Chapter 12, Problem 3STP

Summary Introduction

Introduction:

In animals the sexual life cycle involves meiosis. Meiosis involves two consecutive cell divisions called meiosis I and meiosis II. The two divisions produce a set of four haploid daughter cells that are not identical. Meiosis is important because it results in genetic variation.

Answer to Problem 3STP

Correct answer :

The correct answer is option C. cytokinesis

Explanation of Solution

Explanation/justification for the correct answer:

Option C. cytokinesis − During anphase I, as can be seen in the given diagram, the homologous chromosomes separate and line up at the equator of the cell. The spindle fibers attach to the centromere of each chromosome. The homologous chromosomes then move towards the opposite poles of the cell. After anaphase I, the cell moves to next stage of telophase I. Cytokinesis occurs during this period during which cell divides to form two daughter cells with haploid nuclei. Hence this is the correct option.

Explanation for incorrect answer:

Option A. change to diploid- After anaphase I, cell divides to form two daughter cells with haploid number of chromosomes. They do not change to diploid. Hence this is not the correct option.

Option B. crossing over- After anaphase I, cell divides to form two daughter cells with haploid number of chromosomes. They do not undergo crossing over. Hence this is not the correct option.

Option D. DNA replication- After anaphase I, cell divides to form two daughter cells with haploid number of chromosomes. They do not undergo DNA replication. Hence this is not the correct option.