Question

4. The CFTR gene is carried on human chromosome 7, an autosome. A recessive mutation in CFTR is the cause of cystic fibrosis disease. A female who doesn't have cystic fibrosis has a child with a man who also doesn't have the disease. Their first child has cystic fibrosis. What is the probability their second child will have the disease?

A. 0 (definitely will not have it)

B. 1/4

C. 1/2

D. 3/4

E. 1 (definitely will have it)

Answer 1

Cystic fibrosis is an autosomal recessive disease so it will inherent in their progeny in the same manner. Both mother and father don't have disease but have a child with diseased condition mean both parents are heterozygous for this gene say both have a genotype of Bb. So if we cross Bb with Bb then the individual with bb genotype will definitely show a diseased phenotypic character. The probability for the diseased child is 1/4 and their non-diseased have 3/4. This probability will remain the same whether its first child or fifth child. Thus the answer will be option B i.e. 1/4.