Final answer:
The unknown genotype for smooth peas that produces half smooth and half wrinkled offspring when crossed with wrinkled (rr) peas must be heterozygous (Rr). If the parent plant is heterozygous, the probability of getting three round pea offspring in a single sample is 27/64.
Step-by-step explanation:
In genetics, a test cross is used to determine the genotype of an individual expressing a dominant trait when crossed with an individual expressing a known recessive trait. In this case, smooth peas have a dominant phenotype and wrinkled peas have a recessive phenotype (rr).
Considering the scenario where the unknown genotype for round peas produces offspring that are half smooth and half wrinkled when crossed with rr peas, it suggests that the unknown plant is heterozygous (Rr). This is because a homozygous dominant plant (RR) would not produce any wrinkled progeny when crossed with a rr plant, as R is dominant over r and all offspring would display the dominant round phenotype. On the other hand, a heterozygous plant (Rr) crossed with a homozygous recessive plant (rr) would produce offspring with a 1:1 ratio of round to wrinkled peas, explaining the observed results.
If the round pea parent plant is heterozygous (Rr), the probability that a random sample of three progeny peas would all be round can be calculated. Since the round pea (R) is dominant over the wrinkled pea (r), each offspring has a 3/4 chance of being round. Therefore, the probability of all three being round is (3/4) × (3/4) × (3/4) = 27/64.