Download Complex Mendelian Genetics Problems

Complex Mendelian Genetics Problems
Incomplete Dominance: (Mixing/Blending)
1. In Japanese four-o’clocks, the gene for red flower color (R) is incompletely dominant over the white
flower color (W). For each of the following situations, predict the genotypic and phenotypic ratios.
a. a red plant and a white plant
RR
x
WW (parent genotypedetermine this first to make the Punnett)
R
R
RW
RW
RW
RW
W
Geno.Ratio:
Phen.Ratio:
4 :
4 :
0 - all RW
0- all pink
W
b. a white plant and a pink plant
WW
x
RW (parent genotypes)
W
W
RW
RW
WW
WW
R
R
RR
RR
RW
RW
R
Geno.Ratio:
Phen.Ratio:
2 : 2- RW : WW
2 : 2- pink:white
W
c. a red plant and a pink plant
RR
x
RW (parent genotypes)
R
Geno.Ratio:
Phen.Ratio:
d. two pink plants
RW
x
2 : 2- RR : RW
2 : 2- red : pink
W
RW (parent genotypes)
R
W
RR
RW
RW
WW
R
Geno.Ratio:
Phen.Ratio:
1: 2 : 1- RR : RW: WW
1 : 2 : 1 red : pink: white
W
2. In some cats the gene for tail length shows incomplete dominance. Cats with long tails and those with
no tails are homozygous for the respective alleles. Cats with one long tail allele and one no-tail allele
have short tails. Predict the genotypic and phenotypic ratios of a cross between:
First, identify an allele for long tail (L), then an allele for no tail (N). Cats with short tails will have one
allele of each.
a. a long tail cat and a cat with no tail
LL
x
NN (parent genotypes)
L
L
LN
LN
LN
LN
N
Geno.Ratio:
Phen.Ratio:
4 : 0- all LN
4 : 0 all short
N
b. a long tail cat and a short tail cat
LL
x
LN
L
L
LL
LL
LN
LN
L
Geno.Ratio:
Phen.Ratio:
2 : 2- LL : LN
2 : 2 long : short
N
c. a short tail cat and a cat with no tail
LN
x
NN
L
N
LN
NN
LN
NN
N
Geno.Ratio:
Phen.Ratio:
2 : 2- LN : NN
2 : 2 short: no tail
N
d. two short tail cats
LN
x
LN
L
N
LL
LN
LN
NN
L
Geno.Ratio: 1: 2 : 1- LL : LN : NN
Phen.Ratio: 2 : 2
long : short: no tail
N
3. In the gene that decodes for uni-brow, no uni-brow is incompletely dominant over full, bushy uni-brow.
The phenotype for a heterozygote is a half uni-brow. What would the probability of having a child with
a full, bushy uni-brow be if we crossed a half uni-brow male with a full, bushy uni-brow female.
First, identify an allele for the NO uni-brow (U-no uni brow), then for full, bushy uni-brow (B). A
half-uni brow would be one of each allele.
UB
x
BB (parent genotypes)
U
Possibility of having a child with a full,bushy uni-brow
(BB) is 2/4, or 50%
B
B
B
UB
BB
UB
BB
4. A cross between round radishes (RR) and long radishes (LL) produces oval radishes (RL). What would
happen if you crossed a round radish and an oval radish.
RR x
RL (parent genotypes)
R
R
R
Offspring would be 2/4 or 50% round &
2/4 or 50% oval
RR
RR
RL
RL
L
Codominance
1. A father has the blood type AB and a mother is heterozygous for blood type A. What are the possible
blood types of their first child? This question is asking for the PHENOTYPE of the first child (type of
blood)-you do not have to specify if it is homo-or heterozygous.
(parents genotypes-do this first to make the Punnett) iAiB x iAi
iA
Answer: Type A, Type B, and Type AB
iB
iAiA
iAiB
iAi
IBi
iA
i
2. What are ALL the possible blood types of children in the following families?
Mother
Father
Genotypes
A A
A
A.
homozygous A i i
heterozygous A i i
iAiA - iAi
Phenotypes
Type A
B.
heterozygous A iAi
O
ii
iAi - i i
Type A, Type O
C.
AB
iAiB
O
ii
iAi – iBi
Type A, Type B
D.
AB
iAiB
AB
iAiB
iAiA- iAiB- iBiB
Type A,Type,B,Type AB
a. For the above question, first write the genotype of the parents.
b. Then, ‘foil’ the parent genotypes to determine the possible children genotypes (since it is not asking
for a ratio, you only need to write each possible genotype one time).
c. Use the genotypes to then determine the possible children phenotypes
d. When writing phenotypes of blood-types, you do not need to specify if it is ‘homozygous or
heterozygous, unless it asks you.
3. The palomino horse is a hybrid (mix) showing a golden coat with a lighter mane and tail. A pair of
codominant alleles, D1 and D2 is known to be involved in this trait. Horses with the D1D1 genotype are
chestnut colored, horses with the D1D2 genotype are palomino, and horses with the D2D2 genotype are
white in color.
A. Two palomino horses are mated by artificial insemination. What types of offspring could be
produced? Question is asking for ‘types’- or phenotypes.
D1D2 x D1D2 (parents genotypes-do this first to make the Punnett)
Answer: Chestnut, palomino, and white horses
D1
D1
D2
D1D1
D1D2
D1D2
D2D2
D2
B. If a palomino horse and a white horse are mated, what types of offspring could be produced?
Question is asking for ‘types’- or phenotypes
D1D2 x D2D2 (parents genotypes-do this first to make the Punnett)
Answer: Palomino and white horses
D1
D2
D2
D1D2
D2D2
D1D2
D2D2
D2
4. If a man with blood type B, one of whose parents had blood type O, marries a woman with blood type
AB, what will be the theoretical percentage of their children with blood type B?
Answer: First, determine the genotype of the father. Since one of his parents is Type O, he must be
heterozygous. iB i x iAiB (these parents genotypes)
50% of the children will be Type B (don’t need to specify homo-or heterozygous)
iB
iAiB
i
iAi
iA
iBiB
iB
iBi
Multiple Alleles
1. A couple have four children. Their first child has type A blood, the second has type O, the third has type
AB, and the fourth has type B. What are the genotypes of the parents?
Answer: Remember, that this problem is talking about 4 pregnancies, which would be 4 Punnetts. The order of
birth doesn’t matter. Because it is not asking for ratios, you can do just one Punnett, to look at the possible
PHENOTYPES. Fill in the Punnett first. HINT: If an offspring is Type 0, always put them in the bottom
right box. This means that each parent is contributing an ‘i’, so fill that in the parent gamete. Because they
produce an AB child, each must contribute an A or a B. You’re finished!
iB
iA
i
iAiB
iB i
iAi
ii
i
2. A woman with type AB blood marries a man with type A blood. His mother had type O blood. What
blood type is not likely to occur among this couple’s children? Why – explain.
First, determine the genotype of the father. Since one of his parents is Type O, he must be heterozygous.
IA i x iAiB (these parents genotypes-man first)
Then, do the Punnett.
The answer is Type O, since both parents do NOT contribute an ‘i’.
iA
i
iAiA
iA i
iAiB
iB i
iA
iB
3. What is the probability that a couple whose blood types are AB and O will have a type A child?
iAiB x i i (parents genotype)
Doesn’t matter how you set up Punnett, because you don’t know which is which parent)
The answer is 50%. Doesn’t matter if it is homo-or heterozygous iA
iB
iAi
iB i
iAi
iB i
i
i