Final answer:
To determine the fraction of offspring predicted to exhibit recessive phenotypes for at least two of three characters from the cross PpYyRr x Ppyyrr, we calculated the probability for each gene separately and combined the results. Three scenarios meet the condition, each with a probability of 1/8, leading to a final probability of 3/8 for offspring to exhibit at least two recessive phenotypes.
Step-by-step explanation:
The question asks what fraction of offspring from the cross PpYyRr x Ppyyrr would exhibit recessive phenotypes for at least two of the three characters. We need to address each gene separately using the probability method, and then combine the probabilities accordingly.
For the first gene (P or p), the cross Pp x pp will produce offspring with a 1/2 chance of being heterozygous (Pp) and a 1/2 chance of being homozygous recessive (pp). Only pp displays the recessive phenotype.
For the second gene (Y or y), the cross Yy x yy will also produce offspring with a 1/2 chance of being heterozygous (Yy) and a 1/2 chance of being homozygous recessive (yy). Again, only yy displays the recessive phenotype.
For the third gene (R or r), the cross Rr x rr will follow the same pattern as the other genes, with a 1/2 chance for heterozygous offspring (Rr) and a 1/2 chance for homozygous recessive offspring (rr), with rr showing the recessive phenotype.
To have at least two recessive phenotypes, we can have the following scenarios: pp yy Rr, pp Yy rr, or Pp yy rr. To find the probability for each scenario, we multiply the probabilities for each individual gene, resulting in 1/2 x 1/2 x 1/2 = 1/8. Since there are three possible scenarios that fit our condition, we multiply 1/8 by 3, and the final probability is 3/8.
Therefore, the fraction of offspring displaying at least two recessive phenotypes is 3/8.