Exercise #1: Monohybrid Crosses

In solving a genetics problem there are five basic steps:

  1. Determine the genotype of each parent.
  2. Determine the possible types of gametes each parent can produce.
  3. Determine all possible gene combinations that may result when these gametes combine.
  4. Determine the various phenotypes possible by analysing the various genotypes.
  5. Determine genotypic and phenotypic ratios as required.
  1. A pure black male cat mates with a white female. Black coat colour is the product of a dominant allele. Show the genotypes and phenotypes of the parental, F1 and F2 generations. Indicate the phenotypic and genotypic ratios of the F2 generation.
  2. In humans, six fingers (F) is the dominant trait and five fingers (f) is the recessive trait. Both parents are heterozygous for six-fingers. Indicate the genotypes and phenotypes of the parents and their possible offspring. What is the probability of producing a five-fingered child?
  3. In cattle the polled (hornless) trait is dominant over the horned trait. A purebred polled bull is bred with a horned cow. Show the genotypes and phenotypes of the parental, F1 and F2 generations. Indicate the phenotypic and genotypic ratios of the F2 generation.
  4. A single pair of genes in which the Siamese pattern is recessive to the ordinary solid coat colour pattern controls the Siamese coat pattern in cats. Predict the phenotypes, genotypes and their probable proportions in the kittens of a homozygous solid coat male with a Siamese female cat.
  5. In a certain species of plant, one purebred variety has hairy leaves and another purebred variety has smooth leaves. A cross of the two varieties produces offspring that all have smooth leaves. Predict the phenotypic and genotypic ratio of the F2 generation.
  6. Melanin pigments range in colour from yellow to reddish-brown to black. The amount and the colour of melanin in the skin account for differences in human skin coloration. Albinism is a genetic disorder that results in unpigmented skin and other tissues. About 1 in 20 000 humans has albinism. In humans, it can be caused by an autosomal recessive allele (a). Its dominant allele (A) results in normal pigmentation.

a) If a homozygous albino marries a homozygous person of normal pigmentation, what would be the expected phenotype and genotype of their children?

b) If an albino married a heterozygous person of normal pigmentation, what phenotypes would you expect in their children?

c) Two parents of normal skin pigmentation had an albino child. How was this possible?

  1. In Drosophila, the common fruit fly, the normal grey body is dominant to the recessive trait of black body. How can you determine the unknown genotype of a grey bodied fly?
  2. Black colour in guinea pigs is dominant over white. Outline a cross that would make it possible to determine if a black male guinea pig is homozygous or heterozygous.
  3. Wild red foxes occasionally have silver-black pups. If two silver-black foxes mate, their offspring are all silver-black. Explain the inheritance of these coat colours in foxes. Use a Punnett square to provide evidence for your answers.
  4. When 20 purebred Himalayan rabbits are mated with a solid gray male of unknown ancestry, 46 of the offspring are Himalayan and 52 are gray. A single pair of genes controls coat colour in rabbits.

a)Is the Himalayan coat pattern controlled by a recessive or a dominant allele?

b)What are the genotypes of the female and male rabbits?

c)How many offspring of each phenotype would you expect from the cross outlined above? Use a Punnett square to provide evidence for your answer.

d)In rabbits, certain short-haired individuals crossed with long-haired ones produce only short-haired individuals. Other short-haired ones when crossed with long-haired ones produce approximately equal numbers of short-haired and long-haired offspring. When long-haired rabbits are crossed, they always produce long-haired offspring like themselves.

  1. Which trait is do you hypothesize is dominant?
  2. Give the genotypes of all the rabbits described:

First cross: short-haired rabbit

  • long-haired rabbit
  • short-haired offspring

Second cross: short-haired rabbit

  • long-haired rabbit
  • short-haired offspring
  • long-haired offspring
  1. In humans, brown eyes are dominant to blue. Both parents of a blue-eyed man are brown eyed. The blue-eyed man, Ed, marries a brown-eyed woman, Sharon, whose mother had brown eyes and whose father had blue eyes. The woman has a brother who has blue eyes. Ed and Sharon marry and have a brown-eyed child. Give the genotypes of all the individuals described.

Ed's mother _____ Sharon's mother _____

Ed's father _____ Sharon's father _____

Ed _____ Sharon's brother _____

Sharon _____

Ed and Sharon's child _____

  1. What are the chances that the first child from a marriage of two heterozygous brown-eyed parents has blue eyes? If the first child has brown eyes, what are the chances that the second child will also have brown eyes? Explain your reasoning.
  2. When two rough-coated guinea pigs are bred, the resulting offspring consisted of 18 rough- and 4 smooth-coated offspring. Which type of coat is dominant? What proportion of the offspring are homozygous for the dominant trait?
  3. In peas, inflated pods are the product of a dominant allele and constricted pods are produced by a recessive allele. Long stems are the product of a dominant allele and short stems are produced by a recessive allele.
    a) What symbols would be used for coding these genes?

b) What are the two possible genotypes of a plant, which has the phenotype of inflated pods and short stems?

c) What is the genotype of a plant, which is a hybrid for pod shape and stem length (heterozygous for both traits)?

  1. The polled or hornless condition is dominant over the horned trait in cattle. A certain polled bull is bred to three cows. Cow A, which is horned, produces a horned calf. Cow B, which is also horned, produces a polled calf. Cow C, which is polled, produces a horned calf. What are the genotypes of the animals described? What further offspring could be expected from future mating of these animals?

Bull _____

Cow A _____ Her horned calf _____

Future calves _____

Cow B _____ Her polled calf _____

Future calves _____

Cow C _____ Her horned calf _____

Future calves _____

  1. In Holstein-Friesian cattle, the two colours black-and-white and red-and-white are controlled by a single pair of alleles. A black-and-white bull was bred with 20 red-and white cows. All of the resulting calves were black-and-white.

a)What allele for the trait of coat colour is dominant?

b)When the F1 calves grew to maturity, the ones that had the most desirable traits were mated. What would be the expected phenotypic ratio of coat colours of the F2 calves?

  1. Sickle cell anemia is caused by the sickle cell allele (HbS) of a gene that contributes to hemoglobin (Hb) production. The abnormal hemoglobin (hemoglobin-S) produced causes red blood cells to become deformed and block capillaries. Tissue damage results. Affected individuals homozygous for the sickle cell gene rarely survive to reproductive age. Heterozygous individuals produce both normal hemoglobin and a small percentage of hemoglobin-S. These individuals are more resistant to malaria than individuals who are homozygous for the allele for normal hemoglobin (HbA). Their red blood cells are prone to sickling when there is a deficiency of oxygen.If a man and a woman who are both heterozygous for the alleles HbA and HbS have a child, what is the probability that the child would not be heterozygous?
  2. In Drosophila, the common fruit fly, the normal gray body is dominant to the recessive trait of black body. A gray-bodied male fly was allowed to breed with many black-bodied females. Some of the offspring had gray bodies and some had black bodies.

a)What were the genotypes of the parents and the offspring?

b)Two gray-bodied flies were mated, and all the offspring had gray bodies. Can you conclusively determine the genotypes of the parents? Explain your answer.

c)Two gray bodied flies were mated, and both gray and black bodies were observed in their offspring. What were the genotypes of the parents? What were the genotypic and phenotypic ratios of the offspring?

  1. Marfan syndrome is an autosomal-dominant disorder of humans. Affected individuals tend to be tall and thin. They have defects in the lens of the eye and weak connective tissue around the aorta. Often, affected individuals excel in sports like volleyball or basketball, but it is not uncommon for people with this syndrome to die suddenly.A man, heterozygous for Marfan syndrome and a homozygous recessive woman have a child. What is the probability that the child will be affected by Marfan syndrome? If the couple’s first child has Marfan, what is the probability of their second child also having this disorder?

Exercise #2: Other Patterns of Inheritance

Codominance/Incomplete Dominance/Multiple Alleles

  1. In the four o’clock flower, similar to a petunia, the allele for red flower colour is incompletely dominant over the allele for white flower colour. A heterozygous plant has pink flowers showing intermediate inheritance.
  1. Show the genotypes of the parents and F1 generation of a cross between a red flowered and a white flowered four o’clock plant.
  2. What would be the anticipated offspring if an F1 plant were back-crossed to the red flowered parent?
  3. What would be the anticipated offspring if an F1 plant were back-crossed to the white flowered parent?
  1. In the radish plant three shapes are observed in the root - round, long and oval. Different crosses of radishes gave the following results:
  2. long x oval produces 52 long and 48 oval
  3. long x round produces 98 oval
  4. oval x round produces 51 oval and 50 round
  5. oval x oval produces 24 long, 53 oval, and 27 round

Explain the inheritance of root shape in radishes. Using symbols, demonstrate that your hypothesis is true for all crosses.

  1. In a certain strain of chickens, a mating between a black chicken and a white chicken always produces offspring which have a distinctive feather appearance called blue Andalusian. A cross between two blue Andalusians produces blue, black and white offspring.
  1. Using Punnett squares, illustrate the two crosses described.
  2. What are the phenotypic and genotypic ratios of the F2 generations?
  3. What would be the expected phenotypic ration if a blue Andalusian hen were bred with a black rooster?
  1. Blood typing:The alleles for A (IA) and B (IB) are codominant, and both are dominant to O (i).The alleles for M and N are codominant. The allele for RH+ (R) is dominant to the allele for RH- (r).
  1. Complete the table below to indicate all possible genotypes for the phenotypes indicated.

Phenotypes / Genotypes
Type A
Type B
TypeAB
Type O
  1. A woman having type A blood claims that her former husband who has type B blood is the father of her baby. His mother has type B blood and his father has type A blood. The baby has type O blood. The man denies that he is the father of the child and refuses to pay child support. Show how you would determine if the man is in fact the father of the child.
  2. If a woman with the genotype IAIBRr and a man with the blood type O Rh- have a child, what is the probability that the child will have blood type A Rh-.
  3. Which children could belong to a couple in which the woman has blood type A, N, Rh+ and the man has blood type O, M, Rh+.

ChildBlood Types

One O MN Rh+

Two A N Rh+

Three A MN Rh-

Four AB MN Rh-

  1. Coat colour in mice is dependent on members of a series of multiple alleles. The hierarchy of dominance is : C+ (full colour) > Cch (chinchilla) > Cd (blonde) > c (albino)
  1. Complete the table below to indicate all possible genotypes for the phenotypes indicated.

Phenotypes Genotypes

Full colour

Chinchilla

Blonde

Albino

  1. A chinchilla female which was heterozygous for albino was mated with a full colour male which was heterozygous for blonde. What phenotypes could be expected in their offspring?
  1. Multiple alleles control the intensity of pigment in mice. The gene D1 designates full colour, D2 designates diluted colour, and D3 is deadly when homozygous. The order of dominance is D1>D2>D3. When a full colour male is mated to a dilute colour female, the offspring are produced in the following ratio: two full colour to one dilute to one dead. Indicate the genotypes of the parents.

Exercise #3 Dihybrid Crosses

  1. Horses which are black in colour and which have a trotting gait carry autosomal dominant genes, while horses, which are chestnut in colour and have a pacing gait, carry autosomal recessive genes. A purebred black trotting stallion is bred with a chestnut-pacing mare.

a)What are the parental genotypes and what are the possible sperm and eggs produced?

b)What would the genotype and phenotype of the F1 offspring be?

c)What kind of gametes could the F1 offspring produce at maturity?

d)Construct a Punnett square to show the possible F2 offspring if two F1 horses were bred maturity.

e)What is the expected phenotypic ratio of the F2 generation if many of these breeding took place?

  1. In summer squash, white fruit is dominant over yellow fruit while disc is dominant over sphere shape. A plant which homozygous for white fruit and sphere shape is cross-pollinated with one which is homozygous for yellow fruit colour and disc shape.

a)What will be the genotype and phenotype of the F1?

b)What is the phenotypic ratio of the F2 generation?

c)If a F1 plant were crossed with pollen from its yellow disc parent, what would be the expected phenotypes of the offspring and in what ratio?

  1. In certain breeds of dogs, black colour is determined by an autosomal dominant gene and red colour is due to an autosomal recessive gene. Solid coat is dominant trait and spotted coat is recessive. A male which is heterozygous black and homozygous white spotted is bred with a female which red and white spotted.

a)What is the probability of producing a pup which is red and white spotted?

b)What is the probability of these two dogs producing a solid black puppy?

  1. In garden peas, the allele for tall plant height (T) is dominant over the allele for short plant height (t), and the autosomal allele for axial flower position (A) is dominant over the allele for terminal flower position (a). A plant heterozygous for both traits was crossed with plant homozygous recessive for both traits.

a)What percentage of the offspring produced would be expected to display at least one of the dominant traits?

b)Predict what fraction of the offspring will be heterozygous?

  1. In pea plants, tall (T) dominant over short (t), and round seed (R) is dominant over wrinkled seed (r). A heterozygous tall-heterozygous round –seed pea plant and a short-heterozygous round-seed pea plant were crossed.

a)What is the probability of producing an offspring that is homozygous recessive for both traits?

b)What portion of the offspring are tall and round?

  1. R is the allele for red flesh fruit colour in tomatoes, while r is the allele for yellow fruit. P is the allele which gives tomato stem a purplish colour, and p shows up as a greenish stem. Suppose a cross was made between two tomato plants which resulted in the following kinds of offspring:

247 red fruit, purple stem

261 red fruit, green stem

253 yellow fruit, purple stem

256 yellow fruit, green stem

What would be the phenotypes and genotypes of the parent plants?

  1. A homozygous dominant snapdragon having red flowers and narrow leaves was crossed with one having white flowers and broad leaves. Indicate the genotypes and phenotypes of the parental, F1 and F2 generations.

Genetics Exercise #4

Polygenic characteristics: Epistatic Genes

One Gene Many Effects: Pleiotropic Genes

  1. Assume that there are two gene pairs involved in determining eye colour: one codes for pigment in the front of the iris and other codes for pigment in the back of the iris.

Genotype Eye Colour

AABB black-brown

AABb dark brown

AAbb brown

AaBB brown-green flecked

AaBb light brown

Aabb grey-blue

aaBB green

aaBb dark blue

aabb light blue

  1. A man has gray-blue eyes and a woman has green eyes. Which eye colour phenotypes would be possible for children born to this man and woman?
  2. If one parent has light brown eyes and the other has dark brown eyes, what is the probability that they would have an offspring with gray-blue eyes?
  1. In humans, normal pigmentation dominates no pigmentation (albino). Black hair dominates blonde hair. An albino person will have white hair color even though they may also have the genes for black or blonde hair colour. An albino male who is homozygous for black hair marries a woman who is heterozygous for normal pigmentation and who has blond hair. What colours of hair can their children have and what is the probability for each hair colour?
  2. 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 colour. The recessive gene, b, causes incomplete breakdown of the pigment, and a tan or light-brown colour is produced. The genes that produce black or tan coat colour rely on the gene C, which produces the pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F1 generations from the following crosses:
  3. CCBB x Ccbb
  4. ccBB x CcBb
  5. CcBb x ccbb
  6. CcBb x CcBb
  7. 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?

  1. 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 gene (rrpp) produces birds with single combs. However, when the rose and pea comb genes are present together, they interact to produce a walnut comb (R_P_). Indicate the phenotypes of the parents and give the genotypic ratios of the F1 generations from the following crosses:
  2. rrPP x RRpp
  3. RrPp x RRPP
  4. RrPP x rrPP
  5. RrPp x RrPp

Chromosome Mapping