Problem

Solutions For An Introduction to Genetic Analysis Chapter 2 Problem 57P

Step-by-Step Solution

Solution 1

Step #1 of 3

a. To construct the pedigree of the information given in the question, let us consider the alleles as follows:

C as normal allele and c as the allele responsible for cystic fibrosis.

C- c/ c cystic fibrosis (homozygous recessive)

Cc c-

Step #2 of 3

b. The probability of man having 'c' allele = 100%

The probability of the woman having 'c' allele = 1/50

If both are carriers of the allele (heterozygous), then the probability of both the parents passing the allele to the children is 1/4.

So, the probability of their first child having the disease is

Step #3 of 3

c. If the first child has cystic fibrosis, both the mother and the father should be heterozygous. So, the probability of the first child having disease (cc) is .

In this case, the first child is has the disease. So, the probability of having unaffected child is 1-= and the probability of having an unaffected child is The cross between yields three genotypes, CC, Cc, cc in the ratio 1:2:1. the child with the genotype, cc will be affected and the children with other genotypes will not be affected.