Final answer:
The pea plant RrYy can produce sperm with RY, Ry, rY, or ry alleles. For a cross between RrYY and rrYy plants, a 16-square Punnett square is needed. If all offspring from a test cross have round peas, the round pea parent may be homozygous RR but further breeding tests are needed for confirmation.
Step-by-step explanation:
The question pertains to genetic inheritance patterns in pea plants and entails the analysis of potential genotypic and phenotypic results from specified genetic crosses. In the given scenario where a pea plant has a genotype RrYy, the possible alleles in its sperm cells could include RY, Ry, rY, and ry. Each combination represents a potential allele pairing that the sperm can contribute to the offspring upon fertilization. These correspond to the characteristics described in Mendel's inheritance patterns: R (round seeds) is dominant over r (wrinkled seeds), and Y (yellow peas) is dominant over y (green peas).
When considering the crosses, in the case of a cross between the genotypes RrYY and rrYy, the possible genotypic outcomes include RrYy (yellow and round seeds), Rryy (yellow and round seeds), rrYy (yellow and wrinkled seeds), and rryy (green and wrinkled seeds). The Punnett square analysis for this cross would require 16 squares since each parent can produce four different types of gametes (2 for each trait).
In case of a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype with round peas, if all three offspring have round peas, it suggests that the unknown parent is likely to be homozygous dominant (RR) as no wrinkled peas appeared in the offspring. However, if the round pea parent plant was heterozygous (Rr), the probability that a random sample of three progeny peas would all be round is 0.5^3, since each offspring has a 1/2 chance of being round from a heterozygous parent. But since the sample size is small, it is not conclusive.