Read Protein Structure and Stability

Rigid proline backbone disrupts α-helical structure. Proline's cyclic side chain restricts backbone rotation and its peptide-bond nitrogen lacks an H that would normally donate an α-helical hydrogen bond. Consequently proline often disrupts regular α-helices and creates kinks, making it less common in transmembrane helical segments. Statements B, C and D capture these structural consequences.

Hydrogen-deuterium exchange probes conformational flexibility. At low temperature, fewer amide protons exchange in the thermophilic enzyme, indicating stronger/more persistent hydrogen-bond protection: A. The thermophilic protein remains unusually rigid at low temperature and needs very high temperature to attain exchange/flexibility comparable with the active mesophilic enzyme; hence D. Statements claiming lesser rigidity are inconsistent with the graph.

Removing urea and reductant permits native conformation recovery. Urea disrupts non-covalent interactions including hydrogen bonds, while β-mercaptoethanol reduces disulfide bonds. These treatments disrupt secondary and tertiary structure while leaving the peptide-bond sequence (primary structure) intact. Recovery of activity after removal demonstrates spontaneous refolding.

Choose the method from the evidence needed: X-ray crystallography resolves a protein's three-dimensional atomic structure, while mass spectrometry, chromatography and gels do not. α-helix hairpins, β-hairpins and β-α-β units are supersecondary motifs. Denaturation disrupts the folded, functional globular state and commonly makes proteins insoluble and inactive. When comparing fluorescence curves before and after urea treatment, use the wavelength with the largest separation between the curves: it gives the most sensitive signal for the conformational change.

Arginine has a positively charged guanidinium side chain. A salt bridge needs an opposite charge, so glutamic acid's negatively charged carboxylate is the correct partner. Lysine is also positive, while valine and methionine are nonpolar. Choose B.

CNBr cleaves on the carbonyl side of methionine. First mark every methionine, then ask whether a cut after it actually separates another piece of chain. A methionine at the C-terminus creates no additional fragment; three internal cuts create four fragments. Choose B.