Solutions to Incomplete Dominance
1. In dwarf rabbits, the alleles for coat colour display incomplete dominance. Pure bred dwarf
rabbits can be black or white, but there are also some gray coloured individuals in the
population. Show the F1 and F2 generations when a white rabbit is mated with a black rabbit.
B = black allele
B’ = white allele
Genotype of a white rabbit = B’B’
Genotype of a black rabbit = BB
Crossing B;B’ x BB for F1 generation:
B’
B’
B
BB’
BB’
B
BB’
BB’
Crossing BB’
x
B
B’
B
BB
BB’
B’
BB’
B’B’
100% of the offspring have genotype BB’. This means that 100% of the
offspring are gray.
BB’ for F2 generation:
50% of the offspring have genotype BB’, 25% are B’B’, and 25% are BB.
This means that 50% of the offspring are gray, 25% are white, and 25%
are black.
2. When a purple pansy is crossed with a white pansy, the offspring are all lilac coloured. If pansies
display incomplete dominance show the cross and give the genotypic and phenotypic ratios of:
a) a cross between two lilac flowers
P = purple allele
P’ = white allele
Genotype of a white flower = P’P’
Genotype of a purple flower = PP
Genotype of a lilac flower = PP’
Crossing PP’ x
P
P’
P
PP
PP’
P’
PP’
P’P’
PP’:
50% of the offspring have genotype PP’, 25% are P’P’, and 25% are PP.
This means that 50% of the offspring are lilac, 25% are white, and 25% are
purple.
b) a cross between a lilac flower and a purple flower
Crossing PP’ x
P
P’
P
PP
PP’
P
PP
PP’
PP:
50% of the offspring have genotype PP’, 50% are PP. This means that 50%
of the offspring are lilac and 50% are purple.
3. In horses, the alleles for chestnut colour and white colour exhibit incomplete dominance.
a) If a baby horse is a palomino (heterozygous), then what are the possible genotypes and
phenotypes of its parents?
C = chestnut allele
C’ = white allele
Genotype of chestnut = CC
Genotype of white = C’C’
Genotype of palomino = CC’
Since there is a white allele, both parents cannot be chestnut coloured. Instead, one parent
could be chestnut (CC) and one parent could be white (C’C’) OR both parents could be
palomino (CC’) OR one parent could be white (C’C’) and one palomino (CC’) OR one could be
chestnut (CC)and one palomino (CC’). This is shown in these punnet squares:
C’
C’
C
CC’
CC’
C
CC’
CC’
If one parent is chestnut and one is white, 100% of the offspring are
palomino.
C
C’
C
CC
CC’
C’
CC’
C’C’
C’
C’
C
CC’
CC’
C’
C’C’
C’C’
C
C
C
CC
CC
C’
CC’
CC’
If both parents are palomino, 50% of the offspring are palomino, 25% are
white, and 25% are chestnut.
If one parent is white and one is palomino, 50% of the offspring are
palomino and 50% are white.
If one parent is chestnut and one is palomino, 50% of the offspring are
palomino and 50% are chestnut.
b) If a baby horse is white, then what are the possible genotypes and phenotypes of its
parents?
Since the baby is white, neither of the parents can be chestnut coloured. Instead, both
parents could be white (C’C’), one parent could be palomino (CC’) and one parent could be
white (C’C’) OR both parents could be palomino (C’C’). This is shown in these punnet
squares:
C’
C’
C’
C’ C’
C’ C’
C’
C’
C’
If both parents are white, 100% of the offspring are white.
C
C’
C’
C C’
C’ C’
C’
C C’
C’ C’
C
C’
C
CC
CC’
C’
CC’
C’ C’
If one parents is white and one is palomino, 50% of the offspring are
palomino and 50% are white
If both parents are palomino, 50% of the offspring are palomino, 25% are
white, and 25% are chestnut.
4. A heterozygous blood type A man and an AB type woman have a child. What is the probability
that the child will be type B?
Blood type A allele = IA
Blood type B allele = IB
Blood type O allele = O
Heterozygous blood type A genotype = IAi
Type AB genotype = IAIB
Crossing
IA i
x
IA
IB
IA
IA IA
IA IB
i
IA i
IB i
IAIB
The genotypes are 25% IA IA, IA IB, IAi, and IBi. The probability that the child
will be blood type B is 25%.
5. A homozygous blood type A man and a heterozygous blood type B woman have 12 children.
How many children will have blood type b? Blood type A?
Blood type A allele = IA
Blood type B allele = IB
Blood type O allele = O
Homozygous blood type A genotype = IAIA
Heterozygous blood type B genotype = IBi
Crossing
IB i
x
IA
IA
IB
IA IB
IA IB
i
IA i
IA i
IAIA
The genotypes are 50% IA IB, and 50% IAi. The probability that the children
will be blood type B is 0% and the probability that the children will be
blood type A is 50%. Therefore, 0 out of 12 children will be blood type B
and 6 out of 12 children will be blood type A.
6. A man with blood type O and his wife (blood type unknown) have 2 children.
a) If one of the children has blood type A and one of them has blood type B, what must the
mother’s genotype be?
Blood type A allele = IA
Blood type B allele = IB
Blood type O allele = O
Blood type O genotype = ii
Since the father is blood type O, each child will always inherit a i allele from him. If one child
has blood type A and another has blood type B, this means the children must have inherited
the A and B alleles from the mother. Therefore, the mother must have both the A and B
alleles. Therefore, the mother must have blood type AB and her genotype must be IAIB.
b) If one of the children has blood type A and the other has blood type O, what must the
mother’s genotype be?
Since the father is blood type O, each child will always inherit a i allele from him. If one child
has blood type A and another has blood type O, this means the children must have
inherited the A and extra O alleles from the mother. Therefore, the mother must have both
the A and O alleles. Therefore, the mother must be heterozygous for blood type A and have
a genotype of IAi.