IB Biology Quiz: Understand Cell And Nuclear Division
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Understand Cell And Nuclear DivisionQuestion 1 of 20

A single somatic cell in a diploid organism (2n) undergoes a complete mitotic division, but the sister chromatids of one chromosome fail to separate during anaphase. What will be the chromosome number in the two resulting daughter cells?

One cell will be 2n+1 and one cell will be 2n-1.
Both cells will have a normal diploid number of 2n.
One cell will be 4n and the other cell will be anucleated.
One cell will be 2n and one cell will be 2n+1.
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IB Biology Quiz

IB Biology Quiz: Understand Cell And Nuclear Division

Practice Understand 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.

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This quiz focuses on Understand 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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Question 1

A single somatic cell in a diploid organism (2n) undergoes a complete mitotic division, but the sister chromatids of one chromosome fail to separate during anaphase. What will be the chromosome number in the two resulting daughter cells?

  1. One cell will be 2n+1 and one cell will be 2n-1. (correct answer)
  2. Both cells will have a normal diploid number of 2n.
  3. One cell will be 4n and the other cell will be anucleated.
  4. One cell will be 2n and one cell will be 2n+1.
Explanation: This event is mitotic non-disjunction. When the sister chromatids of one chromosome fail to separate, one pole receives both chromatids (which become individual chromosomes) while the other pole receives none from that homologous pair. After cytokinesis, the daughter cell that received both chromatids will have an extra chromosome (2n+1, trisomy), and the daughter cell that received none will be missing a chromosome (2n-1, monosomy).

Question 2

A mutation in a gene encoding a specific cyclin-dependent kinase (CDK) results in a protein that is permanently active, even without binding to its corresponding cyclin. What is the most likely direct outcome for a cell with this mutation?

  1. The cell would be permanently arrested at a specific checkpoint in the cell cycle.
  2. The cell would bypass a specific cell cycle checkpoint, increasing the rate of cell division. (correct answer)
  3. The process of DNA replication during the S phase would be completely inhibited.
  4. Apoptosis, or programmed cell death, would be immediately triggered by the abnormal protein.
Explanation: Cyclin-CDK complexes act as signals to progress through cell cycle checkpoints. A permanently active CDK would continuously signal 'go', causing the cell to bypass the checkpoint it regulates. This leads to uncontrolled progression through the cell cycle. A) is incorrect because active CDKs promote, not arrest, the cell cycle. C) is incorrect because specific CDKs initiate DNA replication; permanent activity would likely promote, not inhibit it. D) is incorrect because while checkpoint failure can eventually lead to apoptosis, it is not the immediate and direct effect of an active CDK; the direct effect is progression.

Question 3

A researcher observes a tissue sample and calculates a mitotic index of 0.05 for healthy tissue and 0.30 for a nearby tumorous tissue. What is the most valid conclusion that can be drawn from this data?

  1. The cells in the healthy tissue have a much longer M phase than the cells in the tumour.
  2. The cell cycle is completed more rapidly in the tumorous tissue compared to the healthy tissue.
  3. A higher proportion of cells in the tumorous tissue are actively undergoing mitosis at any given time. (correct answer)
  4. The tumorous tissue contains six times as many cells as the healthy tissue.
Explanation: The mitotic index is the ratio of cells in mitosis to the total number of cells. A higher index (0.30 vs 0.05) means that at the moment of observation, a larger fraction of the tumour cells were in M phase. This is the direct interpretation of the data. While this suggests a higher rate of proliferation (B), it's an inference, not a direct conclusion from the index alone. The index doesn't provide information on the length of specific phases (A) or the total number of cells (D).

Question 4

Meiosis II is mechanistically similar to mitosis. However, there is a fundamental difference in the genetic status of the cells entering each process. Which statement correctly identifies this difference?

  1. Cells entering meiosis II are diploid, whereas cells entering mitosis are haploid.
  2. The chromosomes in cells entering meiosis II are unreplicated, unlike in mitosis.
  3. Cells entering meiosis II are genetically diverse, whereas the cell entering mitosis is identical to its parent.
  4. The cells entering meiosis II are haploid regarding chromosome number, whereas cells entering mitosis are diploid. (correct answer)
Explanation: Following meiosis I, the cells are haploid (n) because the homologous chromosome pairs have been separated. These haploid cells then enter meiosis II. In contrast, somatic cells entering mitosis are diploid (2n). While the cells entering meiosis II are genetically diverse due to crossing over (C), the most fundamental difference defining the process is the ploidy level.

Question 5

An organism has a diploid number of 10 (2n = 10). Assuming independent assortment and no crossing over, how many genetically distinct types of gametes can this organism produce?

  1. 10
  2. 25
  3. 32 (correct answer)
  4. 1024
Explanation: The number of possible combinations of chromosomes in the gametes due to independent assortment is calculated by the formula 2^n, where n is the haploid number of chromosomes. If the diploid number (2n) is 10, then the haploid number (n) is 5. Therefore, the number of combinations is 2^5 = 32. Distractor D (1024) is 2^10, using the diploid number. Distractor B is 5^2, an incorrect formula.

Question 6

Non-disjunction of a single chromosome pair occurs during meiosis I in a human primary spermatocyte. If the four resulting sperm cells all fertilize normal ova, what is the expected distribution of karyotypes in the resulting zygotes?

  1. Two zygotes will be normal (46 chromosomes), one will be trisomic (47), and one will be monosomic (45).
  2. Two zygotes will be trisomic (47 chromosomes) and two will be monosomic (45 chromosomes). (correct answer)
  3. All four zygotes will be abnormal, with two being triploid (69) and two being haploid (23).
  4. One zygote will be normal (46 chromosomes), two will be trisomic (47), and one will be monosomic (45).
Explanation: Non-disjunction in meiosis I means a pair of homologous chromosomes fails to separate. This results in two secondary spermatocytes that are abnormal: one has an extra chromosome (n+1) and the other is missing one (n-1). Meiosis II then proceeds, resulting in two sperm that are n+1 and two sperm that are n-1. When these fertilize normal ova (n), the zygotes will be 2n+1 (trisomic) and 2n-1 (monosomic), respectively. Therefore, two trisomic and two monosomic zygotes are produced. Distractor A describes the outcome of non-disjunction in meiosis II.

Question 7

A cell undergoes a normal mitotic division of its nucleus, but cytokinesis fails to occur. What is the immediate and most likely result?

  1. A single cell with two genetically identical nuclei. (correct answer)
  2. Two separate cells, one with a nucleus and one without.
  3. A single cell with one large, tetraploid (4n) nucleus.
  4. Four separate, genetically distinct haploid cells.
Explanation: Mitosis (karyokinesis) is the division of the nucleus. At the end of telophase, two distinct and genetically identical nuclei are formed at opposite poles of the cell. Cytokinesis is the subsequent division of the cytoplasm. If cytokinesis fails, the result is a single large cell containing both of these newly formed nuclei. Such a cell is called a binucleate cell. A tetraploid nucleus (C) might form if the nuclear envelopes fused or if the cell re-entered the cell cycle from this state, but the most immediate result is two separate nuclei in one cytoplasm.

Question 8

Which statement correctly distinguishes the process of cytokinesis in a typical plant cell from that in a typical animal cell?

  1. Plant cells form a cleavage furrow by pinching inwards, while animal cells build a cell plate from the inside out.
  2. Plant cells must synthesize a new plasma membrane, while animal cells reuse the existing membrane for both daughter cells.
  3. In animal cells, cytokinesis begins during prophase, while in plant cells, it begins during telophase.
  4. Animal cells divide by forming a contractile ring of actin, while plant cells transport vesicles to form a cell plate. (correct answer)
Explanation: The primary mechanistic difference is the formation of a contractile ring made of actin and myosin filaments in animal cells, which creates a cleavage furrow. In plant cells, the rigid cell wall prevents this pinching. Instead, vesicles from the Golgi apparatus align at the cell's equator and fuse to form a cell plate, which develops into a new cell wall and plasma membrane. A reverses the two mechanisms. C is incorrect regarding the timing. D is incorrect as both cell types must synthesize new plasma membrane to accommodate the new cells.

Question 9

A cell has a DNA content arbitrarily measured as 8x during prophase I of meiosis. What would be the expected DNA content of a single gamete produced from this cell?

  1. 1x
  2. 2x (correct answer)
  3. 4x
  4. 8x
Explanation: If the DNA content during prophase I (after replication) is 8x, this represents a tetraploid amount of DNA (4n DNA content) for the diploid cell. The cell in G1 would have had 4x DNA content. Meiosis I halves the chromosome number and DNA content, resulting in cells with 4x DNA content. Meiosis II separates sister chromatids, halving the DNA content again. Therefore, the final gamete will have a DNA content of 2x. This represents the haploid (1n) amount of DNA.

Question 10

In the life cycle of a flowering plant (angiosperm), meiotic division directly results in the formation of which structures?

  1. Haploid gametes (sperm and egg cells).
  2. Diploid zygotes that develop into embryos.
  3. Haploid spores that develop into gametophytes. (correct answer)
  4. Diploid somatic cells for the growth of the plant.
Explanation: Flowering plants exhibit an alternation of generations. The diploid sporophyte undergoes meiosis not to produce gametes directly, but to produce haploid spores (microspores and megaspores). These spores then undergo mitosis to develop into the multicellular haploid gametophytes (pollen grain and embryo sac), which then produce the gametes. A is the pattern in animals. B is the result of fertilization. D is the result of mitosis.

Question 11

During prophase I, homologous chromosomes pair up in synapsis. What are the structures called that represent the physical points of crossing over between non-sister chromatids?

  1. Centromeres
  2. Kinetochores
  3. Chiasmata (correct answer)
  4. Centrosomes
Explanation: Chiasmata (singular: chiasma) are the X-shaped points of contact between paired homologous chromosomes during prophase I where crossing over and the exchange of genetic material occurs. Centromeres (A) are the constricted regions of a chromosome where sister chromatids are joined. Kinetochores (B) are protein structures on centromeres where spindle fibers attach. Centrosomes (D) are the organelles that organize the microtubules for the spindle.

Question 12

Genes A and B are linked on the same chromosome. A dihybrid individual with genotype Ab/aB undergoes meiosis. Which statement accurately describes the resulting gametes?

  1. Four types of gametes (Ab, aB, AB, ab) will be produced in equal proportions.
  2. Only two types of gametes (Ab and aB) will be produced.
  3. The parental gametes (Ab and aB) will be more frequent than the recombinant gametes (AB and ab). (correct answer)
  4. The recombinant gametes (AB and ab) will be more frequent than the parental gametes (Ab and aB).
Explanation: When genes are linked, they tend to be inherited together. The original combinations on the chromosomes (Ab and aB) are the parental types. Crossing over between the two gene loci can create new combinations (AB and ab), which are the recombinant types. Because crossing over is a relatively infrequent event at any given location, the parental combinations will always be more frequent than the recombinant combinations. A describes independent assortment. B describes complete linkage with no crossing over.

Question 13

A diploid cell has a chromosome number of 2n = 46. What will be the chromosome number and number of DNA molecules in one of its daughter cells immediately following meiosis I?

  1. 23 chromosomes, 23 DNA molecules
  2. 46 chromosomes, 92 DNA molecules
  3. 46 chromosomes, 46 DNA molecules
  4. 23 chromosomes, 46 DNA molecules (correct answer)
Explanation: Meiosis I separates homologous chromosomes, thus halving the chromosome number from diploid (2n) to haploid (n). So, the cell will have n = 23 chromosomes. However, at the end of meiosis I, each of these chromosomes still consists of two sister chromatids because sister chromatids do not separate until meiosis II. Since each chromatid is one DNA molecule, there are 23 chromosomes x 2 DNA molecules/chromosome = 46 DNA molecules in total.

Question 14

The G1 checkpoint is a critical decision point for a cell. What is the primary condition that must be met for a cell to pass this checkpoint and enter S phase?

  1. The chromosomes must be correctly attached to the mitotic spindle.
  2. DNA must be fully replicated without significant errors.
  3. The cell must have reached a sufficient size and have adequate resources for DNA synthesis. (correct answer)
  4. The homologous chromosomes must be paired and ready for separation.
Explanation: The G1 checkpoint, also known as the restriction point, primarily assesses the cell's internal and external conditions to decide whether to commit to division. Key factors include cell size, nutrient availability, and the presence of growth factors. If conditions are favorable, the cell proceeds to S phase. A describes the metaphase checkpoint. B describes the G2 checkpoint. D describes an event in meiosis I, not a general cell cycle checkpoint.

Question 15

Which event is essential for generating genetic variation during meiosis but would be detrimental if it occurred during mitosis?

  1. Condensation of chromatin into visible chromosomes.
  2. Separation of sister chromatids at the centromere.
  3. Pairing of homologous chromosomes to form bivalents. (correct answer)
  4. Formation of a spindle apparatus from the centrosomes.
Explanation: The pairing of homologous chromosomes (synapsis) occurs in prophase I and is a prerequisite for crossing over, a major source of genetic variation. If homologous chromosomes paired up during mitosis, it could lead to improper segregation and result in aneuploid daughter cells, which would be detrimental to the organism. The other processes (A, B, D) are essential and occur normally in both mitosis and meiosis (though sister chromatid separation occurs at different stages overall).

Question 16

What is a fundamental difference in the state of the chromosomes migrating to opposite poles during anaphase of mitosis compared to anaphase I of meiosis?

  1. Chromosomes in anaphase of mitosis are unreplicated, while chromosomes in anaphase I are replicated.
  2. Chromosomes moving to the poles in anaphase I consist of two chromatids, while those in anaphase of mitosis consist of a single chromatid. (correct answer)
  3. Chromosomes in anaphase I are not attached to spindle fibres, whereas they are attached in anaphase of mitosis.
  4. Chromosomes in anaphase of mitosis are maternal in origin only, while in anaphase I they are a random mix of paternal and maternal.
Explanation: In anaphase of mitosis, sister chromatids separate, so the structures moving to the poles are individual, single-chromatid chromosomes. In anaphase I of meiosis, homologous chromosomes separate, but the sister chromatids remain attached. Therefore, the structures moving to the poles are replicated chromosomes, each still composed of two sister chromatids.

Question 17

Which process directly contributes to the principle of independent assortment as described by Mendel?

  1. The random exchange of genetic material between homologous chromosomes during prophase I.
  2. The semi-conservative replication of DNA during the S phase before meiosis begins.
  3. The separation of sister chromatids into daughter cells during anaphase II.
  4. The random orientation of homologous pairs at the metaphase plate during metaphase I. (correct answer)
Explanation: Independent assortment refers to the fact that alleles for different traits are inherited independently of one another. The physical basis for this is the random orientation of each pair of homologous chromosomes at the metaphase plate during metaphase I. The orientation of one pair (e.g., maternal left, paternal right) does not influence the orientation of any other pair. This leads to various combinations of paternal and maternal chromosomes in the resulting gametes. A describes crossing over, which creates new allele combinations on a single chromosome, but not the assortment of different chromosomes.

Question 18

Colchicine is a chemical that inhibits the formation of microtubules. If a eukaryotic cell is treated with colchicine during mitosis, at which stage will the cell cycle be arrested?

  1. Prophase, because chromosomes will not be able to condense.
  2. Metaphase, because the mitotic spindle cannot form to align the chromosomes. (correct answer)
  3. Anaphase, because sister chromatids will not be able to separate.
  4. Telophase, because the nuclear envelope will not be able to reform.
Explanation: Microtubules are the primary components of the mitotic spindle. The spindle is responsible for attaching to chromosomes and aligning them at the metaphase plate. Without microtubules, a functional spindle cannot form, and the cell will arrest at the spindle assembly checkpoint in metaphase. A is incorrect as condensation is independent of microtubules. C is incorrect because the cell would not reach anaphase if chromosomes are not aligned. D is incorrect as reformation of the nuclear envelope is also downstream of metaphase and anaphase events.

Question 19

A cell in the G2 phase is exposed to ionizing radiation, causing multiple double-strand breaks in its DNA. Which is the most likely immediate response mediated by the cell cycle control system?

  1. The cell cycle will be arrested at the G2/M checkpoint, preventing entry into mitosis until repairs are made. (correct answer)
  2. The cell will immediately trigger apoptosis as the damage is irreparable.
  3. The cell will reverse its cycle to re-enter S phase to use replication machinery for repair.
  4. The cell will proceed into mitosis but will be arrested at the metaphase checkpoint.
Explanation: The G2/M checkpoint is a critical control point that assesses whether all DNA has been replicated and if any replicated DNA is damaged. If damage is detected, as in this case, the checkpoint machinery will arrest the cell cycle to provide time for DNA repair mechanisms to work. Only after the damage is repaired will the cell be allowed to enter M phase. While severe, irreparable damage might lead to apoptosis (B), the primary, immediate response is arrest for repair. Reversing the cycle (C) is not a known mechanism. Arresting in metaphase (D) is a response to spindle attachment errors, not DNA damage detected in G2.

Question 20

What is the primary role of the protein complex cohesin during cell division?

  1. It pulls homologous chromosomes apart during anaphase I.
  2. It holds sister chromatids together from S phase until anaphase. (correct answer)
  3. It condenses chromatin into tightly packed chromosomes during prophase.
  4. It forms the contractile ring that divides the cytoplasm during cytokinesis.
Explanation: Cohesin is a protein complex that regulates the separation of sister chromatids. It forms rings that hold the two chromatids together after DNA replication in S phase. The degradation of cohesin by the enzyme separase at the anaphase transition allows the sister chromatids to be pulled apart. A describes the action of the spindle. C describes the role of condensins. D describes the action of actin and myosin.