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

This problem involves a dihybrid cross. The parent generation has genotypes \( RRyy \) (round, green) and \( rryy \) (wrinkled, yellow). The \( F_1 \) generation is \( RrYy \) (round, yellow).

The \( F_2 \) generation phenotypic ratio is:

• 9 round yellow
• 3 round green
• 3 wrinkled yellow
• 1 wrinkled green

Therefore, to observe at least one plant with each phenotype, you would need a minimum of 16 plants. The question asks for the minimum number to observe wrinkled green seeds. Therefore, at a minimum, you would need 1 plant.