Biochemistry — Proteins & Enzymes
Metabolic pathways, enzymes, proteins
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Showing 1–10 of 100 questions in Proteins & Enzymes
An enzyme exhibits a Km of 2 mM and Vmax of 100 μmol/min. When substrate concentration is 6 mM and an allosteric inhibitor is added, the Vmax decreases to 50 μmol/min while Km remains unchanged. What type of inhibition is occurring?
A Competitive inhibition
B Non-competitive inhibition
C Uncompetitive inhibition
D Mixed inhibition
Correct Answer:  B. Non-competitive inhibition
EXPLANATION

Non-competitive inhibition decreases Vmax while keeping Km constant. This occurs when an inhibitor binds to a site other than the active site (allosteric site), preventing product formation regardless of substrate concentration. The Km value remains unchanged because substrate binding affinity is unaffected.

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A researcher studying protein folding observes that a newly synthesized polypeptide chain contains multiple disulfide bonds between cysteine residues. Which cellular compartment is most likely responsible for facilitating the formation of these disulfide bonds?
A Rough endoplasmic reticulum and Golgi apparatus
B Mitochondrial matrix
C Cytoplasm
D Lysosomal lumen
Correct Answer:  A. Rough endoplasmic reticulum and Golgi apparatus
EXPLANATION

Disulfide bonds are formed in oxidizing environments. The rough endoplasmic reticulum (RER) and Golgi apparatus maintain oxidizing conditions suitable for disulfide bond formation, unlike the reducing environment of the cytoplasm. The enzyme protein disulfide isomerase (PDI) facilitates this process in the ER lumen.

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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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Which proteolytic enzyme is responsible for activating trypsinogen to trypsin in the small intestine?
A Enteropeptidase
B Chymotrypsin
C Elastase
D Carboxypeptidase A
Correct Answer:  A. Enteropeptidase
EXPLANATION

Enteropeptidase (enterokinase), secreted by the duodenal mucosa, cleaves a specific peptide bond in trypsinogen to produce active trypsin, initiating the cascade of pancreatic protease activation.

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In a temperature vs. enzyme activity graph, why does enzyme activity decrease above the optimal temperature?
A The tertiary structure denatures and active site geometry is lost
B Substrate concentration decreases
C Cofactors are oxidized
D The enzyme converts to its inactive zymogen form
Correct Answer:  A. The tertiary structure denatures and active site geometry is lost
EXPLANATION

Above optimal temperature, increased thermal energy disrupts hydrogen bonds and hydrophobic interactions maintaining the 3D structure, causing denaturation and loss of catalytic activity.

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Q.10 Medium Proteins & Enzymes
Which prosthetic group is found in cytochrome c oxidase?
A Heme a and copper centers
B Only nicotinamide adenine dinucleotide
C Flavin adenine dinucleotide exclusively
D Iron-sulfur clusters only
Correct Answer:  A. Heme a and copper centers
EXPLANATION

Cytochrome c oxidase (Complex IV) contains heme a, heme a3, and copper centers (CuA and CuB) essential for electron transfer and oxygen reduction to water.

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