This document discusses sex-linked genes and traits. It explains that females have two X chromosomes and can be homozygous or heterozygous for genes on the X chromosome, while males only have one X chromosome and will always express the phenotype of the single allele they carry for X-linked genes. It uses haemophilia, a bleeding disorder caused by deficient coagulation factor VIII, as an example of a sex-linked recessive trait. The document presents a Punnett square showing that if a healthy carrier woman and healthy man have children, there is a 25% chance of having a hemophilic son since he would receive the defective allele from his mother.

1. Lesson #11
Sex-Linked genes

2. Sex-linked genes
Inheritance linked to the X chromosome

3. Sex-Linked genes Women Men
X
X
X
Y
Male gametes
Femalegametes
Inheritance of genes in the X chromosome has some pecularities:
• Women carry 2 alleles for genes in the X chromosome, therefore they can
be homozygous or heterozygous for those genes
• Men can only carry one allele for the genes in the X chromosome,
therefore they will express the phenotype of that single allele they carry
for genes on the X chromosome.
We represent the allele on the sex chromosome, the Y chromosomo would
not carry an allele, because it has different genes.
1 1
2 2
3 3
. .
. .
. .
X X
1 1
2 2
3 3
. .
. .
. .
X Y
X X X Y

4. Haemophilia (Sex-Linked trait)
Normal Factor VIII
Healthy (H)
Deficient Factor VIII
Haemophilia (h)
Healthy > HaemophiliaAlleles
Gene: Coagulation Factor VIII
Linked to the X chromosome
Disorder mainly caused by a deficient coagulation Factor VIII which limits the ability of
the blood to clot or coagulate. This deficiency is due to genetic reasons. The gene that
codifies for this coagulation Factor VIII is on the X chromosome.
Females Males
Genotype XHXH XHXh XhXh XHY XhY
Phenotype Healthy Healthy Haemophilic Healthy Haemophilic

5. Sex-Linked
XHXh
Problem solving exercise
A healthy woman who carries an allele for a deficient Factor VIII has offspring with a healthy man.
What are the chances of this couple of having a hemophilic child?
P: Healthy carrier woman Healthy man
XHY
x
x
Gametes: p(XH)= p(Xh) = ½
Punnet F1:
Answer: There is a 25% probability of having a hemophilic child.
F1 Genotype F1 Phenotype
p(XHXH) ¼ 25% Healthy girl
p(XHXh ) ¼ 25% Carrier healthy girl
p(XHY) ¼ 25% Healthy boy
p(XhY) ¼ 25% Hemophilic boy
p(XH) = p(Y) = ½