CBSE Previous Year Questions Sexual Reproduction in Flowering Plants - Biology Class

CBSE Previous Year Questions 2025

Multiple Choice Questions (MCQs) - 1 Mark

Q.1. The diagram given below shows labelling of four parts of a dicot embryo during its development as P, Q, R and S. 
Choose the option that indicates correct labelling of ‘P’, ‘Q’, ‘R’ and ‘S’ of embryo in different stages of its development. (1 Mark)
 

Answer: (D) Zygote, Suspensor, Cotyledon, Radicle

Solution: The process of embryogeny begins with the diploid Zygote. The zygote develops into the proembryo stage, which is supported by the Suspensor cells. The mature embryo (dicot) consists of an embryonal axis and two Cotyledons. The lower end of the embryonal axis terminates in the Radicle or root tip. Option (D) lists these four major components or structures encountered during the early and mature stages of dicot embryo development. 
(NCERT Reference: 1.4.2 Embryo)

Q.2. How many pollen grains and ovules are likely to be formed in the anther and the ovary of an angiosperm bearing 50 microspore mother cells and 50 megaspore mother cells respectively? (1 Mark)
(A) 100, 25 
(B) 200, 50 
(C) 50, 50 
(D) 200, 100

Answer: (B) 200, 50
Solution: 

1. Pollen Grains (Male Gametophytes): Each Microspore Mother Cell (MMC) or Pollen Mother Cell (PMC) undergoes meiosis to form four haploid microspores. These develop into pollen grains. Therefore, 50 MMCs will produce $50 \times 4 = \mathbf{2 00}$ pollen grains.
2. Ovules/Embryo Sacs: The Megaspore Mother Cell (MMC) differentiates within the ovule. While each MMC yields four megaspores via meiosis, only one functional megaspore typically develops into the embryo sac (female gametophyte) within the ovule (monosporic development). Assuming one MMC per ovule is differentiated, 50 MMCs would lead to the development of $\mathbf{50}$ functional ovules (each containing one embryo sac). 

(NCERT Reference: 1.2.1 Microsporogenesis and 1.2.2 Megasporogenesis)

Q.3. Endosperm is completely consumed by the developing embryo in which of the following: (1 Mark)
(A) Maize and Castor 
(B) Castor and Groundnut 
(C) Maize and Pea 
(D) Pea and Groundnut

Answer: (D) Pea and Groundnut

Solution: Seeds where the endosperm is completely consumed during embryo development are termed non-albuminous or ex-albuminous seeds. Examples of seeds where the endosperm is fully consumed include pea, groundnut, and beans. 
(NCERT Reference: 1.4.3 Seed)

Q.4. Given below is a diagram of T.S. of a monocot seed with parts I, II & III labelled. 
Choose the option where parts I, II and III are identified correctly. (1 Mark)
(A) Pericarp, Endosperm, Scutellum 
(B) Pericarp, Endosperm, Coleorhiza 
(C) Scutellum, Pericarp, Coleorhiza 
(D) Coleorhiza, Scutellum, Pericarp

Answer: (B) Pericarp, Endosperm, Coleorhiza

Solution: In a monocot seed (like grass), the outer covering is the pericarp, the large storage tissue is the endosperm, and the protective covering around the radicle region is the coleorhiza.
(NCERT Reference: 1.4.3 Seed)

Q.5. Some flowers are unisexual, this property of unisexuality of flowers prevents which kind of pollination? (1 Mark)
(A) Both Autogamy and Geitonogamy 
(B) Both Geitonogamy and Xenogamy 
(C) Geitonogamy but not Xenogamy 
(D) Autogamy but not Geitonogamy

Answer: (D) Autogamy but not Geitonogamy

Solution: Unisexual flowers possess either male (staminate) or female (pistillate) parts, thereby preventing the transfer of pollen within the same flower. This prevents Autogamy. However, if both male and female flowers are borne on the same plant (monoecious condition, e.g., castor, maize), pollen transfer between the flowers on the same plant (Geitonogamy) can still occur. 
(NCERT Reference: 1.2.3 Pollination (Outbreeding Devices))

Q.6. Select the statements that are true for a typical dicotyledonous embryo from the given options. (1 Mark)
(i) It consists of an embryonal axis and scutellum. 
(ii) The portion of embryonal axis above the level of cotyledon is epicotyl. 
(iii) The portion of embryonal axis below the level of cotyledon is coleorhiza. 
(iv) The lower end of the embryo has radicle covered with a root cap. 
Choose the correct answer: 
(A) (i) and (ii) 
(B) (i) and (iii) 
(C) (iii) and (iv) 
(D) (ii) and (iv)

Answer: (D) (ii) and (iv)

Solution: (i) False: Dicot embryos have two cotyledons; scutellum is characteristic of monocots. 
(ii) True: The epicotyl is the embryonal axis portion above the cotyledons. 
(iii) False: The portion below the cotyledons is the hypocotyl; coleorhiza is a monocot sheath. 
(iv) True: The hypocotyl terminates at the radicle/root tip, which is covered by a root cap.
(NCERT Reference: 1.4.2 Embryo)

Q.7. Select the statements that are true for a typical megasporangium of flowering plants. (1 Mark)
(i) It is attached to the placenta by hilum. 
(ii) It has a chalaza end that represents the basal part of the ovule. 
(iii) It has nucellus enclosed by integuments. 
(iv) Its micropylar end is known as chalaza. Choose the correct answer: 
(A) (i) and (ii)
(B) (ii) and (iii) 
(C) (iii) and (iv) 
(D) (i) and (iv)

Answer: (B) (ii) and (iii)

Solution: (i) False: The ovule is attached to the placenta by the funicle; the hilum is the junction between the ovule and the funicle. 
(ii) True: Chalaza is located opposite the micropylar end and represents the basal part of the ovule. 
(iii) True: The nucellus, a mass of cells containing food reserves, is enclosed by protective integuments.
(iv) False: The micropylar end is the small opening at the tip; chalaza is the basal part.
(NCERT Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac)

Q.8. Select the statements that are true for outbreeding devices in the flowering plants. (1 Mark) 
(i) They discourage cross-pollination and ensure self-pollination. 
(ii) They are advantageous as they lead to inbreeding depression. 
(iii) Self-incompatibility prevents self-pollen from fertilising the ovules. 
(iv) To prevent self-pollination in some species pollen release and stigma receptivity are not synchronised. 
Choose the correct answer: 
(A) (i) and (ii) 
(B) (ii) and (iii) 
(C) (iii) and (iv) 
(D) (i) and (iv)

Answer: (C) (iii) and (iv)

Solution: (i) False: Outbreeding devices are meant to discourage self-pollination and encourage cross-pollination. 
(ii) False: Outbreeding devices prevent the disadvantageous outcome of inbreeding depression. 
(iii) True: Self-incompatibility is a genetic mechanism that prevents self-pollen from fertilizing ovules by inhibiting pollen tube growth or germination. 
(iv) True: Lack of synchrony between pollen release and stigma receptivity (dichogamy) is a mechanism evolved to prevent autogamy.
(NCERT Reference: 1.2.3 Pollination (Outbreeding Devices)

Q.9. Select the following statements that are true for insect pollinated flowers from the given options. (1 Mark)
(i) Majority of these flowers are large, colourful and rich in nectar. 
(ii) Nectars and seeds are usual floral rewards to attract insects for pollination. 
(iii) Pollen grains of these flowers are generally mucilaginous so as to stick to the body of the insects. 
(iv) A foul odour is emitted by some flowers to attract flies and beetles. 
Choose the correct answer: 
(A) (i) and (ii)
(B) (ii) and (iii) 
(C) (iii) and (iv)
 (D) (i) and (iv)

Answer: (D) (i) and (iv)
Solution: (i) True: Majority of insect-pollinated flowers are large, colourful, fragrant, and rich in nectar, as these traits attract animals. 
(ii) False: The usual floral rewards are nectar and pollen grains, not seeds. 
(iii) False: Pollen grains in animal-pollinated flowers are generally sticky. Mucilaginous covering is typical of water-pollinated species. 
(iv) True: Flowers pollinated by flies and beetles specifically emit foul odours to attract them.
(NCERT Reference: 1.2.3 Pollination (Agents of Pollination)

Q.10. Select the statements that are true for the seed of angiosperm from the given options : (1 Mark)
(i) Non-albuminous seeds have no residual endosperm. 
(ii) Residual, persistent nucellus in wheat is known as perisperm. 
(iii) Integuments of ovules harden as tough protective seed coat.
(iv) Metabolic activity of the embryo slows down in dormancy. 
Choose the correct option: 
(A) (i) and (ii) 
(B) (ii) and (iii) 
(C) (iii) and (iv) 
(D) (i) and (iv)

Answer: (D) (i) and (iv)

Solution: (i) True: Non-albuminous seeds (e.g., pea, groundnut) have the endosperm completely consumed during embryo development. 
(ii) False: Residual nucellus (perisperm) is found in black pepper and beet, not wheat. Wheat is albuminous (retains endosperm). 
(iii) True: The integuments of the ovules harden to form the tough protective seed coat(s). 
(iv) True: As the seed matures, its water content is reduced, metabolic activity slows down, and the embryo may enter dormancy.
(NCERT Reference: 1.4.3 Seed)

Q.11. Choose the option that correctly describes the gynoecium of Michelia (1 Mark)
(A) Multicarpellary, Apocarpous 
(B) Bicarpellary, Apocarpous 
(C) Multicarpellary, Syncarpous 
(D) Bicarpellary, Syncarpous

Answer: (A) Multicarpellary, Apocarpous

Solution: The gynoecium of Michelia is an example of a structure consisting of more than one pistil (multicarpellary) where the pistils are found free (unfused) from one another, which is the definition of the apocarpous condition.
(NCERT Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac)

Q.12. Given below are the events that are observed in an artificial hybridisation programme. Arrange them in the correct sequential order and select the correct option. (1 Mark)
(i) Re-bagging 
(ii) Selection of parents 
(iii) Bagging 
(iv) Dusting the pollens on stigma
(v) Emasculation
(vi) Collection of pollens from male parent
(A) (ii), (iii), (v), (vi), (iv), (i) 
(B) (ii), (v), (iii), (vi), (iv), (i) 
(C) (v), (ii), (iii), (vi), (i), (iv) 
(D) (ii), (iii), (vi), (iv), (v), (i)

Answer: (B) (ii), (v), (iii), (vi), (iv), (i)
Solution: The correct steps for artificial hybridisation involving bisexual flowers are:

1. Selection of parents (ii).
2. Emasculation (v) (removal of anthers from the female parent flower bud).
3. Bagging (iii) (covering the emasculated flower).
4. Collection of pollens from male parent (vi) (when the stigma of the female flower is receptive).
5. Dusting the pollens on stigma (iv).

6. Re-bagging (i).

(NCERT Reference: 1.2.3 Pollination (Pollen-pistil Interaction)

Q.13. In which one of the following floral plants are many embryos formed in the seeds without fertilisation of the egg cell ?
(A) Black pepper
(B) Mustard
(C) Groundnut
(D) Citrus

Answer: (D) Citrus
Solution: In Citrus, many embryos are formed in a single seed without fertilisation of the egg cell due to polyembryony associated with apomixis. As described in the NCERT chapter Sexual Reproduction in Flowering Plants, nucellar cells surrounding the embryo sac divide and protrude into the embryo sac to form additional embryos, resulting in multiple embryos within one ovule. This phenomenon is characteristic of Citrus.
(NCERT Reference: 1.5 APOMIXIS AND POLYEMBRYONY)

Assertion and Reason Questions - 1 Mark

Q.14. Assertion (A): Perisperm is a diploid tissue. 
Reason (R): Perisperm is the remains of nucellus which surrounds the embryo in certain seeds. (1 Mark)
(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.

Answer: (A) Both A and R are true and R is the correct explanation of A.

Solution: Assertion (A) is True: The nucellus, from which perisperm is derived, is the diploid sporophytic tissue of the ovule. Therefore, perisperm (the residual nucellus) is diploid.
Reason (R) is True: Perisperm is defined as the residual, persistent nucellus found in certain seeds such as black pepper and beet.
Explanation: Reason (R) directly explains Assertion (A) because the tissue of origin for perisperm is the nucellus, which is derived from the diploid maternal sporophytic generation.
(NCERT Reference: 1.4.3 Seed)

Q.15. Assertion (A): In some species of asteraceae and grasses, seeds are formed without fertilization. 
Reason (R): Formation of fruit without fertilization is called parthenocarpy. (1 Mark)
(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.

Answer: (B) Both A and R are true, but R is not the correct explanation of A.

Solution: Assertion (A) is True: Apomixis, the formation of seeds without fertilization, occurs in species like Asteraceae and grasses.
Reason (R) is True: Parthenocarpy is the development of fruits without fertilization.
Explanation: Both statements describe true biological phenomena related to reproduction without fertilization. However, Reason (R) defines parthenocarpy (fruit formation), and thus does not explain the mechanism of apomixis (seed formation) mentioned in Assertion (A).  (NCERT Reference: 1.5 Apomixis and Polyembryony)

Q.16. Assertion (A) : The zygote is present at the micropylar end of the embryo sac and develops into an embryo. 
Reason (R) : The zygote gives rise to heart-shaped embryo and subsequently proembryo in most angiosperms.
(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.

Answer:  (C) A is true, but R is false.

Solution: The assertion is correct because the zygote is formed at the micropylar end of the embryo sac after fertilisation and it develops into the embryo.

The reason is incorrect because the developmental sequence of the embryo in angiosperms is: zygote → proembryo → globular stage → heart-shaped stage → mature embryo. The statement wrongly places the heart-shaped stage before the proembryo.

Therefore, the assertion is true, but the reason is false.
(NCERT Reference: 1.4 Post-fertilisation: Structures and Events – Embryo Development)

SA (I) – Short Answer Type I (2 Marks)

Q.17. (i) How is apomixis different from parthenocarpy ? (1 Mark) 
(ii) Why are apples and cashews not called true fruits ? (1 Mark)

Answer: 

(i) Apomixis is the production of seeds without fertilization, essentially an asexual reproduction mimicking sexual reproduction. Parthenocarpy is the formation of fruits without fertilization; such fruits are typically seedless (e.g., banana).
(ii)  Apples and cashews are called false fruits because in their formation, the thalamus (or other floral parts) contributes significantly to the fruit development, besides the ovary. True fruits develop only from the ovary.
(NCERT Reference: 1.4.3 Seed (False fruits) and 1.5 Apomixis and Polyembryony (Parthenocarpy)

SA (II) – Short Answer Type II (3 Marks)

Q.18.:
(i) Write two crucial changes the seed undergoes while reaching maturity that enable them to be in a viable state until the onset of favourable conditions. (2 Marks) 
(ii) Name the oldest viable seed excavated from Arctic Tundra as per the records. (1 Mark) 

Answer: 
(i) 1. Reduction in water content – Seeds become relatively dry with 10-15% moisture, allowing them to remain dormant and viable.

2. Metabolic slowdown and dormancy – The embryo enters a state of inactivity (dormancy) to survive until favourable conditions for germination arise.

(ii) Lupine (Lupinus arcticus) – Germinated after about 10,000 years of dormancy.
NCERT Reference: 1.4.3 Seed

Q.19. (i) Pea flower produce assured seed sets. Give reason. (1 Mark) 
(ii) In case of Polyembryony, an embryo 'P' develops from a synergid and the embryo 'Q' develops from the nucellus. State the ploidy of embryo 'P' and 'Q'. (2 Marks) (Total 3 Marks for question position)

Answer:  (i) Pea flowers achieve assured seed setting because they possess adaptations that ensure self-pollination (autogamy). This form of pollination is successful because the anthers and stigma are positioned close to each other, allowing pollen grains to contact the stigma and effect pollination internally.

(ii) In case of Polyembryony, an embryo 'P' develops from a synergid and the embryo 'Q' develops from the nucellus. State the ploidy of embryo 'P' and 'Q'. (2 Marks)

Embryo	Origin	Ploidy
P	Synergid	Haploid (n)
Q	Nucellus	Diploid (2n)

NCERT Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain

Q.20. (a) A bilobed dithecous anther has 200 microspore mother cells per microsporangium. How many male gametophytes can be produced by this anther? (1 Mark) 
(b) Write the composition of intine and exine layers of a pollen grains. (2 Marks) (Total 3 Marks)

Answer: (a) A typical angiosperm anther is bilobed and dithecous, meaning it contains four microsporangia (or pollen sacs). Total Microspore Mother Cells (MMCs) = 4 microsporangia x 200 MMCs/microsporangium = 800 MMCs. 
Since each MMC undergoes meiosis to form four male gametophytes (pollen grains): 
Total male gametophytes produced = 800 MMCs x 4 pollen grains/MMC = 3200 male gametophytes.

(b) 

1. Exine: The hard outer layer of the pollen grain wall is made up of sporopollenin.
2. Intine: The inner wall of the pollen grain is a thin and continuous layer made up of cellulose and pectin.

NCERT Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain

Q.21. Mention any four strategies adopted by flowering plants to prevent self-pollination. (3 Marks) 

Answer:  Flowering plants employ various outbreeding devices to discourage self-pollination and promote cross-pollination:

1. Lack of Synchrony (Dichogamy): Pollen release and stigma receptivity are separated in time. Either the pollen is released before the stigma becomes receptive, or vice versa.
2. Heterostyly/Herkogamy: The anther and stigma are placed at different physical positions within the flower, preventing pollen from coming into contact with the stigma of the same flower.
3. Self-Incompatibility: A genetic mechanism that recognizes and prevents self-pollen (from the same flower or plant) from fertilizing the ovules, typically by inhibiting pollen germination or pollen tube growth.
4. Unisexual Flowers (Dicliny): The production of male and female flowers separately prevents autogamy. If the plants are dioecious (male and female flowers on different plants, e.g., papaya), both autogamy and geitonogamy are prevented. 
   (NCERT Reference: 1.2.3 Pollination (Outbreeding Devices))

Q.22. Flowering plants with hermaphrodite flowers have developed many reproductive strategies to ensure cross-pollination. Study the given outbreeding devices adopted by certain flowering plants and answer the questions that follow. 
Note :  All plants belong to the same species.
x - No pollen tube growth/inhibition of pollen germination on stigma. 
✔ - Pollen germination on stigma. 
 (a) Name and define the outbreeding device described in the above table. (1 Mark) 
(b) Explain what would have been the disadvantage to the plant in the absence of the given strategy. (2 Marks)

Answer: 

(a) Self-incompatibility is “a genetic mechanism and prevents self-pollen (from the same flower or other flowers of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.”

(b)  In the absence of outbreeding devices, the plant would predominantly undergo continuous self-pollination. The major disadvantage of continuous self-pollination is that it results in inbreeding depression, which reduces vigor and fitness over successive generations.
NCERT Reference: 1.2.3 Pollination (Outbreeding Devices)

Q.23. Comment upon the mode of pollination in Vallisneria and Zostera. (3 Marks) 

Answer:  Both Vallisneria and Zostera utilize water as an abiotic agent for pollination, a method that is relatively rare in flowering plants.

The specific modes of pollination for each are as follows:

• Vallisneria (Freshwater pollination at the surface):

    ◦ In this plant, the female flowers reach the surface of the water using a long stalk.

    ◦ The male flowers or pollen grains are released onto the surface of the water.

    ◦ These pollens are carried passively by water currents; some eventually reach the female flowers and their stigmas to achieve pollination.

• Zostera (Marine pollination submerged):

    ◦ As a marine seagrass, its female flowers remain submerged in the water.

    ◦ The pollen grains are released inside the water rather than on the surface.

    ◦ The pollen grains in such species are characterized as being long and ribbon-like.

    ◦ They are carried passively inside the water until they reach the stigma.

Common Adaptations: In both cases, the pollen grains are typically protected from wetting by a mucilaginous covering. Additionally, because they rely on water currents rather than animals, these flowers are not very colourful and do not produce nectar.
NCERT Reference: 1.2.3 Pollination (Agents of Pollination)

Q.24. Distinguish between the two cells enclosed in a mature male gametophyte of an angiosperm. (3 Marks)

Answer: In a mature male gametophyte (pollen grain) of an angiosperm, the two enclosed cells are the vegetative cell and the generative cell. They can be distinguished based on the following characteristics:

• Size and Content: The vegetative cell is bigger and contains abundant food reserves, whereas the generative cell is small.

• Physical Appearance: The vegetative cell has a large, irregularly shaped nucleus. In contrast, the generative cell is spindle-shaped with dense cytoplasm and its own nucleus.

• Position and Function: The generative cell floats in the cytoplasm of the vegetative cell. While the vegetative cell provides nourishment, the generative cell undergoes mitotic division to give rise to two male gametes.

In over 60 percent of angiosperms, the pollen is shed at this 2-celled stage, while in others, the generative cell divides before shedding to reach a 3-celled stage.
NCERT Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain

Q.25. Draw a longitudinal section of pistil of a flower showing growth of the pollen tube. Label the part: 
(a) through which the pollen tube moves down 
(b) the cell wherein the pollen tube releases its contents.

Answer: 

(NCERT Reference: 1.2.3 Pollination)

Case-Based Questions (4 Marks)

Q.26. Given below is a flower with its characteristic features specialised for the most common type of abiotic pollination.
Answer the following questions based on the above diagram: 
(a) Name the mode of abiotic pollination that will be adopted by the given plant species in the above picture. (1 Mark) 
(b) State the need of exposed large feathery stigmas for the flower. (1 Mark) 
(c) What will be the two important adaptations in the pollen grains of the flowers pollinated by the above mode of pollination ? (1 Mark) (d) What could be the probable reason for the petals being small and non-green ? (1 Mark) 

Answer: (a) The most common abiotic pollination mode involving exposed floral parts is Wind Pollination (Anemophily).
(b) The large, feathery stigmas are an adaptation necessary to easily trap air-borne pollen grains floating in the wind currents.
(c) The pollen grains must be light and non-sticky so they can be easily transported over long distances in the wind currents.
(d) Wind-pollinated flowers do not require attractive mechanisms because they do not use biotic agents. Therefore, they are typically not colourful and do not produce nectar.
NCERT Reference: 1.2.3 Pollination (Agents of Pollination)

Q.27. The following question is based on pollination. Study the figures carefully and answer the questions that follow. 
(a) Give the scientific terms for the processes taking place in Figures A and B respectively. (1 Mark) 
(b) Mention two conditions necessary for the process occurring in Figure B. (2 Marks) 
(c) (i) State one advantage and one disadvantage of the process occurring in Figure B. (1 Mark) 
OR
(c) (ii) Name one plant where, in some flowers only, the process in Figure B takes place and give the reason responsible for it. (1 Mark)

Answer: (a) Figure A: Geitonogamy It shows the transfer of pollen grains from the anther of one flower to the stigma of another flower on the same plant. Figure B: Autogamy It shows the transfer of pollen grains from the anther to the stigma of the same flower.

(b)  Two conditions necessary for the process occurring in Figure B are: 

1. The pollen must be compatible and germinate on the stigma.

“If it is of the right type, the pistil accepts the pollen and promotes post-pollination events that lead to fertilisation.”

2. The pollen tube must grow through the style to reach the ovule.

“Pollen tube grows through the tissues of the stigma and style and reaches the ovary.”

(c) (i) Advantage: Xenogamy ensures genetic variation because it brings genetically different pollen grains to the stigma.

Disadvantage: The success of the process depends on external agents (wind, water or animals), so pollination may fail if these agents are absent. 

OR
(ii) Plant Name: Viola (common pansy), Oxalis, or Commelina.
Reason: These plants produce cleistogamous flowers (closed flowers) alongside chasmogamous ones. In cleistogamous flowers, the anthers and stigma are positioned close together and the flowers do not open, guaranteeing that pollen contacts the stigma to effect pollination, leading to assured seed-set.
(NCERT Reference: 1.2.3 Pollination (Pollen-pistil Interaction) and 1.3 Double Fertilisation)

Long Answer Type Questions (5 Marks)

Q.28. Study the diagram given below showing the modes of pollination. Answer:  the questions that follow.
(i) The given diagram shows three methods of pollen transfer in plants. Examine them carefully and write the technical terms used for pollen transfer methods ‘1’, ‘2’ and ‘3’. (1 Mark) 
(ii) How do the following plants achieve pollination successfully? 
(a) Water lily (1 Mark)
(b) Vallisneria (1 Mark) 
(iii) Write advantages of pollen transfer method ‘3’. (2 Marks)

Answer:

(ii)
(a) Water lily: Pollination occurs by insects or wind, because the flowers emerge above water; therefore water is not the pollinating agent.
(b) Vallisneria: Pollination occurs through water currents. Male flowers or pollen grains float on the surface and are carried by water to female flowers.

(iii) 1. It brings genetically different pollen to the stigma, which increases variation. 
2. It promotes genetic diversity and better adaptation in offspring.
NCERT Reference: 1.2.3 Pollination (Agents of Pollination)

Q.29. (i) Describe the process of megasporogenesis in an angiosperm. (3 Marks) 
(ii) Draw a diagram of a mature embryo sac of an angiosperm. Label its any four parts. (2 Marks) 
Answer: 

(i) Megasporogenesis is the process of the formation of megaspores from the megaspore mother cell (MMC).

1. A single Megaspore Mother Cell (MMC) differentiates in the micropylar region of the nucellus. The MMC is a large cell with dense cytoplasm and a prominent nucleus.
2. The MMC undergoes meiotic division.
3. Meiosis results in the production of four megaspores.
4. In most flowering plants, three of these megaspores degenerate, and only one functional megaspore develops into the female gametophyte (embryo sac).

(ii) NCERT Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac

Q.30. Study the figures given below showing initial stages in the formation of female gametophyte and answer the questions that follow. 

(i) Identify (P) and (Q). (1 Mark) 
(ii) I. What kind of division does cell (P) undergo to form (Q)? (1 Mark) 
II. How many (Q) cells form the embryo sac? What is the name given to such kind of development? (1 Mark) 
III. How many free nuclear mitotic divisions will the functional megaspore undergo to form the embryo sac? (1 Mark) 
IV. Describe the structure of a mature female gametophyte. (1 Mark) 

Answer: (i) P: Megaspore Mother Cell (MMC)  Q: Megaspores (or Megaspore Tetrad)

(ii) Cell (P) undergoes Meiotic division.
II. Only one functional megaspore (Q cell) develops into the embryo sac. The method of embryo sac formation from a single megaspore is called monosporic development.
III. The nucleus of the functional megaspore undergoes three sequential mitotic nuclear divisions (leading to 2-, 4-, and then 8-nucleate stages).
IV.  A mature female gametophyte (embryo sac) is typically 7-celled and 8-nucleate. The cells are distributed such that three form the egg apparatus (one egg cell, two synergids) at the micropylar end, three are antipodals at the chalazal end, and one large central cell contains two polar nuclei.
NCERT Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac

Q.31. What is meant by ‘right’ and ‘wrong type of pollen when pollen lands on the stigma. A two-celled right type pollen has landed on the stigma. Explain the events taking place till fertilization in angiosperms. Why is pollen pistil interaction said to be dynamic ? (5 Marks) (Total 5 Marks for question position)

Answer: Events taking place till fertilization in angiosperms

1. Pollen Germination: Following compatible pollination (right type pollen), the pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores.
2. Tube Growth and Male Gamete Formation: The pollen tube grows through the tissues of the stigma and style and proceeds toward the ovary. Since the pollen was shed at the two-celled stage, the generative cell divides mitotically during this growth period to form the two male gametes.
3. Entry into Embryo Sac: The pollen tube enters the ovule (usually through the micropyle) and enters one of the synergids of the embryo sac, guided by the specialized cellular thickenings known as the filiform apparatus.
4. Double Fertilization: The two male gametes are discharged into the cytoplasm of the synergid. One male gamete fuses with the egg cell (syngamy) to form the diploid zygote. The other male gamete fuses with the two polar nuclei in the central cell (triple fusion) to form the triploid Primary Endosperm Nucleus (PEN).

Pollen-pistil interaction is described as dynamic because it is a complex process involving a continuous "dialogue" between the pollen and the pistil. This dialogue is mediated by the interaction of chemical components between the two entities. This chemical recognition process determines whether the pistil accepts the pollen (if compatible) and promotes post-pollination events, or rejects it (if incompatible) by inhibiting tube growth or germination.  
(NCERT Reference: 1.2.3 Pollination (Pollen-pistil Interaction) and 1.3 Double Fertilisation)

CBSE Previous Year Questions 2024

Multiple Choice Questions (MCQs)  - 1 Mark

Q.1. Study the following diagram of Transverse Section of a young anther of an angiosperm: Select the option where parts ‘A’, ‘B’ and ‘C’ are correctly identified. 

(A) A – Connective, B – Endothecium, C – Pollen grain. 
(B) A – Endothecium, B – Connective, C – Pollen grain. 
(C) A – Pollen grain, B – Connective, C – Endothecium. 
(D) A – Endothecium, B – Pollen grain, C – Connective.

Answer: (A) A – Connective, B – Endothecium, C – Pollen grain. 
Solution: In the transverse section of a young anther, the central tissue joining the two lobes is the connective, the wall layer beneath the epidermis is the endothecium, and the structures inside the microsporangium are pollen grains.
(NCERT Topic Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain.)

Q.2. Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below: 

(A) A – zygote, B – degenerating synergids, C – degenerating antipodals, D – PEN. 
(B) A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals. 
(C) A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote. 
(D) A – degenerating synergids, B – zygote, C – degenerating antipodals, D – PEN.

Answer: (D) A – degenerating synergids, B – zygote, C – degenerating antipodals, D – PEN

Solution: After double fertilisation, the zygote is located at the micropylar end, the synergids degenerate, the antipodals degenerate at the chalazal end, and the central cell becomes the Primary Endosperm Nucleus (PEN).
(NCERT Topic Reference: 1.3 Double Fertilisation.)

Q.3. In which of the following plants are both male and female flowers born on the same plant and the mode of pollination can be geitonogamy or xenogamy? 
(A) Papaya 
(B) Date Palm 
(C) Maize 
(D) Spinach

Answer: (C) Maize
Solution: Maize is a monoecious plant, meaning both male and female flowers are present on the same plant. This condition prevents autogamy (pollination within the same flower) but allows geitonogamy. Additionally, xenogamy can occur as it is common in wind-pollinated grasses like maize. Dioecious plants like Papaya and Date Palm prevent geitonogamy as male and female flowers are on different plants.
(NCERT Topic Reference: 1.2.3 Pollination.)

Q.4. In a fertilized ovule of an angiosperm, the cells in which n, 2n and 3n conditions respectively occur are: 
(A) antipodal, zygote and endosperm 
(B) zygote, nucellus and endosperm 
(C) endosperm, nucellus and zygote 
(D) antipodals, synergids and integuments

Answer: (A) antipodal, zygote and endosperm
Solution: In a typical fertilized embryo sac, antipodals are haploid (n). The zygote is diploid (2n) because it is formed by the fusion of one haploid male gamete with a haploid egg cell (syngamy). The endosperm is triploid (3n) resulting from triple fusion (one male gamete + two polar nuclei).
(NCERT Topic Reference: 1.3 Double Fertilisation.)

Q.5. The part of the ovule that develops into protective coats of a seed after fertilization in a typical flowering plant is: 
(A) embryo sac 
(B) nucellus 
(C) integuments 
(D) megaspore

Answer: (C) integuments
Solution: Following fertilization, the integuments of the ovule harden to become tough, protective seed coats.
(NCERT Topic Reference: 1.4.3 Seed.)

Q.6. The functional megaspore of an angiosperm develops into: 
(A) Embryo sac 
(B) Endosperm 
(C) Embryo 
(D) Ovule

Answer: (A) Embryo sac
Solution: In a majority of flowering plants, three megaspores degenerate and only one functional megaspore remains. This functional megaspore undergoes mitotic divisions to develop into the female gametophyte, also known as the embryo sac. 
(NCERT Topic Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac.)

Q.7. Refer to the Venn diagram given below. Select the option with correct examples P, Q, and R: 

Answer: B – Bean, Castor, Maize

Solution: P (Dicotyledonous, non-endospermic): Bean

Dicot seeds like bean and groundnut consume the endosperm completely.

“Non-albuminous seeds… e.g., pea, groundnut.” (Bean is same category

Q (Endospermic seed in dicots): Castor

“Albuminous seeds retain a part of endosperm… e.g., castor.”

R (Monocotyledonous): Maize

Monocot seeds like maize retain endosperm.

“Albuminous seeds… e.g., wheat, maize, barley.”

(NCERT Topic Reference: 1.4.3 Seed.)

Q.8. An angiosperm embryo sac is located within the : 
(A) Placenta
(B) Megasporangium
(C) Nucellus
(D) Ovary

Answer: (C) Nucellus
Solution: In angiosperms, the embryo sac (female gametophyte) is situated within the nucellus of the ovule. The nucellus is a mass of nutritive tissue enclosed by the integuments, and it houses the embryo sac where the egg apparatus, polar nuclei, and antipodal cells are present.

The ovule itself is the megasporangium and is attached to the placenta inside the ovary, but the immediate tissue that contains the embryo sac is the nucellus. Hence, the correct option is (C) Nucellus.
(NCERT Topic Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac)

Q.9. Endosperm development precedes embryo development. Justify. (1 Mark)

Answer: Endosperm development precedes embryo development because the endosperm serves as a nutritive tissue for the developing embryo. After double fertilisation, the primary endosperm cell divides rapidly to form endosperm, which stores reserve food material. The zygote usually remains dormant initially and starts dividing only after a sufficient amount of endosperm has formed. This ensures an assured supply of nutrition for the growing embryo during its early stages of development.
(NCERT Topic Reference: 1.4 Post-fertilisation: Structures and Events)

SA (I) – Short Answer Type I (2 Marks)

Q.10. Why does endosperm development precede embryo development in angiosperm seeds? State the role of endosperm in mature albuminous seeds.

Answer: Endosperm development precedes embryo development as an adaptation to provide assured nutrition to the developing embryo. The cells of the endosperm tissue are filled with reserve food materials essential for the growth of the zygote as it matures into an embryo. In mature albuminous seeds (e.g., wheat, maize, castor), the endosperm is not completely consumed during embryo development and persists. Its role is to provide nutrition during the process of seed germination to support the young seedling until it can photosynthesize. 
(NCERT Topic Reference: 1.4.1 Endosperm.)

Expert Tip: Remember that the primary endosperm cell divides repeatedly to form a triploid endosperm tissue before the zygote significantly develops into an embryo to ensure food is ready.

Q.11. How is geitonogamy different from xenogamy?

Answer: Geitonogamy involves the transfer of pollen grains from the anther to the stigma of another flower on the same plant. While it often requires a pollinating agent, it is genetically similar to autogamy because the pollen grains come from the same plant. Xenogamy is the transfer of pollen grains from the anther to the stigma of a different plant. This is the only type of pollination that brings genetically different types of pollen to the stigma. 
(NCERT Topic Reference: 1.2.3 Pollination.)

Q.12. Name the type of flowers that are always (invariably) autogamous. Explain why.

Answer: Cleistogamous flowers are invariably autogamous. These flowers do not open at all, ensuring that the anthers and stigma remain in close contact. This prevents cross-pollen from landing on the stigma and allows for pollination and assured seed-set even in the absence of pollinators. 
(NCERT Topic Reference: 1.2.3 Pollination.)

Q.13. Explain the development of male gametophyte in an angiosperm.

Answer: The development of the male gametophyte (pollen grain) involves:

• Microsporogenesis: Inside the microsporangium, diploid pollen mother cells undergo meiosis to form microspore tetrads.
• Pollen Maturation: As anthers mature and dehydrate, microspores dissociate and develop into pollen grains.
• Cellular Organization: A mature pollen grain contains a large vegetative cell (abundant food reserves) and a small generative cell. In 60% of angiosperms, pollen is shed at this 2-celled stage. In others, the generative cell divides mitotically to form two male gametes before shedding (3-celled stage). 

(NCERT Topic Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain.)

Q.14. Identify and define A, B, C and D regarding seed parts: 

Answer:

• A (Pericarp): The wall of the fruit that develops from the wall of the ovary.
• B (Scutellum): The single cotyledon specifically found in the grass family.
• C (Epicotyl/Hypocotyl): Portions of the embryonal axis; the part above the level of cotyledons is the epicotyl, and the cylindrical portion below is the hypocotyl.
• D (Perisperm): The residual, persistent remnants of the nucellus found in some seeds like black pepper. 

(NCERT Topic Reference: 1.4.2 Embryo and 1.4.3 Seed.)

Q.15. “Farmers prefer apomictic seeds to hybrid seeds.” Justify giving two reasons.

Answer:

1. Avoidance of Character Segregation: Hybrid seeds have a disadvantage where their progeny segregates, meaning the plants do not maintain desirable hybrid traits in the next generation. Apomixis produces seeds without fertilization, ensuring no segregation.
2. Cost-Effectiveness: Hybrid seeds are expensive and must be bought every year. If hybrids are made into apomicts, farmers can reuse the seeds year after year, significantly reducing costs. 

(NCERT Topic Reference: 1.5 Apomixis and Polyembryony.)

Q.16. Name any two outbreeding devices that flowering plants have developed and explain how they help in encouraging cross-pollination.

Answer:  (any two)

1. Unisexuality - production of unisexual flowers, prevent self pollination, 
2. Release and receptivity of stigma are not synchronized, either pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen,
3. Anther and stigma are placed at different positions, so that the pollen cannot come in contact with stigma,
4. Self-incompatibility/genetic incompatibility, genetic mechanism and prevents selfpollen from fertilizing the ovule by inhibiting pollen germination or pollen tube growth in the pistil.
   (NCERT Topic Reference: 1.2.3 Pollination (Outbreeding Devices)).

SA (II) – Short Answer Type II (3 Marks)

Q.17. If the cells in the leaves of a maize plant contain 10 chromosomes each, write the number of chromosomes in its endosperm and zygote. Name and explain the process by which an endosperm and a zygote are formed in maize.

Answer:

Chromosome Numbers:Leaf cells are diploid (2n=10), so the haploid number (n) is 5.

• Zygote (2n): 10 chromosomes.
• Endosperm (3n): 15 chromosomes.

Processes of Formation:

• Syngamy: The zygote is formed when one haploid male gamete fuses with the haploid egg cell nucleus.
• Triple Fusion: The endosperm is formed when the second haploid male gamete fuses with two haploid polar nuclei in the central cell to produce a triploid Primary Endosperm Nucleus (PEN). 

(NCERT Topic Reference: 1.3 Double Fertilisation.)

Concept Applied: Ploidy levels. Somatic cells (leaves) are 2n, gametes are n. Zygote (n+n=2n) and Endosperm (n+n+n=3n).

Q.18. State why plant breeders are interested in artificial hybridisation programme. How do they carry out this process?

Answer:

1. Interest: Breeders use artificial hybridization to cross different species or genera to combine desirable traits from both parents, creating commercially superior and high-yielding varieties.
2. Process:
   • Emasculation: Removal of anthers from a bisexual female parent's flower bud before they dehisce using forceps.
   • Bagging: Covering the emasculated flower with a bag (usually butter paper) to prevent stigma contamination by unwanted pollen.
   • Pollination and Rebagging: When the stigma becomes receptive, mature desired pollen is dusted onto it; the flower is rebagged to allow fruit development.  

(NCERT Topic Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac.)

Long Answer Type Questions (5 Marks)

Q.19. Explain the process of double fertilization in an angiosperm starting from the germination of pollen grains on the stigma, mentioning the ploidy of the end products formed at the end. State the role of synergids during the course of the process.

Answer:

• Process:
  1. Pollen germinates on the stigma, producing a pollen tube that grows through the style to the ovary.
  2. The tube enters the ovule via the micropyle and penetrates one synergid.
  3. Two male gametes are released into the synergid's cytoplasm.
  4. Syngamy: One gamete fuses with the egg nucleus to form a diploid (2n) zygote.
  5. Triple Fusion: The other gamete fuses with two polar nuclei in the central cell to form a triploid (3n) Primary Endosperm Nucleus (PEN).
• End Products: Zygote (2n) and PEN (3n).
• Role of Synergids: Synergids have a filiform apparatus at the micropylar tip that guides the pollen tube entry into the embryo sac. 

(NCERT Topic Reference: 1.3 Double Fertilisation.)

Q.20. Explain any four devices that flowering plants have developed to encourage cross-pollination. Why do they discourage self-pollination?

Answer:

• Outbreeding Devices (Any Four):
  1. Non-Synchronization: Pollen release and stigma receptivity are not timed together.
  2. Different Positioning: Anthers and stigma are placed so pollen cannot physically reach its own stigma.
  3. Self-Incompatibility: A genetic mechanism inhibiting pollen germination or tube growth from the same plant.
  4. Unisexual Flowers: Production of male and female flowers on different plants (dioecy), ensuring only cross-pollination.
• Reason for Discouraging Self-Pollination: Continued self-pollination leads to inbreeding depression, reducing genetic vigor and fitness.

(NCERT Topic Reference: 1.2.3 Pollination (Outbreeding Devices)).

Q.21. Draw a labelled diagram of a three-celled male gametophyte.

Answer: 
(NCERT Topic Reference: 1.2.1 Stamen, Microsporangium and Pollen Grain.)

Q.22. With the help of labelled diagram only, show the different stages of embryo development in a dicot plant.

Answer:  (NCERT Topic Reference: 1.4.2 Embryo.)

Q.23. Show the development of megaspore mother cell upto the formation of mature embryo sac in flowering plants with the help of labelled diagrams only.

Answer: 

NCERT Topic Reference: 1.2.2 The Pistil, Megasporangium (ovule) and Embryo sac.

Q.24. Observe the picture of Commelina plant bearing two types of flowers given above. 

(i) Identify the two types of flowers labelled ‘A’ and ‘B’ in the picture. 
(ii) Compare the two types of flowers with reference to: 
(1) Characteristic feature
(2) modes of pollination
(iii) List any two ‘out breeding devices’ in flowering plants. Explain why do plants develop such devices.

(i) A- chasmogamous flower  B- cleistogamous flower 

(ii) 

(iii) 

• In some species, pollen release and stigma receptivity are not synchronized. Either the pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen.
• The anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.
• Self-incompatibility. This is a genetic mechanism and prevents selfpollen from fertilizing the ovules.
• Production of unisexual flowers. 
  (NCERT Topic Reference: 1.2.3 Pollination (Outbreeding Devices)).