The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (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. Their 4 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 5th child 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 the parents.) O PoFF (female) x ppfF (male) O Poft (female) x ppft (male) ppft female) x PPFI (malel O PPF female) x ppF (malel 2. What is/are the target genotypes? O pof. OPF. O pptt OP.M 3. 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 Sth child will have at least one of these conditions? Answer to two decimal places (es. 0.88).

Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
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Chapter10: From Proteins To Phenotypes
Section: Chapter Questions
Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...
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The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant
trait (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. Their
4 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 5th
child 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 the
parents.)
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 pof
OP.F.
O pott
O P.M
3. 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 to
two decimal places (eg. 0.88).
Transcribed Image Text:The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (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. Their 4 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 5th child 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 the parents.) 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 pof OP.F. O pott O P.M 3. 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 to two decimal places (eg. 0.88).
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