1. Today: compute the experimental electron density map of proteinase K Fourier synthesis  (xyz)=  |F hkl | cos2  (hx+ky+lz -  hkl ) hkl

2. 3 Crystals 5 Measured Quantities H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| ) 1 1 1 681.4 725.8 722.4 730.8 707.6 1 1 2 752.8 733.6 695.3 813.9 805.3 1 1 3 332.1 444.5 456.2 296.1 312.5 1 1 4 526.9 575.8 564.7 527.4 518.3 1 1 5 719.2 827.8 805.4 759.6 766.3 1 1 6 358.4 349.8 354.2 375.6 358.9 1 1 7 273.3 359.4 390.8 300.5 286.6 1 1 8 400.7 362.5 411.2 396.7 411.5 1 2 0 162.5 73.8 132.3 149.8 159.8 Native PCMBS (Hg) EuCl 3 (Eu)

3. Measureable differences 2) Anomalous Differences (i.e. differences between Friedel pairs): 3) Dispersive differences (differences due to changing the wavelength) : F P (hkl) =F PH (hkl) - f H (hkl) 1) Isomorphous differences (between native and derivative) F P (hkl) =F PH (hkl) n - f H (hkl) n F P (hkl) =F PH (-h-k-l) * - f H (-h-k-l) *

4. 3 Crystals 5 Measured Quantities H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 1 1 1 681.4 725.8 722.4 730.8 707.6 1 1 2 752.8 733.6 695.3 813.9 805.3 1 1 3 332.1 444.5 456.2 296.1 312.5 1 1 4 526.9 575.8 564.7 527.4 518.3 1 1 5 719.2 827.8 805.4 759.6 766.3 1 1 6 358.4 349.8 354.2 375.6 358.9 1 1 7 273.3 359.4 390.8 300.5 286.6 1 1 8 400.7 362.5 411.2 396.7 411.5 1 2 0 162.5 73.8 132.3 149.8 159.8 Native PCMBS (Hg) EuCl 3 (Eu)

5. Vector equations for this experiment Isomorphous and Anomalous Differences Isomorphous and Anomalous Differences F P (hkl) =F PHg (hkl) - f Hg (hkl) F P (hkl) =F PHg (-h-k-l) * - f Hg (-h-k-l) * F P (hkl) =F PEu (hkl) - f Eu (hkl) F P (hkl) =F PEu (-h-k-l) * - f Eu (-h-k-l) * PCMBS (Hg) For EuCl 3 (Eu)

6. Vector equations for this experiment Isomorphous Differences F P (hkl) =F PHg (hkl) - f Hg (hkl) We have collecting data on the native and derivative crystals. We know the coordinates of Hg. How many unknown quantities remain?

7. SIR Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Real axis Imaginary axis |F P | F P =F P·Hg (+) - f Hg (+)

8. SIR Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Imaginary axis f Hg =f Hg +f’+if[e 2  i*(h(x)+k(y)+l(z)) + e 2  i*(h(-x)+k(-y)+l(½+z)) + e 2  i*(h(½-y)+k(½+x)+l(¾+z) + e 2  i*(h(½+y)+k(½-x)+l(¼+z) + e 2  i*(h(½-x)+k(½+y)+l(¾-z) + e 2  i*(h(½+x)+k(½-y)+l(¼-z) + e 2  i*(h(y)+k(x)+l(-z) + e 2  i*(h(-y)+k(-x)+l(½-z) ] |F P | F P =F P·Hg (+) - f Hg (+) Real axis f’ and f” are anomalous scattering corrections specific for wavelength used. x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935) f Hg is a real number proportional to the number of e- in Hg

9. SIR Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Imaginary axis No! f Hg =f Hg [e 2  i*(h(x)+k(y)+l(z)) + e 2  i*(h(-x)+k(-y)+l(½+z)) + e 2  i*(h(½-y)+k(½+x)+l(¾+z) + e 2  i*(h(½+y)+k(½-x)+l(¼+z) + e 2  i*(h(½-x)+k(½+y)+l(¾-z) + e 2  i*(h(½+x)+k(½-y)+l(¼-z) + e 2  i*(h(y)+k(x)+l(-z) + e 2  i*(h(-y)+k(-x)+l(½-z) ]+f’+if” |F P | F P =F P·Hg (+) - f Hg (+) Real axis f Hg ≠ |F P |-|F PH (+)| f’ and f” are anomalous scattering corrections specific for wavelength used. x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935) f Hg is a real number proportional to the number of e- in Hg

10. SIR Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 |f Hg |=282  Hg =58° Imaginary axis |F P | F P =F P·Hg (+) - f Hg (+) Real axis 282 58°

11. SIR Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Imaginary axis |F P | F P =F P·Hg (+) - f Hg (+) Real axis Let’s look at the quality of the phasing statistics up to this point.

12. SIR Phasing H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 0 90 180 270 360 Which of the following graphs best represents the phase probability distribution, P(  )? a) b) c) 500 -500 -250 -500 250-250 250 Imaginary axis |F P | Real axis

13. SIR Phasing H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 500 -500 -250 -500 250-250 250 Imaginary axis |F P | Real axis 0 90 180 270 360 The phase probability distribution, P(  ) is sometimes shown as being wrapped around the phasing circle. 90 0 180 270

14. SIR Phasing H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Which of the following is the best choice of F p ? a) b) c) Radius of circle is approximately |F p | 500 -500 Imaginary axis |F P | Real axis 90 0 180 270 90 0 180 270 90 0 180 270 90 0 180 270

15. 500 SIR Phasing H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 500 -500 Imaginary axis |F P | Real axis best F P = |F p |e i  P(  )d   Sum of probability weighted vectors F p Usually shorter than F p 0 90 180 270

16. 500 SIR Phasing H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 500 -500 Imaginary axis |F best | Real axis best F P = |F p |e i  P(  )d   Sum of probability weighted vectors F p Usually shorter than F p 0 90 180 270

17. 90 0 180 270 SIR Sum of probability weighted vectors F p Best phase best F = |F p |e i  P(  )d  

18. 90 0 180 270 SIR a)1.00 b)2.00 c)0.50 d)-0.10 Which of the following is the best approximation to the Figure Of Merit (FOM) for this reflection? FOM=|F best |/|F P | Radius of circle is approximately |F p |

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

20. F best |F PH | Imaginary axis SIR Real axis |F p | a)2.50 b)1.00 c)0.50 d)-0.50 Which of the following is the best approximation to the phasing power for this reflection? Lack of closure = |F PH |-|F P +F H | = 0.5 (at the  P of F best ) |F p | fHfH |f H ( h k l) | = 1.4 Phasing Power = |f H | Lack of closure

21. F best |F PH | Imaginary axis SIR Real axis |F p | a)2.50 b)1.00 c)0.50 d)-0.50 Which of the following is the most desirable phasing power? |F p | fHfH What Phasing Power is sufficient to solve the structure? >1 Phasing Power = |f H | Lack of closure

22. fHfH SIR a)-0.5 b)0.5 c)1.30 d)2.00 Which of the following is the R Cullis for this reflection? R Cullis = Lack of closure isomorphous difference From previous page, LoC=0.5 Isomorphous difference= |F PH | - |F P | 1.0 = 2.8-1.8 |F P ( hkl) | = 1.8 |F PHg (hkl) | = 2.8 F best Imaginary axis Real axis |F p | |F PH |

23. SIRAS Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Imaginary axis |F P | F P =F P·Hg (-)* - f Hg (-)* Real axis |f Hg |=282  Hg *=48° 282 48°

24. SIRAS Phasing 500 -500 -250 -500 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 Imaginary axis |F P | F P =F P·Hg (-)* - f Hg (-)* Real axis

25. Isomorphous differences Anomalous differences SIRAS 0 90 180 270 360 Which P(  ) corresponds to SIR? Which P(  ) corresponds to SIRAS?

26.  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  -helices).  If anomalous data is included, then one map will appear significantly better than the other. Note: Inversion of the space group symmetry (P4 3 2 1 2 →P4 1 2 1 2) accompanies inversion of the coordinates (x,y,z→ -x,-y,-z) Center of inversion ambiguity Patterson map

27. 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?

28. Use a Cross difference Fourier to resolve the handedness ambiguity With newly calculated protein phases,  P, a protein electron density map could be calculated. The amplitudes would be |F P |, the phases would be  P.  xyz  =1/V*  |F P |e -2  i(hx+ky+lz-  P ) Answer: If we replace the coefficients with |F PH2 -F P |, the result is an electron density map corresponding to this structural feature.

29.  x  =1/V*  |F PH2 -F P |e -2  i(hx-  P ) What is the second heavy atom, Alex. When the difference F PH2 -F P is taken, the protein contribution is cancelled and we are left with only the contribution from the second heavy atom. This cross difference Fourier will help us in two ways: 1)It will resolve the handedness ambiguity - high peak in difference map calculated with  P in correct hand -only noise in difference map calculated with  P in incorrect hand. 2)It will improve our electron density map of the protein -identify the position of the 2nd heavy atom -include 2 new vector equations for Eu (more accurate  P )

30. Phasing Procedures 1)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. 2)Negate x,y,z of Hg and invert the space group from P4 3 2 1 2 to P4 1 2 1 2. 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. 3)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.

31. |F P | = 486 ± 2 MIRAS Phasing 500 -500 -250 -500 F P =F P·Hg (+) - f Hg (+) 250-250 250 H K L |F P | |F PH (+)| |F PH (-)| |F PH (+)| |F PH (-)| 9 2 1 486 586 611 536 499 H K L fH+f’ f”(-)  H fH+f’ f”  H 9 2 1 281 27 53° 100 24 -114° F P =F P·Hg (-)* - f Hg (-)* Real axis Imaginary axis

32. SIR SIRAS MIRAS

34. Density modification A) Solvent flattening. Calculate an electron density map. If  solvent If  >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

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

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

37. 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

38. HOMEWORK

39. Barriers to combining phase information from 2 derivatives 1)Initial Phasing with PCMBS 1)Calculate phases using coordinates you determined. 2)Refine heavy atom coordinates 2)Find Eu site using Cross Difference Fourier map. 1)Easier than Patterson methods. 2)Want to combine PCMBS and Eu to make MIRAS phases. 3)Determine handedness (P4 3 2 1 2 or P4 1 2 1 2 ?) 1)Repeat calculation above, but in P4 1 2 1 2. 2)Compare map features with P4 3 2 1 2 map to determine handedness. 4)Combine PCMBS and Eu sites (use correct hand of space group) for improved phases. 5)Density modification (solvent flattening & histogram matching) 1)Improves Phases 6)View electron density map