Question

If the number of bivalent are $8$ in metaphase – I, what shall be the number of chromosomes in daughter cells after meiosis – I and meiosis – II respectively.
A. $8$ and $4$
B. $4$ and $4$
C. $8$ and $8$
D. $16$ and $8$

Answer 1

The first metaphase of meiosis I encompasses the alignment of paired chromosomes that takes place in the first metaphase of meiosis I along the center of a cell. It ensures that $2$ copies of complete chromosomes will be present in the resulting $2$ daughter cells of meiosis I. The Metaphase I is followed by prophase I and is preceded by anaphase I.

Complete answer:
Before the process of meiosis begins, at the S phase of the cell cycle, the DNA of each chromosome is replicated so that it has two identical sister chromatids which are attached at the centromere. The homologous chromosomes in meiosis I pairs with each other and they exchange genetic material in a process known as chromosomal crossover. The homologous chromosomes are then pulled apart into two new separate daughter cells, in which each of the daughter cells contains half the number of chromosomes as that of the parent cell. At the end stage of meiosis, I, the sister chromatids remain attached and may differ from each other if crossing-over has occurred. In the phase of meiosis II, the two cells produced during meiosis I get divided again. At this phase of division, the sister chromatids get detached from one another and get separated into a total of four daughter cells. So, the number of chromosomes decreases in meiosis I and not in meiosis II. So, if the bivalents are $8$ in number at metaphase I, the number of chromosomes in daughter cells after meiosis I and meiosis II will also be $8$ and $8$ respectively.
A single bivalent has $2$ chromosomes. So, $8$bivalents will have $16$ chromosomes. Meiosis I is a reduction division so after meiosis I, there will be $8$ daughter cells. Meiosis II is an equational division so after that there will be the same number of daughter cells i.e., $8$.
$8$ and $4$: Here, after meiosis I, there will be $8$ daughter cells and after meiosis II there will be $8$ daughter cells. In the given option, it is given that after meiosis II there are $4$ daughter cells. Hence option A is wrong.
$4$ and $4$: After meiosis I there will be $8$ daughter cells and after meiosis II there will be $8$ daughter cells. Here, both the daughter cells are given as $4$. Hence option B is wrong.
$8$ and $8$: After the process of meiosis, I and II, there will be $8$ and $8$ daughter cells respectively. Hence option C is the correct answer.
$16$ and $8$: Here the given option says that after Meiosis I there will be $16$ daughter cells. But it is a reduction division hence only $8$ are formed. Hence option D is wrong.

So, Option D. $8$ and $8$ is the correct answer.

Note:
The association of two replicated homologous chromosomes which have exchanged DNA strands in at least one site called the chiasmata, is known as a bivalent. Each of the bivalent consists of a minimum of one chiasma and rarely forms more than three. The less number is due to crossover interference, which is a poorly understood phenomenon that limits the number of resolutions of repair events into crossover.