Using the Patterson Map for Isomorphous Replacement

A Patterson map, different for each space group, is a unique puzzle that must be solved to gain a foothold on the phase problem. It is by finding the absolute atomic coordinates of a heavy atom, for both small molecule and macromolecular crystals, that initial estimates (later to be improved upon) can be obtained for the phases of the structure factors needed to calculate an electron density map. 

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. 

For each symmetry relationship (and corresponding difference between equivalent positions), it will be found that one of the three Patterson coordinates is a constant, 0 or j or and so on. These define the Harker sections of the Patterson map. For example, if one of the Patterson relationships is u = 2x, v = and w = 2z, then the Harker section , j, w is first calculated and investigated. On this section, like the real unit cell of the crystal whose periodicity it reflects, u and w will vary from 0 to 1 and v will be constant, j. It is generally useful to separately calculate the individual Harker sections, review them, and only then calculate the full map. The appearance of the Harker sections and their content 

When the unit cell contains several symmetry elements, then each of these will each give rise to a Harker section. The additional Harker sections will in turn yield information about one of the coordinates already known (but not both), as well as information about the missing coordinate. A second Harker section will not only fill out the x, y, z coordinate set for the heavy atom, but will confirm, or serve as a cross check on one of the coordinates deduced from the first Harker section. A third Harker section will, in its turn, provide a cross check on heavy atom coordinates derived from both of the others. 

Another problem that frequently arises with multiple isomorphous derivatives is that of handedness. In space group P2i2i2i, Patterson maps for two independent derivatives may be interpreted to yield a set of symmetry related sites for one derivative and, independently, a second set for the other. Because handedness is completely absent in a Patterson map (because it contains a center of symmetry), there is an equal chance that the heavy atom constellation for the first will be right handed, and the constellation for the other will be left handed, and vice versa. This won t do. The two heavy atom sets will not cooperate when used to obtain phase information. There are ways of unraveling this problem too, and once again, it involves difference Patterson maps between the two derivative data sets and cross vectors. This case can also be resolved by calculating phases based on only one derivative and then computing a difference Fourier map (see Chapter 10) for the other. 

---