Preparations for Patterson Lab
Structural Molecular Biology Laboratory, ChemM230D
|
Michael R. Sawaya, Duilio Cascio
|
1) Login as sawaya.
cd ~sawaya/HTML/m230d/Patterson/pymol. Open two windows, run pymol
screw4a.pml. Print out ~sawaya/HTML/m230d/Patterson/handout_p2example.ppt.
Print out ~sawaya/HTML/m230d/Patterson/handout_p4example.ppt.
Mike's note for next year. Present the p2 plane group then p43212 example
given in the two powerpoint presentations. Gotta include somewhere
the equation for the Patterson function and the meaning of convolution
of the electron density function with the inverted density and how
this leads to peaks located at positions given by the vectors between atoms.
2) You will be calculating difference Patterson maps. Your assignment
is to use the Patterson peaks for solving the xyz coordinates of the heavy
atom in the PCMBS data set. This is important for obtaining the phases (though
you might not realize yet why this is true). Solving heavy atoms in low symmetry
space groups is easy. P43212 is fairly complicated. The calculations
are simple, involving simplpe algebra, but there are many steps. We
will walk you through the process. Before beginning the algebraic deluge,
lets gain a graphical understanding of the nature of the Patterson map.
3) Suppose our structure is in P41 space group. There is one molecule
in the asymmetric unit. They are arranged around a 4-fold screw. Their
positions are described by the symmetry operators shown. They are operators,
not coordinates. Click on each symmetry copy, one at a time.
I am showing more than one asymmetric unit. The heavy atom is bound
as shown. (which heavy atom am I trying to solve?) The structure factors
will contain information from both the protein and the heavy atom. But
if we subtract the native structure factors, from the derivative FP-FPH, then
we are left with structure factors which describe only the positions of the
heavy atoms. Think of them as heavy atom structure factors. FH. Turn off
the cartoons in the graphic display.
4) We dont have phases to put with the heavy atom structure factors
so we cant calculate the electron density map of the heavy atoms. We
have to calculate a Patterson synthesis where we set the phases to zero. This
gives us the convolution of the heavy atom structure with its inverted image.
In other words the Patterson map contains peaks located at positions
described by interatomic vectors. (type @screw4b.pml). The difference
vectors are shown here. Notice that there are families of vectors with
the same length. Neighbors on the screw produce blue vectors of 45
Angstroms. They have differen x and y directions, but they share the same
z-component (one quarter of the c axis). This is due to the crystallographic
symmetry. You can see it in the symmetry operator expression. Neighbors
two positions down the screw differ by 1/2 unit cell in c.
5) In the real Patterson map, the peaks are not arranged like this, however.
Each vector is translated to the origin. In the second window run
pymol screw4c.pml. Both ends of the difference vector are placed at the
origin. The vectors look more orderly now. You can see
that certain vectors lie on sections z=1/4 and z=1/2.
In the second window type @screw4d.pml. These planes
are Harker planes. No matter where the heavy atom is located, we expect
peaks on these planes. The coordinates of the peak is given by subtracting
symmetry operators. We call Patterson space u,v,w. The
contour maps represent these sections.
6) Now we will try working backward from uvw to xyz. Not as straight
forward. There are degeneracies in going from crystal coordinates to
Patterson coordinates. Go through Eu example.
7)Explain the files they are working with. The anomalous and isomorphous
difference Pattersons.
|
Back
to data reduction experiment
Back to CHEM
M230D course syllabus
[Overview] ·[Facilities] ·
[People]
· [Services] ·[Lectures] ·
[BioLinks] ·
[Stats]
·[Search]
|