AP Biology Quiz: Cell Cycle

What this quiz covers

This quiz focuses on 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. They delay in G1 until all breaks are repaired before DNA replication begins
2. They enter S phase and replicate DNA that still contains break sites (correct answer)
3. They skip S phase and enter mitosis with 2C DNA and uncondensed chromatin
4. They undergo homolog pairing and reduction division to form haploid cells
5. They increase spindle attachment, preventing any chromosome condensation in M phase

Explanation: This question assesses the skill of analyzing the cell cycle, focusing on checkpoint roles in response to DNA damage. The stimulus explains that inhibiting the G1 checkpoint allows cells with radiation-induced double-strand breaks to enter S phase without repair. In AP Biology, the G1 checkpoint normally delays progression to prevent replication of damaged DNA, but when inhibited, replication proceeds over break sites, leading to duplicated chromosomes with inherited damage in G2. Consequently, these cells enter S phase and replicate the damaged DNA templates. A tempting distractor is A, suggesting delay in G1 until repairs, but this is incorrect due to a misunderstanding of inhibition effects, specifically a teleological misconception that cells inherently prioritize repair over progression. To approach similar questions, distinguish between normal and disrupted checkpoint functions, then trace the cycle's progression and potential genomic consequences.

Question 2

1. G1 checkpoint failure allowing entry into S phase before DNA replication enzymes assemble.
2. G2 checkpoint failure allowing mitosis to start before chromosomes condense.
3. Spindle checkpoint failure allowing anaphase with incorrect kinetochore attachments. (correct answer)
4. G1 checkpoint failure allowing cytokinesis before metaphase alignment occurs.
5. G2 checkpoint failure allowing homologous chromosomes to separate instead of sister chromatids.

Explanation: This question assesses the skill of analyzing the cell cycle, focusing on mitotic fidelity and checkpoint failures. The stimulus depicts chromosomes with monopolar kinetochore attachments proceeding to anaphase without delay, leading to aneuploid daughter cells. This indicates a failure of the spindle checkpoint, which normally delays anaphase until all kinetochores are bipollarly attached, as in choice C, aligning with AP Biology's role of the checkpoint in preventing missegregation. The observed imbalance directly results from premature chromatid separation with incorrect attachments. A tempting distractor is choice B, claiming G2 checkpoint failure allows mitosis before condensation, but this involves a structure-function confusion, as G2 checks replication completeness, not attachment during metaphase. To handle similar questions, identify the checkpoint that monitors the specific defect and evaluate effects of its failure on daughter cells.

Question 3

1. Spindle checkpoint, because microtubules cannot attach until DNA damage is repaired.
2. G2 checkpoint, because damaged DNA is detected after replication is complete. (correct answer)
3. G1 checkpoint, because sister chromatids must separate before S phase can begin.
4. G1 checkpoint, because chromosomes are already condensed before DNA replication.
5. Spindle checkpoint, because homologous chromosomes are not synapsed in prophase.

Explanation: This question assesses the skill of analyzing the cell cycle, highlighting checkpoint responses to DNA damage. The stimulus describes a cell progressing from cytokinesis through S phase to G2, where an unrepaired double-strand break on one chromosome prevents entry into mitosis. This is because the G2 checkpoint detects DNA damage post-replication and blocks mitotic entry to allow repair, matching choice B and AP Biology's concept of checkpoints preserving genome stability. The uncondensed chromosomes and intact envelope support that the cell remains in G2 without proceeding. A tempting distractor is choice A, claiming the spindle checkpoint blocks due to unrepaired damage affecting attachments, but this reflects a teleology misconception, assuming checkpoints anticipate future issues rather than responding to current conditions like attachments in mitosis. For these questions, link the timing of damage or issues to the checkpoint that monitors that phase's completion.

Question 4

1. Sister chromatids could fail to segregate properly, producing daughter cells with unequal chromosome sets. (correct answer)
2. Homologous chromosomes would pair and recombine, producing genetically identical daughter cells.
3. DNA content would drop to 2C before mitosis begins, preventing spindle formation.
4. Chromosomes would replicate again during prophase, doubling DNA content to 8C.
5. The nuclear envelope would remain intact throughout mitosis, preventing chromosome alignment.

Explanation: This question assesses the skill of analyzing the cell cycle, evaluating G2 checkpoint importance. With intertwined chromatids at G2, delay allows resolution before mitosis; without it, entry would cause improper segregation and unequal chromosome sets in daughters, as in choice A, per AP Biology's checkpoint ensuring decatenation for faithful division. The uncondensed state and intact envelope during delay support G2 arrest. This prevents aneuploidy. A tempting distractor is choice B, suggesting homolog recombination leading to identical daughters, but this arises from a teleology misconception, assuming entanglement promotes meiosis-like events rather than disrupting mitosis. When considering checkpoint bypass, forecast risks to segregation accuracy based on the monitored condition.

Question 5

1. Segregation occurs with some chromosomes lacking fully duplicated sister chromatids (correct answer)
2. Cells remain permanently in G1 with 2C DNA and never initiate replication
3. DNA replication resumes only after anaphase, restoring identical chromatids at poles
4. Homologous chromosomes pair and separate, yielding four nuclei after cytokinesis
5. Spindle fibers fail to form because DNA replication is required for microtubule polymerization

Explanation: This question assesses the skill of analyzing the cell cycle, exploring effects of incomplete replication in checkpoint-disabled cells. The stimulus describes cells entering mitosis with unreplicated genome sections due to a disabled G2 checkpoint after inhibitor removal. In AP Biology, the G2 checkpoint ensures replication completion before M phase; without it, mitosis proceeds with partially duplicated chromosomes, leading to flawed segregation. Therefore, segregation occurs with some chromosomes lacking fully duplicated chromatids. A tempting distractor is B, claiming permanent G1 arrest, but this is incorrect due to a structure-function confusion, assuming replication blocks prevent all progression rather than allowing faulty mitosis. To approach similar questions, compare normal checkpoint functions to experimental disruptions, forecasting impacts on chromosome integrity during division.

Question 6

1. G1 checkpoint, allowing entry into S phase without sufficient DNA content.
2. G2 checkpoint, allowing entry into mitosis without completed DNA replication. (correct answer)
3. Spindle checkpoint, allowing DNA replication to begin before chromosome alignment.
4. Spindle checkpoint, allowing cytokinesis to occur before DNA synthesis begins.
5. G2 checkpoint, allowing homologous chromosomes to pair before metaphase.

Explanation: This question assesses the skill of analyzing the cell cycle, assessing checkpoint bypass effects. The treatment induces cells with 2C DNA to form spindles and condense, leading to mitotic entry without replication and aneuploid daughters, indicating G2 checkpoint bypass allowing mitosis with incomplete DNA synthesis, as in choice B, consistent with AP Biology's G2 role in verifying replication. The unaffected attachments post-treatment pinpoint the entry defect. This results in segregation of unreplicated chromosomes. A tempting distractor is choice A, claiming G1 bypass allows S without sufficient DNA, but this involves a level-of-organization error, as 2C is normal for G1, and the issue is skipping replication entirely. When treatments alter progression, identify the checkpoint overridden and link to observed abnormalities.

Question 7

1. Sister chromatids remain paired and fail to move to opposite poles (correct answer)
2. Chromosomes decondense immediately and the nuclear envelope reforms at metaphase
3. DNA content doubles again as the cell repeats S phase during anaphase
4. Homologous chromosomes separate, producing two haploid nuclei at telophase
5. Kinetochores detach and the cell returns to G1 with 2C DNA without division

Explanation: This question assesses the skill of analyzing the cell cycle, highlighting mechanisms of chromatid separation in anaphase. The stimulus details a toxin blocking proteolysis of cohesin at centromeres, preventing sister chromatid separation despite a satisfied spindle checkpoint. In AP Biology, anaphase requires cohesin degradation to release chromatids, allowing microtubule pulling to opposite poles; without it, paired chromatids cannot separate. Thus, during anaphase, sister chromatids remain paired and fail to move apart. A tempting distractor is B, proposing immediate decondensation and envelope reformation, but this is incorrect due to a structure-function confusion, mistaking cohesion's role for overall mitotic exit signals. To approach similar questions, identify key molecular events in each mitotic phase and predict effects of targeted disruptions on chromosome behavior.

Question 8

1. G1 checkpoint, because DNA content is too high to begin replication.
2. Spindle checkpoint, because kinetochores are not properly attached to microtubules. (correct answer)
3. G2 checkpoint, because DNA replication has not initiated.
4. Spindle checkpoint, because homologous chromosomes are not paired at the equator.
5. G1 checkpoint, because cytokinesis cannot occur before chromosome condensation.

Explanation: This question assesses the skill of analyzing the cell cycle, focusing on mitotic checkpoints and spindle dynamics. The stimulus shows cells treated with a drug preventing microtubule polymerization, resulting in condensed chromosomes, absent metaphase plate, and unattached kinetochores, with no progression to anaphase. This indicates the spindle checkpoint is activated, halting the cell in metaphase until all kinetochores are properly attached to microtubules from opposite poles, as in choice B, consistent with AP Biology's emphasis on the checkpoint ensuring accurate chromosome segregation. The post-S phase DNA content and condensed state confirm the cells are in mitosis, where the spindle checkpoint operates. A tempting distractor is choice C, suggesting the G2 checkpoint due to uninitiated replication, but this involves a phase confusion misconception, as G2 checks occur before mitosis, not after chromosome condensation has begun. To solve similar problems, correlate cellular observations like condensation and attachments to specific mitotic stages and identify the relevant checkpoint.

Question 9

1. DNA replication begins, increasing DNA content as sister chromatids are synthesized (correct answer)
2. Sister chromatids separate and move to opposite poles of the cell
3. The nuclear envelope breaks down and chromosomes condense into visible structures
4. Kinetochores attach to spindle fibers and align chromosomes at the metaphase plate
5. Cytokinesis divides the cytoplasm, producing two daughter cells with 2C DNA

Explanation: This question assesses the skill of analyzing the cell cycle, examining G1 checkpoint responses to nutrients. The stimulus describes a cell stalled in G1 under low nutrients, then progressing upon nutrient addition, initiating S phase. In AP Biology, passing the G1 checkpoint triggers DNA replication, increasing content from 2C to 4C via sister chromatid synthesis. Thus, DNA replication begins immediately after passing the checkpoint. A tempting distractor is B, proposing chromatid separation, but this is incorrect due to a teleological misconception that checkpoints directly control division rather than replication entry. To approach similar questions, identify environmental cues affecting checkpoints and predict immediate downstream processes like replication.

Question 10

1. Cell X skipped S phase, so chromosomes lacked sister chromatids for equal segregation. (correct answer)
2. Cell Y underwent meiosis, so homologs separated and reduced DNA content in daughters.
3. Cell X repeated G2, so DNA content increased beyond 4C before mitosis began.
4. Cell Y bypassed the spindle checkpoint, so chromatids separated before alignment occurred.
5. Cell X remained in telophase, so DNA replication continued during nuclear reformation.

Explanation: This question assesses the skill of analyzing the cell cycle, comparing outcomes of mitosis with and without replication. Cell X, entering from G1 without S phase, lacks sister chromatids, leading to improper segregation and daughters with fewer DNA molecules, while Cell Y proceeds normally post-G2, supporting choice A and AP Biology's requirement for duplication before mitotic division. The stimulus emphasizes Cell X's unreplicated state causing atypical outcomes. This highlights S phase's necessity for equal partitioning. A tempting distractor is choice B, suggesting Cell Y underwent meiosis with homolog separation, but this stems from a structure-function confusion, misapplying meiotic reduction to mitotic contexts. For comparative questions, trace each cell's path and identify deviations from standard cycle that explain differences.

Question 11

1. 1C DNA in each daughter cell because chromatids are lost during anaphase
2. 2C DNA in each daughter cell because sister chromatids segregate equally (correct answer)
3. 3C DNA in each daughter cell because replication was incomplete in S phase
4. 4C DNA in each daughter cell because DNA content doubles during cytokinesis
5. 8C DNA in each daughter cell because chromosomes replicate again in G2

Explanation: This question assesses the skill of analyzing the cell cycle, particularly DNA content changes during mitosis. The stimulus describes a cell progressing from G1 with 2C DNA through S phase to 4C in G2, then completing mitosis with proper chromatid segregation and cytokinesis. In AP Biology, mitosis divides duplicated chromosomes equally, so each daughter cell receives one chromatid per chromosome, restoring 2C DNA content post-cytokinesis. Therefore, after cytokinesis, each daughter cell has 2C DNA due to equal segregation of sister chromatids. A tempting distractor is D, claiming 4C DNA because content doubles during cytokinesis, but this is incorrect due to a level-of-organization error, confusing cellular division with DNA replication events. To approach similar questions, track DNA content across phases and remember that mitosis halves the content per cell without altering total genomic copies.

Question 12

1. It remains in G1 without initiating DNA replication. (correct answer)
2. It proceeds directly into mitosis because DNA damage triggers chromosome condensation.
3. It enters G2 because DNA content is already sufficient for mitosis.
4. It completes cytokinesis, producing two cells each with 1C DNA content.
5. It enters meiosis I and aligns homologous chromosomes at the metaphase plate.

Explanation: This question assesses the skill of analyzing the cell cycle, examining checkpoint responses to DNA damage in G1. The damaged cell at the G1 checkpoint with 2C DNA does not enter S phase, remaining in G1 to allow repair, as per choice A and AP Biology's G1 checkpoint function in blocking replication of damaged DNA. In contrast, the undamaged cell proceeds normally, highlighting the checkpoint's role. No manipulation of other phases supports this arrest. A tempting distractor is choice B, suggesting direct entry into mitosis triggered by damage, but this reflects a teleology misconception, assuming damage accelerates progression rather than halting it for repair. For damage-related questions, recall which checkpoint detects issues and the typical cellular response of arrest.

Question 13

1. Sister chromatids separate and move toward opposite poles, producing two identical chromosome sets.
2. Cells remain arrested in metaphase because anaphase onset is blocked despite spindle attachment. (correct answer)
3. DNA is replicated again, doubling the DNA content without chromosome separation.
4. Homologous chromosomes pair and segregate, reducing chromosome number by half.
5. Nuclear envelopes reform around uncondensed chromosomes before any spindle forms.

Explanation: This question assesses the skill of analyzing the cell cycle by evaluating the impact of a microtubule-disrupting drug on mitotic progression and checkpoint activation. The correct answer indicates that cells remain arrested in metaphase because the drug prevents microtubule shortening, which is necessary for generating the tension that satisfies the spindle assembly checkpoint, thus blocking anaphase onset even with bipolar kinetochore attachments. This aligns with the observation of condensed chromosomes aligned at the metaphase plate without proceeding to anaphase, as the checkpoint detects improper microtubule dynamics despite attachment. The nuclear envelope breakdown and prior DNA replication confirm the cells are in M phase, where the checkpoint halts progression to ensure accurate chromosome segregation. A tempting distractor is choice A, which suggests sister chromatids separate anyway, reflecting a teleology misconception by assuming the cell cycle proceeds purposefully without checkpoint regulation despite mechanical impediments. For similar questions, identify how disruptions to cellular structures like microtubules affect specific checkpoints, focusing on the requirements for tension and attachment in mitosis.

Question 14

1. Daughter cells contain half the chromosome number because G1 controls homolog separation.
2. Daughter cells are smaller but have normal DNA content per nucleus. (correct answer)
3. Daughter cells have doubled DNA content because S phase repeats before mitosis.
4. Daughter cells lack nuclei because the nuclear envelope cannot reform after G1 bypass.
5. Daughter cells show paired homologs because meiosis is triggered by reduced cell size.

Explanation: This question assesses the skill of analyzing the cell cycle, examining the consequences of bypassing growth-related checkpoints. The stimulus involves forcing a cell past the G1 checkpoint despite insufficient size, allowing normal DNA replication and mitosis, resulting in smaller daughter cells with complete chromosome sets. Choice B is correct as the G1 checkpoint typically ensures adequate cell size before S phase, but bypassing it does not affect DNA content or segregation, per AP Biology's separation of growth and replication controls. The normal progression through subsequent phases confirms that only size is impacted. A tempting distractor is choice A, suggesting halved chromosome number due to G1 controlling homolog separation, but this arises from a level-of-organization error, confusing mitosis with meiosis where homologs separate. When approaching such scenarios, distinguish between checkpoint functions and predict outcomes based on which processes remain intact.

Question 15

1. Population 1, because uncondensed chromatin indicates DNA synthesis has completed.
2. Population 2, because DNA content increases during replication between 2C and 4C. (correct answer)
3. Population 3, because metaphase alignment requires ongoing DNA replication.
4. Population 1, because 2C DNA indicates chromosomes have duplicated and separated.
5. Population 3, because condensed chromosomes indicate the G1 checkpoint has not been passed.

Explanation: This question assesses the skill of analyzing the cell cycle by interpreting observations of DNA content and chromatin state to identify the S phase. The correct answer identifies Population 2 as being in S phase because its DNA content is between 2C and 4C, which directly corresponds to the ongoing DNA replication that doubles the genetic material from the diploid 2C amount post-G1 to 4C before mitosis in the cell cycle. Additionally, the uncondensed chromatin in Population 2 aligns with the interphase characteristic of S phase, where DNA is accessible for replication enzymes without the condensation seen in mitotic phases. In contrast, Population 3 has 4C DNA and condensed chromosomes aligned at the metaphase plate, indicating it is in metaphase of mitosis after replication has completed, not actively replicating. A tempting distractor is choice A, which incorrectly assumes that uncondensed chromatin in Population 1 with 2C DNA means replication is finished, reflecting a structure-function confusion by misattributing chromatin state solely to post-replication without considering DNA content changes. To approach similar questions, always correlate DNA content levels with cell cycle phases, remembering that S phase is marked by increasing DNA from 2C to 4C during interphase.

Question 16

1. Both daughter cells receive identical chromosome sets because chromatids separate evenly.
2. Each daughter cell is haploid because homologous chromosomes segregate in anaphase.
3. One daughter cell may gain chromosomes while the other loses them due to missegregation. (correct answer)
4. Daughter cells contain twice the normal DNA because S phase repeats during mitosis.
5. The cell remains arrested in G2 because DNA replication is incomplete.

Explanation: This question requires analysis of the cell cycle, highlighting consequences of impaired chromosome attachment during mitosis. The toxin causes monopolar attachments for some chromosomes, and disabling the spindle checkpoint allows premature anaphase, leading to improper segregation. In AP Biology, bipolar kinetochore-microtubule attachments ensure equal chromatid distribution; without them, missegregation results in aneuploidy, as no additional replication occurs. Thus, daughter cells may have unequal chromosome numbers, with one gaining and the other losing chromosomes post-cytokinesis. A tempting distractor is choice A, suggesting identical sets from even separation, but this is incorrect due to a teleology misconception assuming perfect outcomes despite disrupted mechanisms. To solve similar questions, evaluate how perturbations affect segregation and predict genomic imbalances.

Question 17

1. Chromosomes condense and a mitotic spindle begins forming while DNA remains 4C. (correct answer)
2. DNA content decreases from 4C to 2C while the nuclear envelope remains intact.
3. Homologous chromosomes pair and align as tetrads while DNA remains 2C.
4. DNA content increases from 4C to 8C with no change in chromosome condensation.
5. Two daughter nuclei form before sister chromatids separate at the centromeres.

Explanation: This question requires analysis of the cell cycle, tracking DNA content changes to identify phase transitions. From post-mitosis 2C in G1, DNA increases to 4C by 6 hours via S phase, remaining 4C at 8 hours in G2 with no spindle, indicating preparation for M phase. In AP Biology, entry into M phase involves chromosome condensation and mitotic spindle formation while DNA stays at 4C, marking prophase. This is the next expected observation after G2. A tempting distractor is choice B, suggesting DNA decrease to 2C, but this is incorrect due to a teleology misconception assuming reduction without division. Use time-course data to predict progression and hallmark features of each phase in such problems.

Question 18

1. The cell remains in metaphase with sister chromatids joined at centromeres. (correct answer)
2. The cell enters G1 with unreplicated chromosomes and a single centrosome.
3. The cell completes cytokinesis, producing four haploid cells with duplicated DNA.
4. The cell initiates S phase again, doubling DNA content without division.
5. The cell separates sister chromatids despite unattached kinetochores, yielding equal genomes.

Explanation: This question requires analysis of the cell cycle, focusing on the mitotic spindle checkpoint. The drug prevents microtubule attachment to kinetochores, leaving chromosomes unattached at metaphase, which activates the spindle assembly checkpoint to halt progression to anaphase. In AP Biology, this checkpoint ensures all chromosomes are properly aligned and attached before sister chromatid separation, preventing aneuploidy. Since the checkpoint is not satisfied, the cell remains arrested in metaphase with sister chromatids still joined at centromeres, as indicated by the condensed chromosomes aligned near the center and sustained M-phase cyclin-dependent kinase activity. A tempting distractor is choice E, which suggests sister chromatids separate despite unattached kinetochores, but this is incorrect due to a structure-function confusion where the role of kinetochore attachment in triggering anaphase is overlooked. To approach similar questions, always identify the specific checkpoint involved and recall its criteria for cell cycle progression.

Question 19

1. The cell enters anaphase on schedule, and sister chromatids separate simultaneously
2. The cell remains in metaphase with sister chromatids still paired at centromeres (correct answer)
3. The cell returns to S phase to re-replicate DNA until attachments are correct
4. The cell completes telophase first, then aligns chromosomes at the equator
5. The cell undergoes synapsis of homologs and forms tetrads at the equator

Explanation: This question requires analysis of the cell cycle, specifically how the spindle checkpoint responds to improper chromosome attachments during metaphase. The spindle checkpoint monitors whether all kinetochores are properly bi-oriented (attached to opposite poles), and even one improperly attached chromosome prevents anaphase onset. With persistent monotelic attachment, the cell remains in metaphase with sister chromatids still paired at centromeres (answer B), as the checkpoint continuously blocks separase activation and cohesin cleavage until all attachments are corrected. Answer A incorrectly suggests normal anaphase progression, representing a quantitative error where students think most correct attachments are sufficient when actually all chromosomes must be properly attached. The strategy is to remember that the spindle checkpoint operates on an all-or-nothing principle—even one error blocks progression.

Question 20

1. Anaphase onset is delayed and sister chromatids remain paired at centromeres (correct answer)
2. DNA replication resumes to replace the severed microtubules with new chromatids
3. The cell completes cytokinesis, then aligns chromosomes at the metaphase plate
4. Homologous chromosomes segregate first, followed by sister chromatid separation
5. The cell returns to S phase and reduces DNA content from 4C to 2C

Explanation: This question assesses the skill of analyzing the cell cycle, assessing checkpoint activation from experimental disruption. The stimulus details laser severance leaving one chromosome unattached, triggering the spindle checkpoint to delay anaphase. In AP Biology, the checkpoint halts progression until all kinetochores reattach, keeping chromatids paired. Thus, anaphase is delayed with chromatids remaining paired. A tempting distractor is D, implying meiotic-like separation, but this is incorrect due to a level-of-organization error, mixing mitotic and meiotic mechanisms. To approach similar questions, consider how targeted interventions activate specific checkpoints and their effects on timing.