Cell Division — Hard Practice Quiz

A Biology cheat sheet for Cell Division — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Practice quiz

1. A cell experiences a significant decrease in its surface area to volume ratio, making nutrient and waste exchange inefficient. Simultaneously, internal regulatory proteins, such as cyclins, are found to be non-functional, leading to a bypass of cell cycle checkpoints. Which of the following is the most likely long-term consequence for this cell, and which specific phase of the cell cycle would be most directly impacted by the cyclin dysfunction?

   • The cell would undergo apoptosis due to inefficient exchange, and the S phase would be unregulated.
   • The cell would become cancerous due to uncontrolled division, primarily affecting the G1 and G2 checkpoints.
   • The cell would differentiate into a specialized tissue, with the M phase being prematurely initiated.
   • The cell would enter a quiescent G0 state, and DNA replication in the S phase would be inhibited.

   Answer: The cell would become cancerous due to uncontrolled division, primarily affecting the G1 and G2 checkpoints.

2. An organism has a diploid number of $2N=6$ chromosomes. If nondisjunction of a homologous pair occurs during Anaphase I of meiosis, and subsequently, Meiosis II proceeds normally, what would be the chromosome number in the resulting gametes, and how many of these gametes would be viable?

   • Two gametes with $N+1=4$ chromosomes and two gametes with $N-1=2$ chromosomes; all four would be viable.
   • Two gametes with $N+1=4$ chromosomes and two gametes with $N-1=2$ chromosomes; only the $N$ gametes would be viable.
   • All four gametes would have $N=3$ chromosomes, but two would be genetically identical.
   • Two gametes with $2N=6$ chromosomes and two gametes with $0$ chromosomes; none would be viable.

   Answer: Two gametes with $N+1=4$ chromosomes and two gametes with $N-1=2$ chromosomes; all four would be viable.

3. A prokaryotic bacterium undergoes binary fission, a form of asexual reproduction. Compare this process to the M phase (mitosis and cytokinesis) in a eukaryotic animal cell. Which of the following statements accurately highlights a fundamental difference in their division mechanisms and genetic outcomes?

   • Both processes involve spindle fiber formation and chromosome condensation, but only eukaryotic cells produce genetically identical offspring.
   • Eukaryotic cells utilize internal and external regulators for precise cell cycle control, whereas bacterial division is unregulated and always leads to genetic diversity.
   • Bacterial binary fission lacks distinct prophase, metaphase, anaphase, and telophase stages, and typically results in genetically identical daughter cells, unlike the genetically diverse products of eukaryotic mitosis.
   • Eukaryotic mitosis involves nuclear envelope breakdown and reformation, while bacterial division involves direct partitioning of the nucleoid region, both producing genetically identical cells unless mutations occur.

   Answer: Eukaryotic mitosis involves nuclear envelope breakdown and reformation, while bacterial division involves direct partitioning of the nucleoid region, both producing genetically identical cells unless mutations occur.

4. A pluripotent stem cell, characterized by rapid division and an unspecialized state, is induced to differentiate into a terminally differentiated, non-dividing neuron. Which of the following best describes the changes in cell cycle regulation and potential involvement of apoptosis during this process?

   • The cell would increase expression of cyclins and growth factors, promoting entry into the M phase, while apoptosis would be inhibited to ensure cell survival.
   • The cell would likely exit the cell cycle and enter a G0 phase, with increased expression of genes for specialized neuronal functions and potential activation of apoptosis for improperly differentiated cells.
   • The cell would undergo multiple rounds of meiosis to reduce its ploidy, followed by a rapid increase in surface area to volume ratio to support its new function.
   • The cell's internal regulators would be suppressed, leading to uncontrolled division and the formation of a benign tumor, with apoptosis being irrelevant to differentiation.

   Answer: The cell would likely exit the cell cycle and enter a G0 phase, with increased expression of genes for specialized neuronal functions and potential activation of apoptosis for improperly differentiated cells.

5. A patient is diagnosed with a malignant tumor caused by a mutation in the $p53$ gene, a critical internal regulator. How does this specific gene defect contribute to uncontrolled cell proliferation, and which phases of the cell cycle are most directly impacted by the loss of functional $p53$?

   • The mutated $p53$ gene causes cells to prematurely enter Meiosis I, leading to aneuploidy and tumor formation.
   • Loss of functional $p53$ allows cells with damaged DNA to bypass the G1 and G2 checkpoints, leading to uncontrolled progression through the S and M phases.
   • The $p53$ mutation enhances the activity of external regulators, causing cells to differentiate rapidly and form benign tumors.
   • Defective $p53$ prevents cytokinesis, resulting in multinucleated cells that are unable to divide, thus inhibiting tumor growth.

   Answer: Loss of functional $p53$ allows cells with damaged DNA to bypass the G1 and G2 checkpoints, leading to uncontrolled progression through the S and M phases.

6. In an organism with $2N=4$ chromosomes, consider a scenario where crossing over completely fails to occur during Prophase I of meiosis. Assuming all other meiotic events proceed normally, what would be the ploidy level of the cells immediately after Meiosis I and Meiosis II, and how would the genetic diversity of the resulting gametes compare to a normal meiotic process?

   • After Meiosis I, cells would be $N$; after Meiosis II, gametes would be $N$. Genetic diversity would be significantly reduced due to the absence of new allele combinations.
   • After Meiosis I, cells would be $2N$; after Meiosis II, gametes would be $N$. Genetic diversity would be unaffected as independent assortment still occurs.
   • After Meiosis I, cells would be $N$; after Meiosis II, gametes would be $N$. Genetic diversity would be limited to independent assortment, lacking recombination.
   • After Meiosis I, cells would be $2N$; after Meiosis II, gametes would be $2N$. Genetic diversity would be completely eliminated, resulting in clones.

   Answer: After Meiosis I, cells would be $N$; after Meiosis II, gametes would be $N$. Genetic diversity would be limited to independent assortment, lacking recombination.

7. An early embryonic cell is characterized by a high surface area to volume ratio and rapid, continuous cell cycles. As this cell differentiates into a mature, specialized cell (e.g., a large muscle cell), which of the following changes in its cell cycle and metabolic efficiency would typically occur?

   • The cell would maintain a high surface area to volume ratio to support increased metabolic demands, and its cell cycle would accelerate to produce more specialized cells.
   • The cell's surface area to volume ratio would decrease, potentially limiting efficient nutrient exchange, and its cell cycle would likely slow down or exit into G0.
   • The cell would increase its volume disproportionately to its surface area, enhancing nutrient uptake, and would enter a prolonged S phase for extensive DNA replication.
   • The cell would undergo apoptosis to make way for more efficient, smaller cells, and its cell cycle would become highly regulated by external growth factors.

   Answer: The cell's surface area to volume ratio would decrease, potentially limiting efficient nutrient exchange, and its cell cycle would likely slow down or exit into G0.

8. A karyotype analysis confirms a diagnosis of Trisomy 21 (Down Syndrome) in an individual. Given the typical mechanisms of chromosomal disorders, which of the following meiotic errors is the most common cause, and how does it relate to the normal outcome of meiosis?

   • Nondisjunction of sister chromatids during Anaphase II in the father, resulting in two gametes with $N+1$ and two with $N-1$ chromosomes.
   • Nondisjunction of homologous chromosomes during Anaphase I in the mother, leading to gametes with either $N+1$ or $N-1$ chromosomes.
   • Crossing over occurring too frequently in Prophase I, leading to an extra chromosome in the zygote.
   • Failure of cytokinesis after Meiosis II, resulting in diploid gametes that then fuse with a normal haploid gamete.

   Answer: Nondisjunction of homologous chromosomes during Anaphase I in the mother, leading to gametes with either $N+1$ or $N-1$ chromosomes.

9. Consider a population of single-celled organisms that reproduce exclusively through asexual reproduction. If a mutation arises in one individual that disables a key cell cycle checkpoint, leading to uncontrolled cell division (analogous to cancer), how would the spread and impact of this mutation differ from a similar mutation occurring in a sexually reproducing multicellular organism?

   • In the asexual population, the mutation would rapidly spread to all offspring, potentially leading to a population crash due to resource depletion, whereas in sexual reproduction, it might be diluted or selected against.
   • In both populations, the mutation would be quickly eliminated by apoptosis, preventing its spread and impact.
   • In the asexual population, the mutation would be confined to the original individual, as there is no genetic exchange, while in sexual reproduction, it would spread rapidly through gamete fusion.
   • The asexual population would develop increased genetic diversity due to the mutation, making it more resilient, while the sexual population would become more susceptible to environmental changes.

   Answer: In the asexual population, the mutation would rapidly spread to all offspring, potentially leading to a population crash due to resource depletion, whereas in sexual reproduction, it might be diluted or selected against.

10. Compare the processes of cytokinesis in a typical animal cell and a typical plant cell following mitosis. If an external growth factor is introduced to both cell types, how might their responses differ in terms of cell division, and what specific structures are involved in their respective cytokinesis processes?

    • Animal cells form a cell plate, while plant cells form a cleavage furrow. Growth factors would primarily stimulate DNA replication in both.
    • Animal cells form a cleavage furrow by pinching the membrane, while plant cells form a cell plate from vesicles. Growth factors would stimulate division in both, but plant cells might also be influenced by hormones for cell wall formation.
    • Both cell types use spindle fibers to divide the cytoplasm. Growth factors would only affect animal cells, as plant cells have rigid cell walls.
    • Plant cells undergo cytokinesis during anaphase, while animal cells undergo it during telophase. Growth factors would inhibit division in both to prevent overgrowth.

    Answer: Animal cells form a cleavage furrow by pinching the membrane, while plant cells form a cell plate from vesicles. Growth factors would stimulate division in both, but plant cells might also be influenced by hormones for cell wall formation.