Question

Two pea plants, one round green (\(RRyy\)) seeds and another with wrinkled yellow (\(rrYY\)) seeds produce \(F_1\) progeny that has round yellow (\(RrYy\)) seeds. If the \(F_1\) progeny plants are selfed, then the minimum number of plants of the \(F_2\) progeny will have the following observable characters:

• A. Wrinkled green
• B. Wrinkled yellow
• C. Round green
• D. Round yellow

Answer 1

The Correct Option is A

Parental Genotypes:

Parent 1: \( RRyy \) (round, green)
Parent 2: \( rryy \) (wrinkled, yellow)
Dominance: R>r (round>wrinkled), Y>y (green>yellow).

F1 Generation Genotype:

Cross of parents yields F1 generation with genotype \( RrYy \). This results in round, green phenotype due to dominant alleles.

F2 Generation Phenotypic Ratio:

Self-cross of F1 (\( RrYy \times RrYy \)) produces the F2 generation with the following phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Minimum Plants for All Phenotypes:

To observe at least one plant of each of the four phenotypes, a minimum of 16 plants are required, reflecting the total parts in the 9:3:3:1 ratio.

Minimum Plants for Wrinkled Green Seeds:

The wrinkled green phenotype appears with a frequency of 1 in 16. Therefore, a minimum of 1 plant is needed to observe this specific phenotype.

Conclusion:

Based on the F2 phenotypic ratio, 1 plant is the minimum required to observe at least one wrinkled green seed.