Final answer:
Expanding (R+W)^2 in a genetic context results in R^2 + 2RW + W^2, representing possible genotypes. Given all offspring from a test cross have round peas, it suggests the round pea parent is more likely RR than Rr. A male cannot be a carrier for red-green color blindness, and a Punnett square between RrYY and rrYy requires 4 squares.
Step-by-step explanation:
When expanding the expression (R+W)^2, which represents the possible pairings of two genes (R for round peas and W for wrinkled peas), we use the binomial expansion method. The expansion of this expression results in R^2 + 2RW + W^2. This is analogous to a Punnett square, where each term in the expansion represents the probability of a particular genotype in the offspring if these genes were independently assorting.
Applying this to a test cross scenario, if a plant with wrinkled peas (rr) is crossed with a plant that has round peas (genotype either RR or Rr), and all the offspring have round peas, it is likely—but not certain—that the round pea parent plant is homozygous dominant (RR). If the round pea parent plant is heterozygous (Rr), the probability that all three progeny will display the round pea phenotype is (3/4)^3, since each offspring has a 3/4 chance of being round if round is dominant.
A human male cannot be a carrier of red-green color blindness because males have only one X chromosome. Red-green color blindness is an X-linked recessive condition, meaning that if a male has the allele for red-green color blindness, he will be color blind.
In the cross between RrYY and rrYy pea plants, the possible genotypes for the offspring will be RrYy, RrYy, rrYy, and rrYy, leading to phenotypes of round yellow peas and wrinkled yellow peas. A Punnett square for this cross will require 4 squares, as there are two possible alleles for round/wrinkled (R or r) being combined with two possible alleles for yellow/green (Y or y).