Question

A recessive allele inherited in a Mendelian manner causes the disease cystic fibrosis. A phenotypically normal man whose father had cystic fibrosis, and a mother who comes from a long line of non-affected people, marries a phenotypically normal woman from outside the family. If the frequency in the general population of heterozygotes for cystic fibrosis is 1 in 25, what is the chance that the couple's first child will have cysitc fibrosis

Answer 1

The chance that the couple's first child will have cystic fibrosis = 1/100

Step-by-step explanation:

Since both parents are phenotypically normal, it means that they are both heterozygous for cystic fibrosis.

The probability of a cross of two heterezygotes for a recessive allele to produce an offspring having two recessive alleles is 1 : 4 in accordance with Mendelian inheritance.

The initial probability of the couple's first child to be have two allelic pairs for cystic fibrosis = 1 : 4.

However, the frequency in the general population of heterozygotes for cystic fibrosis is 1 in 25.

Therefore, the chance that the couple's first child will have cystic fibrosis = 1/4 * 1/25 = 1/100