Kevin Cowtan, York The Patterson Function. Kevin Cowtan

Transcription

1 Kevin Cowtan

2 Outline Doing without phases. Interatomic vectors. Harker vectors.

3 We cannot measure the phases associated with the spots in the diffraction pattern and as a result, we cannot calculate an electron density map. However, the diffraction pattern still contains a lot of information, in the form of the distribution of strong and weak structure factors. This information should be able to tell us something about the structure. The Patterson is one means to access some of that information.

4 A two-atom structure: With the structure factors and phases.

5 The (3,1) reflection is strong: What happens if we set its phase to zero? Lose the positions, keep the spacing.

6 The (1,4) reflection is strong: What happens if we set its phase to zero? Lose the positions, keep the spacing.

7 What happens if we zero all of the reflections? We lose the position of the atoms, but retain information of their spacing.

8 What happens if we zero all of the reflections? We lose the position of the atoms, but retain information of their spacing. There is a big peak at the origin, and two smaller peaks either side of it, at positions given by the vector from each atom to the other.

9 The Patterson has a peak for every interatomic vector:

10 The Patterson has a peak for every interatomic vector: 4 atoms = 12 peaks N atoms = N(N-1) peaks (not including origin)

11 Can be solved by hand for up to ~10 atoms. Can be solved by computer for up to ~100 atoms. Can't be solved for proteins Although you may be able to solve a heavy atom substructure.

12 Theory: Mathematically, if we set the phases to zero, it also makes sense to square the amplitudes: P(h) = F(h) 2 = F(h)F(h)* Then when we take the Fourier transform, the resulting map is the auto-correlation function of the electron density: p(x) = y ρ(y) ρ(x+y) i.e. the Patterson function has a peak at every position y which moves one density peak onto another.

13 What about symmetry? Note: there are lots of peaks on the v=0 line, corresponding to vectors between corresponding atoms in the two molecules.

14 In P2, the symmetry equivalent positions are: (u, v, w); (-u, v, -w) There will always be vectors in the Patterson from an atom to its symmetry equivalent. These will be at positions: (u, v, w) - (-u, v, -w ) = ( 2u, 0, 2w ) i.e. there will be many peaks in the v=0 plane. If we see a Patterson peak at ( u 1, 0, w 1 ), then there must be an atom at ( ½u 1,?, ½w 1 ).

15 What about symmetry? Note: there are lots of peaks on the v=1/2 line, corresponding to vectors between corresponding atoms in the two molecules.

16 In P2 1, the symmetry equivalent positions are: (u, v, w); (-u, v+½, -w) There will always be vectors in the Patterson from an atom to its symmetry equivalent. These will be at positions: (u, v, w) - (-u, v+½, -w ) = ( 2u, ½, 2w ) i.e. there will be many peaks in the y=½ plane. If we see a Patterson peak at ( u 1, ½, w 1 ), then there must be an atom at ( ½u 1,?, ½w 1 ).

17 Examples from other spacegroups: For P6 1 : w=1/6 is a Harker section. For P422: v=u is a Harker section. For P23: u+v+w=0 is a Harker section.

18 The Patterson Gives a map of inter-atomic vectors. These include shorter intra-molecular vectors and longer inter-molecular vectors. Symmetry elements lead to special Harker sections, with greater concentrations of peaks. Harker peaks involve vectors between symmetry equivalent atoms, and so may be simpler to interpret.