Sex-Linked Traits
… and sex-influenced traits
Definitions
 Sex-linked trait - controlled by gene
located on the X or Y chromosome.
 Sex-influenced trait - trait controlled by
a gene on the autosomes, but influenced
by sex hormones.
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Sex Influenced Traits
(on autosomes; influenced by sex hormones)
 One allele is dominant for males and
the other allele is dominant for
females. Written like co-dominant
alleles. (BB’)
 Examples: human baldness, sheep
horns, some colors in cattle.
Baldness
(androgenic alopecia)
 A sex-influenced trait.
 B = bald, B’ = not bald
 For men, B is dominant.
2
Baldness
(androgenic alopecia)
 A sex-influenced trait.
 B = bald, B’ = not bald
 For men, B is dominant.
 For women, B’ is dominant.
Key:
BB - bald
B’B’ - not bald
BB’ - (m) bald
(w) not bald
(wearing a wig)
Key:
BB - bald
B’B’ - not bald
BB’ - (m) bald
(w) not bald
Q. A heterozygous bald man and a
heterozygous woman have kids.
Find the probabilities of offspring.
 Parent phenotypes: Bald dad, not bald mom
 Parent genotypes: BB’
BB’
B
B’
B
BB
BB’
B’
BB’ B’B’
 Offspring genotypes: 1/4 BB, 1/2 BB’, 1/4 B’B’
Calculate separate phenotypes - one for males, one for females.
 Son phenotypes: 3/4 bald, 1/4 not bald
 Daughter phenotypes: 1/4 bald, 3/4 not bald
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Q1. What is the probability that
their daughter will be bald?
Q2. What is the probability of having
a daughter who will be bald?
  Remember:
 Son phenotypes: 3/4 bald, 1/4 not bald
 Daughter phenotypes: 1/4 bald, 3/4 not bald
  A1 - They already have the daughter, just look
at the phenotypes for daughters: 1/4 bald
  A2 - They have a 1/2 chance of having a
daughter, then a 1/4 chance of her becoming
bald. So …
1/2 x 1/4 = 1/8.
They have a 1/8 chance of having a girl who will go bald.
Sex-Linked Traits
1.
These are located on the X or Y chromosome. The few on the
Y have to do with male-only traits. Very few of either kind
are dominant.
2.
Usually X-linked recessive, we will be studying only the Xlinked recessive traits. These traits are more common in
males. (Males can’t have the dominant gene on a second X to
protect them.)
3.
These are written with the X chromosomes. If the gene is
“linked” to the X, then it is NOT on the Y. (XhY)
4.
Always include the person’s sex in the phenotype.
5.
Examples: hemophilia, some cat colors, body colors in fruit
flies, color-blindness, barred-pattern in chickens, also
Duchenne Muscular Dystrophy and ALD.
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Hemophiliaa sex-linked gene (h) is responsible for the “Bleeders disease”.
XHY
XHXh
XHY
XH X-
XHX-
 1. Make a key:
 XHY
- male, normal
 XhY - male hemophiliac
 XHXH - female, normal
 XHXh - female, normal (carrier)
 XhXh - female hemophiliac
XhY
2. Then, compute the
genotype of each person.
a.  Males
b.  Females with the trait
c.  Other females XH
d.  Other females second X
Hemophiliaa sex-linked gene (h) is responsible for the “Bleeders disease”.
 XHY - male, normal
 XhY - male hemophiliac
 XHXH - female, normal
 XHXh - female, normal (carrier)
 XhXh - female hemophiliac
XHY
XHXh
XHY
XH X-
XHX-
XhY
 If the hemophiliac male above married a normal, noncarrying female, predict the usual.
H




Parents phenotypes: Male hemo, female normal
h
H H
Parents genotypes: X Y, X X
1/2 XHXh and 1/2 XHY
Offspring genotypes:
Offspring phenotypes: 1/2 normal (carrier females)
X
XH
Xh
XHXh
XHXh
Y
XHY
XHY
1/2 normal males
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Try some problems!
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