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Cystic fibrosis is an autosomal recessive genetic disorder resulting from the absence of a functional transmembrane...

Cystic fibrosis is an autosomal recessive genetic disorder resulting from the absence of a functional transmembrane conductance regulator (CFTR) protein. A woman whose older sister has a son with the disorder is concerned that she will also have a child with cystic fibrosis. The woman’s husband is healthy but he had a paternal uncle who died of cystic fibrosis as a teenager.
   A. ​Based on the information given above, draw a pedigree for the couple and the extended families described above. Here is some additional information about both families: Woman - has two siblings, the sister with the affected son and a younger brother who does not have children. Neither of the woman’s parents or her siblings have the disease. The sister has a healthy daughter and the sister’s husband is also healthy. Husband – has one unaffected older brother who has two unaffected daughters. Both of his parents are unaffected. His father’s older brother is the uncle that died from cystic fibrosis. Include as many generations and individuals as you can in your pedigree. For simplicity, assume that individuals homozygous for a non-functional CFTR protein do not live long enough to reproduce.   
B. What is the probability that this woman and her husband will have a child with cystic fibrosis?
Assume that only one of the woman’s parents is a carrier of the recessive allele for cystic fibrosis. Also assume that her husband’s mother has two normal alleles of the CFTR gene.

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Expert Solution

Ans A) the Pedigree of given family from the given information is as below :

Alleles = F = Dominant = normal or unaffected ; f = recessive = affected in ff (Homozygous Recessive Condition).

Ans B) Probability that this woman and her husband will have a child with Cystic Fibrosis ( ff). In the question the information about their parents Genotype is also given as either of woman's Parents have a recessive allele (f) and the husband's mother is Homozygous for Dominant allele of CFTR gene = FF. The husband's Father could be FF or Ff as he is healthy and has an elder brother who died of Cystic Fibrosis. So, he can be FF or Ff and hence with wife FF he can have child with Genotypes FF or FF and Ff. We have to find the Probability of having affected child of woman and husband then husband will be a carreir by Probability 1/2 as ( 1 = Ff ; 2 = FF + Ff).

Mother carrier by = 2/4 or 1/2

Father = 1/2

Child can be affected by = 1/4

Probability of affected child = 2/4 * 1/2 * 1/4

= 1/16.

gladiator answered 1 year ago
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