The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominanttrait (polydactyly - P; wildtype - p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis- f; wildtype - F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD.You will walk through a series of steps to answer this question: What is the probability that their 5thchild will have at least one of these conditions? Here is the first step:1. What is the cross? (Hint: You can use the 4 existing children to determine the genotypes of theparents.)O PpFF (female) x ppFF (male)O PoFf (female) x ppft (male)O pof female) x PPFI (male)O PPFF Ifemale) x ppFf (male)2. What is/are the target genotypes?O pofOP.F.O pottO P.M3. What is the probability the child will have PD AND cystic fibrosis? Answer to two decimal places(eg. 0.88).4. What is the probability that their 5th child will have at least one of these conditions? Answer totwo decimal places (eg. 0.88).

Name: The next four questions are all related to this problem: Polydactyly
Uploaded: 2024-03-20T06:42:55.000Z
Channel: Bartleby
Description: The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominanttrait (polydactyly - P; wildtype - p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis- f; wildtype - F). A PD woman, otherwise n