Answers:
a.) draw the pedigree as far as described? Â
Pedigree:
C/–    c/c
    C/c    C/–
       ?
b.) If the frequency in the population of heterozygotes for cystic fibrosis is 1 in 50, what is the chance the couples first child will have cystic fibrosis? Â
Man: has the disease
Wife: 1/50 chance to have the c allele
First child: 1.0 x 1/50 x 1/4 = 1/200 = 0.005 Â
c.) If the first child does have cystic fibrosis, what is the probability that the second child will be normal?
If the first child has the disease, then the mother is a carrier of the
c allele. In consequence, the probability is 3/4