Page 1
Page 2
Week 3, Lesson 1
Genetics


Explain the basic rules and processes associated with the transmission of genetic characteristics
Describe the evidence for dominance, segregation and the independent assortment of genes on different
chromosomes, as investigated by Mendel
• Heredity
– _____________________________________________________
• Genetics
– is the study of the patterns of _______________________ as hereditary
characteristics or traits
• Gregor Mendel
– the "father of modern genetics“
• simple theory
• Mendel was the first person to realize that genetic traits are inherited as
__________________________________
• He proposed that organisms have a pair of "factors" for each trait
– __________________________________________
• We NOW know that the particles of inheritance are segments of __________
– which we call ________________
Mendel's Laws
•
•
•
Inherited characteristics are controlled by pairs of factors
– genes
– one from each parent.
One gene may “mask” the effect of another.
– The gene which is expressed is ______________________, while the one which
is masked is __________________________.
Pairs of genes ____________________________ during gamete formation
– each sex cell contains only one member of a pair of genes.
Terms
• Alleles
– Two or more alternate forms of a _____________, which produce contrasting effects for
a certain trait
– e.g. red (R) and white (r) for flower colour of peas
• Homozygous
– having two of the _______________ allele… eg. red/red (RR)
• _______________________
Page 3
• Heterozygous
– having two __________________________ alleles eg. red/white (Rr)
• Genotype
– the ______________________ makeup of an individual
• Phenotype
– the expression of the genes, or ____________________ of an individual
• Purebred
– an organism having ________ ______________________ gene pairs
• Hybrid
– an organism having at least _______ _____________________ gene pair
• Monohybrid
– an organism having only _______ _____________________ gene pair
• Dihybrid
– an organism having _______ _____________________ gene pair
Monohybrid Cross
• The phenotypic ratio for a monohybrid cross is always ____: ____
dominant trait : recessive trait
Page 4
MONOHYBRIDS
____ = ____________
____ = ____________
Page 5
***If you have any questions or difficulties, let me know so we can work through a couple
together. It is super important that you get these basics down!***
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. (Tells us that yellow is dominant)
Example
P: yellow x green
YY x yy
F1: yellow
Yy
**How do I know? Do a Punnet Square
b. 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. Pea plants having axial flowers (produced along the stem of the plant) were
cross-pollinated with plants having terminal flowers (produced at the tip of
the stems). The offspring all had axial flowers.
f. A brown-eyed man and a blue-eyed woman had 6 children. All the
children had brown eyes.
g. A rough-coated guinea pig was crossed with a smooth-coated guinea pig.
All of the litter were rough-coated.
h. Constricted-pod pea plants were cross-pollinated with pollen from inflated-pod
plants. There were no constricted-pod pea plants among the offspring.
Page 6
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 five-fingered 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. The Siamese coat pattern in cats is controlled by a single pair of genes in which
the Siamese pattern is recessive to the ordinary solid coat colour pattern. Predict
the phenotypes, genotypes and their probable proportions in the kittens of a
homozygous solid coat male with a Siamese female cat.
Page 7
6.
In Drosophila, the common fruit fly, the normal gray body is dominant to the
recessive trait of black body.
a)
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.
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?
7. In mice the genotype GG is gray, Gg is yellow and gg dies as a small embryo.
a. What offspring would be expected from a cross between a yellow mouse and
a gray mouse?
b. What offspring would be expected from a cross between two yellow mice?
c. Which cross would produce the largest size litter: gray X gray,
gray X yellow, or yellow X yellow?
Page 8
8. 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.
Challenge
9. Black colour in guinea pigs is dominant over white. Outline a cross which would
make it possible to determine if a black male guinea pig is homozygous or
heterozygous.
Page 9
Genetics Exercise #2
1. 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.
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
short-haired offspring
Second cross:
short-haired rabbit
long-haired rabbit
short-haired offspring
long-haired offspring
c) How could you test your hypothesis of which trait is dominant?
Use Punnett squares to show the results of the crossed you would use.
Page 10
2. 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 Punnett square to provide evidence for
your answers.
3. 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.
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.
Page 11
4. Humans who have normal skin pigmentation have the genotype of either AA
or Aa, while those who lack pigmentation (albinos) have the genotype aa.
a) Which trait is dominant?
b) If a homozygous albino marries a homozygous person of normal pigmentation,
what would be the expected phenotype and genotype of their children?
c) If an albino married a heterozygous person of normal pigmentation, what
phenotypes would you expect in their children? Use a punnet square to prove.
d) Two parents of normal skin pigmentation had an albino child. How was this possible? Show a
punnet square as proof.
5. 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 _____
Ed _____
Sharon's father _____
Sharon's brother _____
Sharon _____
Ed and Sharon's child _____
Page 12
6. 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?
7. What are the chances that the first child from a marriage of two heterozygous
brown-eyed parents have 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.
8. 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?
Page 13
9. 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 _____
Cow B _____
Cow C _____
Her horned calf _____
Future calves _____
Her polled calf _____
Future calves _____
Her horned calf _____
Future calves _____
CHALLENGE QUESTION:
10. 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)?
Page 14
TEST CROSS
____ = ____________
____ = ____________
Page 15
If an organism has the dominant phenotype,
how can you determine whether it is homozygous or heterozygous?
•
•
You conduct a test cross.
• ________________________________________________________________
________________________________________________________________
________________________________________________________________
____________________________________
For example:
– determine whether a red-flowered pea plant is homozygous or heterozygous
P: Red flowered X _____________________________
R?
F1:
Scenario
#1
___________
Suppose the cross produces ________________________
________________________
•
If no offspring showing the recessive phenotype are produced, then the unknown
parent must be …__________________________________
P: Red flowered X White-flowered
R?
rr
F1: Suppose the cross produces __________ red flowers and ______ white flowers
• The only way a white flower could appear is if it ______________________
____________________________________________________________
– The unknown parent must be …_________________________________
End of Lesson 1
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 16
Scenario
#2
DIHYBRIDS
____ = ______________________
____ = ______________________
____ = ______________________
____ = ______________________
Page 17
Week 3, Lesson 2
Dihybrid Cross
•
Mendel performed dihybrid crosses of plants with ______ different pairs of contrasting
alleles
• In one experiment, Mendel crossed plants homozygous for seeds that were both
___________ and ______________ with plants homozygous for ________________,
____________________ seeds
• All the F1 offspring had round, yellow seeds (which traits were dominant?)
• Self-fertilization of the F1 plants produced and F2 generation of seeds with the
following phenotypes:
______ round yellow
______ round green
______ wrinkled yellow
______ wrinkled green
To find the ratio among the F2 phenotypes…
– Take the number of offspring in the _____________ category - 32 and divide it into the number of offspring in the other categories:
– The quotient is then rounded to the nearest whole number
•
Thus Mendel determined the phenotypic ratio in the F2 generation to be
_____:______:_____:______
• Mendel explained these data by assuming that the genes governing seed color and
seed shape move ________________ during gamete formation
• In the process of independent assortment, each pair of alleles behaves as it would in a
monohybrid cross - independently of the other pair
• A dihybrid can produce ______ possible gene combinations (with equal probability)
• If the alleles are:
R = round
Y = yellow
r = wrinkled
y = green
• The possible gamete combinations are
_______ _______ _______ _______
One of the hardest things with two gene combination questions is determining the gamete
combinations and setting your punnet square up properly. You can FOIL the gene pairs to make
sure you get all of the combinations. Check out the tutorial link for more help.
Page 18
Genetics Exercise #3 Dihybrid Genetics
1.
Horses which are black in colour and which have a trotting gait carry dominant
genes, while horses which are chestnut in color and have a pacing gait carry
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 at maturity.
Parent Genotype
_________
Parent Genotype
_________
Page 19
____ = ____________
____ = ____________
____ = ____________
____ = ____________
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 cross-pollinated with one which is homozygous for yellow fruit
color and disc shape.
a)
What will be the genotype and phenotype of the F1?
Page 20
____ = ____________
____ = ____________
____ = ____________
____ = ____________
b) Construct a Punnett square to show the F2 generation. What is the
phenotypic ratio of the F2 generation?
Parent Genotype
Parent Genotype
c)
If an F1 plant were crossed with pollen from its yellow disc parent, what
would be the expected phenotypes of the offspring and in what ratio?
Parent Genotype
Parent Genotype
Page 21
3.
In certain breeds of dogs, black color is dominant and red color is recessive; solid
coat is dominant and spotted coat is recessive. A male which is homozygous black
and white spotted is bred with a female which is red and white spotted. What is
the probability of these two dogs producing a solid black puppy?
____ = ____________
____ = ____________
____ = ____________
____ = ____________
Parent Genotype
Parent Genotype
4.
5.
In Holstein-Friesian cattle, the two colors 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 color 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 colors
of the F2 calves?
Normal skin pigmentation (A) dominates lack of pigmentation (albino = a).
Brown eyes (B) dominate 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 albino. What are the
possible genotypes of the parents?
____ = ____________
____ = ____________
____ = ____________
____ = ____________
End of Lesson 2
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 22
MULTIPLE ALLELES & INCOMPLETE DOMINANCE


Compare ratios and probabilities of genotypes and phenotypes for dominant and recessive, multiple,
incomplete dominant, and codominant alleles
Explain the relationship between variability and the number of genes controlling a trait
____ = ____________
____ = ____________
Page 23
Week 4, Lesson 1
Multiple Alleles & Incomplete Dominance
Multiple Alleles
•
•
Many genes actually exist in more than two allelic forms,
Although only two genes control coat colour in rabbits it is controlled by a series of
four alleles for the same gene:
1. ___
Full colour
2. ___
Chinchilla
3. ___
Himalayan
4. ___
Albino
•
The dominance hierarchy of these alleles is ___ > ___> ___> ___
•
Determine the genotypes for the following phenotypes:
Phenotype
Possible Genotype
• Full color
• Chinchilla
• Himalayan
• Albino
Incomplete Dominance
• In some cases, a heterozygous organism shows a _____________
of genes because
___________ gene is dominant:
– this is termed incomplete dominance
– for example, in snapdragons, neither the red nor the white allele is dominant…
•
•
P:
•
F1:
•
F2:
If two different alleles each contribute to a phenotype, they are termed codominant
Page 24
•
•
One of the best-known example of codominant genes occurs in humans, and
determine ABO blood types
There are three possible alleles:
– ______
–
______
– ______
Phenotype
A
B
AB
O
Possible Genotypes
Page 25
Genetics Exercise #4 Incomplete Dominance & Codominance
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
a)
Complete the table below to indicate all possible genotypes for the
phenotypes indicated.
Phenotypes
full colour
Genotypes
chinchilla
blonde
albino
b)
2.
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?
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.
Page 26
3.
b)
What are the phenotypic and genotypic ratios of the F2 generations?
c)
What would be the expected phenotypic ratio if a blue Andalusian hen were
bred with a black rooster? Use a punett square.
In the “four o'clock”- a 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?
Page 27
c)
4.
What would be the anticipated offspring if an F1 plant were back-crossed to
the white flowered parent?
The best established series of multiple alleles in humans is associated with the
different blood types - A, B, AB and O. The alleles for type A and B antigens are
codominant, and the allele for type O (no antigen) is recessive to both the A and B
alleles. The coding usually used to represent these alleles is:
type A antigen
type B antigen
no antigen
a)
Complete the table below to indicate all possible genotypes for the
phenotypes indicated.
Phenotypes
Type A
Type B
Type AB
Type O
b)
IA
IB
i
Genotypes
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.
Page 28
Exercise 5
Polygenic characteristics: Epistatic Genes One Gene Many Effects: Pleiotropic Genes
1. 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.
2. 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 color 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?
Page 29
3. 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?
4. In humans, normal pigmentation dominates no pigmentation (albino). Black hair dominates
blonde hair. An albino person will have white hair collar even though they may also have the
genes for black or blonde hair color. An albino male who is homozygous for black hair
marries a woman who is heterozygous for normal pigmentation and who has blond hair. What
colors of hair can their children have and what is the probability for each hair color?
Page 30
Challenge Question:
5. 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?
6.
In snapdragons, the characteristics of flower color and leaf shape are both
determined by genes which show incomplete dominance. The genotypes and
corresponding phenotypes are given below:
rr = red flower
ww = white flower
rw = pink flower
nn = narrow leaves
bb = broad leaves
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.
End of Lesson 1
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 31
Week 4, Lesson 2
CHROMOSOME MAPPING
 Explain the influence of gene linkage and crossing over on variability
Chromosomal Theory of Inheritance
1.
2.
3.
4.
•
•
Genes are located on ______________________
Chromosomes undergo _____________________ during meiosis
Chromosomes assort ___________________ during meiosis
Each chromosome contains many different ______________
Each chromosome contains hundreds or thousands of genes
Genes located on the same chromosome are inherited _______________,
– they are part of a single chromosome that is passed along as a _________
– such genes are said to be ________________
• during meiosis, chromosomes may exchange segments of DNA by _______
____________
A A
a a
B B
C C
b b
c c
B B
C C
b b
c c
D D
d d
D D
d d
E E
F F
e e
f f
Page 32
• The ___________ two genes are on a chromosome, the fewer the possible points of
crossover are between them, and the ________ frequently such a cross-over will
occur
• In other words, if two genes are ________________ on a chromosome it is likely they
will _________________ and not be exchanged between chromatids during meiosis
 To determine the location of genes along a chromosome is called MAPPING a
chromosome
•
A chromosome map indicates
–
–
The ____________ in which specific genes occur on a chromosome
The ______________ between the genes
• Example:
•
– In Drosophila, the following data was obtained from genetic crosses:
• 13% recombination between bar eye and garnet eye
– High percentage recombination indicates that these two genes are ________
______________________
– High likelihood that crossing over _________________________ these two genes.
• 7% recombination between garnet eye and scalloped wings
– These two genes are ___________________ than bar eye and garnet eye
• 6% recombination between scalloped wings and bar eye
This data can be used to map the chromosome:
There is a good tutorial on this in the textbook, starting on pg 636.
Page 33
Biology 30
Genetics Exercise #6
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. 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
A
-
GENE
B
8
C
12
D
4
E
1
B
C
D
E
8
12
4
1
4
12
9
4
16
13
12
16
3
9
13
3
-
Recombinations
per 100 fertilized eggs
End of Lesson 2
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 34
Week 4, Lesson 3
Sex Determination
• There are two chromosomes involved in the determination of sex of most animals,
– the sex chromosomes (_______)
• Any other chromosome not involved in sex determination is called an
________________
– for example, humans have _______ pairs of autosomes and ______ of sex
chromosomes
• In most mammals, females are homozygous and have 2 X chromosomes, while males
are heterozygous and carry an X and a Y chromosome
• Female = _____
• Male = _______
• Females can produce eggs carrying only an ____chromosome,
• Males produce sperm carrying __________________ chromosome (___% of each)
• Thus, it is the ________ who determines the sex of his offspring!
Page 35
SEX LINKAGE

Compare the pattern of inheritance produced by genes on the sex chromosomes to that produced by genes on
autosomes, as investigated by Morgan and others.
Draw an X and a Y chromosome here.
____ = ____________
____ = ____________
Page 36
Sex Linkage
•
•
•
Like other chromosomes, the sex chromosomes carry many __________
Some of the regions of the X-chromosome have a ___________________ region on
the Y- chromosome
– There are also large __________________ portions:
• That is, the X chromosome carries some genes that have ________
_______________________ on the Y chromosome
Genes for _______________ are carried on the ___ chromosome but not on the ____
chromosome
– Therefore in a male, the gene on the ___ chromosome is _____________
____________________
– In a female, she must have two recessive alleles to have the recessive phenotype
Fly Solution
•
•
•
•
•
•
•
R = Red
r = white
XR XR = Red eyed Female
XR Xr = Red eyed Female
Xr Xr = White eyed Female
XR Y = Red eyed Male
Xr Y = White eyed Male
Page 37
•
Some sex-linked traits in humans are
– Colour vision
– Hemophilia
– Duchenne's Muscular Dystrophy
Biology 30
Genetics 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?
Page 38
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 Lisa, what type of colour vision would be expected in their children?
Prove with a punnet square.
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?
Page 39
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?
Page 40
5.
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?
Page 41
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?
End of Lesson 3
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 42
Week 5, Lesson 1
SEX INFLUENCED GENES
•
The main role of sex hormones is to influence the ________________ and its
________________________,
– These hormones, however, also affect many other parts of the body
• Genes that are expressed to a ______________________ as a result of the level of
sex hormones are called __________________________.
• These genes are usually located on the ____________________
• Males and females with the __________________ may differ greatly in phenotype
because the differing levels of ________________
• For example:
– A ______ may have a gene for high ______________, but he will not
produce milk because he has low levels of ___________________.
• In humans, the gene for male pattern baldness is ________________ and
___________________
• A man will still become bald if he has _____________ allele for baldness, because the
____________________________ somehow stimulate the expression of the allele
• In a woman, however, the allele acts as a recessive allele so that she must have
________________________ before she loses her hair
Page 43
Lethal Alleles
• If an organism has a ____________ that destroys the _________________ for a
•
•
•
•
•
_______________________________, the organism will often die prematurely.
This gene that fails to code for a functional protein is called a _____________
It is possible for lethal alleles to be _________________, but most are rapidly
eliminated from a population because they cause _______________ before the
individual carrying the allele ___________________.
An exception of a lethal dominant allele that remains in a population is the one
responsible for __________________ in humans, because this allele is not expressed
until later in life (_________ years of age)
An example of a recessive lethal allele in humans is the one for Brachydactyly:
– Heterozygotes have a short ____________________ that makes the fingers
appear to have only two bones instead of three
– Homozygous babies _____________ and have abnormal development of the
skeleton that result in death in infancy
Some human examples of lethal alleles are:
– Sickle cell anemia
– _________________________
– Cystic fibrosis
– _________________________
Page 44
PEDIGREES
Autosomal Recessive
AA =
A=
a=
Aa =
aa =
Autosomal Dominant
AA =
A=
a=
Aa =
aa =
X-Linked Recessive
XA =
Xa =
XA XA =
XA Xa =
Xa Xa =
XA Y =
Xa Y=
X-Linked Dominant
XA =
Xa =
XA XA =
XA Xa =
Xa Xa =
XA Y =
Xa Y=
Page 45
Page 46
Pedigrees
• Because geneticists are unable to manipulate the mating patterns of people, they must
•
•
•
•
analyze the results of matings that have already occurred.
As much information as possible is collected about a _________________ for a particular
trait, and this information is assembled into a _______________ ___________ describing
the interrelationships of parents and children across the generations
– This is called a ___________________
From a pedigree you should be able to determine if a particular trait is ______________ or
__________, ______________ or _________________
A pedigree not only helps us understand the past but also helps us ______
_________________________
Geneticists, physicians, and genetic counselors use pedigrees
– for analysis of genetic disorders
– to advise prospective parents of genetic risks involved
The rest of the booklet is practice with pedigrees, which combines every other previous topic in
here. You use the same techniques with your punnet squares and inheritance rules but now you
apply them to a family tree where generally you have even less information than before. 
Pedigrees are terrific logic puzzles using genetics. One of the hardest things with pedigrees, is
when you have to figure out what the inheritance rules are. If they just give you a pedigree and no
other information, you have to use common patterns and inheritance laws to determine whether the
trait is autosomal or X-linked, dominant or recessive. Often you have to do guess and check with
each possibility, until you find one that cannot be true. Despite the Pedigree Key and some
examples, you will never be told which people are carriers; you will have to figure that out by
looking at the phenotypes of their parents and/or children.
Page 47
Biology 30
HUMAN PEDIGREES
BACKGROUND
Geneticists often draw pedigree diagrams to show the inheritance of a particular genetic trait within a family.
These diagrams can help them determine whether a phenotype is controlled by a dominant, recessive, or sex-linked
allele. For example, if both parents show a trait but some of their children do not, then the trait is controlled by an
autosomal (not on a sex chromosome) dominant allele. If neither parent shows the trait, but it appears in one or
more of their children, then the trait is controlled by an autosomal recessive allele. If the trait appears primarily in
males and it "skips a generation", it is probably a sex-linked recessive trait. Many of the traits examined using
pedigree diagrams are hereditary disorders.
In a pedigree diagram, each generation is numbered using Roman numerals, with the oldest generation always
number 1. Each individual within each generation is numbered with an Arabic numeral, so that each individual is
known by the combination of the generation and the individual numbers, e.g. III-4.
A carrier is an individual with a normal phenotype, but who has the gene in question and can pass it on to
offspring.
The following is a key to how the pedigrees are coded:
Page 48
Biology 30 Pedigree Analysis
Autosomal Dominant Inheritance
As discussed above, conditions due to autosomal dominant conditions result from an individual carrying one changed gene.
Examples of autosomal dominant forms of dwarfism include achondroplasia and hypochondroplasia. A person with
achondroplasia for example, has a 50% or 1 in 2 chance of passing the gene on to his/her offspring.
Clues that geneticists use when looking at pedigrees to determine autosomal dominant inheritance include the following
facts:
1. Autosomal dominant conditions are seen in every generation (vertical pattern).
2. Males and females have the condition with equal frequency and severity.
3. Unaffected individuals do not have children with the condition.
4. Each child of an affected individual has a 50% chance of being affected, regardless of sex or birth order.
5. Homozygotes for autosomal dominant conditions (individuals with two changed genes) have a more severe form
of the disease. An example of this is the "double-dominant" form of achondroplasia that can occur when two
people with achondroplasia have children (see below).
Autosomal Dominant Pedigree
Autosomal Recessive Inheritance
Autosomal recessive inheritance is seen in conditions that are due to having two changed genes.. Dwarfing conditions
inherited in an autosomal recessive manner include diastrophic dysplasia and cartilage hair hypoplasia (metaphyseal
chrondrodysplasia, McKusick type). Characteristics of autosomal recessive traits include:
1. The condition typically appears in one generation (siblings) and notin their parents or offspring ("horizontal
inheritance").
2. Males and females are equally affected.
3. Both parents are asymptomatic heterozygotes (carriers) meaning they only have one changed gene.
4. Two carrier parents have a 1 in 4 (or 25%) chance of having an affected child.
5. Each unaffected full sibling of an affected individual has a 2/3 chance of being a carrier.
6. Offspring of an affected individual will be a carrier and therefore be unaffected unless the other parent is a carrier
or is affected with the same condition.
Autosomal Recessive Pedigree
Page 49
X-linked Recessive Inheritance
X-linked dwarfing conditions are rare. These include a form of chondrodysplasia punctata (X-linked recessive type) and
Hunter's syndrome which is a mucopolysaccharide disease. Criteria for this mode of inheritance include:
1. Males are almost exclusively affected.
2. No father to son transmission occurs but all daughters of affected fathers will be carriers.
3. Depending on the condition, examination of a carrier female may reveal some mild manifestation.
4. Carrier females have a 50% or 1 in 2 chance of having an affected son and a 50% or 1 in 2 chance of having a
carrier daughter.
5. The first affected son in a family may inherit the changed gene from a carrier mother or may be the result of a new
genetic change.
X-Linked Recessive Pedigree
X-Linked Dominant Inheritance
There are very few X-linked dominant traits. Dwarfing conditions due to X-linked dominant conditions include another
form of chondrodysplasia punctata (X-linked dominant type) and X-linked hypophoshatemic rickets. Characteristics of this
mode of inheritance include:
1. Affected males transmit the trait to all of their daughters and none of their sons.
2. Affected females transmit the trait to half of their sons and half of their daughters.
3. There are usually twice as many affected females as affected males but affected females often express the
condition to a milder degree.
X-Linked Dominant Pedigree
Part 1 - Tay-Sachs Disease
Page 50
Tay-Sachs disease is caused by a lethal autosomal recessive gene (t). Individuals with Tay-Sachs disease first
shown the condition at the age of 6 months and die at about 4 years old.
Pedigree One: Tay-Sachs Disease
I
1
2
II
2
1
4
3
5
III
2
1
5
4
3
6
Pedigree diagram for a family with a history of Tay-Sachs disease
a) Write the genotype for each individual by the circle or square.
b) What is the probability that individuals II-4 and II-5 could produce another Tay-Sachs child?
Part 2 - Sickle-cell Anaemia
Sickle-cell anaemia is a disease that occurs in about 1 in 500 black children born in North America. Individuals with this
disease can suffer from "sickle cell crises" when they are deprived of oxygen by exertion or respiratory ailment. When this
happens, red blood cells collapse into a "sickle" shape that can block capillaries. This causes severe pain, and the lack of
blood flow can result in oxygen deprivation that can make the condition even worse.
Pedigree Two: Sickle-cell Anaemia
I
1
I
I
1
2
2
3
4
3
5
6
4
7
III
1
2
4
3
5
6
a) How is this trait inherited?
b) Write the genotype for each individual below the circle or square.
c) About 1 in 10 North American blacks are heterozygous for sickle-cell anaemia. This number is very high for a
trait that can be so detrimental for homozygous individuals. However, research has shown that heterozygous
individuals have enhanced resistance to malaria. What effect might this factor have had on the frequency of
sickle-cell gene in populations of North American blacks?
d) Under what condition is the genotype of heterozygous individuals known?
Page 51
Part 3 - Haemophilia
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
I
2
1
I
I
1
2
4
3
5
III
1
3
2
a) Write in the genotype for each individual, using:
XH
= normal haemoglobin
Xh
5
4
= abnormal haemoglobin
b) Why do more males have haemophilia than females?
c) Explain how the female (III-3) can have haemophilia.
Part 4 - Problems
1.
Answer the following questions for Pedigrees Four and Five, Six and Seven.
How is this trait inherited? Place a check mark in the box below to indicate which modes
a)
“work” and which modes “don’t work .
b)
Write in the genotype for each individual.




Pedigree Four
I
2
1
I
I
1
2
3
4
5
7
6
8
Autosomal Recessive
Autosomal Dominant
X-Linked Recessive
X-Linked Dominant
9
III
1
2
3
4
Page 52
5
6
7
8
9




Pedigree Five
I
Autosomal Recessive
Autosomal Dominant
X-Linked Recessive
X-Linked Dominant
2
1
II
1
2
4
3
5
6
8
7
III
1
2
3
Pedigree Six
Page 53
5
4




Autosomal Recessive
Autosomal Dominant
X-Linked Recessive
X-Linked Dominant
Pedigree Seven




Autosomal Recessive
Autosomal Dominant
X-Linked Recessive
X-Linked Dominant
2. Follow the directions below and draw the pedigree and label appropriate genotypes for all
individuals below.
a) Generation I : A male with blood type O marries a female with blood type A.
b) Generation II : They have 4 children. The first born is a male with A type blood, the
second born is a girl with O type blood, and the last two children were identical twin
boys.
c) Generation III : The last boy married a women with type B blood. They had two girls,
one with AB blood type and one with B blood type.
Page 54
3. 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) What evidence is there that colour blindness is a sex-linked trait?
b) Give the possible genotypes of all the individuals.
Analysis and Interpretation
1.
Suppose you discovered that both you and your spouse are heterozygous for a lethal recessive gene.
Would you have children? Explain. Would your decision change if the trait involved were debilitating but
not lethal? Explain.
2.
Many genetic diseases extract a high cost in terms of health care dollars. Since Medicare is funded by
government funds, should the government be allowed to legislate who may and who may not have
children? Defend your position.
End of Lesson 1
When you have completed the practice in this section, complete the formative quiz linked in
PowerSchool.
Page 55
Paternity Investigation
OBJECTIVES:
BACKGROUND INFORMATION:
A three year old girl named Lucy was lost at sea and presumed dead when her family fishing boat
capsized during a weekend outing. Miraculously, Lucy was discovered unconscious two days later by
the Pena family. The Pena family raised Lucy with their other children.
At the age of 20 Lucy contracted a rare form of kidney disease. In order to live Lucy needed to find a
suitable donor. Although the Penas offered, Lucy's tissue type did not match the tissue type of her
adoptive parents. As a result, the family turned to the coast guard in hopes of finding the identity of
Lucy's natural family. The Pena's search found five families that had lost a three year old girl in a
boating accident around the time that Lucy was found.
Help Lucy determine which is her family by proving with 5 pedigrees.
Lucy Pena
Jackson Family
Phenotype
Genotype
Dolores
Ramirez Family
Leon
Maria
Raul
lactose intolerant
LL
Lactose intolerant
lactose intolerant
lactose tolerant
lactose intolerance
does not taste PTC
pp
non taster
non taster
taster
non taster
connected eyebrows
nn
not connected
not connected
not connected
not connected
straight hair
cc
straight
straight
straight
wavy
attached earlobes
ff
attached
attached
not attached
not attached
Castillo family
Xochitl
Davis Family
Juan
Karina
Fuentes Family
Marcus
Claudia
Raul
lactose intolerant
lactose
intolerant
lactose intolerant
Lactose intolerant
lactose intolerant lactose intolerant
taster
taster
non taster
taster
taster
taster
connected
not connected
not connected
not connected
not connected
not connected
curly
wavy
curly
curly
straight
straight
not attached
not attached
not attached
attached
not attached
not attached
Page 56
Review Assignments
Biology 30
Genetics Exercise #9: Human Pedigree
Use the following information to answer this pedigree question.
The Saguenay River community, located about 200 km from Quebec City, has world's highest incidence of myotonic
dystrophy - 1 in 500 people. It is an inherited disorder that affects the muscles and nerves of its victim. Typically,
myotonic dystrophy affects 1 in 7 500 Canadians.
An international research team has isolated the gene that causes the disease. The
myotonic dystrophy gene has an interesting characteristic: the gene size expands with each generation through the addition
of extra nitrogen bases. The greater the degree of expansion, the greater the severity of symptoms. In addition, greater
gene expansion increases the likelihood of offspring being born with myotonic dystrophy. Diagnosis of carriers can now
be done with a simple blood test.
A twenty-two-year-old woman and a twenty-three-year-old man from the Saguenay
River community were both unaffected by myotonic muscular dystrophy. They sought genetic counselling before
marriage. The counsellor collected data from their living relatives and recorded the information in a table.
Woman's
Family
Man's
Relative
Age
maternal
grandmother
69
Myotonic
Dystrophy
no
maternal
grandfather
71
mother
Family
Relative
Age
paternal
grandmother
77
yes
paternal
grandfather
74
no
48
no
paternal
grandfather's
brother
78
yes
father
49
no
maternal
uncle
44
yes
sister
24
no
mother
46
no
brother
23
yes
father
51
yes
sister
22
yes
brother
21
no
sister
19
no
* maternal means "on the mother's side of the family"
* paternal means "on the father's side of the family"
Page 57
Myotonic
Dystrophy
no
1.
Construct a pedigree to illustrate the data in the table and the information given
about the man and the woman who are seeking genetic counselling. Use the
symbols given on the Biology 30 Data Sheet when constructing a pedigree.
2.
Based on the data, identify the mode (type) of inheritance of this family's
myotonic dystrophy i.e. is it dominant or recessive, autosomal or sex-linked?
Give evidence to support your choice.
3.
If the man and woman who are seeking genetic counselling married and had a
child, what would be the probability that their child would have myotonic
dystrophy. Show your method and/or calculations.
4.
To what degree would your pedigree be useful for predicting the severity of
myotonic dystrophy in offspring? Explain your answer.
5.
Provide an hypothesis to explain why the incidence of this disease is higher
in the Saguenay River community than in the rest of Canada.
6.
As the genetic counsellor for the man and woman considering marriage, you are
required to prepare a report for the couple.
•
Outline the information you would give the couple regarding the probablilty
of them producing a child affected by myotonic dystrophy and the severity
of the disorder.
•
Outline three recommendations and/or cautions you would give this couple.
Your recommendations and/or cautions must be based on the data and on
scientific information.
Page 58
Biology 30
1.
2.
3.
Genetics Exercise #10: Mixed Problems
How many different types of gametes could be produced by the individuals
having the following genotypes, and what are the gametes?
a. XRY
f. WwDd
b. LlGG
g. XrYLL
c. ssBB
h. XNXnBb
d. XHXh
i. rwTT
e. AATtRr
j. RrYyAa *challenge
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
In peas, yellow seed color is dominant to green. What will be the colors and
their proportions in the offspring of the following crosses?
a. homozygous yellow x green
Page 59
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 hemophiliac 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 hemophilia.
c. What are the chances that if they have a son he will be normal for blood clotting?
d. Rarely a female hemophiliac is born. How could this happen?
Page 60
5.
In cats, short hair is dominant over long and solid black coat is dominant over
the Siamese coat pattern of black and tan. A cat breeder wants to know if a
certain black, short-haired 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 color inherited in sweet peas?
b. What is the phenotype of the parents? Explain how you determined this.
7.
8.
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
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.
a.
M rM r x M rM
Page 61
b. MrMr x Mrm
c. MrM x Mrm
d. 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? Why or why not?
10. Normal pigmentation (A) dominates no pigmentation (albino = aa). Browneyed 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 S, and long hair to its
recessive allele s. 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 short-haired black and 29 long-haired black offspring. What are
the probable genotypes of the parents?
Page 62
Biology 30
Genetics Exercise #11: 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
b = white
D = dull
d = sharp
R = rough
S = smooth
RS = wrinkled
Xt = thick skin
XT = normal skin
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?
7.
XTXt x XtY
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
10. BbRS x BbRS
What fraction of the offspring will be black & sharp?
What fraction of the offspring will be black & wrinkled?
Page 63
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 coloring 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.
Construct a chromosome map to show the position of these genes.
Page 64
Use this information to answer questions 14 to 16.
In Drosophila melanogaster gray body is dominant to black body, and long
wings is dominant to short wings.
14. Name the characteristics 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)
gray body, long wings
____________
b)
black body, short wings
____________
c)
black body, long wings
____________
d)
gray 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?
Page 65
19. Hemophilia is a sex-linked, recessive blood clotting disorder. Below is a pedigree showing inheritance of hemophilia
in a family.
Give the genotypes of all the individuals indicated in this pedigree. (Write neatly please)
Page 66
Challenge Question:
20. A man, whose hobby is the rearing of Mexican swordtail fish, crosses a strain of true-breeding fish which are stippled
and patternless (lacking spots at the base of the tail fin) with true-breeding fish which 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
crescent, nonstippled
twin spot, stippled
twin spot, nonstippled
patternless, stippled
patternless, nonstippled
83
27
29
9
35
12
Explain how these results would be possible.
Biology 30
1.
Genetics Exercise #10 Answers
a. 2 gametes: XR and Y
b. 2 gametes: LG and lG
c. 1 gamete: sB
d. 2 gametes: XH and Xh
e. 4 gametes: ATR, ATr, Atr, and AtR
f. 4 gametes: WD, Wd, wD and wd
g. 2 gametes: XrL and YL
h. 4 gametes: XNB, XnB, XNb and Xnb
i. 2 gametes: rT and wT
j. 8 gametes! you figure them out.....
2.
a. all offspring will have round seeds and tall stems
b. 9 round & tall: 3 round & short: 3 wrinkled & tall: 1 wrinkled & short
c. 1 round & tall: 1 round & short: 1 wrinkled & tall: 1 wrinkled & short
3.
a.
b.
c.
d.
all will be yellow
1 yellow: 1 green
all will be yellow
3 yellow: 1 green
4.
a. There is a 50% chance the son will be hemophiliac (inherited from
mother).
b. There is a 50% chance the son will be hemophiliac (inherited from
mother again, same as in part a).
c. If a hemophiliac man and a woman who is a carrier for hemophilia
have a daughter, there is a 50% chance she will have hemophilia.
i.e. XhY x XHXh
6.
a.
b.
7.
a.
b.
c.
8.
a.
9.
Flower colour is inherited by incomplete dominance
(codominance).
Both parents are rw (heterozygous). Show a Punnett square to
prove this answer.
1 white: 1 pink
all red
1 red : 2 pink : 1 white
phenotypes - all restricted mallard/ genotypes 1 MrMr : 1 MrM
b.
c.
phenotypes - all restricted mallard/ genotypes 1 MrMr : 1 Mrm
phenotypes 3 restricted mallard : 1 mallard
genotypes 1 MrMr : 1 MrM : 1 Mrm : 1 Mm
d.
phenotypes 2 restricted mallard : 1 mallard : 1 dusky mallard
genotypes 1 MrM : 1 Mrm : 1 Mm : 1 mm
He could be the father of the child if he is heterozygous (IAi). It
cannot be proven on this evidence alone.
10.
AaBb and Aabb (based on probaility this is the best answer however both parents could have been AaBb)
11.
SsBB and Ss__ (We can't predict the colour genotype for the second
parent, because the first was homozygous black.)
12.
The allele producing myopia appears to be recessive from this
pedigree. Two parents having normal vision produced a myopic
child. It is autosomal because an unaffected father produced and
affected daughter
5.
Breed the cat several times with a long haired Siamese.
If any Siamese coat kittens result, the unknown cat is heterozygous
for coat colour.
If all black kittens are produced, the cat is homozygous black.
If any long haired kittens result, the unknown cat is heterozygous for
length of fur.
If all short haired kittens are produced, the cat is homozygous for
short hair.
If any long haired, siamese kittens result, the unknown cat is
heterozygous for both traits.
(You could do 2 separate breedings and follow each trait separately)
Page 67
Bio. 30 Genetics Worksheet 11 Answers
1.
2.
3.
4.
5.
100% black
0% smooth
25% sharp
50% black
50% wrinkled
6.
7.
8.
9.
10.
18.75%
25% thick skinned & male
1/16 white & thick-skinned & male
3/16 black and sharp
3/8 black and wrinkled
11.
adopted child - type AB, because the type O woman cannot produce a type AB child
child from the previous marriage - type O, because the type AB man cannot
produce a type O child
12.
S = spotted green
s = plain green
P: Ss x ss
F1: 1 Ss : 1 ss / 50% spotted and 50% plain
K S  L  B
13.
14.
The genetic characteristics are body colour and wing length.
The genetic traits are gray and black body, long and short wing.
15.
The alleles are G and g for body colour; L and l for wing length.
16
a)
b)
17.
P: XnYIAIB x XNXnIAIB
Chance of a son who is colourblind and has type B blood = 1/16 (6.25%).
18.
IA = A allele
IB =
B allele
i=
19.
20.
4 genotypes
1 genotype
O allele
c)
d)
2 genotypes
2 genotypes
R = Rh-positive
r = Rh-negative
M = M allele
N= N allele
P: i i r r N N x IA IB R _ M M
a) Yes, if the man is heterozygous for Rh factor.
b) No.
A. XhY
F. XHY
B. XHXh
C. XHY
G. XhY
H. XHXh
D. XHXh
E. XhY
I. XHY
J. XHY
S=
s=
pp = patternless tail fin
tt = twin spot tail fin
pt = crescent spot tail fin
stippled
nonstippled
The stippled allele is dominant over the and non-stippled allele;
and the patternless tail fin and twin spot tail fin alleles are codominant.
The dihybrid cross of Sspt x Sspt produced a 3 : 1 ratio of stippled to nonstippled;
and a 1 : 2 : 1 ratio of patternless to crescent spot to twin spot tail fin.
Page 68
K. XHY
L. XHXM. XhY
N. XHX-