Learning Genetics Can Be Fun

1. For Labrador retrievers, black fur is dominant to yellow.

a) Explain how two black dogs can have different genotypes.

b) Could a black dog have the same genotype as a yellow dog?

c) How could two black dogs have a yellow pup?

2. A pea plant with round seeds is crossed with a pea plant that has wrinkled seeds. For the cross, indicate each of the following:

a) the genotype of each of the parents if the round seed plant is heterozygous.

b) the gametes produced by the parents

c) the genotypes and phenotypes of the F₁ generation

d) the F₂ generation if two round plants from the F₁ generation were crossed

3. A pea plant with a tall phenotype is pollinated by a short plant, and the seeds of the first generation hybrid produce 327 tall plants and 321 short plants. Give the genotypes of all the plants. (parents, Tt and tt; all tall were Tt; dwarfs, tt)

4. In a certain species of plant, one purebred variety has hairy leaves and another pure variety has smooth leaves. A cross of the 2 varieties produces offspring that all have smooth leaves. Predict the ratio of phenotypes in the F₂ generation.

(3:1, smooth to hairy)

5. In summer squash, white fruit color (yep, it’s a fruit) is dominant and yellow is recessive. A squash plant that is homozygous for white is crossed with a homozygous yellow one. Predict the appearance of

a) the F₁ generation (F₁ = all white)

b) the F₂ (F₂ = 3/4 white and 1/4 yellow)

c) the offspring of a cross between an F₁ individual and a homozygous white individual. (A crossbetween an F₁ individual and a homozygous white individual wouldproduce only white offspring.)

6. For Dalmatian dogs, the spotted condition is dominant to non-spotted.

a) Using a Punnett square, show the cross between two heterozygous parents.

b) A spotted female Dalmatian dog is mated to an unknown male. From the appearance of the pups, the owner concludes that the unknown male was a Dalmatian. The owner notes that the female had six pups, three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?

7. In horses, the trotter characteristic is dominant to the pacer characteristic. A male trotter mates with three different females, and each female produces a foal. The first female, a pacer, gives birth to a foal that is a pacer. The second female, also a pacer, gives birth to a foal that is a trotter. The third female, a trotter, gives birth to a foal that is a pacer. Determine the genotypes of the male, all three females, and the three foals sired.

8. Imagine for hair color that B gives brown hair and b gives blonde hair. Use a Punnett square to determine the following in a cross of two heterozygous parents.

a) What are the chances of the offspring being homozygous brown haired?

b) What are the chances of the offspring having blonde hair?

c) What are the chances of the offspring being heterozygous brown haired?

d) What is the genotypic ratio?

e) What is the phenotypic ratio?

f) Is there a heterozygous blonde haired offspring? Why?

g) If curly hair is dominant to straight hair, what letters will we use to show these genes?

h) A heterozygous curly haired male marries a straight haired female. What are their genotypes?

i) What would be the gametes for the male parent?

j) What would be the gametes for the female parent?

k) What are the chances of the offspring being homozygous curly haired?

l) What are the chances of the offspring having straight hair?

m) What are the chances of the offspring being heterozygous curly haired?

n) What is the genotypic ratio?

o) What is the phenotypic ratio?

p) Is there a heterozygous straight haired offspring? Why?

9. Thalassemia is a serious human genetic disorder that causes severe anemia. The homozygous condition (T^mT^m) leads to severe anemia. People with thalassemia die before sexual maturity. The heterozygous condition (T^mTⁿ) causes a less serious form of anemia. The genotype TⁿTⁿ causes no symptoms of the disease. Indicate the possible genotypes and phenotypes of the offspring if a male with the genotype T^mTⁿ marries a female of the same genotype.

10. In pea plants, tall is dominant to short and yellow is dominant to green. Show the F₁ and F₂ results of a cross between a homozygous tall pea plant that produces yellow seeds, and a short plant that produces green seeds.

11. In guinea pigs, black coat color (B) is dominant to white (b), and short hair length (S) is dominant to long (s). Indicate the genotypes and phenotypes from the following crosses:

a) Homozygous for black, heterozygous for short hair guinea pig crossed with a white, long hair guinea pig.

b) Heterozygous for black and short hair guinea pig crossed with a white, long hair guinea pig.

c) Homozygous for black and long hair crossed with a heterozygous black and short hair guinea pig.

d) For each of these crosses, give the probability that an offspring will have:

            (i) black coat, long hair

            (ii) black coat, short hair

            (iii) white coat, long hair

12. For guinea pigs, black fur is dominant to white fur color. Short hair is dominant to long hair. A guinea pig that is homozygous for white and homozygous for short hair is mated with a guinea pig that is homozygous for black and homozygous for long hair. Indicate the phenotype(s) of the F₁ generation. If two hybrids from the F₁, generation are mated, determine the phenotypic ratio of the F₂ generation.

13 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. Indicate the genotypes of the parents.

a) The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid.

b) The mating with female B, which is black, solid, produces a single pup, which is white, spotted.

c) The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted.

14. T = Tall, t = short B = brown hair, b = blonde hair

Cross a homozygous tall, heterozygous brown haired male with a heterozygous tall, blonde haired female. Use a Punnett square to show parents and gametes.

a) What is the phenotypic ratio?

b) How many offspring are homozygous for both characteristics?

c) How many offspring are heterozygous for both characteristics?

15. Assume that curly hair (C) is dominant to straight hair (c). Albinism is a condition in which cells which normally produce pigment do not do so. The allele for skin albinism (a) is recessive to the normal allele (A). A woman with curly hair and albinism and a man with straight hair and normal pigmentation have a child that has straight hair and is an albino. What are the genotypes of the parents?

16. PKU is a recessive disorder. Suppose two people who were heterozygous for PKU (i.e., are carriers) married. What is the probability of each of the following?
a) All three of their children will be of normal phenotype.
b) One or more of the children will have the disease.
c) All three children have the disease.
d) At least one child will be phenotypically normal.
(Remember that the probabilities of all possible outcomes always ad up to 1)

17. For chickens, the gene for rose comb (R) is dominant to that for single comb (r). The gene for feather legged (F) is dominant to that for clean legged (f). Four feather legged, rose combed birds mate. Rooster A and hen C produce offspring that are all feather legged and mostly rose combed. Rooster A and hen D produce offspring that are feathered and clean, but all have rose combs. Rooster B and hen C produce birds that are feathered and clean. Most of the offspring have rose combs, but some have single combs. Determine the genotypes of the parents.

18. In a disputed paternity case, a woman with blood type B has a child with type O, and she claimed that it had been fathered by a man with type A. What can be proved from these facts? (man could be father but this is not proof)

19. Multiple alleles control the intensity of pigment in mice. The gene D¹ designates full color, D² designates dilute color and D³ is deadly when homozygous. The order of dominance is D¹ > D² > D³. When a full color male is mated to a dilute color female, the offspring are produced in the following ratio: two full color to one dilute to one dead. Indicate the genotypes of the parents.

20. Multiple alleles control the coat color of rabbits. A gray color is produced by the dominant allele C. The C^ch allele produces a silver-gray color when present in the homozygous condition, C^chC^ch, called chinchilla. When C^ch is present with a recessive gene, a light silver-gray color is produced. The allele C^h is recessive to both the full color allele and the chinchilla allele. The C^h allele produces a white color with black extremities. This coloration pattern is called Himalayan. An allele C^a is recessive to all the other alleles. The C^a allele results in a lack of pigment, called albino. The dominance hierarchy is C > C^ch > C^h > C^a. The table below provides the possible genotypes and phenotypes for coat color in rabbits. Notice that four genotypes are possible for full color but only one for albino.

Phenotypes     Genotypes

Full color        CC, CC^ch, CC^h, CC^a

Chinchilla       C^chC^ch

Light gray       C^chC^h, C^chC^a

Himalaya        C^hC^h, C^hC^a

Albino                        C^aC^a

a) Indicate the genotypes and phenotypes of the F₁ generation from the mating of a heterozygous Himalayan coat rabbit with an albino coat rabbit.

b) The mating of a full color rabbit with a light gray rabbit produces two full color offspring, one light gray offspring, and one albino offspring. Indicate the genotypes of the parents.

c) A chinchilla color rabbit is mated with a light gray rabbit. The breeder knows that the light grey rabbit had an albino mother. Indicate the genotypes and phenotypes of the F₁ generation from this mating.

d) A test cross is performed with light gray rabbit, and the following offspring are noted: five Himalayan color rabbits and five light gray rabbits. Indicate the genotype of the light-gray rabbit.

21. A geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. If oval radishes are crossed with oval radishes, the following phenotypes are noted in the F₁ generation: 100 long, 200 oval, and 100 round radishes. Use symbols explain the results obtained for the F₁ and F₂ generations.

22. Palomino horses are known to be caused by the interaction of two different genes. The allele C^r in the homozygous condition produces a chestnut, or reddish color, horse. The allele C^m produces a very pale cream color, called cremello, in the homozygous condition. The palomino color is caused by the interaction of both the chestnut and cremello alleles. Indicate the expected ratios in the F₁ generation from mating a palomino with a cremello.

23 In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. The recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light-brown, color is produced. The genes that produce black or tan coat color rely on the gene C, which produces pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) CCBB x Ccbb (b) ccBB x CcBb (c) CcBb x ccbb (d) CcBb x CcBb

24. The mating of a tan mouse and a black mouse produces many different offspring. The geneticist notices that one of the offspring is albino. Indicate the genotype of the tan parent. How would you determine the genotype of the black parent?

25. The gene R produces a rose comb in chickens. An independent gene, P, which is located on a different chromosome, produces a pea comb. The absence of the dominant rose comb gene and pea comb allele (rrpp) produces birds with single combs. However, when the rose and pea comb alleles are present together, they interact to produce a walnut comb (R_P_).

Indicate the phenotypes of the parents and give the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) rrPP x RRpp (b) RrPp x RRPP (c) RrPP x rrPP (d) RrPp x RrPp

26. For shorthorn cattle, the mating of a red bull and a white cow produces a calf that is described as roan. Roan results from intermingled red and white hair. Many matings between roan bulls and roan cows produce cattle in the following ratio: 1 red, 2 roan, 1 white. Is this a problem of codominance or multiple alleles? Explain your answer.

27. For ABO blood groups, the A and B genes are codominant but both A and B are dominant over type O. Indicate the blood types possible from the mating of a male who is blood type O with a female of blood type AB. Could a female with blood type AB ever produce a child with blood type AB? Could she ever have a child with blood type O?

28. For mice, the allele C produces color. The allele c is an albino. Another allele, B, causes the activation of the pigment and produces black color. The recessive allele, b, causes the incomplete activation of pigment and produces brown color. The alleles C and B are located on separate chromosomes and segregate independently. Determine the F₁ generation from the cross CcBb x CcBb.

29. Baldness (H^B) is dominant in males but recessive in females. The normal gene (Hⁿ) is dominant in females, but recessive in males. Explain how a bald offspring can he produced from the mating of a normal female with a normal male. Could these parents ever produce a bald girl? Explain your answer.

30. Use the phenotype chart (pedigree) at the right to answer the following questions.

a) How many children do the parents A and B have?

b) Indicate the genotypes of the parents.

c) Give the genotypes of M and N.