Final answer:
The cross between a pea plant with genotype RrYy and one with genotype rrYy can produce offspring with four potential genotypes and three possible phenotypes. A Punnett square with 16 squares is used to analyze the outcome of the dihybrid cross.
Step-by-step explanation:
When a pea plant with genotype RrYy is crossed with a plant with genotype rrYy, we can predict the outcome of the seed phenotypes using a Punnett square analysis. To do this, we need to account for the alleles for seed shape and color.
For each trait, the parent plant RrYy can produce gametes with alleles RY, Ry, rY, and ry, while the parent plant rrYy can produce gametes with alleles rY and ry. Combining these gametes produces offspring with four potential genotypes: RrYy, Rryy, rrYy, and rryy. This results in three possible phenotypes: round and yellow, round and green, and wrinkled and yellow, since there are no homozygous dominant RR alleles in the offspring, and hence no chance of wrinkled and green peas (rryy).
To analyze this cross using a Punnett square, we need 16 squares, which result from the combination of the four possible gametes from the RrYy parent with the two possible gametes from the rrYy parent. This elaboration in phenotype possibilities is typical in a dihybrid cross.