AP Biology: Unit 4, Cell Cycle & Regulation (Drill 15)

Question 1. Based on the table, which conclusion about Line T is most directly supported by the data?

• A) Line T cells divide faster than Line N cells because a higher percentage are in M phase.
• B) Line T cells are more likely to be actively replicating DNA at any given time than Line N cells. ✓
• C) Line T cells are less likely to enter S phase and replicate their DNA than Line N cells as described in the passage.
• D) Line T cells cannot enter G0 because tumor cells lack the machinery for quiescence.

Explanation: The table shows 47% of Line T cells are in S phase (DNA replication) compared to 20% of Line N cells. At any given snapshot, nearly half of Line T cells are actively replicating DNA, directly supporting the conclusion that Line T is more likely to be replicating at any given time. A is incorrect because a lower percentage of Line T cells are in M phase (6% vs. 10%). C is incorrect because the data show phase distribution at a single time point, not the duration of each phase or the total cycle length. D overstates the data -- Line T has 1% of cells in G0, meaning some cells can still enter quiescence.

Question 2. A researcher proposes that Line T cells are less responsive to internal checkpoint signals that would normally halt cell cycle progression. Which finding from the table is most consistent with this proposal?

• A) Line T has fewer cells in M phase than Line N, indicating that its cells are pausing at internal checkpoints rather than progressing through division.
• B) Line T has more cells in S and G2 phases combined, suggesting cells are progressing through the cycle without pausing.
• C) Line T has only 1% of cells in G0, indicating that most cells are cycling rather than responding to quiescence signals.
• D) Line T has only 18% of cells in G1 compared to 45% in Line N, consistent with cells bypassing the G1 checkpoint and committing to S phase without appropriate signals. ✓

Explanation: The G1 checkpoint (restriction point) is the primary decision point at which normal cells integrate growth signals and DNA integrity checks before committing to S phase. Line T's low G1 occupancy (18% vs. 45%) is consistent with cells bypassing this checkpoint. D is superior to C because the G1 checkpoint is specifically an internal cell cycle control mechanism, whereas G0 exit reflects external quiescence signals -- making D more directly relevant to the proposal. B describes the outcome of checkpoint bypass but does not identify a specific checkpoint mechanism. A does not suggest checkpoint bypass.

Question 3. CDK inhibitors reduce cell proliferation by blocking cyclin-dependent kinases. Which mechanism best explains why Line T shows a greater reduction in cell number than Line N after CDK inhibitor treatment?

• A) Line T cells are more sensitive to all drugs because tumor cells have thinner membranes that allow greater drug uptake.
• B) Line T cells have a higher proportion of cells actively cycling, so more cells are dependent on CDK activity at the time of treatment. ✓
• C) Line T cells lack G0 cells entirely, so every single cell is simultaneously affected by CDK inhibition at the moment of treatment in the treatment scenario described.
• D) CDK inhibitors preferentially target tumor cells because tumor CDKs have a different molecular structure than normal CDKs.

Explanation: CDK activity is required for cell cycle progression, particularly at the G1/S and G2/M transitions. Line T has far more actively cycling cells (47% in S, 28% in G2, only 1% in G0). Cells in G0 are less dependent on CDK activity and therefore less affected by CDK inhibition. The greater proportion of actively cycling cells in Line T explains the greater reduction. A is biologically unsupported. C overstates the data -- 1% of Line T cells are in G0, not zero. D is incorrect -- no molecular selectivity for tumor CDKs is indicated.

Question 4. A student argues that the CDK inhibitor would be a more effective cancer treatment if it could be modified to exclusively target cells in S phase. Which reasoning best evaluates this argument?

• A) The argument is well-supported because S-phase cells are undergoing DNA replication, which makes them uniquely and exclusively vulnerable to CDK inhibition.
• B) The argument is flawed because CDK inhibitors work at the G1/S checkpoint, not within S phase itself, so targeting S phase cells would have no effect.
• C) The argument has merit because Line T has 47% of cells in S phase, meaning an S-phase-specific drug would affect nearly half the tumor cells at any given time -- but cells in other active phases would be spared, limiting effectiveness. ✓
• D) The argument is flawed because all cycling cells are equally dependent on CDK activity regardless of phase, so phase-specific targeting offers no advantage.

Explanation: The argument has partial merit -- targeting S phase cells would affect a large fraction of Line T cells (47%) given the phase distribution in the table. However, cells in G2 (28%) and other active phases would not be targeted, potentially allowing them to continue cycling. C correctly identifies both the strength of the argument and its limitation. A overstates the case by calling S phase cells "uniquely vulnerable" without mechanistic support. B is incorrect because CDKs are active during S phase as well as at checkpoints. D is incorrect because CDK activity varies by phase -- different cyclin-CDK complexes are active at different points in the cycle.

Question 5. In normal cells, the retinoblastoma protein (Rb) acts as a tumor suppressor by inhibiting progression from G1 into S phase. Which molecular event would most likely result in a cell cycle profile similar to Line T?

• A) Overexpression of Rb, causing cells to remain in G1 indefinitely.
• B) Loss-of-function mutation in Rb, allowing cells to enter S phase without appropriate growth signals. ✓
• C) Overexpression of a CDK inhibitor protein, preventing cyclin-CDK complex formation at the G1/S boundary.
• D) Deletion of S phase cyclins, preventing DNA replication initiation.

Explanation: Rb normally acts as a brake at the G1/S transition -- when hypophosphorylated, it binds and inhibits E2F transcription factors needed for S phase entry. A loss-of-function mutation in Rb removes this brake, allowing cells to enter S phase without appropriate signals, producing fewer cells in G1 and more in S phase -- consistent with Line T. A would produce the opposite: cells accumulating in G1. C would also cause G1 accumulation by preventing the cyclin-CDK phosphorylation of Rb needed for S phase entry. D would block cells at the G1/S boundary rather than producing high S phase occupancy.