1. Lecture 3 Patterson functions

2. 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 inter- atomic 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 R iso range of about 10 -28 %).

3. 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!

4. Here is a simple 2D Patterson function for a set of three atoms. What does a Patterson function look like ? The blue dots are atoms, linked by their inter-atomic vectors. Real space Patterson space The red dots plot the inter-atomic vectors.

5. 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?).

6. 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.

7. 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?

8. 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.

9. 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.

10. 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.