Amino acids and protein chemistry sample question 2

contains nearby negative charges that stabilize the protonated imidazolium form.

Amino acids and protein chemistry sample question 3

competing with the histidine side chains for coordination sites on the immobilized Ni²⁺.

Amino acids and protein chemistry sample question 4

It has partial double-bond character, holding the six atoms around it in one plane.

Amino acids and protein chemistry sample question 5

The tail has no stable folded structure to lose, so heating has no cooperative transition to disrupt.

MCAT · Chemical and Physical Foundations · Biochemistry

Amino acids and protein chemistry

Amino acids and protein chemistry is part of Biochemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include amino acid and peptide charge versus pH, isoelectric point and migration, side-chain classification and placement and peptide bond and sequence notation.

Common traps: giving the side chain the wrong charge; forgetting the terminal groups of a peptide; taking protonated as charged for every group.

Sample questions

Question 1
Which amino acid is most likely to be found on the lipid-facing surface of a transmembrane α-helix?
- A.
  Leucine
- B.
  Aspartate
- C.
  Lysine
- D.
  Serine
Answer: A. Leucine
The lipid-facing surface of a membrane-spanning helix contacts hydrocarbon chains, so nonpolar residues such as leucine, isoleucine, valine, and phenylalanine predominate there. Aspartate and lysine are charged and would be enormously costly to bury in the bilayer, and serine's hydroxyl prefers water or hydrogen-bonding partners.
Question 2
A histidine in an enzyme's active site has an apparent side-chain pKa of 7.5, compared with about 6.0 for free histidine. Its microenvironment most likely:
- A.
  contains nearby negative charges that stabilize the protonated imidazolium form.
- B.
  contains nearby positive charges that stabilize the protonated form.
- C.
  is nonpolar, which favors the charged form of the side chain.
- D.
  is fully exposed to water, which raises every pKa.
Answer: A. contains nearby negative charges that stabilize the protonated imidazolium form.
A higher pKa means the protonated (cationic) form persists to a higher pH, so something stabilizes the positive charge: nearby carboxylates (Asp or Glu) do this electrostatically. Nearby positive charges would repel the proton and lower the pKa, a nonpolar pocket disfavors the charged form and also lowers it, and full exposure to water simply returns the pKa toward the free-histidine value.
Question 3
A recombinant protein carrying a hexahistidine tag is purified on a nickel-affinity column and eluted with a buffer containing imidazole. The imidazole works by:
- A.
  denaturing the protein so that it no longer fits the column's pores.
- B.
  competing with the histidine side chains for coordination sites on the immobilized Ni²⁺.
- C.
  raising the pH so that the histidines become negatively charged.
- D.
  cleaving the tag from the rest of the protein.
Answer: B. competing with the histidine side chains for coordination sites on the immobilized Ni²⁺.
The tag binds because the imidazole rings of histidine coordinate the immobilized Ni²⁺; free imidazole at high concentration occupies those sites and displaces the protein, which then washes off intact. The protein is neither denatured nor cleaved, and histidine's imidazole never becomes negatively charged in aqueous buffer.
Charge of a Lysine-Rich Histone Tail
Histone tails extend from the nucleosome core and make contact with DNA. A synthetic peptide models one such tail. It is 17 residues long and has the sequence Ser-Lys-Arg-Gly-Lys-Ala-Asp-Lys-Pro-Lys-Thr-Arg-Lys-Gly-Lys-Ala-Arg, with a free amino terminus and a free carboxyl terminus. It therefore contains 6 lysines, 3 arginines and 1 aspartate. In solution the free peptide has no stable secondary structure, and a circular dichroism scan shows no change between 20 °C and 90 °C.
Table 1 Ionizable groups of the tail peptide
Group Number pKa
α-Carboxyl (C-terminus) 1 2.0
Aspartate side chain 1 3.9
α-Amino (N-terminus) 1 9.0
Lysine side chain 6 10.5
Arginine side chain 3 12.5
At pH 7.4 each group whose pKa is well above 7.4 is essentially fully protonated and each group whose pKa is well below 7.4 is essentially fully deprotonated. Protonated amino and guanidinium groups carry +1 and deprotonated carboxyl groups carry –1. The researchers bind the peptide to double-stranded DNA and find that the dissociation constant increases about 50-fold when NaCl is raised from 0.05 M to 0.5 M. They also plan to digest the peptide with trypsin and to run the products on a cation-exchange column of carboxymethyl cellulose, which carries fixed negative charges, at pH 7.4.
Question 4
Which statement correctly describes the peptide bond between Lys-2 and Arg-3 of the tail peptide?
- A.
  It rotates freely, which allows the tail to adopt any backbone shape.
- B.
  It has partial double-bond character, holding the six atoms around it in one plane.
- C.
  It is formed between the side-chain amino group of Lys-2 and the carboxyl group of Arg-3.
- D.
  It is formed by the addition of a water molecule across the bond between the two residues.
Answer: B. It has partial double-bond character, holding the six atoms around it in one plane.
Resonance between the amide nitrogen lone pair and the carbonyl gives the C–N bond partial double-bond character, so it is planar and rotation is restricted. A contradicts this. C is wrong because backbone bonds join α-amino and α-carboxyl groups. D is wrong because formation releases water, whereas hydrolysis adds it.
Question 5
The globular core of a histone shows a sharp, cooperative loss of structure on heating, whereas the isolated tail peptide shows no change in its circular dichroism signal between 20 °C and 90 °C. Which explanation is most consistent with these observations?
- A.
  The tail peptide contains disulfide bonds that hold its structure at high temperature.
- B.
  The strong positive charge of the tail prevents any thermal denaturation of its peptide bonds.
- C.
  The tail peptide's peptide bonds are stronger than those in the globular core.
- D.
  The tail has no stable folded structure to lose, so heating has no cooperative transition to disrupt.
Answer: D. The tail has no stable folded structure to lose, so heating has no cooperative transition to disrupt.
A melting transition reports loss of folded secondary and tertiary structure. A disordered peptide has none, so no signal changes. The sequence has no cysteine (A is wrong), charge does not protect structure (B) and peptide bonds are identical covalent amides in both (C).

Table 1 Ionizable groups of the tail peptide
Group	Number	pKa
α-Carboxyl (C-terminus)	1	2.0
Aspartate side chain	1	3.9
α-Amino (N-terminus)	1	9.0
Lysine side chain	6	10.5
Arginine side chain	3	12.5

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Amino acids and protein chemistry

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Charge of a Lysine-Rich Histone Tail

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