CHEM 210 Biochemistry Module 6 Exam
Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: METABOLIC PATHWAYS & ENERGY PRODUCTION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 19-year-old athlete collapses during preseason conditioning and is found to have McArdle
disease (myophosphorylase deficiency). During a brief exercise tolerance test, her serum
lactate fails to rise despite vigorous muscle contraction. Which enzyme deficiency directly
explains the absence of the normal exercise-induced lactate surge?
A. Phosphofructokinase-1 deficiency would block glycolysis distal to glucose-6-phosphate
but would not abolish lactate production entirely.
B. Pyruvate kinase deficiency would reduce lactate generation but still permits some
anaerobic glycolysis through alternative pathways.
C. Muscle glycogen phosphorylase deficiency prevents glycogenolysis, so
glucose-1-phosphate cannot enter glycolysis to produce pyruvate and lactate under anaerobic
conditions. ✓ CORRECT
D. Glucose-6-phosphatase deficiency causes glycogen accumulation in liver but does not
impair muscle glycogen breakdown or lactate formation.
Correct Answer: C
Rationale: Muscle glycogen phosphorylase catalyzes the rate-limiting step of glycogenolysis,
releasing glucose-1-phosphate for entry into glycolysis; without this activity, no substrate
reaches pyruvate kinase and lactate dehydrogenase during anaerobic exercise.
Phosphofructokinase-1 deficiency (choice A) would indeed impair glycolysis but is not the
defect in McArdle disease, and some lactate could still form from blood glucose uptake. For
CHEM 210, always distinguish tissue-specific enzyme isoforms—muscle phosphorylase
deficiency affects exercising skeletal muscle, whereas liver phosphorylase deficiency
presents with fasting hypoglycemia, not exercise intolerance.
,Question 2 of 50
A patient with type 2 diabetes presents with fasting hyperglycemia and elevated hepatic
glucose output. Her physician explains that gluconeogenesis is inappropriately active. Which
allosteric regulator most directly accelerates the flux through the gluconeogenic pathway by
relieving inhibition of a key bypass enzyme?
A. High citrate levels activate phosphofructokinase-1 and thereby stimulate glycolysis rather
than gluconeogenesis.
B. Fructose-2,6-bisphosphate strongly activates phosphofructokinase-1 and inhibits
fructose-1,6-bisphosphatase, thereby suppressing gluconeogenesis.
C. Acetyl-CoA accumulation activates pyruvate carboxylase, the first committed step of
gluconeogenesis, while simultaneously inhibiting pyruvate dehydrogenase. ✓ CORRECT
D. AMP binding activates fructose-1,6-bisphosphatase and promotes glycogenolysis in the
well-fed state.
Correct Answer: C
Rationale: Acetyl-CoA serves as a potent allosteric activator of pyruvate carboxylase and
signals abundant energy availability for anabolic glucose synthesis, while its inhibition of
pyruvate dehydrogenase diverts pyruvate away from oxidation toward carboxylation.
Fructose-2,6-bisphosphate (choice B) actually suppresses gluconeogenesis by inhibiting
fructose-1,6-bisphosphatase, making it the classic trap answer for fasting-state questions. In
the fasting state, glucagon lowers fructose-2,6-bisphosphate, and acetyl-CoA from
β-oxidation drives pyruvate carboxylase to initiate gluconeogenesis in the liver.
Question 3 of 50
During a study of the pentose phosphate pathway in red blood cells, researchers observe that
glucose-6-phosphate dehydrogenase (G6PD) activity is essential for maintaining reduced
glutathione levels. In a patient with G6PD deficiency exposed to oxidative stress, which
metabolic consequence most directly explains the hemolytic anemia?
A. Loss of NADPH production impairs the regeneration of reduced glutathione, leaving
erythrocytes unable to neutralize reactive oxygen species and vulnerable to membrane lipid
peroxidation. ✓ CORRECT
B. Defective ribose-5-phosphate synthesis prevents nucleotide biosynthesis and causes
premature red cell apoptosis due to impaired DNA repair.
C. Impaired glycolytic flux through the oxidative branch reduces ATP generation and causes
osmotic fragility of the erythrocyte membrane.
D. Accumulation of 6-phosphogluconate inhibits hexokinase and blocks glucose entry into
the red blood cell.
Correct Answer: A
,Rationale: G6PD catalyzes the committed step of the oxidative phase of the pentose
phosphate pathway, generating NADPH required by glutathione reductase to maintain reduced
glutathione for detoxifying hydrogen peroxide and other oxidants. Choice B confuses the
non-oxidative phase with the oxidative phase; mature erythrocytes lack nuclei and do not
require ribose for DNA synthesis, making nucleotide deficiency an implausible mechanism
for hemolysis. On CHEM 210 exams, remember that G6PD deficiency is the most common
human enzyme deficiency and presents with hemolytic anemia under oxidative stress
because red blood cells rely exclusively on the pentose phosphate pathway for NADPH.
Question 4 of 50
A biochemistry student is tracing the fate of a glucose molecule during the fed state. After
entering the liver via GLUT2, glucose is phosphorylated and ultimately stored as glycogen.
Which sequence correctly identifies the regulatory enzymes that control the transition from
glucose uptake to glycogen synthesis in hepatocytes?
A. Hexokinase IV → phosphofructokinase-1 → pyruvate kinase → glycogen synthase
B. Glucokinase → phosphofructokinase-1 → pyruvate carboxylase → glycogen
phosphorylase
C. Glucokinase → phosphofructokinase-1 → pyruvate kinase → glycogen synthase ✓
CORRECT
D. Hexokinase I → phosphofructokinase-2 → aldolase → glycogen branching enzyme
Correct Answer: C
Rationale: In the fed state, hepatocytes use glucokinase (hexokinase IV) to phosphorylate
glucose, phosphofructokinase-1 commits it to glycolysis, pyruvate kinase completes
glycolysis to pyruvate, and glycogen synthase (activated by insulin-stimulated
dephosphorylation via protein phosphatase 1) polymerizes glucose into glycogen. Choice A
incorrectly substitutes hexokinase IV for glucokinase in liver and includes pyruvate kinase
without connecting to glycogen synthesis; hepatocytes express glucokinase, not hexokinase
I, because its low affinity and lack of product inhibition allow glucose phosphorylation
proportional to portal blood glucose concentration. For exam strategy, trace the pathway: fed
state = high insulin = glycogen synthesis active, glycogenolysis inactive.
Question 5 of 50
A researcher is studying pyruvate dehydrogenase complex (PDC) regulation in isolated
mitochondria. When she adds a high concentration of acetyl-CoA and NADH to the reaction
mixture, she observes a marked decrease in pyruvate oxidation. Which mechanism best
explains this feedback inhibition?
A. Acetyl-CoA and NADH activate pyruvate dehydrogenase kinase, which phosphorylates and
inactivates the E1 component (pyruvate dehydrogenase) of the PDC. ✓ CORRECT
, B. Acetyl-CoA directly competes with pyruvate for the active site of the E2 dihydrolipoyl
transacetylase component.
C. NADH allosterically activates pyruvate dehydrogenase phosphatase, thereby stimulating
PDC activity and increasing pyruvate flux.
D. High energy charge inhibits the E3 component by reducing the binding affinity of FAD for
the dihydrolipoyl dehydrogenase active site.
Correct Answer: A
Rationale: Pyruvate dehydrogenase kinase is activated by the products of the PDC
reaction—acetyl-CoA and NADH—and phosphorylates three specific serine residues on the E1
α-subunit, converting the complex to its inactive form. Choice B misidentifies the mechanism;
product inhibition of PDC occurs through covalent modification by kinase activation, not
direct competitive inhibition at the E2 active site. Remember that PDC is the critical junction
between glycolysis and the citric acid cycle, and its regulation by energy charge (ATP/ADP,
NADH/NAD⁺, acetyl-CoA/CoA) determines whether glucose is oxidized or diverted to
gluconeogenesis or lipogenesis.
Question 6 of 50
A patient with von Gierke disease (glucose-6-phosphatase deficiency) develops severe
fasting hypoglycemia and lactic acidosis. Which metabolic block most directly explains why
lactate accumulates in this glycogen storage disease?
A. Glucose-6-phosphate cannot be dephosphorylated to free glucose, so it is shunted into
glycolysis and converted to lactate, which cannot be recycled back to glucose via
gluconeogenesis. ✓ CORRECT
B. Pyruvate dehydrogenase is constitutively activated due to lack of glucose-6-phosphate,
causing excessive pyruvate oxidation and lactate buildup.
C. The urea cycle is impaired because glucose-6-phosphate inhibits carbamoyl phosphate
synthetase I, leading to accumulation of nitrogenous waste that somehow interferes with
lactate clearance.
D. Glycogen phosphorylase is permanently inactivated, trapping glucose within glycogen and
preventing any glycolytic intermediates from forming.
Correct Answer: A
Rationale: In von Gierke disease, the absence of glucose-6-phosphatase in the endoplasmic
reticulum prevents hepatic glucose release, so glucose-6-phosphate accumulates and is
diverted through glycolysis to lactate; simultaneously, the block prevents gluconeogenic
lactate clearance, causing both fasting hypoglycemia and lactic acidosis. Choice D is
incorrect because glycogen phosphorylase remains functional and actually generates
glucose-6-phosphate from glycogen, but the downstream block prevents glucose release. For
CHEM 210, von Gierke disease is the classic example of a glycogen storage disease
Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: METABOLIC PATHWAYS & ENERGY PRODUCTION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 19-year-old athlete collapses during preseason conditioning and is found to have McArdle
disease (myophosphorylase deficiency). During a brief exercise tolerance test, her serum
lactate fails to rise despite vigorous muscle contraction. Which enzyme deficiency directly
explains the absence of the normal exercise-induced lactate surge?
A. Phosphofructokinase-1 deficiency would block glycolysis distal to glucose-6-phosphate
but would not abolish lactate production entirely.
B. Pyruvate kinase deficiency would reduce lactate generation but still permits some
anaerobic glycolysis through alternative pathways.
C. Muscle glycogen phosphorylase deficiency prevents glycogenolysis, so
glucose-1-phosphate cannot enter glycolysis to produce pyruvate and lactate under anaerobic
conditions. ✓ CORRECT
D. Glucose-6-phosphatase deficiency causes glycogen accumulation in liver but does not
impair muscle glycogen breakdown or lactate formation.
Correct Answer: C
Rationale: Muscle glycogen phosphorylase catalyzes the rate-limiting step of glycogenolysis,
releasing glucose-1-phosphate for entry into glycolysis; without this activity, no substrate
reaches pyruvate kinase and lactate dehydrogenase during anaerobic exercise.
Phosphofructokinase-1 deficiency (choice A) would indeed impair glycolysis but is not the
defect in McArdle disease, and some lactate could still form from blood glucose uptake. For
CHEM 210, always distinguish tissue-specific enzyme isoforms—muscle phosphorylase
deficiency affects exercising skeletal muscle, whereas liver phosphorylase deficiency
presents with fasting hypoglycemia, not exercise intolerance.
,Question 2 of 50
A patient with type 2 diabetes presents with fasting hyperglycemia and elevated hepatic
glucose output. Her physician explains that gluconeogenesis is inappropriately active. Which
allosteric regulator most directly accelerates the flux through the gluconeogenic pathway by
relieving inhibition of a key bypass enzyme?
A. High citrate levels activate phosphofructokinase-1 and thereby stimulate glycolysis rather
than gluconeogenesis.
B. Fructose-2,6-bisphosphate strongly activates phosphofructokinase-1 and inhibits
fructose-1,6-bisphosphatase, thereby suppressing gluconeogenesis.
C. Acetyl-CoA accumulation activates pyruvate carboxylase, the first committed step of
gluconeogenesis, while simultaneously inhibiting pyruvate dehydrogenase. ✓ CORRECT
D. AMP binding activates fructose-1,6-bisphosphatase and promotes glycogenolysis in the
well-fed state.
Correct Answer: C
Rationale: Acetyl-CoA serves as a potent allosteric activator of pyruvate carboxylase and
signals abundant energy availability for anabolic glucose synthesis, while its inhibition of
pyruvate dehydrogenase diverts pyruvate away from oxidation toward carboxylation.
Fructose-2,6-bisphosphate (choice B) actually suppresses gluconeogenesis by inhibiting
fructose-1,6-bisphosphatase, making it the classic trap answer for fasting-state questions. In
the fasting state, glucagon lowers fructose-2,6-bisphosphate, and acetyl-CoA from
β-oxidation drives pyruvate carboxylase to initiate gluconeogenesis in the liver.
Question 3 of 50
During a study of the pentose phosphate pathway in red blood cells, researchers observe that
glucose-6-phosphate dehydrogenase (G6PD) activity is essential for maintaining reduced
glutathione levels. In a patient with G6PD deficiency exposed to oxidative stress, which
metabolic consequence most directly explains the hemolytic anemia?
A. Loss of NADPH production impairs the regeneration of reduced glutathione, leaving
erythrocytes unable to neutralize reactive oxygen species and vulnerable to membrane lipid
peroxidation. ✓ CORRECT
B. Defective ribose-5-phosphate synthesis prevents nucleotide biosynthesis and causes
premature red cell apoptosis due to impaired DNA repair.
C. Impaired glycolytic flux through the oxidative branch reduces ATP generation and causes
osmotic fragility of the erythrocyte membrane.
D. Accumulation of 6-phosphogluconate inhibits hexokinase and blocks glucose entry into
the red blood cell.
Correct Answer: A
,Rationale: G6PD catalyzes the committed step of the oxidative phase of the pentose
phosphate pathway, generating NADPH required by glutathione reductase to maintain reduced
glutathione for detoxifying hydrogen peroxide and other oxidants. Choice B confuses the
non-oxidative phase with the oxidative phase; mature erythrocytes lack nuclei and do not
require ribose for DNA synthesis, making nucleotide deficiency an implausible mechanism
for hemolysis. On CHEM 210 exams, remember that G6PD deficiency is the most common
human enzyme deficiency and presents with hemolytic anemia under oxidative stress
because red blood cells rely exclusively on the pentose phosphate pathway for NADPH.
Question 4 of 50
A biochemistry student is tracing the fate of a glucose molecule during the fed state. After
entering the liver via GLUT2, glucose is phosphorylated and ultimately stored as glycogen.
Which sequence correctly identifies the regulatory enzymes that control the transition from
glucose uptake to glycogen synthesis in hepatocytes?
A. Hexokinase IV → phosphofructokinase-1 → pyruvate kinase → glycogen synthase
B. Glucokinase → phosphofructokinase-1 → pyruvate carboxylase → glycogen
phosphorylase
C. Glucokinase → phosphofructokinase-1 → pyruvate kinase → glycogen synthase ✓
CORRECT
D. Hexokinase I → phosphofructokinase-2 → aldolase → glycogen branching enzyme
Correct Answer: C
Rationale: In the fed state, hepatocytes use glucokinase (hexokinase IV) to phosphorylate
glucose, phosphofructokinase-1 commits it to glycolysis, pyruvate kinase completes
glycolysis to pyruvate, and glycogen synthase (activated by insulin-stimulated
dephosphorylation via protein phosphatase 1) polymerizes glucose into glycogen. Choice A
incorrectly substitutes hexokinase IV for glucokinase in liver and includes pyruvate kinase
without connecting to glycogen synthesis; hepatocytes express glucokinase, not hexokinase
I, because its low affinity and lack of product inhibition allow glucose phosphorylation
proportional to portal blood glucose concentration. For exam strategy, trace the pathway: fed
state = high insulin = glycogen synthesis active, glycogenolysis inactive.
Question 5 of 50
A researcher is studying pyruvate dehydrogenase complex (PDC) regulation in isolated
mitochondria. When she adds a high concentration of acetyl-CoA and NADH to the reaction
mixture, she observes a marked decrease in pyruvate oxidation. Which mechanism best
explains this feedback inhibition?
A. Acetyl-CoA and NADH activate pyruvate dehydrogenase kinase, which phosphorylates and
inactivates the E1 component (pyruvate dehydrogenase) of the PDC. ✓ CORRECT
, B. Acetyl-CoA directly competes with pyruvate for the active site of the E2 dihydrolipoyl
transacetylase component.
C. NADH allosterically activates pyruvate dehydrogenase phosphatase, thereby stimulating
PDC activity and increasing pyruvate flux.
D. High energy charge inhibits the E3 component by reducing the binding affinity of FAD for
the dihydrolipoyl dehydrogenase active site.
Correct Answer: A
Rationale: Pyruvate dehydrogenase kinase is activated by the products of the PDC
reaction—acetyl-CoA and NADH—and phosphorylates three specific serine residues on the E1
α-subunit, converting the complex to its inactive form. Choice B misidentifies the mechanism;
product inhibition of PDC occurs through covalent modification by kinase activation, not
direct competitive inhibition at the E2 active site. Remember that PDC is the critical junction
between glycolysis and the citric acid cycle, and its regulation by energy charge (ATP/ADP,
NADH/NAD⁺, acetyl-CoA/CoA) determines whether glucose is oxidized or diverted to
gluconeogenesis or lipogenesis.
Question 6 of 50
A patient with von Gierke disease (glucose-6-phosphatase deficiency) develops severe
fasting hypoglycemia and lactic acidosis. Which metabolic block most directly explains why
lactate accumulates in this glycogen storage disease?
A. Glucose-6-phosphate cannot be dephosphorylated to free glucose, so it is shunted into
glycolysis and converted to lactate, which cannot be recycled back to glucose via
gluconeogenesis. ✓ CORRECT
B. Pyruvate dehydrogenase is constitutively activated due to lack of glucose-6-phosphate,
causing excessive pyruvate oxidation and lactate buildup.
C. The urea cycle is impaired because glucose-6-phosphate inhibits carbamoyl phosphate
synthetase I, leading to accumulation of nitrogenous waste that somehow interferes with
lactate clearance.
D. Glycogen phosphorylase is permanently inactivated, trapping glucose within glycogen and
preventing any glycolytic intermediates from forming.
Correct Answer: A
Rationale: In von Gierke disease, the absence of glucose-6-phosphatase in the endoplasmic
reticulum prevents hepatic glucose release, so glucose-6-phosphate accumulates and is
diverted through glycolysis to lactate; simultaneously, the block prevents gluconeogenic
lactate clearance, causing both fasting hypoglycemia and lactic acidosis. Choice D is
incorrect because glycogen phosphorylase remains functional and actually generates
glucose-6-phosphate from glycogen, but the downstream block prevents glucose release. For
CHEM 210, von Gierke disease is the classic example of a glycogen storage disease
