Question

out of a population of 800 f2 individuals of a cross between yellow round and green wrinkled p returns the number of yellow wrinkled seeds would be 150 to 200 402 800

Answer 1

150 to 200 as the phenotype ratio will be 9:3:3:1

Answer 2

Answer:

On following the mendel’s laws of inheritance for a dihybrid cross we can conclude that the answer would be 150 to 200 pea plants of yellow wrinkled pea plants in the f2 generation.  

Assuming that they are homozygous, we would see that the f1 population of pea plants will have pea plants in the ratio of 3:1 for yellow and round seed: green and wrinkled seed. In the f2 progeny, we will see that the ratio is 9:3:3:1 for the phenotypes, yr:yw:gr:gw. All this can be understood on making a simple punnet square.