In garden peas, the yellow seed trait is dominant over green seeds. Also, the round seed trait is dominant over the wrinkled-seed trait. These genes are on separate chromosomes. Suppose that two plants that are heterozygous for both traits are crossed and 720 offspring are produced. About how many progeny will be yellow with round seeds? Select one:

a. 405

b. 180

c. 360

d. 60

e. 540

405

Explanation:

Gregor Mendel worked with pea plant and observed the inheritance of seed colour and shape in the plant. This is a dihybrid cross involving two genes; one coding for seed colour and the other for seed shape. The allele for yellow seed (Y) is dominant over green (y) in the first gene, while the allele for round seed (R) is dominant over wrinkled seed (r) in the second gene.

When Mendel crossed truebreeds i. e. homozygous parents, he got a F1 generation which is heterozygous for both traits (YyRr). He crossed two heterozygous plants, and according to the law of independent assortment, four possible chromosome combinations of gametes were produced viz: YR, Yr, yR, yr.

Mendel discovered that from a cross involving two hybrid genes, 16 variations were possible. In the genetic make-up, the kinds of F2 offsprings possible from this dihybrid cross are; Yellow round, Yellow wrinkled, Green round and Green wrinkled in a theoretical ratio 9:3:3:1 respectively.

According to the question, 720 offsprings were produced. Therefore, to get the number of progeny with yellow round seeds, the ratio will be used.

The ratio of yellow round seeds in Mendel's dihybrid cross is 9/16, hence, among 720 offsprings, we will have about - 9/16 * 720 = 405 pea plants with yellow and round seeds.