- 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?
Answer: The cell cycle will be arrested at the G2/M checkpoint, preventing entry into mitosis until repairs are made.
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.
- In the life cycle of a flowering plant (angiosperm), meiotic division directly results in the formation of which structures?
Answer: Haploid spores that develop into gametophytes.
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.
- A cell undergoes a normal mitotic division of its nucleus, but cytokinesis fails to occur. What is the immediate and most likely result?
Answer: A single cell with two genetically identical nuclei.
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.
- 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?
Answer: The cells entering meiosis II are haploid regarding chromosome number, whereas cells entering mitosis are diploid.
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.
- 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?
Answer: Chiasmata
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.
- 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?
Answer: The cell must have reached a sufficient size and have adequate resources for DNA synthesis.
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.
- 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?
Answer: 2x
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.
- 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?
Answer: One cell will be 2n+1 and one cell will be 2n-1.
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).
- What is the primary role of the protein complex cohesin during cell division?
Answer: It holds sister chromatids together from S phase until anaphase.
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.
- 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?
Answer: 23 chromosomes, 46 DNA molecules
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.
- 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?
Answer: The parental gametes (Ab and aB) will be more frequent than the recombinant gametes (AB and ab).
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.
- 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?
Answer: Chromosomes moving to the poles in anaphase I consist of two chromatids, while those in anaphase of mitosis consist of a single chromatid.
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.
- 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?
Answer: A higher proportion of cells in the tumorous tissue are actively undergoing mitosis at any given time.
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).
- 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?
Answer: 32
The number of possible combinations of chromosomes in the gametes due to independent assortment is calculated by the formula 2n, 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 25 = 32. Distractor D (1024) is 210, using the diploid number. Distractor B is 52, an incorrect formula. - 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?
Answer: Metaphase, because the mitotic spindle cannot form to align the chromosomes.
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.
- Which process directly contributes to the principle of independent assortment as described by Mendel?
Answer: The random orientation of homologous pairs at the metaphase plate during metaphase I.
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.
- 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?
Answer: The cell would bypass a specific cell cycle checkpoint, increasing the rate of cell division.
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.
- 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?
Answer: Two zygotes will be trisomic (47 chromosomes) and two will be monosomic (45 chromosomes).
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.
- Which statement correctly distinguishes the process of cytokinesis in a typical plant cell from that in a typical animal cell?
Answer: Animal cells divide by forming a contractile ring of actin, while plant cells transport vesicles to form a cell plate.
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.
- Which event is essential for generating genetic variation during meiosis but would be detrimental if it occurred during mitosis?
Answer: Pairing of homologous chromosomes to form bivalents.
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).
