AP Biology
AP Bio Unit 3 Free Response
Shared by Ashton · Nov 12, 2026 · 0 plays
11 questions. Write your answer, then check it against the rubric and score yourself.
Sample questions
Open an item to see its answer.
(a) Describe why yeast in these sealed tubes produce CO₂ and ethanol, and explain how this lets them keep making ATP without O₂.
Sample answer: Without O₂ the yeast carry out alcoholic fermentation: pyruvate from glycolysis is converted to ethanol and CO₂. The step that makes ethanol oxidizes NADH back to NAD⁺. Glycolysis needs NAD⁺ to accept electrons, so regenerating it lets glycolysis keep running and keep making 2 ATP per glucose even with no O₂ to accept electrons at an ETC.
(b) Identify the independent and dependent variables, and identify an appropriate negative control for this experiment. Justify your control.
Sample answer: The independent variable is sucrose concentration (%). The dependent variable is the volume of CO₂ released in 30 minutes. The 0% sucrose tube is an appropriate negative control because it shows how much CO₂ the yeast release from their own stored reserves without sucrose, so any extra CO₂ in the other tubes can be credited to fermentation of the added sucrose.
(c) Construct an appropriate graph of the data, including error bars. (Describe your graph in words: type, axes, scale, points, error bars.)
Sample answer: A line graph with sucrose concentration (%) on the x-axis, scaled 0 to 40 in steps of 5, and CO₂ released (mL) on the y-axis, scaled 0 to 10 in steps of 1. Points: (0, 0.4), (5, 3.1), (10, 6.0), (20, 8.8), (40, 4.2), connected with straight segments. Error bars: 0%: 0.2–0.6; 5%: 2.7–3.5; 10%: 5.5–6.5; 20%: 8.2–9.4; 40%: 3.7–4.7.
(d) Calculate the mean rate of CO₂ production (mL/min) at 10% sucrose. Then use the error bars to determine whether CO₂ production at 20% and 40% sucrose differs significantly. Justify your answer.
Sample answer: Rate at 10% = 6.0 mL ÷ 30 min = 0.20 mL CO₂/min. The 20% and 40% results differ significantly: the ±2 SE ranges (8.2–9.4 mL and 3.7–4.7 mL) don't overlap.
(e) Propose an explanation for why CO₂ production decreases at 40% sucrose, even though more sugar is available.
Sample answer: A 40% sucrose solution is strongly hypertonic to the yeast cells, so water leaves them by osmosis. The dehydrated cells can't keep up normal metabolism: the cytoplasm becomes too concentrated for the glycolysis and fermentation enzymes to work efficiently, so less sucrose is fermented and less CO₂ is released, even though more sugar is available.