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Quiz: The Cell Cycle, Mitosis, and Cancer

Test your understanding of the cell cycle, mitotic stages, checkpoint control, and the genetic basis of cancer with these review questions.

1. What is the correct order of phases within interphase, before a cell enters mitosis?

  1. S → G1 → G2
  2. G1 → S → G2
  3. G2 → S → G1
  4. S → G2 → G1
Show Answer

The correct answer is B. Interphase proceeds through G1 (initial growth and signal integration), then S phase (DNA replication, producing sister chromatids), then G2 (further growth and preparation for mitosis). This ordered sequence ensures the cell only replicates its DNA once it has grown sufficiently and only enters mitosis after replication is complete.

Concept Tested: Cell cycle phases (G1, S, G2)

2. Which checkpoint verifies that every kinetochore is properly attached to spindle microtubules from opposite poles before allowing anaphase to begin?

  1. G1/S checkpoint
  2. G2/M checkpoint
  3. Restriction point
  4. Spindle assembly checkpoint
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The correct answer is D. The spindle assembly checkpoint monitors kinetochore-microtubule attachment during metaphase. Any unattached kinetochore generates the mitotic checkpoint complex, which inhibits the anaphase-promoting complex/cyclosome (APC/C) and prevents cohesin cleavage, blocking anaphase until every chromosome is correctly bioriented on the spindle.

Concept Tested: Spindle assembly checkpoint

3. Why must a cyclin-dependent kinase (CDK) bind to a cyclin in order to become catalytically active?

  1. CDKs are constitutively present in the cell but are catalytically inactive unless bound to a cyclin, whose concentration rises and falls at specific points in the cycle
  2. Cyclins are enzymes that directly replicate DNA without any CDK involvement
  3. CDK activity is entirely independent of cyclin levels
  4. Cyclins degrade CDKs to terminate the cell cycle
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The correct answer is A. CDK protein levels remain roughly constant throughout the cell cycle, but CDKs have no catalytic activity on their own. Binding to a cyclin — whose concentration oscillates because it is synthesized and then targeted for degradation at specific points — activates the CDK, allowing cyclin-CDK complexes to phosphorylate target proteins only during the appropriate phase of the cycle.

Concept Tested: Cyclins and CDKs

4. Why are tumor suppressor gene mutations generally considered recessive at the cellular level, while oncogene mutations are considered dominant?

  1. Oncogene mutations require loss of both alleles, while tumor suppressor mutations require only one
  2. Both oncogene and tumor suppressor mutations require loss of both gene copies to affect the cell
  3. Tumor suppressor mutations generally require loss of both gene copies to eliminate the braking function, while a single gain-of-function oncogene mutation is sufficient to drive proliferation
  4. Oncogenes and tumor suppressors are regulated identically and show no difference in dominance
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The correct answer is C. Tumor suppressor genes act as brakes on the cell cycle, and one functional copy is usually enough to maintain that brake, so both copies must be inactivated (the "two-hit hypothesis") before control is lost. Oncogenes behave like a stuck accelerator: a single gain-of-function mutation produces a constitutively active protein that drives proliferation even in the presence of a normal allele.

Concept Tested: Proto-oncogenes vs. tumor suppressor genes

5. What role does the retinoblastoma protein (Rb) normally play in cell cycle regulation?

  1. Rb binds and inhibits E2F transcription factors, preventing S-phase entry until it is phosphorylated by Cyclin D-CDK4/6
  2. Rb directly synthesizes cyclins to drive the cell into S phase
  3. Rb degrades p53 to prevent apoptosis
  4. Rb forms the mitotic spindle during prophase
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The correct answer is A. Rb normally holds E2F transcription factors inactive, preventing the transcription of genes needed for S-phase entry. When Cyclin D-CDK4/6 phosphorylates Rb in response to growth signals, Rb releases E2F, allowing S-phase genes to be transcribed. Loss of both RB1 alleles removes this brake entirely, as seen in retinoblastoma.

Concept Tested: Rb tumor suppressor function

6. A researcher treats dividing cells with a drug that prevents tubulin polymerization. At which stage would these cells most likely arrest, and why?

  1. Anaphase, because cohesin cannot be cleaved
  2. Prophase, because chromatin cannot condense
  3. Telophase, because nuclear envelopes cannot reform
  4. Metaphase, because unattached kinetochores activate the spindle assembly checkpoint and block the transition to anaphase
Show Answer

The correct answer is D. Without tubulin polymerization, spindle microtubules cannot form, so kinetochores cannot attach properly. Unattached kinetochores trigger the spindle assembly checkpoint, which inhibits APC/C and prevents separase-mediated cohesin cleavage. The cell therefore arrests at metaphase, unable to proceed into anaphase until proper kinetochore attachment occurs — which cannot happen without functional spindle microtubules.

Concept Tested: Spindle assembly checkpoint and mitotic arrest

7. A tumor cell carries a p53 mutation that prevents the protein from binding DNA. What is the most likely effect on the cell's response to DNA damage?

  1. The cell would undergo apoptosis more readily than normal
  2. The cell would fail to arrest the cycle or trigger apoptosis in response to DNA damage, allowing damaged cells to continue dividing and accumulate further mutations
  3. DNA replication would become more accurate
  4. The G1/S checkpoint would become more stringent
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The correct answer is B. p53 functions as a transcription factor that must bind DNA to activate genes such as p21 (which halts the cycle) and pro-apoptotic genes like Bax. A mutation that prevents DNA binding disables these downstream responses, so the cell can no longer arrest or self-destruct in response to DNA damage, allowing it to continue dividing while accumulating additional mutations.

Concept Tested: p53 tumor suppressor function

8. A drug locks the APC/C (anaphase-promoting complex/cyclosome) in an inactive state, preventing it from ubiquitinating Cyclin B. What is the most likely effect on the cell cycle?

  1. Cyclin B would be degraded more rapidly, accelerating mitotic exit
  2. The spindle assembly checkpoint would be permanently satisfied
  3. Cyclin B levels would remain high, MPF activity would persist, and the cell would fail to transition from metaphase/anaphase into telophase
  4. DNA replication in the next S phase would begin prematurely
Show Answer

The correct answer is C. Normally, active APC/C ubiquitinates Cyclin B for proteasomal degradation once the spindle assembly checkpoint is satisfied, and the resulting drop in MPF (CDK1-Cyclin B) activity allows the cell to exit mitosis. If APC/C cannot perform this degradation, Cyclin B and MPF activity remain high, and the cell cannot proceed past metaphase/anaphase into telophase and cytokinesis.

Concept Tested: Cyclin B degradation and mitotic exit

9. How does the "two-hit hypothesis" for tumor suppressor gene inactivation change cancer risk for someone who inherits one mutant tumor suppressor allele, compared to the mechanism underlying an activating Ras mutation?

  1. Inherited carriers of one mutant tumor suppressor allele face the same cancer risk as the general population because two independent hits are still required
  2. A single Ras mutation requires a second hit in the other allele before it can drive proliferation, just like a tumor suppressor gene
  3. Individuals who inherit one mutant oncogene allele are at dramatically increased risk because oncogenes always require two-hit inactivation
  4. Individuals who inherit one mutant tumor suppressor allele need only a single additional somatic mutation to lose the remaining functional copy, dramatically increasing their cancer risk compared to those who must acquire two independent hits; an activating Ras mutation, by contrast, needs only one dominant gain-of-function change regardless of inheritance
Show Answer

The correct answer is D. Someone who inherits one mutant tumor suppressor allele already has one "hit" present in every cell, so they need only a single additional somatic mutation to inactivate the gene entirely, greatly raising their lifetime cancer risk compared to someone starting with two normal alleles. Oncogenes like Ras require no such inheritance pattern — a single somatic gain-of-function mutation is sufficient on its own to drive constitutive signaling.

Concept Tested: Two-hit hypothesis and oncogene dominance

10. Why would loss of contact inhibition combined with an activating Ras mutation be expected to produce a more aggressive tumor phenotype than either change alone?

  1. The Ras mutation drives continuous proliferation signaling independent of growth factors, while loss of contact inhibition removes the normal density-dependent brake on division, so together the cells proliferate continuously and pile up without restraint
  2. Loss of contact inhibition alone would completely reverse the effects of the Ras mutation
  3. The two changes would cancel each other out, restoring normal growth control
  4. Ras mutations only affect apoptosis, not proliferation, so contact inhibition loss would have no combined effect
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The correct answer is A. An activating Ras mutation continuously drives the MAPK proliferation cascade regardless of external growth factor signals, while loss of contact inhibition removes the normal restraint that halts division once cells become densely packed. Together, these changes remove two independent layers of growth control, allowing cells to proliferate without limit and pile up into disorganized masses characteristic of aggressive tumors.

Concept Tested: Contact inhibition and oncogene cooperation