Heavy atoms with Patterson maps

The large number of electrons associated with the heavy atom will produce large scattering amplitudes from this atom and, as a result, many of the intensities will be dominated by scattering from this heavy atom. In order to locate the position of the heavy atom, a Patterson function (Patterson, 1935) is calculated which represents a three-dimensional vector space map, also known as an Fp map. The Patterson function is... 

Since protein BLl 1 is nearly globular its location may be determined in a Patterson map with coefficients of [F(wild)-F(mutant)] and may serve, by itself, as a giant heavy-atom derivative. At preliminary stages of structure determination this approach may provide phase information and reveal the location of the lacking protein. 

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... 

Determination of Structure. Analysis of a Patterson map indicated two sets of heavy atoms in genera) positions, a result incompatible with the preconceived opinion of the composition of the materia). For this reason the subsequent analysis was carried out using the diffraction data to establish the composition. The peaks for the heavy atoms were of appropriate relative height to correspond to Xe and As, and other peaks were found which corresponded to six F atoms around the As atom. Fourier maps phased with these eight atoms revealed the seventh fluorine atom. Another smaller peak was tested as a possible fluorine atom, but it was rejected by the least-squares refinement. A later electron density map, prior to the absorption correction and with R - = O.IO, showed no other peaks... 

FIGURE 8.12. The Patterson map represents all interatomic vectors, moved so that one end of each vector is at the origin of the map. (a) An interatomic vector, between two atoms A and B is represented in the Patterson map by a vector with one end at the origin of the Patterson map. (b) In a centrosymmetric triclinic crystal structure with one heavy atom per asymmetric unit, if the Patterson map has a peak at u, v, then the heavy atom lies at i = u/2, y = v/2. If u and v are measured, then x and y will be found. 

With a knowledge of the space-group symmetry, strong vectors can be analyzed to give the fractional atomic coordinates of those atoms in the structure that have the highest atomic numbers. If there is only one such heavy atom in the asymmetric unit, the interpretation of the Patterson map is simplified because the map is dominated by heavy-atom-hea ... 

If the replaceable atoms are heavy, they can be located by a Patterson map. Then, since Fmi and Fm2 are calculated for identical positions of Ml and M2, their relative phases are identical. The value of Fmi - F, / is compared with that of Fi and IF21 obtained from experimental data. This process is shown in Figure 8.23. 

Homometric structure A structure with a uniquely different arrangement of atoms from another, but having the same sets of interatomic vectors, and hence the same Patterson map. Examples to date consist of homometric heavy-atom positions in crystal structures that also contain lighter atoms. The total crystal structure is not, however, homometric. 

The Patterson synthesis (Patterson, 1935), or Patterson map as it is more commonly known, will be discussed in detail in the next chapter. It is important in conjunction with all of the methods above, except perhaps direct methods, but in theory it also offers a means of deducing a molecular structure directly from the intensity data alone. In practice, however, Patterson techniques can be used to solve an entire structure only if the structure contains very few atoms, three or four at most, though sometimes more, up to a dozen or so if the atoms are arranged in a unique motif such as a planar ring structure. Direct deconvolution of the Patterson map to solve even a very small macromolecule is impossible, and it provides no useful approach. Substructures within macromolecular crystals, such as heavy atom constellations (in isomorphous replacement) or constellations of anomalous scattered, however, are amenable to direct Patterson interpretation. These substructures may then be used to solve the phase problem by one of the other techniques described below. 

It may not be obvious how we would locate the x, y, z coordinates of the heavy atom in the unit cell. Indeed it is sometimes not a simple matter to find those coordinates, but as for the heavy atom method described above, it can be achieved using Patterson methods (described in Chapter 9). As we will see later, Patterson maps were used for many years to deduce the positions of heavy atoms in small molecule crystals, and with only some modest modification they can be used to locate heavy atoms substituted into macromolecular crystals as well. Another point. It is not necessary to have only a single heavy atom in the unit cell. In fact, because of symmetry, there will almost always be several. This, however, is not a major concern. Because of the structure factor equation, even if there are many heavy atoms, we can still calculate Juki, the amplitude and phase of the ensemble. This provides just as good a reference wave as a single atom. The only complication may lie in finding the positions of multiple heavy atoms, as this becomes increasingly difficult as their number increases. 

Unless direct methods are used to locate heavy atom positions, an understanding of the Patterson function is usually essential to a full three-dimensional structure analysis. Interpretation of a Patterson map has been one of two points in a structure determination where the investigator must intervene with skill and experience, judge, and interpret the results. The other has been the interpretation of the electron density map in terms of the molecule. Interpretation of a Patterson function, which is a kind of three-dimensional puzzle, has in most instances been the crucial make or break step in a structure determination. Although it need not be performed for every isomorphous or anomalous derivative used (a difference Fourier synthesis using approximate phases will later substitute see Chapter 10), a successful application is demanded for at least the first one or two heavy atom derivatives. 

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. 

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. 

All methods of deduction of the relative phases for Bragg reflections from a protein crystal depend, at least to some extent, on a Patterson map, commonly designated P(uvw) (46, 47). This map can be used to determine the location of heavy atoms and to compare orientations of structural domains in proteins if there are more than one per asymmetric unit. The Patterson map indicates all the possible relationships (vectors) between atoms in a crystal structure. It is a Fourier synthesis that uses the indices, l, and the square of the structure factor amplitude f(hkl) of each diffracted beam. This map exists in vector space and is described with respect to axes u, v, and w, rather than x,y,z as for electron-density maps. 

Apart from the high peak at the origin (the sum of the vectors from each atom to itself), there should be one high peak in the Patterson map at v = A, and the position of this peak will give values forx and z of the heavy atom in the unit cell (with the screw axes at x = 0, z = 0). They coordinate of one atom is arbitrary in this particular space group. This is evident in the general positions of the space group (listed above) since if there is an atom aty there is also one at Vi + y, but no infor-... 

The unit cell of a protein is assigned with respect to a right-handed system of axes. Once a heavy atom has been located, its phase angle may be +a or —a for FH, since it is not known whether the interpreted peak in the Patterson map is from atom 1 to atom 2 or atom 2 to atom 1. Several methods have been developed to remove this ambiguity of which the most decisive are those that involve the preparation of a derivative containing both heavy atoms and/or anomalous dispersion measurements (5). 

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