expand icon
book Human Heredity 9th Edition by Michael Cummings cover

Human Heredity 9th Edition by Michael Cummings

النسخة 9الرقم المعياري الدولي: 978-0538498821
book Human Heredity 9th Edition by Michael Cummings cover

Human Heredity 9th Edition by Michael Cummings

النسخة 9الرقم المعياري الدولي: 978-0538498821
تمرين 1
Jane, a healthy woman, was referred for genetic counseling because she had two siblings, a brother Matt and a sister Edna, with cystic fibrosis who died at the ages of 32 and 16, respectively. Jane's husband, John, has no family history of cystic fibrosis. Jane wants to know the probability that she and John will have a child with cystic fibrosis. The genetic counselor used the Hardy-Weinberg model to calculate the probability that this couple will have an affected child.
The counselor explained that there is a two-in-three chance that Jane is a carrier for the mutant CFTR allele; she used a Punnett square to illustrate this. The probability that John is a carrier is equal to the population carrier frequency (2pq). The probability that John and Jane will have a child who has cystic fibrosis equals the probability that Jane is a carrier (2/3), multiplied by the probability that John is a carrier (2pq), multiplied by the probability that they will have an affected child if they are both carriers (1/4).
Using the heterozygote frequency for cystic fibrosis among white Americans to estimate the probability that John is a carrier, what is the likelihood that their child would have the disease?
Jane, a healthy woman, was referred for genetic counseling because she had two siblings, a brother Matt and a sister Edna, with cystic fibrosis who died at the ages of 32 and 16, respectively. Jane's husband, John, has no family history of cystic fibrosis. Jane wants to know the probability that she and John will have a child with cystic fibrosis. The genetic counselor used the Hardy-Weinberg model to calculate the probability that this couple will have an affected child.  The counselor explained that there is a two-in-three chance that Jane is a carrier for the mutant CFTR allele; she used a Punnett square to illustrate this. The probability that John is a carrier is equal to the population carrier frequency (2pq). The probability that John and Jane will have a child who has cystic fibrosis equals the probability that Jane is a carrier (2/3), multiplied by the probability that John is a carrier (2pq), multiplied by the probability that they will have an affected child if they are both carriers (1/4).  Using the heterozygote frequency for cystic fibrosis among white Americans to estimate the probability that John is a carrier, what is the likelihood that their child would have the disease?
التوضيح
موثّق
like image
like image

The probability that JA is a carrier is ...

close menu
Human Heredity 9th Edition by Michael Cummings
cross icon