z=2/3. But All others will also be pretty nicely arranged on planes
(i don't remember it correctly but these are something like u,v,w = x,2x,any z anyway these are of
less importance). Now you check the positions of the peaks in the Harker plane z=1/3. These have a
certain (u,v,1/3) now go back to the equations you have calculated by subtracting all the symmetry
operators which gave the harker section and start calculating. Remember that the (u,v,1/3) can also
be (-u,-v,1/3) or (1-u,1-v,1/3). So it gives you a lot of equations that will give you certain x,y
for the heavy atom site. Now try and see if these sites give all the other peaks. If so you have
got your self a solution. For more than one heavy atom this becomes pretty complicated, but if you
have one heavy atom bound it is do-able and very illustrative :o)
Also several people suggested the program SOLVE. It worked nicely with one
single solution, which I refined using MLPHARE for both hands.
Once again thank u very much