MCAT practice · Biological and Biochemical Foundations · 1A Proteins and enzymes · 5D Biomolecule structure
Point mutations and the stability of a small protein
An original practice passage written for this site, not an AAMC item. Exam pace is about 8 minutes for a passage like this.
Small single-domain proteins usually unfold in one cooperative step, so the folded (F) and unfolded (U) states are the only species present in appreciable amounts. The stability of such a protein is the standard free energy of unfolding, ΔG°unf, for the reaction F ⇌ U. Because ΔG°unf is typically only 20–60 kJ/mol, a single amino acid substitution can shift it substantially.
Researchers studied P110, a 110-residue single-domain protein from a soil bacterium. Its crystal structure shows that Leu36 is fully buried in the hydrophobic core, Lys72 lies on the solvent-exposed surface with its side chain making no contacts with other residues, and Cys22 and Cys88 are joined by the protein's only disulfide bond. The team expressed wild-type P110 and three variants: L36A, K72A, and the double mutant C22S/C88S. Each protein was purified and dialyzed into 20 mM phosphate buffer, pH 7.0.
Unfolding was monitored by circular dichroism (CD) at 222 nm, a wavelength at which the signal reports α-helical secondary structure. In thermal experiments, samples (10 µM) were heated from 30 °C to 90 °C at 1 °C/min, and the CD signal was converted to the fraction of molecules unfolded (Figure 1). The melting temperature, Tm, was taken as the midpoint of each transition. Heating and then cooling wild-type P110 restored more than 95% of its original CD signal. In chemical experiments, samples at 25 °C were equilibrated overnight in 0–8 M urea. The urea concentration at the midpoint of each transition (Cm) was recorded, and ΔG°unf in the absence of urea was obtained by linear extrapolation of ΔG°unf versus urea concentration to 0 M. All four proteins gave the same slope of that line, −4.6 kJ·mol⁻¹·M⁻¹ (stability falls by 4.6 kJ/mol for each molar of urea). Results are summarized in Table 1; each value is the mean of three independent preparations, and standard deviations were below 0.4 °C for Tm and below 0.8 kJ/mol for ΔG°unf.
| Protein | Tm (°C) | Cm, urea (M) | ΔG°unf (kJ/mol) |
|---|---|---|---|
| Wild type | 68.4 | 5.2 | 24 |
| L36A | 54.1 | 3 | 13.8 |
| K72A | 67.9 | 5.1 | 23.5 |
| C22S/C88S | 57.2 | 3.7 | 17 |
- Wild type
- L36A
- C22S/C88S
Data are illustrative, constructed for practice.
1. Based on Table 1, by how much does the L36A substitution change the free energy of unfolding of P110 at 25 °C?
- A. Wrong: 0.5 kJ/mol is the change caused by K72A (24.0 − 23.5), the surface substitution, not L36A.
- B. Wrong: 7.0 kJ/mol is the change caused by removing the disulfide (24.0 − 17.0), a different variant's row.
- C. Right: ΔΔG°unf = ΔG°unf(wild type) − ΔG°unf(L36A) = 24.0 − 13.8 = 10.2 kJ/mol; the variant is less stable by that amount.
- D. Wrong: 13.8 kJ/mol is the stability of the L36A variant itself, not the change relative to wild type.
2. Which explanation best accounts for the finding that L36A destabilizes P110 far more than K72A does?
- A. Wrong: this reverses a known property; alanine is among the strongest helix-forming residues, and the passage gives no evidence that the helix is disrupted.
- B. Wrong: the passage states that the Lys72 side chain makes no contacts with other residues, so it forms no salt bridge to lose.
- C. Wrong: leucine's side chain is a nonpolar hydrocarbon with no hydrogen-bond donor; its contribution is hydrophobic packing.
- D. Right: truncating a buried leucine to alanine removes nonpolar contacts that exist only in the folded state and leaves a cavity in the core. A solvent-exposed lysine is surrounded by water whether the protein is folded or not, so removing it barely changes ΔG°unf.
3. The C22S/C88S variant could be less stable because the disulfide is lost or because serine itself is poorly tolerated at positions 22 and 88. Which additional experiment would best distinguish these possibilities?
- A. Wrong: changing the temperature shifts every protein's stability and does not separate the effect of the cross-link from the effect of the serine side chains.
- B. Wrong: the single mutant still lacks the disulfide and still carries a serine, so both explanations change together and the result cannot separate them.
- C. Right: reducing the wild-type disulfide removes the cross-link without changing the sequence. If reduced wild type melts near 57 °C, like C22S/C88S, the loss of the disulfide explains the destabilization; if it melts near 68 °C, the serines are responsible.
- D. Wrong: changing the spectroscopic probe changes what structure is reported, not the cause of the lower stability; both explanations predict the same result.
4. The researchers plan one more variant at position 36. Which replacement for Leu36 would be expected to destabilize P110 the least?
- A. Wrong: glycine removes the entire side chain, leaving an even larger cavity than alanine, and it also gives the unfolded chain more backbone freedom, which favors unfolding.
- B. Right: isoleucine is a nonpolar, branched aliphatic residue of the same size as leucine, so it can fill the buried site and keep most of its hydrophobic contacts.
- C. Wrong: aspartic acid would bury a negative charge in the nonpolar core, away from water, which is strongly destabilizing.
- D. Wrong: lysine would bury a positive charge in the nonpolar core; a buried charge that cannot be hydrated costs a great deal of free energy.
5. A sample of the L36A variant in buffer without urea is held at 54.1 °C until it reaches equilibrium. Which value is closest to ΔG°unf for this sample at that temperature?
- A. Right: 54.1 °C is the Tm of L36A, where half the molecules are unfolded, so K = [U]/[F] = 1 and ΔG°unf = −RT ln K = 0.
- B. Wrong: 10.2 kJ/mol is the destabilization caused by the mutation at 25 °C, not the stability of the variant at its melting temperature.
- C. Wrong: 13.8 kJ/mol is the stability of L36A at 25 °C; ΔG°unf falls as temperature rises toward Tm.
- D. Wrong: 24.0 kJ/mol is the stability of wild-type P110 at 25 °C, a different protein at a different temperature.
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