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MCAT · Biological and Biochemical Foundations · Biochemistry
Protein structure and enzyme function is part of Biochemistry in the MCAT Biological and Biochemical 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 michaelis-Menten and reciprocal plots, inhibition type identification, allosteric and covalent regulation and residue substitution and structure.
Common traps: reading the x-intercept as -Km instead of -1/Km; mixing Vmax with the rate at a single substrate concentration; equating low Km with low affinity.
Question 1
At a substrate concentration of 2 mM, an enzyme with a Vmax of 80 μmol/min has a velocity of 40 μmol/min. What is its Km?
1 mM
2 mM
40 mM
80 mM
Answer: B. 2 mM
B: Km is the substrate concentration that gives half of Vmax, and 40 is half of 80, so Km is 2 mM. A halves the concentration for no reason, and C and D confuse velocity values with concentrations.
Question 2
Replacing a residue in the middle of an α-helix with proline most directly disrupts which interaction?
The hydrophobic contacts among R groups that pack the helix against the rest of the protein, because proline is polar.
The hydrogen bonds between backbone C=O and N–H groups four residues apart, because proline has no amide hydrogen.
The disulfide bonds between cysteine side chains, because proline is a cysteine analog.
The covalent peptide bonds along the chain, because proline cannot form an amide link.
Answer: B. The hydrogen bonds between backbone C=O and N–H groups four residues apart, because proline has no amide hydrogen.
Secondary structure is held by backbone hydrogen bonds, and proline's ring nitrogen has no N–H to donate and also restricts rotation, so it breaks a helix. A mislocates the effect in tertiary packing and mislabels proline as polar, C invents a cysteine analogy, and D is wrong because proline forms normal peptide bonds.
Question 3
Which statement correctly contrasts the concerted (MWC) and sequential (KNF) models of cooperativity?
In MWC all subunits share one state at a time, so hybrid forms do not occur; in KNF subunits change one at a time, so hybrids occur.
In MWC each subunit switches state on its own when it binds ligand, whereas in KNF all of the subunits switch state together in a single step.
MWC can explain positive cooperativity only, whereas KNF can explain negative cooperativity only.
In both models every subunit must be in the same conformation at all times.
Answer: A. In MWC all subunits share one state at a time, so hybrid forms do not occur; in KNF subunits change one at a time, so hybrids occur.
The MWC model allows only symmetric T4 and R4 forms, with ligand shifting the T/R balance; KNF lets binding induce a change in the bound subunit that is passed on to neighbors, so mixed forms occur. B reverses the models; C assigns exclusive effects wrongly (KNF can give either kind); D describes only MWC.
Question 4
A valine buried in the hydrophobic core of a globular protein is replaced by a different amino acid. Which replacement is most likely to disrupt folding?
Isoleucine, a slightly larger nonpolar residue that packs into the core with little change
Leucine, a nonpolar residue of similar size
Aspartate, whose side chain is negatively charged at physiological pH
Alanine, a smaller nonpolar residue
Answer: C. Aspartate, whose side chain is negatively charged at physiological pH
A charged side chain buried in a nonpolar core cannot make favorable interactions and pays a large desolvation penalty, so Asp is the most destabilizing. Ile and Leu are similar hydrophobic residues that fit the core with little change (A, B), and Ala only leaves a small cavity, a milder change (D).
Question 5
An enzyme has a bell-shaped activity curve with an optimum at pH 7.5. Activity requires one group (pKa 6.0) to be deprotonated and another (pKa 9.0) to be protonated. A mutation lowers the pKa of the second group to 7.0. What is the new approximate pH optimum?
6.5
7.5
8.0
9.0
Answer: A. 6.5
The optimum lies midway between the two pKa values: (6.0 + 7.0)/2 = 6.5 (A). 7.5 is the original optimum (B); 8.0 pairs the new 7.0 with the old 9.0 (C); 9.0 is the old pKa of the second group, not an optimum (D).
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