IB Biology Quiz: Apply Cell And Nuclear Division
14 questions · exam conditions
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Apply Cell And Nuclear DivisionQuestion 1 of 14

A failure of the G2/M checkpoint to function correctly would have which of the following as the most likely and immediate consequence?

The cell would become arrested in the G2 phase and would not proceed to mitosis.
The cell would initiate mitosis even if its DNA was not completely replicated or was damaged.
The cell would replicate its DNA a second time, resulting in a tetraploid state before division.
The cell would fail to synthesize the necessary proteins, such as cyclins, required for mitosis.
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IB Biology Quiz

IB Biology Quiz: Apply Cell And Nuclear Division

Practice Apply Cell And Nuclear Division in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Cell And Nuclear Division, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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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.

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Question 1

A failure of the G2/M checkpoint to function correctly would have which of the following as the most likely and immediate consequence?

  1. The cell would become arrested in the G2 phase and would not proceed to mitosis.
  2. The cell would initiate mitosis even if its DNA was not completely replicated or was damaged. (correct answer)
  3. The cell would replicate its DNA a second time, resulting in a tetraploid state before division.
  4. The cell would fail to synthesize the necessary proteins, such as cyclins, required for mitosis.
Explanation: The G2/M checkpoint serves as a quality control step to ensure that DNA replication is complete and that any DNA damage has been repaired before the cell enters mitosis. If this checkpoint fails, the cell loses this control and may proceed into M phase with significant genetic errors, which can lead to mutations or cell death.

Question 2

A cell biologist determines that in a tissue sample of 500 cells, 420 cells are in interphase and the entire cell cycle lasts 24 hours. Assuming all cells are dividing, what is the approximate duration of mitosis for this cell type?

  1. 3.84 hours (correct answer)
  2. 4.80 hours
  3. 20.16 hours
  4. 0.16 hours
Explanation: First, calculate the number of cells in mitosis: 500 total cells - 420 interphase cells = 80 cells in mitosis. The mitotic index is the proportion of cells in mitosis: 80 / 500 = 0.16. The duration of mitosis is this proportion multiplied by the total cell cycle time: 0.16 * 24 hours = 3.84 hours. Choice C represents the duration of interphase (420/500 * 24). Choice D is the mitotic index itself, not the duration. Choice B is a calculation error.

Question 3

Colchicine is a substance that inhibits the formation of microtubules. It is often used in labs to arrest cells during mitosis for karyotyping.

If a dividing animal cell is treated with colchicine, at which stage would the cell cycle be arrested and why?

  1. Prophase, because the chromosomes would be unable to condense without microtubules.
  2. Anaphase, because the centromeres would not be able to divide and release the sister chromatids.
  3. Telophase, because the nuclear envelope could not reform around the decondensing chromosomes.
  4. Metaphase, because the mitotic spindle could not form to align the chromosomes at the cell's equator. (correct answer)
Explanation: Microtubules are the polymers that form the mitotic spindle. The spindle is essential for attaching to chromosomes at their kinetochores and aligning them at the metaphase plate. Without a functional spindle, the cell cannot pass the spindle assembly checkpoint and will be arrested in metaphase.

Question 4

If non-disjunction of a single chromosome pair occurs during meiosis II in a human, what would be the expected chromosome numbers in the four gametes produced from that single meiotic event?

  1. 24, 24, 22, 22
  2. 24, 22, 23, 23 (correct answer)
  3. 24, 24, 23, 23
  4. 24, 22, 22, 22
Explanation: If meiosis I proceeds normally, two cells with the haploid number (n=23) enter meiosis II. If non-disjunction occurs in one of these two cells, its sister chromatids fail to separate. This produces one gamete with an extra chromosome (n+1 = 24) and one gamete missing a chromosome (n-1 = 22). The other cell from meiosis I divides normally in meiosis II, producing two normal haploid gametes (n=23). Thus, the final count is 24, 22, 23, 23.

Question 5

An organism has a diploid chromosome number of 2n=12. One specific pair of homologous chromosomes is homozygous for all its alleles. How many genetically unique gametes can this organism produce through independent assortment alone?

  1. 12
  2. 32 (correct answer)
  3. 64
  4. 144
Explanation: The number of unique gametes from independent assortment is calculated as 2^n, where n is the haploid number. For 2n=12, n=6. Normally, this would be 2^6 = 64. However, one pair of homologous chromosomes is homozygous, meaning it does not contribute to genetic variation through assortment. Therefore, we only consider the pairs that are heterozygous, which is n-1 = 5. The calculation is 2^5 = 32.

Question 6

A single crossing-over event occurs between two non-sister chromatids of a homologous pair. Which statement accurately predicts the genetic makeup of the four gametes produced at the end of meiosis?

  1. All four gametes will have a unique, recombinant combination of alleles.
  2. One gamete will be parental, and three will be recombinant.
  3. Two gametes will have the parental combination of alleles, and two will be recombinant. (correct answer)
  4. All four gametes will have the parental combination of alleles, as crossing over is repaired.
Explanation: A single crossover event involves only two of the four chromatids in a bivalent (a homologous pair). The other two chromatids are not involved in that specific exchange. Therefore, after meiosis II, two of the resulting cells will contain the unchanged, parental chromatids, and two will contain the chromatids that underwent recombination. This results in two parental and two recombinant gametes.

Question 7

In which phase of division does the principle of independent assortment have its physical basis?

  1. Metaphase I, due to the random orientation of homologous pairs at the cell's equator. (correct answer)
  2. Prophase I, when homologous chromosomes pair up and form bivalents.
  3. Anaphase II, when sister chromatids separate to opposite poles of the cell.
  4. Metaphase II, due to the random orientation of individual chromosomes at the cell's equator.
Explanation: Independent assortment refers to the fact that the orientation of one homologous pair at the metaphase plate does not influence the orientation of any other pair. This random alignment happens during metaphase I, and the subsequent separation in anaphase I leads to different combinations of maternal and paternal chromosomes in the daughter cells.

Question 8

Imagine a hypothetical organism where crossing over does not occur during meiosis. What would be the main source of genetic variation among its gametes?

  1. Spontaneous mutations arising during the S phase of interphase.
  2. The independent assortment of homologous chromosomes in meiosis I. (correct answer)
  3. The segregation of sister chromatids during meiosis II.
  4. There would be no genetic variation; all gametes would be identical.
Explanation: Meiosis has two main sources of genetic variation: crossing over and independent assortment. If crossing over is absent, the random orientation of homologous pairs at the metaphase plate in meiosis I (independent assortment) is the sole remaining meiotic mechanism for shuffling parental genes into new combinations in the gametes. Spontaneous mutation (A) is a source of new alleles but is not a meiotic mechanism of variation.

Question 9

A male bee develops from an unfertilized egg and is haploid. Its somatic cells must divide for the bee to grow. Which statement accurately describes cell division in a male bee's somatic tissues?

  1. The haploid cells undergo mitosis, producing two genetically identical haploid daughter cells. (correct answer)
  2. The haploid cells undergo meiosis to produce more haploid cells for growth.
  3. The haploid cells first undergo replication without division to become diploid, then divide by mitosis.
  4. The haploid cells divide by binary fission, as they lack homologous chromosome pairs.
Explanation: Mitosis is a process of nuclear division that produces genetically identical daughter cells, regardless of the starting ploidy. A haploid (n) cell can undergo mitosis to produce two identical haploid (n) cells. This allows for growth in haploid organisms. Meiosis (A) requires homologous pairs and reduces ploidy, which is not possible or desirable for somatic growth. Binary fission (D) is for prokaryotes.

Question 10

A researcher observes cytokinesis in an unknown eukaryote. They note the formation of a cleavage furrow that deepens until the cell is pinched into two. This observation strongly suggests the cell is:

  1. a plant cell, because the furrow organizes the material for the cell plate.
  2. a fungal cell, because the chitinous cell wall pinches inward.
  3. a prokaryotic cell, as this process is identical to binary fission.
  4. an animal cell, as it lacks a rigid cell wall and divides by constriction. (correct answer)
Explanation: Cytokinesis by forming a cleavage furrow is characteristic of animal cells and some protists. A contractile ring of actin and myosin filaments forms under the plasma membrane and constricts, pinching the cell in two. Cells with rigid cell walls, like plants and fungi, cannot do this and instead form a cell plate or septum.

Question 11

Budding in yeast is a form of asexual reproduction. Which process of nuclear division is directly involved and why is it evolutionarily advantageous in a stable environment?

  1. Meiosis, because it ensures the daughter cell is haploid and can fuse with other cells.
  2. Meiosis, because it produces genetic variation that helps adaptation to changing conditions.
  3. Mitosis, because it produces a genetically identical daughter cell that is well-adapted to the current conditions. (correct answer)
  4. Mitosis, because it allows rapid cell division without the need for chromosome pairing.
Explanation: Asexual reproduction in eukaryotes like yeast involves mitosis, which creates a daughter nucleus that is genetically identical to the parent. In a stable, favorable environment, this is advantageous because the parent is already well-adapted, and producing genetically identical offspring ensures continuation of this successful genotype.

Question 12

A karyotype analysis reveals the notation 47, XXY. Which statement correctly identifies the syndrome and a possible meiotic error causing it?

  1. Turner syndrome, caused by non-disjunction of autosomes in meiosis I.
  2. Down syndrome, caused by non-disjunction of the X chromosome in meiosis II.
  3. Klinefelter syndrome, caused by non-disjunction of sex chromosomes in meiosis I or II. (correct answer)
  4. Patau syndrome, caused by non-disjunction of the Y chromosome in mitosis.
Explanation: 47, XXY corresponds to Klinefelter syndrome, where an individual has an extra X chromosome. This can result from non-disjunction of sex chromosomes during meiosis I (both X chromosomes go to the same gamete) or meiosis II (sister chromatids of X fail to separate), producing gametes with abnormal numbers of sex chromosomes.

Question 13

Which of these events is a key distinguishing feature of meiosis I, but not mitosis or meiosis II?

  1. The condensation of chromatin into visible chromosomes.
  2. The alignment of chromosomes along the cell's equatorial plate.
  3. The separation of sister chromatids to opposite poles.
  4. The pairing of homologous chromosomes and the separation of these pairs. (correct answer)
Explanation: The pairing of homologous chromosomes to form bivalents (synapsis) and the subsequent separation of these homologous pairs during anaphase I are unique to meiosis I. Condensation (A) and alignment (B) occur in all three processes, although alignment differs (pairs vs. individual). Separation of sister chromatids (C) occurs in mitosis and meiosis II, but not meiosis I.

Question 14

Which of the following describes the most significant challenge during meiosis for a sterile triploid (3n) organism, such as a seedless watermelon?

  1. DNA replication in the S phase is inhibited due to the odd number of chromosome sets.
  2. Cytokinesis cannot divide the cytoplasm into three equal parts following telophase.
  3. The three homologous chromosomes for each type cannot form stable pairs during prophase I. (correct answer)
  4. Sister chromatids fail to separate correctly during anaphase II due to uneven tension.
Explanation: The critical step of meiosis I is the pairing (synapsis) of homologous chromosomes. In a triploid organism, there are three homologues for each chromosome type. They cannot pair up properly (e.g., two might pair, leaving one out), which leads to incorrect segregation during anaphase I. This results in aneuploid (genetically imbalanced) gametes, which are typically non-viable, causing sterility.