Making sense of a MAD experiment.

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

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

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

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

10. DE is a function of the periodic table. DE 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 l=1.54Å). At a synchrotron, the energy of the incident radiation can be tuned to match DE (accurately). Importantly, DEs 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 For these elements, anomalous scattering is significant only at a synchrotron.

11. Choose an element with DE 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. 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. (-15,0,6) · (0,0,0) (15,0,-6) 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 f(hkl)=-f(-h-k-l)

15. e- 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.

16. A single heavy atom per protein can produce a small but measurable difference between FPH(hkl) and FPH(-h-k-l). Differences between Ihkl 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. f-h-k-l |FP-h,-k,-l| FP(-h,-k,-l) In the complex plane, FP(hkl) and FP(-h-k-l) are reflected across the real axis. FP(h,k,l) imaginary |FPh,k,l| FP(hkl)=FP(-h-k-l) and f(hkl)=-f(-h-k-l) fhkl real True for any crystal in the absence of anomalous scattering. Normally, Ihkl and I-h-k-l are averaged together to improve redundancy

18. |FPh,k,l| fhkl f-h-k-l |FP-h,-k,-l| But not FPH(hkl) and FPH(-h-k-l) FPH(h,k,l) imaginary |FPH(hkl)|≠|FPH(-h-k-l)| real FPH(-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.

19. Hey! Look at that! We have two phase triangles now; we only had one before. FPH(hkl) =FP(hkl) +FH(hkl) FPH(-h-k-l) =FP(-h-k-l) +FH(-h-kl-) 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)|and |FPH(-h-k-l)| and... imaginary FPHh,k,l |FPh,k,l| fhkl real f-h-k-l FPH-h,-k,-l |FP-h,-k,-l| we have to be able to calculate the effect of anomalous scattering on the values of FHhkl and FH-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.

20. Q: How do we correct for anomalous scattering effects in our calculation of FH?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. • The oscillator is characterized by • mass=m • position =x,y • natural circular frequency=nB • Characteristic of the atom • Bohr frequency • From Bohr’s representation of the atom e- + nucleus

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

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

27. nucleus f= Amplitude of scattered radiation from the forced e- Amplitude of scattered radiation by a free e- + e- + + + Knowing the amplitude of the e- leads to a definition of the scattering factor, f. 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. Incident photon with n=nB Scattered photon

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

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

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

31. 90o phase shift analogy to a child on a swingForced Oscillator Analogy Maximum negative displacement Zero force Maximum positive 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 -1 0 1 force : displacement incident photon : re-emitted photon. 2 4 1 3

33. f” f’ Construction of FHunder conditions of anomalous scattering Imaginary axis FH( H K L) f fo Real axis FH=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH) scattering factor for H Assume we have located a heavy atom, H, by Patterson methods. Gives f real Positive number imaginary 90° out of phase real 180° out of phase

34. f” f” f’ f’ FH(-h-k-l)is constructed in a similar way as FH(hkl) except f is negative. Imaginary axis Imaginary axis FH( H K L) FH(-H-K-L) FH(-H-K-L) f fo Real axis Real axis -f fo FH( H K L)=[fo + Df’(l) + iDf”(l)] e2pi(+hxH+kyH+lzH) FH(-H-K-L)=[fo + Df’(l) + iDf”(l)] e2pi(-hxH-kyH-lzH)

35. f” f” fo f f’ f’ -f fo fH(+H+K+L) fH(-H-K-L) Again, we see howFriedel’s Law is broken Imaginary axis FH(H K L) Real axis FH(-H-K-L) fH(-h-k-l)≠ -fH(-h-k-l)

36. 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)| 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. Begin by graphing the measured amplitude of FP for (HKL)and (-H-K-L).Circles have equal radius by Friedel’s law Imaginary axis Imaginary axis |FP(HKL) | |FP(-H-K-L) | FH(-H-K-L) Real axis Real axis FH(HKL) Graph FH(hkl)andFH(-h-k-l)using coordinates of H. Place vector tip at origin. FH(-H-K-L)=[fo + Df’(l) + iDf”(l)] e2pi(-hxH-kyH-lzH) FH( H K L)=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH) Structure factor amplitudes and phases calculated using equations derived earlier.

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

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

40. Imaginary axis |FP(-H-K-L) | FH(-H-K-L)* Real axis 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. Imaginary axis |FP(HKL) | Real axis FH(HKL) FPH(-h-k-l)* = FP (-h-k-l)* + FH (-h-k-l)* FPH(hkl) = FP (hkl) +FH (hkl) Friedel’s law, FP (-h-k-l)*=FP (hkl). FPH(-h-k-l)* = FP (hkl) + FH (-h-kl-)*

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

42. Q: How can measurements of |FPH(l1)|,|FPH(l2)|, and |FPH(l3)|be combined to solve the phase of FP in a MAD experiment? A: Again, analogous to MIR, using:measured amplitudes |FPH(l1)|,|FPH(l2)|, and|FPH(l2)| calculated amplitudes & phases of FH(l1),FH(l2), &FH(l3) three phasing triangles FPH(ll) = FP (ll) + FH (ll) FPH(l2) = FP (l2) + FH (l2) FPH(l3) = FP (l3) + FH (l3)and Friedel’s law but no measured amplitude, |FP|

43. After dampening correction Correction factors are largest near n=nB . f = fo+ Df’ + iDf” when n>nB Else, 0 n=frequency of external force (incident photon) nB=natural frequency of oscillator (e-) The REAL COMPONENT becomes negative near v= nB. The IMAGINARY COMPONENT becomes large and positive near n= nB. n-> Df’ Df’ n=nB n=nB

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

45. Df” Df” Df” Df” Df” fo fo fo fo fo Similarly,Df”, varies strongly near the absorption edge, becoming most positive at energies > DE. Se when n>nB Else, 0 Df” is the component of scattered radiation 90° out of phase with the normally scattered component fo DE

46. Four wavelengths are commonly chosen to give the largest differences in FH. FH(low remote) FH(high remote) FH(inflection) FH(peak) f” f” f’ fo f’ fo f’ f’ fo fo FH( l)=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH)

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/l) of the incident radiation. Imaginary axis Imaginary axis Imaginary axis FH (l3) FH (l2) FH (ll) FH( l1)=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH) FH( l2)=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH) FH( l3)=[fo + Df’(l) + iDf”(l)] e2pi(hxH+kyH+lzH) 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(l) amplitudes are graphed as circles centered at the beginning of the FH(l) vectors (as in MAD & SIRAS). Imaginary axis Imaginary axis Imaginary axis FPH(ll) = FP (ll) + FH (ll) FPH(l2) = FP (l2) + FH (l2) FPH(l3) = FP (l3) + FH (l3) No measurement available for FP, but it can be assumed that its value does not change with wavelength because it contains no anomalous scatterers. Hence, FP (ll) = FP (l2) = FP (l3)and all three circles intersect at FP.

49. A three wavelength MAD experiment solves the phase ambiguity. Imaginary axis FPH(ll) = FP + FH (ll) FPH(l2) = FP + FH (l2) FPH(l3) = FP + FH (l3) Real axis 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 l Poor choice of l Imaginary axis Imaginary axis Real axis Real axis Point of intersection clearly defined. Point of intersection poorly defined.