Early in the semester, we discussed the idea that a Patterson function provides us with information about the vectors in 3-space that characterize the displacement of one scatterer in the unit cell from another one. Among the strongest peaks in a Patterson synthesis are the so-called self-peaks, which define the vectors relating an atom to itself. These will always lie right at the origin of the Patterson synthesis, since the displacement of a scatterer at (x,y,z) from itself is (x,y,z) - (x,y,z) = (0,0,0). But if there is rotational symmetry in the space-group, there will be more than one asymmetric unit in the unit cell. For every scatterer at a position (x,y,z) there will be an identical scatterer at another position (x',y',z'), related to the original position by the rotation matrix and translation vector that describes the rotational symmetry of the space-group. We will find strong peaks corresponding to vectors relating a scatterer at a position (x,y,z) and the corresponding scatterer at (x',y',z'). These will appear as strong peaks in the Patterson map at (x-x',y-y',z-z').
For any specific space group there will be particular portions of the overall Patterson map that will contain these strong peaks. If we have strong scatterers (such as the ones contributed by a heavy atom like mercury within a protein), we can search for the peaks relating that mercury atom's position to its symmetry-related position in a different asymmetric unit. If we then look for those peaks within those special portions of the Patterson map where these symmetry-related peaks are expected, we can, in general, learn enough to define where the mercury atom is, or at least two of its three coordinates.
Let us take a simple example. The spacegroup P21
has two asymmetric units within the unit cell, related by a
two-fold screw axis along the y direction.
Therefore the only spatial relationship that defines this unit cell
(other than the trivial one that we have an scatterer at a position
(x,y,z)),
is that any scatterer at (x,y,z) will have a corresponding
symmetry-related position (-x,y+½,-z).
If we have a strong scatterer at (x,y,z) there will also
be one at (-x,y+½,-z).
We say, then, that the equivalent positions in spacegroup
P21 are
(x,y,z) and (-x,y+½,-z).
There will therefore be a strong peak in the Patterson function at
(u,v,w)
= (x,y,z)-(-x,y+½,-z)
= (x-(-x),y-(y+½),z-(-z))
= (2x, -½, 2z).
But what does that -½ really mean?
We can always translate any peak by a full length of a unit cell axis,
so -½ will be exactly the same as -½+1=½.
Therefore we know there should be a big peak at
(u,v,w) = (2x, ½, 2z).
We therefore are primed to look for peaks in the part of the Patterson map where v = ½. Any big peaks we find there are likely to be real ones, and if we find a peak at a particular (u,½,w) value, then it suggests to us that there has to be a peak in the original electron density at x = u / 2, z = w / 2. Unfortunately, that's only two of the three coordinates associated with the peak: we know x and z, but not y. Initially that doesn't matter, because we can set the origin of our coordinate system so that our first atom can be at any y value we want: so we might as well pick it to be y = 0. We can't get away with that for the second or third or subsequent peaks that we find: we need to find ways to determine their y coordinates relative to our first peak. There are ways to do that, but they go beyond the scope of this simple document.
Anyway: the idea here is that we should concentrate our search for interesting peaks in our Patterson map by looking for peaks for which v = ½. We describe these places where we can concentrate our search as Harker sections, in honor of the late Prof. David Harker, who figured out that they would be good places to look for information. In P21, the v = ½ section is the only Harker section. In some other space-group, we may find that there are more than two equivalent positions, and that may mean there is more than one Harker section. By analyzing all of the Harker sections, we may be able to arrive at fully-determined positions for our strong scatterers: that is, we may be able to calculate all three coordinates (x,y,z) for our heavy atoms.