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19 Feb 2008 Biology 555: Crystallographic Phasing II p. 1 of 38 ProteinDataCrystalStructurePhases Overview of the Phase Problem John Rose ACA Summer School.

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Presentation on theme: "19 Feb 2008 Biology 555: Crystallographic Phasing II p. 1 of 38 ProteinDataCrystalStructurePhases Overview of the Phase Problem John Rose ACA Summer School."— Presentation transcript:

1 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 1 of 38 ProteinDataCrystalStructurePhases Overview of the Phase Problem John Rose ACA Summer School 2006 Reorganized by Andy Howard, Biology 555, Spring 2008 Part 2 of 2 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

2 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 2 of 38 From Glusker, Lewis and Rossi Finding the Heavy Atoms or Anomalous Scatterers The Patterson function - a F 2 Fourier transform with  = 0 - vector map (u,v,w instead of x,y,z) - maps all inter-atomic vectors - get N 2 vectors!! (where N= number of atoms)

3 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 3 of 38 The Difference Patterson Map SIR : |  F| 2 = |F nat - F der | 2 SAS : |  F| 2 = |F hkl - F -h-k-l | 2 Patterson map is centrosymmetric - see peaks at u,v,w & -u, -v, -w Peak height proportional to Z i Z j Peak u,v,w’s give heavy atom x,y,z’s - Harker analysis Origin (0,0,0) maps vector of atom to itself

4 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 4 of 38 Harker analysis Certain relationships apply in Patterson maps that enable us to determine some of the coordinates 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

5 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 5 of 38 Example: space group P2 1 P2 1 has peaks at R 1 =(x,y,z) and R 2 =(-x,y+1/2,-z) Therefore we’ll get Patterson (difference) peaks at R 1 -R 1, R 1 -R 2, R 2 -R 1, R 2 -R 1 : (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

6 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 6 of 38 How do we actually use this? Compute difference Patterson map, i.e. map with coefficients derived from F hkl PH - F hkl P or F hkl - 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

7 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 7 of 38 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.

8 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 8 of 38 O N M L  OLM =  OLN Q F PH = F P + F H Need value of F H From Glusker, Lewis and Rossi The Phase Triangle Relationship F P, F PH, F H and -F H are vectors (have direction) F P <= obtained from native data F PH <= obtained from derivative or anomalous data F H <= obtained from Patterson analysis

9 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 9 of 38 O N M L Q From Glusker, Lewis and Rossi The Phase Triangle Relationship 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.

10 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 10 of 38 X 1  true or  false X 2  true or  false From Glusker, Lewis and Rossi Note: F P = protein F H = heavy atom F P1 = heavy atom derivative The center of the F P1 circle is placed at the end of the vector -F H1. Single Isomorphous Replacement 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.

11 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 11 of 38 X 1  true or  false X 2  true or  false From Glusker, Lewis and Rossi Note: F P = protein F H = heavy atom F P1 = heavy atom derivative The center of the F P1 circle is placed at the end of the vector -F H1. Resolving the Phase Ambugity Add more information: (1)Add another derivative (Multiple Isomorphous Replacement) (2)Use a density modification technique (solvent flattening) (3)Add anomalous data (SIR with anomalous scattering)

12 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 12 of 38 X   true X   false X   fals Exact overlap at X 1 dependent on data accuracy dependent on HA accuracy called lack of closure From Glusker, Lewis and Rossi Note: F P = protein F H1 = heavy atom #1 F H2 = heavy atom #2 F P1 = heavy atom derivative F P2 = heavy atom derivative The center of the F P1 and F P1 circles are placed at the end of the vector -F H1 and -F H2, respectively. Multiple Isomorphous Replacement 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.

13 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 13 of 38 B.C. Wang, 1985 From Glusker, Lewis and Rossi 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!

14 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 14 of 38 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

15 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 15 of 38 Using the structure to solve the phase ambiguity Thus, in essence, the phase ambiguity is resolved by the protein image itself! This solvent-flattening process was made practical by the introduction of the ISIR/ISAS program suite (Wang, 1985) and other phasing programs such DM and PHASES are based on this approach.

16 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 16 of 38 Handedness from solvent flattening The ISAS process is performed twice, once with heavy atom sites @ refined locations, once in their inverted locations DataFOM 1 Handed- ness FOM 2 RfactorCorr. Coeff. RHE0.54Correct0.820.260.958 0.54Wrong0.800.300.940 NP + I 3 0.54Correct0.800.270.955 0.54Wrong0.760.360.919 NP+I+S 4 0.56Correct0.820.240.964 0.56Wrong0.780.350.926

17 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 17 of 38 Notes on the handedness table 1 : Figure of merit before solvent flattening 2 : Figure of merit after one filter and four cycles of solvent flattening 3 : Four Iodine were used for phasing 4 : Four Iodine and 56 Sulfur atoms were used for phasing Heavy Atom Handedness and Protein Structure Determination using Single-wavelength Anomalous Scattering Data, ACA Annual Meeting, Montreal, July 25, 1995.

18 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 18 of 38 Does the correct hand make a difference? Yes! The wrong hand will give the mirror image!

19 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 19 of 38 Anomalous Dispersion Methods All elements display an anomalous dispersion (AD) effect in X-ray diffraction For light elements (H, C, N, O), anomalous dispersion effects are negligible; they’re small even for S and P at typical X-ray energies For heavier elements, especially when the X-ray wavelength approaches an atomic absorption edge of the element, these AD effects can be very large.

20 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 20 of 38 Scattering power when anomalous scattering exists The scattering power of an atom exhibiting AD effects is: f AD = f n +  f' + i  f” where: f n is the normal scattering power of the atom in absence of AD effects  f' arises from the AD effect and is a real factor (+/- signed) added to f n  f" is an imaginary term which also arises from the AD effect  f" is always positive and 90° ahead of (f n +  f') in phase angle

21 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 21 of 38  f’ and  f” The values of  f' and  f" are highly dependent on the wavelength of the X-radiation. In the absence of AD effects, I hkl = I -h-k-l (Friedel’s Law). With AD effects, I hkl ≠ I -h-k-l (Friedel’s Law breaks down). Accurate measurement of Friedel pair differences can be used to extract starting phases if the AD effect is large enough.

22 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 22 of 38 f’ Breakdown of Friedel’s Law (F hkl Left) F n represents the total scattering by "normal" atoms without AD effects, f’ represents the sum of the normal and real AD scattering values (f n +  f'),  f" is the imaginary AD component and appears 90° (at a right angle) ahead of the f’ vector and the total scattering is the vector F +++. (F -h-k-l Right) F -n is the inverse of F n (at -  hkl ) and f’ is the inverse of f’, the  f" vector is once again 90° ahead of f’. The resultant vector, F --- in this case, is obviously shorter than the F +++ vector.

23 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 23 of 38 Collecting Anomalous Scattering Data Anomalous scatterers, such as selenium, are generally incorporated into the protein during expression of the protein or are soaked into the crystals in a manner similar to preparing a heavy atom derivative. Bromine, iodine, xenon and traditional heavy atom compounds are also good anomalous scatterers.

24 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 24 of 38 How strong is the signal? The anomalous signal, the difference between |F +++ | and |F --- | is generally about one order of magnitude smaller than that between |F PH (hkl)|, and |F P (hkl)|. Thus, the signal-to-noise (S/n) level in the data plays a critical role in the success of anomalous scattering experiments, i.e. the higher the S/n in the data the greater the probability of producing an interpretable electron density map.

25 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 25 of 38 Why does it work at all? The lack of isomorphism problem is much milder for anomalous data than for isomorphous replacement: One sample, not two or more Unit cell is by definition (?) identical Molecule is in the same place within that unit cell That partly compensates for the low S/N

26 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 26 of 38 Why is selenium a good choice? Methionine is a relatively rare amino acid: 2.4% (vs. average of 5%) So there aren’t a huge number of mets in a typical protein, but there generally are a few It’s possible to make E.coli auxotrophic for methionine and then feed it selenomethionine in its place This incorporates SeMet stoichiometrically and covalently, which is definitely good!

27 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 27 of 38 Anomalous data collection The anomalous signal can be optimized by data collection at or near the absorption edge of the anomalous scatterer. This requires a tunable X-ray source such as a synchrotron. The S/n of the data can also be increased by collecting redundant data. The two common anomalous scattering experiments are Multiwavelength Anomalous Dispersion (MAD) and single wavelength anomalous scattering/diffraction (SAS or SAD) The SAS technique is becoming more popular since it does not require a tunable X-ray source.

28 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 28 of 38 Increasing Number of SAS Structures MAD SAD

29 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 29 of 38 Increasing S/n with Redundancy

30 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 30 of 38 From Glusker, Lewis and Rossi. Multiwavelength Anomalous Dispersion Note: F P = protein F H1 = heavy atom F + PH = F +++ F - PH = F --- F + H” =  f” +++ F - H” =  f” --- The center of the F + PH and F - PH circles are placed at the end of the vector -F + H” and -F - H” respectively In the MAD experiment a strong anomalous scatterer is introduced into the crystal and data are recorded at several wavelengths (peak, inflection and remote) near the X-ray absorption edge of the anomalous scatterer. The phase ambiguity resolved a manner similar to the use of multiple derivatives in the MIR technique

31 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 31 of 38 Single Wavelength Anomalous Scattering The SAS method, which combines the use of SAS data and solvent flattening to resolve phase ambiguity was first introduced in the ISAS program (Wang, 1985). The technique is very similar to resolving the phase ambiguity in SIR data. The SAS method does not require a tunable source and successful structure determination can be carried out using a home X-ray source on crystals containing anomalous scatterers with sufficiently large  f” such as iron, copper, iodine, xenon and many heavy atom salts.

32 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 32 of 38 Sulfur S-SAS: experimental realities The ultimate goal of the SAS method is the use of S- SAS to phase protein data since most proteins contain sulfur. However sulfur has a very weak anomalous scattering signal with  f” = 0.56 e - for Cu X-rays. The S-SAS method requires careful data collection and crystals that diffract to 2Å resolution. A high symmetry space group (more internal symmetry equivalents) increases the chance of success. The use of soft X-rays such as Cr K  (  = 2.2909Å) X-rays doubles the sulfur signal (  f” = 1.14 e - ). There over 20 S-SAS structures in the Protein Data Bank.

33 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 33 of 38 What is the Limit of the SAS Method? Electron density maps of Rhe by Sulfur-ISAS Calculated using simulated data in 1983  f” = 0.56e - using Cu K  X-rays Wang (1985), Methods Enzymol. 115: 90-112

34 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 34 of 38 Molecular Replacement Molecular replacement has proven effective for solving macromolecular crystal structures based upon the knowledge of homologous structures. The method is straightforward and reduces the time and effort required for structure determination because there is no need to prepare heavy atom derivatives and collect their data. Model building is also simplified, since little or no chain tracing is required.

35 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 35 of 38 Molecular Replacement: Practical Considerations The 3-dimensional structure of the search model must be very close (< 1.7Å r.m.s.d.) to that of the unknown structure for the technique to work. Sequence homology between the model and unknown protein is helpful but not strictly required. Success has been observed using search models having as low as 17% sequence similarity. Several computer programs such as AmoRe, X- PLOR/CNS PHASER are available for MR calculations.

36 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 36 of 38 px.cryst.bbk.ac.uk/03/sample/molrep.htm How Molecular Replacment works Use a model of the protein to estimate phases Must be a structural homologue (RMSD < 1.7Å) Two-step process: rotation and translation Find orientation of model (red  black) Find location of oriented model (black  blue)

37 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 37 of 38 Using a protein model to estimate phases: the rotation function We need to determine the model’s orientation in X 1 ’s unit cell We use a Patterson search approach in ( , ,  ), which are Euler angles associated with the rotational space

38 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 38 of 38 Euler angles for rotation function The coordinate system is rotated by: an angle  around the original z axis; then by an angle  around the new y axis; and then by an angle  around the final z axis. zyz convention

39 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 39 of 38 Using a protein model to estimate phases: translation function We need to determine the oriented model’s location in X1’s unit cell We do this with an R-factor search, where

40 19 Feb 2008 Biology 555: Crystallographic Phasing II p. 40 of 38 Translation functions Oriented model is stepped through the X 1 unit cell using small increments in x, y, and z (e.g. x  x+ step) The point where R is lowest represents the correct location There exists an alternative method that uses maximum likelihood to find the translation peak; this notion is embodied in the software package PHASER by Randy Read


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