Final answer:
Both parents must have the homozygous recessive genotype (ss) for all their offspring to express the recessive rough kernel phenotype in corn. In the context of pea plants, a Punnett square with 16 squares would be needed to analyze the cross of RrYY and rrYy, and the genotype of a round pea plant can't be determined from three round pea offspring without further crossing.
Step-by-step explanation:
If we want progeny (offspring) that display only rough kernel phenotype in corn, which is recessive, both parents must have the genotype that provides this trait. In genetics, a recessive trait, like the rough kernel in corn, is only expressed when the individual has two copies of the recessive allele. Therefore, if 100% of the offspring have rough kernels, it means both parents carried two copies of the recessive allele, making their genotypes homozygous recessive (ss).
In a different example, considering pea plants where the round seed shape (R) is dominant to wrinkled seed shape (r), a cross between RrYY and rrYy would require a Punnett square with 16 squares to analyze the possible genotypes and phenotypes of the offspring. When conducting a test cross where one parent has a known recessive genotype (rr) and the other parent has an unknown genotype but a dominant phenotype (round seeds), the appearance of any offspring with a recessive phenotype would indicate that the unknown parent was heterozygous (Rr). If all offspring have round peas, you cannot confirm the genotype of the unknown parent without further crossing.