Biology by Bradford
Unit 2: Cell Structure and Function. Practice questions with scoring guidelines.
105 original AP-style questions on Unit 2, written to the CED learning objectives: 80 multiple-choice, 25 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
- 2.1 Cell Structure: Subcellular Components 15 questions
- 2.2 Cell Structure and Function 5 questions
- 2.3 Cell Size 11 questions
- 2.4 Plasma Membranes 14 questions
- 2.5 Membrane Permeability 11 questions
- 2.6 Membrane Transport 15 questions
- 2.7 Facilitated Diffusion 7 questions
- 2.8 Tonicity and Osmoregulation 24 questions
- 2.9 Mechanisms of Transport 10 questions
- 2.10 Compartmentalization 10 questions
- 2.11 Origins of Cell Compartmentalization 10 questions
In the bank for Unit 2
- 80 multiple-choice
- 25 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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Within a mitochondrion, some energy-releasing reactions occur in the fluid matrix, while the reactions that transfer energy to ATP occur on the inner membrane. Which feature of the mitochondrion increases the inner-membrane surface available for these energy-transferring reactions?
- The presence of a single circular chromosome in the matrix.
- The folding of the inner membrane into cristae.
- The smooth, unfolded outer membrane.
- The large volume of the intermembrane space.
A researcher ranks four substances by how quickly they cross a protein-free artificial bilayer: fastest is a small nonpolar gas, then a small uncharged polar molecule, then a large polar molecule, and slowest is a charged ion. Which membrane property best explains this ranking?
- The bilayer has channels admitting only ions.
- Larger, charged substances cross the fastest.
- The hydrophobic core favors nonpolar solutes.
- Permeability depends only on molecular size.
The nuclear envelope is a double membrane perforated by nuclear pores. mRNA made in the nucleus passes through these pores into the cytoplasm, while the DNA remains inside. Which statement best describes how this structure supports the cell's function?
- The pores let the DNA itself leave the nucleus.
- The double membrane is where mRNA is translated.
- It compartmentalizes DNA and regulates exchange.
- It prevents all molecules from entering or leaving.
In animal cells, materials taken in by endocytosis are delivered to lysosomes, membrane-bound organelles that contain hydrolytic enzymes. A proton pump (H⁺ pump) in the lysosomal membrane uses ATP to move H⁺ from the cytosol into the lysosome, so the interior of the lysosome is acidic (about pH 4.5) while the surrounding cytosol remains at about pH 7.2. Researchers propose that the lysosomal enzymes are active only at the acidic pH maintained inside the lysosome, so that confining the enzymes in a separate acidic compartment is required for the digestion of endocytosed material. The researchers have cultured mammalian cells; compound BQ-1, which blocks the lysosomal H⁺ pump; a protein carrying a fluorescent tag that cells take up by endocytosis and deliver to lysosomes, and whose fluorescence is lost when the protein is digested (so the amount of undigested labeled protein remaining in the cells can be measured); and a dye that reports the pH inside lysosomes.
A student investigated osmosis in carrot tissue. Cores of equal size were cut from a fresh carrot, blotted, weighed, and placed in sucrose solutions of different molarities for 60 minutes. Each core was then removed, blotted, and reweighed, and the mean percent change in mass was calculated. The results are shown in Table 1.
| Sucrose concentration (M) | Mean percent change in mass (%) |
|---|---|
| 0.0 | +14.0 |
| 0.2 | +6.0 |
| 0.4 | -2.0 |
| 0.6 | -10.0 |
| 0.8 | -18.0 |
Researchers prepared artificial phospholipid vesicles (liposomes) that contained no transport proteins. They measured the rate at which four substances crossed the vesicle membrane by simple diffusion. The relative permeability of each substance is shown in Table 1.
The following information applies to parts B, C, and D.
The following information applies to parts B, C and D. The researchers next inserted aquaporin channel proteins into a second batch of identical vesicles and remeasured the water permeability, obtaining a value of 500 μm/s.
| Substance | Molecular property | Relative permeability (μm/s) |
|---|---|---|
| Water (H₂O) | Small, polar | 34 |
| Urea | Small, polar | 4.0×10⁻⁴ |
| Glucose | Large, polar | 1.0×10⁻⁷ |
| Oxygen (O₂) | Small, nonpolar | 23 |
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.
- B — cristae are folds that increase the inner-membrane surface area available for the electron transport chain and ATP synthesis.
- C — the nonpolar interior most readily admits small, nonpolar solutes and increasingly excludes larger and charged particles.
- C — the envelope isolates the genetic material while its pores control which molecules move between the two compartments.
- If lysosomal digestion requires the acidic interior produced by the H⁺ pump, then cells treated with BQ-1 will digest the labeled protein more slowly, so more undigested (fluorescent) protein will remain in BQ-1-treated cells than in untreated cells after the same time.
- Blocking the lysosomal H⁺ pump with BQ-1 will raise the pH inside lysosomes and reduce the rate at which the fluorescent protein is broken down, compared with cells that receive no BQ-1.
Total for part (A): 1 point
- Independent variable = whether the cells are treated with BQ-1 (presence/absence or concentration of BQ-1); dependent variable = the amount of undigested labeled protein remaining in the cells after a fixed time (or the rate at which the fluorescence is lost); control = cells given the same volume of the solvent used to dissolve BQ-1 but no BQ-1, then given the labeled protein and handled identically.
- IV is BQ-1 treatment, DV is the fluorescence (undigested labeled protein) remaining per cell at a set time, and the control is an otherwise identical culture that receives solvent only, so that any difference is due to blocking the H⁺ pump rather than to the handling or the solvent.
Total for part (B): 1 point
- Grow replicate dishes of cells; treat half with BQ-1 and half with solvent only; add the same amount of labeled protein to every dish for the same uptake period, then wash away protein that was not taken up; at set times (for example 0, 1, 2 and 4 hours) measure the fluorescence remaining in the cells of each dish; use the pH dye to confirm that lysosomal pH is higher in the BQ-1-treated cells; compare the mean fluorescence remaining (±SE) in treated and control dishes at each time.
- Set up at least five BQ-1-treated and five control cultures, load all with the fluorescent protein for the same time, then follow the loss of fluorescence over several hours under identical temperature and medium conditions; also measure lysosomal pH in both groups with the dye; plot mean undigested protein against time for each treatment.
Total for part (C): 1 point
- Support: the dye shows lysosomal pH rising toward the pH of the cytosol in BQ-1-treated cells, and much more undigested labeled protein remains in these cells (fluorescence is lost slowly or not at all), whereas in control cells lysosomal pH stays about 4.5 and the fluorescence disappears within a few hours. This shows that the enzymes only work in the acidic conditions the lysosomal membrane and H⁺ pump create, so separating the enzymes into a compartment with conditions different from the cytosol is what allows digestion (and keeps the enzymes inactive if they leak into the cytosol).
- If the hypothesis is correct, BQ-1 cells retain most of the fluorescent protein while control cells digest it; the enzymes are still present, so the loss of digestion must be due to the loss of the acidic environment, demonstrating that the membrane-bound compartment maintains the conditions (low pH) the hydrolytic enzymes require. If instead digestion proceeded at the same rate in both treatments, the hypothesis would be refuted.
Total for part (D): 1 point
Total for question 4: 4 points
- Axes labelled with the given quantities and units; the y-axis includes negative values; points plotted at the tabulated values and joined with a line or best-fit curve.
- Correctly scaled axes with units; all five points plotted accurately and connected.
Total for part (A): 1 point
- As sucrose concentration increases, the percent change in mass decreases, going from a mass gain to a mass loss.
- Percent change falls as concentration rises, crossing from positive to negative values.
Total for part (B): 1 point
- They will lose mass.
- Mass will decrease (the value at 0.6 M is below 0% change).
Total for part (C): 1 point
- The tissue is isotonic at about 0.35 M, so a 0.6 M solution has a lower (more negative) water potential; water leaves the cells by osmosis down the water potential gradient, so the cores lose mass.
- Because 0.6 M is hypertonic to the tissue, water moves out of the cells by osmosis and the cores lose mass.
Total for part (D): 1 point
Total for question 5: 4 points
- As size and/or polarity increases, permeability decreases
- Smaller and less polar substances cross faster
- O₂ is small and nonpolar so it dissolves through the hydrophobic core, while large polar glucose cannot
- The hydrophobic interior admits nonpolar O₂ but blocks polar glucose
Total for part (A): 2 points
- Facilitated diffusion
- Passive (channel-mediated) transport
- Aquaporins form hydrophilic channels that let polar water bypass the hydrophobic core
- They provide a protein pathway for water to cross the membrane
Total for part (B): 2 points
- The vesicle will shrink / decrease in volume
- It will lose water and shrink
- Water moves by osmosis down its gradient out of the vesicle toward the higher external solute concentration, and aquaporins speed this so the vesicle shrinks
- Because outside solute is higher, water exits through aquaporins, reducing volume
- Aquaporins are passive channels, so the direction is still set by the osmotic/concentration gradient
- They only speed diffusion; water still moves down its gradient
Total for part (C): 3 points
- IV = presence/dose of the drug; DV = water permeability (rate of vesicle volume change)
- IV drug concentration, DV rate of water movement
- Aquaporin vesicles with no drug added (solvent only)
- Vesicles treated identically but without the drug
Total for part (D): 2 points
Total for question 6: 9 points
Other AP Biology units
- Unit 1 Chemistry of Life 96
- Unit 3 Cellular Energetics 112
- 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 →