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Phasing I

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Phasing I

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  1. This presentation will probably involve audience discussion, which will create action items. Use PowerPoint to keep track of these action items during your presentation • In Slide Show, click on the right mouse button • Select “Meeting Minder” • Select the “Action Items” tab • Type in action items as they come up • Click OK to dismiss this box This will automatically create an Action Item slide at the end of your presentation with your points entered. Phasing I Biology 555 Andrew J. Howard Crystallographic Phasing I

  2. Overview of the Phase Problem John RoseACA Summer School 2006 Reorganized by Andy Howard, Biology 555, Spring 2008 Remember We can measure reflection intensities We can calculate structure factors from the intensities We can calculate the structure factors from atomic positions We need phase information to generate the image Protein Crystal Data Phases Structure Crystallographic Phasing I

  3. Phasing Topics Crystallographic Phasing I

  4. What is the Phase Problem? X-ray Diffraction Experiment In the X-ray diffraction experiment photons are reflected from the crystal lattice (planes) in different directions giving rise to the diffraction pattern. Using a variety of detectors (film, image plates, CCD area detectors) we can estimate intensities but we lose any information about the relative phase for different reflections. All phase information is lost x,y.z Fhkl [Real Space] [Reciprocal Space] Crystallographic Phasing I

  5. Phases • Let’s define a phase fj associated with a specific plane [hkl] for an individual atom:fj = 2p(hxj + kyj + lzj) • Atom at xj=0.40, yj=0.05, zj=0.10 for plane [213]:fj = 2p(2*0.40 + 1*0.05 + 3*0.10) = 2p(1.35) • If we examine a 2-dimensional case like k=0, then fj = 2p(hxj + lzj) • Thus for [201] (a two-dimensional case):fj = 2p(2*0.40 + 0*0.05 + 1*0.10) = 2p(0.90) • Now, to understand what this means: Crystallographic Phasing I

  6. 0° 360° 0 c A A 2p 201 planes B B D D C C 720° 0.4, y, 0.1 E E H H 4p F F G G I I 1080° a 6p 201 Phases fD = 2p[ 2•(0.40) + 1•(0.10)] = 2p(0.) Crystallographic Phasing I

  7. In General for Any Atom (x, y, z) a dhkl Remember:We express any position in the cell as (1) fractional coordinates: pxyz = xja+yjb+zjc (2) the sum of integral multiples of the reciprocal axes shkl = ha* + kb* + lc* 6π dhkl 4π Atom (j) at x,y,z dhkl 2π φ 0 c Plane hkl Crystallographic Phasing I

  8. Diffraction vector for a Bragg spot • We set up the diffraction vector shkl associated with a specific diffraction direction hkl:shkl = ha* + kb* + lc* • The magnitude of this diffraction vector is the reciprocal of our Bragg-law plane spacing dhkl:|shkl| = 1/ dhkl Crystallographic Phasing I

  9. Phase angle for a spot • The phase angle fj associated with our atom is 2p times the projection of the displacement vector pj onto shkl: fj = 2p shkl • pj • But that displacement vector pj is related to the real-space coordinates of the atom at position j:pj = xja + yjb + zjcwhere the fractional coordinates of our atom within the unit cell are(xj, yj, zj) • Thus fj = 2p (ha* + kb* + lc*) • (xja + yjb + zjc) Crystallographic Phasing I

  10. Real-space and reciprocal space • But these real-space and reciprocal-space unit cell vectors (a,b,c) and (a*,b*,c*) are duals of one another; that is, they obey: a•a* = 1, a•b* = 0, a•c* =0 b•a* = 0, b•b* = 1, b•c* =0 c•a* = 0, c•b* = 0, c•c* = 1 • … even when the unit cell isn’t all full of 90-degree angles! Crystallographic Phasing I

  11. Matrix formulation of this duality • If we construct the 3x3 reciprocal-space unit cell matrix A = (a* b* c*) • And the 3x3 real-space unit cell matrixR = (abc)for a specific position of the sample, then • A and R obey the simple relationshipA = R-1, i.e. AR = I • Where I is a 3x3 identity matrix Crystallographic Phasing I

  12. How to use this in getting phases • fj = 2p (ha* + kb* + lc*) •(xja + yjb + zjc) • But using those dual relationships,e.g. a*•a = 1, b*•c = 0, we getfj = 2p (hxj + kyj + lzj) • Note that this is true even if our unit cell angles aren’t 90º! Crystallographic Phasing I

  13. Fourier transform Inverse Fourier transform Why Do We Need the Phase? • In order to reconstruct the molecular image (electron density) from its diffraction pattern both the intensity and phase, which can assume any value from 0 to 2, of each of the thousands of measured reflections must be known. Structure Factor Electron Density Crystallographic Phasing I

  14. Importance of Phases Hauptman amplitudes with Hauptman phases Karle amplitudes with Karle phases Phases dominate the image! Phase estimates need to be accurate Hauptman amplitudes with Karle phases Karle amplitudes with Hauptman phases Crystallographic Phasing I

  15. Understanding the Phase Problem • The phase problem can be best understood from a simple mathematical construct. • The structure factors (Fhkl) are treated in diffraction theory as complex quantities, i.e., they consist of a real part (Ahkl) and an imaginary part (Bhkl). • If the phases, hkl, were available, the values of Ahkl and Bhkl could be calculated from very simple trigonometry: • Ahkl = |Fhkl| cos (hkl) • Bhkl = |Fhkl| sin (hkl) • This leads to the relationship:(Ahkl)2 + (Bhkl)2 = |Fhkl|2 = Ihkl Crystallographic Phasing I

  16. Argand Diagram (ahkl)2 + (bhkl)2 = |Fhkl|2 = Ihkl The above relationships are often illustrated using an Argand diagram (right). From the Argand diagram, it is obvious that ahkl and bhkl may be either positive or negative, depending on the value of the phase angle, hkl. Note: the units of ahkl, bhkl and Fhkl are in electrons. Crystallographic Phasing I

  17. f0 sinq/l The Structure Factor Atomic scattering factors • The scattering factor for each atom type in the structure is evaluated at the correct sinq/l. That value is the scattering ability for that atom. • Remember sinq/l = 1/(2dhkl) • We now have an atomic scattering factor with magnitude f0 and direction fj Here fj is the atomic scattering factor Crystallographic Phasing I

  18. imaginary Resultant Fhkl Individual atom fjs Bhkl real Ahkl The Structure FactorSum of all individual atom contributions Crystallographic Phasing I

  19. Electron Density • Remember the electron density (image of the molecule) is the Fourier transform of the structure factor Fhkl. Thus Here V is the volume of the unit cell Crystallographic Phasing I

  20. How to calculate r(x,y,z) • In practice, the electron density for one three-dimensional unit cell is calculated by starting at x, y, z = (0, 0, 0) and stepping incrementally along each axis, summing the terms as shown in the equation above for all hkl (as limited by the resolution of the data) at each point in space. Crystallographic Phasing I

  21. What do we mean by the phase problem? • We have seen that for each reflection h=(h,k,l), the Fourier coefficient Fh is complex (hence boldface F) • The real part of that Fourier coefficient |Fh| is determinable directly from the intensity: |Fh| = Ih1/2 • But the phase angle h is not directly determinable from a single diffraction measurement • This inaccessibility of h is what we mean by the phase problem Crystallographic Phasing I

  22. Solving the Phase Problem • Small molecules • Direct Methods • Patterson Methods • Molecular Replacement • Macromolecules • Multiple Isomorphous Replacement (MIR) • Multi Wavelength Anomalous Dispersion (MAD) • Single Isomorphous Replacement (SIR) • Single Wavelength Anomalous Scattering (SAS) • Molecular Replacement • Direct Methods (special cases) Crystallographic Phasing I

  23. Solving the Phase Problem SMALL MOLECULES: • The use of Direct Methods has essentially solved the phase problem for well diffracting small molecule crystals. MACROMOLECULES: • Today, anomalous scattering techniques such as MAD or SAS are the most common techniques used for de novo structure determination of macromolecules. Both techniques require the presence of one or more anomalous scatterers in the crystal. Crystallographic Phasing I

  24. Direct methods • Karle, Hauptman, David Sayre, and others determined algebraic relationships among phase angles of groups of reflections. • The simplest are triplet relationships:For three reflectionsh1=(h1,k1,l1), h2=(h2,k2,l2), h3=(h3,k3,l3),they showed that if h3= -h1- h2, then F1 + F2 + F3 ≈ 0 • Thus if F1 and F2 are known then we can estimate that F3 ≈ -F1 - F2 David Sayre Crystallographic Phasing I

  25. When do triplet relations hold? • Note the approximately zero value in that relationship F1 + F2 + F3 ≈ 0. • The stronger the Bragg reflections are, the closer this condition is to being exact. • For very strong Bragg reflections that sum will be very close to zero • For weaker ones it may differ significantly from zero Crystallographic Phasing I

  26. Phase probabilities • This notion of relationships among phases obliges us to think of phases probabilistically rather than deterministically. This is a key to the direct-methods approach and has a huge influence on how we think about phase determination. • I’m introducing all of this mostly to get you accustomed to the notion of phase probability distributions! Crystallographic Phasing I

  27. Probability distributions P(x) • In general if we can measure or derive information about a variable x but do not know its value to arbitrary accuracy, we can assign a probability distribution P(x) that is highest at its most probable value and lower at other values. • The convention is the integral of the probability distribution over all possible values of x is 1 x ∫-∞ ∞ P(x)dx = 1 Crystallographic Phasing I

  28. Multiple sources of information • If we can derive information about our variable x from more than one experiment or source, then we can derive a phase probability distribution P(x) for x from each source. • If the information sources are 1 and 2, then we describe the two distributions as P1(x) and P2(x) • If the sources of information are truly independent, then the joint (net) probability distribution will be the product of the individual probability distributions:P1&2(x) = P1(x) * P2(x) Crystallographic Phasing I

  29. Probabilities for angles P() • Probability distributions for angular quantities like phase angles are simpler, in a way, than other distributions because the domain is limited to - to .  -  ∫-P()d = 1 Crystallographic Phasing I

  30. Phase probabilities • Any phase has a value between 0 & 2p(or 0 & 360, if we’re using degrees) • If we know it’s close to 2p*0.42, then: • If it’s 2p*(0.42 0.01), it’s a sharp phase probability distribution • If it’s 2p*(0.42 0.32), it’s a much broader phase probability distribution Crystallographic Phasing I

  31. Plots of phase probability P() • Integral of probability must be 1, since every phase has to have some value. Sharp distribution Broad distribution  0 2π Crystallographic Phasing I

  32. How can we use this? • Obviously if we don’t know f1+f2, we can’t use this to calculate f3, even if the intensities of all three are large. • But we could guess what f1 and f2 are and use this to compute f3. • Then we guess f4 and use the triplet relationship to compute f5 and f6,where h5 = -h1-h4 and h6 = -h2-h4 …assuming that reflections 5 and 6 are strong, too! Crystallographic Phasing I

  33. Can we make this work? • We start with guessed phases for a 10-100 strong reflections and use the triplet relationships to determine the phases for another 1000 reflections • Any particular calculated phase can be determined by several different triplet relationships, so if they’re self-consistent, the initial guessed 10-100 are correct; if they aren’t self-consistent, the guess was wrong! • In the latter case, we try a different set of guesses for our 10-100 starting phases and keep going Crystallographic Phasing I

  34. This actually works, provided: • The data are correctly measured • The data are strong enough that we can pick 1000 strong reflections to use in this process • The data extend to high enough resolution that atomicity (separable atoms) is really found • There are ways to do direct methods without assuming atomicity, but they’re more complicated Crystallographic Phasing I

  35. Is this relevant to macromolecules? • Not directly: • Atomicity rarely present • Systematic errors in data • Indirectly yes, because it can be used in conjunction with other methods for locating heavy atoms in the SIR, MIR, and SAS methods • It also helps introduce the notion of phase probability distributions (sneaky!) Crystallographic Phasing I

  36. So how do we get phase estimates for macromolecular data? • Well, it’s more complicated for the user than direct methods, but there are approaches that work. • We’ll give you a sketch of how those work in the next 2 lectures • Hint: most of them involve Patterson functions, one way or another. Crystallographic Phasing I

  37. Isomorphous replacement • Underlying idea: add 1-4 strong scatterers to a macromolecule: they change scattering by ~15% • So: carefully measure the differences between the intensities from the modified molecule and the intensities from the unmodified molecule • Use those differences to determine where the strong scatterer(s) is or are. • Once we know where the strong scatterer is, we can use that knowledge to estimate the phase angles for the starting structure. Crystallographic Phasing I

  38. Finding the Heavy Atomsor Anomalous Scatterers The Patterson function - a F2 Fourier transformwith f = 0 - vector map(u,v,w instead of x,y,z) - maps all inter-atomic vectors - get N2 vectors!!(where N= number of atoms) From Glusker, Lewis and Rossi Crystallographic Phasing I

  39. The Difference Patterson Map SIR : |DF|2 = |Fnat - Fder|2 SAS : |DF|2 = |Fhkl - F-h-k-l|2 Patterson map is centrosymmetric - see peaks at u,v,w & -u, -v, -w Peak height proportional to ZiZj Peak u,v,w’s give heavy atom x,y,z’s - Harker analysis Origin (0,0,0) maps vector of atom to itself Crystallographic Phasing I

  40. Harker analysis • Certain relationships apply inPatterson maps that enable usto determine some of thecoordinates of our heavy atoms • They depend on looking at differences between atomic positions • These relationships were worked out by Lindo Patterson and David Harker • Patterson space is centrosymmetric but otherwise similar to original symmetry; but Patterson symmetry has no translations David Harker Crystallographic Phasing I

  41. Example: space group P21 • P21 has peaks atR1=(x,y,z) and R2=(-x,y+1/2,-z) • Therefore we’ll get Patterson (difference) peaks at R1-R1, R1-R2, R2-R1, R2-R1: • (0,0,0), (2x,-1/2,2z), (-2x,1/2,-2z),(0,0,0) • So if we look at the section of the map at Y=1/2, we can find peaks at (-2x,1/2,-2z) and thereby discern what the x and z coordinates of a real atom are Crystallographic Phasing I

  42. How do we actually use this? • Compute difference Patterson map,i.e. map with coefficients derived from FhklPH - FhklP or Fhkl - F-h-k-l • Examine Harker sections • Peaks in Harker sections tell us where the heavy atoms or anomalous scatterers are • Automated programs like BNP, SOLVE, SHELX can do the heavy lifting for us Crystallographic Phasing I

  43. A Note About Handedness • We identify each reflection by an index, hkl. • The hkl also tells us the relative location of that reflection in a reciprocal space coordinate system. • The indexed reflection has correct handedness if a data processing program assigns it correctly. • The identity of the handedness of the molecule of the crystal is related to the assignment of the handedness of the data, which may be right or wrong! • Note: not all data processing programs assign handedness correctly! • Be careful with your data processing. Crystallographic Phasing I

  44. The Phase Triangle Relationship DOLM = DOLN M FP, FPH, FH and -FH are vectors (have direction) FP <= obtained from native data FPH <= from derivative or anomalous data FH <= obtained from Patterson analysis Q L FPH = FP + FH O Need value of FH From Glusker, Lewis and Rossi N Crystallographic Phasing I

  45. The Phase Triangle Relationship M • In simplest terms, isomorphous replacement finds the orientation of the phase triangle from the orientation of one of its sides. It turns out, however, that there are two possible ways to orient the triangle if we fix the orientation of one of its sides. Q L O From Glusker, Lewis and Rossi N Crystallographic Phasing I

  46. Single Isomorphous Replacement Note: FP = protein FH = heavy atom FP1 = heavy atom derivative The center of the FP1circle is placed at the end of the vector -FH1. • The situation of two possible SIR phases is called the “phase ambiguity” problem, since we obtain both a true and a false phase for each reflection. Both phase solutions are equally probable, i.e. the phase probability distribution is bimodal. X1 = ftrueorffalse X2 = ftrueorffalse From Glusker, Lewis and Rossi Crystallographic Phasing I

  47. Resolving the Phase Ambiguity Note: FP = protein FH = heavy atom FP1 = heavy atom derivative The center of the FP1circle is placed at the end of the vector -FH1. Add more information: • Add another derivative (Multiple Isomorphous Replacement) • Use a density modification technique (solvent flattening) • Add anomalous data (SIR with anomalous scattering) X1 = ftrueorffalse X2 = ftrueorffalse From Glusker, Lewis and Rossi Crystallographic Phasing I

  48. Multiple Isomorphous Replacement Note: FP = protein FH1 = heavy atom #1 FH2 = heavy atom #2 FP1 = heavy atom derivative FP2 = heavy atom derivative The center of the FP1 and FP1 circles are placed at the end of the vector -FH1 and -FH2, respectively. • We still get two solutions, one true and one false for each reflection from the second derivative. The true solutions should be consistent between the two derivatives while the false solution should show a random variation. X1 = ftrue X2 = ffalse X3 = ffalse From Glusker, Lewis and Rossi Exact overlap at X1 dependent on data accuracy dependent on HA accuracy called lack of closure Crystallographic Phasing I

  49. Solvent Flattening Similar to noise filtering Resolve the SIR or SAS phase ambiguity Electron density can’t be negative Use an iterative process to enhance true phase! From Glusker, Lewis and Rossi B.C. Wang, 1985 Crystallographic Phasing I

  50. How does solvent flattening resolve the phase ambiguity? • Solvent flattening can locate and enhance the protein image—viz., whatever is not solvent must be protein! • From the protein image, the phases of the structure factors of the protein can be calculated • These calculated phases are then used to select the true phases from sets of true and false phases Crystallographic Phasing I

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