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Phasing

Phasing. Today’s goal is to calculate phases ( a p ) for proteinase K using PCMBS and EuCl 3 (MIRAS method) . What experimental data do we need? 1) from native crystal we measured |F p | for native crystal 2) from PCMBS derivative we measured |F P•Hg (h k l) |

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Phasing

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  1. Phasing Today’s goal is to calculate phases (ap) for proteinase K using PCMBS and EuCl3 (MIRAS method) . What experimental data do we need? 1) from native crystal we measured |Fp| for native crystal 2) from PCMBS derivative we measured |FP•Hg(h k l)| 3) also from PCMBS derivative we measured |FP•Hg(-h-k-l)| 4) fHg for Hg atom of PCMBS 5) from EuCl3derivative we measured |FP•Eu(h k l)| 6) also from EuCl3derivative we measured |FP•Eu(-h-k-l)| 7) FEu for Eu atom of EuCl3 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)=FP•Hg ( h k l) - fHg ( h k l)isomorphous differences FP ( h k l)=FP•Hg(-h-k-l)*-fHg (-h-k-l)*from Friedel mates (anomalous) EuCl3 FP ( h k l)=FP•Eu( h k l)-fEu ( h k l)isomorphous differences FP ( h k l)=FP•Eu (-h-k-l)*- fEu (-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. FH 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 | Real axis Harker construction 1) |FP ( h k l)| –native measurement 2) FH (h k l) calculated from heavy atom position.

  5. |FPH| |FPH| 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 | Real axis fH Harker construction 1) |FP ( h k l)| –native measurement 2) fH1 (h k l) calculated from heavy atom position. 3)|FPH1(hkl)|–measured from derivative. Let’s look at the quality of the phasing statistics up to this point.

  6. |FPH| |FPH| Which of the following graphs best represents the phase probability distribution, P(a)? SIR Imaginary axis |Fp | a) b) c) 0 90 180 270 360 Real axis fH 0 90 180 270 360 0 90 180 270 360

  7. |FPH| 90 |FPH| 0 180 270 The phase probability distribution, P(a) is sometimes shown as being wrapped around the phasing circle. SIR Imaginary axis |Fp | 0 90 180 270 360 Real axis fH

  8. 90 90 90 0 0 0 180 180 180 |FPH| 270 270 270 90 |FPH| 0 180 270 Which of the following is the best choice of Fp? SIR Imaginary axis a) b) c) |Fp | Real axis fH Radius of circle is approximately |Fp|

  9. best F = |Fp|eia•P(a)da a |FPH| 90 |FPH| 0 180 270 SIR Imaginary axis |Fp | Real axis Sum of probability weighted vectors Fp Usually shorter than Fp fH

  10. best F = |Fp|eia•P(a)da a 90 0 180 270 SIR Sum of probability weighted vectors Fp Best phase

  11. 90 0 180 270 SIR Which of the following is the best approximation to the Figure Of Merit (FOM) for this reflection? • 1.00 • 2.00 • 0.50 • -0.10 FOM=|Fbest|/|FP| Radius of circle is approximately |Fp|

  12. 90 0 180 270 Which phase probability distribution would yield the most desirable Figure of Merit? b) a) 0 + + 270 90 180 0 c) + 270 90 + 180

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

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

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

  16. fH(-h-k-l)* FPH(-h-k-l)* SIRAS Isomorphous differences Anomalous differences To Resolve the phase ambiguity Imaginary axis FP ( h k l)=FPH(-h-k-l)*-fH (-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.

  17. Imaginary axis 0 90 180 270 360 Fp (hkl) fH(-h-k-l)* Real axis fH(hkl) FPH(-h-k-l)* SIRAS Isomorphous differences Anomalous differences Which P(a) corresponds to SIR? Which P(a) corresponds to SIRAS?

  18. Imaginary axis 90 90 0 0 180 180 Fp (hkl) 270 270 fH(-h-k-l)* Real axis fH(hkl) FPH(-h-k-l)* SIRAS Isomorphous differences Anomalous differences Which graph represents FOM of SIRAS phase? Which represents FOM of SIR phase? Which is better? Why?

  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 Eu data (derivative 2) in phase calculation. • I can determine the Eu 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 Eu site. Note the height of the peak and Eu coordinates. • Negate x,y,z of Hg and invert the space group from P43212 to P41212. Calculate a second set of phases and run a second cross difference Fourier to find the Eu site. Compare the height of the peak with step 1. • Chose the handedness which produces the highest peak for Eu. Use the corresponding hand of space group and Hg, and Eu coordinates to make a combined set of phases.

  26. fH1(hkl) Isomorphous Deriv 2 FPH1(hkl) MIRAS fH1(-h-k-l)* FPH1(-h-k-l)* Imaginary axis fH2(hkl) FPH2(hkl) Fp (hkl) Anomalous Deriv 1 Real axis Isomorphous Deriv 1 EuCl3FPH = FP+fH for isomorphous differences

  27. fH1(hkl) EuCl3FP ( h k l)=FPH1 (-h-k-l)*- fH1 (-h-k-l)* FPH1(hkl) Harker Construction for MIRAS phasing (Multiple Isomorphous Replacement with Anomalous Scattering) fH1(-h-k-l)* FPH1(-h-k-l)* Imaginary axis Isomorphous Deriv 2 fH2(hkl) FPH2(hkl) fH2 (-h-k-l)* FPH2(-h-k-l)* Fp (hkl) Anomalous Deriv 1 Real axis Isomorphous Deriv 1 Anomalous Deriv 2

  28. Density modification • 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

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

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

  31. 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

  32. HOMEWORK

  33. Barriers to combining phase information from 2 derivatives • Initial Phasing with PCMBS • Calculate phases using coordinates you determined. • Refine heavy atom coordinates • Find Eu site using Cross Difference Fourier map. • Easier than Patterson methods. • Want to combine PCMBS and Eu 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 Eu sites (use correct hand of space group) for improved phases. • Density modification (solvent flattening & histogram matching) • Improves Phases • View electron density map

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