Question

A genetics counselor explains that cystic fibrosis is an autosomal recessive disease, requiring two copies of a gene to express disease. If both parents are heterozygous for the trait, what are the odds that they would produce a child with cystic fibrosis?

A). 4 in 4
B). 2 in 4
C). 1 in 4
D). 3 in 4

Answer 1

The odds that heterozygous carrier parents will produce a child with cystic fibrosis are 1 in 4, according to Mendelian genetics and the Punnett square method.

Step-by-step explanation:

If both parents are heterozygous for cystic fibrosis, the odds that they would produce a child with cystic fibrosis are C). 1 in 4.

When considering autosomal recessive diseases like cystic fibrosis, two carrier parents have a 25% chance to have a child affected by the disease. This is evidenced by the traditional Mendelian genetics using a Punnett square, where each parent contributes one allele.

The genotypes of the parents (Ff x Ff) can result in the following combinations: FF (normal), Ff (carrier), Ff (carrier), and ff (affected with cystic fibrosis). Since each of these outcomes is equally likely, there is a 1 in 4 chance or 25% probability the child will inherit cystic fibrosis (ff). On the contrary, there is a 75% chance the child will not show the disease. Of the non-affected children, 50% will be carriers (Ff) and 25% will be completely unaffected (FF).