Unit 2: Cell Structure and Function. Practice questions with scoring guidelines.
260 original AP-style questions on Unit 2, written to the CED learning objectives: 178 multiple-choice, 82 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 and Function 46 questions
- 2.2 Cell Size 27 questions
- 2.3 Plasma Membrane 28 questions
- 2.4 Membrane Permeability 33 questions
- 2.5 Membrane Transport 27 questions
- 2.6 Facilitated Diffusion 30 questions
- 2.7 Tonicity and Osmoregulation 44 questions
- 2.8 Mechanisms of Transport 25 questions
- 2.9 Cell Compartmentalization 26 questions
- 2.10 Origins of Cell Compartmentalization 26 questions
In the bank for Unit 2
- 178 multiple-choice
- 82 free-response
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Bacterium S has channel proteins in its plasma membrane that open only when the membrane is stretched; when open, each channel lets ions and other small solutes through. Cells of the normal strain and of a mutant strain that lacks these channels, all with intact cell walls, were grown in a medium with a high solute concentration and then moved suddenly into distilled water or into fresh growth medium. In some cells, ATP synthesis was blocked for 10 minutes before the move. The table shows the percentage of cells surviving 10 minutes after the move.
Which explanation of how the normal strain survives the move into distilled water is best supported by the data?
| Strain | ATP synthesis | Moved into | Cells surviving (%) |
|---|---|---|---|
| Normal | not blocked | fresh growth medium | 99 |
| Mutant (lacks the channels) | not blocked | fresh growth medium | 98 |
| Normal | not blocked | distilled water | 92 |
| Mutant (lacks the channels) | not blocked | distilled water | 8 |
| Normal | blocked | distilled water | 90 |
| Mutant (lacks the channels) | blocked | distilled water | 7 |
- The channels pump solutes out of the cells using energy from ATP, lowering the solute concentration inside
- Stretching opens the channels, so solutes leave down their gradients without ATP and less water enters
- The cell wall alone prevents bursting; the channels matter only for taking up solutes after the move
- Open channels let water out as fast as it enters, so the cells take up no water and never swell at all
The figure shows a mitochondrion in section, with the measured area of each of its two membranes. The mitochondrion is placed in a solution with a very low solute concentration, so water enters the matrix, which swells and becomes rounded. A membrane tears if it is stretched by more than a few percent. For a sphere of radius r, surface area = 4πr² and volume = (4/3)πr³. Which outcome is most likely by the time the matrix has swollen to a volume of 5 µm³?
- Both membranes stay intact, because together they have 43 µm² of membrane, more than enough to enclose a 5 µm³ sphere
- The inner membrane tears, because it presses directly on the swelling matrix, while the smooth outer membrane stays intact
- Both membranes tear, because neither of them can stretch, and a 5 µm³ sphere needs more area than either one of them has
- The outer membrane tears, while the inner membrane stays intact as its folds open out to enclose the swollen matrix
A student cuts equal-sized cylinders from a fresh potato, blots and weighs each one, then submerges each cylinder in a sucrose solution of known molarity for 60 minutes. The cylinders are removed, blotted, and reweighed, and the mean percent change in mass is calculated. The results are shown in the table.
Predict what would happen if a fresh cylinder were left in distilled water (0.0 M) for a much longer period of time. Which outcome is most likely?
| Sucrose concentration (M) | Mean percent change in mass (%) |
|---|---|
| 0.0 | +12.4 |
| 0.2 | +5.1 |
| 0.4 | -2.3 |
| 0.6 | -8.7 |
| 0.8 | -14.0 |
- It would gain mass until turgor pressure prevents further water entry.
- It would lose mass as water exits the cells into the pure water.
- It would show 0.0% change because distilled water is isotonic to the cells.
- It would burst, because plant cells lack a cell wall.
The Ca²⁺ concentration in the cytosol of an animal cell is about 0.0001 mM, and the fluid around the cell contains about 2 mM Ca²⁺. Ca²⁺ enters the cytosol continually through open channel proteins in the plasma membrane, while Ca²⁺ pumps in the plasma membrane move it back out. When the cells were treated with a drug that closes the channels, the cytosolic Ca²⁺ concentration stayed at about 0.0001 mM, but the cells' total rate of ATP use fell by 4 percent. Which statement best explains these results?
- Less Ca²⁺ leaked in, so the pumps had to move less Ca²⁺ out to hold the same level, and used less ATP.
- The open channels had used ATP to let Ca²⁺ in, so closing them saved the ATP that the channels used.
- The pumps had moved Ca²⁺ down its gradient, so fewer open channels left less Ca²⁺ for them to move.
- Closing the channels reversed the Ca²⁺ gradient, so Ca²⁺ could then leave the cytosol without the pumps.
A species of bacterium releases toxin T, a protein that kills animal cells by making large holes in their plasma membranes. Researchers made two kinds of artificial vesicle, each filled with a dye: one whose membrane contained phospholipids only, and one whose membrane contained the same phospholipids together with cholesterol, in about the proportion found in animal plasma membranes. Each kind of vesicle was kept for 10 minutes in a solution with or without toxin T, and the percent of its dye that leaked out was measured in eight replicate samples. The results are shown in Table 1.
The bacterium that makes toxin T is enclosed by a plasma membrane but is not harmed by the toxin. Which explanation is best supported by the data?
| Solution | Phospholipids only (% of dye released; mean ±2SE) | Phospholipids and cholesterol (% of dye released; mean ±2SE) |
|---|---|---|
| Without toxin T | 4 ± 2 | 3 ± 1 |
| With toxin T | 7 ± 3 | 81 ± 6 |
- T makes small holes in any phospholipid bilayer, as it raised dye release from vesicles of phospholipids only from 4 to 7 percent; the bacterium repairs them.
- Cholesterol itself makes a bilayer leaky, since with T present far more dye leaked from the vesicles with cholesterol, so the bacterium's membrane lacks it.
- The bacterium's cell wall shields it, since T damages both kinds of bilayer: with T, dye release from each (4 to 10 and 75 to 87 percent) is above zero.
- T makes holes only in membranes that contain cholesterol, a steroid found in the membranes of eukaryotic cells but not in the bacterium's own membrane.
Two species of single-celled alga, P and Q, each have one chloroplast. In both species, the proteins that capture light energy and release O₂ are embedded in the thylakoid membranes inside the chloroplast. Researchers measured the volume of the chloroplast, the total area of its thylakoid membranes, and the rate at which it released O₂ in bright light. Table 1 shows the means for 30 cells of each species. A student studying the table writes: "Per µm³ of chloroplast, Q releases O₂ four times as fast as P. So each light-capturing protein in Q must work four times as fast as each one in P."
Which statement best identifies the flaw in the student's reasoning?
| Species | Chloroplast volume (µm³) | Thylakoid membrane area (µm²) | Rate of O₂ release (relative units) |
|---|---|---|---|
| P | 60 | 900 | 18 |
| Q | 30 | 1,800 | 36 |
- Q's smaller chloroplast has the larger outer surface of the two, so it can absorb more light through its outer membranes than P's chloroplast can.
- Dividing by thylakoid area is fairer, and it shows that Q's rate per µm² of membrane is twice P's, so each of Q's proteins works twice as fast as P's.
- The reactions take place in the thylakoid membranes, and the rate per µm² of membrane is the same in both species, so Q's proteins need not be faster.
- There is no flaw: the reactions that release O₂ take place throughout the chloroplast's volume, so the rate per µm³ compares the proteins fairly.
The figure shows part of a leaf cell. In the chloroplast, O₂ is released into the space inside the thylakoids. Some of this O₂ diffuses to the matrix of the mitochondrion in the same cell. How many membranes, each a single phospholipid bilayer, must an O₂ molecule cross on this path, and how does it cross them?
- Five, crossing each one through channel proteins, since O₂ cannot pass through a bilayer unaided
- Six, counting the plasma membrane as well, crossing each one by simple diffusion between phospholipids
- Five, counting the thylakoid membrane as well, crossing each by simple diffusion between phospholipids
- Four, the two membranes that surround each of the organelles, crossing each by simple diffusion
The figure is a model of a cube-shaped cell from the lining of an earthworm's gut. Its top face, which faces the food in the gut, carries 2,000 finger-like projections of the plasma membrane (microvilli), each filled with cytoplasm. Model each projection as a cylinder of radius r and height h whose curved side adds 2πrh of membrane area and πr²h of volume to the cell; its flat tip only replaces the patch of top face it stands on. By about how many times do the projections raise the cell's surface area-to-volume ratio, compared with the same cube without them?
- About 7 times: the membrane area of the top face rises from 100 µm² to about 730 µm² in total
- About 2 times: the projections add about 630 µm² of membrane but only about 16 µm³ of cytoplasm
- About 3 times: the projections add about 1,260 µm² of membrane but only about 63 µm³ of cytoplasm
- About 1 time: the projections add volume as well as membrane area, so the ratio stays about the same
A biologist compares a bacterial cell with a plant cell using electron microscopy. Which structural difference best reflects the greater compartmentalization of the eukaryotic cell?
- The bacterium contains a membrane-bound nucleus, while the plant cell does not.
- Only the plant cell contains membrane-bound organelles such as a nucleus.
- The plant cell lacks a plasma membrane, relying on its wall to enclose the cytoplasm.
- Only the bacterium possesses ribosomes, which are needed for protein synthesis.
Researchers tried to remove the cell walls of a plant, a fungus and a bacterium without killing the cells, in a medium with the same solute concentration as their cytoplasm. Each of three different polysaccharide-digesting enzymes removed the walls of just one of the three kinds of cell, a different kind for each enzyme, leaving the cells alive and bounded by their plasma membranes. An enzyme that digests proteins removed none of the walls, and an enzyme that digests phospholipids killed all the cells but left their walls in place. Which conclusion is best supported?
- The walls are made mainly of protein, which the polysaccharide-digesting enzymes loosen until they come away.
- All three walls are made of the same polysaccharide, so any one of the three enzymes would remove all of them.
- The walls of all three are made mainly of polysaccharides, but of a different kind of polysaccharide in each.
- The walls are made of phospholipids like those of the membrane, so the phospholipid enzyme broke them down.
Figure 1 is a model of carrier protein T in the plasma membrane of a cell. Solute S is a polar molecule that does not cross the phospholipid bilayer at a measurable rate. T switches back and forth between shape 1, in which its binding site for S faces the outside of the cell, and shape 2, in which the site faces the cytosol. T switches at random, whether or not S is bound, and uses no ATP. The site binds S equally well in either shape, and a bound molecule of S is released on whichever side the site faces after a switch. S is 5 mM outside the cell and 1 mM in the cytosol.
Which statement best explains why there is net movement of S into the cell through T?
- Facing outward, the site meets more S molecules, so it is filled more often and carries more S in than out.
- T switches to face the cytosol only while S is bound, so it moves S inward in one direction only.
- T holds S more tightly when its site faces outward, so it releases S only into the cytosol.
- T carries equal numbers of S each way, so the net entry must come from S crossing the bilayer itself.
Cylinders of beetroot tissue of equal size were rinsed and then placed in separate tubes of water held at different temperatures for 20 minutes. Beetroot cells store a red pigment inside their vacuoles; when the cell membranes are damaged, the pigment leaks out into the surrounding water. The absorbance of the surrounding water (a measure of how much pigment leaked) was recorded. Table 1 shows the results.
A student predicts that adding a lipid-dissolving solvent such as ethanol to a beetroot cylinder held at 20 °C would increase pigment leakage. Which reasoning best supports this prediction?
| Temperature (°C) | Absorbance of surrounding water |
|---|---|
| 20 | 0.06 |
| 30 | 0.08 |
| 40 | 0.15 |
| 50 | 0.34 |
| 60 | 0.72 |
| 70 | 0.95 |
- Ethanol supplies ATP that pumps the pigment out.
- Ethanol cools the tissue, freezing the membrane solid.
- Ethanol makes the membrane thicker and less permeable.
- Ethanol dissolves the phospholipids of the membrane.
Researchers investigated how a drug that blocks aquaporin channels affects the movement of water into cells. Cells were exposed to one of four treatments, and the rate of osmotic water uptake was measured in eight replicate trials for each treatment. Table 1 shows the mean rate of water uptake ± standard error (SE). All cells were placed in the same hypotonic solution.
Which statement best describes the effect of the aquaporin blocker on the rate of water uptake, based on Table 1?
| Treatment | Mean water uptake rate ± SE (μm³/s) |
|---|---|
| Control (no drug) | 18.4 ± 0.9 |
| Low-dose blocker | 13.1 ± 1.0 |
| High-dose blocker | 6.2 ± 0.7 |
| Cells lacking aquaporins | 5.8 ± 0.6 |
- As the dose of the blocker increases, the mean rate of water uptake decreases.
- The blocker increases the rate of water uptake in a dose-dependent way.
- The blocker has no measurable effect on the rate of water uptake.
- Only cells that lack aquaporins are affected by the blocker.
Red blood cells were placed in a series of sodium chloride (NaCl) solutions of different concentrations for 10 minutes. The relative volume of the cells, compared with their normal volume in blood plasma (taken as 100%), was then measured. Red blood cells lack a cell wall. Table 1 shows the results.
Which NaCl concentration is closest to being isotonic to the cytoplasm of the red blood cells?
| NaCl concentration (%) | Relative cell volume (%) |
|---|---|
| 0.00 | lysed |
| 0.45 | 132 |
| 0.90 | 100 |
| 1.50 | 79 |
| 3.00 | 58 |
- 0.45%
- 0.90%
- 1.50%
- 3.00%
Researchers investigated whether the uptake of large particles by cultured macrophage-like cells requires the cytoskeletal protein actin. Cells drawn from a single culture were divided between two treatments and incubated for 30 minutes at 37 °C with fluorescent latex beads supplied at the same concentration to both treatments. One treatment received buffer alone; the other received cytochalasin D, a compound that prevents actin filaments from assembling. Each cell was then examined under a fluorescence microscope and scored as either 'beads internalized' or 'no beads internalized', and 200 cells were scored in each treatment.
Which of the following states the null hypothesis for this investigation?
| Treatment | Cells with beads internalized | Cells with no beads internalized | Total cells scored |
|---|---|---|---|
| Buffer only | 152 | 48 | 200 |
| Cytochalasin D | 46 | 154 | 200 |
- Cytochalasin D reduces the number of cells that internalize latex beads, relative to untreated cells.
- Fewer than half of the cells treated with cytochalasin D will internalize beads during the assay.
- Assembled actin filaments are required for the internalization of latex beads by these cells.
- The proportion of cells that internalize beads is the same whether or not cytochalasin D is present.
Researchers made two fluorescent probe molecules, F1 and F2. Each has two long nonpolar hydrocarbon tails joined to a small head group. The head of F1 carries a full negative charge; the head of F2 is polar but uncharged. Each probe was added to the solution around cultured cells, where within a minute it entered the outer layer of phospholipids of the plasma membrane. At intervals, compound Q was added to samples of the cells. Q cannot cross membranes, and it dims the glow of any probe head that it touches. No probe left the plasma membrane during the experiment. Table 1 shows the percent of each probe's glow that Q dimmed.
Which of the following best explains the results?
| Time after probe was added (minutes) | F1, charged head (% dimmed) | F2, uncharged polar head (% dimmed) |
|---|---|---|
| 1 | 97 | 96 |
| 20 | 96 | 71 |
| 60 | 95 | 55 |
| 120 | 96 | 50 |
- Both probes cross to the inner layer at the same rate, but Q binds more strongly to F1's charged head and dims it more.
- F2's head is less polar than F1's, so F2 sinks into the nonpolar middle of the bilayer, where Q cannot reach its head.
- F2's uncharged head can pass the bilayer's nonpolar interior, so F2 spreads evenly into both layers; F1's head cannot.
- F1's charged head is attracted to the charged phosphate heads of the inner layer, which holds F1 out of the reach of Q.
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.
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- B: water entering the cells stretches the membrane and opens the channels; solutes then leave passively, down their concentration gradients, which lowers the solute concentration inside so that less water enters and the cells do not burst. Blocking ATP synthesis did not change survival (90 versus 92 percent), so no pumping is involved. A is ruled out by the ATP result, C by the channels opening only when the membrane is stretched, which needs the cells to swell, and D by the mutant, which has a wall yet mostly dies in water.
- D: a sphere of 5 µm³ has r = (3 × 5 ÷ 4π)^(1/3) ≈ 1.06 µm and a surface area of 4πr² ≈ 14 µm². The smooth outer membrane has only 8 µm², so it must tear before the matrix reaches that size, whereas the folded inner membrane has 35 µm², so its folds can open out and it still encloses the swollen matrix. B has the order backwards: the folds give the inner membrane spare area; C adds the two areas, but each membrane must enclose the matrix on its own; D overlooks that the inner membrane's 35 µm² is more than the 14 µm² needed.
- A: In hypotonic distilled water, water keeps entering until the cell wall's turgor pressure balances the osmotic gradient.
- A: the pumps move Ca²⁺ out against a steep gradient, which uses ATP, and to hold a steady concentration they must remove as much Ca²⁺ as leaks in; closing the channels cuts the leak, so less pumping, and less ATP, keeps the same low level. B is wrong because channels allow passive movement down a gradient, C because the concentrations did not change, and D because the pumps work against the gradient.
- D: for vesicles of phospholipids only, the ±2SE intervals are 2 to 6 percent without T and 4 to 10 percent with T, which overlap, so T has no significant effect on a bilayer without cholesterol; for vesicles with cholesterol they are 2 to 4 percent without T and 75 to 87 percent with T, which do not overlap. So T needs cholesterol to make its holes. Cholesterol is a steroid embedded among the phospholipids of eukaryotic membranes, such as the plasma membranes of animal cells; the plasma membrane of a bacterium, a prokaryote, lacks it, so T cannot damage the bacterium's own membrane. B compares means only: 7 against 4 percent is not significant. C uses only the row with T: without T, the two kinds of vesicle released the same small amount of dye (2 to 4 and 2 to 6 percent overlap), so cholesterol alone does not make a bilayer leaky. D compares the release with T against zero instead of against the vesicles kept without T, and phospholipids only gave no significant rise.
- C: dividing each rate by the thylakoid area gives 18 / 900 = 0.02 for P and 36 / 1,800 = 0.02 for Q, so the rate per µm² of the membrane that holds the light-capturing proteins is the same in both; Q releases twice as much O₂ because it packs twice the membrane area into half the volume, so its internal membranes, not faster proteins, account for the difference. Dividing by volume, as the student did, treats the reactions as if they took place throughout the chloroplast. A contradicts the context, which places the proteins in the thylakoid membranes; C miscalculates the rate per µm²; D is wrong because the smaller chloroplast has the smaller outer surface, and light is captured at the thylakoids.
- C: the O₂ crosses the thylakoid membrane, the inner and then the outer chloroplast membrane, and then the outer and inner mitochondrial membranes: five bilayers in all; it never leaves the cell, so it does not cross the plasma membrane, and as a small nonpolar molecule it passes through each bilayer by simple diffusion. A leaves out the thylakoid membrane; B adds the plasma membrane, which lies off the path; C wrongly says that O₂ needs channel proteins.
- B: without projections the cube has 6 × 10² = 600 µm² of membrane and 10³ = 1,000 µm³ of volume, a ratio of 0.6 per µm. Each projection (r = 0.05 µm, h = 1.0 µm) adds 2π(0.05)(1.0) = 0.31 µm² of membrane and π(0.05)²(1.0) = 0.008 µm³ of volume, so 2,000 of them add about 630 µm² and 16 µm³, and the ratio becomes 1,230 ÷ 1,016 = 1.2 per µm, about twice 0.6. A misses that the projections add far more area than volume; C uses the diameter, 0.1 µm, as the radius; D compares the top face alone and leaves out the other five faces and the volume of the cell.
- B: Membrane-bound organelles create separate internal compartments, a hallmark of eukaryotic cells absent in prokaryotes.
- C: each wall was removed by an enzyme that digests a polysaccharide, so the walls are made of these complex carbohydrates, and since each enzyme removed only one kind of wall, the polysaccharide differs among the three. B is contradicted by the protein-digesting enzyme removing no walls, C by the walls staying in place when the phospholipids were digested (killing the cells by breaking down their membranes), and D by each enzyme removing only one kind of wall.
- A: T switches at random and binds S equally well on either side, so the chance that it carries a molecule of S on any one switch depends only on how many molecules of S its site meets. With 5 mM S outside and 1 mM inside, the site is filled more often when it faces outward, so more S is carried in than out: net movement follows the concentration gradient, with no ATP used. A and B contradict the model, in which binding is equally strong on both sides and switching happens whether or not S is bound; D is wrong because T carries more S inward while the concentrations differ, and S does not cross the bilayer at a measurable rate.
- D: because the membrane's permeability barrier is its lipid bilayer, a lipid solvent disrupts it and lets pigment leak out even at low temperature.
- A: higher blocker doses give progressively lower mean water uptake, so increasing the dose decreases uptake.
- B: at 0.90% the cells stay at their normal volume (100%), indicating no net water movement and therefore equal water potential inside and out.
- D: the null hypothesis states that the treatment makes no difference to the measured outcome; the other three choices assert or predict that an effect exists.
- C: Q reaches only heads in the outer layer. F2's dimming falls and levels off at 50%, so half of F2 has moved to the inner layer, which its small uncharged head can do by passing through the nonpolar interior of the bilayer; F1's charged head cannot cross the nonpolar interior, so F1 stays in the outer layer and stays about 96% dimmed. A would lower F2's dimming toward 0%, not level it at 50%. B contradicts F1's steady 95% to 97% dimming. C contradicts the equal dimming of the two probes at 1 minute, when both were in the outer layer.
- 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 17: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 196
- Unit 3 Cellular Energetics 195
- Unit 4 Cell Communication and Cell Cycle 187
- Unit 5 Heredity 203
- Unit 6 Gene Expression and Regulation 219
- Unit 7 Natural Selection 316
- Unit 8 Ecology 215
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