A dihybrid plant on self pollination, produced 400 phenotypes with 9 types of genotype. How many seeds will have genotype TtRr?
1. 200
2. 100
3. 50
4. 150
Dear student,
F1 generation: TtRr * TtRr
Gametes:
The genotypic ratio of a dihybrid cross is 1TTRR:2 TTRr :1 TTrr: 2 tTRR: 4tTrR: 2 tTrr :1 ttRR : 2ttRr :1 ttrr, the total of which is 16.
The ratio of genotype TtRr is 4.
therefore, number of seeds having TtRr genotype in 400 phenotypes would be 4/16*400=100 seeds
Hence, the correct answer to this question would be option B- 100 seeds.
Regards
F1 generation: TtRr * TtRr
Gametes:
| TtRr | Tr | TR | tR | tr |
|---|---|---|---|---|
| TR | TTRr | TTRR | TtRR | TtrR |
| tR | tTrR | TtRR | ttRR | ttrR |
| tr | Ttrr | TtRr | ttRr | ttrr |
The genotypic ratio of a dihybrid cross is 1TTRR:2 TTRr :1 TTrr: 2 tTRR: 4tTrR: 2 tTrr :1 ttRR : 2ttRr :1 ttrr, the total of which is 16.
The ratio of genotype TtRr is 4.
therefore, number of seeds having TtRr genotype in 400 phenotypes would be 4/16*400=100 seeds
Hence, the correct answer to this question would be option B- 100 seeds.
Regards