Lecture 7 Video 10

Protein structure

๐Ÿงฌ Lecture Summary โ€” Patterson Function & Heavy Atom Substructure

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.


๐Ÿ“Œ 1. What is the Patterson Function?

The Patterson function is:

๐Ÿ” A convolution of the electron density with itself

Mathematically:

  • It integrates electron density at position (x,y,z) with density at (x+u, y+v, z+w).
  • If both positions contain atoms โ†’ large Patterson value (peak).
  • If not โ†’ small value.

๐Ÿšจ Super important takeaway

โœ… Patterson maps can be calculated using only diffraction intensities. โŒ No phase information is needed.

This is possible because when electron density is multiplied by itself:

  • The phase terms cancel out (square out).

Thus:

Patterson = Fourier transform of intensities (|F|ยฒ) only.


๐Ÿงญ 2. What Does a Patterson Map Represent?

A Patterson map is:

๐ŸŽฏ An inter-atomic vector map

It does NOT show atomic positions directly. Instead it shows:

  • Vectors between atoms
  • Each peak = vector from atom i โ†’ atom j

Properties

  • Large origin peak = all self-vectors (atom โ†’ itself)
  • Total peaks:

n^2 - n + 1

where n = number of atoms

  • Map always has inversion symmetry

P(u,v,w) = P(-u,-v,-w)

This symmetry exists even if the real crystal is not centrosymmetric.


๐Ÿ”ฌ 3. Simple Examples

๐Ÿงช Two-atom structure

Vectors:

  • Atom 1 โ†’ Atom 2
  • Atom 2 โ†’ Atom 1
  • Two self vectors โ†’ origin peak

Total peaks:

2^2 -2 +1 = 3

These appear:

  • One central peak
  • Two symmetric vector peaks

๐Ÿงช Three-atom structure

Vectors:

  • 1โ†”2
  • 2โ†”3
  • 1โ†”3

Total peaks:

3^2 -3 +1 = 7

So Patterson maps grow very crowded quickly โ†’ interpretation becomes harder.


โšก 4. Difference Patterson Map (Heavy Atom Method)

To solve phases using isomorphous replacement, we want:

โžก 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‰ˆ Patterson of heavy atoms (scaled)

This is hugely important because it allows:

๐Ÿ‘‰ Determination of heavy atom positions from intensity differences.


โš–๏ธ 5. Peak Heights Depend on Electron Numbers

Vector peak weight:

ext{Weight} propto Z_i imes Z_j

Example (bromobenzene):

VectorWeight
Hโ€“H1
Hโ€“C6
Hโ€“Br35
Cโ€“C36
Cโ€“Br210
Brโ€“Br1225

๐Ÿ’ก Heavy atom vectors dominate Patterson maps โ†’ makes them easier to detect.


๐Ÿงฉ 6. Harker Sections (Symmetry Helps!)

If the space group has screw axes, Patterson peaks concentrate in special planes:

โญ Called Harker sections

Example:

  • 2โ‚ screw axis along Y โ†’ vectors lie in plane:

v = rac{1}{2}

Thus instead of searching the full 3D map โ†’ we inspect specific planes.

This dramatically simplifies heavy atom localization.


๐Ÿงฎ 7. Determining Heavy Atom Coordinates (Stepwise Logic)

Procedure shown for space group P212121:

Step 1 โ€” Identify Patterson peaks in one Harker section

โ†’ Gives relations like:

u = pm 2x v = -2z + rac{1}{2}

Result:

  • Some coordinates fixed (e.g., z)
  • Others ambiguous (ยฑx)

Step 2 โ€” Use second Harker section

โ†’ Fix another coordinate (e.g., y)

Step 3 โ€” Combine constraints

โ†’ Possible heavy atom positions found.

Step 4 โ€” Remove spurious peaks

  • Some peaks belong to another Harker section (overlap)
  • Expect one strong peak per heavy atom

Thus number of heavy atoms determined.


๐Ÿ” 8. Origin Ambiguity & Inversion Problem

Because Patterson maps are centrosymmetric:

โš ๏ธ Cannot distinguish:

  • True heavy atom solution
  • Inverted solution

So crystallographers:

  • Build both phase sets
  • Calculate electron density maps
  • Choose the interpretable one

Correct map โ†’ continuous protein density Wrong map โ†’ fragmented nonsense density.


๐Ÿ”ง 9. Symmetry & Origin Shifts

Heavy atom coordinates may initially look incorrect.

But applying:

  • Symmetry operations
  • Origin shifts (e.g., ยฑยฝ cell translation)

Can transform them into the true crystallographic positions.

Thus solving Patterson maps involves:

๐Ÿง  Mathematics ๐Ÿง  Symmetry reasoning ๐Ÿง  Trial refinement


โญ Final Big Picture

This lecture shows the practical workflow for phase determination:

  1. Measure diffraction intensities
  2. Compute difference Patterson map
  3. Locate heavy atom peaks (often in Harker sections)
  4. Determine heavy atom coordinates
  5. Calculate heavy atom structure factors
  6. Use Harker construction โ†’ obtain protein phases
  7. Build interpretable electron density map

๐Ÿง  Ultra-Short Exam Takeaways

  • Patterson = electron density autocorrelation
  • Uses intensities only (no phases!)
  • Shows inter-atomic vectors
  • Peaks = ( n^2 - n +1 )
  • Heavy atom peaks are strongest (high Zยฒ weighting)
  • Screw axes create Harker sections
  • Patterson always has inversion symmetry
  • Heavy atom solution may be origin-shifted or inverted

Quiz

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