Previous year questions: Sexual reproduction in flowering plants
Sexual Reproduction in Flowering Plants28 min readLesson 8 of 8
Previous year questions: sexual reproduction in flowering plants
This set mixes actual CBSE board paper questions from the 2023, 2024, 2025 and 2026 exams with two official sample paper questions from the 2022-23 and 2025-26 sessions. Each heading names the paper and the marks, and each answer is worked the way a board examiner expects. Attempt the question first.
Q1 (CBSE Board 2023, 1 mark)
Choose the option that gives the correct number of pollen grains formed after 325 microspore mother cells undergo microsporogenesis.
Answer: 1300 pollen grains. Each microspore mother cell undergoes meiosis and produces a tetrad of four microspores, and every microspore of the tetrad matures into a pollen grain. So 325 mother cells give 325 x 4 = 1300 pollen grains.
Q2 (CBSE Board 2023, 1 mark)
Assertion (A): Endosperm is completely consumed during the development of the embryo in ex-albuminous seeds. Reason (R): Castor, pea and beans are all examples of ex-albuminous seeds.
Answer: The assertion is true and the reason is false, so the correct option is the one that says A is true but R is false. A seed in which the endosperm is fully used up by the developing embryo is called non-albuminous or ex-albuminous, and pea and beans are examples. Castor does not belong in that list, because castor is an albuminous seed that keeps its endosperm in the mature seed. One wrong example makes the whole reason false. The prefix ex tells you the endosperm is gone from the mature seed.
Q3 (CBSE Board 2023, 2 marks)
(a) From which end of the ovule, and how, does the pollen tube gain entry into the embryo sac of a Hibiscus flower? (b) State the fate of the male nuclei present in the pollen tube.
Answer: (a) The tube reaches the ovule at its micropylar end, travels through the micropyle and then pushes into a synergid. The filiform apparatus at the tip of that synergid is what steers the tube in. (b) Inside the synergid the tube sets free two male nuclei. The first travels to the egg and joins with the egg nucleus, so syngamy gives a diploid zygote. The second travels to the central cell, joins both polar nuclei, and triple fusion gives a triploid primary endosperm nucleus.
Q4 (CBSE Board 2024, 1 mark)
In a fertilised ovule of an angiosperm, the cells in which the 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: Option (A), antipodal, zygote and endosperm. The antipodal cells are haploid products of the embryo sac, and so are the synergids and the egg. The zygote is diploid because it forms by the fusion of two haploid nuclei. The endosperm is triploid because its primary nucleus forms by triple fusion, from one male gamete and two polar nuclei. Nucellus and integuments are diploid parts of the ovule, which rules out the other options.
Q5 (CBSE Board 2024, 3 marks)
Draw a well labelled diagram of a sectional view of the male gametophyte or microspore of an angiosperm and write the functions of any two labelled parts. Any four labels are required.
Answer: The male gametophyte is the pollen grain. A labelled section shows the hard outer exine with a germ pore, the inner intine, the large vegetative cell with its nucleus and stored food, and the small generative cell floating in the cytoplasm of the vegetative cell. Two functions: the vegetative cell grows into the pollen tube and carries the male gametes to the ovule, and it also provides stored food and cytoplasm for the grain. The generative cell divides by mitosis to give the two male gametes, which carry out syngamy and triple fusion. The exine protects the gametophyte, and the germ pore is the point from which the pollen tube emerges.
Q6 (CBSE Board 2025, 3 marks)
Flowering plants with hermaphrodite flowers have developed reproductive strategies to ensure cross-pollination. A table shows pollen grains of plant A on the stigma of plant A failing to germinate, while pollen of plant B and plant C germinates on that stigma. All plants belong to the same species. (a) Name and define the outbreeding device described. (b) Explain the disadvantage to the plant if this strategy were absent.
Answer: (a) The device is self-incompatibility. It is a genetic mechanism. The pistil recognises pollen that came from the same flower or from a different flower on the same plant, and it turns that pollen away. The block can act at the stigma, by stopping the grain from germinating, or inside the style, by stopping the tube from growing. Either way the self pollen never reaches the ovule. (b) Without this device the hermaphrodite flower would self-pollinate repeatedly. Continued selfing causes inbreeding depression, reduces genetic variation in the offspring and leaves the population less able to adapt to a changing environment.
Q7 (CBSE Board 2025, 5 marks)
(a) (i) Explain the structure of a mature embryo sac of a typical flowering plant. (ii) How is triple fusion achieved in these plants?
Answer: (a)(i) A mature embryo sac carries eight nuclei in seven cells. Three cells sit at the micropylar pole: two synergids and one egg. Those three together are the egg apparatus, and each synergid has a thickened patch at its tip called the filiform apparatus. Three more cells, the antipodals, sit at the chalazal pole. The middle holds one large central cell with two polar nuclei inside it. Count them and the total is eight nuclei in seven cells. (ii) After the pollen tube enters a synergid and discharges the two male gametes, one male gamete moves to the central cell and fuses with the two polar nuclei. Three haploid nuclei join in this triple fusion, so the product is the triploid primary endosperm nucleus. The central cell becomes the primary endosperm cell, which develops into the endosperm. Because syngamy also occurs in the same sac, the whole event is double fertilisation.
Q8 (CBSE Board 2026, 3 marks)
(a) Explain the different ways by which apomictic seeds can be developed. (b) Mention one advantage of apomictic seeds for farmers.
Answer: (a) Apomictic seeds develop without fertilisation, in two main ways. In some species the diploid egg cell forms without reduction division, that is without meiosis, and develops into an embryo on its own. In others, as in many Citrus and mango varieties, cells of the nucellus surrounding the embryo sac divide, push into the embryo sac and develop into embryos. This second route also produces polyembryony, with several embryos in one seed. (b) Apomictic seeds carry the same genotype as the parent, so a hybrid variety raised from them does not segregate. A farmer can save seed from the hybrid crop and use it year after year instead of buying fresh hybrid seed every season.
Q9 (CBSE SQP 2022-23, 1 mark)
Assertion: Apomictic embryos are genetically identical to the parent plant. Reason: Apomixis is the production of seeds without fertilisation.
Answer: Both statements are true and the reason explains the assertion. Apomixis avoids both meiosis and fertilisation, so the genes of the parent are not reshuffled in the embryo. The absence of fertilisation is exactly why the embryos stay genetically identical to the parent.
Q10 (CBSE SQP 2025-26, 1 mark)
The male gametes in flowering plants are formed by: mitotic division of the nucleus of the vegetative cell, meiotic division of the nucleus of the vegetative cell, mitotic division of the nucleus of the generative cell, or meiotic division of the nucleus of the generative cell?
Answer: Mitotic division of the nucleus of the generative cell. The generative cell is already haploid, and a mitotic division of its nucleus produces two haploid male gametes. A meiotic division would halve the chromosome number again and would not give functional gametes.
Three of these questions are printed board questions with only small rewording, and the rest keep the original data while the wording is made self-contained. Sort them by mark weight when you revise. The one-mark items test a single fact, such as the number 1300 or the two nuclei that join in triple fusion. The three-mark items want a labelled structure plus a function. The five-mark item wants the full embryo sac and the sequence of fusion. Write the ploidy beside every nucleus you name, because that is where most answers lose a mark.