1. Making sense of a MAD experiment. Chem M230B Friday, February 3, 2006 12:00-12:50 PM Michael R. Sawaya http://www.doe-mbi.ucla.edu/M230B/

2. What is the anomalous scattering phenomenon?What is the anomalous scattering phenomenon? How is the anomalous scattering signal manifested?How is the anomalous scattering signal manifested? How do we account for anomalous scattering effects in its form factor f H ?How do we account for anomalous scattering effects in its form factor f H ? How does anomalous scattering break the phase ambiguity in aHow does anomalous scattering break the phase ambiguity in a –SIRAS experiment? –MAD experiment? Topics Covered

3. Q: What is anomalous scattering? A: Scattering from an atom under conditions when the incident radiation has sufficient energy to promote an electronic transition.

4. An electronic transition is an e - jump from one orbital to another – from quantum chemistry. Orbitals are paths for electrons around the nucleus. Orbitals are organized in shells with principle quantum number, n= 1= Kshell 2= L shell 3= M shell 4= N shell 5= O shell 6= P shell etc. And different shapes possible within each shell. s,p,d,f

5. Orbitals have quantized energy levels. Outer shells are higher energy.

6. Normally, electrons occupy the lowest energy orbitals –ground state. Selenium atom Ground state

7. But, incident radiation can excite an e - to an unoccupied outer orbital if the energy of the radiation (h ) matches the  E between orbitals. h  E Excites a transition from the “K” shell For Se,  E=12.65keV = = 0.9795 Å Selenium atom Electronic transition

8. Under these conditions when an electron can transit between orbitals, an atom will scatter photons anomalously. Selenium atom Excited state Incident X-ray Anomalously Scattered X-ray -90° phase shift diminished amplitude

9. If the incident photon has energy different from the  E between orbitals, then there is little anomalous scattering (usual case). Scattered x-rays are not phase shifted. Selenium atom Ground state h  E No transition possible, Insufficient energy usual case

10. For these elements, anomalous scattering is significant only at a synchrotron.  E is a function of the periodic table.  E is near 8keV for most heavy and some light elements, so anomalous signal can be measured on a home X-ray source with CuKa radiation ( 8kev =1.54Å). At a synchrotron, the energy of the incident radiation can be tuned to match  E (accurately). Importantly,  Es for C,N,O are out of the X-ray range. Anomalous scattering from proteins and nucleic acids is negligible. K shell transitions L shell transitions

11. Choose an element with  E that matches an achievable wavelength. Green shading represents typical synchrotron radiation range. Orange shading indicates CuKa radiation, typically used for home X-ray sources.

12. How is the anomalous scattering signal manifested in a crystallographic diffraction experiment? Q: How is the anomalous scattering signal manifested in a crystallographic diffraction experiment? A: Anomalous scattering causes small but measurable differences in intensity between the reflections hkl and –h-k-l not normally present.

13. Under normal conditions, atomic electron clouds are centrosymmetric. For each point x,y,z, there is an equivalent point at –x,-y,-z. Centrosymmetry relates points equidistant from the origin but in opposite directions.

14. Centrosymmetry in the scattering atoms is reflected in the centrosymmetry in the pattern of scattered x-ray intensities. The positions of the reflections hkl and –h-k-l on the reciprocal lattice are related by a center of symmetry through the reciprocal lattice origin (0,0,0). Pairs of reflections hkl and –h-k-l are called Friedel pairs. Friedel’s law is a consequence of an atom’s centrosymmetry. I (hkl) =I (-h-k-l) and   hkl) =-   -h-k-l) (15,0,-6) (-15,0,6) · (0,0,0)

15. But, under conditions of anomalous scattering, electrons are perturbed from their centrosymmetric distributions. Electrons are jumping between orbits. By the same logic as before, the breakdown of centrosymmetry in the scattering atoms should be reflected in a loss of centrosymmetry in the pattern of scattered x-ray intensities. e-e-

16. A single heavy atom per protein can produce a small but measurable difference between FPH (hkl) and FPH (-h-k-l). Differences between I hkl and I -h-k-l are small typically between 1-3%. Keep I(hkl) and I(-h-k-l) as separate measurements. Don’t average them together. Example taken from a single Hg site derivative of proteinase K (28kDa protein) h k l Intensity sigma 5 3 19 601.8 +/- 15.4 -5 -3 -19 654.8 +/- 15.7 Anomalous difference = 53 Anomalous signal is about 3 times greater than sigma

17. In the complex plane, F P (hkl) and F P (-h-k-l) are reflected across the real axis. real imaginary  hkl |FP h,k,l | F P (h,k,l)  -h-k-l |FP -h,-k,-l | F P (-h,-k,-l) True for any crystal in the absence of anomalous scattering. Normally, I hkl and I -h-k-l are averaged together to improve redundancy F P (hkl) =F P (-h-k-l) and   hkl) =-   -h-k-l)

18. But not F PH (hkl) and F PH (-h-k-l) real imaginary  -h-k-l |FP -h,-k,-l |  hkl |FP h,k,l | F PH( h,k,l) F PH (-h,-k,-l) The heavy atom structure factor is not reflected across the real axis. Hence, the sum of FH and FP=FPH is not reflected across the real axis. Hence, an anomalous difference. |F PH (hkl) |≠|F PH (-h-k-l) |

19. Hey! Look at that! We have two phase triangles now; we only had one before. real imaginary  -h-k-l |FP -h,-k,-l |  hkl |FP h,k,l | FPH h,k,l FPH -h,-k,-l we have to be able to calculate the effect of anomalous scattering on the values of F H hkl and F H -h-k-l precisely given the heavy atom position. So far we just have a very faint idea of what the effect of anomalous scattering is. The form factor f, is going to be different. In isomorphous replacement method, we get a single phase triangle, which leaves an either/or phase ambiguity. Anomalous scattering provides the opportunity of constructing a second triangle that will break the phase ambiguity. We just have to be sure to measure both. FPH (hkl) | |FPH (-h-k-l) | and... | FPH (hkl) | and |FPH (-h-k-l) | and... F PH (hkl) =F P (hkl) +F H (hkl) F PH (-h-k-l) =F P (-h-k-l) +F H (-h-kl-)

20. How do we correct for anomalous scattering effects in our calculation of F H ? Q: How do we correct for anomalous scattering effects in our calculation of F H ? A: The correction to the atomic scattering factor is derived from classical physics and is based on an analogy of the atom to a forced oscillator under resonance conditions.

21. Examples of forced oscillation: A tuning fork vibrating when exposed to periodic force of a sound wave. The housing of a motor vibrating due to periodic impulses from an irregularity in the shaft. A child on a swing

22. The Tacoma Narrows bridge is an example of an oscillator swaying under the influence of gusts of wind. Tacoma Narrows bridge, 1940

23. But, when the external force is matched the natural frequency of the oscillator, the bridge collapsed. Tacoma Narrows bridge, 1940

24. An atom can also be viewed as a dipole oscillator where the electron oscillates around the nucleus. nucleus e-e- + The oscillator is characterized by mass=m position =x,y natural circular frequency= B Characteristic of the atom Bohr frequency From Bohr’s representation of the atom

25. An incident photon’s electric field can exert a force on the e -, affecting its oscillation. e-e- e-e- + E=h + What happens when the external force matches the natural frequency of the oscillator (a.k.a resonance condition)? +

26. In the case of an atom, resonance (  B ) leads to an electronic transition (analogous to the condition h  E from quantum chemistry discussed earlier). The amplitude of the oscillator (electron) is given by classical physics: m=mass of oscillator e=charge of the oscillator c-=speed of light Eo= max value of electric vector of incident photon =frequency of external force (photon) B=natural resonance frequency of oscillator (electron) + nucleus + + + + + e-e- + + Incident photon with  B

27. Knowing the amplitude of the e - leads to a definition of the scattering factor, f. Scattered photon The amplitude of the scattered radiation is defined by the oscillating electron. The oscillating electron is the source of the scattered electromagnetic wave which will have the same frequency and amplitude as the e-. Keep in mind, the frequency and amplitude of the e- is itself strongly affected by the frequency and amplitude of the incident photon as indicated on the previous slide. Amplitude of scattered radiation from the forced e- Amplitude of scattered radiation by a free e- f=f= + nucleus + e-e- + + Incident photon with  B

28. We find that the scattering factor is a complex number, with value dependent on. =frequency incident photon B =Bohr frequency of oscillator (e - ) (corresponding to electronic transition) f = f o +  f’ + i  f” fofo Normal scattering factor REAL correction factor REAL correction factor IMAGINARY

29. Physical interpretation of the real and imaginary correction factors of f. imaginary component,  f” A small component of the scattered radiation is 90°out of phase with the normally scattered radiation given by f o. Bizarre! Any analogy to real life? f = f o +  f’ + i  f” real component,  f” A small component of the scattered radiation is 180° out of phase with the normally scattered radiation given by f o. Always diminishes f o. Absorption of x-rays

30. 90 o phase shift analogy to a child on a swing Forced Oscillator Analogy Maximum positive displacement Zero force Maximum negative displacement Zero force Zero displacement Maximum +/- force Swing force is 90 o out of phase with the displacement.

31. 90 o phase shift analogy to a child on a swing Forced Oscillator Analogy Maximum positive displacement Zero force Maximum negative displacement Zero force Zero displacement Maximum +/- force time- > force : displacement incident photon : re-emitted photon.

32. Displacement of block, x, is 90° behind force applied ForceDisplace- ment 10 (relaxed spring)-1(max negative displacement) 2-1 (max compress)0 30 (relaxed spring)1(max positive displacement) 41 (max expand)0 10 (relaxed spring)-1 (max negative displacement) -1 0 1 force : displacement incident photon : re-emitted photon. 1 2 3 4

33.  Real axis Imaginary axis Construction of F H under conditions of anomalous scattering F H = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) Assume we have located a heavy atom, H, by Patterson methods. Gives  scattering factor for H real Positive number real 180° out of phase imaginary 90° out of phase fofo f’ f” FH ( H K L)

34. Real axis Imaginary axis fofo f’ f”  FH (-H-K-L) FH (-h-k-l) is constructed in a similar way as FH (hkl) except  is negative. Real axis Imaginary axis f’ f” FH ( H K L) FH ( H K L) = [f o +  f’( ) + i  f”( )] e 2  i(+hx H +ky H +lz H ) FH (-H-K-L) = [f o +  f’( ) + i  f”( )] e 2  i(-hx H -ky H -lz H ) FH (-H-K-L) fofo 

35. Again, we see how  Friedel’s Law is broken Real axis Imaginary axis FH (-H-K-L)  fofo f’ f” fofo  FH (H K L)  H (-H-K-L)  H (+H+K+L) f’  H (-h-k-l) ≠ -  H (-h-k-l)

36. How can measurements of |FPH (hkl) |, and |FPH (-h-k-l) | be combined to solve the phase of FP in a SIRAS experiment? Q: How can measurements of |FPH (hkl) |, and |FPH (-h-k-l) | be combined to solve the phase of FP in a SIRAS experiment? A: Analogous to MIR, using: measured amplitude, |FP| | measured amplitudes | FPH (hkl) | and |FPH (-h-k-l) | FPH (hkl) = FP (hkl) + FH (hkl) FPH (-h-k-l) =FP (-h-k-l) + FH (-h-kl-) calculated amplitudes & phases of FH (HKL), FH (-H-K-L), two phasing triangles FPH (hkl) = FP (hkl) + FH (hkl) and FPH (-h-k-l) =FP (-h-k-l) + FH (-h-kl-) and Friedel’s law.

37. Imaginary axis Begin by graphing the measured amplitude of FP for (HKL) and (-H-K-L). Circles have equal radius by Friedel’s law Real axis FH (HKL) Imaginary axis Real axis |FP (-H-K-L) | |FP (HKL) | FH ( H K L) = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) FH (-H-K-L) = [f o +  f’( ) + i  f”( )] e 2  i(-hx H -ky H -lz H ) Structure factor amplitudes and phases calculated using equations derived earlier. Graph FH (hkl) and FH (-h-k-l) using coordinates of H. Place vector tip at origin. FH (-H-K-L)

38. Imaginary axis Graph measured amplitudes of FPH for (H K L) and (-H-K-L). Real axis FH (HKL) Imaginary axis Real axis |FP (-H-K-L) | |FP (HKL) | |FPH (hkl) | |FPH (-h-k-l) | FPH (hkl) = FP (hkl) +FH (hkl) FPH (-h-k-l) =FP (-h-k-l) +FH (-h-kl-) There are two possible choices for FP (HKL) and two possible choices for FP (-H-K-L)

39. FH (-H-K-L) Imaginary axis To combine phase information from the pair of reflections, we take the complex conjugate of the –h-k-l reflection. Real axis FH (HKL) Imaginary axis Real axis |FP (-H-K-L) | |FP (HKL) | |FPH (hkl) | |FPH (-h-k-l) | Complex conjugation means amplitudes stay the same, but phase angles are negated. FPH (hkl) = FP (hkl) +FH (hkl) FPH (-h-k-l) =FP (-h-k-l) +FH (-h-kl-) FPH (-h-k-l) * =FP (-h-k-l) * +FH (-h-kl-) * Reflection across real axis.

40. Imaginary axis FP (-h-k-l) * FP (hkl) Complex conjugation allows us to equate FP (-h-k-l) * and FP (hkl) by Friedel’s law and thus merge the two Harker constructions into one. Real axis FH (HKL) |FP (HKL) | FH (-H-K-L) * Imaginary axis Real axis |FP (-H-K-L) | FPH (hkl) = FP (hkl) +FH (hkl) FPH (-h-k-l) * = FP (-h-k-l) * + FH (-h-k-l) * FPH (-h-k-l) * = FP (hkl) + FH (-h-kl-) * FP (-h-k-l) * FP (hkl) Friedel’s law, FP (-h-k-l) * = FP (hkl).

41. Imaginary axis Phase ambiguity is resolved. Real axis FH (HKL) FH (-H-K-L) FP (HKL) Three phasing circles intersect at one point. Now repeat process for 9999 other reflections FPH (hkl) = FP (hkl) + FH (hkl) FPH (-h-k-l) * = FP (hkl) + FH (-h-k-l) *

42. How can measurements of |FPH ( 1) |, |FPH ( 2) |, and |FPH ( 3) | be combined to solve the phase of FP in a MAD experiment? Q: How can measurements of |FPH ( 1) |, |FPH ( 2) |, and |FPH ( 3) | be combined to solve the phase of FP in a MAD experiment? |FPH ( 1) |, |FPH ( 2) |, |FPH ( 2) | A: Again, analogous to MIR, using: measured amplitudes |FPH ( 1) |, |FPH ( 2) |, and |FPH ( 2) | FH ( 1), FH ( 2), FH ( 3) calculated amplitudes & phases of FH ( 1), FH ( 2), & FH ( 3) FPH ( l) = FP ( l) + FH ( l) three phasing triangles FPH ( l) = FP ( l) + FH ( l) FPH ( 2) = FP ( 2) + FH ( 2) FPH ( 3) = FP ( 3) + FH ( 3) FPH ( 3) = FP ( 3) + FH ( 3) and Friedel’s law but no measured amplitude, |FP|

43. Correction factors are largest near = B. The IMAGINARY COMPONENT becomes large and positive near = B. =frequency of external force (incident photon) B =natural frequency of oscillator (e - ) The REAL COMPONENT becomes negative near v= B. f = f o +  f’ + i  f” when > B Else, 0  f’  BB BB  f’ After dampening correction

44. As the energy of the incident radiation approaches the  E of an electronic transition (absorption edge),  f’, varies strongly, becoming most negative at  E.  f’ is the component of scattered radiation 180° out of phase with the normally scattered component fo  f’ fofo fofo fofo fofo fofo EE Se

45. Similarly,  f”, varies strongly near the absorption edge, becoming most positive at energies >  E.  f” is the component of scattered radiation 90° out of phase with the normally scattered component fo fofo  f” fofo fofo fofo fofo EE Se when > B Else, 0

46. Four wavelengths are commonly chosen to give the largest differences in F H. f” f’ fofo F H ( peak ) F H ( inflection ) FH ( FH ( ) = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) f” f’ fofo fofo fofo F H ( high remote ) F H ( low remote )

47. The basis of a MAD experiment is that the amplitude and phase shift of the scattered radiation depend strongly on the energy (or wavelength, E=hc/ ) of the incident radiation. Imaginary axis FH ( l) FH ( 2) FH ( 3) FH ( 1 FH ( 1) = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) FH ( 2 FH ( 2) = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) FH (  3 FH (  3) = [f o +  f’( ) + i  f”( )] e 2  i(hx H +ky H +lz H ) Hence, the amplitude and phase of FH varies with wavelength. Same heavy atom coordinate, but 3 different structure factors depending on the wavelength.

48. FPH( ) amplitudes are graphed as circles centered at the beginning of the FH( ) vectors (as in MAD & SIRAS). Imaginary axis FPH ( l) = FP ( l) + FH ( l) FPH ( 2) = FP ( 2) + FH ( 2) FPH ( 3) = FP ( 3) + FH ( 3) No measurement available for FP, but it can be assumed that its value does not change with wavelength because it contains no anomalous scatterers. FP ( l) = FP ( 2) = FP ( 3) Hence, FP ( l) = FP ( 2) = FP ( 3) and all three circles intersect at FP.

49. Imaginary axis A three wavelength MAD experiment solves the phase ambiguity. Real axis FPH ( l) = FP + FH ( l) FPH ( 2) = FP + FH ( 2) FPH ( 3) = FP + FH ( 3) Anomalous differences between reflections hkl and –h-k-l could also be measured and used to contribute additional phase circles. In principle, one could acquire 2 phase triangles for each wavelength used for data collection. Let’s examine more closely how FH changes with wavelength.

50. Good choice of Poor choice of Imaginary axis Real axis Imaginary axis Real axis Point of intersection poorly defined. Point of intersection clearly defined.

51. Technological Advances Leading to the Routine use of MAD phasing Appearance of synchrotron stations capable of protein crystallography. Cryo protection to preserve crystal diffraction quality during long 3-wavelength experiment. Production of selenomethionyl derivatives in ordinary E.coli strains. Production of selenomethionyl derivatives in ordinary E.coli strains. Fast, accurate data collection software.

52. Anomalous electrons Need to mention that length of correction factors, f’ and f” are 10 at most, compared to mercury at 80e. Need perfect isomorphism to see signal. Anomalous signal is smaller for lighter elements compared to heavier elements.

53. Accuracy of measurement is extremely important to a successful AS experiment. The anomalous signal from a derivative is sufficient to phase if it produces a 2-5% difference between Friedel related pairs. Useful anomalous signals range from a minimum of f”=4e- (for selenium (requires 1SeMet/100 residues bare minimum to yield a sufficient signal for phasing) to a maximum of about f”=14e- for Uranium). In comparison with isomorphous differences, anomalous differences are much smaller. For example, the maximal isomorphous difference for a Hg atom is 80 e-, while its anomalous difference can be no bigger than 10e-. But the measurement of the anomalous difference does not suffer from nonisomorphism. Also, the anomalous scattering factors do not diminish at high resolution as do the normal scattering factors. Data collection must be highly redundant to improve the accuracy of the measurements. Anomalous differences are small differences taken between large measurements.

54. How to prepare a selenomethionine derivative Use minimal media for bacterial growth and expression. Use of a methionine auxotroph to express protein. Supplement with selenomethionine. OR use of an ordinary bacterial expression strain, but supress methionine biosynthesis by the addition of T,K,F,L,I,V. See Van Duyne et al., JMB (1993), 229, 105-124. $68 for 1 gram selenomethionine Acros organics.

55. Source of ideas & information Concept of anomalous scattering R.W. James, The Optical Principles of Diffraction of X-rays. 1948. Ethan Merrit’s Anomalous scattering website http://www.bmsc.washington.edu/scatter/AS_index.html And references therein Sherwood, Crystals, X-rays and Proteins. 1976. Out of print Woolfson, X-ray Crystallography. 1970 Halliday & Resnick Physics text book Todd Yeates Crystallographic concepts Stout & Jenesen X-ray structure determination Glusker, Lewis & Rossi, Crystal Structure Analysis for Chemists & Biologists Drenth, Principles of Protein X-Ray crystallography. Hendrickson, Science, 1991, vol 254, p51. Ramakrishnan & Biou, Methods in Enzymology vol 276, p538. Giacavazzo, Fundamentals of Crystallography. others

56. Brief review of MIR method. Perspective Reinforce important concepts for understanding MAD Each point illustrated with a figure

57. A typical electron density map is plotted on a 3D grid containing of 1000s of grid points.

58. Y X Z Each grid point has a value  (x,y,z)

59. Each value  (x,y,z) is the summation of 1000s of structure factors, F hkl Y X Z  (x,y,z)= 1 /v  F hkl e -2  i(hx+ky+lz) h k l

60. Each structure factor F hkl specifies a cosine wave with a certain amplitude and phase shift  (x,y,z  1 /v  |F 0,0,1 |e -2  i(0x+0y+1z-  001 ) + |F 0,0,2 |e -2  i(0x+0y+2z-  002 ) + |F 0,0,3 |e -2  i(0x+0y+3z-  003 ) + |F 0,0,4 |e -2  i(0x+0y+4z-  004 ) + |F 0,0,5 |e -2  i(0x+0y+5z-  005 ) +… |F 50,50,50 |e -2  i(50x+50y+50z-  50 50 50 ) } x x x x x     

61. The value of the cosine waves at the point x,y,z sum up to the value  (x,y,z) Y X Z x x x x x     

62. The task of the crystallographer is to amplitudes and phases of 1000s of F hkl to obtain the electron density map  (x,y,z) Y X Z x x x x x       (x,y,z  1 /v  |F 0,0,1 |e -2  i(0x+0y+1z-  001 ) + |F 0,0,2 |e -2  i(0x+0y+2z-  002 ) + |F 0,0,3 |e -2  i(0x+0y+3z-  003 ) + |F 0,0,4 |e -2  i(0x+0y+4z-  004 ) + |F 0,0,5 |e -2  i(0x+0y+5z-  005 ) +…}  (x,y,z)= 1 /v  |F hkl |e -2  i(hx+ky+lz-  hkl ) h k l

63. Remarkably, the F hkl amplitudes and phases we need are encoded in the radiation scattered by the atoms in the crystal. |F h,k,l | is the square root of the intensity of the scattered radiation which can be measured in a standard diffraction expt.  hkl is the phase shift of the scattered radiation It cannot be measured directly, leaving us with the Phase Problem.

64. In solving the phase problem by MIR, it is important to know that each F hkl, is the sum of individual atomic structure factors contributed by each atom in the crystal. F hkl =  f j e 2  i(hx j +ky j +lz j ) Here we show a crystal with a single amino acid containing 12 atoms; In a protein crystal there would be thousands of atoms; = f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c +ky c +lz c ) + f O e 2  i(hx o +ky o +lz o ) + f OT e 2  i(hx ot +ky ot +lz ot ) + f N e 2  i(hx n +ky n +lz n ) f j is called the scattering factor and is proportional to the number of electrons in the atom j. j

65. Each atomic structure factor can be represented as a vector in the complex plane with length f j and phase angle e 2  i(hx j +ky j +lz j ).. F hkl =  f j e 2  i(hx j +ky j +lz j ) j = f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c +ky c +lz c ) + f O e 2  i(hx o +ky o +lz o ) + f OT e 2  i(hx ot +ky ot +lz ot ) + f N e 2  i(hx n +ky n +lz n ) real imaginary Argand diagram

66. The resultant of the atomic vectors give the amplitude and phase of F hkl for the protein. F hkl =  f j e 2  i(hx j +ky j +lz j ) j = f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c  +ky c  +lz c  ) + f C  e 2  i(hx c +ky c +lz c ) + f O e 2  i(hx o +ky o +lz o ) + f OT e 2  i(hx ot +ky ot +lz ot ) + f N e 2  i(hx n +ky n +lz n ) real imaginary Argand diagram F hkl  hkl |F h,k,l |

67. MIR method By the same reasoning, if a heavy atom is added to a protein crystal then the structure factors of the heavy atom derivative F PH must equal the sum of the component vectors F P +F H. F PH =F P +F H, forms the basis for the MIR method. real imaginary FHFHFHFH FPFPFPFP F PH

68. MIR method: F PH =F P +F H Only the amplitude of F P can be measured, not its phase. The amplitude is represented by a circle in the complex plane with radius= |FP| real imaginary FHFHFHFH |F P | |F PH | Both the phase and amplitude of F H can be plotted assuming the heavy atom position (x H,y H,z H ) can be determined by difference Patterson methods. F H = f H e 2  i(hx H +ky H +lz H ) The amplitude of F PH can be measured and is represented by a circle in the complex plane with radius= |F PH |. The circle is centered at the start of the FH vector. So in effect F P =F PH -F H

69. There are two possible choices of phase angle for F P that satisfy: F PH =F P +F H The phasing ambiguity can be resolved by soaking in a different heavy atom and collecting a new data set. F PH 2 =F P +F H 2 real imaginary

70. The phase ambiguity is resolved by combining F PH 1 =F P +F H 1 and F PH 2 =F P +F H 2 All three circle intersect at only one point. real imaginary

71. In practice, the phase ambiguity can be resolved more easily by taking advantage of anomalous scattering from PH1. Screening for a second derivative, PH2,costs time, money, and nerves for expressing protein growing crystals Soaking heavy atom Collecting and analyzing data. Anomalous scattering from PH1 can be used in combination with native data set (SIRAS) or with other data sets from the same crystal collected at different wavelengths (MAD). MAD is like “In situ MIR in which physics rather than chemistry is used to effect the change in scattering strength at the site”. - Hendrickson, (1991). Two phasing circles can be drawn with each new wavelength used for data collection F PH( ). How many crystallization plates does it take to find a decent heavy atom derivative?