AP Biology Cellular Energetics — Worked Answer Explanations

Unit 3 · 12 questions explained

Below is a complete answer key for our AP Biology Cellular Energetics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Cellular Energetics practice test and come back here to review, or head back to the Cellular Energetics unit overview.

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Question 1 · Easy
An enzyme lowers the activation energy of a reaction. Which of the following best describes how the enzyme accomplishes this?
- A
  By adding energy to the reactants to push the reaction forward
  Why not A: Enzymes do not add energy; they reduce the energy barrier required.
- B
  By stabilizing the transition state, reducing the energy required to reach itCorrect
- C
  By permanently bonding to the substrate and forming a new product
  Why not C: Enzymes are not consumed in the reaction; they are regenerated unchanged.
- D
  By increasing the temperature of the reaction environment
  Why not D: Temperature change is a physical condition, not a mechanism of enzyme action.
Explanation
Enzymes are biological catalysts that function by binding substrates at the active site and stabilizing the transition state — the highest-energy intermediate between reactants and products. This stabilization lowers the activation energy, allowing the reaction to proceed faster without the enzyme being consumed. The enzyme is released unchanged at the end of the reaction.

Key takeaway
Enzymes lower activation energy by stabilizing the transition state, not by adding energy or being consumed.
Question 2 · Easy
During the light-dependent reactions of photosynthesis, which molecule directly receives the electrons energized by absorbed photons?
- A
  NADPH
  Why not A: NADPH is the final electron acceptor in the light reactions, not the immediate recipient of energized electrons from photosystems.
- B
  The primary electron acceptor within the photosystemCorrect
- C
  ATP synthase
  Why not C: ATP synthase uses the proton gradient to synthesize ATP, not electrons directly from photons.
- D
  Carbon dioxide
  Why not D: CO₂ is fixed during the Calvin cycle, not involved in the light-dependent reactions.
Explanation
When a photon is absorbed by a reaction-center chlorophyll molecule (P680 in Photosystem II or P700 in Photosystem I), the energy excites an electron to a higher energy level. This excited electron is immediately captured by the primary electron acceptor within the photosystem. From there, it passes down the electron transport chain, releasing energy used to pump protons and ultimately reducing NADP⁺ to NADPH.

Key takeaway
Absorbed photon energy excites electrons that are captured by the primary electron acceptor before entering the electron transport chain.
Question 3 · Easy
Which of the following is the direct product of glycolysis when one molecule of glucose is completely processed?
- A
  2 acetyl-CoA, 2 ATP, 2 NADH
  Why not A: Acetyl-CoA is produced during pyruvate oxidation, the step after glycolysis.
- B
  2 pyruvate, 2 ATP (net), 2 NADHCorrect
- C
  4 pyruvate, 4 ATP, 4 NADH
  Why not C: Glycolysis cleaves one glucose into exactly 2 three-carbon pyruvate molecules.
- D
  2 pyruvate, 36 ATP, 2 NADH
  Why not D: 36–38 ATP represents total yield from full aerobic respiration, not glycolysis alone.
Explanation
Glycolysis splits one 6-carbon glucose into two 3-carbon pyruvate molecules in the cytoplasm. The energy investment phase uses 2 ATP, and the energy payoff phase produces 4 ATP and 2 NADH, for a net gain of 2 ATP and 2 NADH. Pyruvate then enters pyruvate oxidation before the Krebs cycle, if oxygen is available.

Key takeaway
Glycolysis yields 2 pyruvate, 2 net ATP, and 2 NADH per glucose molecule.
Question 4 · Easy
A mutation eliminates the enzyme RuBisCO from a plant cell. Which stage of photosynthesis would be most directly affected?
- A
  The light-dependent reactions in the thylakoid membranes
  Why not A: RuBisCO functions in the Calvin cycle, not in the light reactions.
- B
  The splitting of water molecules (photolysis)
  Why not B: Water splitting is catalyzed by the oxygen-evolving complex in Photosystem II, not RuBisCO.
- C
  Carbon fixation in the Calvin cycleCorrect
- D
  ATP synthesis via the electron transport chain
  Why not D: ATP synthesis is driven by ATP synthase using the proton gradient, independent of RuBisCO.
Explanation
RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase) is the enzyme responsible for carbon fixation — the first step of the Calvin cycle in which CO₂ is attached to the 5-carbon molecule RuBP to form two 3-carbon molecules (3-PGA). Without RuBisCO, the plant cannot fix atmospheric CO₂, halting the Calvin cycle and ultimately preventing glucose synthesis even if the light reactions proceed normally.

Key takeaway
RuBisCO catalyzes CO₂ fixation in the Calvin cycle; its absence blocks the light-independent reactions.
Question 5 · Medium
A researcher adds an uncoupler to mitochondria. The uncoupler makes the inner mitochondrial membrane freely permeable to protons (H⁺). Which of the following would be the immediate result?
- A
  ATP synthesis increases because more protons flow through ATP synthase
  Why not A: Uncouplers bypass ATP synthase, so proton flow through the synthase decreases.
- B
  The electron transport chain stops because it has no electron acceptors
  Why not B: The ETC still passes electrons to O₂; only the proton gradient is dissipated, not the electron acceptors.
- C
  The proton gradient collapses and ATP synthesis via oxidative phosphorylation haltsCorrect
- D
  NADH accumulates because the Krebs cycle accelerates
  Why not D: NADH might accumulate, but this is a downstream consequence; the primary immediate effect is loss of the proton gradient.
Explanation
Oxidative phosphorylation depends on the proton gradient (proton-motive force) built across the inner mitochondrial membrane by the electron transport chain. This gradient drives H⁺ through ATP synthase, powering ATP production. An uncoupler dissipates the gradient by providing an alternative route for H⁺ to re-enter the matrix, bypassing ATP synthase. As a result, the energy released by the ETC is lost as heat rather than captured in ATP, and ATP synthesis halts even though electron transport continues.

Key takeaway
ATP synthesis via oxidative phosphorylation requires the proton gradient; uncouplers dissipate this gradient and stop ATP production.
Question 6 · Medium
During the Krebs cycle, one turn (per acetyl-CoA) produces which of the following sets of energy carriers?
- A
  2 NADH, 1 FADH₂, 1 ATP (or GTP)
  Why not A: This undercounts the NADH; a single turn of the Krebs cycle produces 3 NADH.
- B
  3 NADH, 1 FADH₂, 1 ATP (or GTP)Correct
- C
  3 NADH, 2 FADH₂, 2 ATP
  Why not C: Only one FADH₂ and one ATP/GTP are produced per turn of the Krebs cycle.
- D
  2 NADH, 2 FADH₂, 1 ATP (or GTP)
  Why not D: Overcounts FADH₂; only the succinate dehydrogenase step produces FADH₂.
Explanation
Each turn of the Krebs (citric acid) cycle processes one 2-carbon acetyl group from acetyl-CoA. The cycle generates: 3 NADH (from isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase steps), 1 FADH₂ (from succinate dehydrogenase), and 1 ATP or GTP (substrate-level phosphorylation via succinyl-CoA synthetase). Two turns occur per glucose molecule, doubling these values.

Key takeaway
One Krebs cycle turn yields 3 NADH, 1 FADH₂, and 1 ATP/GTP per acetyl-CoA.
Question 7 · Medium
A particular organism lives in an environment with abundant light and CO₂ but very little inorganic phosphate (Pi). Which of the following processes would be most severely limited?
- A
  Carbon fixation by RuBisCO
  Why not A: Carbon fixation requires RuBP and CO₂ but not Pi directly as a substrate.
- B
  ATP synthesis and regeneration of RuBP in the Calvin cycleCorrect
- C
  Absorption of photons by chlorophyll
  Why not C: Photon absorption is a physical process that depends on chlorophyll structure, not Pi availability.
- D
  Splitting of water molecules in Photosystem II
  Why not D: Photolysis of water uses light energy and produces O₂, electrons, and H⁺, not Pi.
Explanation
Inorganic phosphate (Pi) is the substrate used by ATP synthase to regenerate ATP from ADP. ATP is required in the Calvin cycle (3 ATP per CO₂ fixed) to convert 3-phosphoglycerate (3-PGA) to glyceraldehyde-3-phosphate (G3P) and to regenerate RuBP. Without adequate Pi, ATP cannot be synthesized, stalling the Calvin cycle even when light and CO₂ are abundant.

Key takeaway
Phosphate availability limits ATP synthesis, which is essential for regenerating RuBP and driving the Calvin cycle.
Question 8 · Medium
An athlete sprinting at maximum effort for 8 seconds relies primarily on which metabolic process to generate ATP?
- A
  Aerobic cellular respiration (oxidative phosphorylation)
  Why not A: Aerobic respiration requires several seconds to ramp up oxygen delivery; it cannot supply ATP fast enough for an 8-second maximal effort.
- B
  Fermentation producing ethanol
  Why not B: Ethanol fermentation occurs in yeast, not in human muscle cells.
- C
  Phosphocreatine hydrolysis and glycolysis (anaerobic)Correct
- D
  Beta-oxidation of fatty acids
  Why not D: Fatty acid oxidation is slow and aerobic; it powers endurance activities, not sprinting.
Explanation
During maximal-intensity exercise lasting under ~10 seconds, muscles rely on stored phosphocreatine (PCr) to rapidly rephosphorylate ADP to ATP, and on anaerobic glycolysis, which generates ATP quickly without oxygen. Aerobic respiration requires a lag time for oxygen delivery via the cardiovascular system. Lactate fermentation (not ethanol) accompanies anaerobic glycolysis in human muscle. This energy system is known in the AP Biology CED as a fitness-related application of cellular respiration.

Key takeaway
Short maximal-intensity activity relies on phosphocreatine and anaerobic glycolysis because aerobic pathways are too slow to ramp up.
Question 9 · Medium
In the light reactions, the electron transport chain between Photosystem II and Photosystem I pumps protons into the thylakoid lumen. What is the primary function of this proton gradient?
- A
  To reduce NADP⁺ to NADPH directly
  Why not A: NADP⁺ reduction is catalyzed by ferredoxin-NADP⁺ reductase at Photosystem I, not by the proton gradient.
- B
  To drive ATP synthesis through ATP synthase (chemiosmosis)Correct
- C
  To split water molecules and release oxygen
  Why not C: Water splitting (photolysis) is driven by the energy of P680, not the proton gradient.
- D
  To absorb additional photons for Photosystem I
  Why not D: Photon absorption depends on chlorophyll pigments, not on the proton gradient.
Explanation
As electrons pass from Photosystem II through plastoquinone, the cytochrome b6f complex, and plastocyanin to Photosystem I, they release energy that pumps H⁺ from the stroma into the thylakoid lumen, creating an electrochemical gradient. This gradient drives H⁺ back across the thylakoid membrane through ATP synthase (CF₁CF₀ complex), powering ATP synthesis via chemiosmosis — the same fundamental mechanism used in mitochondria during oxidative phosphorylation.

Key takeaway
The proton gradient across the thylakoid membrane drives ATP synthesis via chemiosmosis, linking the ETC to photophosphorylation.
Question 10 · Hard
A scientist discovers a novel inhibitor that competes with NAD⁺ at the active site of isocitrate dehydrogenase in the Krebs cycle. If this inhibitor is added to a cell, which of the following best predicts the downstream effects?
- A
  NADH production increases and ATP yield per glucose rises
  Why not A: Blocking isocitrate dehydrogenase halts the Krebs cycle at that step, reducing NADH production.
- B
  Isocitrate accumulates, the Krebs cycle slows, and overall ATP yield from aerobic respiration decreasesCorrect
- C
  The cell shifts entirely to glycolysis with no change in total ATP output
  Why not C: Even if glycolysis continues, losing the ATP from oxidative phosphorylation dramatically reduces total yield.
- D
  Pyruvate oxidation accelerates to compensate for reduced Krebs cycle activity
  Why not D: Pyruvate oxidation produces acetyl-CoA, which enters the Krebs cycle; if the Krebs cycle is blocked, acetyl-CoA would accumulate, not accelerate through.
Explanation
Isocitrate dehydrogenase catalyzes the oxidative decarboxylation of isocitrate to α-ketoglutarate, producing NADH and CO₂ — one of three NADH-generating steps in the Krebs cycle. Competitive inhibition of NAD⁺ binding blocks this step. As a result, isocitrate (and by reversible equilibrium, citrate) accumulates. The Krebs cycle slows or halts, reducing NADH and FADH₂ production. With fewer electron carriers entering the ETC, proton pumping decreases, the proton gradient weakens, and ATP synthesis via oxidative phosphorylation declines sharply, lowering overall cellular ATP yield.

Key takeaway
Inhibiting a Krebs cycle dehydrogenase blocks NADH production, reducing the proton gradient and cutting ATP yield from oxidative phosphorylation.
Question 11 · Hard
In a C₃ plant, the Calvin cycle consumes ATP and NADPH to fix CO₂. Under conditions of high light intensity but a sudden drop in CO₂ concentration (stomata close due to drought), which of the following would be expected?
- A
  RuBP levels decrease because the Calvin cycle continues to consume it at the same rate
  Why not A: With CO₂ unavailable, RuBP cannot be consumed by RuBisCO for fixation, so RuBP accumulates.
- B
  ATP and NADPH accumulate while RuBP builds up, and photorespiration may increaseCorrect
- C
  The light reactions slow down because there is less demand for ATP and NADPH
  Why not C: Although the Calvin cycle slows and demand drops, photon absorption is not regulated by CO₂; excess energy can cause photoinhibition rather than clean slowdown.
- D
  Glucose production increases because the Calvin cycle is no longer limited by light
  Why not D: Without CO₂ to fix, the Calvin cycle cannot produce G3P or glucose regardless of light availability.
Explanation
When stomata close during drought, CO₂ cannot enter the leaf, so RuBisCO has no substrate for carbon fixation. RuBP accumulates because it is not being consumed. Meanwhile, the light reactions continue producing ATP and NADPH as long as light is present, causing these molecules to build up. In the absence of CO₂, RuBisCO's oxygenase activity becomes more competitive: it fixes O₂ instead of CO₂, initiating photorespiration. Photorespiration is energetically wasteful, releasing CO₂ without net sugar synthesis — this is a significant limitation of C₃ plants compared to C₄ or CAM plants.

Key takeaway
When CO₂ drops, the Calvin cycle slows, RuBP and ATP/NADPH accumulate, and RuBisCO's oxygenase activity drives wasteful photorespiration in C₃ plants.
Question 12 · Hard
A cell has a high ratio of AMP to ATP. This condition allosterically activates phosphofructokinase-1 (PFK-1), a key enzyme in glycolysis. Which of the following best explains why this regulation is adaptive?
- A
  It prevents overproduction of pyruvate when ATP is abundant, conserving glucose
  Why not A: This describes the opposite scenario — when ATP is low (AMP high), glycolysis should accelerate, not be inhibited.
- B
  It accelerates ATP production precisely when the cell's energy charge is low, restoring homeostasisCorrect
- C
  It inhibits the Krebs cycle to prevent further energy expenditure
  Why not C: High AMP signals energy deficit; slowing the Krebs cycle would worsen it, not help.
- D
  It signals the cell to begin gluconeogenesis to produce more glucose from amino acids
  Why not D: PFK-1 activation accelerates glycolysis (glucose breakdown), not gluconeogenesis (glucose synthesis).
Explanation
PFK-1 catalyzes an irreversible, committed step in glycolysis: the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate. It is a major control point for glycolytic flux. AMP is a signal of low energy charge — when AMP/ATP is high, the cell has consumed most of its ATP. Allosteric activation of PFK-1 by AMP (and ADP) accelerates glycolysis, increasing the rate of ATP production (via substrate-level phosphorylation and the subsequent Krebs cycle and ETC). This negative-feedback-like mechanism is adaptive because it restores the ATP:AMP ratio toward homeostasis. Conversely, high ATP inhibits PFK-1, slowing glycolysis when energy is already sufficient.

Key takeaway
Allosteric regulation of PFK-1 by AMP/ATP couples glycolytic rate directly to cellular energy demand, maintaining energy homeostasis.