Solutions to X-linked genetic problems
1. Baldness is an x-linked recessive trait. Cross a male who is not bald with a female that is a carrier
(i.e. no bald but heterozygous). What percent of the male offspring will be bald? What percent
of the female offspring will be bald? Give the genotypes of their offspring.
XB = allele for not bald
Xb = allele for bald
Genotype of the father = XBY
Genotype of the mother = XBXb
Crossing XBXb
x
XBY
The genotypes of the offspring are:
25% XBXB, XBXb, XBY, XbY
XB
Xb
XB
XB XB
XB Xb
Y
XB Y
XbY
50% of the male offspring will be bald and 0% of the female offspring will
be bald.
2. Hemophilia is an x-linked disorder. A man who is hemophiliac marries a woman that is a carrier
of the disease. What percent of the male offspring will have the disorder? What percent of the
female offspring will have the disorder? Give the genotypes of their offspring.
XH = allele for not hemophilic
Xb = allele for hemophilia
Genotype of the father = XhY
Genotype of the mother = XHXh
Crossing XHXh
x
XH
Xh
Xh
XH Xh
Xh Xh
Y
XHY
XhY
XhY
The genotypes of the offspring are:
25% XHXh, XhXh, XHY, XhY
50% of the male offspring will have hemophilia and 50% of the female
offspring will have hemophilia.
3. Colour blindness is an x-linked recessive disorder. A normal man marries a colour blind woman.
What percent of the male offspring will have the disorder? What percent of the female offspring
will have the disorder? Give the genotypes of the offspring.
XB = allele for normal vision
Xb = allele for colour blindness
Genotype of the father = XBY
Genotype of the mother = XbXb
Crossing XbXb
x
XBY
The genotypes of the offspring are:
50% XBXb, XbY
Xb
Xb
XB
XB Xb
XB Xb
Y
XbY
XbY
100% of the male offspring will be colour blind and 0% of the female
offspring will be colour blind.
4. Can a mother who is a hemophiliac have a son that does not have the disorder? Explain why or
why not.
Hemophilia is an x-linked recessive trait. If a mother is a hemophiliac, she carries two recessive
alleles and her genotype is XhXh. If she has a son, he will always inherit an allele that has the trait
on it because the mother is homozygous recessive. Since males only inherit one X-chromosome,
he will not have the opportunity to inherit another, dominant allele to mask the recessive allele.
Therefore, it is not possible for a hemophiliac mother to have a son that is not hemophiliac.
5. Red-green colour blindness is an x-linked recessive trait. A colour-blind man and a normal –
vision woman have a colour blind daughter. How did this happen? Explain your answer using a
Punnet Square.
XB = allele for normal vision
Xb = allele for colour blindness
Genotype of the father = XbY
Genotype of the mother = XBXb OR XBXB
Crossing XBXB
x
XbY
AND
XBXb
x
XbY
XB
XB
Xb
XB Xb
XB Xb
Y
XB Y
XB Y
XB
Xb
Xb
XB Xb
Xb Xb
Y
XB Y
XbY
The genotypes of the offspring are:
50% XBXb, XbY
In this case, the 0% of the daughters will be colour blind. Therefore, the
mother’s genotype is NOT XBXB
The genotypes of the offspring are:
25% XBXb, XbXb, XBY, XbY
In this case, 50% of the daughters will be colour blind. This is because the
mother was heterozygous for the colour blindness trait and was a carrier
for the disorder.