Final answer:
When a pea plant of genotype RrYy is crossed with one of genotype rrYy, the offspring are predicted to have a phenotypic ratio of 3 round yellow : 1 round green : 3 wrinkled yellow : 1 wrinkled green, which indicates that the correct answer is option C.
Step-by-step explanation:
The cross between a pea plant with genotype RrYy and one with genotype rrYy will produce offspring with different combinations of these alleles. In pea plants, round seed shape (R) is dominant to wrinkled seed shape (r), and yellow seeds (Y) are dominant to green seeds (y). We can predict the phenotypic outcome by creating a Punnett square.
To solve this cross, we need to understand how each trait segregates independently (Law of Independent Assortment). The RrYy parent can produce gametes that are RY, Ry, rY, and ry. The rrYy parent can produce gametes that are rY and ry. By combining these gametes, we can determine the ratio of phenotypes in the offspring.
Here is the Punnett square:
- Ry x rY = RrYy (round, yellow)
- Ry x ry = Rryy (round, green)
- ry x rY = rrYy (wrinkled, yellow)
- ry x ry = rryy (wrinkled, green)
From the Punnett square we can see that the phenotypes of the offspring will be:
- 6/16 round yellow
- 2/16 round green
- 6/16 wrinkled yellow
- 2/16 wrinkled green
So the expected phenotypic ratio will be:
3 round yellow : 1 round green : 3 wrinkled yellow : 1 wrinkled green
Therefore, the correct answer is option C.