CHEM 210 Biochemistry Module 1 to 8 Exams' & Final Exam
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100% Guaranteed Pass ||Complete A+ Guide - 175 Questions and
Answers Already Graded A+ Premium Exam Tested And
Verified
Subject Area Biochemistry
Description Comprehensive biochemistry exam covering protein structure, enzyme kinetics,
metabolism (carbohydrates, lipids, amino acids, nucleotides), DNA replication,
transcription, translation, signal transduction, and membrane transport.
Emphasizes mechanistic understanding, regulatory logic, and integration of
pathways.
Expected Grade A+
Total Questions 175
Duration 3 hours
Learning Outcomes 1. Predict protein folding outcomes from sequence and cellular conditions
2. Analyze enzyme inhibition using kinetic parameters and graphical data
3. Integrate metabolic pathway regulation to maintain homeostasis
4. Calculate energy yield from catabolic processes
5. Explain proofreading and repair mechanisms in DNA replication
6. Interpret signal transduction cascade outcomes given receptor mutations
Accreditation Designed to meet rigorous standards of top US R1 universities (Harvard, MIT,
Stanford, Yale, Princeton) for advanced biochemistry coursework.
Page 1
,1. In the presence of an allosteric inhibitor of PFK-1, how would the glycolytic
flux change under hypoxic conditions in a hepatocyte simultaneously active in
gluconeogenesis?
A. Glycolytic flux increases due to ATP depletion
B. Glycolytic flux decreases because the inhibitor mimics the effect of high ATP
C. Glycolytic flux remains unchanged because gluconeogenesis supplies
glucose-6-phosphate
D. Glycolytic flux increases due to feed-forward activation by
fructose-2,6-bisphosphate
Answer: B. Glycolytic flux decreases because the inhibitor mimics the effect
of high ATP
PFK-1 is inhibited by ATP and activated by AMP and fructose-2,6-bisphosphate.
An allosteric inhibitor mimics high ATP, reducing PFK-1 activity and glycolytic
flux. Hypoxia would normally increase flux via AMP, but the inhibitor overrides
this. Gluconeogenesis does not directly affect PFK-1 activity.
2. A patient presents with lactic acidosis after a minor exercise. Muscle biopsy
reveals deficient succinate dehydrogenase activity. Which metabolic pathway is
most directly impaired?
A. Glycolysis
B. Tricarboxylic acid cycle
C. Fatty acid oxidation
D. Pentose phosphate pathway
Answer: B. Tricarboxylic acid cycle
Succinate dehydrogenase (SDH) catalyzes the conversion of succinate to fumarate
in the TCA cycle. Its deficiency impairs TCA cycle flux, causing NADH depletion
and compensatory glycolysis leading to lactic acidosis. Glycolysis, fatty acid
oxidation, and PPP are not directly affected.
Page 2
,3. In translation, which of the following best explains why the Shine-Dalgarno
sequence is essential for prokaryotic but not eukaryotic protein synthesis?
A. Eukaryotic ribosomes have a higher affinity for initiator tRNA
B. Eukaryotic mRNAs are monocistronic and contain a 5' cap that directs ribosome
binding
C. Prokaryotic ribosomes lack the ability to scan for start codons
D. The Shine-Dalgarno sequence base-pairs with the 23S rRNA to position the
ribosome
Answer: B. Eukaryotic mRNAs are monocistronic and contain a 5' cap that
directs ribosome binding
Prokaryotes lack a 5' cap; instead, the Shine-Dalgarno sequence base-pairs with
16S rRNA to position the ribosome near the start codon. Eukaryotes use a 5' cap
and scanning mechanism. Option A is false (initiator tRNA is similar), C is
incorrect (prokaryotes do not scan), D is wrong (it pairs with 16S, not 23S).
4. A biochemist measures the Km of hexokinase for glucose as 0.1 mM. In a
liver cell under fasting conditions, glucose concentration is approximately 5
mM. If glucokinase is also present (Km ~ 10 mM), which enzyme dominates
glucose phosphorylation at this glucose level?
A. Hexokinase, because its low Km ensures near-maximal velocity
B. Glucokinase, because its Vmax is higher and it is not inhibited by
glucose-6-phosphate
C. Both contribute equally because the substrate concentration is near the Km of
glucokinase
D. Neither; most glucose is phosphorylated by a third enzyme, glucose-6-phosphatase
Answer: A. Hexokinase, because its low Km ensures near-maximal velocity
At 5 mM glucose, hexokinase is nearly saturated (V ~ Vmax) because its Km is
0.1 mM. Glucokinase operates at half Vmax (Km 10 mM). Although glucokinase
has higher Vmax, the actual rate depends on substrate concentration relative to
Km. Hexokinase dominates at this concentration. Glucose-6-phosphatase
dephosphorylates, not phosphorylates.
Page 3
, 5. In a reconstituted system of isolated mitochondria, succinate is added as an
electron donor, and rotenone is present. Which of the following will occur?
A. Oxygen consumption is completely inhibited
B. ATP synthesis continues at a reduced rate via complex II
C. Electrons flow from succinate to oxygen, but complex I is bypassed
D. The proton gradient collapses, and no ATP is made
Answer: C. Electrons flow from succinate to oxygen, but complex I is
bypassed
Succinate feeds electrons via complex II (succinate dehydrogenase) into the
electron transport chain, bypassing complex I. Rotenone inhibits complex I, but
since succinate enters at complex II, electron flow continues through complexes
III and IV to oxygen, allowing ATP synthesis. Option A is false; oxygen is still
consumed. Option B incorrect because ATP synthesis can continue normally.
Option D false; gradient is maintained.
Page 4
() Portage Learning Questions and Verified Answers,
100% Guaranteed Pass ||Complete A+ Guide - 175 Questions and
Answers Already Graded A+ Premium Exam Tested And
Verified
Subject Area Biochemistry
Description Comprehensive biochemistry exam covering protein structure, enzyme kinetics,
metabolism (carbohydrates, lipids, amino acids, nucleotides), DNA replication,
transcription, translation, signal transduction, and membrane transport.
Emphasizes mechanistic understanding, regulatory logic, and integration of
pathways.
Expected Grade A+
Total Questions 175
Duration 3 hours
Learning Outcomes 1. Predict protein folding outcomes from sequence and cellular conditions
2. Analyze enzyme inhibition using kinetic parameters and graphical data
3. Integrate metabolic pathway regulation to maintain homeostasis
4. Calculate energy yield from catabolic processes
5. Explain proofreading and repair mechanisms in DNA replication
6. Interpret signal transduction cascade outcomes given receptor mutations
Accreditation Designed to meet rigorous standards of top US R1 universities (Harvard, MIT,
Stanford, Yale, Princeton) for advanced biochemistry coursework.
Page 1
,1. In the presence of an allosteric inhibitor of PFK-1, how would the glycolytic
flux change under hypoxic conditions in a hepatocyte simultaneously active in
gluconeogenesis?
A. Glycolytic flux increases due to ATP depletion
B. Glycolytic flux decreases because the inhibitor mimics the effect of high ATP
C. Glycolytic flux remains unchanged because gluconeogenesis supplies
glucose-6-phosphate
D. Glycolytic flux increases due to feed-forward activation by
fructose-2,6-bisphosphate
Answer: B. Glycolytic flux decreases because the inhibitor mimics the effect
of high ATP
PFK-1 is inhibited by ATP and activated by AMP and fructose-2,6-bisphosphate.
An allosteric inhibitor mimics high ATP, reducing PFK-1 activity and glycolytic
flux. Hypoxia would normally increase flux via AMP, but the inhibitor overrides
this. Gluconeogenesis does not directly affect PFK-1 activity.
2. A patient presents with lactic acidosis after a minor exercise. Muscle biopsy
reveals deficient succinate dehydrogenase activity. Which metabolic pathway is
most directly impaired?
A. Glycolysis
B. Tricarboxylic acid cycle
C. Fatty acid oxidation
D. Pentose phosphate pathway
Answer: B. Tricarboxylic acid cycle
Succinate dehydrogenase (SDH) catalyzes the conversion of succinate to fumarate
in the TCA cycle. Its deficiency impairs TCA cycle flux, causing NADH depletion
and compensatory glycolysis leading to lactic acidosis. Glycolysis, fatty acid
oxidation, and PPP are not directly affected.
Page 2
,3. In translation, which of the following best explains why the Shine-Dalgarno
sequence is essential for prokaryotic but not eukaryotic protein synthesis?
A. Eukaryotic ribosomes have a higher affinity for initiator tRNA
B. Eukaryotic mRNAs are monocistronic and contain a 5' cap that directs ribosome
binding
C. Prokaryotic ribosomes lack the ability to scan for start codons
D. The Shine-Dalgarno sequence base-pairs with the 23S rRNA to position the
ribosome
Answer: B. Eukaryotic mRNAs are monocistronic and contain a 5' cap that
directs ribosome binding
Prokaryotes lack a 5' cap; instead, the Shine-Dalgarno sequence base-pairs with
16S rRNA to position the ribosome near the start codon. Eukaryotes use a 5' cap
and scanning mechanism. Option A is false (initiator tRNA is similar), C is
incorrect (prokaryotes do not scan), D is wrong (it pairs with 16S, not 23S).
4. A biochemist measures the Km of hexokinase for glucose as 0.1 mM. In a
liver cell under fasting conditions, glucose concentration is approximately 5
mM. If glucokinase is also present (Km ~ 10 mM), which enzyme dominates
glucose phosphorylation at this glucose level?
A. Hexokinase, because its low Km ensures near-maximal velocity
B. Glucokinase, because its Vmax is higher and it is not inhibited by
glucose-6-phosphate
C. Both contribute equally because the substrate concentration is near the Km of
glucokinase
D. Neither; most glucose is phosphorylated by a third enzyme, glucose-6-phosphatase
Answer: A. Hexokinase, because its low Km ensures near-maximal velocity
At 5 mM glucose, hexokinase is nearly saturated (V ~ Vmax) because its Km is
0.1 mM. Glucokinase operates at half Vmax (Km 10 mM). Although glucokinase
has higher Vmax, the actual rate depends on substrate concentration relative to
Km. Hexokinase dominates at this concentration. Glucose-6-phosphatase
dephosphorylates, not phosphorylates.
Page 3
, 5. In a reconstituted system of isolated mitochondria, succinate is added as an
electron donor, and rotenone is present. Which of the following will occur?
A. Oxygen consumption is completely inhibited
B. ATP synthesis continues at a reduced rate via complex II
C. Electrons flow from succinate to oxygen, but complex I is bypassed
D. The proton gradient collapses, and no ATP is made
Answer: C. Electrons flow from succinate to oxygen, but complex I is
bypassed
Succinate feeds electrons via complex II (succinate dehydrogenase) into the
electron transport chain, bypassing complex I. Rotenone inhibits complex I, but
since succinate enters at complex II, electron flow continues through complexes
III and IV to oxygen, allowing ATP synthesis. Option A is false; oxygen is still
consumed. Option B incorrect because ATP synthesis can continue normally.
Option D false; gradient is maintained.
Page 4
