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Question

If the number of bivalents 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
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B
4 and 4
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C
8 and 8
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D
16 and 8
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Solution

The correct option is C 8 and 8
In meiosis, DNA replication is followed by two rounds of cell division to produce four daughter cells with half the number of chromosomes as the original parent cell. The two meiotic divisions are known as meiosis I and meiosis II. Before meiosis begins, during S phase of the cell cycle, the DNA of each chromosome is replicated so that it consists of two identical sister chromatids attached at a centromere. In meiosis I, homologous chromosomes pair with each other and can exchange genetic material in a process called chromosomal crossover. The homologous chromosomes are then pulled apart into two new separate daughter cells, each containing half the number of chromosomes as the parent cell. At the end of meiosis I, sister chromatids remain attached and may differ from one another if crossing-over occurred. In meiosis II, the two cells produced during meiosis I divide again. During this division, sister chromatids detach from one another and are separated into four total daughter cells. These cells can mature into gametes, spores, pollen, and other reproductive cells. So, the chromosome number is decreased in meiosis I and not in meiosis II. So, If the number of bivalents is 8 in metaphase I, the number of chromosomes in daughter cells after meiosis I and meiosis II respectively are 8 and 8.
So, the correct answer is option C.

Biology

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