Biochemistry — Proteins & Enzymes
Metabolic pathways, enzymes, proteins
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Showing 1–10 of 27 questions in Proteins & Enzymes
Which scenario best describes non-competitive enzyme inhibition kinetically?
A Inhibitor binds to both free enzyme and enzyme-substrate complex with equal or different affinities, reducing Vmax while Km remains unchanged
B Inhibitor only binds to free enzyme, increasing Km
C Inhibitor irreversibly denatures the enzyme
D Inhibitor reduces substrate availability in the reaction mixture
Correct Answer:  A. Inhibitor binds to both free enzyme and enzyme-substrate complex with equal or different affinities, reducing Vmax while Km remains unchanged
EXPLANATION

In non-competitive inhibition, the inhibitor binds to an allosteric site on both E and ES, preventing product formation. This decreases Vmax (fewer active enzymes) while Km remains unchanged (substrate binding affinity unaffected).

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A protein exhibits a β-sheet structure rich in proline residues. Why is this structurally problematic?
A Proline lacks a free NH group to form backbone hydrogen bonds in β-sheets
B Proline causes excessive protein cross-linking
C Proline increases hydrophobicity excessively
D Proline participates in peptide bond cleavage
Correct Answer:  A. Proline lacks a free NH group to form backbone hydrogen bonds in β-sheets
EXPLANATION

Proline is an imino acid with its side chain bonded to the backbone nitrogen, eliminating the NH group needed for β-sheet hydrogen bonding between strands. High proline content disrupts β-sheet formation, commonly found in turns and loops instead.

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A mutation changes a hydrophobic valine residue to a charged aspartate in the hydrophobic core of a globular protein. What is the most likely consequence?
A Protein instability and misfolding due to disruption of hydrophobic interactions
B Enhanced enzyme activity
C Increased protein solubility with maintained function
D Formation of additional disulfide bonds
Correct Answer:  A. Protein instability and misfolding due to disruption of hydrophobic interactions
EXPLANATION

Substituting a nonpolar residue with a charged, hydrophilic one in the protein core disrupts critical hydrophobic interactions that stabilize the tertiary structure, leading to misfolding, aggregation, or degradation.

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How does the proteasome recognize proteins marked for degradation in the ubiquitin-proteasome system?
A By recognizing polyubiquitin chains (Lys48-linked) attached to lysine residues on target proteins
B By scanning for hydrophobic amino acid sequences
C By identifying proteins with exposed disulfide bonds
D By sensing changes in protein charge
Correct Answer:  A. By recognizing polyubiquitin chains (Lys48-linked) attached to lysine residues on target proteins
EXPLANATION

E3 ubiquitin ligases catalyze the attachment of ubiquitin chains (primarily through Lys48 linkages) to lysine residues on target proteins. The 19S proteasomal subunit recognizes these polyubiquitin chains and unfolds the protein for degradation.

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A student observes that an enzyme shows sigmoidal kinetics instead of Michaelis-Menten kinetics. What does this indicate?
A The enzyme has multiple subunits and exhibits cooperative binding
B The enzyme is completely inhibited
C The enzyme lacks specificity
D The substrate concentration is too high
Correct Answer:  A. The enzyme has multiple subunits and exhibits cooperative binding
EXPLANATION

Sigmoidal (S-shaped) kinetics indicate positive cooperativity, typical of allosteric enzymes with multiple subunits (e.g., aspartate transcarbamoylase). Binding of substrate to one subunit increases affinity in others.

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What is the physiological significance of the Cori cycle?
A It transfers reducing equivalents between liver and muscles
B It recycles lactate from anaerobic muscle metabolism back to glucose in liver, maintaining blood glucose during exercise
C It oxidizes fatty acids exclusively in the liver
D It synthesizes glucose from amino acids only
Correct Answer:  B. It recycles lactate from anaerobic muscle metabolism back to glucose in liver, maintaining blood glucose during exercise
EXPLANATION

The Cori cycle (glucose-lactate cycle) allows muscles undergoing anaerobic glycolysis to produce lactate, which is transported to liver and converted back to glucose via gluconeogenesis, maintaining blood glucose homeostasis during intense exercise.

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Which of the following is true regarding enzyme specificity?
A All enzymes show absolute specificity for one substrate
B Enzymes show varying degrees of specificity: absolute, group, and linkage specificity
C Enzyme specificity is determined solely by size of active site
D Specificity can be completely overcome by increasing substrate concentration
Correct Answer:  B. Enzymes show varying degrees of specificity: absolute, group, and linkage specificity
EXPLANATION

Enzyme specificity varies: absolute (one substrate only), group (substrates with similar functional groups), linkage (specific types of bonds), and stereochemical (stereoisomers). Specificity results from 3D active site structure and orientation of catalytic residues.

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What is the structural role of zinc in alcohol dehydrogenase?
A Zinc acts as a cofactor for NAD+ binding
B Zinc coordinates the substrate and facilitates hydride transfer
C Zinc stabilizes the enzyme-NAD+ complex only
D Zinc has no catalytic role, only structural role
Correct Answer:  B. Zinc coordinates the substrate and facilitates hydride transfer
EXPLANATION

Alcohol dehydrogenase contains catalytic zinc that coordinates the hydroxyl group of ethanol/aldehyde substrate, activating it for hydride transfer to NAD+, and structural zinc that maintains protein stability.

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In the urea cycle, which enzyme catalyzes the condensation of carbamoyl phosphate and ornithine?
A Carbamoyl phosphate synthetase I
B Ornithine transcarbamylase
C Argininosuccinate synthase
D Arginase
Correct Answer:  B. Ornithine transcarbamylase
EXPLANATION

Ornithine transcarbamylase (OTC) catalyzes the condensation of carbamoyl phosphate (formed by CPS I) with ornithine to form citrulline, the second step of the urea cycle. OTC deficiency is the most common urea cycle disorder.

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What is the mechanism of allosteric regulation in phosphofructokinase (PFK)?
A AMP and ADP activate; ATP and citrate inhibit through binding at regulatory sites distinct from active site
B Substrate acts as allosteric activator
C Direct competitive inhibition by ATP only
D Covalent modification by phosphorylation
Correct Answer:  A. AMP and ADP activate; ATP and citrate inhibit through binding at regulatory sites distinct from active site
EXPLANATION

PFK exhibits allosteric regulation where AMP/ADP (signals of low energy) activate the enzyme, while ATP/citrate (signals of high energy/biosynthesis) inhibit it by binding to allosteric sites, changing enzyme conformation.

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