AP Biology: Unit 3, Cellular Respiration: Glycolysis & Krebs Cycle (Drill 12)

Question 1. Based on the data in Table 1, what is the relationship between O2 concentration and cellular respiration rate in the germinating seeds?

• A) Respiration rate is independent of O2 concentration because seeds can switch entirely to fermentation.
• B) Respiration rate increases as O2 concentration increases, consistent with O2 serving as the final electron acceptor in aerobic respiration. ✓
• C) Respiration rate decreases as O2 concentration increases due to product inhibition of the electron transport chain.
• D) Respiration rate increases proportionally with O2 concentration only above 10% O2, staying completely flat below that level as the passage notes.

Explanation: Correct answer: B. The mean O2 consumption increases from 0.93 mL at 2% O2 to 2.43 mL at 10% and 3.93 mL at 21%, showing a clear positive relationship. This is expected because O2 is the terminal electron acceptor in the electron transport chain (ETC). At low O2 concentrations, the ETC is substrate-limited -- fewer O2 molecules are available to accept electrons, slowing the chain and reducing ATP production and overall respiration rate. (A) is incorrect because fermentation produces far less ATP and is not fully substituted for aerobic respiration in seeds that have access to any O2; the data show active O2 consumption even at 2%, indicating aerobic respiration is operating. (C) is incorrect; product inhibition does not apply here -- O2 is a reactant, not a product that accumulates. (D) is incorrect; the data show increases at both transitions (2% to 10% and 10% to 21%), not only above 10%.

Question 2. What was the purpose of the control respirometer containing boiled seeds, and what result would falsify the experiment's conclusions?

• A) The boiled seeds control for temperature; if the boiled seeds showed O2 consumption, the experiment would need to be repeated at a different temperature.
• B) The boiled seeds serve as a positive control confirming that KOH absorbs CO2; if the boiled seeds showed no volume change, the KOH would need to be replaced.
• C) The boiled seeds control for non-biological changes in gas volume (e.g., pressure or temperature fluctuations); if the boiled seeds showed significant gas volume changes, those changes could not be attributed to cellular respiration. ✓
• D) The boiled seeds measure background O2 production by non-living chemical reactions in the chamber; if no O2 is produced by them, the entire experiment is invalid.

Explanation: Correct answer: C. Boiling denatures enzymes and kills cells, so boiled seeds cannot perform cellular respiration. Any gas volume change in the boiled-seed respirometer would therefore represent a non-biological source -- such as a change in atmospheric pressure, temperature-induced expansion/contraction of the gas, or a leak. This is a negative control: it establishes the baseline "no respiration" measurement. If the boiled seeds showed substantial gas volume changes, researchers could not confidently attribute the changes seen in the germinating seed respirometers to respiration. (A) is incorrect because the boiled seeds are at the same temperature as the experimental respirometers; they do not independently control temperature. (B) reverses the logic: the boiled seed control not showing volume change is the expected negative result that validates the setup; it does not test KOH function. (D) is incorrect because boiled seeds cannot produce O2 -- dead cells do not perform photosynthesis or any oxygen-generating metabolism.

Question 3. If the experiment were repeated at 4C instead of 22C, which outcome would be most expected, and why?

• A) O2 consumption rates would decrease at all O2 concentrations because lower temperature reduces the kinetic energy of molecules and slows enzyme-catalyzed reactions, including those of cellular respiration. ✓
• B) O2 consumption rates would increase at all O2 concentrations because lower temperature increases gas solubility, providing more dissolved O2 to the mitochondria.
• C) O2 consumption rates would remain unchanged at 4C because cellular respiration is driven entirely by O2 concentration gradients rather than by temperature.
• D) O2 consumption rates would increase at 2% O2 but decrease at 21% O2 because cold temperatures selectively activate fermentation pathways.

Explanation: Correct answer: A. Cellular respiration is a series of enzyme-catalyzed reactions (glycolysis, the citric acid cycle, oxidative phosphorylation). Decreasing temperature reduces molecular kinetic energy, which lowers collision frequency and energy between enzyme and substrate molecules. This slows reaction rates across all steps, producing measurably lower O2 consumption at all O2 concentrations. (B) is incorrect; while gas solubility does increase at lower temperatures, the primary effect of cooling on a living biological system is a reduction in enzyme-catalyzed reaction rates -- this dominates any solubility benefit, and O2 availability is not the limiting factor at these concentrations anyway. (C) is incorrect; enzyme kinetics drive respiration rates -- concentration gradients alone do not maintain metabolic activity independent of temperature. (D) is incorrect; cold temperatures do not selectively activate fermentation. Fermentation is triggered by insufficient O2, not by low temperature.

Question 4. During aerobic cellular respiration at 21% O2, which stage produces the greatest number of ATP molecules per glucose molecule?

• A) Glycolysis, because it occurs in the cytoplasm and does not require a membrane, allowing it to generate the bulk of the cell's ATP per glucose molecule.
• B) The citric acid cycle (Krebs cycle), because it directly produces ATP through substrate-level phosphorylation at every turn.
• C) Pyruvate oxidation, because it converts pyruvate to acetyl-CoA and releases the most free energy.
• D) Oxidative phosphorylation (electron transport chain + chemiosmosis), because it uses the NADH and FADH2 produced in earlier stages to generate the majority of ATP via a proton gradient. ✓

Explanation: Correct answer: D. The electron transport chain (ETC) coupled with chemiosmosis (oxidative phosphorylation) produces approximately 26-28 of the ~30-32 total ATP per glucose -- the vast majority. Glycolysis yields only 2 net ATP (substrate-level phosphorylation). The citric acid cycle yields 2 ATP directly per glucose (1 per turn x 2 turns), but its primary contribution is generating NADH and FADH2 that feed into the ETC. Pyruvate oxidation produces 0 ATP directly; it generates 2 NADH per glucose. The ETC uses the electrons from NADH and FADH2 to pump protons across the inner mitochondrial membrane, creating a gradient that drives ATP synthase. (A) is incorrect; glycolysis yields only 2 net ATP. (B) is incorrect; the citric acid cycle yields only 2 ATP directly per glucose, though it generates electron carriers. (C) is incorrect; pyruvate oxidation produces no ATP directly.

Question 5. The researchers used KOH solution in the respirometers. What would happen to the data if the KOH were accidentally omitted from the experimental respirometers?

• A) O2 consumption would appear higher than actual because CO2 released would add to the gas volume.
• B) O2 consumption would appear lower than actual because CO2 released would partially offset the volume decrease caused by O2 consumption, reducing or eliminating the net volume change measured. ✓
• C) O2 consumption would be unaffected by omitting the KOH because CO2 and O2 are measured independently by separate sensors within each respirometer.
• D) The seeds would switch to fermentation because the CO2 buildup would inhibit aerobic respiration.

Explanation: Correct answer: B. In aerobic respiration, O2 is consumed and CO2 is released in roughly equal volumes (respiratory quotient ~ 1.0 for carbohydrates). The respirometer measures net gas volume change. If KOH is present, it absorbs CO2, so the only gas volume change is the decrease due to O2 consumption -- giving a clean measure of respiration rate. Without KOH, CO2 released by the seeds adds back to the gas volume, partially or fully canceling the volume decrease from O2 consumption. The net result is a much smaller (or near-zero) volume change, making the respiration rate appear much lower than it actually is. (A) is incorrect; CO2 produced would not add to volume because the seeds consume O2 at a similar rate; the net effect is near-zero change, not an increase. (C) is incorrect because the respirometer is a closed physical system measuring total gas volume; there are no separate sensors. (D) is incorrect because CO2 accumulation in a sealed vessel would not directly switch seeds to fermentation; fermentation is triggered by insufficient O2, not by CO2 buildup in this context.