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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 fivefingered 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?
7. 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?
8. 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.
9. 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.
10. 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.
i. Which trait is do you hypothesize is dominant?
ii. 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
11. 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 _____
12. What are the chances that the first child from a marriage of two heterozygous browneyed 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.
13. 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?
14. 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)?
15. 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 _____
16. 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 F 2 calves?
17. 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?
18. 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?
19. 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.
a. Show the genotypes of the parents and F1 generation of a cross between a red
flowered and a white flowered four o’clock plant.
b. What would be the anticipated offspring if an F 1 plant were back-crossed to the red
flowered parent?
c. What would be the anticipated offspring if an F 1 plant were back-crossed to the white
flowered parent?
2. In the radish plant three shapes are observed in the root - round, long and oval. Different
crosses of radishes gave the following results:
a.
b.
c.
d.
long x oval produces 52 long and 48 oval
long x round produces 98 oval
oval x round produces 51 oval and 50 round
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.
3. 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.
a. Using Punnett squares, illustrate the two crosses described.
b. What are the phenotypic and genotypic ratios of the F2 generations?
c. What would be the expected phenotypic ration if a blue Andalusian hen were bred
with a black rooster?
4. 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).
a. Complete the table below to indicate all possible genotypes for the phenotypes
indicated.
Phenotypes
Type A
Type B
Type AB
Type O
Genotypes
b. 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.
c.
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-.
d. 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+.
Child Blood Types
One O MN Rh+
Two A N Rh+
Three A MN RhFour AB MN Rh5. 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)
a. Complete the table below to indicate all possible genotypes for the phenotypes
indicated.
Phenotypes
Genotypes
Full colour
Chinchilla
Blonde
Albino
b. 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?
6. 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 chestnutpacing 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?
2. 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 crosspollinated 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?
3. 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?
4. 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?
5. 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 shortheterozygous 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?
6. 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?
7. 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
a. 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?
b. 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?
2. 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?
3. 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:
1.
2.
3.
4.
CCBB x Ccbb
ccBB x CcBb
CcBb x ccbb
CcBb x CcBb
4. 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?
5. 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:
1. rrPP x RRpp
2. RrPp x RRPP
3. RrPP x rrPP
4. RrPp x RrPp
Chromosome Mapping
1. The recombination frequency between gene A and gene B is 9%; between A and C is
17%; and between B and C is 26%. What is the sequence of genes in the chromosome?
Sketch a map of the chromosome, showing the map unit distances between the genes.
2. The crossing-over frequency between genes A and B is 35%; between B and C 10%;
between C and D 15%; between A and C 25%; between A and D is 10%, and between B
and D 25%. What is the sequence of genes in the chromosome? Sketch a map of the
chromosome, showing the map unit distances between the genes.
3. The crossing-over frequency between genes A and B is 5%; between B and C 17%;
between C and D 18%; between A and C 12%; between B and D 1%; and between A
and D 6%. What is the sequence of genes in the chromosome? Sketch a map of the
chromosome, showing the map unit distances between the genes.
4. In a series of breeding experiments, a group of genes located on the same chromosome
was found to show the recombination frequencies in the chart below. Using this data,
map the chromosome.
GENE
A B
A B
C
D
E
8 12
4
1
8 -
4
12 9
Recombinations
C 12 4 -
16 13
per 100 fertilized eggs
D 4 12 16 - 3
E 1 9
13 3 -
5. For a breeding experiment, a linkage group composed of genes W, X, Y, and Z was
found to show the following gene combinations. Use this data to construct a gene map.
Gene
WXYZ
W - 5 78
X 5 - 23
Y 7 2 -1
Z 8 3 1Crossover Frequencies for Some Genes on an Autosome of Organism Z
P and Q 5%
P and R 8%
P and S 12%
Q and R 13%
Q and S 17%
6. Which chromosome map best represents the sequence of genes on the autosome from
organism Z?
7. The use of marker genes and the analysis of crossover frequencies of genes have
enabled geneticists to map the location of many genes on human chromosomes. Blue
colour vision and blue colour blindness (tritanopia) are both controlled by a gene on
chromosome 7. The gene for the production of trypsin (a digestive enzyme) and the gene
responsible for cystic fibrosis are also found on chromosome 7. Some crossover
frequencies of these genes are shown below.
Pair of Genes Crossover Frequency
Marker gene cystic fibrosis 18%
Marker gene tritanopia 13%
Cystic fibrosis trypsin 6%
Trypsin tritanopia 1%
Which of the following gene maps shows the correct sequence of these genes on
chromosome 7?
Note: For all genetics problems involving sex-linked traits, state the results of the two sexes
separately.
Exercise #6: Sex linkage
1. Colour blindness in humans is a sex-linked recessive trait. A woman of normal colour
vision whose father was colour blind marries a man of normal vision whose father was
also colour blind. What type of colour vision will be expected in their children?
2. Mr. and Mrs. White have normal colour vision. They have three children: Bob, who is
colour blind and who has a daughter of normal colour vision; Joan, who has normal
colour vision and who had one son who is colour blind and one son who is normal; and
Susan, who also has normal colour vision and who has five sons of normal colour vision.
What are the genotypes of the individuals described?
3. Peter's maternal grandmother had normal colour vision and his maternal grandfather was
colour blind. Peter's mother is colour blind, and his father has normal colour vision.
A. What are the genotypes of the individuals described?
B. What type of colour vision does Peter have?
C. What type of colour vision do his sisters have?
D. If Peter were to marry a woman who is genotypically the same as his sister Joan,
what type of colour vision would be expected in their children?
4. Eye colour in Drosophila is sex-linked. It is controlled by a pair of genes in which red is
dominant over white.
A. If a red-eyed male fruit fly is crossed with a white-eyed female, what will be the
phenotypes of the offspring?
B. If a male F1 is mated with a white-eyed female, what will be the appearance of
the offspring produced?
C. If a female F1 is mated with a red-eyed male, what will be the appearance of the
offspring produced?
D. If a white-eyed female is crossed with a red-eyed male and the F1 are allowed to
freely interbreed, what will be the eye colour of the F2 offspring?
5. "Alligator men" or "fish women" were exhibited for their physical abnormalities in fairs or
circuses earlier this century. These people probably suffered from X-linked ichthyosis,
which produces symmetric dark scales on the body. The disease occurs in 1 in 6 000
males and is much more rarely in females. Ichthysosis is likely a recessive disorder. If an
"alligator man" were to marry a woman homozygous for the normal condition, what is the
percentage probability that their children would have ichthyosis?
6. Baldness is a sex-influenced trait. Two autosomal alleles (HB and HN) exist for the pattern
baldness gene. The HB allele is dominant in men but recessive in women. The H N
(normal) allele is dominant in females but recessive in males. A single H B allele seems to
cause pattern baldness only in the presence of the level of testosterone normally found in
adult males. Males will develop pattern baldness with the genotypes H BHN or HBHB;
whereas, for females to develop pattern baldness, they must inherit two H B alleles. It
appears that testosterone destroys or inhibits the production of an enzyme necessary for
hair growth in the hair follicle.
A. What is influencing the difference in expression of the pattern baldness gene in
women and men?
B. Parents who do not display pattern baldness have a son who exhibited pattern
baldness by the age of thirty. What are his possible genotypes?
C. If they also have a daughter, could she also display baldness. Explain.
HUMAN PEDIGREES
1. Haemophilia is a recessive X-linked condition that used to be called the "bleeder's disease"
because the blood of affected individuals takes a long time to clot. Normal blood usually clots
within five minutes after being placed in a test tube, whereas haemophiliac blood may require
several hours to clot. Without treatment, even common bruises often lead to serious internal
bleeding in the haemophiliac individual.
Pedigree Three: Haemophilia
a. Why do more males have haemophilia than females?
b. Explain the conditions required for female (III-3) to have haemophilia.
2. Below is a diagram showing the inheritance of blue-green colour blindness. A black symbol
indicates colour blindness and a white symbol indicates normal colour vision.
a) How is this trait inherited?
b) Write in the genotype for each individual.
c) What evidence is there that colour blindness is a sex-linked trait?
d) Give the possible genotypes of all the individuals.
3. Marfan syndrome is an inherited condition that affects the connective tissue, resulting in
unusually long bones and spinal curvature, as well as vision, cardiac, and respiratory problems.
The syndrome tends to become increasingly severe over time. The following pedigree shows
inheritance of Marfan syndrome in a multigenerational family.
a) How is this syndrome inherited?
b) Can you determine individual II4’s genotype? Explain.
c) Individual II1 and II2 are considering having another child. What is the probability that this
child will have Marfan syndrome? Explain using a Punnett square.
4. Examine the following pedigree showing the inheritance of straight hair in a four-generation
family.
a) Is straight hair a dominant or recessive trait? Explain.
b) Identify the genotype of each individual in the pedigree. Whose genotype can you not be
certain of?
c) If individual V3 marries a man who is heterozygous, what is the probability that they will
have a girl with straight hair? Explain using a Punnett square.
5.
Humans may have a peaked or smooth hairline. If a man and a woman both have a smooth
hairline, none of their children will have peaked hairlines. How is a peaked hairline most likely
inherited? Draw a pedigree for a family where one parent and two of three children have a
peaked hairline. One of the children with a peaked hairline marries an individual with a
smooth hairline. Their children both have a peaked hairline. Identify the genotypes and
phenotypes of each individual.
6. As a top research scientist, you and your colleagues have discovered a new condition. In the
course of your research, you have come across a pedigree (see below) for a family in which
the condition occurs.
a) On which chromosome is the allele for the condition found?
b) Which individuals in the pedigree can you be certain are carriers of the allele? Explain your
reasoning.
c) If individual III3 has a son with a woman who is not a carrier of the allele, what is the
probability that the son will have the condition? Explain.
7. In the pedigree below, different blood types are identified by the letters A, B, AB, and O.
a) Individuals II4 and II5 have just had identical twin girls. List the possible blood types these
infants may have based on the information provided in the pedigree.
b) Individuals II6 and II7 have a second child with blood type O. What does this tell you about
II6’s genotype?
c) Could I1 and I2 have a child with the AB blood type? Explain why or why not.
Genetics Exercise #7: Mixed Problems
1. In pea plants, round seeds and tall stems are dominant to wrinkled seeds and short stems.
Determine the phenotypes and their proportions for the following crosses.
a. RRtt x rrTT
b. RrTt x RrTt
c. Rrtt x rrTt
2. In peas, yellow seed colour is dominant to green. What will be the colours and their proportions
in the offspring of the following crosses?
a.
b.
c.
d.
homozygous yellow x green
heterozygous yellow x green
heterozygous yellow x homozygous yellow
heterozygous yellow x heterozygous yellow
3. In Drosophila wing length is controlled by autosomal genes in which the normal long-wing
allele is dominant over the vestigial-wing allele. Vestigial wings are very small and non-functional.
a) If a female fruit fly having white eyes and homozygous long wings is crossed with
a male having red-eyes and vestigial wings, what will be the appearance of the
F1?
b) What will be the appearance of the F2?
c) If an F1 female is mated with a male, which has the same genotype as her father,
what will be the appearance of the offspring?
d) If an F1 male is mated with a female, which has the same genotype as his
mother, what will be the appearance of the offspring?
4. In humans, the condition for normal blood clotting (H) dominates the condition "bleeder's
disease". The gene controlling blood clotting is carried on the X chromosome.
a. A male haemophiliac marries a woman who is a carrier for this condition. What
are the chances that if they have a male child he will be normal for blood
clotting?
b. A male who has normal blood clotting marries a woman who is a carrier for
haemophilia. What are the chances that if they have a son he will be normal for
blood clotting?
c. Rarely a female haemophiliac is born. How could this happen?
5. In cats, short hair is dominant over long and solid black coat is dominant over he Siamese coat
pattern of black and tan. A cat breeder wants to know if a certain black, shorthaired cat is
homozygous or heterozygous. How can this be determined?
6. A cross between two sweet pea plants produced 41 plants with pink flowers, 18 with white
flowers, and 19 with red flowers.
a. How is flower colour inherited in sweet peas?
b. What is the phenotype of the parents? Explain how you determined this.
7. Using the information from the previous question, determine the expected phenotypes and
their proportions among the offspring of the following crosses.
a. white x pink
b. red x red
c. pink x pink
8. Plumage colour in ducks is dependent on a set of three alleles: Mr produces restricted mallard
plumage, M produces mallard, and m produces dusky mallard. The dominance hierarchy is Mr >
M > m. Determine the genotypic and phenotypic ratios expected in the F 1 from the following
crosses.
a. MrMr x MrM
b. MrMr x Mrm
c. MrM x Mrm
d.
Mrm x
9. A young woman with type O blood gave birth to a baby with type O blood. In a court case she
claims that a certain young man is the father of her child. The man has type A blood. Could he be
the father of her child? Can it be proven on this evidence alone that he is the father?
10. Normal pigmentation (A) dominates no pigmentation (albino = aa). Brown-eyed colouring (B)
dominated blue-eyed colouring (bb). Two people with normal skin pigmentation produce one
brown-eyed, three blue-eyed, and one albino child. What are the parents' possible genotypes?
11. In rabbits, short hair is due to a dominant gene L, and long hair to its recessive allele l. Black
fur is due to a dominant gene B and white hair to its recessive allele b. Two rabbits were crossed
several times and they produced a total of 88 shorthaired black and 29 longhaired black offspring.
What are the probable genotypes of the parents?
12. Below is a pedigree showing the inheritance of the common condition of myopia (shortsightedness). A square represents a male and a circle represents a female. A white symbol
indicates normal vision and a black symbol represents myopia.
a. Is the allele controlling myopia dominant or recessive? Sex-linked or autosomal?
How do you know?
b. Give the genotypes of all individuals.
13. You have seen a pedigree can be used to trace the inheritance pattern of a single-allele trait.
A pedigree can also be used to study multiple-allele traits. In the pedigree below, different blood
types are identified by the letters A, B, AB, and O. Examine the pedigree below, and then answer
the questions that follow.
a. Neither individual I 4 or I 6 has ever had their blood tested. What are their blood types?
b. Write the genotypes of as many individuals as you can. Whose genotypes can you not be
sure of? Explain.
c. Individual III 3 marries a man with blood type AB, and they have four children. Will any of
these children have blood type O? Explain.
d. Individuals II 1 and II 2 have a second child. After you see the child’s blood-test results,
you know that both parents have the genotype IAi. What blood type is the child?
e. Using examples from the pedigree, describe some of the limitations that apply to human
genetic analysis. In what ways could these limitations affect genetic research?
f. Blood typing has often been used as an aid in paternity disputes. Can a blood test ever
prove that a man is definitely the father of a particular child? Explain.