Part E: Sex Linkage
1. A carrier is a female who is heterozygous for the characteristic. She carries the gene for the
trait but does not express it.
A man only has one X-chromosome and since hemophilia is a recessive x-linked trait, men
only need one allele for the trait to express it (therefore they cannot be carriers). It is
impossible for men to be heterozygous for the trait since the X and Y chromosomes are not
homologous.
2. If the recessive mutation occurs in the egg (Xr), there are two types of sperm that can fertilize
the egg (Y or XR). If the “Y” sperm fertilizes the egg, the offspring will be XrY and the male
offspring will express the trait. If the “XR” sperm fertilizes the egg, the offspring will be XRXr and
the female offspring will not express the trait (they will be carriers). Therefore the sperm must
carry the “Y” chromosome for the F1 generation to express the recessive mutation.
3. Colour blindness is a recessive X-linked trait. Since males carry only one X chromosome, they
cannot be heterozygous for any given gene on the X chromosome. Remember, heterozygous
means two different alleles.
4. Woman (XHXH) x man (XhY)
H
X
XH
Xh
XHXh
XHXh
Y
X HY
X HY
Of the male offspring, 0% will be hemophiliac.
5. Man (XbY), woman (XBXb)  woman’s father can only donate an X chromosome…if father was
colour blind, that means he donated the Xb allele.
B
X
Xb
Xb
XBXb
XbXb
Y
X BY
XbY
There is a 25% chance that their first child will be a colour blind boy.
6. Man (XBY) x Woman (XBXb)
XB
Xb
XB
XBXB
XBXb
Y
X BY
XbY
There is a 25% chance that they will have a colour blind son.
Genotype: 25% XBXB : 25% XBXb : 25% XBY : 25% XbY
Phenotype: 50% normal visioned female : 25% normal visioned males : 25%
colour-blind males
7. Chance alone determines which sperm fertilizes the egg.
8. Unicorn  16 chromosomes = 8 pairs
If we assume there is a single pair of sex chromosomes, then:
1 pair sex chromosomes and 7 pairs of autosomes.
9. Man (XbY) x woman (XBXB)
B
X
XB
Xb
XBXb
XBXb
Y
X BY
X BY
Yes it is possible to have children with normal vision.
There is a 100% chance that he will have daughters that are carriers for redgreen colour blindness.
10. Man (XBY) x woman (XBXb)
B
X
Xb
XB
XBXB
XBXb
Y
X BY
XbY
1/3 normal females, 1/3 normal males and 1/3 carrier females.
The child with genotype (XbY) will die since Xb is lethal if expressed.
11. Male (XRY) x female (XrXr)
r
X
Xr
XR
XRXr
XRXr
Y
XrY
XrY
Red-eyed offspring are possible, however they will all be female.
Red-eyed males are not possible – since the mother is homozygous
recessive, the males can only get a recessive allele from their mother and
the Y chromosome from their father.
12. Father & son (XbY)
The son’s Y chromosome must have come from his father (women don’t have Y chromosomes),
therefore the colour blindness trait must have been passed on by his mother.
13. A = normal, a = hemophilia
Woman (XAXa  father was a hemophiliac, XaY, so she must be a carrier) x XAY
A
X
Xa
XA
XAXA
XAXa
Y
X AY
XaY
Of the girls, there is a 0% chance of having a daughter with hemophilia.
Of the boys, there is a 50% chance of having a son with hemophilia.
14. Man (XbY) x Wife #1 (XBX ?)  4 normal sons (XBY)
This wife can either be XBXB or XBXb
Man (XbY) x Wife #2 (XBX ?)  1 colour blind daughter (XbXb)
This wife must be XBXb in order to have normal vision but be able to donate a recessive allele to
her daughter.