Question:medium

A diploid cell which has 8 chromosomes undergoes meiosis and produces 4 daughter cells. What is the number of chromosomes present in each daughter cell formed at the end of meiosis-I?

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Remember that Meiosis I is reductional (halves the chromosome number) while Meiosis II is equational (maintains the haploid chromosome number).
Updated On: Apr 28, 2026
  • 8 chromosomes
  • 4 chromosomes
  • 16 chromosomes
  • 32 chromosomes
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Problem:
Meiosis is a type of cell division that consists of two consecutive stages: Meiosis-I and Meiosis-II. The primary purpose of Meiosis-I is the segregation of homologous chromosomes, which halves the genetic content.
Step 2: Approach and Formula:
Since Meiosis-I is a reductional division, the chromosome number is reduced to half:
\[ n = \frac{2n}{2} \]
Step 3: Detailed Explanation:
The given parent cell is diploid (\( 2n \)) with 8 chromosomes.
During Meiosis-I, homologous pairs separate. This reductional division creates two daughter cells, each containing half the original number of chromosomes.
Applying the formula:
\[ \text{Number of chromosomes} = \frac{8}{2} = 4 \]
Therefore, at the end of Meiosis-I, each daughter cell will have 4 chromosomes.
Meiosis-II will follow as an equational division, keeping the chromosome number at 4 in the final 4 daughter cells.
Step 4: Final Answer:
The number of chromosomes present is 4 chromosomes.
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