13th Edition Campbell Biology Advanced Prep:
Master Cellular Energetics & Metabolism
Practice Questions & Detailed Explanations
Subject: Cell Biology and Biochemistry / Cellular Energetics and Metabolism
Question 1: During the light-dependent reactions of photosynthesis, the movement of electrons
through the electron transport chain (ETC) directly facilitates the generation of a proton-motive
force. If a thylakoid membrane were treated with a mild detergent that increases permeability to
ions, which of the following would be the most immediate physiological consequence?
A) The complete cessation of the Calvin cycle due to lack of G3P production.
B) The inhibition of the photolysis of water at Photosystem II.
C) The immediate decrease in ATP synthesis by ATP synthase despite sustained electron flow.
D) An increase in the reduction rate of NADP+ to NADPH.
Correct Answer: C) The immediate decrease in ATP synthesis by ATP synthase despite
sustained electron flow.
Explanation: ATP synthesis in the thylakoid relies on the chemiosmotic gradient (proton-motive
force) established by the concentration difference of H+ across the membrane. A detergent
increasing membrane permeability would dissipate this gradient. While electron flow (and thus
NADPH production) might briefly continue, the lack of a proton gradient would render ATP
synthase non-functional. Option A is a downstream effect, not an immediate physiological
consequence, and Option B is unrelated to the proton gradient.
Question 2: In the regulation of phosphofructokinase-1 (PFK-1) during glycolysis, how does
high ATP concentration act as an allosteric inhibitor, and what is the biological significance of
this feedback mechanism?
A) ATP binds to the active site, competitively inhibiting fructose-6-phosphate, signaling an
abundance of energy for immediate cellular work.
B) ATP binds to a regulatory allosteric site, decreasing the enzyme’s affinity for fructose-6-
phosphate, ensuring that glucose is not unnecessarily broken down when energy stores are
sufficient.
C) ATP induces a conformational change that prevents the binding of glucose, thereby forcing
the cell to use fatty acids exclusively.
,D) ATP acts as a non-competitive inhibitor by binding to the enzyme-substrate complex, halting
glycolysis even when ADP levels rise.
Correct Answer: B) ATP binds to a regulatory allosteric site, decreasing the enzyme’s
affinity for fructose-6-phosphate, ensuring that glucose is not unnecessarily broken down
when energy stores are sufficient.
Explanation: PFK-1 is the "committed step" of glycolysis. ATP is both a substrate and an
allosteric inhibitor of PFK-1. When ATP levels are high, the cell does not need to perform
glycolysis for ATP production; thus, ATP binds to a regulatory site to slow down the process,
conserving glucose. Option A is incorrect because ATP binds to a regulatory site, not the active
site. Option D is incorrect because PFK-1 inhibition is typically reversible and sensitive to the
ratio of ATP/AMP.
Question 3: Consider the mitochondrial electron transport chain. If a cell is exposed to a
chemical uncoupler that inserts into the inner mitochondrial membrane to facilitate proton
leakage, what will happen to the rates of oxygen consumption and heat production?
A) Both oxygen consumption and heat production will decrease.
B) Oxygen consumption will increase, and heat production will decrease.
C) Oxygen consumption will increase, and heat production will increase.
D) Oxygen consumption will remain constant, but heat production will increase.
Correct Answer: C) Oxygen consumption will increase, and heat production will increase.
Explanation: Uncouplers bypass ATP synthase by allowing protons to move down their
electrochemical gradient without producing ATP. To compensate for the lack of ATP, the ETC
accelerates to re-establish the gradient, leading to increased oxygen consumption (the final
electron acceptor). Because the potential energy of the proton gradient is not captured in ATP
bonds, it is released as heat, increasing thermogenesis.
Question 4: Which statement correctly explains the thermodynamic necessity of the "Energy
Investment Phase" of glycolysis?
A) It is required to destabilize the six-carbon glucose molecule to facilitate its cleavage into two
three-carbon molecules.
B) It serves to decrease the activation energy of the entire metabolic pathway by increasing the
entropy of the system.
C) It stores energy in the form of high-energy phosphate bonds that are directly converted to
ATP in the final steps.
,D) It is an evolutionary relic that reduces the overall efficiency of cellular respiration.
Correct Answer: A) It is required to destabilize the six-carbon glucose molecule to facilitate
its cleavage into two three-carbon molecules.
Explanation: Glucose is a stable molecule. Phosphorylating it (using 2 ATP) adds negative
charge and destabilizes the hexose ring, making the subsequent cleavage by aldolase
thermodynamically favorable. This investment is paid back in the energy payoff phase.
Question 5: During the Calvin Cycle, why is the regeneration of Ribulose-1,5-bisphosphate
(RuBP) considered the most energy-intensive portion of the cycle?
A) Because it requires the fixation of atmospheric CO2.
B) Because it involves the reduction of 3-phosphoglycerate to glyceraldehyde-3-phosphate.
C) Because it requires the rearrangement of multiple G3P molecules using additional ATP to
ensure the cycle can continue.
D) Because it is the step where oxygen is released as a byproduct.
Correct Answer: C) Because it requires the rearrangement of multiple G3P molecules using
additional ATP to ensure the cycle can continue.
Explanation: The Calvin Cycle is a cycle; for it to persist, the CO2 acceptor (RuBP) must be
recreated. This regeneration phase consumes significant ATP to rearrange the carbon skeletons
of G3P molecules to reform RuBP. Option A refers to the fixation phase, and Option B refers to
the reduction phase.
Question 6: An enzyme catalyzes a reaction with a ΔG of -20 kJ/mol. If the enzyme is removed
from the system, what happens to the ΔG of the reaction?
A) The ΔG becomes more negative as the reaction proceeds more slowly.
B) The ΔG becomes positive, making the reaction non-spontaneous.
C) The ΔG remains unchanged, as enzymes only affect the rate of reaction, not the free energy
change.
D) The ΔG becomes zero, as the reaction reaches equilibrium.
Correct Answer: C) The ΔG remains unchanged, as enzymes only affect the rate of reaction,
not the free energy change.
, Explanation: Enzymes lower the activation energy (Ea) of a reaction, thereby increasing the
rate. However, they have no impact on the initial or final energy states of the reactants and
products, meaning the overall ΔG (free energy change) remains constant.
Question 7: During the transition reaction (pyruvate oxidation), which of the following best
describes the fate of the carbons in pyruvate?
A) They are all incorporated into the Citric Acid Cycle as part of the acetyl group.
B) One carbon is released as CO2, and the remaining two form an acetyl group attached to
Coenzyme A.
C) They are completely oxidized to CO2 and H2O before entering the mitochondria.
D) They are converted to oxaloacetate to prime the Citric Acid Cycle.
Correct Answer: B) One carbon is released as CO2, and the remaining two form an acetyl
group attached to Coenzyme A.
Explanation: Pyruvate (3 carbons) undergoes decarboxylation (releasing 1 CO2) and oxidation
to become a 2-carbon acetyl group, which is then bound to CoA to form Acetyl-CoA, the
substrate that enters the Citric Acid Cycle.
Question 8: In C4 photosynthesis, what is the primary structural and functional advantage of the
bundle-sheath cells?
A) They contain Photosystem II, allowing for the independent production of oxygen.
B) They provide a high-CO2 environment, minimizing the oxygenase activity of Rubisco
(photorespiration).
C) They are the primary site of initial CO2 fixation into 3-phosphoglycerate.
D) They transport water from the roots to the leaves to prevent desiccation.
Correct Answer: B) They provide a high-CO2 environment, minimizing the oxygenase
activity of Rubisco (photorespiration).
Explanation: In C4 plants, CO2 is initially fixed in mesophyll cells into a 4-carbon compound,
which is then transported to the bundle-sheath cells. There, it is decarboxylated to release CO2
near Rubisco. This spatial separation keeps CO2 concentrations high, ensuring Rubisco acts as
a carboxylase rather than an oxygenase.
Question 9: Which of the following is true regarding the role of NADH and FADH2 in cellular
respiration?
Master Cellular Energetics & Metabolism
Practice Questions & Detailed Explanations
Subject: Cell Biology and Biochemistry / Cellular Energetics and Metabolism
Question 1: During the light-dependent reactions of photosynthesis, the movement of electrons
through the electron transport chain (ETC) directly facilitates the generation of a proton-motive
force. If a thylakoid membrane were treated with a mild detergent that increases permeability to
ions, which of the following would be the most immediate physiological consequence?
A) The complete cessation of the Calvin cycle due to lack of G3P production.
B) The inhibition of the photolysis of water at Photosystem II.
C) The immediate decrease in ATP synthesis by ATP synthase despite sustained electron flow.
D) An increase in the reduction rate of NADP+ to NADPH.
Correct Answer: C) The immediate decrease in ATP synthesis by ATP synthase despite
sustained electron flow.
Explanation: ATP synthesis in the thylakoid relies on the chemiosmotic gradient (proton-motive
force) established by the concentration difference of H+ across the membrane. A detergent
increasing membrane permeability would dissipate this gradient. While electron flow (and thus
NADPH production) might briefly continue, the lack of a proton gradient would render ATP
synthase non-functional. Option A is a downstream effect, not an immediate physiological
consequence, and Option B is unrelated to the proton gradient.
Question 2: In the regulation of phosphofructokinase-1 (PFK-1) during glycolysis, how does
high ATP concentration act as an allosteric inhibitor, and what is the biological significance of
this feedback mechanism?
A) ATP binds to the active site, competitively inhibiting fructose-6-phosphate, signaling an
abundance of energy for immediate cellular work.
B) ATP binds to a regulatory allosteric site, decreasing the enzyme’s affinity for fructose-6-
phosphate, ensuring that glucose is not unnecessarily broken down when energy stores are
sufficient.
C) ATP induces a conformational change that prevents the binding of glucose, thereby forcing
the cell to use fatty acids exclusively.
,D) ATP acts as a non-competitive inhibitor by binding to the enzyme-substrate complex, halting
glycolysis even when ADP levels rise.
Correct Answer: B) ATP binds to a regulatory allosteric site, decreasing the enzyme’s
affinity for fructose-6-phosphate, ensuring that glucose is not unnecessarily broken down
when energy stores are sufficient.
Explanation: PFK-1 is the "committed step" of glycolysis. ATP is both a substrate and an
allosteric inhibitor of PFK-1. When ATP levels are high, the cell does not need to perform
glycolysis for ATP production; thus, ATP binds to a regulatory site to slow down the process,
conserving glucose. Option A is incorrect because ATP binds to a regulatory site, not the active
site. Option D is incorrect because PFK-1 inhibition is typically reversible and sensitive to the
ratio of ATP/AMP.
Question 3: Consider the mitochondrial electron transport chain. If a cell is exposed to a
chemical uncoupler that inserts into the inner mitochondrial membrane to facilitate proton
leakage, what will happen to the rates of oxygen consumption and heat production?
A) Both oxygen consumption and heat production will decrease.
B) Oxygen consumption will increase, and heat production will decrease.
C) Oxygen consumption will increase, and heat production will increase.
D) Oxygen consumption will remain constant, but heat production will increase.
Correct Answer: C) Oxygen consumption will increase, and heat production will increase.
Explanation: Uncouplers bypass ATP synthase by allowing protons to move down their
electrochemical gradient without producing ATP. To compensate for the lack of ATP, the ETC
accelerates to re-establish the gradient, leading to increased oxygen consumption (the final
electron acceptor). Because the potential energy of the proton gradient is not captured in ATP
bonds, it is released as heat, increasing thermogenesis.
Question 4: Which statement correctly explains the thermodynamic necessity of the "Energy
Investment Phase" of glycolysis?
A) It is required to destabilize the six-carbon glucose molecule to facilitate its cleavage into two
three-carbon molecules.
B) It serves to decrease the activation energy of the entire metabolic pathway by increasing the
entropy of the system.
C) It stores energy in the form of high-energy phosphate bonds that are directly converted to
ATP in the final steps.
,D) It is an evolutionary relic that reduces the overall efficiency of cellular respiration.
Correct Answer: A) It is required to destabilize the six-carbon glucose molecule to facilitate
its cleavage into two three-carbon molecules.
Explanation: Glucose is a stable molecule. Phosphorylating it (using 2 ATP) adds negative
charge and destabilizes the hexose ring, making the subsequent cleavage by aldolase
thermodynamically favorable. This investment is paid back in the energy payoff phase.
Question 5: During the Calvin Cycle, why is the regeneration of Ribulose-1,5-bisphosphate
(RuBP) considered the most energy-intensive portion of the cycle?
A) Because it requires the fixation of atmospheric CO2.
B) Because it involves the reduction of 3-phosphoglycerate to glyceraldehyde-3-phosphate.
C) Because it requires the rearrangement of multiple G3P molecules using additional ATP to
ensure the cycle can continue.
D) Because it is the step where oxygen is released as a byproduct.
Correct Answer: C) Because it requires the rearrangement of multiple G3P molecules using
additional ATP to ensure the cycle can continue.
Explanation: The Calvin Cycle is a cycle; for it to persist, the CO2 acceptor (RuBP) must be
recreated. This regeneration phase consumes significant ATP to rearrange the carbon skeletons
of G3P molecules to reform RuBP. Option A refers to the fixation phase, and Option B refers to
the reduction phase.
Question 6: An enzyme catalyzes a reaction with a ΔG of -20 kJ/mol. If the enzyme is removed
from the system, what happens to the ΔG of the reaction?
A) The ΔG becomes more negative as the reaction proceeds more slowly.
B) The ΔG becomes positive, making the reaction non-spontaneous.
C) The ΔG remains unchanged, as enzymes only affect the rate of reaction, not the free energy
change.
D) The ΔG becomes zero, as the reaction reaches equilibrium.
Correct Answer: C) The ΔG remains unchanged, as enzymes only affect the rate of reaction,
not the free energy change.
, Explanation: Enzymes lower the activation energy (Ea) of a reaction, thereby increasing the
rate. However, they have no impact on the initial or final energy states of the reactants and
products, meaning the overall ΔG (free energy change) remains constant.
Question 7: During the transition reaction (pyruvate oxidation), which of the following best
describes the fate of the carbons in pyruvate?
A) They are all incorporated into the Citric Acid Cycle as part of the acetyl group.
B) One carbon is released as CO2, and the remaining two form an acetyl group attached to
Coenzyme A.
C) They are completely oxidized to CO2 and H2O before entering the mitochondria.
D) They are converted to oxaloacetate to prime the Citric Acid Cycle.
Correct Answer: B) One carbon is released as CO2, and the remaining two form an acetyl
group attached to Coenzyme A.
Explanation: Pyruvate (3 carbons) undergoes decarboxylation (releasing 1 CO2) and oxidation
to become a 2-carbon acetyl group, which is then bound to CoA to form Acetyl-CoA, the
substrate that enters the Citric Acid Cycle.
Question 8: In C4 photosynthesis, what is the primary structural and functional advantage of the
bundle-sheath cells?
A) They contain Photosystem II, allowing for the independent production of oxygen.
B) They provide a high-CO2 environment, minimizing the oxygenase activity of Rubisco
(photorespiration).
C) They are the primary site of initial CO2 fixation into 3-phosphoglycerate.
D) They transport water from the roots to the leaves to prevent desiccation.
Correct Answer: B) They provide a high-CO2 environment, minimizing the oxygenase
activity of Rubisco (photorespiration).
Explanation: In C4 plants, CO2 is initially fixed in mesophyll cells into a 4-carbon compound,
which is then transported to the bundle-sheath cells. There, it is decarboxylated to release CO2
near Rubisco. This spatial separation keeps CO2 concentrations high, ensuring Rubisco acts as
a carboxylase rather than an oxygenase.
Question 9: Which of the following is true regarding the role of NADH and FADH2 in cellular
respiration?
