Patterson space

Figure 12. In order to check the experimental Patterson map for a known structural fragment it must be converted into Patterson space first. This is shown in the above sketch for the fragment of a body centered cube which is projected along the 4-fold axis (a). The corresponding motif in Patterson space is shown in (b). Note, that the atom in the center of the fragment defines a 2-fold axis what reduces the number of different peaks in the Patterson map.

While a proper discussion of this function is beyond the scope of this article, it is sufficient to note that this function allows the splitting of the rotation/translation problem. But this decomposition of the problem in Patterson space entails the rotational seeirch to be carried out first without knowing the correct translation beforehand. Residual noise due to the very approximate nature of the Patterson decomposition ensures that the rotation search produces multiple broad peaks rather than a single sharp peak (Figure 3a). This makes the selection of the correct rotational setting relatively difficult. Nevertheless, the model is then rotated as accurately as possible to the candidate rotational setting. The translation function search is then conducted. If the selected rotational orientation is indeed the correct one the translational search should have relatively httle noise and the correct translational solution can be expected to clearly stand out (Figure... 

It is possible, as shown by Rossmann and Blow (1962), to search for redundancies in Patterson space that correspond to the multiple copies of molecular transforms. Rossmann and Blow show, however, that the Patterson map does not need to be computed and used in any graphical sense, but that an equivalent search process can be carried out directly in diffraction or reciprocal space. Using such a search procedure, called a rotation function, they showed that noncrystallographic relationships, both proper and improper rotations, could be deduced in many cases directly from the X-ray intensity data alone, and in the complete absence of phase information. Translational relationships (only after rotations have been established) can also be deduced by a similar approach. Rotation functions and translation functions constitute what we call molecular replacement procedures. Ultimately the spatial relationships among multiple molecules in an asymmetric unit can be defined by their application. 

In addition to being confusing, this might at first also appear quite useless. The value of P(u, v, w) at any point u,v,w in Patterson space is the sum of products obtained by... 

The Patterson function is yet another example of a convolution function (see Chapter 3), and it maps vector relationships in real space into a second coordinate system, which is Patterson space. It will be instructive here to examine the Patterson function s relationship to a real atom distribution by asking how it may be physically generated if the distribution of atoms in real space is known. Examples are illustrated in Figures 9.3 and 9.4. The Patterson function of a known structure is formed in the following way. 

A Patterson coordinate system is defined based on unit vectors u, v, and w, which are parallel with the axes of the real unit cell of the crystal. Each point (u, v, and w) in Patterson space defines the end point of a vector u having a unique direction and length from the origin of Patterson space to that point, that is, u = (u,v, w). Every point, or vector, in Patterson space (u, v, w) will have associated with it a value P(u). 

The process is now repeated for every vector u that can be drawn within the unit cell of the crystal until the value of P(u) for every point (u, v, w) in Patterson space has been calculated and entered. The collection of all points (u, v, w) with their associated value P(u, v, w) is the Patterson map of the crystallographic unit cell. The Patterson map yields, at least directly, no information regarding the absolute positions of scattering matter in the unit cell, atoms, but it does provide a map of all interatomic vectors in the crystal. The... 

Occasionally more than one pair of atoms in a structure may be related by identical vectors, for example, atoms that lie on opposite sides of a benzene ring. Therefore overlap can occur in Patterson space. For some points u, v, w the Patterson value P(u, v, w) will be the sum of more than one nonzero product. This is what makes interpretation of Patterson maps particularly challenging, and obviously this becomes more so as the number of atoms increases. In general, however, every pair of atoms in the unit cell will give rise to two centrosymmetrically related, nonzero vectors in Patterson space. Hence, if there are N atoms in the unit cell, there will be N(N — 1) peaks in the Patterson map, as well as the vector from every atom to itself, but this vector is always u = (0, 0, 0). 

This alternative way of looking at a Patterson map is illustrated by a four atom and a five atom structure in Figures 9.5 and 9.6. This is sometimes a useful way of considering the Patterson map because it provides the basis for various kinds of Patterson search methods where the objective is to find the image of a known part of a molecule in Patterson space. It is also the basis for the rotation and translation functions used in molecular replacement procedures (see Chapter 8). 

FIGURE 9.9 Two RNase A molecules are related by a twofold axis along y in a crystal. Assume that the points indicated correspond to sites of heavy atoms. The vector between any pair of symmetry-equivalent heavy atom sites is U, 0, W in Patterson space and is Xi — x, yi — yi, Zi — Zi in real space. Thus V = x2 — xt, V = yi — yq = 0, and IT = -2 — z,. U, V, and VT can be obtained by inspection of the Patterson map calculated from the diffraction intensities of the crystal and x and z, two coordinates of the heavy atoms, be determined unambiguously. 

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