Biology by Bradford
Unit 4: Cell Communication and Cell Cycle. Practice questions with scoring guidelines.
187 original AP-style questions on Unit 4, written to the CED learning objectives: 130 multiple-choice, 57 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
- 4.1 Cell Communication 29 questions
- 4.2 Introduction to Signal Transduction 42 questions
- 4.3 Signal Transduction Pathways 46 questions
- 4.4 Feedback 29 questions
- 4.5 Cell Cycle 25 questions
- 4.6 Regulation of Cell Cycle 35 questions
In the bank for Unit 4
- 130 multiple-choice
- 57 free-response
Every question is original and tagged to a learning objective and science practice.
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Researchers used time-lapse microscopy to follow single cells of two cultured human cell lines, J and K, through one complete cell cycle. A fluorescent marker protein whose pattern in the nucleus changes from one phase of interphase to the next let them time G1, S and G2; the start of chromosome condensation and the completion of cytokinesis marked the start and end of mitosis (M). Ten cells of each line were timed. The results are shown in Table 1.
Which claim about the events of the cell cycle in line K, compared with line J, is best supported by the data?
| Phase | Line J (hours; mean ±2SE) | Line K (hours; mean ±2SE) |
|---|---|---|
| G1 | 10.0 ± 2.4 | 8.2 ± 1.6 |
| S | 6.0 ± 0.4 | 7.4 ± 0.6 |
| G2 | 3.0 ± 0.6 | 3.6 ± 0.8 |
| M | 1.0 ± 0.2 | 1.1 ± 0.2 |
- Line K cells complete the whole cell cycle more slowly, because one phase is significantly longer and the means of the two after it are higher.
- Line K cells take significantly longer to replicate their DNA, because the intervals 5.6 to 6.4 and 6.8 to 8.0 hours do not overlap at all.
- No event takes significantly longer in line K, because with twice the ±2SE as the interval, 6.0 and 7.4 span 5.2 to 6.8 and 6.2 to 8.6 hours.
- Line K cells spend less time growing before they copy their DNA, because that phase's mean is 1.8 hours shorter, the largest difference.
In cardiac muscle tissue, adjacent cells are connected by gap junctions that allow ions to flow directly from the cytoplasm of one cell into the next, so an electrical signal spreads rapidly and the cells contract in a coordinated wave. Which type of cell communication does this best illustrate?
- Quorum sensing based on population density
- Long-distance signaling by hormones traveling through the bloodstream
- Direct signaling between adjacent cells through connecting cell junctions
- Signaling that requires secretion of a molecule into the extracellular space
In some cancers, a relay protein that passes growth signals from the growth-factor receptor to a downstream kinase cascade carries a mutation that keeps it permanently switched on. Which treatment is most likely to slow division of these cancer cells?
- Adding a drug that inhibits a kinase acting after the relay protein
- Adding a drug that blocks the growth-factor receptor on the cell surface
- Adding extra growth factor so that every receptor on the cell is occupied
- Removing all of the growth factor from the fluid surrounding the cells
A line of cultured cells divides only when a growth factor is present in the medium; in these cells G1 lasts about 10 hours and S phase about 8 hours. Researchers grew the cells with growth factor, collected cells that had just completed mitosis, and removed the growth factor from separate samples at different times afterward. They recorded the percentage of cells in each sample that had completed S phase 24 hours after mitosis. The results are shown in the table. Which conclusion is best supported by the data?
| Time growth factor was removed (hours after mitosis) | Cells that completed S phase (%) |
|---|---|
| 1 | 3 |
| 3 | 4 |
| 5 | 6 |
| 7 | 91 |
| 9 | 94 |
| Not removed | 95 |
- Cells need growth factor only until 5 to 7 hours into G1, and they then complete S phase without it
- Removing growth factor at 7 or 9 hours had little effect because those cells were already in S phase
- Cells need growth factor only in the first 3 hours after mitosis, and then complete S phase without it
- Cells need growth factor throughout G1, so removing it at any time before S phase prevents S phase
The model shows how the concentration of glucose in human blood is regulated by two hormones from the pancreas.
A drug blocks the glucagon receptors on liver cells. Which result is predicted during a long fast, when blood glucose is falling?
- The α cells stop secreting glucagon because glucose is not detected
- Muscle cells take up more glucose to compensate for the drug
- Blood glucose falls further because glycogen is not broken down
- Blood glucose rises above the set point because insulin stops
Researchers study a signaling pathway in a cultured animal cell line. Hormone H binds a G-protein-coupled receptor on the cell surface, which activates the enzyme phospholipase C. Phospholipase C produces the second messenger IP₃, which causes Ca²⁺ to be released from the endoplasmic reticulum (ER) into the cytosol. The rise in cytosolic Ca²⁺ activates the cellular response. The researchers measured cytosolic Ca²⁺ (relative units) under the conditions shown in the table.
According to the pathway described, IP₃ is produced by phospholipase C and then causes Ca²⁺ to be released from the endoplasmic reticulum. Which role does IP₃ play in this pathway?
| Condition | Cytosolic Ca²⁺ (relative units) |
|---|---|
| Normal cells + hormone H | 9 |
| Normal cells, no hormone H | 1 |
| Normal cells + hormone H + receptor antagonist | 1 |
| Normal cells, no hormone H, IP₃ injected into the cytosol | 9 |
| Normal cells + hormone H, ER Ca²⁺ stores emptied beforehand | 1 |
| Mutant cells (ER Ca²⁺ channel cannot open when IP₃ binds) + hormone H | 1 |
- The surface receptor for hormone H.
- A second messenger that triggers Ca²⁺ release.
- The extracellular ligand binding the receptor.
- The final cellular response of the pathway.
A growth signal binds a receptor protein that crosses the plasma membrane. When the signal binds, two receptor molecules pair up, and the intracellular region of each receptor adds phosphate groups to the intracellular region of its partner; a receptor passes the signal on into the cell only once its partner has phosphorylated it. Researchers make cells that produce a normal receptor and a mutant receptor in equal amounts, with the same total number of receptor molecules as unmodified cells. The mutant receptor binds the signal and pairs normally but has no intracellular region. Assuming receptors pair at random and the response is proportional to the number of pairs that relay the signal, the response of the engineered cells to a high concentration of signal is closest to what percentage of the response of unmodified cells?
- About 50 percent, because half of all the receptor molecules in the engineered cells are normal and can relay the signal
- About 0 percent, because every normal receptor is paired with a mutant receptor, so no pair can add any phosphate groups
- About 100 percent, because the normal receptors still bind the signal, and the mutant receptors simply do nothing at all
- About 25 percent, because only a pair of two normal receptors relays the signal, and 1/2 × 1/2 of all pairs are such pairs
Drug H is used to treat some tumors. It enters all cells, but it kills a cell only if the cell is replicating its DNA while the drug is present; cells in G1, G2, or M are not harmed. Compound R inhibits a cyclin-dependent kinase (CDK) that cells need in order to pass the G1 checkpoint. Researchers studied two lines of cultured human cells that each complete a cell cycle in about 24 hours: line N, made from normal cells, and line T, made from a tumor. Cells of line T keep dividing at their usual rate when R is present.
Each line was given one of four treatments: no drug; R alone for 30 hours; H alone for 6 hours; or R for 30 hours with H added for the last 6 of those hours. The drugs were then washed out, and one day later the researchers determined the percentage of cells in each culture that were alive. The results are shown in Table 1.
| Treatment | Line N cells alive (%) | Line T cells alive (%) |
|---|---|---|
| No drug | 97 ± 2 | 96 ± 3 |
| R alone | 95 ± 3 | 95 ± 2 |
| H alone | 58 ± 5 | 55 ± 6 |
| R, with H for the last 6 hours | 94 ± 4 | 53 ± 6 |
In the 1950s, researchers studied how the hormone epinephrine makes liver cells break down glycogen. The enzyme that breaks down glycogen, glycogen phosphorylase, is found in the cytosol. The researchers tested two hypotheses. Hypothesis 1: epinephrine activates glycogen phosphorylase by acting on the enzyme directly. Hypothesis 2: epinephrine acts on the cell membranes, which then make a substance that activates the enzyme. They broke liver cells open and separated the contents into a membrane fraction (pieces of the plasma membrane and other membranes, with no glycogen phosphorylase) and a soluble fraction (the cytosol, with inactive glycogen phosphorylase and no membranes). For tubes 3 to 5, they first incubated the membrane fraction with or without epinephrine, then removed the membranes and added the fluid that remained to the soluble fraction. Heating to 100 °C for 10 minutes destroys the activity of the enzymes and other proteins in these fluids. They then measured the activity of glycogen phosphorylase in each tube. The results are shown in Table 1.
| Tube | Contents | Glycogen phosphorylase activity (relative units) |
|---|---|---|
| 1 | Soluble fraction only | 4 |
| 2 | Soluble fraction + epinephrine | 5 |
| 3 | Soluble fraction + fluid from membranes incubated without epinephrine | 5 |
| 4 | Soluble fraction + fluid from membranes incubated with epinephrine | 38 |
| 5 | Soluble fraction + fluid as in tube 4, heated to 100 °C for 10 minutes before it was added | 36 |
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- B: DNA is replicated in S phase. The ±2SE intervals for S are 5.6 to 6.4 hours (J) and 6.8 to 8.0 hours (K); they do not overlap, so the longer S phase in K is a significant difference. The G1 intervals (7.6 to 12.4 and 6.6 to 9.8), G2 intervals (2.4 to 3.6 and 2.8 to 4.4) and M intervals (0.8 to 1.2 and 0.9 to 1.3) all overlap, so the 1.8-hour difference in mean G1, though the largest difference in the table (A), is not significant, and neither are K's higher G2 and M means. The table values are already ±2SE; doubling them (B) widens the intervals beyond 95 percent. D generalises from one row: the means add to 20.0 hours for J and 20.3 hours for K, so the whole cycle is not detectably slower.
- C: Gap junctions link the cytoplasm of neighboring cells, allowing signals to pass directly without secreting a molecule into the surroundings.
- A: The mutant relay protein is active without any signal from the receptor, so only blocking a step downstream of it can stop the division signal; removing growth factor or blocking the receptor acts upstream of the defect.
- A: Removing growth factor at 5 hours or earlier stopped almost all cells, but removing it at 7 or 9 hours, while the cells were still in their 10-hour G1, hardly mattered; so the cells pass the growth factor-dependent G1 checkpoint between 5 and 7 hours and are then committed to S phase. Removal at 5 hours still blocked S phase, which rules out a need confined to the first 3 hours.
- C: Glucagon is still secreted, but without its receptors the liver cannot respond by breaking down glycogen and releasing glucose, so the lower loop in the model fails and blood glucose keeps falling below the set point.
- B: IP₃ is a small intracellular second messenger that carries the signal from phospholipase C to the ER, triggering Ca²⁺ release.
- D: in a mixed pair the mutant has no intracellular region, so it cannot phosphorylate its normal partner and has no intracellular region of its own to pass the signal on; only pairs of two normal receptors relay the signal. With random pairing, each receptor has a 1/2 chance of being normal, so 1/2 × 1/2 = 1/4 of the pairs (25 percent) are normal pairs, compared with all of the pairs in unmodified cells. A ignores that the mutant receptors pair with normal ones; B counts receptors rather than working pairs; D forgets that random pairing still brings some normal receptors together.
- S phase is only part of the cell cycle, so at any time only some of the cells are replicating their DNA; only cells that are in S phase, or enter it, during the 6 hours are killed, while cells that spend the whole exposure in G1, G2, or M are not harmed.
- The cells are spread through the phases of the cycle, and in 6 hours (a quarter of a 24-hour cycle) many cells never reach S phase, so H does not affect them.
Total for part (A): 1 point
- A bar graph (or plotted points) with the four treatments on the x-axis and cells alive (%) on the y-axis, both axes labeled, a linear scale that uses most of the grid, and a key that distinguishes line N from line T.
- Correctly labeled axes and scale with the two cell lines shown side by side for each treatment and identified in a legend.
- All eight means plotted correctly (97, 95, 58 and 94 for line N; 96, 95, 55 and 53 for line T), each with an error bar extending the stated 2SE above and below the mean.
- Eight accurately placed bars or points, each with a correct ±2SE error bar.
- Yes, the difference is significant: the ±2SE intervals 53 to 63% (H alone) and 90 to 98% (R with H) do not overlap.
- Significantly higher survival with R, because the error bars of the two treatments do not overlap.
- R raises survival of line N cells given H from 58% to 94%, close to untreated cells (97%), but has no significant effect on line T, whose survival stays at about the level of H alone (53% compared with 55%, overlapping intervals).
- R protects line N cells from H but does not protect line T cells.
Total for part (B): 4 points
- It shows whether R itself affects survival; because R alone gives about the same survival as no drug (95% compared with 97% and 96%), differences in the R with H treatment can be attributed to R changing the effect of H rather than to R killing (or protecting) cells on its own.
- It is a control for the effect of R by itself, so that the survival with R and H can be compared with the right baseline.
- An increase of about 62%: (94 − 58) / 58 × 100 = 62% (62.1%).
- +62%.
Total for part (C): 2 points
- Claim: R holds line N cells in G1, so none of them replicates DNA while H is present, but line T cells pass the G1 checkpoint even when the CDK is inhibited, enter S phase, and are killed. Evidence: line T keeps dividing in R, and its survival with R and H (53 ± 6%) is the same as with H alone (55 ± 6%), whereas line N survival rises to 94 ± 4%.
- R works by stopping cells at the G1 checkpoint; tumor cells that ignore this checkpoint still reach S phase, so H kills them as if R were absent, as shown by the overlapping intervals for line T.
- If R kept H out of cells, R would protect line T cells as well, because H enters all cells; but line T survival with R and H (53%) is no higher than with H alone (55%), so H still enters and kills cells given R. R protects only the cells it holds in G1.
- R gives no protection to line T cells, which are also exposed to R, so R cannot be blocking the entry of H in general.
Total for part (D): 2 points
Total for question 8: 9 points
- Refuted: epinephrine added directly to the soluble fraction, which contains the enzyme (tube 2), gives an activity of 5, no higher than the soluble fraction alone (4, tube 1).
- Refuted, because the enzyme is not activated when epinephrine is with it but no membranes are present (tube 2 about the same as tube 1).
Total for part (A): 1 point
- The membranes make (release) a substance that activates the enzyme only when they are exposed to epinephrine: fluid from membranes incubated with epinephrine raises activity to 38, but fluid from membranes incubated without it gives only 5.
- Epinephrine causes the membranes to produce an activating factor; tube 4 (38) against tube 3 (5) shows the factor appears only when epinephrine is present.
Total for part (B): 1 point
- The substance is not a protein (not an enzyme): heating destroys the activity of proteins, yet the heated fluid still raises the enzyme's activity almost as much as unheated fluid (36 against 38).
- It is a non-protein (small) molecule that survives heating, since its effect is kept after a treatment that destroys the activity of proteins.
Total for part (C): 1 point
- Epinephrine acts at the membrane, not on the enzyme, and the membrane then makes a different, small non-protein molecule that carries the signal to the enzyme in the cytosol, which is the role of a second messenger; the hormone itself never needs to reach the enzyme.
- The signal is received at the membrane and relayed to the enzyme by a second molecule made there (a second messenger), so the first signal does not have to enter the cytosol to change the enzyme's activity.
Total for part (D): 1 point
Total for question 9: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 196
- Unit 2 Cell Structure and Function 260
- Unit 3 Cellular Energetics 195
- Unit 5 Heredity 203
- Unit 6 Gene Expression and Regulation 219
- Unit 7 Natural Selection 316
- Unit 8 Ecology 215
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