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Question
Out of a population of 800 individuals in F2 generation of a cross between yellow round and green wrinkled Pea plants, what would be number of yellow and wrinkled seeds
Out of a population of 800 individuals in F2 generation of a cross between yellow round and green wrinkled Pea plants, what would be number of yellow and wrinkled seeds
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Solution
The correct option is C 150
A cross is made between homozygous yellow round seeds(YYRR) to another homozygous green wrinkled seeds(yyrr). In the F1 generation, all plants are heterozygous yellow round(YyrRr). When this plant is subjected to self-pollination, in the F2 generation, 9 plants are Yellow round, 3 plants are yellow wrinkled, 3 plants are green round and 1 plant is green wrinkled are formed.
The dihybrid phenotypic ratio is 9:3:3:1. So, in the dihybrid cross, 800 individuals are formed, in this 450 are yellow round 150 are yellow wrinkled, 150 are green round, 50 are green wrinkled.
So, the correct option is ‘150’.
A cross is made between homozygous yellow round seeds(YYRR) to another homozygous green wrinkled seeds(yyrr). In the F1 generation, all plants are heterozygous yellow round(YyrRr). When this plant is subjected to self-pollination, in the F2 generation, 9 plants are Yellow round, 3 plants are yellow wrinkled, 3 plants are green round and 1 plant is green wrinkled are formed.
The dihybrid phenotypic ratio is 9:3:3:1. So, in the dihybrid cross, 800 individuals are formed, in this 450 are yellow round 150 are yellow wrinkled, 150 are green round, 50 are green wrinkled.
So, the correct option is ‘150’.Suggest Corrections
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