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 plantRR x WW (parent genotype- R R

determine this first to make the Punnett)

WGeno.Ratio: 4 : 0 - all RW Phen.Ratio: 4 : 0- all pink W

b. a white plant and a pink plant

WW x RW (parent genotypes) W W

RGeno.Ratio: 2 : 2- RW : WW Phen.Ratio: 2 : 2- pink:whiteW

c. a red plant and a pink plantRR x RW (parent genotypes) R R

RGeno.Ratio: 2 : 2- RR : RW Phen.Ratio: 2 : 2- red : pink W

d. two pink plantsRW x RW (parent genotypes) R W

RGeno.Ratio: 1: 2 : 1- RR : RW: WW Phen.Ratio: 1 : 2 : 1 red : pink: white W

RW RW

RW RW

RW RW

WW WW

RR RR

RW RW

RR RW

RW WW

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

NGeno.Ratio: 4 : 0- all LN Phen.Ratio: 4 : 0 all short N

b. a long tail cat and a short tail cat LL x LN L L

LGeno.Ratio: 2 : 2- LL : LN Phen.Ratio: 2 : 2 long : short N

c. a short tail cat and a cat with no tail LN x NN

L N

NGeno.Ratio: 2 : 2- LN : NN Phen.Ratio: 2 : 2 short: no tail N

d. two short tail cats LN x LN L

N

LN LN

LN LN

LL LL

LN LN

LN NN

LN NN

LL LN

LN NN

LGeno.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 B

Possibility of having a child with a full,bushy uni- brow B(BB) is 2/4, or 50%

B

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

ROffspring would be 2/4 or 50% round &

2/4 or 50% oval

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 iB

UB BB

UB BB

RR RR

RL RL

Answer: Type A, Type B, and Type AB

iA

i

2. What are ALL the possible blood types of children in the following families?Mother Father Genotypes Phenotypes

A. homozygous A iAiA heterozygous A iAi iAiA - iAi Type A

B. heterozygous A iAi O i i iAi - i i Type A, Type O

C. AB iAiB O i i 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 phenotypesd. 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 D2

D1

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)

iAiA iAiB

iAi IBi

D1D1 D1D2

D1D2 D2D2

Answer: Palomino and white horses

D1 D2

D2

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 i

iA

iB

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!

iA i

iB

i

D1D2 D2D2

D1D2 D2D2

iAiB iAi

iBiB iBi

iAiB iB i

iAi i 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

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

i

i

iAiA iA i

iAiB iB i

iAi iB i

iAi iB i