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Cell Cycle and Cell Division - Meiosis and its significance

Grade 11CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Meiosis is a specialized cell division that reduces the chromosome number by half, producing four haploid (nn) daughter cells from a single diploid (2n2n) parent cell.

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The process consists of two sequential cycles of nuclear and cell division: Meiosis I (reductional division) and Meiosis II (equational division), but only a single cycle of DNA replication occurs during the SS phase.

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Prophase I is uniquely long and complex, divided into five substages: Leptotene (chromatin condensation), Zygotene (synapsis and formation of the synaptonemal complex), Pachytene (crossing over between non-sister chromatids of homologous chromosomes via the enzyme recombinase), Diplotene (dissolution of synaptonemal complex and appearance of X-shaped Chiasmata), and Diakinesis (terminalisation of chiasmata).

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During Anaphase I, homologous chromosomes separate and move towards opposite poles, while sister chromatids remains associated at their centromeres. This is where the actual reduction of chromosome number occurs (2n→n2n \rightarrow n).

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Meiosis II is similar to Mitosis, where sister chromatids are separated during Anaphase II, resulting in four haploid cells with a DNA content of 1C1C.

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Significance: Meiosis ensures the maintenance of a constant chromosome number across generations in sexually reproducing organisms and introduces genetic variations through crossing over and independent assortment, which are crucial for evolution.

📐Formulae

Chromosome number: 2n→Meiosis In→Meiosis IIn\text{Chromosome number: } 2n \xrightarrow{\text{Meiosis I}} n \xrightarrow{\text{Meiosis II}} n

DNA Content: 2C→S-phase4C→Meiosis I2C→Meiosis II1C\text{DNA Content: } 2C \xrightarrow{\text{S-phase}} 4C \xrightarrow{\text{Meiosis I}} 2C \xrightarrow{\text{Meiosis II}} 1C

Number of recombinations possible∝Frequency of crossing over\text{Number of recombinations possible} \propto \text{Frequency of crossing over}

💡Examples

Problem 1:

An onion root tip cell has 2n=162n = 16 chromosomes. If a cell from the same plant undergoes meiosis, how many chromosomes and what will be the DNA content (CC) at the end of Telophase I and Telophase II, assuming the initial DNA content in G1G_1 is 2C2C?

Solution:

At the end of Telophase I: Chromosomes = 88, DNA content = 2C2C. At the end of Telophase II: Chromosomes = 88, DNA content = 1C1C.

Explanation:

In Meiosis I, the chromosome number is halved from 2n2n (1616) to nn (88). Although the DNA was 4C4C after the SS phase, the first division reduces it back to 2C2C. Meiosis II is equational for chromosome number (8→88 \rightarrow 8) but reduces the DNA content from 2C2C to 1C1C because sister chromatids separate.

Problem 2:

During which specific stage of Prophase I do bivalents or tetrads clearly appear, and at which stage does the enzyme recombinase act?

Solution:

Bivalents/Tetrads are clearly visible in the Pachytene stage (though they begin forming in Zygotene). Recombinase acts during the Pachytene stage.

Explanation:

Synapsis occurs in Zygotene to form bivalents, but they become more distinct as tetrads in Pachytene. Pachytene is also the stage where crossing over (exchange of genetic material) is catalyzed by the enzyme complex known as recombinase.

Meiosis and its significance - Revision Notes & Key Diagrams | CBSE Class 11 Biology