Mitosis vs Meiosis I Anaphase — NEET Biology

Correct answer: B — S2, S3 and S6

This question explores the functional significance of Meiosis II and the consequences of its absence or errors, as well as related meiotic processes. S1: Correct. After Meiosis I, two haploid cells are formed, but each chromosome within these cells still consists of two sister chromatids. If Meiosis II is skipped, these cells will persist in this state. S2: Incorrect. If Meiosis II is skipped, the two cells formed after Meiosis I (which are haploid) would not undergo further division to separate sister chromatids. These cells would be haploid (n), not diploid (2n) as Meiosis I already reduced the chromosome number. Therefore, they cannot form diploid gametes. S3: Incorrect. The primary purpose of Meiosis II is the separation of sister chromatids, which is an equational division. The separation of homologous chromosomes and reduction in chromosome number occur during Meiosis I. S4: Correct. Nondisjunction refers to the failure of homologous chromosomes or sister chromatids to separate. If sister chromatids fail to separate during Anaphase II, it would lead to gametes with an incorrect number of chromosomes (e.g., n+1 or n-1). S5: Correct. Genetic recombination, facilitated by crossing over, occurs during the pachytene stage of Prophase I, significantly contributing to genetic variability. S6: Incorrect. In the absence of Meiosis II, the resulting cells would contain a haploid number of chromosomes (n), but each chromosome would still be duplicated (i.e., composed of two chromatids). Thus, the DNA content per cell would be 2C (relative to a normal haploid gamete with 1C DNA content). Therefore, statements S2, S3, and S6 are NOT correct. Recap: The question tests the understanding of the specific roles of Meiosis I and Meiosis II, their products, and the genetic and chromosomal implications of deviations from the normal meiotic process.