Question

Two true-breeding pea plants were crossed. One parent is round, terminal, violet, constricted, while the other expressed the respective contrasting phenotypes of wrinkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the F1 only round, axial, violet, and full were expressed. In the F2, all possible combinations of these traits were expressed in ratios consistent with Mendelian inheritance. In the F2 generation, how often is either of the P1 phenotypes likely to occur? Express your answer as a fraction

Answer 1

Answer:

9/256

Explanation:

Parental generation: two true-breeding plants (homozygous for all genes).

Round, terminal, violet, constricted X wrinkled, axial, white, full

F1: Homogeneous round, axial, violet, and full

According to Mendel's Law of Dominance, the traits observed in the phenotype of F1 are dominant.

Therefore, the genotypes of the parents are:

P: RRaaVVff x rrAAvvFF

F1: RrAaVvFf

The phenotypes of the P1 individuals that we're looking for in the F2 as a result of crossing F1xF1 are:

• R_aaV_ff
• rrA_vvF_

The genes are independent, so we can use Mendel's laws individually to predict the expected phenotypes for each gene, and then multiply all the probabilities to obtain the probability of having the P1 phenotypes.

Phenotype R_aaV_ff

• Rr x Rr ---> 3/4 R_ and 1/4 rr
• Aa x Aa --> 3/4 A_ and 1/4 aa
• Vv x Vv --> 3/4 V_ and 1/4 vv
• Ff x Ff --> 3/4 F_ and 1/4 ff

Probability of R_aaV_ff  = 3/4 × 1/4 × 3/4 × 1/4 = 9/256

Phenotype rrA_vvF_

• Rr x Rr ---> 3/4 R_ and 1/4 rr
• Aa x Aa --> 3/4 A_ and 1/4 aa
• Vv x Vv --> 3/4 V_ and 1/4 vv
• Ff x Ff --> 3/4 F_ and 1/4 ff

Probability of rrA_vvF_  = 1/4 × 3/4 × 1/4 × 3/4 = 9/256