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Biology 30 Genetics Package
Exercise #1: Monohybrid Crosses
1.
In order to solve any genetics problem, you must be able to code the genotypes of the parents
and offspring. In this question, you will practice coding. For each of the following, indicate
the genotypes and phenotypes of the P and F1 generation. Assume pure traits in the parents.
a) Mendel crossed pea plants producing yellow seeds with plants producing green seeds.
All offspring produced yellow seeds.
b) Homozygous smooth-seed plants were crossed with wrinkled-seed plants. Smooth
seeds are dominant.
c) White-fruit squash plants were crossed with yellow-fruit squash plants. All the
offspring produced white fruit.
d) A yellow-pod pea plant was crossed with a green pod plant. None of the offspring
produced green pods.
e) A brown-eyed man and a blue-eyed woman had 6 children. All the children had
brown eyes.
In solving a genetics problem there are six basic steps:
1. Assign letters to represent the alleles.
2. Determine the genotype of each parent.
3. Determine the possible types of gametes each parent can produce.
4. Determine all possible gene combinations that may result when these gametes combine.
5. Determine the various phenotypes possible by analysing the various genotypes.
6. Determine genotypic and phenotypic ratios as required.
2.
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.
3.
In humans, six fingers (F) is the dominant trait and five fingers (f) is
the recessive trait. Both parent 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?
4.
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.
5.
Scientist believe that a mutant recessive form of an autosomal gene called BRCA1 may be
associated with 5% to 10% of all cases of breast cancer. About 80% of women who inherit
the gene in its defective form are likely to develop a cancerous breast tumour. Men who
carry the faulty BRCA1 gene rarely develop breast cancer, but they may pass the gene to their
offspring. A couple has two children, a girl and a boy. The mother has a single mutant gene
for breast cancer; the father is not a carrier of the mutant BRCA1 gene. What is the
probability that their daughter has inherited the mutant BRCA1 gene?
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Biology 30 Genetics Package
6.
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.
7.
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.
8.
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 an 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?
9.
In cattle, hornless or polled (P) is dominant over the horned (p) condition. It is an autosomal
trait. The semen of a polled bull is used to artificially inseminate three cows. Cow 1
(horned) produced a horned calf, cow 2 (polled) produced a horned calf, and cow 3 (polled)
produces a polled calf. Determine the genotypes of the above animals then answer the
following questions:
a) Which of the cattle must have a heterozygous genotype for this trait?
b) Which of the above cattle could have two possible genotypes?
10. 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.
11. In mice the genotype GG is gray, Gg is yellow and gg dies as a small embryo. Assuming (g)
in the homozygous state is a lethal allele, what offspring would be expected from a cross
between a yellow mouse and a gray mouse? What offspring would be expected from a cross
between two yellow mice?
12. 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?
13. 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.
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Biology 30 Genetics Package
14. When 20 purebred Himalayan rabbits are mated with a solid grey male of unknown ancestry,
46 of the offspring are Himalayan and 52 are grey. A single pair of genes controls coat
colour in rabbits.
Himalayan rabbit
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.
15. In rabbits, certain short-haired individuals crossed with long-haired ones produce only shorthaired 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.
a) Which trait is do you hypothesize is dominant?
b) Give the genotypes of all the rabbits described:
First cross:
short-haired rabbit
long-haired rabbit
Second cross: short-haired rabbit
long-haired rabbit
short-haired offspring
short-haired offspring
long-haired offspring
16. 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 _____
17. 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.
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Biology 30 Genetics Package
18. A brown-eyed man marries a blue-eyed woman. They have eight children, all of whom have
brown eyes. What are the genotypes of all the individuals in the family?
19. When two rough-coated guinea pigs are bred, the resulting offspring consisted of 18 roughand 4 smooth-coated offspring. Which type of coat is dominant? What proportions of the
offspring are homozygous for the dominant trait?
CHALLENGE QUESTION:
20. 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)?
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Biology 30 Genetics Package
Exercise # 2 Dihybrid & Monohybrid 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?
2.
In summer squash, white fruit is dominant over yellow fruit while disc is dominant over
sphere shape. A plant which is 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 white spotted
coat is recessive. A male which is heterozygous black and homozygous white spotted is bred
with a female which is 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 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 for both traits?
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 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?
c) How could you test your hypothesis of which trait is dominant? Use Punnett squares
to show the results of the crosses you would use.
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Biology 30 Genetics Package
6.
In Drosophila, the common fruit fly, the normal grey body is dominant to the recessive trait
of black body. A grey-bodied male fly was allowed to breed with many black-bodied
females. Some of the offspring had grey bodies and some had black bodies.
a) What were the genotypes of the parents and the offspring?
b) Two grey-bodied flies were mated, and all the offspring had grey bodies. Can you
conclusively determine the genotypes of the parents? Explain your answer.
c) Two grey bodied flies were mated, and both grey 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?
7.
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?
8.
Normal skin pigmentation (A) dominates lack of pigmentation (albino = a). Brown eyes (B)
dominant blue eyes (b). Two people with normal pigmentation produce one brown-eyed
child of normal skin pigmentation, two blue-eyed children of normal skin pigmentation, and
one blue-eyed albino. What are the possible genotypes of the parents?
Challenge Question:
9. In snapdragons, the characteristics of flower colour and leaf shape are both determined by
genes which show incomplete dominance. The genotypes and corresponding phenotypes are
given below:
rr = red flower nn = narrow leaves
ww = white flower bb = broad leaves
rw = pink flower bn = intermediate leaves
A 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
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Biology 30 Genetics Package
Exercise #3: Other Patterns of Inheritance
Codominance/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 F1 plant were back-crossed to the red
flowered parent?
c) What would be the anticipated offspring if an F1 plant were back-crossed to the white
flowered parent?
2.
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?
3.
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). The coding usually used to represent these alleles is:
type A antigen
IA
type B antigen
IB
no antigen
i
a) Complete the table below to indicate all possible genotypes for the phenotypes
indicated.
Phenotypes
Type A
Type B
Type AB
Type O
Genotypes
4.
A woman having type A blood claims that her former husband who has type B blood is the
father of her baby. 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.
5.
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-.
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Biology 30 Genetics Package
6.
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 Rh-
7.
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)
Complete the table below to indicate all possible genotypes for the phenotypes indicated.
Phenotypes
Genotypes
Full colour
Chinchilla
Blonde
Albino
a) 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?
8.
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.
a) 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?
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Biology 30 Genetics Package
Genetics Exercise #4
Polygenic characteristics: Epistatic Genes
One Gene Many Effects: Pleiotropic Genes
1.
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) long x oval  52 long and 48 oval
b) long x round  98 oval
c) oval x round  51 oval and 50 round
d) oval x oval  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.
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?
3. In mice, the gene C causes pigment to be produced, while the recessive gene c makes it
impossible to produce pigment (albinism). Another gene B, located on a different
chromosome, causes a chemical reaction with the pigment and produces a black colour. The
recessive gene b causes incomplete breakdown of the pigment and causes a light brown or
tan colour to be 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 genotype 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
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. What is the genotype of the tan parent?
How could you determine the genotype of the black parent?
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Biology 30 Genetics Package
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 or pea
comb allele (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_). Pictures of these
comb shapes are shown on page 487 of your text. Indicate the phenotypes of the parents and
provide 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
6.
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?
7.
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) 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?
b) If the couple’s first child has Marfan, what is the probability of their second child also
having this disorder?
8.
In humans, normal pigmentation dominates no pigmentation (albino). Black hair dominates
blonde hair. An albino person will have white hair colour 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?
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Biology 30 Genetics Package
Genetics Exercise #5 Chromosome Mapping
The probability of a crossover between 2 loci (specific places on a chromosome where genes are
located) in a chromosome is proportional to the distance between the two loci. Since a crossover can
occur at any place in the chromosome, the longer the chromosome length between the loci the greater
the chance that crossover will take place. By knowing recombination frequencies, the sequence of
genes in a chromosome can be determined.
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%; 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. 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.
W
X
Y
Z
W
Gene
X
Y
Z
5
7
8
5
2
3
8
3
1
-
7
2
1
5. 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.
A
B
C
D
E
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A
8
12
4
1
GENE
B
8
4
12
9
C
12
4
16
13
D
4
12
16
3
E
1
9
13
3
-
Recombinations
per 100 fertilized eggs
Biology 30 Genetics Package
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.
Pairs of Genes
Crossover
Frequency
Marker gene - cystic fibrosis
18
Marker gene - tritanopia
13%
Cystic fibrosis- trypsin
6%
Trypsin - tritanopia
1%
6. Which of the following gene maps shows the correct sequence of these genes on chromosome 7?
Crossover Frequencies for Some Genes on an Autosome of Organism Z
Genes Crossover Frequency
P and Q 5%
P and R 8%
P and S 12%
Q and R 13%
Q and S 17%
7. Which chromosome map best represents the sequence of genes on the autosome from organism
Z?
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Biology 30 Genetics Package
Exercise #6: Sex linkage
Note:
For all genetics problems involving sex-linked traits state the results of the two sexes separately.
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 more rarely found in females. Ichthyosis 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?
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Biology 30 Genetics Package
6.
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 nonfunctional.
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?
7.
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 HN (normal)
allele is dominant in females but recessive in males. A single HB 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 HBHN or HBHB; whereas, for
females to develop pattern baldness, they must inherit two HB 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.
8.
In humans, the conditions for blood clotting dominates the condition for non-clotting
(haemophilia). Both genes are linked to the X chromosome. A male haemophilic 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?
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Biology 30 Genetics Package
Biology 30 Genetics Exercise #7: Mixed Problems
1. How many different types of gametes could be produced by the individuals having the
following genotypes, and what are the gametes?
a) XRY
f) XNXnBb
b) WwDd
g) XHXh
c) LlGG
h) rwTT
r
d) X YLL
i) AATtRr
e) ssBB
j) RrYyAa *challenge
2. 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
3. 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) homozygous yellow x green
b) heterozygous yellow x green
c) heterozygous yellow x homozygous yellow
d) heterozygous yellow x heterozygous yellow
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
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Biology 30 Genetics Package
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 F1 from the
following crosses.
MrMr x MrM
MrMr x Mrm
MrM x Mrm
Mrm x mm
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? How do you know?
b) Give the genotypes of all individuals.
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Biology 30 Genetics Package
Biology 30
Genetics Exercise #8: Review
The information below is given to answer questions 1 to 10.
Each letter represents an allele of a gene controlling a character.
B = black
D = dull
R = rough
Xt = thick skin
b = white
d = sharp
S = smooth
XT = normal skin
RS = wrinkled
Parental Genotypes
1. BB x bb
What percentage of the offspring will be black?
2. RR x SS
What percentage of the offspring will be smooth?
3. Dd x Dd
What percentage of the offspring will be sharp?
4. Bb x bb
What percentage of the offspring will be black?
5. RS x RS
What percentage of the offspring will be wrinkled?
6. BbDd x BbDd
What percentage of the offspring will be white & dull?
T
t
t
7. X X x X Y
What percentage of the offspring will be thick-skinned males?
8. BbXTXt x BbXtY What fraction of the offspring will be white thick-skinned males?
9. BbDd x BbDd
What fraction of the offspring will be black & sharp?
10. BbRS x BbRS
What fraction of the offspring will be black & wrinkled?
11. The father of a certain family has type AB blood, and the mother has type O blood. They
have four children, of which one belongs to each of the four blood groups (A, B, AB and O).
One of the 4 children is adopted and another is from a previous marriage of the mother. State
which of the children is adopted and which is from the earlier marriage. Show clearly how
you arrived at your answer.
12. Skin colouring in the common leopard frog can be spotted green or plain green. (When
spotted frogs are crossed with plain frogs, all the offspring are spotted). In the space below,
outline a cross between a heterozygous spotted frog and a green frog.
13. In chickens there is 10% recombination between the genes for brown eye (B) and light down
(L). There is a 26% recombination between brown eye and silver plumage (S), and a 16%
recombination between silver plumage and light down. Another gene, slow feathering (K) is
found to have 11% recombination with S and 27 % recombination with L.
a) Construct a chromosome map to show the position of these genes.
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Biology 30 Genetics Package
Use this information to answer the next two questions.
In Drosophila melanogaster gray body is dominant to black body, and long wings is
dominant to short wings.
14. Name the characteritics and traits involved in this situation.
15. Define the alleles for each trait.
16. For each of the phenotypes below, state how many genotypes could apply. Your answer
should be a number.
a) grey body, long wings ____________
b) black body, short wings ____________
c) black body, long wings_____________
d) grey body, short wings____________
17. Colour blindness is a sex-linked, recessive disorder. A colour-blind man who has type AB
blood marries a woman who is a carrier of colour blindness and who also has type AB blood.
What is the probability of these parents producing a colour-blind son having type B blood?
18. There are many possible factors for which human blood can be typed, including the ABO
factor and the Rh factor, which were previously studied. Another blood factor is the MN
factor, which is controlled by two codominant alleles, M and N. A woman having type O,
Rh-negative, N blood marries a man having type AB, Rh-positive, M blood.
a) Is it possible for these parents to produce a child having type B, Rh-negative, MN blood?
b) Is it possible for these parents to produce a child having the same blood type as either the
mother or the father?
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Biology 30 Genetics Package
19. Haemophilia is a sex-linked, recessive blood clotting disorder. Below is a pedigree showing
inheritance of haemophilia in a family.
Give the genotypes of all the individuals indicated in this pedigree.
Challenge Question:
20. A man, whose hobby is the rearing of Mexican swordtail fish, crosses a strain of truebreeding fish that are stippled and patternless (lacking spots at the base of the tail fin) with
true-breeding fish that are nonstippled and have a twin spot tail fin. He finds all the offspring
of these matings are stippled and possess a tail fin spotting pattern unlike that of either
parent, called crescent spot. When the offspring are allowed to mate at random, he finds that
the offspring fall into six categories in the following numbers:
crescent, stippled
83
crescent, nonstippled
27
twin spot, stippled
29
twin spot, nonstippled
9
patternless, stippled
35
patternless, nonstippled
12
Explain how these results would be possible.
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