1. Announcements Grades (for Mid-term) in Web portal
Review your Mid-term today or tomorrow in 328/326 LLP.
(Worthwhile top review test so you can see what you
should know for final.)
Homework
Read chapter on DNA and X-ray Diffraction. (It was assigned last Tuesday.
There will be a quiz next Tuesday, in class.
Written homework will be assigned by the end of Friday. Do next Thursday.

2. 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

3. 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
The distance between vertices of the diamonds corresponds to the interbase spacing. Then, the fact that there are 10 layer lines (corresponding to periodicity of the helix) per diamond tells you that there are 10 basepairs per helical period.

4. 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)

5. 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

6. Gayle Wittenberg 2003
Notice where blue and red dots totally interfere: only at Layer line 4l/P. Therefore no signal there.

7. 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
The distance between vertices of the diamonds corresponds to the interbase spacing. Then, the fact that there are 10 layer lines (corresponding to periodicity of the helix) per diamond tells you that there are 10 basepairs per helical period.

8. The phase problem: Coherence length
Young’s Double Slit: Diffraction and Interference
The phase problem: Coherence length
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
Recall that the phase between two (or more) points must be well-defined in order to have interference.

9. The phase problem: Coherence length
If the phase of a light wave is well defined at all times (oscillates in a simple pattern with time and varies in a smooth way in space at any instant), then the light is said to be coherent.
If, on the other hand, the phase of a light wave varies randomly from point to point, or from moment to moment (on scales coarser than the wavelength or period of the light) then the light is said to be incoherent.
For example, a laser produces highly coherent light. In
a laser, all of the atoms radiate in phase.
An incandescent or fluorescent light bulb produces incoherent light. All of the atoms in the phosphor of the bulb radiate with random phase. Each atom oscillates for about 1ns, and produces a wave about 1 million wavelengths long.

10. X-ray sources are incoherent!!
They are not a laser!
Only the most sophisticated synchrotrons are slightly coherent.
So how did Franklin et al. produce their diffraction pattern?
Maximum possible resolution ~λ
X-rays: 10 – 0.1 nm ~Interatomic distances
They used a lead and also somewhat monochromatic light.
Coherent over a length of P.
For Copper electrode, 1.54A

11. Or use synchrotron radiation
Equipment for x-ray diffraction data collection
Of 3-D crystals
Or use synchrotron radiation
Notice that you need the ability to rotate the crystal!
The basic equipment necessary to conduct an X-ray diffraction experiment consists of an X-ray source, a means for orienting the crystal and a means for detecting (measuring) the diffraction spots.

12. Structure determination by X-ray diffraction
Not just 1dimensional crystal, but three dimensional
Strategy & objectives
Determine location and intensities of electron densities in crystal
Model the positions of atoms in 3 dimensions.
(You have the 1d amino acid or nucleotide sequence.)
Thread it through the electron density.
In general, can’t see H’s but know heavier atoms: C, O, N, P

13. Detection of X-ray diffraction
(X-ray film or electronic)
Diffraction pattern
Each diffraction spot is the result of interference of all X-rays with the same diffraction angle emerging from all atoms
The stronger the signal, the more intense the scattered beam, and
the greater the electron-density that scattered it

14. What does the diffraction pattern tell us?
Each spot contains a small amount of information about the position of each atom in the unit cell
To produce all possible spots, X-rays must be allowed to strike the crystal from many different directions: crystal is rotated in the X-ray beam
Positions & intensities of each spot are the basic experimental data
The ~ 25,000 diffracted beams diffracted by myoglobin contain scattered
X-rays from each of the protein’s ~1,500 atoms