Yes. To have a colorblind daughter, the man would have to have the colorblind allele himself.
Men get their X chromosome from their mother.
Yes. A colorblind daughter must receive the colorblind allele from each parent.
XBXb
XBY
XBXb
XbY
XbXb
Yes
XHXh x XhY
XH
Xh
Xh
XH Xh
Xh Xh
Y
XH Y
Xh Y
1/4
1/4
1/4
1/4
XHXh
XhXh
XHY
XhY
1/4
1/4
1/4
1/4
normal (carrier) female
hemophiliac female
normal male
hemophiliac male
¼
¼
¼
¼
XHXh
4
BbXbY
bbXBXb
B Xb
BY
b Xb
bY
B XB Bb XBXb Bb XBY bb XBXb bb XB Y
B
Xb
Bb
XbXb
Bb
XbY
bb
XbXb
bb
XbY
b XB Bb XBXb Bb XBY bb XBXb bb XBY
b Xb Bb XbXb Bb XbY bb XbXb bb XbY
2/16
8/16
2/16
2/16
2/16
8/16
2/16
2/16
2/16
2/16
2/16
2/16
2/16
2/16
Bb Xb Xb
Bb XBXb
Bb Xb Y
Bb XBY
bb Xb Xb
bb XB Xb
bb Xb Y
bb XBY
2/16 Brown-eyed colorblind
female
2/16 Brown-eyed carrier female
2/16 Brown-eyed colorblind
male
2/16 Brown-eyed normal male
2/16 Blue-eyed colorblind
female
2/16 Blue-eyed carrier female
2/16 Blue-eyed colorblind male
2/16 Blue-eyed normal male
BBXBY
bb XbXb
Bb XBXb and BbXbY
(Bb XBXB and BbXbY)
B Xb
BY
b Xb
bY
B XB BB XBXb BB XBY Bb XBXb Bb XBY
B XB BB XBXb BB XBY Bb XBXb Bb XBY
b XB Bb XBXb Bb XBY
bb XBXb bb XBY
2/16
2/16
4/16
4/16
2/16
2/16
BB XBXb
BB XBY
Bb XBXb
Bb XBY
bb XBXb
bb XBY
6/16 Brown-eyed carrier
female
6/16 Brown-eyed normal
male
2/16 Blue-eyed carrier
female
2/16 Blue-eyed normal
male
b XB Bb XBXb Bb XBY bb XBXb bb XBY
XBXb
XBY
XB
Y
XB
Xb
XBXB
XBXb
XBY
Xb Y
1/4 XBXB
1/4 XBXb
1/4 XBY
1/4 XbY
2/4 Female with normal vision
1/4 Normal male
1/4 Colorblind male
The woman is most likely XH XH and is not carrying the recessive allele for
colorblindness. If she carried one copy of the recessive allele, it is probable that half of
her sons would have hemophilia. The daughter is most likely XH XH as well.
Man = XHY
Woman = XHXH
There is no chance of the trait appearing in their children.
XBXb
XBY
Xb Xb
Xb Y
XBXb
Xb Y
If the brother has hemophilia, their mother was a carrier for the disease. The daughter
could be XHXH or XHXh. If the daughter is XHXH there is no chance of producing a child
with hemophilia.
If the daughter is XHXh , she might produce a son (25% chance) who has hemophilia.
If the man has normal vision, he could not produce a daughter with the disease. He
could use this as evidence. He is not the child’s father.
XBY
XBXb
Xb Y
XBXb
XBXB
XBY
Xb Y
XBXb
XBY
See last page for these answers set up in genealogy table format.
Xr Y
XR XR
Xr
XR
XR
2
2
2/4
0
2/4
0
XR
Xr
XR Xr
Y
X RY
X RY
2/4 XR Xr
2/4 Red-eyed females
2/4 XRY
2/4 Red-eyed males
XRY
XR Xr
XR Xr or XR XR
XRY
Xr Y
Xr Xr
XRY
XR Xr and Xr Y
XRXr x XRY
XR
Y
XR
XR XR
X RY
Xr
XR Xr
Xr Y
1/4 XR XR
2/4 Red-eyed females
1/4 XR Xr
1/4 Red-eyed males
1/4 XRY
1/4 White-eyed males
1/4 XrY
Nn XrY
N Xr
nn XRXr
NY
n Xr
nY
n XR Nn XR Xr Nn XR Y nn XRXr
nn XRY
r r
r
r r
n Xr Nn X X Nn X Y nn X X
nn XrY
n XR Nn XR Xr Nn XR Y nn XRXr nn XRY
n Xr Nn Xr Xr Nn XrY nn XrXr
nn XrY
2/16
2/16
2/16
2/16
2/16
2/16
2/16
2/16
Nn XR Xr
Nn XRY
Nn Xr Xr
Nn XrY
nn XR Xr
nn XRY
nn Xr Xr
nn XrY
The father of the liter is XbY which is yellow in color.
There are two fathers. One father is XbY and the other father is XBY.
The genotype of the mother is XbXb. The genotype of the father is XBY. The yellow kitten
would have to be male and has the genotype XbY.
Question 12 set up in genealogy table format.
Mom = XBXb
XbY
Dad = XBY
XBXb
XBXb
XBXB
XbY XBY
XBY