Cell-cycle & cell division practice · Mixed difficulty

Option A (Both Statement-I and Statement-II are true) is correct because plants exhibit continuous growth due to the presence of meristematic cells.

Option C (S-Phase) is correct because centriole duplication in animal cells occurs during the S-phase of the cell cycle, along with DNA replication.

Option A (Homologous chromosomes behave independently) is correct because in mitosis, homologous chromosomes do not pair or interact, whereas in meiosis, they pair and may crossover.

Option C (250) is correct because each seed requires one meiotic division in the egg mother cell and four meiotic divisions for each pollen grain that fertilizes the egg, but only one pollen grain is needed. Therefore, 200 eggs require 200 meiotic divisions, and 200 pollen grains require 200/4 = 50 meiotic divisions. Total = 200 + 50 = 250.

Option A (8 pg, 16 chromosomes) is correct because a gamete has half the DNA and half the number of chromosomes as a somatic cell in G1 phase.

Option B (Telophase) is correct because the listed features are characteristic of telophase, the final stage of mitosis.

Option D (Metaphase) is correct because Colchicine inhibits microtubule polymerization, disrupting the spindle apparatus and arresting cells at metaphase.

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.

The correct answer is B (M) because the M phase (mitosis) is a distinct phase of the cell cycle, separate from interphase, which includes G1, S, and G2 phases.

Option B (12000) is correct because the amount of DNA doubles during the S phase, so if there are 6000 nucleotides in G1, there will be 12000 in S phase.

Option B (No change) is correct because mitosis is an equational division where the chromosome number remains the same in daughter cells.

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 B (Four times DNA content but double chromosome number) is correct because after S phase, DNA content is doubled and since the cell is diploid, the chromosome number is double that of a gamete.

Option D (5) is correct because each mitotic division doubles the number of cells, so 32 cells require 5 generations ($2^5 = 32$).

Option A (Prophase) is correct because during prophase, the chromatin condenses, and the tangled chromosomal material starts to untangle.

Option A (64) is correct because in 2 hours (120 minutes), the bacteria will divide 120/20 = 6 times. Therefore, the number of bacteria formed will be 2⁶ = 64.

Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because mitosis is equational division and the chromosome number remains the same.

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 B (II and IV) is correct because in G1 phase, cell organelles increase in number and RNAs, proteins and other biochemicals are synthesized.

Option A (Reduces by half) is correct because meiosis is a reductional division, meaning it reduces the chromosome number from diploid (2n) to haploid (n), which is half the original number.

Option A (Inhibit spindle formation during mitosis) is correct because colchicine disrupts microtubule polymerization, which is essential for spindle formation.

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 A (There is no growth of cells) is correct because cleavage divisions are characterized by rapid cell division without significant cell growth, resulting in smaller cells with each division.

Option A (AB, aB, Ab, ab) is correct because with independent assortment during meiosis, an AaBb genotype will produce all four possible combinations of alleles in the gametes.

Option B (Twice the number of chromosomes and four times the amount of DNA) is correct because after S phase, a somatic cell has twice the number of chromosomes and four times the DNA content compared to a gamete.

Option B is correct because meiosis II results in the formation of four haploid cells from the two haploid cells produced in meiosis I.

Option A (Haploid) is correct because meiosis II results in the separation of sister chromatids, producing haploid cells, each with one set of chromosomes.

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