Amino Acids and Proteins Questions and Answers

Answer: D

The one that is not such a method is pyrolysis.

Pyrolysis means breaking a sample down by heat, so it is a deliberate thermal decomposition and the opposite of what the question asks. It is used in pyrolysis-GC-MS to study polymers from their fragments.

MALDI (matrix-assisted laser desorption/ionisation), electrospray ionisation and plasma desorption are all 'soft' ionisation methods developed to bring large, fragile molecules such as proteins and nucleic acids into the gas phase as intact ions, without thermal decomposition.

Answer: C

Disulfide bonds are broken with β-mercaptoethanol.

β-Mercaptoethanol is a thiol reducing agent: it reduces each S-S bridge between cysteine residues to two free -SH groups, so that subunits held together by disulfide bonds separate and the chains unfold fully. Dithiothreitol does the same job.

SDS is a detergent that denatures proteins and coats them with negative charge, but it does not break covalent disulfide bonds. TEMED catalyses the polymerisation of the gel, and bisacrylamide is the cross-linker that forms the gel network.

Answer: A

The pI of aspartic acid is 2.945.

The isoelectric point is the average of the two pK values on either side of the form with no net charge. Aspartic acid has two acidic groups, the alpha-carboxyl (pK₁ = 1.99) and the side-chain carboxyl (pK_R = 3.90), so the neutral form lies between them:

\[ \begin{aligned} pI &= \frac{1.99 + 3.90}{2} \\ &= 2.945 \end{aligned} \]

5.945 averages pK₁ and pK₂, the rule for an amino acid with no ionisable side chain. 6.9 averages pK_R and pK₂, and 7.895 is half the sum of all three.

Answer: B

Cyanogen bromide gives 2 fragments.

CNBr cleaves the peptide bond on the carboxyl side of methionine residues, converting the methionine to homoserine lactone. The oligopeptide contains one methionine, at position 6, so it is cut once:

Gly-Ala-His-Arg-Asp-Met (as homoserine lactone) and Cys-Lys-Val-Leu.

One cut always gives two pieces, so 1 fragment would mean no cleavage, and 3 or 4 fragments would need two or three methionines. The basic residues His, Arg and Lys are trypsin's targets, not cyanogen bromide's.

Answer: A

The key gives 8.25. The stem does not say what kind of group the third pK belongs to, so this rests on the usual reading, which I could not confirm from a source.

The isoelectric point is the average of the two pK values on either side of the neutral (zwitterion) form. If X has a basic side chain, the groups lose protons in the order alpha carboxyl (2.5), side chain (7.5), alpha amino (9.0). The neutral form sits between the last two, so:

\[ pI = \frac{pK_2 + pK_3}{2} = \frac{7.5 + 9.0}{2} = 8.25 \]

Had the side chain been acidic, the neutral form would lie between the first two pK values and the pI would be (2.5 + 7.5)/2 = 5.0, which is option B. A side chain with a pK of 7.5 is closer to a basic group such as an imidazole than to a carboxylic acid (about 4), which is why 8.25 is the intended answer. Options C and D do not come from averaging any pair of the three values.

Answer: A

Intrinsic protein fluorescence comes from aromatic amino acids.

Tryptophan, tyrosine and, much more weakly, phenylalanine absorb ultraviolet light near 280 nm and emit fluorescence. Tryptophan dominates, emitting around 340 nm, and its emission shifts as its surroundings change, which is why intrinsic fluorescence is used to follow protein folding and ligand binding.

Charged and branched-chain amino acids have no conjugated ring system and do not fluoresce. Disulfide bonds absorb weakly and can quench fluorescence, but they are not fluorophores.

Answer: A

The number of possible sequences is \((20)^{146}\).

Each of the 146 positions can be filled by any of the 20 standard amino acids, independently of the others, so the choices multiply:

\[ 20 \times 20 \times \cdots \times 20 = 20^{146} \]

\(20 \times 146\) adds the choices instead of multiplying them. \((3)^{146}\) confuses amino acids with something like the three nucleotides of a codon, and \((146)^{20}\) inverts the base and the exponent: it would be the count for a 20-residue chain built from 146 kinds of unit.

Answer: D

The amino acid is proline.

In proline the side chain loops back and bonds to the backbone nitrogen, forming a five-membered pyrrolidine ring. The ring locks the backbone angle (phi) near -60 degrees and leaves the nitrogen with no hydrogen to donate, so proline breaks alpha-helices and often sits at turns.

Alanine (a methyl group), leucine (a branched hydrocarbon chain) and serine (a hydroxymethyl group) all have open side chains attached only to the alpha-carbon, so their backbones rotate freely.