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:
Step 1: Understanding the Genotypes of the Parents:
The parent generation consists of two individuals with the following genotypes: - Parent 1: \( RRyy \) (round, green) - Parent 2: \( rryy \) (wrinkled, yellow) The round shape (R) is dominant over wrinkled (r), and the green color (Y) is dominant over yellow (y).
Step 2: Determining the Genotype of the F1 Generation:
When these two parents are crossed, the F1 generation will inherit one allele from each parent for each gene. The F1 genotype will be \( RrYy \), which results in round yellow seeds because round (R) and yellow (Y) are dominant traits.
Step 3: F2 Generation and Phenotypic Ratio:
The F1 generation is crossed with another F1 individual, resulting in the following phenotypic ratio for the F2 generation: - 9 round yellow - 3 round green - 3 wrinkled yellow - 1 wrinkled green
Step 4: Minimum Number of Plants to Observe Each Phenotype:
To observe at least one plant of each phenotype, you would need a minimum of 16 plants, as this is the total number of plants that would provide all four phenotypic combinations based on the ratio (9:3:3:1).
Step 5: Minimum Number of Plants to Observe Wrinkled Green Seeds:
The phenotype "wrinkled green" occurs in 1 out of every 16 plants. Therefore, at a minimum, you would need 1 plant to observe this phenotype. It is expected to appear in the F2 generation in a 1:16 ratio.
Conclusion:
The minimum number of plants required to observe at least one wrinkled green seed is 1, based on the phenotypic ratio of the F2 generation.
