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Sexual Reproduction in Flowering Plants - Flower Structure

Grade 12CBSEBiology

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

🔑Concepts

A flower is the reproductive unit in Angiosperms, often described as a modified shoot. It consists of four whorls: Calyx, Corolla, Androecium, and Gynoecium attached to the thalamus.

The Androecium consists of stamens. A typical stamen has a long filament and a bilobed, dithecous anther. Each lobe contains two microsporangia, making it tetrasporangiate.

The microsporangium wall consists of four layers: the epidermis, endothecium, middle layers, and the Tapetum. The tapetum nourishes the developing pollen grains.

Microsporogenesis: The process of formation of microspores from a Pollen Mother Cell (PMCPMC) through meiosis. Each PMCPMC (2n2n) produces a microspore tetrad (nn).

Pollen Grain: Represents the male gametophyte. It has a two-layered wall: the outer Exine (made of sporopollenin, C90H142O27C_{90}H_{142}O_{27}) and the inner Intine (cellulose and pectin).

The Gynoecium consists of the pistil (stigma, style, and ovary). The megasporangium, or ovule, is attached to the placenta by a funicle. The main body of the ovule is the nucellus (2n2n).

Megasporogenesis: The Megaspore Mother Cell (MMCMMC) undergoes meiosis to form four megaspores. In most plants, three degenerate, and one functional megaspore develops into the female gametophyte (monosporic development).

The mature Embryo Sac is typically 77-celled and 88-nucleate. It contains the egg apparatus (11 egg cell + 22 synergids), 33 antipodals, and 11 central cell with 22 polar nuclei.

📐Formulae

Number of Meiosis for n Microspores=n4\text{Number of Meiosis for } 'n' \text{ Microspores} = \frac{n}{4}

Number of Meiosis for n Megaspores/Seeds=n\text{Number of Meiosis for } 'n' \text{ Megaspores/Seeds} = n

Total Meiosis for n Seeds=n+n4=5n4\text{Total Meiosis for } 'n' \text{ Seeds} = n + \frac{n}{4} = \frac{5n}{4}

Ploidy: Endosperm=3n, Nucellus=2n, Male Gamete=n\text{Ploidy: Endosperm} = 3n, \text{ Nucellus} = 2n, \text{ Male Gamete} = n

💡Examples

Problem 1:

Calculate the number of meiotic divisions required to produce 400400 seeds in a pea plant.

Solution:

500500 meiotic divisions.

Explanation:

To produce 400400 seeds, we need 400400 pollen grains and 400400 ovules. Meiosis for 400400 pollen grains = 4004=100\frac{400}{4} = 100. Meiosis for 400400 ovules = 400400 (since each meiosis produces only one functional megaspore). Total divisions = 100+400=500100 + 400 = 500.

Problem 2:

If the leaf cell of an angiosperm has 2424 chromosomes, find the chromosome count in the Pollen Grain, Tapetum, and Endosperm.

Solution:

Pollen Grain: 1212; Tapetum: 2424 (or multiple); Endosperm: 3636.

Explanation:

Leaf cell is diploid (2n=242n = 24), so n=12n = 12. 1. Pollen grain is haploid (nn), so it has 1212. 2. Tapetum is a somatic layer of the anther, so it is diploid (2n=242n = 24), though it can become polyploid. 3. Endosperm is triploid (3n=3×12=363n = 3 \times 12 = 36).

Flower Structure - Revision Notes & Key Diagrams | CBSE Class 12 Biology