Question

A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.

Answer the following questions giving reasons:

(i) How many ovules are minimally involved ?

(ii) How many megaspore mother cells are involved ?

(iii) What is the minimum number of pollen grains that must land on stigma for pollination ?

(iv) How many male gametes are involved in the above case ?

(v)  How many microspore mother cells must have undergone reduction division prior to dehiscence of another in the above case ?

Answer 1

(i) 360 ovules are involved. One ovule after fertilisation forms one seed.

(ii) 360 MMC are involved. Each MMC forms four megaspores out of which only one remains functional.

(iii) 360 pollen grains. One pollen grains participates in fertilisation of one ovule.

(iv) 720 male gametes are involved. Each pollen grain carries two male gametes (which participate in double fertilisation) (360 × 2 = 720).

(v) 90 MMC undergo reduction division. Each microspore mother cell meiotically divides to form four pollen grains. (360/4 = 90).