This lecture focuses on one of the most important practical tools in macromolecular crystallography:
โญ The Patterson function โ a method that allows us to extract structural information without knowing phases.
This is especially powerful for finding heavy atom positions, which is crucial for solving the phase problem using isomorphous replacement.
The Patterson function is:
๐ A convolution of the electron density with itself
Mathematically:
โ Patterson maps can be calculated using only diffraction intensities. โ No phase information is needed.
This is possible because when electron density is multiplied by itself:
Thus:
Patterson = Fourier transform of intensities (|F|ยฒ) only.
A Patterson map is:
๐ฏ An inter-atomic vector map
It does NOT show atomic positions directly. Instead it shows:
n^2 - n + 1
where n = number of atoms
P(u,v,w) = P(-u,-v,-w)
This symmetry exists even if the real crystal is not centrosymmetric.
Vectors:
Total peaks:
2^2 -2 +1 = 3
These appear:
Vectors:
Total peaks:
3^2 -3 +1 = 7
So Patterson maps grow very crowded quickly โ interpretation becomes harder.
To solve phases using isomorphous replacement, we want:
โก Patterson map of heavy atoms only
But we cannot measure their structure factors directly.
Instead we use:
Delta F_ = |F_| - |F_P|
Using trigonometry:
(Delta F_)^2 approx rac{1}{2} |F_H|^2
So:
โญ Difference Patterson โ Patterson of heavy atoms (scaled)
This is hugely important because it allows:
๐ Determination of heavy atom positions from intensity differences.
Vector peak weight:
ext{Weight} propto Z_i imes Z_j
Example (bromobenzene):
| Vector | Weight |
|---|---|
| HโH | 1 |
| HโC | 6 |
| HโBr | 35 |
| CโC | 36 |
| CโBr | 210 |
| BrโBr | 1225 |
๐ก Heavy atom vectors dominate Patterson maps โ makes them easier to detect.
If the space group has screw axes, Patterson peaks concentrate in special planes:
โญ Called Harker sections
Example:
v = rac{1}{2}
Thus instead of searching the full 3D map โ we inspect specific planes.
This dramatically simplifies heavy atom localization.
Procedure shown for space group P212121:
โ Gives relations like:
u = pm 2x v = -2z + rac{1}{2}
Result:
โ Fix another coordinate (e.g., y)
โ Possible heavy atom positions found.
Thus number of heavy atoms determined.
Because Patterson maps are centrosymmetric:
โ ๏ธ Cannot distinguish:
So crystallographers:
Correct map โ continuous protein density Wrong map โ fragmented nonsense density.
Heavy atom coordinates may initially look incorrect.
But applying:
Can transform them into the true crystallographic positions.
Thus solving Patterson maps involves:
๐ง Mathematics ๐ง Symmetry reasoning ๐ง Trial refinement
This lecture shows the practical workflow for phase determination: