1. Mid-term

2. Mid-term by Question

3. One of the premier techniques of Structural Biology
X-ray Diffraction
One of the premier techniques of Structural Biology
Can get Angstrom resolution of nearly every atom in biological macromolecules—now- a-days get up to (beyond?) 2 MD (Ribosomes).
1st one: Whale myoglobin
--won Nobel Prize, 1962, Max Perutz, Sir John Cowdery Kendrew.

4. Watson & Crick 1953 Minor Groove 1.2 nm = 3P/8 Major Groove
Diameter 2 nm
Interbase distance
0.34 nm = P/10
X pattern
Period
P = 3.4 nm
05_08_major_minor_gr.jpg

5. Fraunhofer X-ray Diffraction
Maximum possible resolution ~λ
X-rays: 10 – 0.1 nm ~Interatomic distances

6. X-ray Diffraction is elastic scattering
86,817 X-ray crystal structures of proteins, nucleic acids and other biological molecules have been determined
The scattering is elastic; the scattered X-rays have the same wavelength as the incoming X-ray. By contrast, inelastic X-ray scattering methods are useful in studying excitations of the sample, rather than the distribution of its atoms.

7. “Photo 51” – Rosalind Franklin 1952
X pattern
Layer Lines
Missing 4th layer
Diamond Pattern

8. Destructive Interference
X-rays behave as electromagnetic waves
They interfere
amplitude
wavelength λ
Constructive Interference
(Bright Spot)
“In Phase”
Destructive Interference
(Dark Spot)
ΔL = λ/2
“Out of Phase”

9. Constructive interference
ΔL = 0
ΔL = λ
ΔL = 2λ
Constructive interference
ΔL = nλ n = 0, 1, 2, ...

10. Destructive interference
ΔL = λ/2
ΔL = 3λ/2
ΔL = 5λ/2
Destructive interference
n’ = 0, 1, 2, ...

11. Constructive Interference (Bright Spot)
Destructive Interference
(Dark Spot)

12. Two-wave Interference Pattern
Interference depends on waves traveling different distances
Constructive interference
Destructive interference
Now, imagine that we have a source of waves; sound waves in this case, coming out of a speaker. The sound waves will travel radially away from the speaker.
Now, let’s imagine that we add another speaker, also generating sound waves of the same wavelength. The sounds waves coming out of these two speakers will interfere with one another! In fact, looking at the resulting pattern, you can visually SEE what’s going to happen. Where the two waves are in phase, the peaks and troughs line up perfectly and you get
Constructive interference. You can see this in the pattern as the lighter parts of the pattern. Alternatively, where the two waves are out of phase, the peaks of one wave line up with the troughs of the other,
And you get destructive interference. This corresponds to the regions of the pattern that look darker. This combination of light and dark regions corresponding to regions of constructive and destructive interference is what’s called an INTERFERENCE PATTERN,
Here you can see a similar sort of pattern generated by water waves. The regions that still look “wavy” are regions of constructive interference, while those regions where destructive interference occur look flat.
You can also see that if we change the relative distance between the speakers
The shape of the interference pattern changes, and the locations of the regions of constructive and destructive interference change. This is because what we are doing by moving the speakers is changing the relative distances that each wave must travel to get to each point,
And the resulting interference pattern depends upon wave traveling different distances to result in different phase changes. We will see exactly how this works in the coming slides!

13. X-rays behave as electromagnetic waves They diffract
Atom
Interference!
What if there are two (or more) atoms?

14. Young’s Double Slit: Diffraction and Interference
Bottom wave travels extra 1λ
Both waves travel same distance
To further study interference, let’s look at an experiment called Young’s double slit. Here we have two screens, the first of which has two holes, or slits, in it. We will hit this screen with planar waves, all in phase and with the same wavelength. In practice, this is usually done with laser light. Now, remember that each point on the wavefronts acts as a point source wavelet.
When the waves hit the screen, most of the wavefront is blocked, except for the portion of the wave that hits the slits; each of the slits then acts as a point source wavelet, radiating spherical waves towards the second screen.
BOTH of the slits will behave like this, and you can see that the two waves from each of the slits will interfere with each other, forming an interference pattern. Again, the areas that appear darker correspond to regions of destructive interference, where the light cancels out. The lighter
Areas correspond to regions of constructive interference, where the light from both waves add together. The result, when projected on the second screen, is what appears as a pattern of light and dark spots. This is because, for each of the light and dark spots, the two light waves travel different distances to get there, resulting in phase changes.
Along the center line, both waves travel the same distance, and the phase difference is zero, resulting in constructive interference.
At the next bright spot above the center line, the bottom wave has had to travel an extra distance to get there that corresponds to an extra wavelength phase shift, lambda.
Similarly, for the next bright spot below the center line, the top wave has had to travel an extra distance to get there that corresponds to an extra wavelength phase shift.
Top wave travels extra 1λ

16. Imagine a simple 1-D crystal...
ΔL = 0 Constructive
Bright Spot
X-rays

17. Destructive Interference (Dark Spots)
ΔL = λ/2 Destructive
Dark Spot θ P θ ΔL
ΔL = λ/2
Destructive Interference (Dark Spots)
Constructive Interference (Bright Spots)

19. Question about Crystal diffraction
In a NaCl crystal, the spacing between atoms is nm. Which of the following wavelengths could be used to see a clear diffraction pattern?
need λ < d
First interference maximum:
0.282 nm
A. λ = 0.1 nm B. λ = 1 nm C. λ = 10 nm
Need very short wavelength light: X-rays!

20. Gayle Wittenberg 2003
Θ, λ, P
Can get P

21. X-ray crystallography
Given X-ray wavelength λ, diffraction angles θ provide information about distance d between atoms in crystal
Crystalline fiber of DNA
“Photo 51” Rosalind Franklin θ
As long as λ < d, small features lead to large θ. BUT need regular ordering of atoms – i.e. a crystal!

22. 2-D crystals... Four simple arrays...
...and their diffraction patterns
Notice the inverse relationship between spacing
in the array and spacing in the diffraction pattern.
The array in (x,y,z) real space and the diffraction space (h,l,m) are Fourier Transform of each other. (Will learn about this.)
Rows of spots are perpendicular to original lines

24. The first three mask patterns
Basic pattern

25. A helix (flattened into 2 dimensions) is very similar to a zig zag
Periodic
Periodic
X-shape = Helix!

26. Question: DNA cross pattern
You discover a new structure of DNA in which the diffraction pattern is the same as the “normal” DNA in every respect EXCEPT that the cross makes a more acute angle α α P
New form of DNA α P
“Normal” DNA α α Δθ Δθ
Which statement regarding the new DNA structure must be true?
A. It cannot be a helix
B. The helix repeat distance P must be different
C. It must be a wider molecule

27. E
Question: What does increasing the diameter of the helix do to slope of
decreases the slope?
Decreases. H

28. Layer Lines: Helix period P Angle α: Helix radius
X pattern: Helix
Layer Lines: Helix period P
Angle α: Helix radius
10 layers lines/diamond: Interbase spacing P/10
Missing 4th Layer line? α

29. Missing lines → Additional Interference
Even if Blue is in phase with Blue, q P
Notice that both the red and blue dots are spaced 1P apart, but because they’re offset from each other by an amount q.
Two helices identical except the starting point, their phase.
Blue may be out of phase with Red
It turns out that 1/8P is
For 4th layer line, n = 4
This is mathematically the correct answer. However, P/8 would make the atoms collide: 10.4bp/8 = 1.3 bases apart. For 5th layer lines you get 5P/8, but this is the same as 3P/8

30. Continued next class

31. Answer, and turn in at the end of class.
Class evaluation
1. What was the most interesting thing you learned in class today?
2. What are you confused about?
3. Related to today’s subject, what would you like to know more about?
4. Any helpful comments.
Answer, and turn in at the end of class.