Biochemistry I Fall Term, 2000

Lecture 23: X-ray Diffraction and NMR Spectroscopy of Proteins

Assigned reading in Campbell: None
Lecture and notes by Dr. Gordon Rule

Take the Review Quiz on Lecture 23 concepts.

X-Ray Diffraction in Structure determination:

1. Proteins must be crystallized in a regular lattice, just like NaCl.
2. No real limitations as to the size of structures (the structures of some viruses are known).
3. X-rays are scattered by electrons, the amount of scattering is proportional to the number of electrons.
4. Interference between x-rays that are scattered from atoms in different locations change the amplitude and the phase of the scattered x-rays.
5. It is only possible to measure the intensity of the scattered X-rays, but both the intensity and phase are important for the calculation of the electron density.
6. Technique produces an electron density map into which atoms are placed. An example is shown below:



7. The resolution is related to the scattering angle. High scattering angles imply the measurement of close spacing between atoms, therefore locating the position of atoms to high resolution.
8. The resolution affects the amount of information that can be obtained:

Resolution and Structural Knowledge:

The effect of Resolution on the electron density (and consequently on the ability to locate atoms) is shown below:

The position of hydrogens can be defined in very high resolution x-ray diffraction studies. For example, the proton that is shared between the His and Asp residues in serine proteases can be seen in this electron density map.

NMR SPECTROSCOPY:

1. Form of Spectroscopy that detects transition of nuclear spin angular momentum.
2. ¹H, ¹³C, ¹⁵N, ³¹P can be detected.
3. Absorption frequency depends on the local environment (electron density)

• Type of bonding (e.g. aromatic versus aliphatic)
• Charge
• Polarity

1. Absorption frequency, f, is converted to chemical shift:
2. Spectra are plotted as intensity versus chemical shift.

pK_a Determination:

Chemical shift of ionizable groups depends on whether the group is protonated or not. Many different nuclei can be observed for this experiment, such as:

A: In the case of Glu or Asp, the ¹³C resonance lines from the sidechain carboxyl group are often observed. Typical chemical shifts for carbonyl carbons are 175 ppm

B: In the case of Histidine, the ring protons (either those on the nitrogen or those attached to the carbon) will show a chemical shift change due to protonation of the ring.

1. The NMR spectrum of unprotonated His would show an absorbance line at 7 ppm (unprotonated)
2. The NMR spectrum of protonated His would show an absorbance line at 8 ppm (protonated)

The NMR spectrum of a mixture of protonated and deprotonated species would show a single NMR line at:

Where f_protonated = fraction protonated, f_unprotonated = fraction unprotonated. An averaged chemical shift is usually observed because the ionizable group (e.g. His sidechain) is rapidly protonated and deprotonated.

The chemical shift of the mixture of protonated and unprotonated species can be used to obtain the fraction of each species (remember f_protonated + f_unprotonated = 1). The pK_a can be determined by measuring the NMR chemical shift at different pH values. The following formula is used for this determination:

Example: His Titration:

The following is a typical graph of the Chemical shift of a His ring proton as a function of pH. At low pH, His is fully protonated and a chemical shift of 8.0 is observed. At high pH, His is fully deprotonated and a chemical shift of 7.0 is observed.

When the chemical shift is half-way between the protonated and deprotonated states (7.5 ppm in this case), then f_protonated = f_unprotonated, = 0.5 (This is the same as saying R = 1, as defined in the lecture on acids and bases.

At this point in the titration:

The important point to remember is: at the halfway point in the titration, the pH equals the pK_a!

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