Sexual reproduction in flowering plants practice · Medium

Option D (Both (A) and (R) are false) is correct because insects visit flowers for both nectar and pollen, not just honey, and the attraction of flowers doesn't necessarily prevent them from damaging other parts of the plant.

Option D (Generative cell) is correct because in angiosperms, the generative cell divides to form two male gametes.

Option D (16) is correct because each anther typically has four microsporangia (pollen sacs), and each microspore mother cell within these sacs undergoes meiosis to produce four pollen grains. If there are four cells in an anther, and each cell is a microspore mother cell, then 4 cells * 4 pollen grains/cell = 16 pollen grains.

The correct answer is A (Fusion of male gamete with diploid secondary nucleus) because vegetative fertilization in angiosperms refers to the fusion of a sperm cell with the secondary nucleus to form the triploid endosperm.

Option B (24) is correct because the aleurone layer is triploid (3n), and if synergids (n) have 8 chromosomes, then the aleurone layer has 3 * 8 = 24 chromosomes.

Option C (100) is correct because each pollen mother cell produces 4 pollen grains, and each seed requires one pollen grain. Therefore, 400 seeds require 400 pollen grains, which are produced by 100 pollen mother cells.

Option D (16) is correct because each microsporocyte undergoes meiosis to produce 4 pollen grains. Therefore, 4 microsporocytes will produce 4 x 4 = 16 pollen grains.

Option A (Dioecy) is correct because dioecy, the condition of having separate male and female plants, strictly prevents autogamy and geitonogamy, thus promoting xenogamy.

Option B (FALSE) is correct because microspores develop into pollen grains, not ova (eggs).

Option A (Polyembryony) is correct because polyembryony is the occurrence of more than one embryo in a seed.

Option D (All of the above) is correct because a micropyle is a small pore for water entry, a gap where integuments are absent, and necessary for seed existence.

Option C (Scutellum) is correct because pollen grain, synergid, and polar nucleus are all haploid (n) structures, while the scutellum is a diploid (2n) structure.

Option B (380) is correct because the meiocyte (diploid cell undergoing meiosis) of a butterfly has 380 chromosomes.

Option B (Mango) is correct because mango trees exhibit a distinct recovery phase after fruiting, where they build up resources for the next season.

Option B (24) is correct because the endosperm is triploid (3n), and if the leaf (2n) has 16 chromosomes, then the endosperm has (3/2) * 16 = 24 chromosomes.

Option C (Both (a) and (b)) is correct because the pericarp is the mature ovary wall, which forms the fruit wall.

Option D is correct because the transfer of pollen grains to the stigma of the same flower is self-pollination (autogamy), not cross-pollination.

Option C (If Assertion is true but Reason is false) is correct because the assertion about nucellar endosperm formation is correct, but the reason is false as coconut initially has free-nuclear endosperm which later becomes cellular.

Option D (None of the above) is correct because apomixis arises due to specific genetic and developmental factors, not simply rapid or slow reproduction.

Option D is correct because wind-pollinated flowers typically have dry pollen grains, not surrounded by a mucilaginous covering which is more common in insect or water-pollinated flowers.

Option D (All of these) is correct because polyembryony, the condition of having multiple embryos in a single ovule, is common in Citrus, Orange, and Mango.

Option B (Capsella) is correct because Capsella exhibits onagrad type embryo development, characterized by specific cell division patterns during embryogenesis.