Explanation
$$\textbf{Correct option:A}$$
$\textbf{Solution:}$$
$$\textbf{Option (A)}$$
$$\bullet$$Leptotene stage is the first step of meiotic division-I,which is also known as bouquet stage ,chromosome become gradually visible ,nuclear membrane initiates to disappear.
$$\textbf{Option (B)}$$
$$\bullet$$ Synapse occurs in the zygotene phase of meiosis I where the homologous chromosomes get attached to each other by means of a nucleoprotein known as synaptonemal protein ,this protein is present in between the chromatids also .The linkage is such that all the genes expressing the same characters falls against each other.
$$\textbf{Option (C)}$$
$$\bullet$$ Terminalisation is the phenomenon of chiasmata shifting towards the ends of the chromosomes that takes place in the diakinesis stage of meiosis I, nucleolus, nuclear envelope finally degenerates.
$$\textbf{Option (D)}$$
$$\bullet$$ Crossing over is the process of exchange of genetic material between two non sister chromatids that takes place in the pachytene stage.
$$\bullet$$ At the pachytene stage, the two homologous chromosomes start condensing and shorten in size such that all the chromatids present in each chromosome become more distinct and these look like four chromatids and it is known as tetrad.
$$\textbf{Hence, the correct option is (A)}$$
Correct Option: C
Explanation:
Hence, if the number of bivalents is 8 in metaphase I, the number of chromosomes in daughter cells after meiosis I and meiosis II respectively will be 8 and 8.
Please disable the adBlock and continue. Thank you.