MCAT 2026 Biochemistry Practice Exam | Enzymes,
Metabolism & Molecular Biology Review | INSTANT PDF
DOWNLOAD
MCAT Biochemistry practice exam. This resource contains 200 high-yield questions covering enzymes,
kinetics, metabolic pathways (glycolysis, TCA, ETC, gluconeogenesis, glycogen metabolism, fatty acid
oxidation, amino acid metabolism), molecular biology (DNA replication, transcription, translation, gene
regulation), and lab techniques. Each question includes the correct answer and rationale. Use this exam to
solidify your biochemistry knowledge for MCAT success.
Key Topics Covered
• Enzymes & Kinetics – Michaelis-Menten, inhibition, catalytic mechanisms, cofactors
• Carbohydrate Metabolism – Glycolysis, gluconeogenesis, TCA cycle, oxidative phosphorylation, glycogen
synthesis/breakdown, pentose phosphate pathway
• Lipid Metabolism – Fatty acid oxidation, ketogenesis, fatty acid synthesis, cholesterol metabolism
• Amino Acid & Nitrogen Metabolism – Transamination, urea cycle, amino acid degradation, one-carbon
metabolism
• Molecular Biology – DNA replication, repair, transcription, RNA processing, translation, genetic code
• Gene Regulation – Operons, transcription factors, epigenetics, non-coding RNAs
• Biochemical Techniques – Chromatography, electrophoresis, centrifugation, spectrometry, blotting
Questions 1–200
1. Which of the following is true about the Michaelis constant (Km)?
A) It is the substrate concentration at which Vmax is reached.
B) It is the substrate concentration at which the reaction rate is half of Vmax.
C) It is a measure of the enzyme’s catalytic efficiency.
D) It increases in the presence of a competitive inhibitor.
Answer B: It is the substrate concentration at which the reaction rate is half of Vmax.
Rationale: Km is the substrate concentration at half-maximal velocity. It reflects affinity: lower Km = higher
affinity (except for certain cases).
2. A competitive inhibitor will:
A) Decrease Vmax and leave Km unchanged.
B) Increase Km and leave Vmax unchanged.
,C) Decrease both Vmax and Km.
D) Increase Vmax and decrease Km.
Answer B: Increase Km and leave Vmax unchanged.
Rationale: Competitive inhibitor binds active site, increasing apparent Km; at high substrate, Vmax is still
achievable. Vmax unchanged.
3. The enzyme enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate in
glycolysis. This reaction is an example of:
A) Isomerization
B) Dehydration
C) Oxidation-reduction
D) Phosphorylation
Answer B: Dehydration
Rationale: Enolase removes a water molecule (dehydration) from 2-phosphoglycerate, forming a double
bond in phosphoenolpyruvate.
4. In the pentose phosphate pathway, the primary purpose is to generate:
A) ATP and FADH2
B) NADPH and ribose-5-phosphate
C) Acetyl-CoA and CO2
D) Glucose-6-phosphate and pyruvate
Answer B: NADPH and ribose-5-phosphate
Rationale: PPP produces NADPH for biosynthesis and antioxidant defense, and ribose-5-phosphate for
nucleotide synthesis. No ATP is directly generated.
5. Which of the following statements about the TCA cycle is correct?
A) It occurs in the cytoplasm.
B) It produces GTP (or ATP), NADH, and FADH2 per turn.
C) It directly consumes oxygen.
D) It is inhibited by high NAD+/NADH ratio.
Answer B: It produces GTP (or ATP), NADH, and FADH2 per turn.
Rationale: TCA cycle in mitochondrial matrix produces 3 NADH, 1 FADH2, 1 GTP (or ATP) per acetyl-CoA.
Oxygen is not directly used.
,6. Which enzyme is responsible for the rate-limiting step of glycolysis?
A) Hexokinase
B) Phosphofructokinase-1 (PFK-1)
C) Glyceraldehyde-3-phosphate dehydrogenase
D) Pyruvate kinase
Answer B: Phosphofructokinase-1 (PFK-1)
Rationale: PFK-1 converts fructose-6-phosphate to fructose-1,6-bisphosphate; it is allosterically regulated
by ATP, AMP, and citrate, making it the key regulatory step.
7. A mutation in the gene encoding pyruvate dehydrogenase (PDH) would most directly affect the
conversion of:
A) Glucose to pyruvate
B) Pyruvate to acetyl-CoA
C) Acetyl-CoA to oxaloacetate
D) Lactate to pyruvate
Answer B: Pyruvate to acetyl-CoA
Rationale: PDH complex converts pyruvate to acetyl-CoA, linking glycolysis to TCA cycle. Deficiency causes
lactic acidosis and neurological problems.
8. In oxidative phosphorylation, the electron transport chain complexes pump protons from the:
A) Cytosol into the mitochondrial matrix
B) Matrix into the intermembrane space
C) Intermembrane space into the matrix
D) Matrix into the cytosol
Answer B: Matrix into the intermembrane space
Rationale: Complexes I, III, and IV pump protons from matrix to intermembrane space, creating a proton
gradient used by ATP synthase.
9. Which of the following is an allosteric activator of glycogen synthase?
A) Epinephrine
B) Glucose-6-phosphate
, C) Glucagon
D) cAMP
Answer B: Glucose-6-phosphate
Rationale: Glucose-6-phosphate allosterically activates glycogen synthase, promoting glycogen synthesis
when glucose is abundant. Epinephrine and glucagon inhibit it.
10. The conversion of glucose-6-phosphate to glucose is catalyzed by:
A) Hexokinase
B) Glucose-6-phosphatase
C) Phosphoglucomutase
D) Glycogen phosphorylase
Answer B: Glucose-6-phosphatase
Rationale: Glucose-6-phosphatase is present in liver and kidney, not muscle, and allows gluconeogenesis to
release free glucose into blood.
11. Which enzyme in fatty acid synthesis is the primary regulatory point?
A) Acetyl-CoA carboxylase
B) Acetyl-CoA carboxylase (same – yes)
C) Fatty acid synthase
D) Carnitine acyltransferase I
Answer A: Acetyl-CoA carboxylase
Rationale: Acetyl-CoA carboxylase converts acetyl-CoA to malonyl-CoA; inhibited by AMPK and glucagon,
activated by insulin and citrate.
12. During β-oxidation of a saturated fatty acid with 16 carbons (palmitate), how many molecules of
acetyl-CoA are produced?
A) 7
B) 8
C) 16
D) 6
Answer B: 8
Rationale: Each β-oxidation cycle removes 2 carbons as acetyl-CoA. Palmitate (C16) yields 8 acetyl-CoA
(16/2 = 8) and 7 FADH2 + 7 NADH.
Metabolism & Molecular Biology Review | INSTANT PDF
DOWNLOAD
MCAT Biochemistry practice exam. This resource contains 200 high-yield questions covering enzymes,
kinetics, metabolic pathways (glycolysis, TCA, ETC, gluconeogenesis, glycogen metabolism, fatty acid
oxidation, amino acid metabolism), molecular biology (DNA replication, transcription, translation, gene
regulation), and lab techniques. Each question includes the correct answer and rationale. Use this exam to
solidify your biochemistry knowledge for MCAT success.
Key Topics Covered
• Enzymes & Kinetics – Michaelis-Menten, inhibition, catalytic mechanisms, cofactors
• Carbohydrate Metabolism – Glycolysis, gluconeogenesis, TCA cycle, oxidative phosphorylation, glycogen
synthesis/breakdown, pentose phosphate pathway
• Lipid Metabolism – Fatty acid oxidation, ketogenesis, fatty acid synthesis, cholesterol metabolism
• Amino Acid & Nitrogen Metabolism – Transamination, urea cycle, amino acid degradation, one-carbon
metabolism
• Molecular Biology – DNA replication, repair, transcription, RNA processing, translation, genetic code
• Gene Regulation – Operons, transcription factors, epigenetics, non-coding RNAs
• Biochemical Techniques – Chromatography, electrophoresis, centrifugation, spectrometry, blotting
Questions 1–200
1. Which of the following is true about the Michaelis constant (Km)?
A) It is the substrate concentration at which Vmax is reached.
B) It is the substrate concentration at which the reaction rate is half of Vmax.
C) It is a measure of the enzyme’s catalytic efficiency.
D) It increases in the presence of a competitive inhibitor.
Answer B: It is the substrate concentration at which the reaction rate is half of Vmax.
Rationale: Km is the substrate concentration at half-maximal velocity. It reflects affinity: lower Km = higher
affinity (except for certain cases).
2. A competitive inhibitor will:
A) Decrease Vmax and leave Km unchanged.
B) Increase Km and leave Vmax unchanged.
,C) Decrease both Vmax and Km.
D) Increase Vmax and decrease Km.
Answer B: Increase Km and leave Vmax unchanged.
Rationale: Competitive inhibitor binds active site, increasing apparent Km; at high substrate, Vmax is still
achievable. Vmax unchanged.
3. The enzyme enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate in
glycolysis. This reaction is an example of:
A) Isomerization
B) Dehydration
C) Oxidation-reduction
D) Phosphorylation
Answer B: Dehydration
Rationale: Enolase removes a water molecule (dehydration) from 2-phosphoglycerate, forming a double
bond in phosphoenolpyruvate.
4. In the pentose phosphate pathway, the primary purpose is to generate:
A) ATP and FADH2
B) NADPH and ribose-5-phosphate
C) Acetyl-CoA and CO2
D) Glucose-6-phosphate and pyruvate
Answer B: NADPH and ribose-5-phosphate
Rationale: PPP produces NADPH for biosynthesis and antioxidant defense, and ribose-5-phosphate for
nucleotide synthesis. No ATP is directly generated.
5. Which of the following statements about the TCA cycle is correct?
A) It occurs in the cytoplasm.
B) It produces GTP (or ATP), NADH, and FADH2 per turn.
C) It directly consumes oxygen.
D) It is inhibited by high NAD+/NADH ratio.
Answer B: It produces GTP (or ATP), NADH, and FADH2 per turn.
Rationale: TCA cycle in mitochondrial matrix produces 3 NADH, 1 FADH2, 1 GTP (or ATP) per acetyl-CoA.
Oxygen is not directly used.
,6. Which enzyme is responsible for the rate-limiting step of glycolysis?
A) Hexokinase
B) Phosphofructokinase-1 (PFK-1)
C) Glyceraldehyde-3-phosphate dehydrogenase
D) Pyruvate kinase
Answer B: Phosphofructokinase-1 (PFK-1)
Rationale: PFK-1 converts fructose-6-phosphate to fructose-1,6-bisphosphate; it is allosterically regulated
by ATP, AMP, and citrate, making it the key regulatory step.
7. A mutation in the gene encoding pyruvate dehydrogenase (PDH) would most directly affect the
conversion of:
A) Glucose to pyruvate
B) Pyruvate to acetyl-CoA
C) Acetyl-CoA to oxaloacetate
D) Lactate to pyruvate
Answer B: Pyruvate to acetyl-CoA
Rationale: PDH complex converts pyruvate to acetyl-CoA, linking glycolysis to TCA cycle. Deficiency causes
lactic acidosis and neurological problems.
8. In oxidative phosphorylation, the electron transport chain complexes pump protons from the:
A) Cytosol into the mitochondrial matrix
B) Matrix into the intermembrane space
C) Intermembrane space into the matrix
D) Matrix into the cytosol
Answer B: Matrix into the intermembrane space
Rationale: Complexes I, III, and IV pump protons from matrix to intermembrane space, creating a proton
gradient used by ATP synthase.
9. Which of the following is an allosteric activator of glycogen synthase?
A) Epinephrine
B) Glucose-6-phosphate
, C) Glucagon
D) cAMP
Answer B: Glucose-6-phosphate
Rationale: Glucose-6-phosphate allosterically activates glycogen synthase, promoting glycogen synthesis
when glucose is abundant. Epinephrine and glucagon inhibit it.
10. The conversion of glucose-6-phosphate to glucose is catalyzed by:
A) Hexokinase
B) Glucose-6-phosphatase
C) Phosphoglucomutase
D) Glycogen phosphorylase
Answer B: Glucose-6-phosphatase
Rationale: Glucose-6-phosphatase is present in liver and kidney, not muscle, and allows gluconeogenesis to
release free glucose into blood.
11. Which enzyme in fatty acid synthesis is the primary regulatory point?
A) Acetyl-CoA carboxylase
B) Acetyl-CoA carboxylase (same – yes)
C) Fatty acid synthase
D) Carnitine acyltransferase I
Answer A: Acetyl-CoA carboxylase
Rationale: Acetyl-CoA carboxylase converts acetyl-CoA to malonyl-CoA; inhibited by AMPK and glucagon,
activated by insulin and citrate.
12. During β-oxidation of a saturated fatty acid with 16 carbons (palmitate), how many molecules of
acetyl-CoA are produced?
A) 7
B) 8
C) 16
D) 6
Answer B: 8
Rationale: Each β-oxidation cycle removes 2 carbons as acetyl-CoA. Palmitate (C16) yields 8 acetyl-CoA
(16/2 = 8) and 7 FADH2 + 7 NADH.
