A young couple has undergone a successful in vitro fertilization procedure. The father has cystic fibrosis and the mother has a sister with cystic fibrosis. The father as well as the mother's sister are both known to have ΔF508 mutations, but the mother's carrier status is unknown. Before making the decision to conceive, the couple underwent extensive genetic counseling regarding the potential risks of having a child with cystic fibrosis. The family pedigree is diagrammed below with the unborn child marked by the red arrow. What is the probability that the unborn child will have cystic fibrosis?

Question

A young couple has undergone a successful in vitro fertilization procedure.  The father has cystic fibrosis and the mother has a sister with cystic fibrosis.  The father as well as the mother's sister are both known to have ΔF508 mutations, but the mother's carrier status is unknown.  Before making the decision to conceive, the couple underwent extensive genetic counseling regarding the potential risks of having a child with cystic fibrosis.  The family pedigree is diagrammed below with the unborn child marked by the red arrow.  What is the probability that the unborn child will have cystic fibrosis?

Quizplus · Accepted Answer

11ec160b_3769_6d67_9249_f1ea1b8fc801_MD0001_00 Cystic fibrosis (CF) results from an autosomal recessive defect in the CF transmembrane conductance regulator (CFTR) gene. Although most men with CF are infertile due to congenital absence of the vas deferens, they are not sterile and can have children via assisted reproductive technology. Calculating the probability that the unborn child will have CF can be done by analyzing the above pedigree as follows: Because the father is homozygous for the mutant CFTR allele, he will always transmit the mutant allele to his offspring. Because the mother has an affected sibling and neither of her parents is affected, she most likely had 2 heterozygous carrier parents. Therefore, the mother's 4 possible genotypes are: homozygous for the normal allele, heterozygous with her mother's mutant allele, heterozygous with her father's mutant allele, and homozygous for the mutant allele. However, the mother does not have CF and therefore is not homozygous for the mutant allele. This leaves 3 possible genotypes for the mother. Two of the 3 remaining genotypes result in her being a carrier for the mutant CFTR allele, while the last one results in her being homozygous normal. Therefore, the mother's probability of being a carrier equals 2/3. If the mother is a carrier (2/3 chance), the probability that she will transmit the mutant allele to the child is 1 in 2. As a result, the probability that the child will inherit a mutant allele from the mother (and therefore have CF as the father will always contribute a mutant allele) is: 2/3 x 1/2 = 1/3. Educational objective: The probability that an autosomal recessive disease will be transmitted to a child can be calculated based on the maternal and paternal pedigrees. An unaffected individual (with unaffected parents) who has a sibling affected by an autosomal recessive condition has a 2/3 chance of being a carrier for that condition. __________ References: Genetic counseling issues in cystic fibrosis. (http://www.ncbi.nlm.nih.gov/pubmed/20416541) Cystic fibrosis and fertility. (http://www.ncbi.nlm.nih.gov/pubmed/23429570)

A Young Couple Has Undergone a Successful in Vitro Fertilization