The goal of Data Processing From a series of diffraction images, obtain the intensity ( I ) and standard deviation ( ( I )) for each reflection, hkl

$Page 1: The goal of Data Processing From a series of diffraction images, obtain the intensity ( I ) and standard deviation ( ( I )) for each reflection, hkl$
The goal of Data ProcessingFrom a series of diffraction images, obtain the intensity (I) and standard deviation (s(I)) for each reflection, hkl.

H K L I s0 0 4 3295.4 174.00 0 8 482.1 28.70 0 12 9691.0 500.70 0 16 1743.9 67.40 0 20 5856.0 221.00 0 24 14066.5 436.20 0 28 9936.3 311.70 0 36 8409.8 273.40 0 40 790.5 32.80 0 44 103.4 18.4. . . . .. . . . .. . . . .37 7 0 28.5 16.237 7 1 110.1 10.937 7 2 337.4 13.337 7 3 98.5 10.637 7 4 25.9 10.7

Set of 360 images Final intensities

1. Index2. Integrate3.Merge

3 x 109 bytes (3Gb) 8 x 105 bytes (800kB)

Mike Sawaya

We divided the diffraction pattern of the crystal into a series of 1 degree slices. One slice on each film. Up to 360 films. We convert this data into a single file containing the integrated intensities of all the reflections on the films and average together the measurements that are related by symmetry.This year (2010) Students looked tired and confused at the end of the lecture.Next year, consider removing the description of autoindexing with mosflm and offer this as a separate powerpoint tutorial.Give a X-rayview demo instead.Lukasz suggests using a movie of the 360 1degree oscillation images and get them to see the lattice inside the annular window.

Indexing sounds like a trivial taskplane L=0 b*

a*

(6,2,0)

How many dimensions?

Proteinase K Crystal

Diffraction

Fourier transform

?Detector

3 3 2

Recorda*

b*

c*

Sawaya

How do we obtain a 3D diffraction pattern from data collected on a 2D detector? Do we have just one image? No, we have a series of them. Each image covers a 1 degree wedge of the reciprocal lattice. So each image covers 3 dimensions. We know how the images are related to each other. Each image is one degree advances over the previous image. All the pieces of information for reconstructing the 3 dimensions of the reciprocal lattice are known. Lets illustrate how the individual images accumulate to complete (reconstruct) a 3D lattice.

The Fourier transform of a 3D crystalis a 3D reciprocal lattice

a

b c

Unit cell lengths a, b, cAtom coordinates x, y, z

Reciprocal cell lengths a*, b*, c*Reflection coordinates h, k, l

a*

b*

c*

3D reciprocal lattice is projected on 2D detector(projection is in direction of X-ray beam).

detector

In an undistorted view of the reciprocal lattice, recorded reflections would reside on the surface of a sphere, not a plane.

A distortion-corrected representation of the reflections

Restored depth of the diffraction pattern is evident from an orthogonal view

Each circle corresponds to a different reciprocal lattice plane

Start indexing process using reflections within the same plane and lying near the origin.

Draw a set of evenly spaced rows

Draw the vector representing this repeat distance, a*, between rows

Draw a set of evenly spaced columns

a*

Draw the vector representing this repeat distance, b*, between columns

a*

What is the angle between a* and b*?

a*b*

Which of 14 Bravais Lattices has a=b and g=90°

Cubic

Rhombohedral

Hexagonal/Trigonal

Tetragonal

Orthorhombic

Monoclinic

Triclinic

Draw a set of evenly spaced rows in orthogonal view

a*b*

Draw the vector representing this repeat distance, a*, between rows

a*b*

Draw a set of evenly spaced columns in orthogonal view

a*b*

a*

Draw the vector representing this repeat distance, c*, between columns

a*b*

a*

Is the length of c* related to a* and b*?What are the angles a and b?

a*b*

a*c*

Which of 14 Bravais Lattices has a=b≠c and = =a b g=90°

Cubic

Rhombohedral

Hexagonal/Trigonal

Tetragonal

Orthorhombic

Monoclinic

Triclinic

What is the index of the lowest resolution reflection?

a*b*

What is the index of the highest resolution reflection in the l=0 plane?

a*b*

(-2,2,0)

(3,6,0)

FILM

X-ray beam

Determine unit cell length “a”

1,0,0 reflection

1,0,0

(0,0,0) origin of reciprocal lattice,Also known asx-beam, y-beam

crystal

DCF=80 mm

Da*=2.0 mma*

1/l

a*=Da/( *l DCF)

a*=2.0mm/80mm*1.54Å

a*=0.1623 Å-1

a=61.6 Å

a*/Da* = 1/ l / DCF

Review which experimental parameters were required to index a spot.

a*b*

(-2,2,0)

Coordinates of the direct beam, (X,Y)

Coordinates (X,Y) for the spot position

Unit cell parametersa,b,c, , ,a b g

The orientation of the unit cell axes with respect to the laboratory axes (fyk).

Crystal-to-detector distance

The wavelength of the incident radiation

What are some reasons why indexing might be inaccurate or unreliable?

The wavelength of the incident radiation

Coordinates (X,Y) of the direct beam

Coordinates (X,Y) for the spot position

Unit cell parametersa,b,c, , ,a b g

The orientation of the unit cell axes with respect to the laboratory axes (fyk).

Need a program that can index spots from multiple lattice planes without manually aligning crystal

Automatic indexing algorithm explained

Acta Cryst. (1999), D55, 1690-1695

Locate reflections positions (peaks of high intensity)

1) Display first image in your data set with

2) Press “Peak Search”. Red circles indicate position of prominent peaks (spots).

3) Evaluate whether you need more or fewer peaks.

4) Press “OK”

5) Spot positions (x,y) are written to a file “peaks.file.”

Peak Search177 peaks found

Peaks.file• 7777 0.0 0.0 1 1 height X Y frame• 13 2695.7 1350.5 1 1• 27 2669.5 1062.4 1 1• 16 2570.6 1143.5 1 1• 26 2569.4 1302.4 1 1• 30 2562.5 1592.5 1 1• 32 2554.5 1902.4 1 1• 32 2524.5 1103.4 1 1• 22 2514.5 1523.8 1 1• 12 2503.4 1316.6 1 1• 21 2494.5 1949.5 1 1• 15 2492.5 1923.4 1 1• 35 2488.5 1721.5 1 1• 17 2483.5 1870.6 1 1• 12 2479.4 1212.5 1 1• 32 2465.5 1452.5 1 1• 15 2456.4 638.4 1 1• 13 2444.7 900.7 1 1• 14 2437.6 1183.4 1 1• 23 2436.4 1969.4 1 1• Etc…………………………………………..

Project vectors onto a line. Measure the length of each projection.

Michael Sawaya

Distribution of vectors lengths would be continuous, non-discrete, non-quantized.It signifies that the vector direction does not correspond to a principal unit cell dimension.What direction does correspond to the principal unit cell dimension?Its obvious by inspection that one of the principal unit cell dimensions is only 15 degrees away, counterclockwise.

Look for incremental differences in lengths.

12

34

Distribution of lengths is not incremental, it is continuous

Michael Sawaya


Rotate, 7300 orientations tested.

Michael Sawaya


Projected vectors for rotated image.

Sort the vector projections by length. Count the number of observations of each length.

2 vectors of length 28.5 mm




et cetera

1) Note: Projected vectors have a quantized values (distribution looks like steps).2) The incremental difference D is proportional to the reciprocal cell length

12

34

D D D D D D D D

D D D D D D

Fourier analysis of length histogram reveals cell dimension.

Unit cell length = 62.5 Å

1-D Fourier Transform

A cosine wave with periodicity of 62.5 Å is a major contributor to the

1-D FT.

Run autoindexing script

The autoindexing script is simply named “a.”

Type “denzo” to start the program.

Then type @a to pass instructions to Denzo.


Select a space group with desired Bravais Lattice (e.g. new space group P4)

Predicted pattern should match observed diffraction pattern.

“go” to refine

Paste parameters into integration script (integ.dat).

Insert refined unit cell and crystal orientation parameters into

integration script (integ.dat).Type “list” to obtain refined parameters..

It is not necessary to index following images from scratch.

1o

Film 1, exposed over 1 to 2 degrees Film 2, exposed over 2 to 3 degrees

Integration

1)Draw boundaries of each reflection2) Sum up the intensities recorded on each pixel within boundary.3) Repeat for each spot on each film.

Integrated intensities are written to .x files

Film 1, exposed over 1 to 2 degrees

h k l flag I(profit) I(prosum) c2 s(I) cos incid. X pix Y pix29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.1……………………………………………………….

One .x file for each film




h k l flag I(profit) I(prosum) c2 s(I) cos incid. X pix Y pix

29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061


29 20 -33 1 52.3 50.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061

h k l flag I(profit) I(prosum) c2 s(I) cos incid. X pix Y pix-

29 -20 33 1 212.3 220.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061

prok_001.img

prok_001.x

prok_002.img prok_360.img

prok_002.x prok_360.x


29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061

360 frames, 1 degree

rotation each

With .x files, we can map intensities onto a reciprocal

lattice

1) Accuracy will improve if we Merge multiple observations of the same reciprocal lattice point

2) But, we must test if rotational symmetry exists between lattice points.


29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061


29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061


29 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 22 -29 1 24.0 25.2 1.29 1.2 0.564 28.0 1489.129 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061.629 20 -33 1 202.3 200.8 1.36 17.4 0.556 6.4 1353.0 29 21 -31 1 102.1 105.0 1.08 7.7 0.560 16.8 1421.5 30 26 -19 1 1291.2 1323.2 1.19 50.0 0.554 23.9 1808.7 31 28 -11 1 1554.0 1618.7 1.26 95.1 0.536 24.2 2061

prok_001 -> 360.x

Choose point group symmetry (4 or 422)

Is it Point group 422 Or Point group 4?

P4 H, K,L-H,-K,L-K, H,L K,-H,L

H, K,-L H,-K,-L K, H,-L-K,-H,-L

P422

Test existenceof 4-fold

symmetry

Test existence of4-fold Symmetry

andPerpendicular

2-fold symmetry

j observations of the reflection 30 22 6

<I>= (550 + 500 + 543) / 3 = 531

j H K L I

1 30 22 6 550

2 -30 -22 6 500

3 30 -22 -6 543

Rmerge= S|Ij-<I>|

SIj

Discrepancy between symmetry related reflections

S|Ij-<I>| = |550-531|+|500-531|+|543-531|

= 19 + 31 + 12 = 62SIj = 550 + 500 + 543

= 1593

Rsym = 62/1593 = 0.126 = 12.6%

?

H, K, L=

-H,-K, L=

H,-K,-L

Merge

Average (merge together) symmetry related reflections.

Plane L=0 b*

a*

?-H,-K, L=-K, H, L= K,-H, L= H, K,-L= H,-K,-L= K, H,-L=-K,-H,-L=-H,-K,-L= K,-H,-L=-K, H,-L=-H, K, L=-K,-H, L= K, H, L

Plane L=0 b*

a*

K,-H,-L

H,K,L

-H,K,-L

-K,-H,L

K,H,L

-H,-K,L

H,-K,-L

-K,H,-L

Discrepancy between symmetry related reflections (Rsym) increases with increasing resolution. Why?

Shell Rsym

100-5.0Å 0.04

5.0-3.0Å 0.06

3.0-2.0Å 0.08

2.0-1.7Å 0.15

Statistics are analyzed as a function of resolution (N shells).

Average I/s decreases with increasing resolution

High resolution shells with I/s <2 should be discarded.

Shell I/sI100-5.0 Å 20.0

5.0-3.0 Å 10.0

3.0-2.0 Å 7.0

2.0-1.7 Å 3.0

SIGNAL TO NOISE RATIO (I/s)

COMPLETENESS?

What percentage of reciprocalLattice was measured for a givenResolution limit?

Better than 90% I hope.

Shell completeness

100-5.0Å 99.9%

5.0-3.0Å 95.5%

3.0-2.0Å 89.0%

2.0-1.7Å 85.3%

Overall 92.5%

Assignment

FILM

X-ray beam

Layer line screen blocks diffracted rays from upper layers ( that is, l≠0)

1,0,0 reflection

1,0,0

1/l

(0,0,0) origin of reciprocal lattice,Also known asx-beam, y-beam

crystal

Da*

a*

1,0,1

Knowing the orientation of the reciprocal lattice allows prediction of the position

of each reflection on the detector

a

b

a

b

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

b*a*

1/l(0,0)

(0,-1)

(0,-2)

(0,-3)

(-1,3)

X-ray beam

crystal

detector

What’s the h,k,l of this spot?

3 lattice p

oin

ts in

a* d

irection

2 lattice points in

b* direction

For a given spot on the film, we just trace the diffracted ray back to the reciprocal lattice point (h,k,l)

The answer is HKL=3,2,2

What parameters must be defined to complete this construction?

2,0,0

3,0,-1

4,0,-25,0,-3

5,0,-45,0,-5

4,0,-63,0,-7

2,0,-7

1,0,-80,0,-8

-1,0,-8

Indexing

Assign an h,k,l coordinate to each reflection of the first image.

Indices h,k,l are coordinates of the reflections, analogous to how atom positions are described by coordinates x,y,z.

X-ray scattering from a 2D crystal

ab

CrystalXrays Detector

Michael Sawaya

Indexing a crystal requires understanding of the relationship between three key elements:1) crystal (Bragg planes)2) the reciprocal lattice3) and the pattern of recorded reflections.Let's take this opportunity to review the diffraction experiment.Here is a crystal. How many unit cells?How many 5's in the asymmetric unit?Plane group?We expect X-rays to diffract from electrons in the fives and intercept the detector.

Diffracted intensities arise as if reflecting from families of planes --

William Bragg.

(0,1) planes

ab

Crystal(1,1) planes

(2,1) planes

(3,1) planes

(0,2) planes

(1,2) planes

(2,2) planes

(3,2) planes

Xrays Detector

(0,1) planesCrystal

(1,1) planes

(2,1) planes

(3,1) planes

(0,2) planes

(1,2) planes

(2,2) planes

(3,2) planes

ab

Xrays Detector

What are the chances of observing reflections from all these planes in a single orientation?

A) ExcellentB) Zero

Crystal(1,1) planes

ab

Xrays Detector

The planes must be oriented w.r.t the X-ray beam such that the difference in path length of each

scattered ray is nl.

Difference in path lengths is not nl.Scattered waves are out of phase. Total destructive interference. (1,1) reflection not observed in this crystal orientation.

2dsinq

Reciprocal Lattice

a*

b*

ab

CrystalXrays Detector1/l

Sphere of reflection shows the relationship between Bragg plane (HKL) and location of reflection (HKL) on

detector

RECIPROCAL LATTICE

a*

b*

(0,1) planes

length=1/d0,1

(0,1)

ab

Crystal DetectorXrays

(0,1) Bragg planes produce (0,1) reciprocal lattice point

RECIPROCAL LATTICE

a*

b*

(1,1) planes

length=1/d1,1

(1,1)

ab

(0,1)Crystal DetectorXrays


RECIPROCAL LATTICE

a*

b*

(2,1) planes

length=1/d2,1

(1,1)(0,1)

ab



RECIPROCAL LATTICE

a*

b*

(3,1) planes

length=1/d3,1

(1,1)(0,1)

ab

(2,1)(3,1)



RECIPROCAL LATTICE

a*

b*

(0,2) planes

length=1/d0,2

(1,1)(0,1)

(2,1)(3,1)

(0,2)

ab



RECIPROCAL LATTICE

a*

b*

(1,2) planes

length=1/d1,2

(1,1)(0,1)

(2,1)(3,1)

(0,2)(1,2)

ab



RECIPROCAL LATTICE

a*

b*

(2,2) planes

length=1/d2,2

(1,1)(0,1)

(2,1)(3,1)

(2,2)

(0,2)(1,2)

ab



RECIPROCAL LATTICE

a*

b*

(3,2) planes

length=1/d3,2

(1,1)(0,1)

(2,1)(3,1)

(2,2)

(0,2)(1,2)

(3,2)

ab



RECIPROCAL LATTICE

a*

b*

(1,1)(0,1)

(2,1)(3,1)

(2,2)

(0,2)(1,2)

(3,2)

ab


And so on... to fill out reciprocal lattice.

(1,0)(2,0)

(3,0)

(1,-2)(2,-2)

(3,-2)

(2,-1)(1,-1)

(3,-1)

(0,-1)

(0,-2)

(-1,1)

(-1,2)

(-1,0)

(-1,-2)

(-1,-1)

(-2,1)

(-2,2)

(-2,0)

(-2,-2)

(-2,-1)

(-3,1)

(-3,2)

(-3,0)

(-3,-2)

(-3,-1)

In this crystal orientation Which reflections will appear on the detector?

RECIPROCAL LATTICE

a*

b*

(1,1) planes

(1,1)

ab


What operation must we perform to the crystal in order to observe the (1,1) reflection?

33°

Michael Sawaya

We want to bring the reciprocal lattice point 1,1 into contact with sphere of reflection.Translate? noRotate? yesHow do we rotate reciprocal lattice?We do it by Rotating the crystal.

RECIPROCAL LATTICE

(1,1) planes

a*

b*


Rotate crystal to observe (1,1) reflection.

ab

(1,1) reflection is located here!

Michael Sawaya

reciprocal lattice rotates around the origin. It is fixed at the point where xrays exit the sphere of reflection.

ab ab(1,1) planes


A crystal rotation that brings the reciprocal lattice point (HKL) in contact with sphere of reflection, satisfies Bragg’s Law for

reflection (HKL).

q1/l

d*

d*/2 = 1/l•sinq

2d•sin q = l

}

(1,1) reflection is located here!

RECIPROCAL LATTICE

ab ab(1,1) planes


Indexing - We observe the location of a reflection on the detector. Which set of Bragg planes produced it?

There is a reflection located here!

What is it’s index?

RECIPROCAL LATTICE

a*

b*

How many increments of a* and b* is this point from the origin?d*=ha*+ kb*+ lc*

Michael Sawaya

I illustrated how the sphere of reflection can be used to determine the location of a reflection H,K,L from a given set of Bragg Planes (H,K,L).Indexing uses the same construct but approching from the opposite direction.In indexing, we start with the location of a reflection on the detector, then use the sphere of reflection to determine which set of Bragg planes produced it.The solution is trivial, if you know the unit cell parameters and orientation of the crystal.Just trace the diffraction vector from the film to the point where it crosses the sphere of reflection. The distance from this point to the origin is d*.How many increments of a* and b* is this point from the origin?d*=ha*+ kb*+lc*How can we use these concepts to index our data set?Let's review our method of data collection.

1/l

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We collected a series of images, each covering 1° rotation. Here we rotate 15° per image.

5°

DCF

X,Y

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We collected a series of exposure while crystal rotates.

13°

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We collected a series of exposure while crystal rotates.

15°

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Map onto reciprocal lattice

Image rotation1 0-15°

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Image rotation1 0-15°2 15-30°

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Image rotation1 0-15°2 15-30°3 30-45°

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°5 60-75°

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°5 60-75°6 75-90°

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°5 60-75°6 75-90°7 90-105°

a*

b*

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°5 60-75°6 75-90°

ab

a*

b*

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Image rotation1 0-15°2 15-30°3 30-45°4 45-60°5 60-75°6 75-90°

(1,-2)(2,-2)

(3,-2)

(0,-2)(-1,-2)

(-2,-2)(-3,-2)

(2,-1)(1,-1)

(3,-1)

(0,-1)(-1,-1)

(-2,-1)(-3,-1)

(1,0)(2,0)

(3,0)

(-1,0)(-2,0)

(-3,0)

(1,1)(0,1)

(2,1)(3,1)

(-1,1)(-2,1)

(-3,1)

(2,2)

(0,2)(1,2)

(3,2)

(-1,2)(-2,2)

(-3,2)

ab

a*

b*

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ab

3D reciprocal lattice is projected on 2D detector.

detector

3 steps of data reduction• Indexing

– Assign H,K,L values to each reflection

– We learn unit cell parameters• a,b,c, a,b,g• Identify which of

14 Bravais Lattices

• Integration– Sum up the number

of X-ray photons that intercepted the detector for each reflection, H,K,L.

– We learn the intensity values for each recorded reflection, H,K,L

• Scale and Merge– Average together

reflections related by symmetry

– Calculate scale factor for each image to minimize discrepancies between measurements of symmetry related reflections.

– We obtain the final symmetry averaged data set.

– We learn the space group symmetry

H K L I s 37 7 1 28.5 9.2-37 -7 1 30.1 10.9 7 37 1 37.4 13.3 37 7 -1 28.7 10.6-37 -7 -1 25.9 9.7 37 7 2 337.4 13.3 37 7 3 98.5 10.6 37 7 4 25.9 10.7

H K L I s 37 7 1 30.1 10.7 37 7 2 337.4 13.3 37 7 3 98.5 10.6 37 7 4 25.9 10.7

Outline 1• Overview –We will process diffraction data in 3 steps. Briefly:

– 1) indexing- assign coordinates (h,k,l) to each reflection in the data set

– 2) integration- extract intensity values from the diffraction images for each reflection, h,k,l

– 3) scaling and merging -average together the multiple intensity measurements related by symmetry.

• Indexing is the most challenging step of the three.– The concept of indexing sounds trivial –locate spots on the

image and assign them coordinates on the reciprocal lattice. But, complexity arises from the fact that the diffraction pattern has 3 dimensions and the detector used for data collection has only 2 dimensions. The 3D reciprocal lattice is projected on a 2D surface in our diffraction experiment.

– Recall that the Fourier transform of a 3D crystal is a 3D reciprocal lattice. What are the names of the three coordinates used to index atoms in crystal space? (x,y,z) What are the names of the three coordinates used to index reflections in diffraction space? (h,k,l) What are the repeating unit dimensions along x,y,z? along h,k,l?

– Indexing would be trivial if our diffraction images captured an undistorted view of the 3D reciprocal lattice like these. Ideally, reflections would be divided neatly into sections, aligned with reciprocal cell axes and without breaks in the pattern, as shown. Does our data come like this? No. It looks like these images: misoriented, distorted from projection, and discontinuous. It is possible to collect undistorted images of the 3D lattice using a precession camera, but unfortunately, it is inefficient and time consuming to collect data with a precession camera.

– Our task of indexing is to assign h,k,l values to spots in our diffraction images. In so doing, we can map them onto an undistorted 3D reciprocal lattice, computationally—as shown here.

• Indexing concepts– We are going to use a program, Denzo, to help us index the

thousands of reflections that we recorded. However, I would also like to show you how to index by inspection so you gain an intuitive feeling for indexing.

– 1) Take one of the many diffraction images that we recorded and eliminate from it the distortion due to projection of the 3D pattern onto a 2D detector. We can do this by taking into consideration the curvature of Ewald’s sphere of reflection. Recall that a reflection is recordable only when the corresponding planes in the crystal are oriented in the beam in such a way that satisfies Bragg’s law. That is, the photons reflected from the planes differ in path length precisely by integer multiples of the wavelength. This condition is satisfied when the reciprocal lattice point crosses the sphere of reflection. So, all the reflections recorded on the film originate from a curved surface. Eliminating the distortion from projecting this curved surface onto a 2D detector involves re-introducing the 3rd dimension and restoring the curvature to the diffraction pattern. Like this. We can now see undistorted, but sparsely populated, 3D reciprocal lattice. Show orthogonal views.

– 2) Identify the sets of evenly spaced rows and columns in the reciprocal lattice. Draw a set of evenly spaced lines through columns of spots. Draw a set of evenly spaced lines through rows of spots.

– 3) Find the reciprocal cell lattice parameters. Draw the vector representing this repeat distance, a*, between columns. Draw the vector representing the repeat distance, b*, between rows.

Outline 2– 4) Note the angle between a* and b*? This is gamma. – 5) Note the relationship between the lengths of a* and b*, if

any.– 6) Narrow down the possibilities of choice of Bravais

lattice. Note, current info suggests either primitive tetragonal or cubic.

– 7) Identify the third unit cell length, c*. Identify the sets of evenly spaced columns in the reciprocal lattice. Draw the vector representing this repeat distance, c*, between columns. What angle does c* make with a* and b*? These are beta and gamma.

– 8) Is the crystal primitive tetragonal or cubic?– 9) What is the index for this reflection?– Review. In our intuitive indexing process, the following

parameters were either measured or derived from measurements. Which of the following parameters were measured directly from the diffraction experiment? Which parameters were derived from the measurements.

– Why might indexing fail?• Indexing in practice

– Autoindexing using Denzo. It uses a computer algorithm to perform the same analysis as we just did here.

– Identify spots. Uses an algorithm, not very good. You use your eyes and pattern recognition.

– Identify rows and columns of spots. It performs a systematic search of all orientations of the image for periodicities among spot locations. It does this by projecting each spot on a line. In certain orientations the lengths of these projections will differ by integer multiples of a constant, corresponding to the reciprocal cell length.

– A one-dimensional Fourier transform of the vector lengths identifies this increment, a*.

– Our task of indexing is to assign h,k,l values to spots in our diffraction images so we can map them onto an undistorted 3D reciprocal lattice, computationally—as shown here.

• Indexing concepts– We are going to use a program, Denzo, to help us

index the thousands of reflections that we recorded. However, I would also like to show you indexing by inspection so you become familiar with the concepts of indexing.

– Distortions due to projection of the 3D pattern onto a

– 2) integration- for each reflection we sum the intensity values for all pixels within the reflection boundary

– 3) scaling and merging –for each unique reflection (h,k,l) average all intensity measurements of (h,k,l) and its symmetry mates. Determine scale factors to obtain the best agreement between I(h,k,l) values measured from different images.





Documents

The goal of Data Processing From a series of diffraction images, obtain the intensity ( I ) and standard deviation ( ( I )) for each reflection, hkl