Science, asked by manikandanmaru69, 9 months ago

A dihybrid plants on self polination produced 400 phenotypes with 9 types of genotypes. the number of seeds having genotype RrYy is

Answers

Answered by mazumdarsuraj42

1

Answer:

The self pollination of a dihybrid plant yields the genotypic ratio of 1 TTRR : 2 TTRr : 1 TTrr : 2 TtRR : 4TtRr : 2 Ttrr : 1ttRR : 2 ttRr : 1 ttrr. The ratio of genotype TtRr is 4/16. Hence, out of total 400 phenotypes, number of seeds having TtRr genotype will be 4/16x400 = 100. Option "B" is correct.

F

1

generation : TtRr x TtRr

Gametes :

TtRr -->

TtRr TR Tr tR tr

TR TTRR TTRr TtRR

TtRr

Tr TTRr TTrr TtRr Ttrr

tR TtRR TtRr ttRR ttRr

tr TtRr Ttrr ttRr ttrr

Explanation:

Answered by guhabhavani

0

Answer:

The self pollination of a dihybrid plant yields the genotypic ratio of 1 TTRR : 2 TTRr : 1 TTrr : 2 TtRR : 4TtRr : 2 Ttrr : 1ttRR : 2 ttRr : 1 ttrr. The ratio of genotype TtRr is 4/16. Hence, out of total 400 phenotypes, number of seeds having TtRr genotype will be 4/16x400 = 100. Option "B" is correct.

F 1 generation : TtRr x TtRr

Gametes :

TtRr -->

TtRr TR Tr tR tr

TR TTRR TTRr TtRR

TtRr

Tr TTRr TTrr TtRr Ttrr

tR TtRR TtRr ttRR ttRr

tr TtRr Ttrr ttRr ttrr

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