Final answer:
The genetic outcomes for two corn plants both heterozygous for yellow seeds and smooth seeds result in a 9:3:3:1 phenotypic ratio. To analyze the pea plants crosses mentioned, a Punnett square with 16 squares and considerations of dominant and recessive traits are needed. With a small sample of three offspring, we cannot conclusively determine the genotype of a plant producing round peas, but if heterozygous, the probability all three offspring will show the dominant trait is 1/8.
Step-by-step explanation:
Predicting the Outcome of a Genetic Cross
When predicting the outcome between two corn plants that are heterozygous for yellow seeds (Y being dominant) and smooth seeds (S being dominant), we need to consider both traits. The Punnett square analysis for a dihybrid cross where both parents are heterozygous (YySs) would show that the genotypic ratio is 9:3:3:1 for the phenotypes yellow smooth, yellow wrinkled, purple smooth, and purple wrinkled, respectively.
In the case of pea plants, for a cross between a plant with the genotype RrYY (round yellow peas) and rrYy (wrinkled yellow peas), the Punnett square would require 16 squares to show all possible genotype outcomes. The possible genotypes for this cross would be RrYY, RrYy, rrYY, and rrYy, resulting in phenotypes of round yellow peas and wrinkled yellow peas.
If you were to perform a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype with round peas, obtaining three offspring all with round peas could suggest that the parent plant with round peas is homozygous dominant, but without more offspring, we cannot be certain. If the parent plant is heterozygous (Rr), the probability of all three offspring having round peas is 1/2 * 1/2 * 1/2 = 1/8.