1. Overview of the Phase Problem
Protein
Crystal
Data
Phases
Structure
John Rose ACA Summer School 2006
Reorganized by Andy Howard, Biology 555, Spring 2008
Remember
We can measure reflection intensities
We can calculate structure factors from the intensities
We can calculate the structure factors from atomic positions
We need phase information to generate the image
Biology 555 Crystallographic Phasing I
14 Feb 2008

2. What is the Phase Problem?
X-ray Diffraction Experiment
All phase information is lost
x,y.z
Fhkl
[Real Space]
[Reciprocal Space]
In the X-ray diffraction experiment photons are reflected from the crystal lattice (planes) in different directions giving rise to the diffraction pattern.
Using a variety of detectors (film, image plates, CCD area detectors) we can estimate intensities but we lose any information about the relative phase for different reflections.

3. Biology 555 Crystallographic Phasing I
Phases
Let’s define a phase fj associated with a specific plane [hkl] for an individual atom: fj = 2p(hxj + kyj + lzj)
Atom at xj=0.40, yj=0.05, zj=0.10 for plane [213]: fj = 2p(2* * *0.10) = 2p(1.35)
If we examine a 2-dimensional case like k=0, then fj = 2p(hxj + lzj)
Thus for [201] (a two-dimensional case): fj = 2p(2* * *0.10) = 2p(0.90)
Now, to understand what this means:
Biology 555 Crystallographic Phasing I
14 Feb 2008

4. Biology 555 Crystallographic Phasing I
201 Phases A B G C H D F I E 0°
720° c a
201 planes 4p
360° 2p
1080° 6p
0.4, y, 0.1
fD = 2p[ 2•(0.40) + 1•(0.10)] = 2p(0.)
Biology 555 Crystallographic Phasing I
14 Feb 2008

5. In General for Any Atom (x, y, z)
dhkl 6π
dhkl 4π
Atom (j) at x,y,z
dhkl 2π φ c
Remember: We express any position in the cell as
(1) fractional coordinates: pxyz = xja+yjb+zjc
(2) the sum of integral multiples of the reciprocal axes
hkl = ha* + kb* + lc*
Plane hkl
Biology 555 Crystallographic Phasing I
14 Feb 2008

6. Diffraction vector for a Bragg spot
We set up the diffraction vector shkl associated with a specific diffraction direction hkl: shkl = ha* + kb* + lc*
The magnitude of this diffraction vector is the reciprocal of our Bragg-law plane spacing dhkl: |shkl| = 1/ dhkl
Biology 555 Crystallographic Phasing I
14 Feb 2008

7. Biology 555 Crystallographic Phasing I
Phase angle for a spot
The phase angle fj associated with our atom is 2p times the projection of the displacement vector pj onto shkl: fj = 2p shkl• pj
But that displacement vector pj is related to the real-space coordinates of the atom at position j: pj = xja + yjb + zjc where the fractional coordinates of our atom within the unit cell are (xj, yj, zj)
Thus fj = 2p (ha* + kb* + lc*) • (xja + yjb + zjc)
Biology 555 Crystallographic Phasing I
14 Feb 2008

8. Real-space and reciprocal space
But these real-space and reciprocal-space unit cell vectors (a,b,c) and (a*,b*,c*) are duals of one another; that is, they obey:
a•a* = 1, a•b* = 0, a•c* =0
b•a* = 0, b•b* = 1, b•c* =0
c•a* = 0, c•b* = 0, c•c* = 1
… even when the unit cell isn’t all full of 90-degree angles!
Biology 555 Crystallographic Phasing I
14 Feb 2008

9. Matrix formulation of this duality
If we construct the 3x3 reciprocal-space unit cell matrix A = (a* b* c*)
And the 3x3 real-space unit cell matrix R = (a b c) for a specific position of the sample, then
A and R obey the simple relationship A = R-1, i.e. AR = I
Where I is a 3x3 identity matrix
Biology 555 Crystallographic Phasing I
14 Feb 2008

10. How to use this in getting phases
fj = 2p (ha* + kb* + lc*) • (xja + yjb + zjc)
But using those dual relationships, e.g. a*•a = 1, b*•c = 0, we get fj = 2p (hxj + kyj + lzj)
Note that this is true even if our unit cell angles aren’t 90º!
Biology 555 Crystallographic Phasing I
14 Feb 2008

11. Biology 555 Crystallographic Phasing I
Why Do We Need the Phase?
Fourier transform
Inverse Fourier transform
Structure Factor
Electron Density
In order to reconstruct the molecular image (electron density) from its diffraction pattern both the intensity and phase, which can assume any value from 0 to 2, of each of the thousands of measured reflections must be known.
Biology 555 Crystallographic Phasing I
14 Feb 2008

12. Importance of Phases Phases dominate the image!
Hauptman amplitudes
with Hauptman phases
Karle amplitudes
with Karle phases
Karle amplitudes
with Hauptman phases
Hauptman amplitudes
with Karle phases
Phases dominate the image!
Phase estimates need to be accurate
Biology 555 Crystallographic Phasing I
14 Feb 2008

13. Understanding the Phase Problem
The phase problem can be best understood from a simple mathematical construct.
The structure factors (Fhkl) are treated in diffraction theory as complex quantities, i.e., they consist of a real part (Ahkl) and an imaginary part (Bhkl).
If the phases, hkl, were available, the values of Ahkl and Bhkl could be calculated from very simple trigonometry:
Ahkl = |Fhkl| cos (hkl)
Bhkl = |Fhkl| sin (hkl)
This leads to the relationship: (Ahkl)2 + (Bhkl)2 = |Fhkl|2 = Ihkl
Biology 555 Crystallographic Phasing I
14 Feb 2008

14. Biology 555 Crystallographic Phasing I
Argand Diagram
(Ahkl)2 + (Bhkl)2 = |Fhkl|2 = Ihkl
The above relationships are often illustrated using an Argand diagram (right).
From the Argand diagram, it is obvious that Ahkl and Bhkl may be either positive or negative, depending on the value of the phase angle, hkl.
Note: the units of Ahkl, Bhkl and Fhkl are in electrons.
Biology 555 Crystallographic Phasing I
14 Feb 2008

15. Biology 555 Crystallographic Phasing I
The Structure Factor
sinq/l f0
Atomic scattering factors
Here fj is the atomic scattering factor
The scattering factor for each atom type in the structure is evaluated at the correct sinq/l. That value is the scattering ability for that atom.
Remember sinq/l = 1/(2dhkl)
We now have an atomic scattering factor with magnitude f0 and direction fj
Biology 555 Crystallographic Phasing I
14 Feb 2008

16. The Structure Factor Sum of all individual atom contributions
real
imaginary
Individual
atom fjs
Resultant
Fhkl
Ahkl
Bhkl
Biology 555 Crystallographic Phasing I
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17. Biology 555 Crystallographic Phasing I
Electron Density
Remember the electron density (image of the molecule) is the Fourier transform of the structure factor Fhkl. Thus
Here V is the volume of the unit cell
Biology 555 Crystallographic Phasing I
14 Feb 2008

18. How to calculate r(x,y,z)
In practice, the electron density for one three-dimensional unit cell is calculated by starting at x, y, z = (0, 0, 0) and stepping incrementally along each axis, summing the terms as shown in the equation above for all hkl (as limited by the resolution of the data) at each point in space.
Biology 555 Crystallographic Phasing I
14 Feb 2008

19. Solving the Phase Problem
Small molecules
Direct Methods
Patterson Methods
Molecular Replacement
Macromolecules
Multiple Isomorphous Replacement (MIR)
Multi Wavelength Anomalous Dispersion (MAD)
Single Isomorphous Replacement (SIR)
Single Wavelength Anomalous Scattering (SAS)
Direct Methods (special cases)
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14 Feb 2008

20. Solving the Phase Problem
SMALL MOLECULES:
The use of Direct Methods has essentially solved the phase problem for well diffracting small molecule crystals.
MACROMOLECULES:
Today, anomalous scattering techniques such as MAD or SAS are the most common techniques used for de novo structure determination of macromolecules. Both techniques require the presence of one or more anomalous scatterers in the crystal.
Biology 555 Crystallographic Phasing I
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21. Biology 555 Crystallographic Phasing I
Direct methods
Karle, Hauptman, David Sayre, and others determined algebraic relationships among phase angles of groups of reflections.
The simplest are triplet relationships: For three reflections h1=(h1,k1,l1), h2=(h2,k2,l2), h3=(h3,k3,l3), they showed that if h3= -h1- h2, then
F1 + F2 + F3 ≈ 0
Thus if F1 and F2 are known then we can estimate that F3 ≈ -F1 - F2
David Sayre
Biology 555 Crystallographic Phasing I
14 Feb 2008

22. When do triplet relations hold?
Note the approximately zero value in that relationship F1 + F2 + F3 ≈ 0.
The stronger the Bragg reflections are, the closer this condition is to being exact.
For very strong Bragg reflections that sum will be very close to zero
For weaker ones it may differ significantly from zero
Biology 555 Crystallographic Phasing I
14 Feb 2008

23. Biology 555 Crystallographic Phasing I
Phase probabilities
This notion of relationships among phases obliges us to think of phases probabilistically rather than deterministically. This is a key to the direct-methods approach and has a huge influence on how we think about phase determination.
I’m introducing all of this mostly to get you accustomed to the notion of phase probability distributions!
Biology 555 Crystallographic Phasing I
14 Feb 2008

24. Biology 555 Crystallographic Phasing I
Phase probabilities
Any phase has a value between 0 and 2p (or 0 and 360, if we’re using degrees)
If we know it’s close to 2p*0.42, then:
If it’s 2p*(0.42 0.01), it’s a sharp phase probability distribution
If it’s 2p*(0.42 0.32), it’s a much broader phase probability distribution
Biology 555 Crystallographic Phasing I
14 Feb 2008

25. Plots of phase probability
Integral of probability must be 1, since every phase has to have some value.
Sharp distribution
Broad distribution  2π
Biology 555 Crystallographic Phasing I
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26. Biology 555 Crystallographic Phasing I
How can we use this?
Obviously if we don’t know f1+f2, we can’t use this to calculate f3, even if the intensities of all three are large.
But we could guess what f1 and f2 are and use this to compute f3.
Then we guess f4 and use the triplet relationship to compute f5 and f6, where h5 = -h1 - h4 and h6 = -h1 - h4 … assuming that reflections 5 and 6 are strong, too!
Biology 555 Crystallographic Phasing I
14 Feb 2008

27. Biology 555 Crystallographic Phasing I
Can we make this work?
We start with guessed phases for a strong reflections and use the triplet relationships to determine the phases for another 1000 reflections
Any particular calculated phase can be determined by several different triplet relationships, so if they’re self-consistent, the initial guessed are correct; if they aren’t self-consistent, the guess was wrong!
In the latter case, we try a different set of guesses for our starting phases and keep going
Biology 555 Crystallographic Phasing I
14 Feb 2008

28. This actually works, provided:
The data are correctly measured
The data are strong enough that we can pick 1000 strong reflections to use in this process
The data extend to high enough resolution that atomicity (separable atoms) is really found
There are ways to do direct methods without assuming atomicity, but they’re more complicated
Biology 555 Crystallographic Phasing I
14 Feb 2008

29. Is this relevant to macromolecules?
Not directly:
Atomicity rarely present
Systematic errors in data
Indirectly yes, because it can be used in conjunction with other methods for locating heavy atoms in the SIR, MIR, and SAS methods
It also helps introduce the notion of phase probability distributions (sneaky!)
Biology 555 Crystallographic Phasing I
14 Feb 2008

30. Biology 555 Crystallographic Phasing I
SIR and SAS Methods
Need a heavy atom (lots of electrons) or a anomalous scatterer (large anomalous scattering signal) in the crystal.
SIR - heavy atoms usually soaked in.
SAS - anomalous scatterers usually engineered in as selenomethional labels. Can also be soaked.
SIR collect a native and a derivative data set (2 sets total). SAS collect one highly redundant data set and keep anomalous pairs separate during processing.
SAS - may want to choose a scatterer or wavelength that enhances the anomalous signal.
Must find the heavy atoms or anomalous scatterers
can use Patterson analysis or direct methods.
Must resolve the bimodal ambiguity.
use solvent flattening or similar technique
Biology 555 Crystallographic Phasing I
14 Feb 2008

31. What’s the bimodal ambiguity?
As we’ll show next time, a single isomorphous derivative or anomalous scatterer enables us to measure each phase apart from an ambiguity
That is, for each phase we get two answers (e.g. 2π*0.12 and 2π*0.55), and we can’t pick one out
A second scatterer will resolve that
Biology 555 Crystallographic Phasing I
14 Feb 2008

32. Phase probabilities with no error
A single derivative with no error gives a phase probability like this:  2π
Biology 555 Crystallographic Phasing I
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33. Biology 555 Crystallographic Phasing I
2 derivatives, no error
P()
Wrong estimate derived from derivative 2
The two distributions overlap at the correct answer, not at the wrong answer
Wrong estimate derived from derivative 1
Correct phase  2π
Biology 555 Crystallographic Phasing I
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34. Biology 555 Crystallographic Phasing I
Errors spread this out
Each phase estimate is not really that sharp
Lack of isomorphism (see below) makes each distribution spread out
Joint probability distribution from 2 or more experiments is the product of the probability distributions of the individual experiments
Biology 555 Crystallographic Phasing I
14 Feb 2008

35. Realistic probability distributions
Joint probability distribution = product of individual ones  2π
Biology 555 Crystallographic Phasing I
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36. Joint probability distribution
Biology 555 Crystallographic Phasing I
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37. Heavy Atom Derivatives
Heavy atom derivatives MUST be isomorphous
Heavy atom derivatives are generally prepared by soaking crystals in dilute ( mM) solutions of heavy atom salts (see Table II below for some examples).
Crystal cracking is generally a good indication that that heavy atom is interacting with the crystal lattice, and suggests that a good derivative can be obtained by soaking the crystal in a more dilute solution.
Biology 555 Crystallographic Phasing I
14 Feb 2008

38. Is the derivative worth using?
Once derivative data has been collected, the merging R factor (Rmerge) between the native and derivative data sets can be used to check for heavy atom incorporation and isomorphism. Rmerge values for isomorphous derivatives range from 0.05 to Values below 0.05 indicate that there is little heavy atom incorporation. Values above 0.15 indicate a lack of isomorphism between the two crystals.
Biology 555 Crystallographic Phasing I
14 Feb 2008

39. Biology 555 Crystallographic Phasing I
What is isomorphism?
Isomorphism for derivatives means that the structure of the derivatized macromolecule is identical to the structure of the underivatized molecule except at the site where the derivative compound has been introduced.
Biology 555 Crystallographic Phasing I
14 Feb 2008

40. What is lack of isomorphism?
A derivative may be nonisomorphous if:
It alters the unit cell lengths or angles significantly (>0.2%?)
It rotates or translates the entire macromolecule within the unit cell
It alters significantly the conformation of a large segment (> 8 amino acids or 4 nucleotides?) of the mcromolecule
Biology 555 Crystallographic Phasing I
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41. Biology 555 Crystallographic Phasing I
Derivative compounds
Biology 555 Crystallographic Phasing I
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42. Finding the Heavy Atoms or Anomalous Scatterers
The Patterson function
- a F2 Fourier transform with f = 0
- vector map (u,v,w instead of x,y,z)
- maps all inter-atomic vectors
- get N2 vectors!! (where N= number of atoms)
From Glusker, Lewis and Rossi
Biology 555 Crystallographic Phasing I
14 Feb 2008