Additional problems for lectures 1, 2, and 3. Most of these are from past years exams. They will be representative of the kinds of problems that will appear on the Spring 1998 exams.

 1) A plant heterozygous for four independently assorting pairs of genes (Aa, Bb, Cc, Dd) and homozygous for a fifth (EE) is crossed to a second plant heterozygous for all five genes (Aa, Bb, Cc, Dd, Ee).

a) Determine the expected frequency of the following genotypes in the progeny of such a plant:

i) aa bb cc dd EE

  (1/4) (1/4) (1/4) (1/4) (1/2) = 1 / 512

ii) aa bb Cc Dd Ee

  (1/4) (1/4) (1/2) (1/2) (1/2) = 1 / 128

 iii) Aa Bb Cc Dd Ee

(1/2) (1/2) (1/2) (1/2) (1/2) = 1 / 32

 b) Determine the expected frequency of the parental phenotype among the progeny.

 Both parents have the same phenotype A B C D E.

So probability = p of A- B- C- D- E- = (3/4) (3/4) (3/4) (3/4) (1) = 81 / 256


2) With which of the following modes of inheritance are the following pedigrees inconsistent: autosomal recessive, autosomal dominant, X-linked dominant, X-linked recessive? Provide a BRIEF justification for each answer.

Pedigree (a)

Is inconsistent with both X-linked and autosomal dominant inheritance since neither parent was affected.


Pedigree (b)

Is inconsistent with X-linked dominant inheritance because the father did not pass the trait to his daughter. Is also inconsistent with recessive X-linked or recessive autosomal inheritance since affected parents did not pass on the trait.


Pedigree (c)

Is inconsistent with X-linked recessive inheritance because the affected mother did not pass the trait to her son.


Pedigree (d)

Is inconsistent with X-linked dominant inheritance since the father would have had to have passed the trait to the son. X-linked recessive is possible since the mother could have been heterozygous.



3) Suppose you were on a jury to decide the following case: The Jones family claims that baby Jane, given to them at Chicago Hope, does not belong to them and that baby Sara, who was presented to the Smith family, really belongs to the Jones family. The Jones allege that the babies were exchanged soon after birth as part of "Sweeps Week" plot to improve ratings. The Smiths deny the allegation. Blood group determinations show the following results:


Ms. Jones, AB

Mr. Jones, O

Mrs. Smith, A

Mr. Smith, O

Baby Jane, A

Baby Sara, O


Which baby belongs to which family?


Sara belongs to the Smith's. Mrs. Smith must have genotype IA/ i.


Jane belongs to the Jones's. Her genotype is IA/ i.



4) In sheep white fleece (W) is dominant over black fleece (w). Horned (H) is dominant over hornless (h) in males, but is recessive in females.


a) If a homozygous white horned male is bred to a homozygous hornless black ewe, what will the appearance of the F1? What about the F2?


The F1 males would be horned and white, and the F1 females, hornless and white. HHWW male X female hh ww So all progeny are HhWw which leads to horned white males and hornless white females.

For the F2 :




appearance of

appearance of



(1/4) (3/4) = 3 / 16

Horned, White

Horned, White


HH ww

(1/4) (1/4) = 1 / 16

Horned, black

Horned, black


Hh W_

(1/2) (3/4) = 6 / 16

Horned, White

hornless, White


Hh ww

(1/2) (1/4) = 2 / 16

Horned, black

hornless, black


hh W_

(1/4) (3/4) = 3 /16

hornless, White

hornless, White


hh ww

(1/4) (1/4) = 1 / 16

hornless, black

hornless, black


Probability of:

male Horned, White 3/16 + 6/16 X (1/2) = 9/32

female Horned, White 3/16 X (1/2) = 3/32

male Horned, black 1/16 + 2/16 X 1/2) = 3/32

female Horned, black 1/16 X (1/2) = 1/32

male hornless, White 3/16 X (1/2) = 3/32

female hornless, White 3/16 + 6/16 X 1/2) = 9/32

male hornless, black 1/16 X (1/2) = 1/32

female hornless, black 1/16 + 2/16 X (1/2) = 3/32





b) A horned white ram was bred to the following four ewes with the following results:




Ewe A hornless, black

1 horned white female

Ewe B hornless, white

1 hornless black female

Ewe C horned, black

1 horned, white female

Ewe D hornless, white

1 hornless, black male + 1 Horned white female

What are the genotypes of the five parents?


Father: Hh Ww

Mother A: Hh ww

Mother B: Hh or hh and Ww

Mother C: HH ww

Mother D: Hh Ww


5) In mammals X chromosome dosage compensation is achieved via the inactivation of one of the two Xs present in females. In eutherian mammals (mice, orangutans, undergraduates) the inactivation process is random: Individual cells inactivate either the paternally-derived or the maternally-derived X chromosome with equal probability. In contrast, in marsupials (opossums, kangaroos, Tasmanian devils) all cells inactivate the same X chromosome. Furthermore, the inactive X is always derived from the same parent.


(a) Given the following pedigree for an X-linked trait in kangaroos, can you determine whether it is the paternal X or the maternal X that is inactivated in marsupials? If so, state your reasons. If you cannot tell, explain why, and describe a simple cross, or set of crosses, that will allow you to distinguish the possibilities.

The paternal X is inactivated. If it were the maternal X both daughters of the affected male would show the trait. The mother must therefore have been heterozygous and non-expressing for the trait. By chance only 1/4 of her offspring express the trait.



(b) Based on your answer to part (a): What can you say about the phenotypes and genotypes of the grandparents? If there are ambiguities briefly explain why.


Since the mother was heterozygous and did not express the trait she must have received the X carrying the trait from her father. Thus, the maternal grandfather must have expressed the trait.

The maternal grandmother must have been either homozygous for the wild-type or heterozygous. Her phenotype cannot be inferred.

For the paternal grandparents. The paternal grandmother must have been a carrier since she passed on an X carrying the trait to her son who expressed it. It cannot be determined if she was homo- or heterozygous and her phenotype cannot be determined. Nothing can be said about the genotype or phenotype of the paternal grandfather.



6) Mary and her brother Paul are both healthy adults. Mary and her live-in boyfriend Mike have a healthy baby girl and Mary is pregnant again. They learn that Mary and Paul's mother has just had a baby by a second marriage, and the baby has Duchenne muscular dystrophy. (DMD is a rare X-linked recessive disorder.)


a. Draw the pedigree of the individuals described and their relevant relatives.




b. Do Mary and Mike have to worry about their new baby having DMD? Explain.


Mary and Paul's mother must have been a heterozygote D/d. Mary has a 1/2 chance of being a carrier, and if so 1/2 of her sons would be affected.


c. If so, what is the probability the baby will have DMD?


P = 1/2 (Mary a carrier) X 1/2 (having a son) X 1/2 (son having DMD. i.e. d/Y).

Overall a 1/8 probability.


d. Why didn't Mary and Paul and the first baby get DMD?


Mary and her daughter are not affected because they are either D/D or D/d. Paul was lucky. His genotype is D/Y.




7) What is the MAXIMUM number of phenotypes that could result from the mating of F1 progeny from a cross of AA BB males and aa bb females (A and B are unlinked)? How could this come about?


The maximum number of phenotypes will result if both A and B expressed in a codominant fashion, such that all possible genotypes present a unique phenotype. For complete codominance there will be 9 phenotypes. This could also occur in the case of incomplete dominance.