1. Phasing
Today’s goal is to calculate phases (ap) for proteinase K using MIRAS method (PCMBS and GdCl3).
What experimental data do we need?
1) from native crystal we measured |Fp| for native crystal
2) from PCMBS derivative we measured |FPH1(h k l)|
3) also from PCMBS derivative we measured |FPH1(-h-k-l)|
4) FH1 for Hg atom of PCMBS
5) from GdCl3 derivative we measured |FPH2(h k l)|
6) also from GdCl3 derivative we measured |FPH2(-h-k-l)|
7) FH2 for Gd atom of GdCl3
How many phasing triangles will we have for each structure factor if we use all the data sets ?
FP ( h k l) = FPH ( h k l) - FH ( h k l) is one type of phase triangle.
FP ( h k l) = FPH(-h-k-l) * - FH (-h-k-l)* is another type of phase triangle.

2. Four Phase Relationships
PCMBS
FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l) isomorphous differences
FP ( h k l) = FPH1(-h-k-l)* - FH1 (-h-k-l)* from Friedel mates (anomalous)
GdCl3
FP ( h k l) = FPH2( h k l) FH2 ( h k l) isomorphous differences
FP ( h k l) = FPH2 (-h-k-l)* - FH2 (-h-k-l)* from Friedel mates (anomalous)

3. FP ( h k l) = FPH1 ( h k l) - FH1 (h k l)
Imaginary axis
|Fp ( h k l) | = 1.8 2 1
Real axis -2 -1 1 2 -1 -2
Harker construction
1) |FP ( h k l) | –native measurement

4. FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
Imaginary axis
|Fp ( h k l) | = 1.8
|FH ( h k l) | = 1.4
aH (hkl) = 45°
|Fp | FH
Real axis
Harker construction
1) |FP ( h k l) | –native measurement
2) FH (h k l) calculated from heavy atom position.

5. FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
Imaginary axis
|FP ( hkl) | = 1.8
|FH1 ( hkl) | = 1.4
aH1 (hkl) = 45°
|FPH1 (hkl) | = 2.8
|Fp |
|FPH|
Real axis
|FPH| FH
Harker construction
1) |FP ( h k l) | –native measurement
2) FH1 (h k l) calculated from heavy atom position.
3) |FP1(hkl)|–measured from derivative.
Let’s look at the quality of the phasing statistics up to this point.

6. SIR
Which of the following graphs best represents the phase probability distribution, P(a)?
Imaginary axis
|Fp |
|FPH| a) b) c)
Real axis
|FPH| FH

7. SIR
The phase probability distribution, P(a) is sometimes shown as being wrapped around the phasing circle.
Imaginary axis
|Fp |
|FPH|
Real axis 90
180
270
|FPH| FH

8. SIR
Which of the following is the best choice of Fp? 90
180
270
Imaginary axis 90
180
270 a) b) c)
|Fp |
|FPH| 90
180
270
Real axis
|FPH| FH 90
180
270
Radius of circle is approximately |Fp|

9. best F = |Fp|eia•P(a)da
SIR
Imaginary axis 90
180
270
|Fp |
best F = |Fp|eia•P(a)da a
|FPH|
Real axis
Sum of probability weighted vectors Fp
Usually shorter than Fp
|FPH| FH

10. best F = |Fp|eia•P(a)da
SIR 90
180
270
best F = |Fp|eia•P(a)da a
Sum of probability weighted vectors Fp
Best phase

11. Radius of circle is approximately |Fp|
SIR
Which of the following is the best approximation to the Figure Of Merit (FOM) for this reflection? 90
180
270
1.00
2.00
0.50
-0.10
FOM=|Fbest|/|FP|
Radius of circle is approximately |Fp|

12. Which phase probability distribution would yield the most desirable Figure of Merit?90
180
270 b) a) + +
270 90
180 c) +
270 90 +
180

13. SIR 2.50 1.00 0.50 -0.50 Phasing Power = |FH| Lack of closure
Which of the following is the best approximation to the phasing power for this reflection?
Imaginary axis
|FPH|
2.50
1.00
0.50
-0.50
|Fp |
Fbest
Real axis FH
|Fp |
Phasing Power = |FH|
Lack of closure
|FH ( h k l) | = 1.4
Lack of closure = |FPH|-|FP+FH| = 0.5
(at the aP of Fbest)

14. SIR 2.50 1.00 0.50 -0.50 Phasing Power = |FH| Lack of closure >1
Which of the following is the most desirable phasing power?
Imaginary axis
|FPH|
2.50
1.00
0.50
-0.50
|Fp |
Fbest
Real axis FH
|Fp |
Phasing Power = |FH|
Lack of closure
>1
What Phasing Power is sufficient to solve the structure?

15. SIR Which of the following is the RCullis for this reflection? -0.5
|FP ( hkl) | = 1.8
|FPHg (hkl) | = 2.8
Which of the following is the RCullis for this reflection?
Imaginary axis
|FPH|
-0.5
0.5
1.30
2.00
|Fp |
Fbest
Real axis FH
|Fp |
RCullis = Lack of closure isomorphous difference
From previous page, LoC=0.5
Isomorphous difference= |FPH| - |FP|
1.0 =

16. To Resolve the phase ambiguity
SIRAS
Isomorphous differences
Anomalous differences
Imaginary axis
FP ( h k l) = FPH(-h-k-l)* - FH (-h-k-l)*
FH(-h-k-l)*
FPH(-h-k-l)*
Fp (hkl)
Real axis
FH(hkl)
We will calculate SIRAS phases using the PCMBS Hg site.
FP –native measurement
FH (hkl) and FH(-h-k-l) calculated from heavy atom position.
FPH(hkl) and FPH(-h-k-l) –measured from derivative. Point to these on graph.
To Resolve the phase ambiguity

17. SIRAS Isomorphous differences Anomalous differences
Real axis
Imaginary axis
FH(hkl)
Fp (hkl)
FH(-h-k-l)*
FPH(-h-k-l)*
Which P(a) corresponds to SIR?
Which P(a) corresponds to SIRAS?

18. SIRAS Isomorphous differences Anomalous differences
Real axis
Imaginary axis
FH(hkl)
Fp (hkl)
FH(-h-k-l)*
FPH(-h-k-l)*
Which graph represents FOM of SIRAS phase?
Which represents FOM of SIR phase?
Which is better? Why? 90
180
270

19. Calculate phases and refine parameters (MLPhaRe)

20. A Tale of Two Ambiguities
We can solve both ambiguities in one experiment.

21. Center of inversion ambiguity
Remember, because the position of Hg was determined using a Patterson map there is an ambiguity in handedness.
The Patterson map has an additional center of symmetry not present in the real crystal. Therefore, both the site x,y,z and -x,-y,-z are equally consistent with Patterson peaks.
Handedness can be resolved by calculating both electron density maps and choosing the map which contains structural features of real proteins (L-amino acids, right handed a-helices).
If anomalous data is included, then one map will appear significantly better than the other.
Note: Inversion of the space group symmetry (P43212 →P41212) accompanies inversion of the coordinates (x,y,z→ -x,-y,-z)
Patterson map

22. Choice of origin ambiguity
I want to include the Gd data (derivative 2) in phase calculation.
I can determine the Gd site x,y,z coordinates using a difference Patterson map.
But, how can I guarantee the set of coordinates I obtain are referred to the same origin as Hg (derivative 1)?
Do I have to try all 48 possibilities?

23. Use a Cross difference Fourier to resolve the handedness ambiguity
With newly calculated protein phases, fP, a protein electron density map could be calculated.
The amplitudes would be |FP|, the phases would be fP. r(x)=1/V*S|FP|e-2pi(hx+ky+lz-fP)
Answer: If we replace the coefficients with |FPH2-FP|, the result is an electron density map corresponding to this structural feature.

24. r(x)=1/V*S|FPH2-FP|e-2pi(hx-fP)
What is the second heavy atom, Alex.
When the difference FPH2-FP is taken, the protein component is removed and we are left with only the contribution from the second heavy atom.
This cross difference Fourier will help us in two ways:
It will resolve the handedness ambiguity by producing a very high peak when phases are calculated in the correct hand, but only noise when phases are calculated in the incorrect hand.
It will allow us to find the position of the second heavy atom and combine this data set into our phasing. Thus improving our phases.

25. Phasing Procedures
Calculate phases for site x,y,z of Hg and run cross difference Fourier to find the Gd site. Note the height of the peak and Gd coordinates.
Negate x,y,z of Hg and invert the space group from P43212 to P Calculate a second set of phases and run a second cross difference Fourier to find the Gd site. Compare the height of the peak with step 1.
Chose the handedness which produces the highest peak for Gd. Use the corresponding hand of space group and Hg, and Gd coordinates to make a combined set of phases.

26. GdCl3 FPH = FP+FH for isomorphous differences
FH1(hkl)
MIRAS
FPH1(hkl)
FH1(-h-k-l)*
FPH1(-h-k-l)*
Imaginary axis
FH2(hkl)
Isomorphous
Deriv 2
FPH2(hkl)
Fp (hkl)
Anomalous
Deriv 1
Real axis
Isomorphous
Deriv 1
GdCl FPH = FP+FH for isomorphous differences

27. GdCl3 FP ( h k l) = FPH1 (-h-k-l)* - FH1 (-h-k-l)*
Imaginary axis
FH2(hkl)
Isomorphous
Deriv 2
FPH2(hkl)
FH2 (-h-k-l)*
FPH2(-h-k-l)*
Fp (hkl)
Anomalous
Deriv 1
Real axis
Isomorphous
Deriv 1
Anomalous
Deriv 2
Harker Construction for MIRAS phasing (Multiple Isomorphous Replacement with Anomalous Scattering)

28. For opposite hand Fp (hkl) FH1(hkl) FPH1(hkl) FH1(-h-k-l)*
Isomorphous
Deriv 1
Imaginary axis
FH2(hkl)
FPH2(hkl)
Anomalous
Deriv 2
FH2 (-h-k-l)*
FPH2(-h-k-l)*
Fp (hkl)
Real axis
Anomalous
Deriv 1
Isomorphous
Deriv 2
Note: in the presence of anomalous scattering, FH(HKL) ≠ FH(-H-K-L)* so if this opposite hand is the incorrect hand, the phase angle (aopposite) will be significantly different than (-aoriginal). The electron density would not just be inverted, but inverted and more noisy.

29. Density modification Histogram matching A) Solvent flattening.
Calculate an electron density map.
If r<threshold, -> solvent
If r>threshold -> protein
Build a mask
Set density value in solvent region to a constant (low).
Transform flattened map to structure factors
Combine modified phases with original phases.
Iterate
Histogram matching

30. MIR phased map + Solvent Flattening + Histogram Matching
MIRAS phased map

31. MIR phased map + Solvent Flattening + Histogram Matching
MIRAS phased map

32. Density modification B) Histogram matching.
Calculate an electron density map.
Calculate the electron density distribution. It’s a histogram. How many grid points on map have an electron density falling between 0.2 and 0.3 etc?
Compare this histogram with ideal protein electron density map.
Modify electron density to resemble an ideal distribution.
Number of times a particular electron density value is observed.
Electron density value

33. Each grid point has a value r(x,y,z)

34. represent r(x,y,z) in shades of gray
How many grid points have density 0?
How many grid points have density 1?
How many grid points have density x?

35. ntotal=130 pixels total r=2 n=17 r=3 n=5 r=4 n=1 r=7 n=2 r=8 n=4

36. Histogram of electron density distribution P(r).
Number of pixels with density (r)
Density (r)

37. Cumulative density distribution function N(r).
Density (r)
Number of pixels with density (r)
Number of pixels with density < r
Cumulative density

38. Histogram matching density distribution function N(r).
Number of pixels with density < r
Number of pixels with density < r
Pixels with r=4 in original map will be modified to r= 2.5 in the histogram matched map.
Cumulative density
observed
Cumulative density
ideal

39. HOMEWORK

40. Barriers to combining phase information from 2 derivatives
Initial Phasing with PCMBS
Calculate phases using coordinates you determined.
Refine heavy atom coordinates
Find Gd site using Cross Difference Fourier map.
Easier than Patterson methods.
Want to combine PCMBS and Gd to make MIRAS phases.
Determine handedness (P43212 or P41212 ?)
Repeat calculation above, but in P41212.
Compare map features with P43212 map to determine handedness.
Combine PCMBS and Gd sites (use correct hand of space group) for improved phases.
Density modification (solvent flattening & histogram matching)
Improves Phases
View electron density map