Patterson maps Harker planes

Unit-cell symmetry can also simplify the search for peaks in a three-dimensional Patterson map. For instance, in a unit cell with a 2X axis (twofold screw) on edge c, recall (equivalent positions, Chapter 4, Section II.H) that each atom at (x,y,z) has an identical counterpart atom at (-x,-y,V2 + z). The vectors connecting such symmetry-related atoms will all lie at (u,v,w) = (2x,2y,V2) in the Patterson map (just subtract one set of coordinates from the other), which means they all lie in the plane that cuts the Patterson unit cell at w = l/2. Such planes, which contain the Patterson vectors for symmetry-related atoms, are called Harker sections or Harker planes. If heavy atoms bind to the protein at... 

FIGURE 9.2 A section from a difference Patterson map calculated between a heavy atom derivative and native diffraction data (known as a difference Patterson map). This map is for a mercury derivative of a crystal of bacterial xylanase. The plane of Patterson density shown here corresponds to all values of u and w for which v =. Because the space group of this crystal is P2, this section of the Patterson map is a Harker section containing peaks denoting vectors between 2t symmetry related heavy atoms. 

FIGURE 9.7 Two molecules of tRNA in (a) are related by a twofold symmetry axis along z in the crystal. A point x, y, z, which could be the site of a heavy atom in one molecule, has an identical corresponding site in the dyad-related molecule at — x, —y, z The vector that connects the two sites will be (x, y, z) — (—x, —y, z) = 2x, 2y, 0. This vector, a Harker vector, must appear on the ro = 0 section of the corresponding Patterson map computed from the intensities of the diffraction pattern. In (b) the heavy atom site on the protein molecule at x, y, z appears on the 2i screw axis (along z) related asymmetric unit at —x, —y, Z—. But —, the unit translation, is the same as +, so the difference vector is 2x, 2y,. This Harker vector would appear on the plane of the Patterson map containing points for which w =. ... 

Thus we see that for space group P222i there are three planes of the Patterson map that must contain Harker peaks, and they are (0, v, w), (u, v, j), and ( , 0, w). 

FIGURE 9.11 The w = j plane of the difference Patterson map for the K2HgI4 heavy atom derivative of the hexagonal crystal form of the protein canavalin. The space group is P6, so w = is a Harker section. The derivative crystal contained two major K2HgI4 substitution sites and one minor substitution site per asymmetric unit. The Patterson peaks corresponding to those sites are marked with crosses. Note that the Patterson peak corresponding to the minor site cannot be discriminated from noise peaks in the Patterson map as is often the case. 

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