Biology by Bradford
Unit 3: Cellular Energetics. Practice questions with scoring guidelines.
112 original AP-style questions on Unit 3, written to the CED learning objectives: 84 multiple-choice, 28 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 3.1 Enzyme Structure 10 questions
- 3.2 Enzyme Catalysis 13 questions
- 3.3 Environmental Impacts on Enzyme Function 27 questions
- 3.4 Cellular Energy 11 questions
- 3.5 Photosynthesis 27 questions
- 3.6 Cellular Respiration 32 questions
- 3.7 Fitness 13 questions
In the bank for Unit 3
- 84 multiple-choice
- 28 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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A student measured the initial reaction rate of an enzyme at increasing substrate concentrations, first with no inhibitor present and then in the presence of a fixed concentration of inhibitor Q. Enzyme concentration, temperature, and pH were held constant. The results are shown in the table.
Based on the pattern of inhibition shown in the data, where does inhibitor Q most likely bind to the enzyme?
| Substrate concentration (mM) | Initial rate, no inhibitor (μmol·min⁻¹) | Initial rate, + inhibitor Q (μmol·min⁻¹) |
|---|---|---|
| 1 | 20 | 8 |
| 2 | 33 | 15 |
| 5 | 50 | 30 |
| 10 | 63 | 46 |
| 20 | 71 | 62 |
| 40 | 76 | 71 |
- To an allosteric site far from the active site, changing the active-site shape so substrate cannot overcome it.
- To the product, reversing the reaction after it occurs.
- To the active site, where it competes with the substrate for binding.
- To the substrate molecules, preventing them from entering the cell.
During intense exercise, oxygen delivery to muscle cells cannot keep pace with demand, yet the cells continue to make ATP by glycolysis for a time by converting pyruvate to lactate. What is the essential role of lactate production under these conditions?
- It regenerates NAD⁺ for glycolysis.
- It fixes CO₂ into glucose for storage.
- It splits water to provide electrons.
- It produces most of the cell's ATP.
Chloroplasts were isolated and mixed with DPIP, a blue dye that becomes colorless when it accepts electrons. As DPIP is reduced, less blue light is absorbed and the solution's percent transmittance rises. Three tubes were prepared: illuminated chloroplasts, chloroplasts kept in the dark, and boiled chloroplasts kept in the light. Percent transmittance was recorded over time.
Why does the boiled-chloroplast tube show almost no change in transmittance over the 15 minutes?
| Time (min) | Illuminated (% transmittance) | Dark (% transmittance) | Boiled + light (% transmittance) |
|---|---|---|---|
| 0 | 20 | 20 | 20 |
| 5 | 45 | 22 | 21 |
| 10 | 62 | 24 | 20 |
| 15 | 78 | 25 | 21 |
- Boiling sped up the Calvin cycle.
- Boiling removed all DPIP from the tube.
- Boiling gave DPIP too much energy to accept.
- Boiling denatured the transport proteins.
Two purified forms of the same enzyme are available: one isolated from human cells and one isolated from a bacterium that lives in a hot spring at 75 °C. Both forms catalyze the same reaction, which converts a colorless substrate into a yellow product whose concentration can be read on a spectrophotometer. A student hypothesizes that the bacterial form of the enzyme retains activity at higher temperatures than the human form. Water baths from 20 °C to 90 °C, buffers of any pH, and a spectrophotometer are available.
Enzymes speed up reactions by lowering the activation energy required. Some molecules slow an enzyme-catalyzed reaction by binding to the enzyme. A competitive inhibitor binds the active site itself. A noncompetitive inhibitor binds a separate site elsewhere on the enzyme.
A researcher measures the rate of oxygen (O₂) evolution by a submerged aquatic plant across a range of light intensities. Carbon dioxide concentration, temperature, and pH are held constant. A positive value indicates net O₂ release; a negative value indicates net O₂ uptake. The results are shown in the table.
The following information applies to parts B, C, and D.
The following information applies to parts B, C, and D. The researcher notes that at the highest light intensities the O₂ evolution rate no longer increases, and that at a light intensity of 0 the plant takes up O₂ rather than releasing it.
| Light intensity (μmol photons·m⁻²·s⁻¹) | O₂ evolution rate (μmol O₂·min⁻¹) |
|---|---|
| 0 | -2 |
| 100 | 6 |
| 200 | 13 |
| 400 | 22 |
| 600 | 27 |
| 800 | 28 |
| 1000 | 28 |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · AP® is a trademark registered by the College Board, which was not involved in and does not endorse this site.
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- C — because excess substrate overcomes the inhibition, Q must be competing with substrate for the same active site.
- A — reducing pyruvate to lactate reoxidizes NADH to NAD⁺, keeping glycolysis running when the electron transport chain is oxygen-limited.
- D — heat denatures the electron transport proteins, so light-driven electron flow to DPIP cannot occur and the dye stays reduced-colored.
- Set up tubes containing the same concentration of substrate and the same concentration of one enzyme form, incubate identical sets of tubes at a series of temperatures (for example 20, 35, 50, 65, 80, and 90 °C) for the same length of time, measure the yellow product formed per minute at each temperature, and repeat the entire series with the other enzyme form for comparison
- Run both enzymes across the same range of temperatures with all other conditions identical, measure the rate of yellow product formation at each temperature, and compare the temperature at which each enzyme's rate falls off
Total for part (A): 1 point
- The rate at which yellow product is formed (change in absorbance per minute)
- Enzyme activity, measured as the amount of yellow product produced in a set time
Total for part (B): 1 point
- Tubes containing substrate and buffer but no enzyme, incubated at each temperature, to show how much yellow product forms without the enzyme
- A no-enzyme blank at every temperature, so that any spontaneous conversion of substrate at high temperature can be subtracted from the treatment values
Total for part (C): 1 point
- pH affects the charges on R groups in the enzyme, which affects folding and the shape of the active site; if pH varied between tubes, differences in rate could not be attributed to temperature alone
- pH is a confounding variable that independently changes enzyme activity, so holding it constant is what allows the temperature effect to be isolated
Total for part (D): 1 point
Total for question 4: 4 points
- The active site is a pocket whose three-dimensional shape and chemical properties (charge, polarity, R groups) are complementary to those of one substrate, so only that substrate binds well enough to be catalyzed
- Folding of the polypeptide creates an active site whose shape and R-group chemistry fit a particular substrate (induced fit), excluding molecules of other shapes
Total for part (A): 1 point
- Binding at the other site changes the enzyme's conformation, which distorts the active site so the substrate no longer binds (or is no longer positioned for catalysis), lowering the reaction rate
- The inhibitor binds an allosteric site and alters the shape of the protein, so the active site is no longer complementary to the substrate
Total for part (B): 1 point
- The reaction rate would increase, returning to (or near) the uninhibited rate
- The inhibition would largely be overcome and the rate would rise toward its maximum
Total for part (C): 1 point
- The competitive inhibitor and the substrate bind the same site, so they compete; raising the substrate concentration raises the proportion of collisions in which substrate rather than inhibitor occupies the active site, so more enzyme-substrate complexes form per unit time
- Because binding is reversible and to the same site, an excess of substrate outcompetes the inhibitor for active sites and restores the rate
Total for part (D): 1 point
Total for question 5: 4 points
- As light intensity increases, the O₂ evolution rate increases and then levels off
- A positive relationship that plateaus at high light intensity
- About 800 μmol photons·m⁻²·s⁻¹
- Around 600–800 units
Total for part (A): 2 points
- Cellular respiration consumes O₂ while no photosynthesis occurs in the dark, giving net O₂ uptake
- With no light there are no light-dependent reactions, so respiration causes net O₂ consumption
- CO₂ availability (or Calvin-cycle capacity/temperature) limits the rate because light is no longer limiting
- A factor other than light, such as CO₂ concentration, becomes limiting once light is saturating
Total for part (B): 2 points
- The rate would increase
- A higher plateau / greater O₂ evolution
- Because CO₂ was the limiting factor, more CO₂ allows faster carbon fixation, which consumes NADPH and ATP and lets the light reactions release O₂ faster
- More CO₂ speeds carbon fixation, so photosynthesis and O₂ release increase
Total for part (C): 2 points
- Temperature (or CO₂) was held constant so its effects do not confound the results, leaving light intensity as the only independent variable
- Controlling it ensures changes in rate are due to light intensity alone
- Measured O₂ is a net value because respiration simultaneously consumes some O₂, so gross photosynthesis is actually higher
- The plant respires at the same time, so O₂ evolution underestimates gross photosynthesis
- Use the magnitude of O₂ uptake at 0 light (the negative value) as an estimate of the respiration rate
- The O₂ consumed in the dark approximates the respiration rate
Total for part (D): 3 points
Total for question 6: 9 points
Other AP Biology units
- Unit 1 Chemistry of Life 96
- Unit 2 Cell Structure and Function 105
- Unit 4 Cell Communication and Cell Cycle 104
- Unit 5 Heredity 130
- Unit 6 Gene Expression and Regulation 116
- Unit 7 Natural Selection 149
- Unit 8 Ecology 131
All eight units → · IB Biology subtopics → · Open the exam maker →