Bragg's law (n-lambda = 2d sin-theta) relates the angle of diffraction to the spacing of lattice planes in the crystal. If a crystal with larger unit cell dimensions is analyzed, how does the diffraction pattern change?
AFewer reflections are observed because larger cells diffract less
BThe diffraction spots are more closely spaced — larger real-space dimensions correspond to finer sampling in reciprocal space, producing more reflections within the same angular range
CThe diffraction spots move farther apart because larger cells scatter at larger angles
DThe diffraction pattern is unchanged because unit cell size does not affect diffraction
This is the reciprocal relationship: large in real space = small in reciprocal space. A crystal with a larger unit cell has more closely spaced lattice planes (more d-values), which according to Bragg's law diffract at more closely spaced angles. The result is more diffraction spots within any given angular range, which means more data (more Fourier components) and ultimately a more finely sampled electron density map. This is why large macromolecular crystals (with unit cells of 50-200 Angstroms) produce many more reflections than small-molecule crystals, but each reflection is weaker because the scattering power is distributed among more spots.
Question 2 True / False
The phase of each diffraction spot carries more information about the structure than the amplitude.
TTrue
FFalse
Answer: True
This seemingly surprising statement has been demonstrated mathematically and computationally. If you compute an electron density map using the phases from protein A and the amplitudes from protein B, the resulting map looks like protein A — not protein B. Phases determine the overall features of the electron density (the positions of atoms and molecules), while amplitudes modulate the contrast and fine detail. This is why the phase problem is so critical: losing the phases loses most of the structural information. It also explains why approximate phases (from molecular replacement or initial experimental phasing) are often sufficient to produce interpretable electron density maps that can be improved by iterative refinement.
Question 3 Short Answer
Explain in simple terms why the electron density and the diffraction pattern are related by a Fourier transform.
Think about your answer, then reveal below.
Model answer: The electron density in a crystal is a periodic function (it repeats with the unit cell). Any periodic function can be decomposed into a sum of sine and cosine waves of different frequencies, amplitudes, and phases — this decomposition is the Fourier series. Each diffraction spot corresponds to one of these component waves: its position in the pattern gives the frequency (the spacing of the corresponding set of lattice planes), its intensity gives the amplitude squared, and its phase (unmeasured) gives the phase of that wave component. The electron density is reconstructed by summing all these waves (inverse Fourier transform). Each reflection adds one wave to the sum, and the more reflections included (higher resolution), the sharper and more detailed the resulting electron density map.
An intuitive analogy: the diffraction pattern is the 'frequency spectrum' of the electron density, just as an audio spectrum shows the frequency components of a sound wave. To reconstruct the sound (electron density), you need both the amplitude and phase of each frequency component (reflection). Crystallography gives you the amplitudes but not the phases — hence the phase problem.