Question

A recessive allele inherited in a Mendelian manner causes the disease cystic fibrosis. A phenotypically normal man whose father had cystic fibrosis marries a phenotypically normal woman from outside the family. If the frequency in the general population of heterozygotes for cystic fibrosis is 1 in 50, what is the chance that the couple's first child will have cysitc fibrosis

Answer 1

Answer:

The chance that the couple's first child will have cysitc fibrosis is equal to  $0.005$

Explanation:

The probability of man having cystic fibrosis is equal to $100$ percent

The probability of his wife having cystic fibrosis is equal to  $\frac{1}{50}$

In case if both have them allele for cystic fibrosis, then the probability of their children having a child with cystic fibrosis

$\frac{1}{4}$

The chance that the couple's first child will have cysitc fibrosis is equal to

$1* \frac{1}{50} *\frac{1}{4} \\= \frac{1}{200}\\= 0.005$