- 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:
Answer: an animal cell, as it lacks a rigid cell wall and divides by constriction.
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.
- 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?
Answer: 24, 22, 23, 23
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.
- 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?
Answer: 3.84 hours
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.
- A karyotype analysis reveals the notation 47, XXY. Which statement correctly identifies the syndrome and a possible meiotic error causing it?
Answer: Klinefelter syndrome, caused by non-disjunction of sex chromosomes in meiosis I or II.
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.
- 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?
Answer: Mitosis, because it produces a genetically identical daughter cell that is well-adapted to the current conditions.
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.
- In which phase of division does the principle of independent assortment have its physical basis?
Answer: Metaphase I, due to the random orientation of homologous pairs at the cell's equator.
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.
- A failure of the G2/M checkpoint to function correctly would have which of the following as the most likely and immediate consequence?
Answer: The cell would initiate mitosis even if its DNA was not completely replicated or was damaged.
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.
- Which of these events is a key distinguishing feature of meiosis I, but not mitosis or meiosis II?
Answer: The pairing of homologous chromosomes and the separation of these pairs.
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.
- If a dividing animal cell is treated with colchicine, at which stage would the cell cycle be arrested and why?
Answer: Metaphase, because the mitotic spindle could not form to align the chromosomes at the cell's equator.
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.
- 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?
Answer: 32
The number of unique gametes from independent assortment is calculated as 2n, where n is the haploid number. For 2n=12, n=6. Normally, this would be 26 = 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 25 = 32. - 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?
Answer: Two gametes will have the parental combination of alleles, and two will be recombinant.
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.
- 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?
Answer: The haploid cells undergo mitosis, producing two genetically identical haploid daughter cells.
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.
- Which of the following describes the most significant challenge during meiosis for a sterile triploid (3n) organism, such as a seedless watermelon?
Answer: The three homologous chromosomes for each type cannot form stable pairs during prophase I.
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.
- Imagine a hypothetical organism where crossing over does not occur during meiosis. What would be the main source of genetic variation among its gametes?
Answer: The independent assortment of homologous chromosomes in meiosis I.
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.
