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X-ray Crystallography-2 (plus extra slides) Math: complex numbers Two and more atoms in a unit cell affect intensity (but not spacing) of spots Structure.

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Presentation on theme: "X-ray Crystallography-2 (plus extra slides) Math: complex numbers Two and more atoms in a unit cell affect intensity (but not spacing) of spots Structure."— Presentation transcript:

1 X-ray Crystallography-2 (plus extra slides) Math: complex numbers Two and more atoms in a unit cell affect intensity (but not spacing) of spots Structure factor Fiber diffraction (especially helices and DNA) (Fourier transform, time permitting) (Phase problem (conceptual), time permitting) (R-factor, time permitting) Reading: van Holde, Physical Biochemistry, Chapter 6; the two Watson & Crick papers, Franklin paper Additional optional reading: Gale Rhodes, Crystallography Made Clear, sections of Chapters 1-4 Remember:Midterm 1: Wednesday, Feb. 26 Outline of biophysics paper due this week! Homework: (see next page), 4.1 due Monday, Feb. 22; 4.2 due Friday, March 7

2 Homework 4.1 (Chapter 6, X-ray diffraction), due Monday, Feb. 24 If not stated otherwise, assume = nm (CuK  -radiation) 1.van Holde NaCl crystals are crushed and the resulting microcrystalline powder is placed in the X-ray beam. A flat sheet of film is placed 6.0 cm from the sample and exposed. Ignoring the possibility of forbidden reflections (which is in fact the case with NaCl, because the lattice is centered), what would be the diameters and indices of the first two (innermost) rings on the photograph? NaCl is cubic with unit cell dimension a = 0.56nm. 3.van Holde 6.6 a-c (Fig. 6.18, don’t need to do 6d) 4.You are working with a linear crystal of atoms (assume to be planes), each spaced 6.28 nm apart. You adjust your x-ray emitter so that it emits nm x-rays along the axis of the array. a.You place a 1 cm 2 spherical detector 1 cm from the sample, centered on the x-axis, on the opposite side of the emitter. Draw the pattern you expect to detect. Clearly mark the expected distances. b.If you performed the experiment on a linear crystal with atoms spaced 0.1 nm apart, what pattern would you detect? Would you have the same pattern if your detector were 1 m 2 ? What does this say about the resolution of your experiment? Homework 4.2 (Chapter 6, X-ray diffraction), due Friday, March 7 If not stated otherwise, assume = nm (CuK  -radiation) 1.van Holde 6.2a (Hint: put one atom at x, y, z,  the other atom at x+1/2, -y, -z) 2.van Holde Extra credit: van Holde 6.9 a-c (in c), real space means on film) 4.Extra credit: van Holde 6.9 d but: Sketch the fiber diffraction pattern expected for A-DNA (not Z- DNA).

3 So far… Bragg’s law and von Laue’s law: By observing the spacing and pattern of reflections (spots) on the diffraction pattern, we can determine the lengths, and angles between each side of the unit cell, as well as the symmetry or space group (from systematic absences) How do we find out what’s inside the unit cell? (i.e. the interesting stuff, like proteins).  The intensity of the spots (square of structure factor) ‘tells us what’s inside a unit cell.’

4 Let’s look at two atoms in a unit cell, that are a distance r 1 and r 2 from the origin. After hit by an incoming X-ray, they both emit a wave. E 2 wave is shifted by a phase factor,   r  is phase difference between the two waves. Point: The intensity of the wave, I = |E| 2, depends on , which, in turn, depends on the wave-vector (scattering direction) and location of the atoms in the unit cell. Example: Two atoms in a unit cell (white board example)

5 The intensity of the spots, I, is equal to the square of the electric field of the wave, E. As shown for two atoms, E depends on the spacing (coordinates),, and number of atoms, and the scattering direction,. So, if we have two or more atoms in a unit cell, the intensity of the spots, I, depends on the spacing (coordinates) and number of atoms, and the scattering direction,  We have a relationship between atomic coordinates, (which we want to know) and the intensity of the spots (which we can measure). Diffraction pattern for T4 lysozyme crystal

6 Solving macromolecular structures by X-ray diffraction If we have two or more atoms in a unit cell, the intensity of the spots changes  the diffraction angle  is not affected by the number of atoms in a unit cell. The diffraction angle is still only related to the unit cell dimension by Bragg’s law. However, with one atom, the phase and amplitude of the resultant wave from each plane (atom) was the same. With more than one atom, the phase and amplitude coming from each plane of different atoms may be different, resulting in different intensities of the reflections.  Deconvolute each reflection into the phase and amplitude contributions from each of the individual reflections from each atom in the molecule. # of atoms in unit cell does NOT affect reflection angle (Bragg’s law). # of atoms (electron density) in unit cell DOES affect intensity of reflection spots.

7 Structure Factor for many atoms inside unit cell: F (hkl) is electric field amplitude for a certain scattering direction (reflection spot h,k,l on film) x, y, z … fractional coordinates describing the position of each atom inside unit cell. f j is the amplitude of the electric field coming from each atom j inside unit cell (amplitude is proportional to electron density). The sum is over all N atoms inside a unit cell. S = (ha *, kb * lc * ) is the scattering vector for each diffraction spot Diffraction pattern for T4 lysozyme crystal The structure factor, F (hkl)

8 Moving from fixed points to electron density inside a cell:  The sum is replaced by an integral over the volume of the unit cell, V.  the amplitude f i is replaced by the electron density  (r) (the more density the larger the amplitude). This is very nice now: If we know the electron density  (r) (i.e. positions of atoms) inside unit cell, we can predict the intensity, of each reflection spot (h, k, l) on the film Knowing the size of the unit cell, and knowing the electron density, we can calculate (predict) the location and intensity of each diffraction spot.

9 The phase of the structure factor The structure factor (a complex number) has a magnitude |(F(hkl)| and a “phase angle”  hkl (remember Euler notation).

10 Example You know that an orthorhombic unit cell has two atoms at 1/12, 1/12, 1/12 and -1/12, 1/12, 1/12 (fractional coordinates). If these are the only unique atom in the unit cell, calculate the structure factor F (hkl) and the phase angle  hkl for the h=1, k=1, l=0 reflection. What is the intensity of that spot, compared to possible maximum intensity? (Assume f j for each atom is 1) Important subpoint: If we know the position and type of all atoms in the unit cell, we can calculate F(hkl), magnitude and phase.

11 Starting with the equation for the structure factor F(hkl), show that the intensity of reflections along the (h00) axis will be zero for all odd values of h, when a two-fold screw axis is aligned along the crystallographic a- axis. h = 4 h = 3 h = 2 h = 1 h = 0 h = -1 h = -3 h = -2 h = -4 k = l = 0 a a a2a2

12 Fiber diffraction If there is inherent symmetry in a molecule (i.e. long helical biopolymers) X-Ray diffraction patterns can be obtained from non-crystalline samples. Like one-dimensional crystal along helix axis. We can treat an exact repeat of helix as a crystalline unit cell. Can get information about helical symmetry, pitch and radius. Meridian reflections: information on pitch and symmetry Equatorial reflections: radius of helix

13 Continuous helix Discontinuous helix Pitch P Rise h Radius r H c… repeat Continuous: “line wrapped around a cylinder, Discontinuous: has c residues per turn Distinguish between continuous and discontinuous helix

14 The Fiber Unit Cell Describe unit cell in cylindrical coordinates. Real lattice: cylindrical coordinates (z, r,  ) Reciprocal lattice: coordinates (Z, R, Ψ) Cylindrical unit cell in real and reciprocal space Position in real space Scattering vector in reciprocal space

15 The structure factor in cylindrical coordinates -Now, we’ll assume a continuous helix: - Helix is a line wrapped around a cylinder (  is zero if not on line) - Unit cell has a length equal to the pitch of helix.

16 The structure factor for continuous helix J n (x), Bessel functions The intensity I ~ J n 2 I -n (x) = I n (x), I n (-x) = I n (x)  Symmetric about n and x  Get x-shaped diffraction pattern

17 Diffraction pattern for continuous helix We get spots on layer lines that are 1/P apart. P is pitch. Each layer line corresponds to a Bessel function for n = 1, 2, 3, …. Lines at Z = n/P We get points on those layer lines that correspond to the maxima of the Bessel functions. This results in a X-pattern. Maxima of J x = 1.85

18 Diffraction pattern for discontinuous helix with integral number of m residues, per repeat c. The X-pattern we got for the continuous helix will just repeat itself on the Z-axis with spacing Z=c/P Examples: –B-DNA: c = 10 residues/turn –A-DNA: c = 11 residues/turn –3 10 -helix: c = 3 residues/turn Each layer line is still spaced at 1/P The pattern is repeated at Z = c/P = 1/h

19 Radius determination from diffraction pattern Maxima along equatorial line (n=0) result from hexagonal crystal packing of unit cells.  Spacing of spots along equatorial line is:

20 Diffraction Pattern or B-DNA In May 1952, Rosalind Franklin took beautiful X-ray diffraction patterns of pictures of B-form DNA. The X-pattern is indicative of a helix!! The bases are 0.34 nm apart; there are ten nucleotides per turn; each turn has a rise of 3.4 nm; the diameter of the helix is 2 nm. This pattern is consistent with the model that Watson and Crick built.

21 Extra material The phase problem Reciprocal space, Bragg’s law and Ewald shere.

22 Moving from the structure factor to the electron density (final goal). We need to do the ‘Fourier transform’ of the structure factor to get the electron density out. Now, the diffraction pattern is actually made up of points, so the integral becomes a sum (Fourier series) over all diffraction spots: And that is what we want!!

23 … but wait, there is a problem …the phase problem We cannot measure F(hkl) (complex number), only |F(hkl)|. We can measure the intensity I(hkl) of the spots We only know the magnitude but not the direction (phase angle  ) of F(S). We have lost the phase. How bad is that? Pretty bad, because we can calculate the “magnitude” of the electron density, but we don’t know where it is in the unit cell.  What can we do??

24 Electron density calculated from the two components of F(hkl). (a)Using only |F(hkl)|. The map does not fit the actual structure. (b)Using only the phase information (|F(hkl)| was set to 1).

25 Phase problem solutions Molecular replacement: –If we are lucky, you can get hold of the structure of a known, isomorphous crystal. –This is often used when one wants to know the effect of a mutation in a crystal, or when one wants to know where a substrate binds (soak substrate into crystal). –We use the structure (  (r)) of the known crystal as a starting point. –  Iterative solution: Make small changes to known structure (  (r)), calculate new diffraction pattern (I(hkl), compare to data (diffraction pattern of new crystal), adjust known structure (  (r)), again, calculate new diffraction pattern (I(hkl)), compare with data,… etc.

26 Multiple Isomorphous Replacement 1.Collect the diffraction data from native crystal. 2.Soak in a heavy atom and collect diffraction data. 3.By comparing those two data sets, the position and phase of the heavy metals can often be figured out (because they are very strong scatterers). 4.Repeat with other heavy atom derivatives. 5.These data sets can then be used to estimate the phases of the native data set. 6.Do iterative improvements (estimate position and phases, calculate diffraction spots, compare with data, estimate improved positions and phases, calculate diffraction spots, etc…)

27 Structure refinement, the R-factor The R-factor is defined as: |F (hkl)| is from measured intensity, F calc is calculated structure factor It’s the error between the real diffraction pattern and the calculated one. A R-factor of ~0.2 (20% error) is considered good enough to give a good model of the structure.

28 Real lattice Reciprocal lattice

29 Relationship between unit cell parameters in real space and in reciprocal space Stout & Jensen (1989) X-Structure Determination. A practical guide. 2 nd ed. John Wiley & Sons, New York

30 Examples of direct and reciprocal lattices (Figures from Jensen and Stout “X-Ray structure determination. A pratical guide” Orthorhombic direct and reciprocal cells

31 X-ray beam Sphere of reflection (Ewald sphere) Reciprocal lattice O 1/  Visualization of Bragg’s law A B L 

32 X-ray beam Reciprocal lattice Visualization of Bragg’s law Sphere of reflection (Ewald sphere) O Rotating beam (or crystal) we get reflections from all points within sphere.

33 Sphere of reflection in 3D reciprocal space  hkl Scattering direction k S hkl

34 Definition of reciprocal lattice (hand out) Bragg’s law in reciprocal space Von Laue condition in reciprocal space (orthogonal lattices): Von Laue condition in reciprocal space for all crystal symmetries:

35 “Resolution in X-Ray Crystallography” 1.Theoretical limit: smallest d in Bragg’s law, 2sin  = /d,  d hkl, min ~0.094 nm for CuK  -radiation (  = nm) and largest practical angle 2  = 110° (good: can get atomic resolution, ~ 1Å) 2.The “resolution” of X-ray diffraction is also limited by being able to separate spots on film. Reflection spots must be separated by at least 2° (because of their own size, ~ 1° at half-height). 2sin  = /d unit cell. For larger unit cells, need to use larger (CuK  over MoK  ). 3.How many data points (reflections) are required to obtain atomic resolution? We need (at least) 4N equations to solve for 4N parameters (x, y, z and temperature factor B for each of the N atoms within unit cell).  Need at least 4N good reflection spots on film.  Need at least 4N reciprocal lattice points inside sphere of reflection. There are several issues:  = 110° # of points inside limiting sphere: Where d is resolution; N # of reflections needed for that resolution; V real is volume of real lattice.

36 # of unique reflections on film If we get N * reflections on film, some of them are redundant because of symmetry. Friedel’s law: I(hkl) = I(–h –k –l) Each symmetry operator reduces # of unique reflections (For P , only 1/8 the reflections are unique). Good news: because of symmetry, not the whole unit cell needs to be solved.


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