Transcript Intro to Patterson functions

Lecture 3 Patterson functions
Patterson functions
The Patterson function is the auto-correlation function of the
electron density ρ(x) of the structure. The key points here are
(i) The Patterson function is independent of the structure factor
phases and can thus be computed directly from the reduced
intensity data for the reflections.
(ii) The peaks in the Patterson function of the difference electron
density (derivative-native) correspond approximately to the interatomic distances between the heavy atoms themselves, provided
that the derivative is isomorphous and that the differences in
structure factor amplitudes are small (i.e. in the Riso range of
about 10 -28 %).
Calculation of the Patterson
function
The Patterson (auto-correlation) function P of ρ(x) is
P(x) = ∫ρ(u)ρ(u-x)du
Peaks in P(x) correspond to the vectors between peaks in ρ(x).
P(x) is easily computed using the fact that correlation in real space
corresponds to multiplication in Fourier space.
Remember, no phases are needed!
What does a Patterson function
look like ?
Here is a simple 2D Patterson function for a set of three atoms.
Real space
The blue dots are atoms,
linked by their inter-atomic
vectors.
Patterson
space
The red dots plot the
inter-atomic vectors.
Properties of a Patterson
Properties of the Patterson function:
1. It is centrosymmetric i.e. for every pair of heavy atoms the
displacement between them can be considered in either the positive or
negative sense. Hence, for a given heavy atom set within the crystal,
the Patterson cannot distinguish between the set and its mirror image.
2. The height of a peak in the Patterson is proportional to the product of
the atomic numbers of the atoms responsible for the peak.
3. The symmetry of the Patterson function is that of the Laue group of the
diffraction pattern (i.e. screw axes become non-screw axes and a
centre of symmetry is added)
4. Symmetry elements in real space give rise to peaks in special planes or
lines in Patterson space, termed Harker planes or lines.
5. The Patterson function has a large origin peak (why?).
How do we deconvolute the
Patterson function ?
Deconvolution has traditionally be done by hand, starting with an
inspection of the Harker sections of the Patterson function.
Peaks in these sections arise from vectors between a particular
heavy atom and its symmetry-related mates and allow coordinates
to be assigned with limited ambiguity to all the heavy atoms.
However, peaks between one heavy atom and another (not a
symmetry mate) are not constrained to be in any special position.
Interpretation of these peaks allows one to check the coordinates
assigned from Harker sections and to resolve the ambiguity in the
peak coordinates.
Simple Harker sections
Consider the space group P222 with a heavy atom at (x,y,z). The
symmetry related heavy atoms are therefore at (-x,y,-z), (x,-y,-z)
and (-x,-y,z). Interatomic vectors are therefore of the form
(2x,0,2z), (2x,2y,0) and (0,2y,2z), i.e. they all lie on the axial
planes.
The axial planes are termed the Harker planes corresponding to
P222, and one need look only in these planes for the vectors
between atoms and their symmetry-related partners.
Given peak coordinates of the form (u,v,w) within the Harker planes,
simple algebra allows the determination of (x,y,z). But note that
one can always add or subtract 1/2 from any one or more of the
final coordinates! Why?
How do we deconvolute the
Patterson function ?
Thus the manual search process is as follows.
1. List all the peaks in the Harker section. Try to find self consistent
sets that yield trial coordinates of each and every heavy atom.
Note that in practice the peaks may be of varying height, some
peaks may be entirely absent, some peaks may be the
superposition of more than one vector. Note further that there
can be both a hand and origin ambiguity associated with the
vectors. Try to account for all the peaks present.
2. Then take the possible coordinates of each pair of atoms and
search for all the cross peaks, in an endeavour to resolve the
ambiguities and to check the assignment.
In practice...
Even once the ambiguity in the heavy atom coordinates has been
resolved, one may still be left with overall ambiguity in the hand
of the space group or in the hand of the heavy atom cluster.
Clearly the process is complicated but it can be automated to some
extent. A particularly useful semi-automated search program in
CCP4 is RSPS.
Automated Patterson does not start from the list of Harker peaks.
Instead it considers every possible coordinate (x,y,z) in the
crystallographic asymmetric unit, computes the corresponding
Harker coordinates (uvw) and then evaluates a score function of
the Patterson values at these positions (say the sum of the
Patterson values). The (xyz) coordinates with the highest score
function in Patterson space are then retained as potential heavy
atom sites.
In practice...
One may then generate all the possible cross vector sets, allowing
for ambiguity and use a score function to check these, finally
taking the set of heavy atom positions that have the highest
overall score.
Alternatively one may perform cross-vector searches directly. From
a starting set of heavy atom positions one may search for a
further site by simply computing the coordinates of all possible
cross-vectors between it and the starting set and scoring these
positions in Patterson space. Then check for the corresponding
This technique is sometimes a better way to proceed than Harker
searches as there are more vectors involved and that leads to
better averaging of scores.