Question

A woman is a carrier for the a severe Cystic Fibrosis mutation (Ff1) and her husband is a carrier for a mild Cystic Fibrosis mutation (Ff3). What is the probability that their children will have the disease?

a) 25%
b) 50%
c) 75%
d) 0%

Answer 1

The probability that the children of two carrier parents for Cystic Fibrosis will have the disease is 25%, as they must inherit two recessive alleles. According to Mendelian genetics, there's a 25% chance for each outcome: healthy and non-carrier, healthy carrier, or having Cystic Fibrosis. The correct option is a) 25%.

Step-by-step explanation:

The probability that the children of a woman who is a carrier for a severe Cystic Fibrosis mutation (Ff1) and her husband who is a carrier for a mild Cystic Fibrosis mutation (Ff3) will have the disease is based on classic Mendelian genetics. Since Cystic Fibrosis is an autosomal recessive disorder, a child must inherit two recessive alleles (one from each parent) to express the disease.

The Punnett square demonstrates that there is a 25% chance the child will inherit two recessive alleles (ff - one from each parent), and hence have Cystic Fibrosis. As each parent carries one copy of the recessive allele (Ff), and one copy of the dominant allele (FF or Ff), it is a 25% probability each, that the child will inherit ff, FF, or Ff from each parent.

Applying this understanding:

• 25% chance to be healthy and non-carrier (FF)
• 50% chance to be a healthy carrier (Ff)
• 25% chance to have Cystic Fibrosis (ff)

Therefore, the correct option is a) 25%, which is the probability that their children will have Cystic Fibrosis.