AP Biology Quiz: Regulation Of Cell Cycle

What this quiz covers

This quiz focuses on Regulation Of Cell Cycle, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Increase kinase activity that adds inhibitory phosphate onto CDK to block mitotic entry
2. Decrease APC/C activity so cyclin persists and CDK signaling remains elevated
3. Reduce cyclin availability so CDK-dependent phosphorylation occurs at a lower rate (correct answer)
4. Stabilize cohesin to prevent chromatid separation and accelerate mitotic exit
5. Increase DNA replication origin firing to ensure rapid completion of S phase

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on CDK-cyclin dynamics and phosphorylation balance during mitosis. The phosphatase inhibitor sustains high phosphorylation of mitotic substrates by preventing dephosphorylation, leading to delayed mitotic exit in intact cells. Reducing cyclin availability decreases CDK-dependent phosphorylation rates, allowing existing phosphatase activity to lower phosphorylation levels more quickly. This shortens mitotic duration by reducing the overall signaling strength opposing exit. A tempting distractor is B, which suggests decreasing APC/C activity to persist cyclin, but this would prolong high CDK activity, stemming from the misconception that inhibiting degradation accelerates exit rather than delays it. To approach similar problems, consider the balance between kinase and phosphatase activities and select changes that tip it toward the desired outcome.

Question 2

1. A slower, less abrupt rise in CDK activity, delaying entry into mitosis (correct answer)
2. Immediate anaphase onset because APC/C is activated earlier by the phosphatase
3. Permanent G1 arrest because separase cannot cleave cohesin without phosphatase activation
4. Faster mitotic entry because inhibitory phosphorylation accumulates more quickly on CDK
5. Failure of DNA replication because kinetochore attachment signaling is reduced in G2

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on positive feedback in CDK activation at G2/M. The mutation prevents CDK from activating the phosphatase that removes its inhibitory phosphate, disrupting the feedback loop for rapid CDK activation. This results in a slower, less abrupt rise in CDK activity as inhibitory phosphate removal is impaired, delaying mitotic entry. The remaining inhibition of the kinase provides some activation but lacks the switch-like rapidity. A tempting distractor is D, which suggests faster entry due to phosphorylation accumulation, but accumulation inhibits, stemming from the misconception that losing positive feedback accelerates rather than slows transitions. To approach similar problems, dissect feedback mechanisms and evaluate how partial disruptions affect timing and sharpness of phase changes.

Question 3

1. Earlier sister chromatid separation because APC/C remains active longer at metaphase.
2. Delayed anaphase onset because APC/C stays inhibited and securin is not degraded. (correct answer)
3. Increased DNA replication because cohesin cleavage directly initiates origin firing in S phase.
4. Immediate cytokinesis because unattached kinetochores activate myosin phosphorylation pathways.
5. Normal anaphase timing because checkpoint signaling only monitors DNA damage, not attachment.

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically the metaphase checkpoint's control of APC/C activity. Normally, unattached kinetochores signal to inhibit APC/C, preventing securin degradation until all attachments occur, after which APC/C activates to allow chromatid separation. The drug maintains the inhibitory signal post-attachment, keeping APC/C inhibited and securin stable, thus delaying anaphase onset and affecting segregation, as stated in choice B. This mimics a persistent checkpoint activation despite proper attachments. A tempting distractor is choice A, proposing earlier separation from prolonged APC/C activity, but this reverses the drug's effect on inhibition, misconstruing signal persistence as activation. A transferable strategy is to map checkpoint signals to their targets and predict outcomes when signals are artificially sustained or disrupted.

Question 4

1. Add a small molecule that prevents the blocker from binding cyclin A–CDK2 (correct answer)
2. Prevent kinetochore attachment so cells remain longer in metaphase
3. Inhibit cyclin B–CDK1 dephosphorylation to extend G2 phase
4. Increase cyclin A synthesis to overwhelm the blocker through gene regulation
5. Activate APC/C early to degrade securin before chromosomes align

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically S-phase progression requiring cyclin A–CDK2 substrate phosphorylation. The blocker binds cyclin A–CDK2, preventing substrate access and reducing cells with intermediate DNA content. Adding a small molecule that prevents blocker binding, as in choice A, would restore substrate access and S-phase progression. This directly counters the blocker's mechanism in the assay. A tempting distractor is choice D, which proposes increasing cyclin A synthesis, but this misconceives that higher levels overcome binding without displacing the blocker. A transferable strategy is to use inhibitors that target accessory proteins rather than core complexes in signaling disruptions.

Question 5

1. Cells accumulate in G2 because Cdk1 remains inhibited despite cyclin B binding. (correct answer)
2. Cells bypass the G2 checkpoint and enter mitosis early due to increased MPF activity.
3. Cells accelerate S phase because cyclin B directly activates DNA polymerase at replication forks.
4. Cells arrest in G1 because cyclin B degradation prevents activation of G1/S cyclin-Cdk complexes.
5. Cells complete mitosis faster because inhibitory phosphorylation stabilizes spindle microtubule attachments.

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, emphasizing the activation of mitosis-promoting factor (MPF) for G2/M transition. Normally, MPF forms when cyclin B binds Cdk1, but full activity requires removal of inhibitory phosphates by a phosphatase to trigger mitosis entry. The small molecule prevents this dephosphorylation, keeping Cdk1 inhibited even with cyclin B present, causing cells to accumulate in G2 without progressing to mitosis, as in choice A. This blocks the sharp rise in MPF activity needed for mitotic entry. A tempting distractor is choice B, suggesting early mitosis due to increased MPF, but this ignores the necessity of dephosphorylation for activation, confusing binding with full enzymatic function. A transferable strategy is to identify key activation steps in cyclin-CDK complexes and evaluate how inhibitors affect checkpoint passage.

Question 6

1. Earlier cohesin cleavage because separase is not inhibited by securin binding (correct answer)
2. Delayed cyclin B degradation because separase must cleave cohesin first
3. Failure to activate cyclin E–CDK2 because securin normally activates CDK2
4. G1 arrest because securin is required to phosphorylate the G1 checkpoint protein
5. Reduced APC/C activation because securin binding to separase is required for APC/C signaling

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically securin's role in inhibiting separase until APC/C-mediated degradation. In the mutant, securin cannot bind separase, leaving separase uninhibited. This allows earlier cohesin cleavage, as in choice A, even with normal APC/C and cyclin B dynamics. The prediction fits the binding defect during the transition. A tempting distractor is choice B, which links cyclin B degradation to separase, but this misconceives the independent regulation of securin and cyclin B by APC/C. A transferable strategy is to isolate the functional impact of binding mutations on downstream effectors.

Question 7

1. Cells more readily enter mitosis because inhibitory signaling to Cdc25-like phosphatase is reduced (correct answer)
2. Cells arrest in S phase because APC/C is directly inhibited by the checkpoint kinase
3. Cells delay cytokinesis because cohesin cleavage requires checkpoint kinase activation
4. Cells prevent DNA damage by increasing repair enzyme transcription before mitosis
5. Cells permanently inactivate Cdk1 because M-cyclin cannot bind without phosphorylation

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how DNA damage checkpoints prevent mitotic entry with unrepaired DNA. The second drug blocks the checkpoint kinase, preventing inhibition of the Cdc25-like phosphatase, so inhibitory phosphates are removed from Cdk1, activating M-cyclin–Cdk1 and allowing mitotic entry despite damage. This overrides the normal G2 arrest signal. The damaged DNA persists, but the treatment removes the inhibitory block on phosphatase activity. A tempting distractor is choice B, suggesting S-phase arrest because APC/C is inhibited, but this confuses G2 checkpoint mechanisms with mitotic regulators, a misconception about phase-specific controls. To approach similar problems, trace how inhibiting an inhibitor affects the pathway's net output.

Question 8

1. More cells enter mitosis because inhibitory phosphorylation is not reinforced despite damage signaling (correct answer)
2. More cells arrest in G1 because securin cannot be degraded by APC/C
3. More cells complete cytokinesis early because kinetochores remain unattached
4. More cells remain in S phase because separase cleaves cohesin before replication
5. More cells halt in anaphase because cyclin binding prevents CDK activation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on DNA damage response and inhibitory phosphorylation of CDK in G2. The mutant cannot increase inhibitory-kinase activity after DNA damage, failing to reinforce CDK inhibition despite normal phosphatase regulation. This allows more cells to enter mitosis because the inhibitory phosphorylation is not sufficiently maintained, reducing the effectiveness of the G2 arrest signal. Consequently, CDK activation proceeds more readily despite damage. A tempting distractor is B, which suggests G1 arrest due to securin issues, but securin is mitotic, stemming from the misconception that G2 damage signals affect unrelated downstream proteins. To approach similar problems, evaluate how mutations disrupt feedback in damage responses and anticipate effects on cell cycle progression.

Question 9

1. Increase activity of the kinase that adds the inhibitory phosphate to the CDK
2. Add a CDK variant that cannot be inhibited by phosphorylation at the inhibitory site (correct answer)
3. Block spindle microtubule attachment to kinetochores to silence the checkpoint
4. Increase cyclin breakdown to promote CDK activation at the G2 checkpoint
5. Enhance DNA replication initiation to ensure cells pass the restriction point

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on CDK activation for mitotic entry. The small molecule prevents the phosphatase from removing the inhibitory phosphate on CDK, causing cells to arrest in G2 with duplicated DNA and intact nuclear envelopes despite cyclin presence. Adding a CDK variant that cannot be inhibited by phosphorylation at the inhibitory site bypasses the need for dephosphorylation, allowing cyclin-bound CDK to remain active. This restores progression into mitosis by circumventing the block in the activation pathway. A tempting distractor is A, which suggests increasing the inhibitory kinase activity, but this would further inhibit CDK, stemming from the misconception that enhancing negative regulation could promote activation. To approach similar problems, identify the blocked regulatory step and select interventions that directly bypass it without reinforcing inhibition.

Question 10

1. Sustained Cdk1 signaling because M-cyclin is tagged but not degraded, preventing activity decline (correct answer)
2. Reduced Cdk1 signaling because proteasome inhibition removes cyclins from cyclin–Cdk complexes
3. Normal exit because nuclear envelope reformation depends only on kinetochore attachment signals
4. Early G1 arrest because proteasome inhibition activates origin firing kinases at centrosomes
5. Mitotic exit occurs because APC/C tagging alone is sufficient to inactivate Cdk1

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how degradation is essential for activity shifts, like mitotic exit. Impaired proteasome function allows APC/C tagging but prevents M-cyclin degradation, sustaining Cdk1 signaling and high fluorescence, blocking envelope reformation. This matches the persistent phosphorylation observed. Tagging alone isn't sufficient; removal is needed. A tempting distractor is choice B, claiming reduced signaling from cyclin removal, but inhibition stabilizes cyclins, misconstruing blockage as enhancement of degradation. To approach similar problems, differentiate between preparatory tagging and the degradative step in regulation.

Question 11

1. Add an inhibitor of DNA helicase to ensure replication does not begin prematurely
2. Introduce a cyclin E–CDK2 complex that cannot bind the inhibitor protein (correct answer)
3. Strengthen the spindle checkpoint by blocking APC/C activation in metaphase
4. Reduce cyclin B–CDK1 activity to prevent entry into mitosis
5. Increase transcription of cyclin E to outcompete inhibitor binding sites

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically the G1/S transition where an inhibitor blocks cyclin E–CDK2 until phosphorylated and dissociated. In mutants, the unphosphorylatable inhibitor keeps cyclin E–CDK2 inactive, preventing S-phase entry. Introducing a cyclin E–CDK2 complex that cannot bind the inhibitor, as in choice B, would maintain kinase activity and promote S-phase entry. This bypasses the mutation's effect on the signaling event described. A tempting distractor is choice E, which suggests increasing cyclin E transcription, but this misconceives that higher levels overcome binding without preventing inhibitor interaction. A transferable strategy is to engineer variants that evade inhibitory binding when normal signaling fails.

Question 12

1. Treatment 2, because increased inhibitory phosphorylation reduces active M-cyclin–Cdk1 (correct answer)
2. Treatment 1, because higher M-cyclin always blocks Cdk1 activation at the G2 checkpoint
3. Treatment 1, because M-cyclin binding prevents inhibitory phosphorylation of Cdk1
4. Treatment 2, because inhibitory kinase activity directly activates APC/C at metaphase
5. Neither, because nuclear envelope breakdown depends only on chromosome attachment signals

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how molecular concentrations and modifications set thresholds for events like nuclear envelope breakdown. Treatment 2 increases inhibitory kinase activity, adding more inhibitory phosphates to Cdk1 and reducing active M-cyclin–Cdk1 below the threshold, delaying breakdown. Treatment 1 raises M-cyclin but doesn't address phosphorylation, while treatment 2 directly lowers activity. Both allow binding, but inhibition dominates in treatment 2. A tempting distractor is choice C, claiming treatment 1 delays because M-cyclin prevents phosphorylation, but this confuses cyclin binding with inhibition relief, a misconception about complex dynamics. To approach similar problems, compare how each intervention affects the net activity crossing a threshold.

Question 13

1. Provide an alternative kinase that phosphorylates the same S-phase substrates using normal ATP (correct answer)
2. Increase spindle checkpoint signaling to ensure proper chromosome attachment
3. Inhibit APC/C to keep cyclin B levels high during mitosis
4. Block separase activation so cohesin is not cleaved during anaphase
5. Increase cyclin levels through transcription to overcome reduced catalytic activity

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically ATP-dependent phosphate transfer in CDK catalysis for S-phase entry. The nonhydrolyzable ATP analog binds but blocks transfer, reducing substrate phosphorylation and causing pre-S accumulation. Providing an alternative kinase phosphorylating the same substrates with normal ATP, as in choice A, would restore progression. This bypasses the CDK block in the stimulus. A tempting distractor is choice E, which suggests increasing cyclin levels, but this misconceives that more complex overcomes ATP site occupancy. A transferable strategy is to introduce redundant effectors when core catalytic mechanisms are impaired.

Question 14

1. Delayed S-phase entry because the inhibitor persists and continues blocking cyclin–Cdk (correct answer)
2. Accelerated S-phase entry because phosphorylation alone fully removes inhibition
3. Premature anaphase because the inhibitor directly activates APC/C at kinetochores
4. Failure of cytokinesis because the inhibitor blocks actin polymerization in telophase
5. No effect because ubiquitin ligases only function during transcriptional responses

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how degradation relieves inhibition at transitions like G1/S. The mutation prevents ubiquitin ligase binding, so the phosphorylated inhibitor persists, continuing to block cyclin–Cdk and delaying S-phase entry. Phosphorylation occurs but doesn't lead to removal. This sustains inhibition despite signaling. A tempting distractor is choice B, claiming accelerated entry because phosphorylation removes inhibition, but this ignores the need for degradation, a misconception about modification versus clearance. To approach similar problems, check if the pathway requires both tagging and actual removal for relief.

Question 15

1. Anaphase begins with missegregation because securin is degraded despite unattached kinetochores (correct answer)
2. Cells arrest in G1 because cyclin D is not produced at sufficient levels
3. Cells arrest in S phase because DNA polymerase cannot bind origins
4. Cells remain in metaphase because APC/C is fully inhibited by unattached kinetochores
5. Cells delay cytokinesis because the contractile ring cannot assemble without transcription

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically the spindle checkpoint where unattached kinetochores inhibit APC/C via recruited proteins. The mutant sensor fails to recruit inhibitors, allowing APC/C activity despite unattached kinetochores. This leads to securin degradation and anaphase with missegregation, as in choice A, causing errors. The outcome reflects the loss of checkpoint signaling in the stimulus. A tempting distractor is choice D, which predicts metaphase arrest, but this misconceives that failed recruitment strengthens rather than weakens inhibition. A transferable strategy is to predict checkpoint failures by tracing signal generation to effector inhibition.

Question 16

1. Prolonged high cyclin B–CDK1 activity and delayed exit from mitosis (correct answer)
2. Earlier G1 checkpoint passage because cyclin D is degraded more quickly
3. Immediate anaphase because securin is degraded without APC/C activity
4. Reduced DNA content because replication forks initiate multiple times per cycle
5. Increased cyclin B gene expression to compensate for decreased APC/C signaling

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically how APC/C-mediated ubiquitination controls cyclin levels for mitotic exit. The mutation reduces APC/C activity, preventing sharp cyclin B degradation at the metaphase-to-anaphase transition. This results in prolonged high cyclin B–CDK1 activity and delayed mitotic exit, as in choice A, despite normal kinetochore attachment. The effect stems from impaired ubiquitination in the stimulus. A tempting distractor is choice C, which predicts immediate anaphase, but this misconceives that reduced APC/C accelerates rather than hinders securin degradation. A transferable strategy is to consider how ligase mutations disrupt degradation timing and extend phase-specific activities.

Question 17

1. Cells arrest in metaphase because APC/C remains inhibited and securin persists (correct answer)
2. Cells enter S phase early because securin activates replication origin firing
3. Cells complete anaphase faster because cohesin is stabilized by APC/C activation
4. Cells bypass mitosis because M-cyclin is destroyed before nuclear envelope breakdown
5. Cells restore division by increasing securin transcription to overcome inhibition

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how the spindle checkpoint ensures alignment before anaphase. Continuous inhibitor production keeps APC/C inhibited even after attachment, preventing securin ubiquitination and causing metaphase arrest with persistent securin. Chromatids cannot separate without securin removal. The mutation overrides the normal shutdown of signaling upon attachment. A tempting distractor is choice D, suggesting bypassing mitosis because M-cyclin is destroyed early, but this assumes APC/C activation, misconstruing persistent inhibition as premature activity. To approach similar problems, identify if the defect sustains or abolishes checkpoint signaling.

Question 18

1. Introduce a CDK4/6 variant that cannot bind the inhibitor protein (correct answer)
2. Inhibit APC/C so cyclin B is not degraded at anaphase
3. Increase spindle checkpoint signaling to prevent chromosome separation
4. Block DNA ligase to prevent completion of DNA replication
5. Increase transcription of the E3 ligase to promote inhibitor degradation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically ubiquitination controlling inhibitor degradation for cyclin D–CDK4/6 activation. The mutation prevents ubiquitination, allowing the inhibitor to persist and bind CDK4/6, causing G1 arrest. Introducing a CDK4/6 variant that cannot bind the inhibitor, as in choice A, would free CDK4/6 and promote S-phase entry. This evades the persistent inhibitor in the mutants. A tempting distractor is choice E, which suggests increasing E3 ligase transcription, but this misconceives that more ligase overcomes a ubiquitination-defective mutation. A transferable strategy is to alter binding interfaces to prevent inhibitory interactions when degradation fails.

Question 19

1. Shortened G2 phase because cyclin B–CDK1 activates earlier (correct answer)
2. Shortened S phase because APC/C degrades cyclin A more rapidly
3. Delayed anaphase because increased CDK1 activity inhibits APC/C permanently
4. Extended G1 because cyclin D cannot bind CDK4/6 when CDK1 is active
5. No change because phosphatases only function after transcriptional activation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically phosphatase-mediated removal of inhibitory phosphates at checkpoints. The mutation increases phosphatase activity, accelerating CDK1 dephosphorylation without changing cyclin B. This shortens G2 via earlier cyclin B–CDK1 activation, as in choice A, advancing mitosis. The change reflects faster inhibitory removal in the stimulus. A tempting distractor is choice C, which predicts delayed anaphase, but this misconceives that higher CDK1 inhibits APC/C rather than promoting entry. A transferable strategy is to predict timing shifts by evaluating enzyme activity changes on key regulators.

Question 20

1. Sister chromatids separate despite unattached kinetochores, increasing segregation errors (correct answer)
2. Cells arrest in G1 because cyclin D–CDK4/6 requires microtubules to function
3. Cells remain in metaphase because APC/C activation strengthens checkpoint inhibition
4. DNA replication increases because APC/C activation directly activates DNA polymerase
5. Cyclin B levels rise because APC/C activity promotes cyclin accumulation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically the spindle checkpoint inhibiting APC/C in response to unattached kinetochores. The microtubule drug creates unattached kinetochores, but the second compound forces APC/C activity, overriding inhibition. This causes chromatid separation with errors, as in choice A, due to missegregation. The outcome stems from checkpoint bypass in the stimulus. A tempting distractor is choice C, which predicts metaphase arrest, but this misconceives that forced APC/C sustains rather than overcomes inhibition. A transferable strategy is to combine perturbations to evaluate checkpoint robustness and failure modes.