Data Collection and Reflection Sphere- Data completeness

- Rsym

Structure factors and Phase Problems- Structure factor

- Fourier transform

- Electron density maps

- Temperature factor

Data Collection and Reflection Sphere

Rotation method and data completeness

Diffraction pattern of Im7Orthorhombic unit cell I222: a = 75.1Å, b = 50.5Å , c = 45.4Å , 90o

H0L plane 6 0 0 295587.00 3564.27 8 0 0 476981.00 3687.43 10 0 0 77658.04 1728.93 12 0 0 76207.25 1882.20 14 0 0 54967.11 1862.80 16 0 0 661011.00 1368.47 18 0 0 28076.70 3147.65 20 0 0 129816.00 3182.27 22 0 0 87852.19 2770.71 24 0 0 165364.00 3898.54 26 0 0 42694.30 2164.06 28 0 0 6260.15 1092.76 30 0 0 2112.36 924.42 32 0 0 3942.09 1122.29 1 1 0 157984.00 1667.29 2 1 0 32571.87 836.24 3 1 0 28649.71 855.31 4 1 0 21480.39 767.46 5 1 0 8201.62 590.49 6 1 0 272377.00 3366.99 7 1 0 137746.00 2150.82 8 1 0 389.05 393.24 9 0 3 3219.17 1215.68 10 1 0 45167.83 1463.33 . . . . . . . . . .

H K L I (or F) I (or F)

Rsym of a data set

For n independent reflections and i observations of a given reflection.

is the average intensity of the i observation.

Summary of reflections intensities and R-factors by shells R-linear = SUM ( ABS(I - <I>)) / SUM (I) R-square = SUM ( (I - <I>) ** 2) / SUM (I ** 2) Chi**2 = SUM ( (I - <I>) ** 2) / (Error ** 2 * N / (N-1) ) )

In all sums single measurements are excluded Shell Lower Upper Average Average Norm. Linear Square limit Angstrom I error stat. Chi**2 R-fac R-fac 99.00 5.37 16608.9 410.3 204.5 1.867 0.035 0.039 5.37 4.26 15793.6 343.3 122.8 1.720 0.034 0.037 4.26 3.72 12598.2 279.4 109.3 1.885 0.038 0.038 3.72 3.38 11247.7 276.9 113.6 2.278 0.040 0.033 3.38 3.14 7337.3 206.6 107.0 2.437 0.050 0.038 3.14 2.95 2931.1 109.7 77.7 2.603 0.079 0.062 2.95 2.81 1669.0 89.0 75.0 2.658 0.117 0.102 2.81 2.68 1311.6 86.1 76.5 2.610 0.137 0.125 2.68 2.58 934.0 87.0 81.8 2.644 0.190 0.176 2.58 2.49 722.5 90.0 86.2 2.771 0.241 0.208All reflections 7312.6 201.3 106.2 2.273 0.045 0.038

Example of data scaling from DENZO-Scale packMonoclinic P21 unit cell

Data collection and processing statistics for nuclease-ColE7/DNA complex

Space group P21

Cell dimensions a = 59.4 Å, b = 46.5 Å, c =78.5 Å = 90.0o, = 90.4o, = 90.0o Resolution (Å) 2.5 Observed reflections 61,443 Unique reflections 15,097 Completeness- all data (%) 96.7 Completeness- last shell (%) 83.2 (resolution range, Å) (2.58 – 2.49) Rsyma- all data (%) 4.0 Rsyma- last shell (%) 23.6 I/(I)- all data (50.0 – 2.49 Å) 36.4 I/(I)- last shell (2.58 – 2.49 Å) 7.8

aRsym =ΣhklΣi∣Ii(hkl)﹣ ＜ I(hkl) ＞∣ / Ii(hkl).

Examples of diffraction statistics

X-Ray diffraction statistics for the nuclease-ColE7 

Data collection and processing Space group P1Cell dimensions a = 37.7 Å, b = 39.2 Å, c =53.2 Å

= 96.8o, = 103.9o, = 91.4o Resolution (Å) 2.1Observed reflections 53,594Unique reflections 16,335Completeness- all data (%) 95.0Completeness- last shell (%) 92.7(resolution range, Å) (2.18-2.10)Rsym- all data (%)a 4.4

Rsym- last shell (%) 17.3

I/(I)- all data (50.0-2.1 Å) 22.5I/(I)- last shell (2.18- 2.10 Å) 7.0

Structure Factor F(h,k,l)

f(j): Scattering factor of each atom

(x, y, z): Atom position Phase contribution from each atom

Vector diagram of structure factor

F = A + iB = |F| cos + i |F| sin |F| (cos +i sin ) = |F| exp i

|F|2 = (A2 + B2)1/2

= I

Friedel’s law: Ihkl = I-h-k-l |Fhkl| = |F-h-k-l| hkl = --h-k-l

Fourier Transform

Back Fourier Transform

F(h) is the Fourier transform of f(x)Units of h are reciprocals of the units of x, i.e, h(sec) and x(sec-1)

h(1/Å) and x(Å).

Now we will try and represent this function in terms of sine waves:

Fourier Transforms in Crystallography

Let us consider an imaginary one-dimensional crystal. There are three atoms in the unit cell; two carbons and an oxygen. The electron density in the unit cell looks like this:

Note that the sum of the three sine-waves is a good approximation to the original unit cell. Thus we can see that the unit cell can be represented quite well using only three sine-waves, given the correct choice of frequency, amplitude and phase.

Now we will look at the Fourier Transform of the same unit cell. Note that the result consists of a series of peaks, the largest of which are at 2, 3 and 5 on the x-axis. These correspond exactly to the sine-wave frequencies which we used to reconstruct the unit cell. If you look carefully you will also see that the heights of the peaks correspond to the amplitudes of the three waves:

An atom, and its Fourier Transform:

A molecule, and its Fourier Transform:

Fourier Transform

An lattice, and its Fourier Transform:

A crystal, and its Fourier Transform:

The Fourier transform of the crystal is thus the product of the molecular transform and the reciprocal lattice. This is the diffraction pattern.

Crystal (real space) Diffraction pattern (reciprocal space)

Back Fourier Transform:

hkl cannot be measured Phase Problem

A duck, and its Fourier

Transform:

Electron density Diffraction pattern

Amplitude is represented by colour saturation and brightness, while phase is given by hue. This is illustrated in the following figure.

Note:In the above figure

Real Space Reciprocal Space

The Fourier Duck

If we only have the low resolution terms of the diffraction pattern, we only get a low resolution duck

Crystallographic Interpretation:There is considerable loss of detail. At low resolution, your atomic model may reflect more what you expect to see than what is actually there. Note the ripples around the duck. These could be mistaken for solvation shells.

Resolution:Minimum interplanar spacing of the real lattice for the corresponding reciprocal lattice point (reflection) that is being measured. It is directly related to the optical definition in which it is the minimum distance that two objects can be apart and still be seen as two separate objects. Resolution is normally quoted in Ångstroms (Å).

Resolution makes differenceThere is no question that a model of phenylalanine (the 6-ring structure) can be correctly placed into the 1.2 Å data. This still can be done with confidence in the 2 Å case, but at 3 Å we already observe a deviation of the centroid of the ring from the correct model.

Resolution limitation Upper limit of = 90o

2d sin = n The maximum d =

If we only have the high resolution terms of the diffraction pattern, we see only the edges of the duck.

Crystallographic Interpretation:It may be difficult to distinguish the solvent and protein regions. This will handicap both building and density modification. You may need to contour at a lower level inside the protein. There are prominent artifact peaks in the solvent.

Do not omit your low resolution data. Collect it and use it.

If a segment of data is missing, features perpendicular to that segment will be blurred.

Crystallographic Interpretation:Helices parallel to the missing data axis will become cylinders. Beta sheets parallel may merge into a flat blob. Beta sheets perpendicular to the missing data may be very weak. You could get into a lot of trouble with anisotropic temperature factors in this case.

Try not to omit any data. Collect it and use it.

Here is our old friend; the Fourier Duck, and his Fourier transform.

And here is a new friend; the Fourier Cat and his Fourier transform.

Now we will mix them up. Let us combine the the magnitudes from the Duck transform with the phases from the Cat transform. (You can see the brightness from the duck and the colours from the cat).

Fduck + cat

Using the magnitudes from the Cat transform and the phases from the Duck transform.

Fcat + duck

Crystallographic Interpretation:In X-ray diffraction experiments, we collect only the diffraction magnitudes, and not the phases. Unfortunately the phases contain the bulk of the structural information. That is why crystallography is difficult.

A Tail of Two CatsSuppose we are trying to reconstruct the image of a cat, and have the fourier magnitudes for it.Only have Fcat but no cat

Since for this experiment we only have the magnitudes of the transform it is represented in monochrome, and we cannot reconstruct the image

We have an image which we know is similar to the missing cat. This image is of a Manx (tailless) cat. Since we have the image, we can calculate both the Fourier magnitudes and phases for the manx cat.

Fc and c

One simple method to try and restore the image of the cat is to simply calculate an image using the known Fourier magnitudes from the cat transform with the phases from the manx cat.

Fo + c

Despite the fact that the phases contain more structural information about the image than the magnitudes, the missing tail is restored at about half of its original weight. This occurs only when the phases are almost correct. The factor of one half arises because we are making the right correction parallel to the estimated phase, but no correction perpendicular to the phase (and <cos2>=1/2). There is also some noise in the image.

Fourier Map (|Fo| map)

This suggests a simple way to restore the tail at full weight: apply double the correction to the magnitudes. An image is therefore calculated with twice the magnitude from the desired image minus the magnitude from the known image. The resulting image shows the tail at full weight. However the noise level in the image has also doubled.

2Fo-Fc

c

Crystallographic Interpretation:Often in crystallography we have an incomplete model. Thus we have observed structure factor magnitudes for a complete molecule, but a model (from which we can calculate both magnitudes and phases) for only part of it. In this case the missing portion of the model may be reconstructed by use of the appropriate Fourier coefficients. The first attempt to reconstruct the cat's tail above corresponds to an |Fo| map, the second to a 2|Fo|-|Fc| map.

Electron density around the Ca2+ binding site of TFs-glucanase. Stereo views of Fourier map (2Fo-Fc) calculated using the final model phases contoured at 1.5 σ above the mean density at 1.7 Å resolution. The calcium ion is bound with nearly perfect decahedral geometry coordinated to the O1 atom of Asn164, three backbone carbonyl oxygen atoms (Asn164, Asn189 and Gly222) and three water molecules.

Electron density maps- Fourier maps

|F(hkl)| = |Fobs|; or 2|Fobs| - |Fcal|

(hkl) Calculated phases from model

Stereo views of the omitted (2Fo-Fc) electron density map of Fis mutants R71L superimposed onto the final model in the BC turn region. The map was calculated by omitting the mutated residue and all the atoms within a 3 Å spherical shell, and contoured at 1.0 of the average electron density.

Omit Maps (or Difference maps)

Electron density maps and models demonstrating the quality and resolution of the initial phases. Maps were calculated using solvent-flattened MAD phases at 2.5 Å resolution, contoured at 1 σ.

Maps calculated by experimental phases

|F(hkl)| = |Fobs| (hkl) Observed phases from different methods (MAD, MIR, etc)

Scattering Factor and Temperature Factor

The normalized scattering curves have been fitted to a 9-parameter equation by Don Cromer and J. Mann. Knowing the 9 coefficients, a(i), b(i) and c, and the wavelength, we can calculate the scattering factor of each atom at any given scattering angle.

In actual cases there will be an additional weakening of the scattering power of the atoms by the so called Temperature-, B- , or Debye-Waller factor. This exponential factor is also angle dependent and effects the high angle reflections substantially

The B-factor can be related to the mean displacement of a vibrating atom <u> by the Debye-Waller equation B=8pΠ2<u>2

As u is given in Å the unit for B must be Å2. In most protein structures it suffices to assume the isotropic average displacement.

1. In space group P212121, the equivalent positions are (x, y, z), (-x+1/2, -y, z+1/2), (-x, y+1/2, -z+1/2) and (x+1/2, -y+1/2, -z). Show that the reflections of (h, 0, 0), (0, k, 0) and (0, 0, l) are absent if h = odd, k = odd, and l = odd, respectively.

Why there are systematic absences in the diffraction pattern?

For (h, 0, 0) reflections:

F(hkl) = ∑ fj exp [2i (hx + ky + lz)]

= ∑ fj [ exp(2ihx) + exp(-2ihx+ih) + exp(-2ihx) + exp(2ihx+ih)]

Because exp(a+b) = exp(a) exp(b), therefore,

= ∑ fj { exp(2ihx) [1 + exp(ih)] + exp (-2ihx) [1+ exp(ih)]}

exp(ih) = cos(h) + i sin(h) = -1, if h = odd integer, cos(h) = -1 and sin(h) = 0

= ∑ fj {exp(2ih) [(1 + (-1)] + exp (-2ihx) [1+ (-1)]}

= 0