Punjab State Board PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants Important Questions and Answers.
PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants
Very short answer type questions
Question 1.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give any one reason.
Answer:
A malfunctioning tapetum does not provide enough nourishment to the developing male gametophytes and thus fail to produce viable male gametophytes.
Question 2.
Complete the following flow chart
Pollen mother cell → Pollen tetrad
[NCERT Exampler]
Answer:
Generative cell.
Question 3.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an example each.
Answer:
Gynoecium of a flower may be apocarpous means the carpels are free from one another and there is no fusion of any part e.g., Ranunculus, Rose. Gynoecium of a flower is syncarpous, means the carpels are fused by their ovaries. The number of fusing carpels may vary from 2 (Petunia) to ∞ (Hibiscus).
Question 4.
Name the parts of the gynoecium which develop into fruit and seeds. [NCERT Exemplar]
Answer:
Ovary develops into fruit and ovules develop into seeds.
Question 5.
How many haploid cells are present in a mature female gametophyte of a flowering plant? Name them.
Answer:
One dikaryotic polar cell with two haploid nuclei and six haploid cells, viz, 3 antipodal, 2 synergids and 1 egg.
Question 6.
Name the type of pollination in self-incompatible plants. [NCERT Exemplar]
Answer:
Xenogamy.
Question 7.
How do flowers of Vallisneria get pollinated?
Answer:
In Vallisneria, the female flower stalk is coiled to reach the water surface to receive the pollen grains carried by water currents.
Question 8.
What is pollen-pistil interaction and how is it mediated?
Answer:
The pistil accepts the right type (compatible) of pollen and promotes fertilisation and rejects the pollen of other species and incompatible pollen of the same species. It is the result of interaction between the chemicals of the pollen and those of stigma.
Question 9.
State the function of filiform apparatus found in mature embryo sac of an angiosperm.
Answer:
Filiform apparatus plays an important role in guiding the path of pollen tubes into the synergids.
Question 10.
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened?
Answer:
An orange seed has many embryos because of polyembryony.
Question 11.
Name the component cells of the ‘egg apparatus’ in an embryo sac. [NCERT Exemplar]
Answer:
Two synergids and an egg.
Question 12.
Name the common function that cotyledons and nucellus perform. [NCERT Exemplar]
Answer:
Cotyledons and nucellus provide nourishment.
Question 13.
In the embryos of a typical dicot and a grass, which are the true homologous structures? [NCERT Exemplar]
Answer:
Cotyledons and scutellum.
Question 14.
In a case of polyembryony, if an embryo develops from the synergid and *another from the nucellus, which is haploid and which is diploid? [NCERT Exemplar]
Answer:
Synergid embryo is haploid and nucellar embryo is diploid.
Short answer type questions
Question 1.
Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer:
Exine is made up of sporopollenin and intine is made up of cellulose and pectin.
Due to the sporopollenin, exine can withstand high temperature and strong acids. It is also not affected by enzymes. It is because of this reason that pollen grains are well preserved as fossils.
Question 2.
Differentiate between geitonogamy and xenogamy in plants. Which one between the two will lead to inbreeding depression and why?
Answer:
| Geitonogamy | Xenogamy |
| 1. It is transfer of pollen grains from the anther to the stigma of another flower of same plant. | It is transfer of pollen grains from the anther to the stigma of a different plant. |
| 2. The pollen grains are genetically similar to the plant. | The pollen grains are genetically different from the plant. |
Geitonogamy will lead to inbreeding depression because the pollen grains are genetically similar, which results in inbreeding. Continued inbreeding will thus reduce fertility and productivity.
Question 3.
Double fertilisation is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilisation events that are responsible for it.
Answer:
The development of endosperm (preceding the embryo) takes place from primary endosperm nucleus (PEN) in both, castor and groundnut. The developing embryo derives nutrition from endosperm.
PEN undergoes repeated division to give free nuclei. Subsequently cell wall is formed and endosperm becomes cellular. At this stage endosperm is retained in castor or is not fully consumed but in groundnut endosperm is consumed by growing embryo.
Question 4.
Differentiate between albuminous and non-albuminous seeds, giving one example of each.
Answer:
- Albuminous seeds have residual endosperm ip them. For example, maize.
- Non-albuminous seeds do not have any residual endosperm. For example, pea.
Question 5.
A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person?
Answer:
In apple only the thalamus (along with ovary) portion contributes to fruit. Therefore, it is a false fruit. Mango develops only from the ovary, therefore it is a true fruit.
Banana develops from ovary but without fertilisation. The method is known as parthenocarpy. Since there is no fertilisation, no seeds are formed.
Question 6.
Why are some seeds referred to as apomictic seeds? Mention one advantage and one disadvantage to a farmer who uses them.
Answer:
Seeds produced without fertilisation are referred to as apomictic.
Advantage: Desired characters are retained in offspring (progeny) as there is no segregation of characters in offspring (progeny). Seed production is assured in absence of pollinators.
Disadvantage: Cannot control accumulation of deleterious genetic mutation. These are usually restricted to narrow ecological niches and lack ability to adapt to changing environment.
Long answer type questions
Question 1.
A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
Answer the following questions giving reasons:
(a) How many ovules are minimally involved?
(b) How many megaspore mother cells are involved?
(c) What is the minimum number of pollen grains that must land on stigma for pollination?
(d) How many male gametes are involved in the above case?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of another in the above case?
Answer:
(a) 360 ovules are involved. One ovule after fertilisation forms one seed.
(b) 360 megaspore mother cells are involved. Each megaspore mother cell forms four megaspores out of which only one remains functional.
(c) 360 pollen grains. One pollen grain participates in fertilisation of one ovule.
(d) 720 male gametes are involved. Each pollen grain carries two male gametes (which participate in double fertilisation) (360 × 2 = 720).
(e) 90 microspore mother cells undergo reduction division. Each microspore mother cell meiotically divides to form four pollen grains (360/4 = 90).