Genetics Problems – Pedigree Charts and Multiple Alleles
1. Tay-Sachs disease is a fatal genetic disorder caused by a recessive allele. The pedigree chart below
shows the appears of the disease in three generations of a family. Tell the genotype or possible
genotypes of each member of the family. Be sure to define your variables.
Key:
1
2
(from a family
with no history
of TaySachs)
= Normal male
= Tay-Sachs male
= Normal female
3
4
5
6
7
= Tay-Sachs female
8
9
2. Draw a pedigree chart for the family described below and use the chart to determine the genotype or
possible genotypes of each member of the family. Be sure to define your variables.
In humans, having freckles is dominant over not having them. Both Mr. And Mrs. Chambers have
freckles. Of their three children, Robin and Daniel have freckles but Alex does not. Robin marries
Stephen, who does not have freckles. They have two children, Kevin and Beth, who do not have
freckles. Daniel marries Diane who also has freckles. They have one son, Phillip, who also has
freckles. Alex marries Mary, who has freckles. They have one daughter, Norma, who does not
have freckles.
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3. On a certain day in a local hospital, two male babies were born. Shortly after the mothers returned home
with their infants, Mrs. Robinson discovered a tag marked “K” on her baby. The other mother, Mrs.
Koffer, insisted she had her own baby and refused to exchange. The matter came to court where all
concerned agreed to submit to blood test. Here are the results:
Mrs. Robinson – Type O
Mr. Robinson – Type O
Baby taken home by the Robinsons – Type A
Mrs. Koffer – Type O
Mr. Koffer – Type AB
Baby taken home by the Koffers – Type O
Did a switch occur? Explain using Punnett Squares to support your answer.
4. A man married a woman of blood type O. They had 4 children. Two of the children had blood type A
and two had blood type B. Draw a pedigree chart showing the phenotypes and genotypes of every one in
the family.
5. For each of the following determine the genotype of the parents. Show a Punnet square to back up your
answer.
A) One parent is type A and the other is type B. Each of their four children has a different blood type.
B) Both parents are type A. Three fourths of their children are type A while the other one fourth is type
O.