Question

A cross was made between pure breeding pea plants one with round and green seeds and other with wrinkled a And yellow seeds a) write the phenotype of F1 progeny. Give reason for your answer b) write the different types of F2 progeny obtained along with their ratio when F1 progeny was served

Answer 1

A) The F1 progeny would express the phenotype of round and green seeds. This is because the cross consists of a homozygous dominant ["pure breeding pea plant" means it is homozygous dominant and has the genotype of RRYY] and a homozygous recessive plant [This can be concluded since it is a cross test which mixes a plant with a homozygous recessive plant which has a geneotype of rryy] which would create a heterozygous offspring. This heterozygous offspring will express the traits of the homozygous dominant parent. [This is because dominant traits always show their characteristics over recessive traits.]

B) The F2 progeny would consist of 16 offsprings which express a 9:3:3:1 ratio which means there would be 9 round and green seeds, 3 round and yellow seeds, 3 wrinkled and green seeds, 1 wrinkled and yellow seeds.

[Note to solve this you would have to determine the ratio for the first trait and the second trait and multiply them to determine the phenotypic ratio of the cross. In this case it was 3 green : 1 yellow, which was determined by crossing Yy x Yy in a punnet square, multiplied by 3 round : 1 wrinkled, which was determined by crossing Rr x Rr, would give you the ratio of 9:3:3:1. In case you do not know how to multiply ratios: Think of it like foil where you do (3:1) x (3:1), you would multiply 3 and 3 to get 9 green and round, 3 and 1 to get 3 green and wrinkled, 1 and 3 to get 3 yellow and round, 1 to 1 to get 1 yellow and wrinkled.]