47th one
Q47. 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 mega spore mother cells are involved?
(c) What is the minimum number of pollen grains that must land on the stigma for pollination?
(d) How many male gametes are involves in above case?
(e) How many micro spore mother cells must have under reduction division prior to dehiscence of anther in the above case?
Dear student,
The flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
a)The minimal number of ovules involved will be 360 as the number of viable seeds is 360 because after the process of fertilization ovules form seeds and ovary forms fruit.So, the number of ovules formed will be equal to the number of seeds.
b)The number of megaspore mother cells involved will be 360 because during the process of gametogenesis only one megaspore of the tetrad is functional and remaining degenerate.
c)The minimum number of pollen grains that must land on stigma will be 360 because of each pollen grain contains 2 gametes in which one fuses with polar nuclei and form endosperm and other fuses with the egg cell to form the zygote which ultimately forms the seed.
d)The number of male gametes involved in this case will be 360 as each male gamete fuses with the one egg nuclei to form the zygote, which will further give rise to seed.
e)In this case, 90 microspore mother cell must have undergone reduction division prior to the dehiscence of the anther. This is because each microspore mother cell will give rise to 4 microspores. Thus, for 360 viable seed formation, we require 90 microspore mother cell.
Regards
The flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
a)The minimal number of ovules involved will be 360 as the number of viable seeds is 360 because after the process of fertilization ovules form seeds and ovary forms fruit.So, the number of ovules formed will be equal to the number of seeds.
b)The number of megaspore mother cells involved will be 360 because during the process of gametogenesis only one megaspore of the tetrad is functional and remaining degenerate.
c)The minimum number of pollen grains that must land on stigma will be 360 because of each pollen grain contains 2 gametes in which one fuses with polar nuclei and form endosperm and other fuses with the egg cell to form the zygote which ultimately forms the seed.
d)The number of male gametes involved in this case will be 360 as each male gamete fuses with the one egg nuclei to form the zygote, which will further give rise to seed.
e)In this case, 90 microspore mother cell must have undergone reduction division prior to the dehiscence of the anther. This is because each microspore mother cell will give rise to 4 microspores. Thus, for 360 viable seed formation, we require 90 microspore mother cell.
Regards