Genetics Problems – Sex Linkage
1. Color blindness is a sex-linked recessive trait in humans. A normal woman whose father was color blind
marries a normal man. Predict the phenotypes of their possible offspring and the probability of each.
XN = Normal allele
XnY
n
X = Colorblind allele
XNY
XN Xn
The woman must be heterozygous since her father could only give her an Xn. Therefore, the cross is:
XN Xn
x
XNY
The Punnett Square would be:
XN
Xn
XN
Y
XNXN
XNY
XnY
XN Xn
Normal males
Colorblind males
½ Normal Females
¼ Normal Males
¼ Colorblind Males
Normal females
2. What parental genotypes could produce a colorblind daughter?
Her father would need to be colorblind. Her mother could either be colorblind or be a carrier:
XnY x XN Xn
OR
n
X Y x Xn Xn
3. White eye color in Drosophila (fruit flies)is inherited as a sex-linked recessive trait. What phenotypes
would be found in the offspring of a cross between a white-eyed female and a normal (red-eyed) male?
XR = Red allele
White-eyed Female x Red-eyed Male
r
X = white allele
XRY
XrXr
Xr
Xr
XR
Y
XR Xr
Xr Y
XR Xr
Xr Y
Red-eyed
females
White-eyed
males
½ Red-eyed Females
½ White-eyed Males
4. Hemophilia is a sex-linked recessive disease in humans in which the blood does not clot properly. A
man whose father was a hemophiliac but whose own blood clotting is normal marries a normal woman
with no record of hemophilia in her family. What kinds of offspring would you predict they may have?
Since the man received only a Y chromosome from his father, the fact that his father had hemophilia will
have no effect on the man’s genotype. Since his wife has no family history of the disease, she is most
likely homozygous for the normal allele. Therefore, all of their children should be normal.
XN = Normal allele
Xn = hemophilia allele
XNY x XNXN
½ Normal Females
½ Normal Males
5. A hemophiliac father has a hemophiliac son. The boy’s mother does not have the disease. Give the
genotypes of the father, the mother and the son.
N
X = Normal allele
Xn = hemophilia allele
XnY
N _
X X
XnY
This must be XNXn
since her son received his
X chromosome from her.
The son received his Y chromosome from his father. Therefore, he did not receive the hemophilia allele
from his father. His X chromosome came from his mother. Therefore, her second X chromosome mus carry
the hemophilia allele. Her genotype must be XNXn.
6.A) Refer to the pedigree chart below. Predict the genotype of each individual if the mutant trait
(represented by a colored in circle or square) is autosomal dominant.
DEFINE YOUR VARIABLES:
M = mutant allele
m = normal allele
Key:
= Normal male
mm
1
2 Mm
= Mutant male
= Normal female
mm 3
4 mm
5 Mm
6 Mm
7 mm
= Mutant female
8 mm
9 mm
10 mm
B) Refer to the pedigree chart below. Predict the genotype of each individual if the mutant trait (represented
by a colored in circle or square) is sex-linked recessive.
DEFINE YOUR VARIABLES:
Key:
XM= normal allele
Xm = mutant allele
XMXm 1
= Normal male
2 XmY
= Mutant male
= Normal female
XMY 3
4 XMXm
5 XmXmm
6 XmY
7 XMX_
= Mutant female
8 XMY
9 XMXm
10 XMY