Cell-cycle & cell division practice · Medium

Option B (III, II, I and IV) is correct because it represents the correct sequence of events in prophase I of meiosis: Synapsis, Crossing over, Terminalization, and Disjunction of genomes.

Option D (Prophase) is correct because chromatin condenses and the nuclear envelope breaks down during prophase.

Option A ($64 \times 10^7$) is correct because the bacteria divide every 40 minutes, so in 4 hours (240 minutes), there are 6 divisions. Thus, the population increases by a factor of 2⁶ = 64.

Option D ($A=30, B=20 p g$) is correct because in the S phase, the DNA content doubles, but the number of chromosomes remains the same. After meiosis I, the DNA content is halved, and the chromosome number is also halved.

Option B (353) is correct because if 'n' is the number of mitotic divisions, then the number of cells formed is 2^n. To form 354 cells starting from 1 cell, we need to find n such that 2^n = 354, which is approximately 353 divisions.

Option D (256) is correct because after 4 hours (240 minutes), there will be 240/30 = 8 divisions. Therefore, the number of bacteria will be 2⁸ = 256.

Option D (I, II and III) is correct because all the statements about plant cytokinesis are accurate.

Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because Meiosis I is a reductional division where the chromosome number is halved, and Meiosis II is equational, maintaining the chromosome number.

Option C (Separation of chromatids and centromere) is correct because meiosis II is essentially a mitotic division of haploid cells, leading to the separation of sister chromatids at the centromere.

Option C (${G_1}$-phase) is correct because RNA and non-histone proteins are synthesized during the ${G_1}$ phase of the cell cycle.

Option D (All of the above) is correct because mitosis results in diploid daughter cells, growth of multicellular organisms, and cell repair.

Option D is correct because during pachytene of meiosis-I, the bivalent chromosomes clearly appear as tetrads.

Option C (A is true, but R is false) is correct because the chromosome number remains the same after S phase, but the DNA content doubles.

Option A ($150$) is correct because to form 120 seeds, 120 fertilizations are required. This means 120 eggs and 120 pollen grains are needed. To produce 120 eggs, 120/1 = 120 meiotic divisions are required. To produce 120 pollen grains, since one meiosis produces 4 pollen grains, 120/4 = 30 meiotic divisions are required. Therefore, the total number of meiotic divisions is 120 + 30 = 150.

Option A (Interphase and M-phase/divisional phase) is correct because the cell cycle is broadly divided into interphase (growth and DNA replication) and the M-phase (mitosis and cytokinesis).

Option C ($1\frac{1}{2}hour$) is correct because the yeast cell cycle duration is approximately 90 minutes, which is equal to 1.5 hours.

Option D (All the above) is correct because amitosis is a direct cell division observed in unicellular organisms where the nucleus constricts followed by cytoplasmic division.

Option B is correct. Mitosis does play a role in gamete formation in haploid organisms, and haploid organisms do divide by mitosis, but the reason doesn't explain why mitosis plays a role in gamete formation.

Option C (40 $pg\, DNA$ and 16 chromosomes) is correct because the microspore mother cell is diploid and has twice the DNA content and chromosome number of the male gamete.

The correct answer is D (${G_0}$-phase) because mature nerve cells are incapable of cell division and are therefore in the ${G_0}$ phase, a quiescent state outside the cell cycle.

Option A (least mitotic activity in the root apex) is correct because the quiescent center in the root apex is characterized by cells with low mitotic activity, serving as a reserve of cells that can replace damaged or differentiated cells in the surrounding region.

Option C (Splitting of centromere) is correct because splitting of the centromere occurs in anaphase, not metaphase.

Option A (16) is correct because if there are 8 bivalents in prophase I, the diploid number is 16. During anaphase II, sister chromatids separate, but the chromosome number remains the same as in the cell after meiosis I, which is half of the original diploid number. Therefore, the number of chromosomes during anaphase II is 16.

Option D (20pico gram) is correct because the DNA content doubles during the S phase and remains doubled until the end of G2 phase. Since the egg cell has 5 pg, the G2 phase will have 4 times that amount.

Option D (A-telophase, B-polyploidy) is correct because failure of cytokinesis after telophase results in a multinucleated cell, leading to polyploidy.

Option B (10 Chromosomes, 20 pg DNA) is correct because the spore mother cell is diploid (2n) and in G2 phase, the DNA content is doubled.

Option D (20 and 60 pg) is correct because in G2 phase, the chromosome number remains the same, but the DNA content doubles due to replication during S phase.

Option C (1 and 4) is correct because the diplotene stage *is* recognized by the dissolution of the synaptonemal complex, and in anaphase II, sister chromatids separate, not homologous chromosomes.

Option B (Chromatin) is correct because during interphase, chromosome material exists as loosely coiled chromatin threads.

The correct answer is B (${\mathrm{G}_0}$-phase) because cells that do not divide further exit the cell cycle and enter the ${\mathrm{G}_0}$ phase, where mitotic division is arrested.