SBI3UName: ______

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

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)0%

B)25%

C)50%

D)75%

E)100%

iiIndividuals II-1 and II-4 can be classified as ______for the trait that is being followed in the pedigree.

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 = XHXhMan = XhY

Mother = XHX-Mother = XHXh

Father = XhYFather = XhY

b) What is the probability that their son will be colourblind? Use a Punnet square to determine the answer.

XH / Xh
Xh / XHXh / XhXh
Y / XHY / XhY

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 AS 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 DogsGenotypes for Sandy Dogs

AS ASay ay

AS ayay at

AS at

The parents have to be AS at crossed with ay atto produce a dark, sandy and spotted puppy.

ay / at
AS / AS ay / AS at
at / ay 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. 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 – A-

Sam – A-

Son #1 – Aa

Son #2 - Aa

8. A naturalist visiting an island in the middle of a large lake observes a species of small bird 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 / BL
BS / BSBS / BSBL
BS / BSBS / BSBL

GenotypesPhenotypes

50% BSBS50 % short, crushing beaks

50% BSBL50% intermediate beaks