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SBI3U
Name: ________________
Mendelian Genetics – Test #2 Review
1. You are to be the judge in a divorce case involving a man and his wife. The man has
accused his wife of infidelity claiming that the child she has given birth to is not his. The child
has blood type B, the woman has blood type A, and the man is type O. Should he be granted
the divorce? Justify your answer using a Punnet square(s).
Child = IBIB or IBi
Woman = IAIA or IAi
Man = ii
The man is justified. There is no way that child can be his as he and the woman
would not be able to produce a type B baby. They could only have a type A or O
baby together.
2. When a white chicken is crossed with a black chicken, the offspring are all ’Blue
Andulusians’, which show a combination of black and white feathers.
a) What are the alleles and genotypes of the different types of chickens that can be
produced?
CW = white feather allele
CB = black feather allele
CWCW = white feathered chicken
CBCB = black feathered chicken
CBCW = Blue Andulusians
SBI3U
Name: ________________
b) Could a farmer have a flock of only Blue Andulusian chickens? Explain using Punnet
squares.
The only way a farmer could have a flock of Blue Andulusian chickens would be
to cross a white feathered chicken with a black feathered chicken. That would
produce 100% CBCW (Blue Andulusians)
3. Use the following to answer questions i-iii:
i. Examine the trait that is represented by the shaded symbols in the pedigree. What
inheritance pattern does this trait illustrate?
A) incomplete dominance
B) multiple alleles
C) co-dominance
D) sex-linked inheritance
E) dihybrid cross
ii. Analyze the pedigree above. Suppose that individual III-2 marries someone with the
same genotype as individual I-1. What is the chance that one of their children will have
hemophilia?
A)
B)
C)
D)
E)
0%
25%
50%
75%
100%
ii Individuals II-1 and II-4 can be classified as __________________ for the trait that is being
followed in the pedigree.
SBI3U
Name: ________________
A) homozygous dominant
B) homozygous recessive
C) carriers
D) siblings
E) mutants
4. Red-Green colourblindness in humans is a sex-linked recessive trait found on the X
chromosome.
a) If a normal-sighted woman whose father was colourblind but mother was not, marries a
colourblind man whose father was also colourblind but mother was not, what are the
genotypes of the parents of the man and woman?
H = normal
h = hemophillia
Woman = XHXh
Mother = XHXFather = XhY
Man = XhY
Mother = XHXh
Father = XhY
b) What is the probability that their son will be colourblind? Use a Punnet square to
determine the answer.
Xh
Y
XH
XHXh
XHY
Xh
XhXh
XhY
SBI3U
Name: ________________
There is a 50% chance that their son will be colourblind.
c) What is the probability that the couple above will have a colourblind daughter?
The probability that the couple will have a colourblind daughter is 25%.
d) In a large family in which all the daughters have normal vision and all of the sons are
colourbind, what are the probable genotypes of the parents? Use Punnet Square(s) to justify
your answer.
XhXh and XHY
5. In dogs, coat colour is determined by the interaction between three alleles. The allele A S
produces a dark coloured dog, ay produces a sandy coloured dog, and at produces a spotted
dog. The order of dominance is AS > ay > at . A dark coloured dog is mated with a sandy
coloured dog. The litter of puppies includes a dark puppy, a sandy puppy, and a spotted
puppy. Use a Punnett Square to determine the possible genotypes of the offspring and the
parents.
Genotypes Dark Coloured Dogs
AS AS
AS ay
AS at
Genotypes for Sandy Dogs
ay ay
ay at
The parents have to be AS at crossed with ay at to produce a dark, sandy and
spotted puppy.
SBI3U
AS
at
Name: ________________
ay
AS ay
ay at
at
AS at
at at
6. A woman with blood type is A has a child with blood type B. She has three
candidates for fatherhood. Their blood types are: Man #1 is type B; Man #2 is type AB; Man #3
is
type O. Based on blood types, the mother says it must have been Man #1.
a. Do you agree? Why or why not?
Woman = IAi
Child = IBIB or IBi
Man #1 = IBIB or IBi
Man #2 = IAIB
Man#3 = ii
No – it could be Man #1 or Man #2
b. This child, a son, is also colorblind. Colorblindness is a x-linked recessive condition. The only
one of the men in question to share this characteristic is #2. The mother is not colorblind. Can
you now determine who the father of the little boy is, assuming it must be one of these men?
Explain your answer.
No – her son would receive his affected X from mom. She must be a carrier. Still can’t
determine who the dad is for sure.
7. Cystic fibrosis is a genetic trait that results in the secretion of abnormal amounts of body
fluids, including unusual sweat and a thick mucus, which prevents the body from properly
cleansing the lungs. The mucus interrupts the function of vital organs and leads to chronic
infections. John exhibits the cystic fibrosis characteristic. He is married to Marie, who does
not. They have three children: Alice, May, and Ken. Only May exhibits the trait for cystic
fibrosis. Both Alice and May are married and have children. Alice married Jack, who does not
exhibit the trait. Alice and Jack have three children: Ralph, Kerri, and Sue. Of the three, Ralph
and Kerri exhibit the trait for cystic fibrosis. May married Sam, who does not exhibit the trait.
SBI3U
Name: ________________
They have two boys, neither of whom have the cystic fibrosis trait.
a. Design and make a pedigree that reflects the information. Include the names of each
individual on your pedigree.
Come see me to check the pedigree
b. What type of inheritance is being expressed (autosomal recessive, autosomal
dominant or sex-linked recessive)?
Autosomal recessive
c. Using the alleles A = normal and a = abnormal, assign genotypes to all individuals
in the pedigree.
John – aa
Marie – Aa
Alice – Aa
May – aa
Ken – Aa
Jack – Aa
Ralph – aa
Kerri – aa
Sue – ASam – ASon #1 – Aa
Son #2 - Aa
8. A naturalist visiting an island in the middle of a large lake observes a species of small bird
SBI3U
Name: ________________
with three distinct types of beaks. Those with short, crushing beaks (BSBS) consume hard
shelled nuts, those with long, delicate beaks (BLBL) pick the seeds from pine cones, and those
with intermediate beaks (BSBL), consume both types of seeds though they are not as good at
either. Assume that this difference in beak morphology is the result of incomplete
dominance. State the phenotypic and genotypic ratios that would result from a cross
between a bird with a short, crushing beak and a bird with an intermediate beak.
BSBS x BSBL
BS
BS
Genotypes
50% BSBS
50% BSBL
BS
BSBS
BSBS
BL
BSBL
BSBL
Phenotypes
50 % short, crushing beaks
50% intermediate beaks