Chapter 02
Mendelian Genetics

Multiple Choice Questions
 
1. The first generation of offspring from the parents is called 
A. P.
B. F1.
C. F2.
D. testcross.
E. backcross.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
2. Which of the following terms is not a type of mating cross? 
A. reciprocal
B. testcross
C. monohybrid
D. dihybrid
E. dominant

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation

3. Individuals having two different alleles for a single trait are called ______. 
A. monohybrid
B. dihybrid
C. dominant
D. recessive
E. dizygotic

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
4. If an individual has 10 gene pairs, how many different gametes can be formed if three of the gene pairs are homozygous and the remaining seven gene pairs are heterozygous? 
A. 49
B. 100
C. 128
D. 1024
E. 131,072

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-07 Mendel's results reflect basic rules of probability
 
5. If the parents of a family already have two boys, what is the probability that the next two offspring will be girls? 
A. 1
B. 1/2
C. 1/3
D. 1/4
E. 1/8

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-07 Mendel's results reflect basic rules of probability

6. In some genetically engineered corn plants, a dominant gene (BT) produces a protein that is lethal to certain flying insect pests that eat the corn plants. It was also found that the pollen could cause death in some flying insects. If the corn plant is heterozygous for BT, what proportion of the pollen would carry the dominant gene? 
A. all pollen
B. 1/2
C. 1/3
D. 1/4
E. 1/8

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
7. A late onset genetic trait description can be used in which of the following? 
A. Cystic fibrosis
B. Sickle-cell anemia
C. Huntington disease
D. Tay-Sachs disease
E. Hurler's disease

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
 
8. The gene responsible for the defective protein in cystic fibrosis is located on which of the following chromosomes? 
A. 11
B. 7
C. 15
D. 4
E. X

Blooms: Remember
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-12 Mendelian Inheritance in Humans

9. When a trait is determined by two or more genes and their interaction with the environment, this is referred to as? 
A. Polygenic
B. Environmental polygenic
C. Recessive
D. Multifactorial
E. Dominant

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-16 Extensions to Mendel for a Single-Gene Inheritance
 
10. Most single-gene diseases in humans that are not of late-onset are caused by which of the following? 
A. Recessive alleles
B. Dominant alleles
C. Horizontal pattern of inheritance
D. Vertical pattern of inheritance
E. Reciprocal allele

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
11. Phenylketonuria (PKU) is caused by _____? 
A. Recessive allele
B. Dominant allele
C. Polygenic
D. Multifactorial
E. Monohybrid allele

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-12 Mendelian Inheritance in Humans

12. Suppose that in plants, smooth seeds (S) are dominant to wrinkled seeds (s) and tall plants (T) are dominant to short plants (t). A tall plant with smooth seeds was backcrossed to a parent that was short and wrinkled. Assuming independent assortment, what proportion of the progeny is expected to be homozygous for short and wrinkled? 
A. 1/2
B. 1/4
C. 1/8
D. 1/16
E. 0

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
13. A rare recessive trait in a pedigree is indicated by which pattern of inheritance? 
A. vertical
B. horizontal
C. diagonal
D. both vertical and horizontal
E. father to daughter inheritance

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-08 Further crosses verify the law of segregation
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

14. Sickle cell anaemia is a recessive trait in humans. The gene that causes this disease is not located on the sex chromosomes. In a cross between a father who has sickle cell anaemia and a mother who is heterozygous for the gene, what is the probability that their first three children will have the normal phenotype? 
A. 1/4
B. 1/2
C. none
D. 1/8
E. 1/16 will be albino

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
15. A dominant trait, Huntington disease, causes severe neural/brain damage at approximately age 40. The gene that causes this disease is not located on the sex chromosomes. A female whose mother has Huntington disease marries a male whose parents are normal. It is not known if the female has the disease. Assuming the female's mother was a heterozygote, and her father was normal, what is the probability that their firstborn will inherit the gene that causes Huntington disease? 
A. 25%
B. 50%
C. 75%
D. 100%
E. 0%

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation

16. In a monohybrid cross AAaa, what proportion of homozygotes is expected among the F2 offspring? 
A. 1/4
B. 1/2
C. 3/4
D. All are homozygotes.
E. None are homozygotes.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
17. An allele that expresses its phenotype even when heterozygous with a recessive allele is called _______ 
A. recessive.
B. recombinant.
C. dominant.
D. parental.
E. independent.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
18. Assume that in guinea pigs, dark brown fur (B) is dominant to black fur (b). If you mate a black guinea pig with a homozygous brown guinea pig, what proportion of the progeny will be homozygous? 
A. none
B. 1/4
C. 1/2
D. 3/4
E. all

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation

19. In the dihybrid cross AaBbaabb, what proportion of individuals are expected to be homozygotic for both genes in the F1 generation? 
A. 1/4
B. 1/2
C. 3/4
D. All are homozygotes.
E. None are homozygotes.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
20. ______ is/are a cross(es) between parents that differ in only one trait. 
A. Self-fertilization
B. Cross fertilization
C. Monohybrid crosses
D. Artificial selection
E. Reciprocal crosses

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
21. Assuming independent assortment, which of the crosses below will produce a 1:1 phenotypic ratio among the F1 progeny? 
A. AABBaabb
B. AaBbAaBb
C. AaBbaabb
D. AaBBaaBB
E. AAbbaaBB

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

22. The actual alleles present in an individual make up the individual's 
A. recombinant types.
B. zygote.
C. dominant allele.
D. allele.
E. genotype.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-08 Further crosses verify the law of segregation
 
23. In a dihybrid cross AAbbaaBB, what proportion of the F2 offspring is expected to be homozygotic for at least one gene? 
A. 1/4
B. 1/2
C. 3/4
D. All are homozygotes.
E. None are homozygotes.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
24. A phenotype reflecting a new combination of genes occurring during gamete formation is called 
A. a recombinant type.
B. an independent assortment.
C. heterozygous.
D. homozygous.
E. a multihybrid cross.

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

25. Assume that in guinea pigs, dark brown fur (B) is dominant to black fur (b). If you mate a homozygous black guinea pig with a heterozygous brown guinea pig, what proportion of the progeny will be black? 
A. none
B. 1/4
C. 1/2
D. 3/4
E. all

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
26. The diploid cell formed by the fertilization of the egg by the sperm during sexual reproduction is a _____ 
A. reciprocal.
B. zygote.
C. dihybrid.
D. gamete.
E. monohybrid.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
27. A gamete is ________. 
A. Only an egg
B. Only a sperm
C. Either an egg or a sperm
D. A zygote
E. Only a sex chromosome

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation

28. In a dihybrid cross for which the parental cross is AABBaabb, what proportion of F2 offspring will be heterozygous for both genes? Assume independent assortment. 
A. 1/4
B. 1/2
C. 3/4
D. All are heterozygotes.
E. None are heterozygotes.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
29. An alternative form of a single gene is known as 
A. parental.
B. dihybrid.
C. reciprocal.
D. allele.
E. recessive.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Topic: 02-04 Mendel devised a new experimental approach
 
30. Assume that in guinea pigs, dark brown fur (B) is dominant to black fur (b). If you mate a homozygous black guinea pig with a homozygous brown guinea pig, what proportion of the progeny will be heterozygous? 
A. none
B. 1/4
C. 1/2
D. 3/4
E. all

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation

31. Which of the crosses listed below will give a 1:1:1:1 genotypic ratio in the F1 generation? Assume independent assortment. 
A. AABBaabb
B. AaBbAaBb
C. AaBbaabb
D. AaBBaaBB
E. AAbbaaBB

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
32. For the cross AaBbaabb, what proportion of F1 offspring will be heterozygous for both gene pairs? Assume independent assortment. 
A. 1/4
B. 1/2
C. 3/4
D. All are heterozygotes.
E. None are heterozygotes.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
33. If a dog breeder chooses the parents for a desired next generation, the dog breeder is using a process called _____________. 
A. artificial selection
B. natural selection
C. mutation
D. evolution
E. random selection

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-01 Background: The Historical Puzzle of Inheritance

34. When both egg and pollen from the same plant produce a zygote, the process is called 
A. cross-pollination
B. outcrossing
C. self-fertilization
D. recombination
E. trans-pollination

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-03 Desirable traits sometimes disappear and reappear
 
35. Which of the following was not involved in the rediscovery of Mendel's work? 
A. Correns
B. de Vries
C. Tschermak
D. Morgan
E. Watson

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
36. What does a vertical pattern of inheritance in a pedigree likely indicate? 
A. rare recessive trait
B. rare dominant trait
C. mulitgenic inheritance
D. environmental impact
E. common recessive trait

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

37. Calculate the probability of either all-dominant or all-recessive genotypes for the alleles A, B, E, and F in the following cross: AaBbccddEeFfAaBbCcddEeFf 
A. 1/32
B. 1/16
C. 1/64
D. 1/128
E. 1/256

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
38. In some plants, a purple pigment is synthesized from a colourless precursor. In a cross between two plants, one purple and the other colourless, an F1 generation was produced that was all-purple. The F2 produced from the F1 had 775 purple, 200 red, and 65 colourless. What is the genotype of the parents? 
A. AABBAABB
B. AABBaabb
C. aabbaabb
D. aaBBaabb
E. AAbbaabb

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

39. Lines that produce offspring carrying specific parental traits that remain constant from generation to generation are called 
A. maternal
B. indeterminate
C. heterozygous
D. True-breeding
E. wild-type

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-04 Mendel devised a new experimental approach
 
40. After a cross between two corn plants, the F1 plants all had a dwarfed phenotype. The F2 consisted of 1,207 dwarf plants and 401 tall plants. Identify the phenotypes and genotypes of the two parents. 
A. DD (dwarf), dd (tall)
B. DD (tall), dd (dwarf)
C. dd (dwarf), dd (tall)
D. DD (dwarf), DD (tall)
E. dd (dwarf), Dd (tall)

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-04 Mendel devised a new experimental approach

41. Rosy coloured eyes and forked bristles are unlinked, recessive traits in Drosophila. A rosy-eyed Drosophila with wild-type bristles was crossed with a forked Drosophila with wild-type eyes. All of the F1 were phenotypically wild-type for both traits, whereas the F2 consisted of 306 wild-type, 94 rosy-eyed, 102 fork-bristled, and 33 forked-bristled and rosy-eyed flies. Infer the genotypes of the parents. 
A. rrff, RRFF
B. Rrff, rrFf
C. RRFF, RRFF
D. rrff, rrff
E. RRff, rrFF

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
42. Which of the following is not a phenotypic description of allele interactions affecting the expression of traits? 
A. incomplete dominance
B. codominance
C. polymorphic
D. multifactorial
E. pleiotropic

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics

43. An interaction between non-allelic genes that results in the masking of expression of a phenotype is 
A. epistasis.
B. epigenetic.
C. dominance.
D. codominance.
E. incomplete dominance.

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
44. Which of the following diseases show pleiotropism? 
A. albinism
B. muscular dystrophy
C. colour blindness
D. sickle cell anaemia
E. male pattern baldness

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics
 
45. A deviation from normal Mendelian ratios, which may be resolved by counting and/or controlled crosses, is seen in which of the following terms? 
A. pleiotropy
B. codominance
C. incomplete dominance
D. complete dominance
E. penetrance and expressivity

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype

46. Which of the following phenotypic ratios show incomplete dominance? 
A. 2:1
B. 3:1
C. 1:2:1
D. 1:1
E. 4:1

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
47. Which of the following ratios show codominance? 
A. 2:1
B. 3:1
C. 1:2:1
D. 1:1
E. 4:1

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
48. Which of the following ratios indicates a lethal gene? 
A. 2:1
B. 3:1
C. 1:2:1
D. 1:1
E. 4:1

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics

49. A person who has type O blood has 
A. anti-A antibodies.
B. anti-B antibodies.
C. anti-AB antibodies.
D. both anti-A and -B antibodies.
E. no surface antigens.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles
 
50. If two or more forms of the same gene exist, the different forms are called. _______ 
A. incomplete dominance.
B. penetrance and expressivity.
C. pleiotropic.
D. alleles.
E. dihybrid.

Blooms: Remember
Difficulty: Medium
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Topic: 02-04 Mendel devised a new experimental approach
 
51. The blood groups A, B, and O are different types of 
A. incomplete dominance.
B. penetrance and expressivity.
C. pleiotropy.
D. alleles.
E. heterozygotes.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles

52. The blood groups A, B, and O show 
A. complete dominance.
B. recessiveness.
C. codominance.
D. complete dominance, recessiveness, and codominance.
E. Corecessiveness.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles
 
53. Which of the following monohybrid ratios can describe incomplete dominance and codominance? 
A. 2:1
B. 3:1
C. 1:3
D. 1:2:1
E. 4:1

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
54. Which of the following ratios demonstrate gene interaction? 
A. 2:1
B. 3:1
C. 1:2:1
D. 9:3:4
E. 1:3

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics

55. A _____________ results whenever the nucleotide sequence is changed. 
A. phenotype
B. genotype
C. mutation
D. trait
E. character

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-19 Mutations are the source of new alleles
 
56. When the same gene is related to respiratory problems and sterility, it can be described as 
A. pleiotropy.
B. codominance.
C. incomplete dominance.
D. complete dominance.
E. penetrance and expressivity.

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics
 
57. Another name for a normal gene is 
A. wild-type.
B. pleiotropy.
C. dominant.
D. codominant.
E. recessive.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-20 One gene may contribute to several characteristics

58. The phenotypic ratio 1:2:1 may indicate 
A. complete dominance.
B. codominance.
C. epistasis.
D. recessive lethal.
E. codominance and epistasis.

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
59. The phenotypic ratio 3:1 may indicate 
A. complete dominance.
B. codominance.
C. epistasis.
D. incomplete dominance.
E. codominance and epistasis.

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
60. The phenotypic ratio 2:1 may indicate 
A. complete dominance.
B. codominance.
C. epistasis.
D. recessive lethal.
E. codominance and epistasis.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics

61. The phenotypic ratio 9:7 may indicate 
A. complete dominance.
B. codominance.
C. epistasis.
D. recessive lethal.
E. complementary gene action.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics
 
62. The phenotypic ratio 9:3:4 may indicate 
A. complete dominance.
B. codominance.
C. epistasis.
D. recessive lethal.
E. codominance and epistasis.

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics
 
63. Which of the following phenotypic ratios show independent assortment? 
A. 9
B. 7
C. 4
D. 13:3
E. 5

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

64. Temperature sensitive (ts) alleles of the Drosophila shibire gene were isolated by David Suzuki. Under permissive conditions, what is the phenotype of flies homozygous for the ts alleles? 
A. lethal
B. conditional on other factors
C. indistinguishable from wild-type
D. continuously variable
E. co-dominant

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
65. People may inherit a specific genotype that predisposes them to cancer. However, not everyone with this genotype develops cancer; the occurrence of cancer in these individuals is dependent on environment. This is an example of: 
A. epistasis
B. incomplete penetrance
C. variable expressivity
D. incomplete dominance
E. complementation

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
66. If a mother is phenotype A and her child is phenotype B then the father's genotype is _____? 
A. IAIB
B. ii
C. IAIA
D. IAi
E. Cannot be determined

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles

67. Which of the following options is considered the universal donor of blood? 
A. IAIB
B. ii
C. IAIA
D. IAi
E. IBi

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles
 
68. Which of these is not an example of a continuous trait? 
A. age at death
B. occurrence of phenylketonuria (PKU)
C. human skin colour
D. birth weight of mice
E. plant height

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation
 
69. Which of the following statements about continuous traits is not true? 
A. they do not obey Mendel's laws
B. they are called complex traits
C. they are also called quantitative traits
D. they are relevant to medicine
E. they are relevant to agriculture

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation

70. Several alleles at several different loci all contribute additively to the same trait. Therefore, for this trait: 
A. homozygotes cannot exist
B. heterozygotes cannot exist
C. only one phenotypic class is possible
D. only two phenotypic classes are possible
E. continuous variation may be observed

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation
 
71. How does penetrance differ from expressivity? 
A. expressivity is dependent on environment; penetrance is not
B. expressivity is qualitative (presence or absence); penetrance is quantitative
C. penetrance involves multiple genes; expressivity involves a single gene
D. penetrance is qualitative (presence or absence); expressivity is quantitative
E. penetrance is dependent on environment; expressivity is not

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
72. When a gene has a more subtle and secondary effect on the phenotype, the gene is usually called _____? 
A. Lethal
B. Modifier
C. Recessive
D. Conditional
E. Permissive

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype

73. When a certain condition stimulates a particular allele to be lethal, this allele is referred to as _____? 
A. Permissive
B. Restrictive
C. Lethal
D. Modifier
E. Conditional

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
74. Wild-type pea flowers are purple. You find spontaneous, white-flowered mutants growing nearby in five different locations (numbered a-e). You establish pure breeding lines of each and perform crosses between them, and record the F1 phenotype in the table below. Based on the data in the table, how many different genes in the pathway for purple flowers have been identified by mutation?
   
 	a	b	c	d	e	
a	white	purple	purple	white	purple	
b	purple	white	purple	purple	purple	
c	purple	purple	white	purple	white	
d	white	purple	purple	white	purple	
e	purple	purple	white	purple	white	
 
A. 1
B. 2
C. 3
D. 4
E. 5

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-10 Evaluate the significance of the complementation test as a tool for genetic analysis.
Topic: 02-24 Breeding studies help decide how a trait is inherited

75. Which of the following is not useful in a complementation test? 
A. recessive alleles
B. sexual reproduction
C. pure breeding lines
D. F1 progeny
E. alleles dominant to wild-type

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-10 Evaluate the significance of the complementation test as a tool for genetic analysis.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
76. If two homozygous recessive mutants show the same phenotype, but are caused by mutations at different loci, what will be the phenotype ratio among their F1 progeny? 
A. 1 wild-type: 0 mutant
B. 1 wild-type: 2 mutant
C. 1 wild-type: 1 mutant
D. 2 wild-type: 1 mutant
E. 0 wild-type: 1 mutant

Blooms: Analyze
Difficulty: Easy
Learning Objective: 02-10 Evaluate the significance of the complementation test as a tool for genetic analysis.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
77. AA and Aa make red flowers, and aa makes white flowers. BB and Bb make tall plants, and bb makes short plants. What would be the expected ratios of phenotypes among the offspring of the cross of AaBb x aaBb? Note the genotypes in the cross carefully. Assume independent assortment of each gene. 
A. 9 (red & tall): 3 (red & short): 3 (white & tall): 1 (white & short)
B. 3 (red & tall): 1 (white & tall)
C. 3 (red & tall): 1 (red & short): 3 (white & tall): 1 (white & short)
D. all (red & tall)
E. 1 (red & tall): 1 (red & short): 1 (white & tall): 1 (white & short)

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation

78. Seeds of some lentils are speckled. A true breeding strain with small speckles is crossed with a true breeding strain with large speckles. All of the F1 progeny have both large and small speckles. Which of the following is true? 
A. The trait is controlled by two genes and the alleles are co-dominant
B. The trait is controlled by one gene and the alleles are co-dominant
C. The trait is controlled by two genes and the alleles are incompletely dominant
D. The trait is controlled by one gene and the alleles are incompletely dominant
E. The trait is controlled by one gene and both alleles are dominant

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles

True / False Questions
 
79. Phenotype for a given trait can be influenced by an environmental factors such as temperature? 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
80. The mating of parents with antagonistic traits produces hybrids. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-04 Mendel devised a new experimental approach

81. Mendel's law of segregation states that two alleles for each trait unite in a specific, predictable manner during gamete formation. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
82. Dihybrid crosses helped reveal the law of independent assortment. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
83. The Punnett square was introduced in 1906 by Reginald Punnett and provides a simple and convenient method of tracking possible combinations of gametes that might be produced in a given cross. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-07 Mendel's results reflect basic rules of probability
 
84. Using the product rule, one would calculate the probability of parents having six children who are all boys as (½)6. 
TRUE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-07 Mendel's results reflect basic rules of probability

85. The sum rule states that the probability of both of two mutually exclusive events occurring is the sum of their individual probabilities. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-07 Mendel's results reflect basic rules of probability
 
86. If you know the phenotype and the dominance relation of the alleles you can predict the genotype. 
FALSE

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
87. An individual can be a heterozygote for one trait and a homozygote for another. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
88. A testcross is a cross between two heterozygotes. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability

89. At fertilization, in the mating of dihybrids, four different kinds of eggs can combine with four different kinds of pollen, producing a total of sixteen different genotypes. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
90. When examining a pedigree, a father to son transmission for a disease that manifests itself in every generation is an indication that the pattern of inheritance is likely to be autosomal dominant. 
TRUE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
91. If a 4 generation family pedigree shows that the disease manifests for the first time in the 4th generation then it's likely that the pedigree would show consanguinity? 
TRUE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait

92. A 3 generation pedigree of Huntington's disease would show a skip generation? 
FALSE

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
93. During gamete formation, different pairs of alleles on different chromosomes segregate independently of each other. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
94. If yellow and round phenotypes in peas are dominant, and pea shape and colour are each controlled by a single gene, you know the genotype of all peas that are green and wrinkled. 
TRUE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
95. Several single-gene disorders are more common in some populations of people than in others. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

96. When examining a dominant trait, affected children always have at least one affected parent. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
97. Two affected parents can produce unaffected children in a recessive trait? 
FALSE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
98. Consanguineous mating increase the likelihood of a dominant trait? 
FALSE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait
 
99. Incomplete dominance means that the hybrid does not resemble either pure-breeding parent? 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete

100. A lethal disorder does not include the inheritance of traits that cause death in adulthood. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics
 
101. Cross-fertilization is the same as reciprocal cross. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-04 Mendel devised a new experimental approach
 
102. Traits such as human height are considered as a type of discrete traits. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation
 
103. When a sperm cell fertilizes an egg cell the result is called zygote. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation

104. The following genotype: Gg is called heterozygote. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
105. Parental generation is designated as (P) and the progeny of the parental generation is designated as F1 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-06 Monohybrid crosses reveal units of inheritance and the law of segregation
 
106. The law of segregation is a Mendelian law that states that both alleles must separate during gamete formation. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-03 Explain Mendel's law of segregation.
Topic: 02-08 Further crosses verify the law of segregation
 
107. Multifactorial inheritance is when a phenotype arises as a result of multiple genes interacting with each other and/or the environment. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-16 Extensions to Mendel for a Single-Gene Inheritance

108. The flower colours white, pink, and red indicate codominant inheritance. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
109. A phenotype that is expressed in 87% of individuals with the same genotype shows complete penetrance. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
110. When a late blooming pea and an early blooming pea are crossed and an intermediate phenotype occurs, this result would suggest incomplete dominant inheritance. 
TRUE

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
111. In codominance, F1 hybrids show the traits of both parents. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete

112. Different alleles indicate unique genes. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-02 Distinguish between the terms gene and allele and contrast dominant alleles with recessive alleles.
Learning Objective: 02-10 Evaluate the significance of the complementation test as a tool for genetic analysis.
Topic: 02-04 Mendel devised a new experimental approach
 
113. Mutations are the source of new alleles. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-19 Mutations are the source of new alleles
Topic: 02-20 One gene may contribute to several characteristics
 
114. A wild-type allele is any allele whose frequency is closest to 100%. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-19 Mutations are the source of new alleles
Topic: 02-20 One gene may contribute to several characteristics
 
115. A measurable traits such as the length of a tobacco flower in millimeters is often considered a form of a discontinuous trait and is polygenic. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation

116. A mutant allele has a rare occurrence in a population. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-19 Mutations are the source of new alleles
Topic: 02-20 One gene may contribute to several characteristics
 
117. Genes with more than one wild-type allele are termed polymorphic. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-19 Mutations are the source of new alleles
Topic: 02-20 One gene may contribute to several characteristics
 
118. The mouse agouti gene has one wild-type allele and several mutant alleles. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-07 Explain the terms wild-type allele, mutant allele, monomorphic, and polymorphic.
Topic: 02-19 Mutations are the source of new alleles
Topic: 02-20 One gene may contribute to several characteristics
 
119. The phenomenon of a single gene determining a number of distinct and seemingly unrelated characteristics is known as pleiotropy. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-20 One gene may contribute to several characteristics

120. HbßsHbßshomozygous are resistant to Plasmodium falciparum 
TRUE

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-08 Describe pleiotropy and how it arises.
Topic: 02-21 A comprehensive example: Sickle-cell disease illustrates many extensions to Mendel's analysis
 
121. In epistasis, one gene's alleles mask the effects of another gene's alleles. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
122. A gene interaction in which the effects of an allele at one gene hide the effects of alleles at another gene is known as dominance. 
FALSE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait
 
123. Epistasis in which a dominant allele of one gene hides the effects of another gene is called recessive epistasis. 
FALSE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait

124. When an organism has two genes that perform the same function, these genes are called redundant genes? 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait
 
125. In complementary gene action, dominant alleles of two or more genes are required to generate a particular trait. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
126. Mutant alleles at one of two or more different genes can result in the same phenotype? 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait
 
127. Dominant epistasis II is also known as dominant suppression. 
TRUE

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait

128. To produce a particular normal phenotype, the dominant allele of two interacting genes can both be necessary. 
TRUE

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-23 Two genes can interact to determine one trait

Short Answer Questions
 
129. You are a judge in a civil trial where a young man is attempting to prove that he is the illegitimate child of a very wealthy man who has recently died. He wishes to be included in the distribution of the wealth. After considering all the testimony about how this person was conceived, the key evidence seems to come down to two main facts. The wealthy man and the mother of the young man are both deaf but the young man is not. Therefore the lawyer of the family suggests that the wealthy man is not the father. The mother, wealthy man, and young man all have O, MM, and Rh Blood Type at the phenotypic level but a genotyping screen indicates that the wealthy man is actually IAIAhh blood type. How do you interpret the evidence presented and how does it influence your decision in this case? 
The fact that the young man can hear is not evidence against his being the son of the wealthy man. Two deaf individuals can, via complementation, give rise to hearing offspring if the mutation they carry is on different genes (hearing is a polygenic trait.) The blood type evidence is definitive in favour of the wealthy man not being the father of the young man. Although both putative parents and the son in question have O blood type, the wealthy man is genetically type A and phenotypically type O because of recessive homozygosity of the h allele which leads to Bombay phenotype; the protein to which the A sugar attaches is missing thereby making the wealthy man phenotypically type O. Any son of his would be highly likely to have A-antigen, as the h allele is very rare in humans, making homozygous recessive offspring extremely unlikely except in consanguineous matings.

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited

130. Can a phenotype O be the father of a child who is phenotype B if the mom is phenotype A? 
NO

Blooms: Understand
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-18 A gene may have more than two alleles
 
131. Calculate the probability of the production of a homozygous recessive genotype for the following cross: AaBbccddEeFfAaBbCcddEeFf 
¼  ¼  ½  1  ¼  ¼ = 1/512

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
132. A phenotypically normal man who has two siblings died from an autosomal recessive disease before the age of 5. What is the risk that this man is heterozygous carrier for the autosomal recessive mutation? 
2/3

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

133. Karen, a 35-year-old woman affected by an autosomal dominant disease that has 80% penetrance marries Jon, a 40 year old man who is similar to his wife (a heterozygous) for the same autosomal dominant disease. If they decide to have a child, what is the probability that the child is going to be phenotypically normal? 
40%

Blooms: Apply
Difficulty: Hard
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
134. In Drosophila, forked (fk) bristles are recessive to normal (fk+) and glassy eyes (gls) are recessive to normal (gls+). If an F1 heterozygous female is backcrossed to the homozygous wild-type male parent, predict the genotypes and phenotypes of the offspring. 

Genotype	Phenotype	
fk+fk+gls+gls+	Wild type	
fk+fk+gls+gls	Wild type	
fk+fk gls+gls+	Wild type	
fk+fk gls+gls	Wild type	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans

135. A science teacher is attempting to convince her class that alcoholism, which has long been known to be a disease of polygenic inheritance, really is partially genetically determined. You are asked to assist in the design of an experiment that will help show eighth graders genetic transmission of differences in alcohol drinking. You have been given outbred rats as your experimental model. Set up a quantitative experiment that would test the hypothesis that alcoholism, as determined by amount of alcohol drunk, is a quantitative trait. 
Set up a selective breeding experiment. Provide rats with water and with a solution of water and alcohol in a low concentration. Measure the consumption of the alcohol-containing solution per day for all rats. Breed the high-drinking male rats with the high-drinking females, and the low-drinking males with low-drinking females. Test the offspring for alcohol solution consumption, and do the same in subsequent generations. If the rats bred for high drinking continue to increase their drinking levels from generation to generation, and the low drinkers decrease their drinking levels in the same way, this is evidence that alcohol consumption is genetically determined. Your data will also show that the individual rats differ in amount of consumption, and when plotted together the data will show a continuous distribution, indicating a quantitative trait (interactions of more than one gene and interactions with the environment contribute to the alcohol drinking trait).

Blooms: Create
Difficulty: Medium
Learning Objective: 02-12 Explain the inheritance of continuous traits.
Topic: 02-27 Mendelian principles can also explain continuous variation

136. In corn, liguleless (l) is recessive to ligules (L) and a green leaf (G) is dominant to the normal non-green (g). If a testcross is performed with a plant that is a dihybrid for both of these genes, what would be the phenotypes and genotypes of the progeny? Assume independent assortment. 

Genotype	Phenotype	
LlGg	Ligules/Green	
Llgg	Ligules/Non-green	
llGg	Liguleless/Green	
llgg	Liguleless/Non-green	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans

137. Short hair in rabbits is produced by a dominant allele (l+) and long hair by its recessive allele (l). Black hair results from the action of a dominant allele (b+) and brown hair from its recessive allele (b). Determine the genotypic and the corresponding phenotypic ratios of the F2 offspring, beginning with a parental cross of a rabbit with brown, short hair to a rabbit with long, black hair. Assume that the parent with short hair is homozygous for that allele, and that the parent with black hair is homozygous for that allele. Assume independent assortment. 

#	Genotype	Phenotype	
1	l+l+ b+b+	Short Black	
2	l+l b+b+	Short Black	
2	l+l+ b+b	Short Black	
4	l+l b+b	Short Black	
1	l+l+ bb	Short Black	
2	l+l bb	Short Black	
1	ll b+b+	Long Black	
2	ll b+b	Long Black	
1	llbb	Long Black	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
 
138. What does a diamond symbol à in a pedigree indicate? 
Sex unspecified

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families

139. You wish to know the genotype of some carrot plants that you have grown in your garden so that you might grow more of them. They have reddish orange flesh, are sweet in taste, long in root, and short in leaf. Using classical genetic techniques how would you determine the genotype? 
You need to determine the dominant/recessive nature of each trait. Set up crosses between reddish orange, sweet tasting, long in root, and short in leaf carrot plants and true orange, plain tasting, short in root, and long in leaf carrot plants to determine each dominant trait. Then create a "tester plant" that is recessive for all four traits. Cross your favourite carrot plants with the tester and observe the offspring. The traits shown in the offspring are indicative of the genotype of your original carrot plant.

Blooms: Create
Difficulty: Easy
Learning Objective: 02-04 Differentiate between the terms homozygous, heterozygous, genotype, and phenotype.
Topic: 02-07 Mendel's results reflect basic rules of probability
Topic: 02-08 Further crosses verify the law of segregation
 
140. List 3 criteria to recognize dominant traits? 
Affected children always have at least one affected parent, there is vertical pattern of inheritance, the trait shows up in every generation, two affected parents can produce unaffected children (if the parents are heterozygous).

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait

141. In Drosophila, forked (fk) bristles are recessive to normal (fk+) and glassy eyes (gls) are recessive to normal (gls+). If a homozygous wild-type male is mated to a forked-bristled, glassy-eyed female, predict the genotypes and phenotypes of the F2. Assume independent assortment. 

# Genotype	Phenotype	
1 fk+fk+ gls+gls+	Wild type	
2 fk+fk+ gls+gls	Wild type	
2 fk+fk gls+gls+	Wild type	
4 fk+fk gls+gls	Wild type	
1 fk+fk+ gls gls	Glassy eyes	
2 fk+fk gls gls	Glassy eyes	
1 fkfk gls+gls+	Forked bristles	
2 fkfk gls+gls	Forked bristles	
1 fkfk gls gls	Forked bristles and glassy eyes	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
 
142. In Drosophila, forked (fk) bristles are recessive to normal (fk+) and glassy eyes (gls) are recessive to normal (gls+). If a homozygous wild-type male is mated to a forked-bristle, glassy-eye female, predict the genotypes and phenotypes of the F1. 

Genotype	Phenotype	
fk+fk gls+gls	Wild type	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

143. Short hair in rabbits is produced by a dominant allele (l+) and long hair by its recessive allele (l). Black hair results from the action of a dominant allele (b+) and brown hair from its recessive allele (b). Determine the genotypic and the corresponding phenotypic ratios of the F1 offspring, beginning with a parental cross of a rabbit with brown, short hair to a rabbit with long, black hair. Assume that the parent with short hair is homozygous for that allele, and that the parent with black hair is homozygous for that allele. Assume independent assortment. 

Genotype	Phenotype	
l+l b+b	short, black	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
144. Stem colour of tomato plants is known to be under the genetic control of at least one pair of alleles such that A_ results in the production of anthocyanin pigment (purple stem). The recessive genotype aa lacks this pigment and hence is green. The production of two locules (seed chambers) in the tomato fruit is controlled by the dominant allele M, and multiple locules is determined by mm. Determine the genotypic and phenotypic ratios of the F1 from a cross between an inbred tomato plant with a purple stem and fruit with two locules crossed to a tomato plant with a green stem and fruit with multiple locules. 

Genotype	Phenotype	
AaMm	purple, 2 locules	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment

145. In corn liguleless, (l) is recessive to ligules (L) and a green leaf (G) is dominant to the normal non-green (g). If a plant homozygous for liguleless and green leaves is crossed to one homozygous for non-green with ligules, predict the phenotypes and genotypes of the F1. Assume independent assortment. 

Genotype	Phenotype	
LlGg	Ligules/Green	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
 
146. If a scientist performs a cross in which the male parent traits and the female parent traits are reversed, the cross is referred to as _____? 
Reciprocal cross

Blooms: Understand
Difficulty: Easy
Learning Objective: 02-01 Explain how monohybrid crosses led Mendel to infer the law of segregation.
Topic: 02-04 Mendel devised a new experimental approach

147. You are out on a nature walk up in the mountains and you find a pretty wildflower in the lower altitude that is short and bushy with small, fragrant, bright purple flowers. In the higher altitude you find what seems to be the same plant, yet it is tall and sparse with larger flowers of the same colour and fragrance. A) Set up an experiment to test the hypothesis that the plants are different due to genetic but not environmental influences. B) Is it possible to tell if both genetic and environmental effects occur? 
A) Assuming these are not endangered plants and you are not in a protected area, obtain several specimens from each location. Plant seeds of both types of plants in both low- and high-altitude locations. Observe the offspring. If the offspring look the same as their parental stock, then the differences are simply genetic in nature. If the offspring look short and bushy with small fragrant, bright purple flowers in the lower altitude, but tall and sparse with larger flowers of the same colour and fragrance in the higher altitude, then the differences are due to environmental influences. B) Yes, a combination of the traits would indicate that both environmental and genetic influences play a role in the differences you have identified.

Blooms: Create
Difficulty: Medium
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
148. List two diseases that are caused by a dominant allele? 
Hypercholesterolaemia, Huntington

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-12 Mendelian Inheritance in Humans
 
149. List two diseases that are caused by a recessive allele? 
Sickle-cell anemia, cystic fibrosis, Tay-Sachs Phenylketonuria, Thalassemia.

Blooms: Remember
Difficulty: Easy
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-12 Mendelian Inheritance in Humans

150. In corn liguleless, (l) is recessive to ligules (L) and a green leaf (G) is dominant to the normal non-green (g). If a plant homozygous for liguleless and green leaves is crossed to one homozygous for non-green with ligules, predict the phenotypes and genotypes of the F2. 

# Genotype	Phenotype	
1 LLGG	Ligules/Green	
2 LLGg	Ligules/Green	
2 LlGG	Ligules/Green	
4 LlGg	Ligules/Green	
1 LLgg	Ligules/Non-green	
2 Llgg	Ligules/Non-green	
1 llGG	Liguleless/Green	
2 llGg	Liguleless/Green	
1 llgg	Liguleless/Non-green	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families

151. Stem colour of tomato plants is known to be under the genetic control of at least one pair of alleles such that A_ results in the production of anthocyanin pigment (purple stem). The recessive genotype aa lacks this pigment and hence is green. The production of two locules (seed chambers) in the tomato fruit is controlled by the dominant allele M, and multiple locules is determined by mm. Determine the genotypic and phenotypic ratios of the F2 offspring beginning with a parental cross between an inbred tomato plant that has a purple stem and fruit with two locules, and a tomato plant that has a green stem and fruit with multiple locules. Assume independent assortment. 

#	Genotype	Phenotype	
1	AAMM	Purple, 2 locules	
2	AaMM	Purple, 2 locules	
2	AAMm	Purple, 2 locules	
4	AaMm	Purple, 2 locules	
1	aaMM	Green, 2 locules	
2	aaMm	Green, 2 locules	
1	AAmm	Purple, Multi locules	
2	AAMm	Purple, Multi locules	
1	Aamm	Green, Multi locules	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families

152. Below is a pedigree for a human trait. Shaded symbols are for individuals exhibiting the trait. Identify the mode of inheritance of the trait and apply the laws of probability to calculate the probability that individual #4 is a heterozygous carrier of the trait.

Mode of inheritance is recessive. The probability that #4 is a carrier is 1/4, since both of his parents are carriers, and since he does not have the trait himself (i.e. 3 Aa: 1 AA).

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait

153. In corn, three dominant genes are necessary for aleuronecolour. The genotype B_D_R_ is coloured. Any homozygous recessive for one gene is colourless. Predict the genotypes and phenotypes of the offspring of the cross BbDdRrBbDdRr
Phenotype: 27 coloured; 37 colourless 

Ratio of Genotypes	
1	BBDDrr	
2	BBDdrr	
2	BbDDrr	
4	BbDdrr	
1	BBddrr	
2	Bbddrr	
1	bbDDrr	
2	bbDdrr	
1	Bbddrr	
2	BBDDRr	
4	BBDdRr	
4	BbDDRr	
8	BbDdRr	
2	BBddRr	
4	BbddRr	
2	bbDDRr	
4	bbDdRr	
2	bbddRr	
1	BBDDRR	
2	BbDDRR	
2	BBDdRR	
4	BbDdRR	
1	bbDDRR	
2	bbDdRR	
1	BBddRR	
2	bbDdRR	
1	bbddRR	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families

154. In corn, three dominant genes are necessary for aleuronecolour. The genotype B_D_R_ is coloured. Any homozygous recessive for one gene is colourless. Predict the genotypes and phenotypes of the offspring of the cross BbDdRRBbDdRR 
Phenotype: 9 colour; 7 colourless
   
Ratio of Genotypes	
1	BBDDRR	
2	BbDDRR	
2	BBDdRR	
4	BbDdRR	
1	bbDDRR	
2	bbDdRR	
1	BBddRR	
2	bbDdRR	
1	bbddRR	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families

155. In corn, three dominant genes are necessary for aleuronecolour. The genotype B_D_R_ is coloured. Any homozygous recessive for one gene is colourless. Predict the genotypes and phenotypes of the offspring of the cross BbDdRRBbDdrr 
Phenotype: 9 colour; 7 colourless
   
Ratio of Genotypes	
1	BBDDRr	
2	BBDdRr	
2	BbDDRr	
4	BbDdRr	
1	BBddRr	
2	BbddRr	
1	bbDDRr	
2	bbDdRr	
1	bbddRr	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-09 Dihybrid crosses reveal the law of independent assortment
Topic: 02-12 Mendelian Inheritance in Humans
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
 
156. In rats, the gene for the pigment (P) is dominant to no pigment (p). The gene for black (B) is dominant to the gene for cream (b). If a pigment gene (P) is absent, genes B and b are inoperative. Predict the genotypes and phenotypes of the F1 of a cross between a homozygous black rat and an albino homozygous for cream. 

Genotype	Phenotype	
PpBb	Black	

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited

157. In rats, the gene for the pigment (P) is dominant to no pigment (p). The gene for black (B) is dominant to the gene for cream (b). If a pigment gene (P) is absent, genes B and b are inoperative. Predict the genotypes and phenotypes of the F2 of a parental cross between a homozygous black rat and an albino homozygous for cream. 
9 Black; 3 cream; 4 colourless
   
 	Genotype	Phenotype	
1	PPBB	Black	
2	PPBb	Black	
2	PpBB	Black	
4	PpBb	Black	
1	ppBB	colourless	
2	ppBb	colourless	
1	PPbb	cream	
2	Ppbb	cream	
1	ppbb	colourless	

Blooms: Analyze
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
158. In the common daisy, the genes A and a and B and b represent two pairs of alleles acting on flower colour. A and B are required for colour. The alleles of these two genes show recessive epistasis. The two gene pairs together thus show duplicate recessive epistasis. Predict the genotypes and phenotypes of the F1 of a cross between two colourless plants, one homozygous for A and the other homozygous for B. 

Genotype	Phenotype	
AaBb	Colour	

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited

159. In the common daisy, the genes A and a and B and b represent two pairs of alleles acting on flower colour. A and B are required for colour. The alleles of these two genes show recessive epistasis. The two gene pairs together thus show duplicate recessive epistasis. Predict the genotypes and phenotypes of the F2 of a cross between two colourless plants, one homozygous for A and the other homozygous for B. 
9 Black; 7 colourless
   
 	Genotype	Phenotype	
1	AABB	Colour	
2	AABb	Colour	
2	AaBB	Colour	
4	AaBb	Colour	
1	aaBB	Colourless	
2	aaBb	Colourless	
1	AAbb	Colourless	
2	Aabb	Colourless	
1	Aabb	Colourless	

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
160. In poultry, if a Black Longshank male with feathered shanks is crossed with a Buff Rock female with unfeathered shanks the F1 are all feathered and the F2 show 90 feathered to 6 unfeathered. Infer the genotypes of the parents. 
AABBaabb; The ratio is a 15:1 which is a dihybrid ratio; therefore the parents are homozygous and produce a heterozygous F1.

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited

161. In a certain breed of plants, dark green is determined by the dominant gene G and light green is determined by the recessive gene g. The heterozygote shows 75% penetrance for the dominant phenotype. If the parental cross is GGgg, what phenotype distribution would be expected in a population of 400 F2 plants? 
250 dark green (GG + 75% Gg); 150 light green (gg + 25% Gg)

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-11 Distinguish between penetrance and expressivity.
Topic: 02-26 The same genotype does not always produce the same phenotype
 
162. A man with blood type A whose father was blood type O married a woman of blood type B whose mother was blood type O. What are the possible blood types of their offspring? 
Blood types A, B, AB, and O are possible.

Blooms: Apply
Difficulty: Easy
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
 
163. What phenotypes and genotypes would you expect from the following cross of blood-related genotypes?
IBirh+rh+IAirh+rh 

IBIArh+rh	AB positive	
IBIArh+rh+	AB positive	
IBirh+rh	B positive	
IBirh+rh+	B positive	
IAirh+rh	A positive	
IAirh+rh+	A positive	
ii rh+rh	O positive	
ii rh+rh+	O positive	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete

164. Coat colour in a certain species of rabbit is governed by multiple alleles. The dominance series for these alleles is as follows: coloured (c+), chinchilla, (cch), himalayan (ch) and albino (c). Give the phenotypes and ratios from the following crosses: (A) c+cchch (B) c+c+chcch (C) c+cchc (D) ccchcch (E) c+chchcch (F) c+cchchcch (G) ccc+cch. 
(A) 2 coloured: 2 himalayan (B) all coloured (C) 2 coloured: 1 himalayan: 1 albino (D) 2 himalayan: 2 chinchilla (E) 2 coloured: 1 himalayan: 1 chinchilla (F) 2 coloured: 2 chinchilla (G) 2 coloured: 2 chinchilla.

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete

165. Affected individuals in the following pedigree are homozygous for the allele that causes the trait. What are the possible genotypes of persons 1, 2, 3 and 4?

Persons 1, 2, 3 are Aa. Person 4 is AA.

Blooms: Apply
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait

166. The pedigree shown is for a human genetic disease in which solid colour indicates affected individuals. Affected individuals in the pedigree are homozygous for the allele that causes the trait. Apply the laws of probability and calculate the probability, the offspring of the cousin marriage (individual 2  individual 3) will exhibit the disease.

The trait is a recessive trait. Individual #2 and individual #3 are both carriers, therefore, there is a ¼ chance their offspring will be homozygous for the recessive allele.

Blooms: Analyze
Difficulty: Medium
Learning Objective: 02-05 Analyze Mendel's law of independent assortment, and explain how Mendel proposed the law from dihybrid crosses.
Topic: 02-13 Pedigrees aid the study of hereditary traits in human families
Topic: 02-14 A vertical pattern of inheritance indicates a rare dominant trait
Topic: 02-15 A horizontal pattern of inheritance indicates a rare recessive trait

167. The following five mothers, (a) through (e), with phenotypes given, each produced one child whose phenotype is described as to blood group (A, B, O), M or N antigens, and Rh factor. For each child, select as the father, one of the five males whose genotypes are given. For some children, more than one male may be a possible father.
 (ii = Type O blood, rr = rh&R = rh+]
   
 	Maternal Phenotype	Child Phenotype	Genotype of Male	
(a)	A M R	O M R	1. IAi MN rr	
(b)	B N r	O N r	2. IBi MN RR	
(c)	O M r	A MN R	3. iiNNrr	
(d)	A N R	AB MN R	4. iiMMrr	
(e)	AB MN r	A MN r	5. IAIA MNRR	
 
For the child of mother (a), the father could be 1 or 4. For the child of mother (b), the father could be 1 or 3. For the child of mother (c), the father could be 5. For the child of mother (d), the father could be 2. For the child of mother (e), the father could be 1 or 3 or 4.

Blooms: Evaluate
Difficulty: Hard
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
Topic: 02-18 A gene may have more than two alleles

168. You have obtained an interesting flower for your garden from your neighbour. The neighbour has given you two pure lines of the plant, one with red flowers and one with yellow flowers. You decide to cross them and find that you obtain all orange flowers. The curious molecular geneticist in you decides to test two independent hypotheses: Hypothesis 1: Incomplete dominance; Hypothesis 2: Recessive epistasis. The first step in your test is to self the F1 orange plants, which you complete only to find that the results do not statistically distinguish the two hypotheses. a) What ratio of yellow, orange, and red would you expect in the F2 population for each hypothesis and b) what crosses would you complete next to definitively test your two hypotheses? 
a) The expected phenotypic ratio for recessive epistasis is 9:3:4, and for incomplete dominance, 1:2:1. b) Cross the yellow F2 flowers with true breeding red flowers. If the hypothesis for incomplete dominance is correct, the yellow colour will be determined by a single gene and all F2 yellow flowers will be homozygous recessive and give rise to only orange flowers in the F3 population [aaAA = Aa]. However, if the hypothesis for recessive epistasis is correct, a cross of F2 yellow and true breeding red flowers will give rise to some red and some orange flowers [YyrryyRR = either yyRr or YyRr].

Blooms: Create
Difficulty: Hard
Learning Objective: 02-06 Compare and contrast complete dominance, incomplete dominance, and codominance relationships, and demonstrate how a dominance series can be established.
Topic: 02-17 Dominance is not always complete
Topic: 02-18 A gene may have more than two alleles
 
169. Genes A and B are required for colour. If A or B is absent (that is, aa or bb) the result is colourless. Give the genotypes and phenotypes for each F1 and F2 progeny of the cross AAbbaabb 
F1 = Aabb/All colourless; F2 = 1AAbb: 2Aabb: 1aabb/All colourless

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited

170. Genes A and B are required for colour. If A or B is absent (that is, aa or bb) the result is colourless. Give the genotypes and phenotypes for each F1 and F2 progeny of the cross aaBBaabb 
F1 = aaBb/All colourless; F2 = 1aaBB: 2aaBb: 1aabb/All colourless

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited
 
171. Genes A and B are required for colour. If A or B is absent (that is, aa or bb) the result is colourless. Give the genotypes and phenotypes for each F1 and F2 progeny of the cross AAbbaaBB 
F1 = AaBbcoloured; F2 = 9 coloured; 7 colourless
   
Genotype	Phenotype	
For F1:	 	
AaBb	Coloured	
For F2:	 	
1AABB	Coloured	
2AABb	Coloured	
2AaBB	Coloured	
4AaBb	Coloured	
1aaBB	Colourless	
1AAbb	Colourless	
2aaBb	Colourless	
2Aabb	Colourless	
1aabb	Colourless	

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 02-09 Compare and contrast complementary gene action, recessive epistasis, dominant epistasis, and redundancy.
Topic: 02-24 Breeding studies help decide how a trait is inherited