1. Genetics Problems

2. Practice problem sheet

3. Problem 1

4. Problem 1 solution

5. Problem 2

6. Problem 2 solution

7. Problem 3
In mice, a short-tailed mutant was discovered. When it was crossed to a normal long-tailed mouse, 4 offspring were short-tailed and 3 were long-tailed. Two short-tailed mice from the F1 generation were selected and crossed. They produced 8 short-tailed and 3 long-tailed mice. These genetic experiments were repeated three times with approximately the same results. Hypothesize the mode of inheritance and diagram the crosses.

8. Problem 4
A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.
What is the simplest explanation for the inheritance of these colors in chickens?
Incomplete Dominance
What offspring would you predict from mating a gray rooster and a black hen?
Equal numbers of gray and black chicks

9. Problem 5
1/64
1/64
1/8
1/32

10. Homework Assignment

15. 6 Tall Pink 3 Tall Red 3 Tall White 2 Dwarf Pink 1 Dwarf RedTR TR Tr Tr
6 Tall Pink
3 Tall Red
3 Tall White
2 Dwarf Pink
1 Dwarf Red
1 Dwarf White tR tR tr tr

25. More problems

26. George Arlene Sandra Tom Sam Wilma Ann Michael Carla Daniel Alan Tina
The pedigree below traces inheritance of alkaptonuria. Affected individuals (colored in squares and circles) are unable to breakdown alkapton, which discolors urine and body tissues
George
Arlene
Sandra
Tom
Sam
Wilma
Ann
Michael
Carla
Daniel
Alan
Tina
Christopher
Is alkaptonuria a dominant of recessive trait?
What are the genotypes of the individuals shown in this pedigree?

27. Carla is the Key! George Arlene Sandra Tom Sam Wilma Ann Michael Carla
Daniel
Alan
Tina
Christopher
You can’t get a dominant trait from two recessive parents, so…Alkaptonuria must be recessive

28. George Arlene Sandra Tom Sam Wilma Ann Michael Carla Daniel Alan Tina
Now we can fill in a lot of the genotypes! Everyone with the trait must be …aa aa
George
Arlene
Sandra
Tom
Sam
Wilma
Ann
Michael aa aa
Carla aa
Daniel
Alan
Tina
Christopher

29. George Arlene Sandra Tom Sam Wilma Ann Michael Carla Daniel Alan Tina
Everyone without the trait is …A_ A_ aa
George
Arlene A_
Sandra
Tom
Sam
Wilma
Ann
Michael aa A_ aa A_ A_
Carla A_ aa
Daniel
Alan
Tina A_ A_ A_
Christopher

30. George Arlene Sandra Tom Sam Wilma Ann Michael Carla Daniel Alan Tina
George must be Aa to have fathered Tom and Wilma.
Sam and Ann are Aa since their mom had to give them an a allele.
Michael must be Aa to have fathered Carla. Aa aa
George
Arlene
AA or Aa
Sandra
Tom
Sam
Wilma
Ann
Michael aa Aa aa Aa Aa
Carla
AA or Aa aa
Daniel
Alan
Tina Aa Aa
AA or Aa
Christopher
Daniel and Alan must be Aa to have been fathered by Tom (aa)
We can not be sure of Sandra, Tina, or Christopher. Each of them may be either AA or Aa.

32. Key cross. Two recessives can’t have both recessives and dominants, so…Polydactyly is Dominant

33. Can’t be sex-linked. Since polydactyly is dominant
Can’t be sex-linked. Since polydactyly is dominant. A recessive mother could not have affected sons by a PY father
Key cross. Two recessives can’t have both recessives and dominants, so…Polydactyly is Dominant

37. RR X Rr R r RR Rr R R

38. Still more problems

39. My sexual partner has type A blood. I have type AB
My sexual partner has type A blood. I have type AB. She has a child with type O blood. Can this child be mine?

40. My sexual partner has type A blood. I have type B
My sexual partner has type A blood. I have type B. She has a child with type O blood. Can this child be mine?

41. Kate and Bob want children.
Kate’s dad had hemophilia (an X-linked recessive disorder).
Kate’s brother Ted has hemophilia.
She wants to know the odds of her having a child with hemophilia. What do you tell her?

42. X-linked Problem a
A female MendAlien, Marge, with no ears (the normal condition) and whose male parent had pointed ears pairs with a male MendAlien with no ears and whose male parent also had pointed ears.
Using e for the pointed ear allele and E for the normal no-ear allele, diagram the cross and give the proportions of the phenotypes in the offspring that will have ears. (Give your answer separately for males and females)

43. X-linked Problem a This picture is wrong!
A female MendAlien, Marge, with no ears (the normal condition) and whose male parent had pointed ears pairs with a male MendAlien with no ears and whose male parent also had pointed ears.
Using e for the pointed ear allele and E for the normal no-ear allele, diagram the cross and give the proportions of the phenotypes in the offspring that will have ears . (Give your answer separately for males and females)
This picture is wrong!
Can you spot the problem with the drawings?

44. X-linked Problem b
Assume, instead, that the female MendAlien, in part a, pairs with a male MendAlien with pointed ears as follows:
Diagram the cross, and give the proportions of the phenotypes in the offspring. (give your answer separately for males and females)

45. X-linked Problem b
Assume, instead, that the female MendAlien, in part a, pairs with a male MendAlien with pointed ears.
Diagram the cross, and give the proportions of the phenotypes in the offspring. (give your answer separately for males and females)

46. Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.
A. A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible?
B. A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes?
C. A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes.
What is the man’s genotype?
What is the woman’s genotype?
What are the genotypes of the three children?

47. Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.
A. A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible? Two types of gametes…50% E and 50% e
B. A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes?
C. A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes.
What is the man’s genotype?
What is the woman’s genotype?
What are the genotypes of the three children?

48. Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.
A. A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible? Two types of gametes…50% E and 50% e
B. A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes? Cross is Ee X Ee (children will be 3 hanging:1 attached)
C. A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes.
What is the man’s genotype?
What is the woman’s genotype?
What are the genotypes of the three children?

49. Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.
A. A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible? Two types of gametes…50% E and 50% e
B. A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes? Cross is Ee X Ee (children will be 3 hanging:1 attached)
C. A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes. ee X E_  2E_ and 1 ee
What is the man’s genotype? ee
What is the woman’s genotype? Ee
What are the genotypes of the three children? Ee and Ee and ee

50. Remember Labradors
Genes often interact with one another. The term epistasis is applied to cases in which one gene alters the expression of another gene that is independently inherited. In Labrador retriever dogs one gene locus is involved with production of melanin pigment: B (black) and b (brown) are its two alleles. Another gene locus determines whether the melanin produced is actually deposited in individual hairs as they grow (E) or not deposited (e). Any dog with at least one dominant allele B and one dominant allele E will be a black Labrador. A dog with the homozygous recessive bb and at least one dominant E will be a lighter, chocolate Labrador. If a dog has the homozygous ee genotype it will be a yellow Labrador, regardless of the alleles at the pigment locus.

51. A. Give possible genotypes for a dog with the black Labrador phenotype?
With respect to locus B BB or Bb
With respect to locus E EE or Ee
B. Give possible genotypes for a dog with the chocolate Labrador phenotype?
With respect to locus B bb
C. Give possible genotypes for a dog with the yellow Labrador phenotype?
With respect to locus B BB or Bb or bb
With respect to locus E ee

52. A yellow Labrador with the genotype bbee mates with a black Labrador that is homozygous for both dominant alleles. What are the ratios of genotypes and phenotypes expected in the F1 and F2 generations?
Parental cross: bbee X BBEE
F1 individuals: BbEe
F1 cross: BbEe X BbEe
F1 gametes: BE or Be or bE or be
Setup the 16 box punnett square for the F2 results
F2 results: 9 Black : 3 Chocolate : 4 Yellow

53. In tigers, a recessive allele causes an absence of fur pigmentation ( a “white tiger”) and a cross-eyed condition.
If two phenotypically normal tigers that are heterozygous at this locus are mated what percentage of their offspring will be white?
What percentage will be cross-eyed?