The Mid-Atlantic ridge is a type of
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Answer:
1/4
Explanation:
Cystic fibrosis is a recessive disorder and therefore, is expressed only in homozygous recessive genotype. Let's assume that the allele "c" is responsible for the disease in the homozygous state. Since the couple already has a child with cystic fibrosis (cc), they both are heterozygous carriers for the disease.
The genotype of male and female each is "Cc". A cross between Cc x Cc gives progeny in the following phenotype ratio=1/4 diseased: 1/2 normal carriers: 1/4 normal. Therefore, the probability of getting a diseased child for them is 1/4.
