Home Subjects Biochemistry

Biochemistry

Metabolic pathways, enzymes, proteins

73 Q 3 Topics Take Test
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Difficulty: All Easy Medium Hard 41–50 of 73
Topics in Biochemistry
All Proteins & Enzymes 100 Carbohydrates 100 Lipids 78
Q.41 Hard Carbohydrates
A competitive athlete is found to have a deficiency in muscle phosphorylase (McArdle disease). During intense exercise, which metabolic consequence is PRIMARY?
A Inability to mobilize muscle glycogen causing severe fatigue and cramps
B Excessive glucose accumulation in muscles
C Uncontrolled glycogenolysis leading to hypoglycemia
D Increased ATP production from fatty acid oxidation alone
Correct Answer:  A. Inability to mobilize muscle glycogen causing severe fatigue and cramps
EXPLANATION

Muscle phosphorylase deficiency prevents glycogen breakdown, depriving muscles of glucose-1-phosphate during exercise, causing severe energy crisis, fatigue, cramps, and myoglobinuria.

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Q.42 Hard Carbohydrates
A 6-month-old infant develops hypoglycemia, hepatomegaly, and lactic acidosis after feeding. Genetic testing reveals glucose-6-phosphatase deficiency (Type I Glycogen Storage Disease). Why does this cause lactic acidosis?
A Inability to convert glucose-6-phosphate to free glucose blocks gluconeogenesis and glycogenolysis, leading to pyruvate and lactate accumulation
B Increased glycolysis due to high glucose-6-phosphate levels
C Direct inhibition of lactate dehydrogenase
D Increased fructose metabolism producing excess lactate
Correct Answer:  A. Inability to convert glucose-6-phosphate to free glucose blocks gluconeogenesis and glycogenolysis, leading to pyruvate and lactate accumulation
EXPLANATION

G6Pase is the final enzyme in both gluconeogenesis and glycogenolysis. Its deficiency traps glucose-6-phosphate, forcing it through glycolysis and the pentose phosphate pathway, increasing pyruvate and lactate production.

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Q.43 Hard Carbohydrates
The Warburg effect describes increased glycolysis in cancer cells even in the presence of oxygen. Which enzyme is typically upregulated in cancer cells to support this?
A Phosphofructokinase-2 (PFK-2)
B Pyruvate kinase M2 (PKM2)
C Both A and B
D Pyruvate dehydrogenase
Correct Answer:  C. Both A and B
EXPLANATION

In the Warburg effect, PFK-2 and PKM2 are upregulated. PFK-2 produces fructose-2,6-bisphosphate (a potent PFK-1 activator), while PKM2 (pyruvate kinase isoform) is upregulated in cancer cells. PKM2 also has non-glycolytic functions in cancer metabolism.

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Q.44 Hard Carbohydrates
In glycolysis, which step is irreversible under physiological conditions and requires a different enzyme during gluconeogenesis?
A Glucose to Glucose-6-phosphate
B Fructose-6-phosphate to Fructose-1,6-bisphosphate
C Phosphoenolpyruvate to Pyruvate
D Glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate
Correct Answer:  B. Fructose-6-phosphate to Fructose-1,6-bisphosphate
EXPLANATION

The PFK reaction is highly exergonic (ΔG°' = -14.2 kJ/mol) and is essentially irreversible. During gluconeogenesis, fructose-1,6-bisphosphatase catalyzes the reverse reaction. This is a major control point in carbohydrate metabolism.

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Q.45 Hard Carbohydrates
Pasteur effect refers to the inhibition of glycolysis by oxidative phosphorylation. In terms of ATP and citrate, which statement is correct?
A High ATP and acetyl-CoA inhibit phosphofructokinase
B High ATP and citrate inhibit phosphofructokinase
C Low ATP activates phosphofructokinase
D Both B and C are correct
Correct Answer:  D. Both B and C are correct
EXPLANATION

PFK is inhibited by high ATP, citrate (signal of sufficient acetyl-CoA), and low pH. In aerobic conditions, efficient oxidative phosphorylation produces sufficient ATP, reducing need for glycolysis (Pasteur effect). Citrate is an allosteric inhibitor of PFK.

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Q.46 Hard Carbohydrates
A patient with galactosemia has a deficiency in which enzyme?
A Galactokinase
B Galactose-1-phosphate uridyltransferase
C UDP-galactose-4-epimerase
D Lactase
Correct Answer:  B. Galactose-1-phosphate uridyltransferase
EXPLANATION

Classic galactosemia results from deficiency of galactose-1-phosphate uridyltransferase (GALT), preventing conversion of galactose-1-phosphate to UDP-galactose. Leads to accumulation of galactose-1-phosphate and galactitol (in polyol pathway), causing cataracts, intellectual disability, and liver damage.

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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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