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BiologyCBSE Class 12 · Chapter 1

Sexual Reproduction in Flowering Plants Class 12: Concepts, Diagrams and Board Answers

Updated 14 October 2026 · Fact-checked

Sexual Reproduction in Flowering Plants is Chapter 1 of CBSE Class 12 Biology and part of Unit VI, Reproduction (16 marks). It covers the stamen and pistil, development of pollen and embryo sac, pollination and outbreeding devices, double fertilisation, endosperm, embryo, seed and fruit, and apomixis and polyembryony.

What Sexual Reproduction in Flowering Plants is about

Flowers are the reproductive organs of angiosperms. This chapter follows a flower from bud to seed.

Pre-fertilisation: the stamen has an anther with four microsporangia, whose wall layers (epidermis, endothecium, middle layers, tapetum) protect and nourish developing pollen. Microsporogenesis (meiosis in pollen mother cells) gives tetrads of microspores that become pollen grains with a tough sporopollenin exine. The pistil has stigma, style and ovary; the ovule contains the nucellus, and megasporogenesis gives a functional megaspore that develops into the 7-celled, 8-nucleate embryo sac (Polygonum type).

Pollination: autogamy (including cleistogamy), geitonogamy and xenogamy; agents (wind, water, insects and other animals); outbreeding devices that discourage self-pollination; pollen–pistil interaction; and artificial hybridisation by emasculation and bagging.

Double fertilisation: one male gamete fuses with the egg (syngamy, forming the diploid zygote) and the other with the two polar nuclei (triple fusion, forming the triploid primary endosperm nucleus).

Post-fertilisation: development of endosperm (usually free-nuclear first, as in coconut water), embryo (dicot and monocot), seed (endospermic and non-endospermic, perisperm, dormancy) and fruit (true, false and parthenocarpic fruits).

Special modes: apomixis (seeds without fertilisation) and polyembryony.

Board questions include labelled diagrams (anther wall, embryo sac, ovule, embryos), ploidy and counting questions, pollination types and devices, double fertilisation, and seed and fruit reasoning.

Weightage: This chapter is part of Unit VI, Reproduction, which carries 16 of the 70 theory marks together with Human Reproduction and Reproductive Health.Source: CBSE Biology (044) syllabus 2026-27, course structure, Unit VI Reproduction

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9.3 marks per paper. See the questions

Key concepts

Stamen, microsporangium and microsporogenesis
NCERT section 1.2.1 Stamen, Microsporangium and Pollen Grain

A stamen has a long, slender filament and a terminal anther. A typical angiosperm anther is bilobed, each lobe with two thecae (dithecous), so a transverse section shows a four-sided structure with four microsporangia (pollen sacs).

Wall of the microsporangium (outside to inside): epidermis, endothecium, middle layers and tapetum. The outer three protect the anther and help it split (dehisce) to release pollen. The tapetum nourishes the developing pollen grains; its cells have dense cytoplasm and are often multinucleate.

In a young anther, the centre of each microsporangium has the sporogenous tissue. Each of its cells is a potential pollen mother cell (PMC) or microspore mother cell, which undergoes meiosis to form a microspore tetrad (four haploid cells). This process is microsporogenesis. When the anther matures and dries, the microspores separate and develop into pollen grains.

Worked example

Name the four wall layers of a microsporangium from outside to inside and state the function of the innermost layer. How many meiotic divisions are needed to produce 100 pollen grains? (2 marks)

Try it first, then show the solutionSolution
  1. Epidermis, endothecium, middle layers, tapetum; the tapetum nourishes the developing pollen grains. (1)
  2. Each meiosis of a pollen mother cell gives a tetrad of 4 microspores, each becoming one pollen grain.
  3. 100 ÷ 4 = 25 meiotic divisions. (1)

Answer: Epidermis, endothecium, middle layers, tapetum (nourishes pollen); 25 meiotic divisions.

The pollen grain (male gametophyte)
NCERT section 1.2.1 Stamen, Microsporangium and Pollen Grain (The pollen grain)

Pollen grains are generally spherical, about 25–50 micrometres across, with a two-layered wall.

  • The hard outer exine is made of sporopollenin, one of the most resistant organic materials: it withstands high temperatures and strong acids and alkalis, and no enzyme that degrades it is known. This is why pollen is well preserved as fossils. The exine has prominent apertures called germ pores, where sporopollenin is absent; the pollen tube emerges through them.
  • The inner intine is thin and continuous, made of cellulose and pectin.

A mature pollen grain contains two cells: the large vegetative cell (abundant food reserve, large irregular nucleus) and the small, spindle-shaped generative cell, which floats in the vegetative cell's cytoplasm. In over 60% of angiosperms, pollen is shed at this 2-celled stage. In the rest, the generative cell divides mitotically to give two male gametes before shedding (3-celled stage).

Pollen and people: pollen of some plants (such as Parthenium or carrot grass) causes allergies and respiratory problems; pollen is also eaten as a food supplement. Viability varies: rice and wheat pollen lose viability within 30 minutes of release, while some members of Rosaceae, Leguminosae and Solanaceae stay viable for months. Pollen can be stored for years in liquid nitrogen (−196 °C) in pollen banks, for crop breeding.

Worked example

Why are pollen grains well preserved as fossils? What is the role of germ pores? At what stage are pollen grains shed in most angiosperms? (3 marks)

Try it first, then show the solutionSolution
  1. The exine is made of sporopollenin, which resists high temperature, strong acids, alkalis and enzymes, so pollen does not decay. (1)
  2. Germ pores are apertures in the exine where sporopollenin is absent; the pollen tube comes out through them. (1)
  3. In over 60% of angiosperms, pollen is shed at the 2-celled stage (vegetative cell + generative cell). (1)

Answer: Sporopollenin in the exine; germ pores allow the pollen tube to emerge; 2-celled stage.

Pistil, ovule, megasporogenesis and the embryo sac
NCERT section 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac

The gynoecium may have one pistil (monocarpellary) or more (multicarpellary); multiple pistils may be fused (syncarpous, as in Papaver) or free (apocarpous, as in Michelia). Each pistil has a stigma (landing platform for pollen), a style and a basal ovary, whose cavity (locule) contains ovules on the placenta. The number of ovules ranges from one (wheat, paddy, mango) to many (papaya, watermelon, orchids).

The ovule (megasporangium) is attached to the placenta by the funicle; the point of attachment is the hilum. It has one or two protective integuments, which leave a small opening, the micropyle; opposite it is the chalaza, the basal part. Inside the integuments is the nucellus, and within it the embryo sac (female gametophyte). An ovule generally has one embryo sac.

Megasporogenesis: a single megaspore mother cell (MMC) near the micropylar end of the nucellus divides by meiosis to give four megaspores. In most flowering plants, one megaspore is functional and three degenerate. This method of forming the embryo sac from a single megaspore is monosporic development.

Embryo sac development (female gametophyte): the functional megaspore nucleus divides mitotically three times (free-nuclear divisions) to give 2, 4 and then 8 nuclei. Cell walls then form, giving the typical mature embryo sac, 8-nucleate but 7-celled:

  • Micropylar end: the egg apparatus: two synergids and one egg cell. The synergids have cellular thickenings at the micropylar tip called the filiform apparatus, which guides the pollen tube.
  • Chalazal end: three antipodal cells.
  • Centre: a large central cell with two polar nuclei.

Worked example

A mature embryo sac is described as '8-nucleate and 7-celled'. Explain this, and state how many mitotic divisions form it from the functional megaspore. (3 marks)

Try it first, then show the solutionSolution
  1. The functional megaspore nucleus divides mitotically three times: 1 → 2 → 4 → 8 nuclei. (1)
  2. Cells form around six nuclei: three at the micropylar end (egg + two synergids) and three at the chalazal end (antipodals). (1)
  3. The remaining two polar nuclei share one large central cell, so there are 8 nuclei but 7 cells. (1)

Answer: Three free-nuclear mitoses give 8 nuclei; 6 single-nucleus cells plus 1 binucleate central cell make 7 cells.

Pollination and its agents
NCERT section 1.2.3 Pollination (kinds and agents of pollination)

Pollination is the transfer of pollen from the anther to the stigma of a pistil.

Kinds by source of pollen:

  • Autogamy: pollen goes to the stigma of the same flower. It needs the anther and stigma to be close and to mature at the same time. Some plants, such as Viola (common pansy), Oxalis and Commelina, have two kinds of flowers: chasmogamous flowers (open, with exposed anthers and stigma) and cleistogamous flowers (never open). Cleistogamous flowers are invariably autogamous and produce assured seed set even without pollinators.
  • Geitonogamy: pollen goes to the stigma of another flower of the same plant. It is functionally cross-pollination (needs an agent) but genetically like autogamy, because the pollen is from the same plant.
  • Xenogamy: pollen goes to the stigma of a different plant. This is the only type that brings genetically different pollen to the stigma.

Agents:

  • Abiotic: wind (most common; pollen is light and non-sticky, stamens well exposed, stigmas large and feathery, flowers often with a single ovule, as in grasses and corn cob tassels) and water (rare, mostly monocots: Vallisneria, Hydrilla, and the marine sea-grass Zostera). In Vallisneria, female flowers reach the water surface on long stalks and male flowers or pollen are released onto the surface. Water-pollinated flowers lack nectar and colour; their pollen is protected by mucilage.
  • Biotic: mostly insects (especially bees), also birds, bats and other animals. Insect-pollinated flowers are large, colourful, fragrant and rich in nectar; small flowers are grouped in an inflorescence. Pollen and nectar are the usual floral rewards. Some plants give a safe place to lay eggs: the tallest flower, Amorphophallus (about 6 feet), and the Yucca moth, which cannot complete its life cycle without the plant, and vice versa.
  • Pollen robbers consume pollen or nectar without pollinating.

Worked example

Distinguish between geitonogamy and xenogamy. Why is geitonogamy described as 'functionally cross-pollination but genetically similar to autogamy'? (2 marks)

Try it first, then show the solutionSolution
  1. Geitonogamy: pollen from one flower to the stigma of another flower on the same plant; xenogamy: pollen to the stigma of a flower on a different plant. (1)
  2. In geitonogamy an agent carries pollen between flowers, as in cross-pollination, but the pollen comes from the same plant, so it is genetically the same as self-pollination. (1)

Answer: Geitonogamy: same plant, different flower; xenogamy: different plant. Geitonogamy needs an agent (functional cross) but the pollen is genetically identical (like selfing).

Outbreeding devices, pollen–pistil interaction and artificial hybridisation
NCERT section 1.2.3 Pollination (Outbreeding devices; Pollen–pistil interaction; Artificial hybridisation)

Continued self-pollination causes inbreeding depression. Flowering plants have several outbreeding devices that discourage self-pollination and encourage cross-pollination:
1. Pollen release and stigma receptivity are not synchronised (one matures before the other).
2. The anther and stigma are placed at different positions, so pollen cannot reach the stigma of the same flower.
3. Self-incompatibility: a genetic mechanism that prevents self-pollen from fertilising ovules by inhibiting pollen germination or pollen tube growth in the pistil.
4. Unisexual flowers: if male and female flowers are on the same plant (monoecious, such as castor and maize), autogamy is prevented but not geitonogamy; if they are on different plants (dioecious, such as papaya), both autogamy and geitonogamy are prevented.

Pollen–pistil interaction: the pistil recognises compatible pollen and rejects incompatible pollen (of another species or self-pollen in self-incompatible plants), through chemical interaction between pollen and pistil. A compatible pollen grain germinates on the stigma and sends a pollen tube through a germ pore; the tube grows through the stigma and style to the ovary, enters the ovule through the micropyle and then enters one synergid through the filiform apparatus. All these events, from pollen landing to tube entry, form pollen–pistil interaction.

Artificial hybridisation (used to combine desired characters): only the desired pollen must reach the stigma.

  • Emasculation: removing the anthers from a bisexual flower bud before the anther dehisces (with forceps).
  • Bagging: covering emasculated flowers with a bag of butter paper to prevent contamination by unwanted pollen. When the stigma becomes receptive, desired pollen is dusted on it and the flower is rebagged.
  • In unisexual flowers, emasculation is not needed; the female flower buds are bagged before they open.

Worked example

Explain how a monoecious plant like castor differs from a dioecious plant like papaya in preventing self-pollination. (2 marks)

Try it first, then show the solutionSolution
  1. Castor (monoecious) has separate male and female flowers on the same plant: autogamy is impossible, but pollen can still reach another flower of the same plant (geitonogamy). (1)
  2. Papaya (dioecious) has male and female flowers on different plants, so both autogamy and geitonogamy are prevented. (1)

Answer: Monoecy prevents autogamy only; dioecy prevents both autogamy and geitonogamy.

Double fertilisation
NCERT section 1.3 Double Fertilisation

After entering a synergid, the pollen tube releases two male gametes into the synergid's cytoplasm.

  • One male gamete moves towards the egg cell and fuses with its nucleus: syngamy, giving a diploid zygote (2n).
  • The other male gamete moves to the two polar nuclei in the central cell and fuses with them, giving a triploid primary endosperm nucleus (PEN, 3n). Because three haploid nuclei fuse, this is called triple fusion.

Since two kinds of fusion (syngamy and triple fusion) take place in one embryo sac, the event is called double fertilisation, a feature unique to flowering plants. The central cell becomes the primary endosperm cell (PEC), which develops into the endosperm, while the zygote develops into the embryo.

Ploidy checklist: nucellus, integuments, megaspore mother cell, zygote and embryo: 2n; megaspores, egg, synergids, antipodals, male gametes, pollen nuclei: n; primary endosperm nucleus and endosperm: 3n.

Worked example

The leaf cells of a flowering plant have 24 chromosomes. How many chromosomes are present in (a) a synergid, (b) the primary endosperm nucleus, (c) the zygote, (d) a nucellus cell? (2 marks)

Try it first, then show the solutionSolution
  1. Leaf cells are diploid: 2n = 24, so n = 12.
  2. (a) Synergid is haploid: 12. (½)
  3. (b) PEN is triploid (two polar nuclei + one male gamete): 3n = 36. (½)
  4. (c) Zygote is diploid: 24. (½)
  5. (d) Nucellus is maternal diploid tissue: 24. (½)

Answer: (a) 12 (b) 36 (c) 24 (d) 24

Post-fertilisation events: endosperm, embryo, seed and fruit
NCERT section 1.4 Post-fertilisation: Structures and Events (1.4.1 Endosperm; 1.4.2 Embryo; 1.4.3 Seed)

Endosperm develops before the embryo, because the embryo needs nutrition from it. The PEN usually divides repeatedly without cell walls (free-nuclear endosperm); cell walls form later (cellular endosperm). The water of a tender coconut is free-nuclear endosperm (thousands of nuclei), and the white kernel around it is cellular endosperm. The endosperm may be used up completely before seed maturity (pea, groundnut, beans) or may persist in the mature seed (castor, coconut) to be used during germination.

Embryo: develops at the micropylar end from the zygote, after some endosperm has formed. Stages: proembryo, globular, heart-shaped and mature embryo.

  • Dicot embryo: an embryonal axis with two cotyledons. The part above the cotyledons is the epicotyl, ending in the plumule (shoot tip); the part below is the hypocotyl, ending in the radicle (root tip), covered by a root cap.
  • Monocot embryo (grass family): one cotyledon, the scutellum, on one side of the axis. The radicle and root cap are enclosed in a sheath, the coleorhiza; the shoot apex and a few leaf primordia are enclosed in a hollow foliar structure, the coleoptile.

Seed: the fertilised ovule, with seed coat(s), cotyledon(s) and embryo axis. Seeds are non-albuminous (no residual endosperm: pea, groundnut) or albuminous (endosperm remains: wheat, maize, barley, castor). In a few seeds such as black pepper and beet, remnants of the nucellus persist as perisperm. Integuments harden into the seed coat; the micropyle remains as a small pore that admits oxygen and water during germination. The seed loses water (moisture 10–15%), its metabolism slows and it may become dormant.

Fruit: the ovary wall becomes the pericarp. Fruits may be fleshy (guava, orange, mango) or dry (groundnut, mustard). True fruits develop only from the ovary; in false fruits (apple, strawberry, cashew) the thalamus also contributes. Parthenocarpic fruits (banana) develop without fertilisation and are seedless; parthenocarpy can be induced by growth hormones.

Significance of seeds: pollination and fertilisation are independent of water, so seed formation is more dependable; seeds have adaptations for dispersal to new habitats; they store food for the seedling; the hard coat protects the embryo; they carry new genetic combinations from sexual reproduction; and they are the basis of agriculture. Seeds can stay viable for long periods: a lupine (Lupinus arcticus) germinated after an estimated 10,000 years, and a 2000-year-old date palm (Phoenix dactylifera) seed was found viable.

Worked example

(a) Why does endosperm develop before the embryo? (b) What is perisperm? Give an example. (c) Why is an apple called a false fruit? (3 marks)

Try it first, then show the solutionSolution
  1. (a) The developing embryo depends on the endosperm for nutrition, so the endosperm must form first. (1)
  2. (b) Perisperm is the persistent remnant of the nucellus in the seed, as in black pepper and beet. (1)
  3. (c) In apple, the thalamus also contributes to the fruit, not only the ovary. (1)

Answer: (a) Embryo needs nutrition from endosperm (b) persistent nucellus; black pepper, beet (c) thalamus contributes to the fruit.

Apomixis and polyembryony
NCERT section 1.5 Apomixis and Polyembryony

Apomixis is the formation of seeds without fertilisation; it is a form of asexual reproduction that mimics sexual reproduction. It occurs in some species of Asteraceae and grasses.

Ways it happens:

  • A diploid egg cell forms without reduction division and develops into an embryo without fertilisation.
  • More often (as in many Citrus and mango varieties), some cells of the nucellus around the embryo sac divide, protrude into the embryo sac and develop into embryos.

In such species each ovule may contain many embryos. The occurrence of more than one embryo in a seed is polyembryony (an orange seed squeezed open shows several embryos of different sizes).

Importance for agriculture: hybrid seeds must be produced fresh every year, because in their progeny the hybrid characters segregate and are lost. Producing hybrid seed every year is expensive for farmers. If hybrids are made apomictic, the characters do not segregate, and farmers can keep using the hybrid seed year after year. Scientists are trying to identify the genes for apomixis and transfer them into hybrid varieties.

Note the difference: parthenocarpy gives a fruit without fertilisation (seedless), while apomixis gives a seed without fertilisation.

Worked example

Why would making hybrid crops apomictic benefit farmers? (2 marks)

Try it first, then show the solutionSolution
  1. Seeds of a hybrid, if produced sexually, give offspring in which hybrid characters segregate, so farmers must buy new hybrid seed every year, which is costly. (1)
  2. Apomictic seeds form without fertilisation, so the embryo is genetically identical to the parent and the hybrid characters do not segregate; farmers can reuse their own seed. (1)

Answer: Apomixis prevents segregation of hybrid characters, so the same hybrid seed can be used every year without buying new seed.

Sexual Reproduction in Flowering Plants in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Outbreeding devices, pollen–pistil interaction and artificial hybridisation

Suppose a plant breeder wants to cross a high-yielding wheat variety (female parent) with a disease-resistant variety (male parent).

List the steps she must follow so that only the desired pollen reaches the stigma.

Try it first, then show the solutionSolution
  1. Emasculation: remove the anthers from the bisexual flower buds of the female parent with forceps before the anthers dehisce.
  2. Bagging: cover the emasculated flowers with butter-paper bags to keep out unwanted pollen.
  3. When the stigma becomes receptive, dust it with pollen collected from the disease-resistant variety, then rebag the flowers and let the fruit develop.

Answer: Emasculate, bag, pollinate with the chosen pollen when the stigma is receptive, and rebag.

Memory hook: Cut, cover, dust, cover again.

Post-fertilisation events: endosperm, embryo, seed and fruit

Imagine you drink the water of a tender coconut and later eat the white kernel of a mature coconut.

Explain what the water and the kernel are, in terms of endosperm development.

Try it first, then show the solutionSolution
  1. The primary endosperm nucleus divides many times without forming cell walls: free-nuclear endosperm.
  2. The coconut water of a tender coconut is this free-nuclear endosperm, with thousands of nuclei.
  3. Later, cell walls form from the edges inward, giving the white kernel: cellular endosperm.

Answer: Coconut water is free-nuclear endosperm; the kernel is cellular endosperm.

Memory hook: Coconut: liquid nuclei first, solid cells later.

Apomixis and polyembryony

Suppose you squeeze open an orange seed and find several small embryos of different sizes inside.

Name this condition and explain how the extra embryos arise.

Try it first, then show the solutionSolution
  1. More than one embryo in a seed is called polyembryony.
  2. In many Citrus varieties, some nucellar cells around the embryo sac divide, protrude into it and develop into embryos without fertilisation (apomixis).
  3. So one seed holds the zygotic embryo plus several nucellar embryos.

Answer: Polyembryony: extra embryos develop from nucellar cells (a form of apomixis) in Citrus.

Memory hook: One orange seed, many babies.

Pollination and its agents

Imagine you see a maize field where the top tassels release clouds of pollen, and silky threads hang from the cobs.

Explain why maize is suited to wind pollination.

Try it first, then show the solutionSolution
  1. Wind-pollinated plants produce light, non-sticky pollen in large amounts.
  2. Maize tassels expose the stamens to wind; the silky threads are the long feathery stigmas and styles that trap airborne pollen.
  3. Such flowers often have one ovule each, and in maize the flowers are packed into an inflorescence (the cob).

Answer: Light, abundant pollen from exposed tassels and large feathery stigmas (silks) suit wind pollination.

Memory hook: Light pollen, big feathery stigma: wind's flower.

Topics in this chapter

Sexual Reproduction in Flowering Plants: key facts

Ploidy of structures
PEN = 2 polar nuclei (n + n) + 1 male gamete (n).
Meiosis counting
x pollen need x/4 meioses; x embryo sacs need x meioses.
Embryo sac formation
Polygonum type: monosporic, 8-nucleate, 7-celled.
Double fertilisation
Syngamy plus triple fusion.
Pollination types
Only xenogamy brings genetically different pollen.
Anther wall layers
Outside to inside; tapetum nourishes pollen.

Sexual Reproduction in Flowering Plants: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Transverse section of a mature anther

Draw a labelled T.S. of a mature anther / microsporangium.

Draw

  1. A four-lobed (butterfly-shaped) outline with two lobes on each side of a central connective containing a vascular bundle.
  2. In each lobe, a circular microsporangium filled with pollen grains.
  3. Around one microsporangium, show the wall layers: outer epidermis, endothecium, middle layers and inner tapetum.

Label

  • Epidermis
  • Endothecium
  • Middle layers
  • Tapetum
  • Microsporangium (pollen sac)
  • Pollen grains
  • Connective
  • Vascular bundle

Watch out: Draw the wall layers enlarged on one microsporangium only; label in order from outside to inside.

Anatropous ovule

Draw a labelled diagram of a typical anatropous ovule.

Draw

  1. An inverted ovule attached to the placenta by a funicle, with the micropyle facing the placenta.
  2. Two integuments (outer and inner) surrounding the nucellus, leaving a gap at the micropyle.
  3. The embryo sac inside the nucellus; the chalaza at the opposite end; the hilum at the junction with the funicle.

Label

  • Funicle
  • Hilum
  • Micropyle
  • Outer integument
  • Inner integument
  • Nucellus
  • Embryo sac
  • Chalaza

Watch out: In an anatropous ovule the body is turned so the micropyle lies close to the funicle.

Mature embryo sac

Draw a labelled mature (7-celled, 8-nucleate) embryo sac.

Draw

  1. An oval sac with the micropylar end at the bottom.
  2. At the micropylar end, the egg cell flanked by two synergids with filiform apparatus at their tips.
  3. At the chalazal end, three antipodal cells.
  4. A large central cell with two polar nuclei in the middle.

Label

  • Micropylar end
  • Filiform apparatus
  • Synergids
  • Egg cell
  • Central cell
  • Polar nuclei
  • Antipodals
  • Chalazal end

Watch out: The egg apparatus is always at the micropylar end; antipodals at the chalazal end.

Dicot and monocot (grass) embryos

Draw a labelled L.S. of a dicot embryo or of a grass embryo.

Draw

  1. Dicot: an embryonal axis with two cotyledons attached; plumule at the top (epicotyl), radicle at the bottom with a root cap (hypocotyl).
  2. Monocot: a single shield-shaped scutellum on one side of the axis; the shoot apex and leaf primordia in a coleoptile above; the radicle and root cap in a coleorhiza below; the epiblast where present.

Label

  • Cotyledons
  • Plumule
  • Epicotyl
  • Hypocotyl
  • Radicle
  • Root cap
  • Scutellum
  • Coleoptile
  • Coleorhiza
  • Shoot apex

Watch out: Coleoptile is with the shoot; coleorhiza is with the root (rhiza = root).

How the board asks Sexual Reproduction in Flowering Plants

The question patterns to expect, and the method that works for each.

Question typeHow it is askedMethod
Labelled diagramsDraw and label the T.S. of a mature anther, an anatropous ovule, a mature embryo sac, a pollen grain, or dicot and monocot embryos; or identify labelled parts.Practise each diagram with at least six correct labels. Keep the micropylar and chalazal ends clearly marked.
Ploidy and countingChromosome numbers in endosperm, synergids, nucellus; number of meioses or mitoses needed for a given number of seeds or pollen grains.Find n from the given cell, then apply n, 2n or 3n. For counts: PMC meiosis gives 4 pollen; MMC meiosis gives 1 functional megaspore.
Pollination and outbreedingTypes of pollination, cleistogamy, adaptations for wind, water and insect pollination, outbreeding devices, self-incompatibility.Give the definition plus an NCERT example (Viola, Vallisneria, Yucca, castor, papaya).
Double fertilisation and post-fertilisationExplain double fertilisation; endosperm types; dicot vs monocot embryo; albuminous seeds; true and false fruits; parthenocarpy.Name both fusions and their products; use examples (coconut, castor, pea, apple, banana).
Assertion-reason and case-basedStatements on cleistogamy, sporopollenin, emasculation, apomixis; cases on plant breeding with emasculation and bagging.Check each statement against NCERT wording; in breeding cases, list emasculation, bagging, dusting pollen and rebagging in order.

Tricks to remember Sexual Reproduction in Flowering Plants

Short memory aids. Say each one aloud once, then use it on a question the same day.

  • Every Elephant Must Tiptoe

    Anther wall from outside in: Epidermis, Endothecium, Middle layers, Tapetum.

    Use it when: Labelling the T.S. of an anther.

  • 3-3-2 in the embryo sac

    Micropylar end: 3 cells (egg + 2 synergids); chalazal end: 3 antipodals; middle: 2 polar nuclei in 1 central cell. 8 nuclei, 7 cells.

    Use it when: Drawing and describing the embryo sac.

  • AGX: Auto, Geitono, Xeno

    Same flower, same plant, different plant: the circle widens from A to G to X.

    Use it when: Pollination types.

  • Syngamy makes 2, triple fusion makes 3

    Egg + gamete = 2n zygote; 2 polar + gamete = 3n PEN.

    Use it when: Ploidy questions.

  • Fruit without sex = Parthenocarpy; Seed without sex = Apomixis

    Parthenocarpy: no seeds (banana). Apomixis: seeds but no fertilisation (Citrus, mango).

    Use it when: Telling the two apart.

Sexual Reproduction in Flowering Plants: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Innermost wall layer · 1 mark · MCQBoard level

Which layer of the microsporangium wall nourishes the developing pollen grains?
(a) Endothecium (b) Middle layer (c) Tapetum (d) Epidermis

Try it first, then show the solutionSolution
  1. The innermost layer has dense cytoplasm and is often multinucleate.
  2. It provides nourishment to the developing microspores.

Answer: (c) Tapetum

Example 2: Sporopollenin · 1 mark · Assertion-reasonBoard level

Assertion (A): Pollen grains are well preserved as fossils.
Reason (R): The exine of pollen grains is made of sporopollenin, which resists high temperatures, strong acids and alkalis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution
  1. Both are true.
  2. The resistant exine prevents decay, which is why pollen fossilises well.

Answer: (a)

Example 3: Counting divisions for seeds · 2 marks · Very short answerHarder board level

How many meiotic divisions are needed to produce 100 seeds in a flowering plant? Show your reasoning. (2 marks)

Try it first, then show the solutionSolution
  1. Each seed needs one embryo sac (one egg) and one pollen grain (male gametes). One MMC meiosis gives one functional megaspore, so 100 embryo sacs need 100 meioses. (1)
  2. One PMC meiosis gives 4 pollen grains, so 100 pollen grains need 25 meioses. Total = 100 + 25 = 125. (1)

Answer: 125 meiotic divisions.

Example 4: Chromosome numbers · 2 marks · Very short answerHarder board level

The endosperm of a plant has 27 chromosomes. Find the number of chromosomes in (a) the egg, (b) a root tip cell, (c) the pollen mother cell, (d) an antipodal cell. (2 marks)

Try it first, then show the solutionSolution
  1. Endosperm is 3n = 27, so n = 9 and 2n = 18.
  2. (a) Egg: n = 9; (b) root tip (somatic): 2n = 18. (1)
  3. (c) Pollen mother cell (before meiosis): 2n = 18; (d) antipodal: n = 9. (1)

Answer: (a) 9 (b) 18 (c) 18 (d) 9

Example 5: Outbreeding devices · 3 marks · Short answerBoard level

Explain three devices that flowering plants have evolved to discourage self-pollination. (3 marks)

Try it first, then show the solutionSolution
  1. Pollen release and stigma receptivity are not synchronised, so the stigma is not receptive when the same flower's pollen is shed (or vice versa). (1)
  2. Anther and stigma are placed at different positions, so pollen cannot reach the stigma of the same flower. (1)
  3. Self-incompatibility: the pistil inhibits germination or tube growth of self-pollen (or production of unisexual flowers). (1)

Answer: Non-synchrony, spatial separation of anther and stigma, self-incompatibility (or unisexuality).

Example 6: From pollen landing to two products · 5 marks · Long answerHarder board level

(a) Describe the events from the landing of a compatible pollen grain on the stigma to the entry of male gametes into the embryo sac. (2)
(b) Explain double fertilisation and name the products with their ploidy. (2)
(c) Why does the endosperm develop before the embryo? (1) (5 marks)

Try it first, then show the solutionSolution
  1. (a) The pistil recognises compatible pollen; the pollen grain germinates and a pollen tube emerges through a germ pore and grows through the stigma and style into the ovary. (1)
  2. It enters the ovule through the micropyle and a synergid through the filiform apparatus, releasing two male gametes. (1)
  3. (b) One male gamete fuses with the egg (syngamy): diploid zygote. (1)
  4. The other fuses with two polar nuclei (triple fusion): triploid primary endosperm nucleus; two fusions make double fertilisation. (1)
  5. (c) The embryo needs nourishment from the endosperm, so endosperm forms first. (1)

Answer: Pollen–pistil interaction → tube through micropyle and synergid → syngamy (2n zygote) + triple fusion (3n PEN); endosperm first to feed the embryo.

Example 7: Reasoning about cleistogamy · 3 marks · Short answerHOTS / competency

Plants such as Viola and Commelina produce both chasmogamous and cleistogamous flowers. Explain the advantage to the plant of having both types. (3 marks)

Try it first, then show the solutionSolution
  1. Chasmogamous flowers open, exposing anthers and stigma, so cross-pollination can bring in genetic variation. (1)
  2. Cleistogamous flowers never open; anthers dehisce inside and pollen falls on the stigma, so seed set is assured even without pollinators. (1)
  3. Having both lets the plant gain variation when pollinators are available and still guarantee some seeds when they are not. (1)

Answer: Chasmogamy allows cross-pollination (variation); cleistogamy assures seed set; together they balance variation and reliability.

Example 8: Seed with the wrong ploidy · 1 mark · MCQHOTS / competency

In an apomictic Citrus seed, an embryo develops from a nucellar cell. The leaf cells have 2n = 18. What is the chromosome number of this embryo, and will it be genetically identical to the mother plant?
(a) 9, no (b) 18, yes (c) 27, no (d) 18, no

Try it first, then show the solutionSolution
  1. The nucellus is maternal diploid tissue, so a nucellar embryo has 2n = 18.
  2. It forms without fertilisation, by mitosis, so it is a genetic copy of the mother.

Answer: (b) 18, yes

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Producing hybrid seed

Suppose a seed company wants to produce hybrid tomato seeds every year by crossing two pure lines. Tomato flowers are bisexual. The company's scientists also want to develop a hybrid that farmers could save and resow, without losing its qualities, by making it apomictic.

  1. (i) · 1 mark

    Why must the anthers of the female parent's flowers be removed, and when?

    Answer

    Answer: To prevent self-pollination; before the anthers dehisce (in the bud stage).

    This is emasculation; it ensures only the chosen pollen fertilises the ovules.

  2. (ii) · 1 mark

    Why are the emasculated flowers bagged?

    Answer

    Answer: To prevent contamination of the stigma by unwanted pollen.

    Butter-paper bags keep out pollen carried by wind or insects.

  3. (iii) · 2 marks

    Explain why farmers must normally buy hybrid seed every year, and how apomixis would solve this.

    Answer

    Answer: Hybrid characters segregate in the next generation; apomictic seeds form without fertilisation, so the hybrid characters are kept.

    Seeds from a sexually reproducing hybrid give a mixed, segregating progeny (1). Apomictic embryos are genetically identical to the parent, so saved seed keeps the hybrid traits (1).

Ploidy in the ovule

Suppose a student examines a fertilised ovule of a plant whose root tip cells have 20 chromosomes. She finds a zygote near the micropylar end, a large mass of free nuclei in the central part, and remnants of three degenerated cells at the chalazal end.

  1. (i) · 1 mark

    What is the chromosome number of the zygote?

    Answer

    Answer: 20 (2n).

    Root tip cells are diploid; the zygote is also diploid.

  2. (ii) · 1 mark

    What is the free-nuclear mass, and its chromosome number per nucleus?

    Answer

    Answer: Free-nuclear endosperm; 30 (3n).

    It comes from repeated divisions of the triploid primary endosperm nucleus.

  3. (iii) · 2 marks

    Identify the degenerated cells at the chalazal end and explain why the endosperm forms before the embryo develops much.

    Answer

    Answer: Antipodal cells; the endosperm must nourish the developing embryo.

    The three chalazal cells of the embryo sac are the antipodals (1); the embryo depends on the stored food of the endosperm, so endosperm develops first (1).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Define self-incompatibility. Explain why, in plants showing it, pollen from the same flower fails to produce seeds. (2 marks) (2 marks)

StepMarks
Self-incompatibility is a genetic mechanism that prevents pollen of the same flower or plant from fertilising its ovules.1
The pistil recognises self-pollen and inhibits its germination or pollen tube growth, so fertilisation, and therefore seed formation, does not occur.1

Describe double fertilisation in angiosperms. Mention the products and their ploidy. (3 marks) (3 marks)

StepMarks
The pollen tube enters a synergid through the filiform apparatus and releases two male gametes.1
One male gamete fuses with the egg (syngamy) forming the diploid zygote.1
The other fuses with the two polar nuclei (triple fusion) forming the triploid primary endosperm nucleus; two fusions = double fertilisation.1

(a) Draw a labelled diagram of a mature embryo sac. (b) Describe megasporogenesis and the development of the embryo sac. (c) What is meant by monosporic development? (5 marks) (5 marks)

StepMarks
(a) Diagram: micropylar end with egg and two synergids (filiform apparatus); chalazal end with three antipodals; central cell with two polar nuclei; at least six labels.2
(b) The megaspore mother cell in the nucellus near the micropyle divides by meiosis into four megaspores; three degenerate.1
The functional megaspore nucleus divides mitotically three times (free-nuclear) to give eight nuclei; walls form, giving the 7-celled, 8-nucleate embryo sac.1
(c) Formation of the embryo sac from a single megaspore is monosporic development.1

Where students lose marks in Sexual Reproduction in Flowering Plants

The slips examiners see most often, and what to write instead.

  • Mistake: Calling the embryo sac 8-celled

    Wrong working

    The mature embryo sac has 8 cells: egg, 2 synergids, 3 antipodals and 2 polar cells.

    Right working

    It is 8-nucleate but 7-celled: the two polar nuclei lie in one central cell.

    Do this: Count the central cell once.

    Usually costs: 1 mark.

  • Mistake: Wrong ploidy of endosperm or nucellus

    Do this: Endosperm is 3n (two polar nuclei + one male gamete); nucellus is maternal 2n tissue.

    Usually costs: ½ mark per error.

  • Mistake: Saying monoecy prevents geitonogamy

    Do this: Monoecy (castor, maize) prevents only autogamy; dioecy (papaya) prevents both autogamy and geitonogamy.

    Usually costs: 1 mark.

  • Mistake: Unlabelled micropylar and chalazal ends in diagrams

    Do this: Always mark the micropyle and chalaza; the egg apparatus must be drawn at the micropylar end.

    Usually costs: ½ to 1 mark of diagram marks.

  • Mistake: Confusing apomixis with parthenocarpy

    Do this: Apomixis gives seeds without fertilisation; parthenocarpy gives fruits without fertilisation (and without seeds).

    Usually costs: 1 mark.

NCERT exercise guide: Sexual Reproduction in Flowering Plants

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises on the stamen, pollen and pistil

What it tests: Parts of a flower, microsporangium wall, microsporogenesis, pollen wall, megasporogenesis, embryo sac, ovule parts, ploidy.

How to approach it: Answer in terms of the cell division involved (meiosis or mitosis) and the ploidy; draw where asked.

Meiosis versus mitosis in gametophyte formation

Which type of cell division occurs during (a) microsporogenesis and (b) formation of the 8-nucleate embryo sac? Give the ploidy of the products.

Show the solutionSolution
  1. (a) Microsporogenesis: the diploid pollen mother cell divides by meiosis; the microspores are haploid.
  2. (b) The haploid functional megaspore divides by mitosis three times; all eight nuclei are haploid.

Answer: (a) Meiosis, haploid microspores (b) mitosis, haploid nuclei.

Exercises on pollination and fertilisation

What it tests: Autogamy versus geitonogamy, cleistogamy, outbreeding devices, self-incompatibility, emasculation and bagging, double fertilisation.

How to approach it: Define each term and give an NCERT example; for hybridisation, list the steps in order.

Advantages and disadvantages of cleistogamy

Give one advantage and one disadvantage of cleistogamous flowers to the plant.

Show the solutionSolution
  1. Advantage: seed set is assured even without pollinators, because pollen is shed directly on the stigma in the closed flower.
  2. Disadvantage: only self-pollination occurs, so there is no genetic variation and inbreeding depression may result.

Answer: Assured seed set; but no cross-pollination, so variation is lost.

Exercises on post-fertilisation events and special modes

What it tests: Endosperm, embryo, seed and fruit; perisperm; false fruits; parthenocarpy; apomixis and polyembryony.

How to approach it: Trace each structure back to the part of the flower or ovule it came from.

Origin of fruit and seed parts

From which part of the flower or ovule does each develop: (a) seed coat, (b) pericarp, (c) perisperm, (d) endosperm?

Show the solutionSolution
  1. (a) Integuments.
  2. (b) Ovary wall.
  3. (c) Remnants of the nucellus.
  4. (d) Primary endosperm cell (from triple fusion).

Answer: (a) integuments (b) ovary wall (c) nucellus (d) primary endosperm cell

Sexual Reproduction in Flowering Plants: previous year questions

  • Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9.3 marks per paper; 31 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Sexual Reproduction in Flowering Plants

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 57/1/1, Q1 · 1 mark · MCQ

Identify the plant in which emasculation is not required for artificial hybridization.

  1. (A)Rice
  2. (B)Wheat
  3. (C)Pea
  4. (D)Papaya
Try it first, then show our solutionSolution
  1. Emasculation (removal of anthers) is needed only in bisexual flowers. Papaya is dioecious, so its female flowers are unisexual and carry no anthers; only bagging is needed. Rice, wheat and pea are bisexual.

    1 mark

Answer: (D) Papaya

Where the marks are

  • Identifies papaya, a unisexual/dioecious plant (1 mark)

See it in the full 2026 board paper, set 57/1/1

2026, set 57/1/1, Q18(a) · 2 marks · Very short answerCompetency-based

Answer the following.

  1. (i)

    How does a farmer use dormancy of seed to his advantage ?

    1 mark
  2. (ii)

    Differentiate between pea seed and castor seed.

    1 mark
Try it first, then show our solutionSolution
  1. (i) Dormant seeds are dehydrated and metabolically inactive, so they can be safely stored for long periods. The farmer can therefore keep them as seed to raise the crop in the next season, and also use them as food or sell them commercially.

    1 mark
  2. (ii) Pea seed is non-endospermic (non-albuminous): the endosperm is completely used up during embryo development, so none is left in the seed. Castor seed is endospermic (albuminous): part of the endosperm remains, because it is not fully consumed during embryo development.

    1 mark

Answer: (i) Dormant seeds can be stored and sown next season, or used as food or for trade; (ii) pea is non-albuminous (no residual endosperm), castor is albuminous (residual endosperm present)

Where the marks are

  • States that dormancy allows storage of seeds for the next season's crop, or use as food or for commerce (1 mark)
  • Correct difference: pea is non-endospermic/non-albuminous, castor is endospermic/albuminous (1 mark)

See it in the full 2026 board paper, set 57/1/1

2026, set 57/1/1, Q22 · 3 marks · Short answer

Answer the following.

  1. (a)

    Do all pollen grains remain viable for the same length of time ? Support your answer with two suitable examples. (2½)

    (The question carries 3 marks together with part (b) (½) )

    0 marks
  2. (b)

    How are pollen grains stored in pollen banks ? (½)

    0 marks
Try it first, then show our solutionSolution
  1. (a) No, pollen viability differs from species to species.

    ½ mark
  2. Example 1: in rice and wheat (cereals), pollen grains lose viability within about 30 minutes of their release.

    1 mark
  3. Example 2: in some members of Rosaceae, Leguminosae and Solanaceae, pollen grains stay viable for months.

    1 mark
  4. (b) In pollen banks, pollen grains are stored by cryopreservation in liquid nitrogen at −196 °C, so they remain viable for years for crop-breeding programmes.

    ½ mark

Answer: (a) No: rice/wheat pollen stays viable about 30 min, while pollen of Rosaceae, Leguminosae and Solanaceae stays viable for months; (b) cryopreservation in liquid nitrogen at −196 °C

Where the marks are

  • States 'No' (½ mark)
  • Rice/wheat pollen viable only about 30 minutes (1 mark)
  • Rosaceae/Leguminosae/Solanaceae pollen viable for months (1 mark)
  • Cryopreservation in liquid nitrogen at −196 °C (½ mark)

See it in the full 2026 board paper, set 57/1/1

Every Sexual Reproduction in Flowering Plants question, by year

For 2027

Sample paper 2026-27: Q1 (1 mark, solved), Q17(A) (2 marks, solved), Q17(B) (2 marks, solved), Q23 (3 marks, solved)

2026

Set 57/1/1: Q1 (1 mark, solved), Q18(a) (2 marks, solved), Q22 (3 marks, solved), Q25 (3 marks, solved)

Set 57/5/1: Q1 (1 mark, solved), Q17(a) (2 marks, solved), Q17(b) (2 marks, solved), Q22 (3 marks, solved)

Sample paper 2025-26: Q1 (1 mark, solved), Q2 (1 mark, solved), Q13 (1 mark, solved), Q17(A) (2 marks, solved), Q17(B) (2 marks, solved), Q29 (4 marks, solved)

2025

Set 57/1/1: Q7 (1 mark, solved), Q10 (1 mark, solved), Q32(A) (5 marks, solved)

Set 57/7/1: Q4 (1 mark, solved), Q6 (1 mark, solved), Q30 (4 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2024-25: Q2 (1 mark, solved), Q13 (1 mark, solved), Q22 (3 marks, solved), Q31(B) (5 marks, solved)

2024

Set 57/1/1: Q1 (1 mark, solved), Q17 (2 marks, solved), Q31(a) (5 marks, solved)

Set 57/5/1: Q7 (1 mark, solved), Q8 (1 mark, solved), Q17(a) (2 marks, solved), Q25 (3 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2023-24: Q1 (1 mark, solved), Q2 (1 mark, solved), Q13 (1 mark, solved), Q22 (3 marks, solved)

2023

Set 57/1/1: Q8 (1 mark, solved), Q15 (1 mark, solved), Q18 (2 marks, solved), Q28(b) (3 marks, solved), Q33(a) (5 marks, solved)

Set 57/5/1: Q12 (1 mark, solved), Q25 (3 marks, solved), Q31(a) (5 marks, solved)

Practice questions on Sexual Reproduction in Flowering Plants

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

  1. Q1 · 1 mark · MCQBoard level

    The 7-celled, 8-nucleate embryo sac develops from:
    (a) a microspore (b) the megaspore mother cell directly (c) a functional megaspore (d) the nucellus

    Hint

    Monosporic development.

    Answer

    (c) a functional megaspore

  2. Q2 · 1 mark · Assertion-reasonBoard level

    Assertion (A): Geitonogamy is genetically similar to autogamy.
    Reason (R): In geitonogamy, pollen grains come from the same plant.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Which plant does the pollen come from?

    Answer

    (a)

  3. Q3 · 2 marks · Very short answerBoard level

    Differentiate between a chasmogamous and a cleistogamous flower. Name a plant that bears both. (2 marks)

    Hint

    Open versus closed.

    Answer

    Chasmogamous flowers open and expose their anthers and stigma; cleistogamous flowers never open and are self-pollinated (1). Viola (or Oxalis, Commelina) bears both (1).

  4. Q4 · 2 marks · Very short answerBoard level

    List two adaptations each of wind-pollinated and insect-pollinated flowers. (2 marks)

    Hint

    Light pollen versus colour and nectar.

    Answer

    Wind: light, non-sticky pollen in large amounts; exposed stamens and large feathery stigmas (1). Insect: large, colourful, fragrant flowers rich in nectar; sticky pollen or flowers grouped in inflorescences (1).

  5. Q5 · 3 marks · Short answerHarder board level

    Describe the structure of a dicot embryo and compare it with a monocot (grass) embryo. (3 marks)

    Hint

    Cotyledons, scutellum, coleoptile, coleorhiza.

    Answer

    Dicot: embryonal axis with two cotyledons; epicotyl ending in plumule above them, hypocotyl ending in radicle with root cap below (1½). Grass: one cotyledon (scutellum) on one side; shoot apex and leaf primordia enclosed in the coleoptile; radicle and root cap enclosed in the coleorhiza (1½).

  6. Q6 · 3 marks · Short answerHarder board level

    A plant has 2n = 16. How many chromosomes are present in its (a) microspore, (b) tapetal cell (diploid wall cell), (c) primary endosperm nucleus? How many meiotic divisions produce 400 pollen grains? (3 marks)

    Hint

    n = 8.

    Answer

    (a) 8 (½) (b) 16 (½) (c) 24 (1). 400 ÷ 4 = 100 meiotic divisions (1).

  7. Q7 · 3 marks · Short answerHarder board level

    Explain pollen–pistil interaction. Why is it important? (3 marks)

    Hint

    Recognition, then tube growth.

    Answer

    The pistil recognises whether pollen is compatible (right species, not self-incompatible) through chemical signals between pollen and pistil (1). Compatible pollen germinates; the tube grows through stigma and style, enters the ovule by the micropyle and a synergid by the filiform apparatus; incompatible pollen is inhibited (1). It ensures only suitable pollen achieves fertilisation, and plant breeders can manipulate it to produce desired hybrids (1).

  8. Q8 · 5 marks · Long answerHarder board level

    (a) Describe the development of endosperm and give the fate of the endosperm in pea and castor. (2) (b) Distinguish albuminous and non-albuminous seeds with examples. (1) (c) What is seed dormancy and why is it advantageous? (2) (5 marks)

    Hint

    Free-nuclear first; persists or is used up.

    Answer

    (a) The PEN divides repeatedly without walls (free-nuclear endosperm), later becoming cellular; it is completely used up by the developing embryo in pea but persists in the mature castor seed (2). (b) Albuminous seeds keep endosperm (castor, wheat); non-albuminous do not (pea, groundnut) (1). (c) As the seed matures it loses water (10–15% moisture) and its metabolism slows to an inactive state (dormancy) (1); this lets it survive unfavourable conditions and be dispersed and stored until conditions suit germination (1).

  9. Q9 · 3 marks · Short answerHOTS / competency

    Suppose a farmer notices that his papaya trees never set fruit, though they flower every year. A botanist finds that all the trees are female. Explain the cause and suggest a remedy. (3 marks)

    Hint

    Papaya is dioecious.

    Answer

    Papaya is dioecious: male and female flowers are on different plants (1). With only female trees there is no pollen, so no pollination or fertilisation and no fruit set (1). Plant some male trees among them (or bring pollen from male plants) so that pollinators can carry pollen to the female flowers (1).

  10. Q10 · 2 marks · Very short answerHOTS / competency

    A seedless banana and an apomictic mango seed both form without fertilisation. Explain the difference between them. (2 marks)

    Hint

    Fruit versus seed.

    Answer

    Banana is parthenocarpic: the fruit develops without fertilisation and has no seeds (1). In apomictic mango, a seed with an embryo forms without fertilisation, often from nucellar cells (1).

  11. Q11 · 2 marks · Very short answerHarder board level

    Explain why the Yucca plant and the Yucca moth depend on each other. (2 marks)

    Hint

    Eggs and pollination.

    Answer

    The moth lays its eggs in the locule of the ovary and pollinates the flower while doing so (1). The plant cannot be pollinated without the moth, and the moth's larvae develop on the seeds, so neither can complete its life cycle without the other (1).

  12. Q12 · 1 mark · MCQHarder board level

    Which of the following is a false fruit?
    (a) Mango (b) Strawberry (c) Guava (d) Orange

    Hint

    The thalamus contributes.

    Answer

    (b) Strawberry

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

  1. 1.Which layer of the anther wall nourishes pollen grains?Show answer

    The tapetum.

  2. 2.What is sporopollenin?Show answer

    The highly resistant organic material of the pollen exine; no known enzyme degrades it.

  3. 3.In what percentage of angiosperms is pollen shed at the 2-celled stage?Show answer

    Over 60%.

  4. 4.What is the filiform apparatus and its role?Show answer

    Cellular thickenings at the micropylar tip of synergids that guide the pollen tube into the synergid.

  5. 5.What are cleistogamous flowers?Show answer

    Flowers that never open; they are invariably autogamous and ensure seed set.

  6. 6.Name two water-pollinated plants.Show answer

    Vallisneria and Hydrilla (fresh water), Zostera (marine sea-grass).

  7. 7.What is triple fusion?Show answer

    Fusion of a male gamete with two polar nuclei to form the triploid primary endosperm nucleus.

  8. 8.What are the coleoptile and coleorhiza?Show answer

    Sheaths enclosing the shoot apex (coleoptile) and the radicle and root cap (coleorhiza) of a monocot embryo.

  9. 9.Give two examples of false fruits.Show answer

    Apple and strawberry (also cashew).

  10. 10.What is polyembryony? Give an example.Show answer

    Presence of more than one embryo in a seed, as in Citrus.

Best order to study it

  1. Learn the stamen, anther wall layers, microsporogenesis and the pollen grain, drawing the T.S. of an anther.
  2. Learn the pistil, ovule parts, megasporogenesis and embryo sac development, drawing the ovule and embryo sac.
  3. Study pollination types, agents with examples, outbreeding devices, pollen–pistil interaction and artificial hybridisation.
  4. Learn double fertilisation with ploidy, then endosperm, embryo (dicot and monocot), seed and fruit.
  5. Finish with apomixis and polyembryony and their use in agriculture.

Quick revision checklist

The night before the exam: can you do each of these without looking?

  • Can you draw and label the T.S. of a mature anther?
  • Can you describe the structure of a pollen grain and explain why it fossilises well?
  • Can you draw a typical anatropous ovule and the 7-celled embryo sac?
  • Can you define autogamy, geitonogamy and xenogamy with examples?
  • Can you list four outbreeding devices?
  • Can you describe the steps of artificial hybridisation?
  • Can you explain double fertilisation and give the ploidy of zygote, PEN and endosperm?
  • Can you compare dicot and monocot embryos and explain apomixis and its value?

Sexual Reproduction in Flowering Plants: frequently asked questions

How many marks does this chapter carry in CBSE Class 12 Biology 2026-27?

It is part of Unit VI, Reproduction, which carries 16 of the 70 theory marks together with Human Reproduction and Reproductive Health. In the 12 papers in our question bank (2023-2026 board sets and sample papers), this chapter alone carried between 6 and 12 marks.

Is asexual reproduction part of this chapter?

No. The NCERT chapter 'Reproduction in Organisms', which covered asexual reproduction, was dropped in the rationalisation. The 2026-27 syllabus covers only sexual reproduction in flowering plants, plus apomixis, parthenocarpy and polyembryony.

Which diagrams are most important in this chapter?

The T.S. of a mature anther, an anatropous ovule, the mature embryo sac, the pollen grain, and the dicot and monocot embryos. Practise each with labels.

What is the difference between endospermic and non-endospermic seeds?

Endospermic (albuminous) seeds keep part of the endosperm at maturity, such as wheat, maize, barley and castor. Non-endospermic (non-albuminous) seeds use up the endosperm during development, such as pea and groundnut.

Why is double fertilisation called 'double'?

Two fusions occur in the same embryo sac: one male gamete fuses with the egg (syngamy) and the other with the two polar nuclei (triple fusion).

Based on: CBSE Biology (044) syllabus 2026-27; NCERT Biology, Textbook for Class XII (Reprint 2026-27)