Question

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 anther in the above case?

Answer 1

The number of viable seeds produced by the brinjal plant through sexual reproduction = 360

(a) The number of ovules minimally involved in this process would be 360, as the number of viable seeds is 360. After fertilisation, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules are corresponding to the number of seeds formed.

(b) During the process of gametogenesis, 360 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.

(c) The minimum number of pollen grains that must land on stigma for pollination are 360 because each pollen grain contains two male gametes. Out of theses two gametes, one fuses with polar nuclei and forms endosperm, while, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds. Therefore, in order to obtain 360 seeds, number of pollen grains needed would be 360.

(d) The number of male gametes involved in seed formation would be 360 as each male gamete will fuse with one egg nuclei to form zygote, which will further give rise to the seed.

(e) In the above case, 90 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 360 microspores, there must be 90 microspore mother cells.