assignment for class 12 IPA

Choose the best answer
1. A heritable feature is a ______ and may have two or more variants called ______.
A. trait ... characteristics
B. character ... traits
C. character ... factors
D. trait ... factors
E. characteristic ... alleles

2. A plant with the genotype AABbcc is _____.
A. homozygous at two loci
B. heterozygous at two loci
C. recessive at three loci
D. triploid
E. all of the above

3. GgTt pea plants can produce _____________ type(s) of gametes, but a ggtt plant can produce _____________ type(s) of gametes.
A. one ... two
B. two... one
C. four ... two
D. four ... one
E. one ... one

4. Two organisms with genotype AaBbCcDdEE mate. These loci are all independent. What fraction of the offspring will have the same genotype as the parents?
A. 3/4
B. 1/4
C. 9/64
D. 1/16
E. 4/3

5. Huntington's disorder is due to an autosomal dominant allele. If a heterozygous male marries a normal female, what percentage of the offspring will have Huntington's?
A. 0%
B. 25%
C. 50%
D. 75%
E. 100%

6. An allele is _____.
A. a type of chromosome
B. the dominant form of a gene
C. a variety of pea plant used by Mendel
D. an alternative version of a gene
E. the recessive form of a gene

7. You cross a true-breeding red-flowered snapdragon with a true-breeding white-flowered one. All of the F1 are pink. What does this say about the alleles for the parental traits?
A. Red and white are co-dominant.
B. Red is dominant.
C. Both red and white are recessive.
D. Red and white show incomplete dominance.
E. Pink is dominant, while red and white are recessive.

8. Folk singer Woody Guthrie died of Huntington's syndrome, an autosomal dominant disease. Which statement below must be true?
A. All of his children will die of Huntington's syndrome.
B. His sons will die of Huntington's syndrome but not his daughters.
C. His daughters will die of Huntington's syndrome but not his sons.
D. At least one of Woody Guthrie's parents must have had Huntington's syndrome also.
E. There is not enough information to answer the question.

9. The most common lethal genetic disease in the United States is _____.
A. sickle-cell disease
B. cystic fibrosis
C. Huntington's disease
D. hemophilia
E. PKU

10. Various procedures can be used to detect genetic disorders before birth. Among the tests discussed in this chapter, ____ is the least invasive, while ____ and ____ allow the chromosomes of the fetus to be examined.
A. amniocentesis ... fetoscopy ... chorionic villus sampling
B. ultrasound imaging ... chorionic villus sampling... amniocentesis
C. ultrasound imaging ... fetoscopy... amniocentesis
D. amniocentesis ... ultrasound imaging ... chorionic villus sampling
E. chorionic villus sampling ... fetoscopy ... amniocentesis

11. The result of the following cross indicates the orange eyes are _____ black eyes.

A. incompletely dominant to
B. recessive to
C. co-dominant to
D. dominant to
E. larger than

12. If O represents the allele for black eyes (dominant) and o represents the allele for orange eyes (recessive), what would be the genotypic ratio of a cross between a heterozygous black-eyed MendAlien and an orange-eyed MendAlien?
A. 1 homozygous black (OO): 1 heterozygote (black) (Oo): 1 homozygous orange (oo)
B. 2 homozygous black (OO): 1 heterozygote (black) (Oo): 1 homozygous orange (oo)
C. 1 homozygous black (OO): 0 heterozygote (black) (Oo): 1 homozygous orange (oo)
D. 0 homozygous black (OO): 1 heterozygote (black) (Oo): 1 homozygous orange (oo)
E. 0 homozygous black (OO): 0 heterozygote (black) (Oo): 1 homozygous orange (oo)

13. If O represents the allele for black eyes (dominant) and o represents the allele for orange eyes (recessive), what would be the phenotypic ratio of a cross between a heterozygous black-eyed MendAlien and an orange-eyed MendAlien?
A. 3 black : 1 orange
B. 1 black : 3 orange
C. 0 black : 1 orange
D. 1 black : 1 orange
E. 1 black : 0 orange

14. The result of the following cross indicates that the genotype of the female parent is _____.

A. bb
B. BB
C. Bb
D. b
E. B

15. The result of the following cross indicates that the genotype of the male parent is _____.

A. bb
B. BB
C. Bb
D. b
E. B

16. The result of the following cross indicates that genotypically the offspring _____.

A. are all bb
B. are all BB
C. are all Bb
D. exhibit a 3 BB : 1 bb ratio
E. exhibit a 1 BB : 2 Bb : 1 bb ratio

17. Which of these is a testcross?
A. A? x aa
B. A? x AA
C. A? x Aa
D. A? x A?
E. aa x aa

18. That each gamete contains a single allele of the eye color gene is an illustration of _____.


A. Beadle and Tatum's one gene¬one enzyme hypothesis
B. Mendel's law of segregation only
C. Darwin's theory of natural selection
D. Mendel's law of independent assortment only
E. Both Mendel's law of segregation and Mendel's law of independent assortment.

19. What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.)

A. OoWw
B. ooWW
C. ooWw
D. OOWW
E. ooww

20. What is the genotype of the offspring?


A. OoWw
B. ooWW
C. ooWw
D. OOWW
E. ooww

21. In order to determine the genotype of a MendAlien with black eyes and green skin, you would cross this individual with a(n) _____ individual.
A. ooww
B. OOWW
C. OOww
D. ooWW
E. OoWw

22. Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is a MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted phenotypic ratio of their offspring would be _____.
A. 1 black eyes, green skin : 1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin
B. 3 black eyes, green skin : 3 black eyes, white skin : 9 orange eyes, green skin : 1 orange eyes, white skin
C. 1 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 9 orange eyes, white skin
D. 9 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 1 orange eyes, white skin
E. There is insufficient information to determine Sam's genotype.

23. A cross between an individual with orange eyes and green skin and an individual with black eyes and white skin is an example of a _____ cross.
A. test
B. difficult
C. dihybrid
D. trihybrid
E. monohybrid

24. A phenotypic ratio of 9:3:3:1 in the offspring of a cross indicates that _____.
A. one parent is homozygous dominant and one parent is homozygous recessive
B. one parent is heterozygous and one parent is homozygous recessive
C. one parent is homozygous dominant and one parent is heterozygous
D. both parents are heterozygous for both genes
E. both parents are homozygous dominant

25. The observed distribution of alleles into gametes is an illustration of _____.

A. Mendel's law of segregation only
B. Thienemann's rule
C. Eimer's principle
D. Mendel's law of independent assortment only
E. Mendel's laws of segregation and independent assortment

26. An individual heterozygous for eye color, skin color, and number of eyes mates with an individual who is homozygous recessive for all three characters; what would be the expected phenotypic ratio of their offspring? [Hint: O = black eyes, o = orange eyes; W = green skin, w = white skin; C = two eyes, c = one eye]
A. 9 black eyes, green skin, two eyes : 3 black eyes, green skin, one eye : 3 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye : 9 orange eyes, green skin, two eyes : 3 orange eyes, green skin, one eye : 3 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye
B. 9 black eyes, green skin, two eyes : 3 black eyes, green skin, one eye : 3 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye
C. 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 orange eyes, green skin, two eyes : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye
D. 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye : 1 orange eyes, green skin, two eyes : 1 orange eyes, green skin, one eye : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye
E. 9 orange eyes, green skin, two eyes : 9 orange eyes, green skin, one eye : 9 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye

27. In the following cross the genotype of the female parent is OoWw. What is the genotype of the male parent? [Hint: O = black eyes, o = orange eyes, W = green skin, w = white skin]

A. OoWw
B. OOWW
C. OOWw
D. OoWW
E. ooWW

28. A OoWw x ooww cross yields a phenotypic ratio of approximately 5 black eyes, green skin : 5 orange eyes, white skin : 1 black eyes, white skin : 1 orange eyes, green skin. Which of the following best explains these results?
A. Mendel's law of independent assortment is being violated.
B. Mendel's law of segregation is being violated.
C. The genes for eye color and skin color are co-dominant.
D. The heterozygous individual is male, and the homozygous individual is female.
E. Mendel's laws of segregation and independent assortment are being violated.

29. In a situation in which genes assort independently, what is the ratio of the gametes produced by an AaBB individual?
A. 1 A : 1 B
B. 3 A : 1 B
C. 3 AB : 1 ab
D. 1 AB : 1 aB
E. 3 AA : 1 BB

30. Mendel worked _____.
A. on the Human Genome Project
B. with fruit flies
C. on problems involving characters that were inherited together more often than expected by chance
D. on gene mapping
E. in a monastery

31. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the relationship between the CB and CW alleles?
A. CB is dominant to CW
B. CB is recessive to CW
C. CW is dominant to CB
D. The relationship between the alleles is one of incomplete dominance.
E. CB and CW are co-dominant

32. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected phenotypic ratio of a CBCW x CBCW cross?
A. 1 black : 1 white
B. 3 black : 1 white
C. 1 black : 2 gray : 1 white
D. 3 gray : 1 white
E. 1 black : 1 gray

33. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected genotypic ratio of a CBCW x CBCW cross?
A. 1 CBCB : 1 CBCW
B. 3 CBCB : 1 CWCW
C. 1 CBCB : 2 CBCW : 1 CWCW
D. 2 CBCB : 1 CBCW : 2 CWCW
E. 9 CBCB : 3 CBCW : 3 CWCB : 1 CWCW

34. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected genotypic ratio of a CBCB x CBCW cross?
A. 1 CBCB : 1 CBCW
B. 3 CBCB : 1 CWCW
C. 1 CBCB : 2 CBCW : 1 CWCW
D. 2 CBCB : 1 CBCW : 2 CWCW
E. 9 CBCB : 3 CBCW : 3 CWCB : 1 CWCW

35. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected phenotypic ratio of a CBCB x CBCW cross?
A. 1 black : 1 white
B. 3 black : 1 white
C. 1 black : 2 gray : 1 white
D. 3 gray : 1 white
E. 1 black : 1 gray

36. The basic difference between Mendel's particulate theory of inheritance and the previous blending theory was that the blending theory, but not the particulate theory, maintained that _____.
A. mutation is the major source of new gene combinations
B. many genes are lost in their transmission to offspring, but the remaining genes gradually reproduce the lost genes
C. the traits governed by genes in the egg are different from the traits governed by genes in the sperm
D. after a mating, genes of the two parents are mixed in the offspring and lose their individual identities
E. genes combine with each other to produce new genes

37. It was important that Mendel chose true-breeding varieties for his experiments because _____.
A. true-breeding varieties are ALWAYS heterozygous for the trait of interest
B. true-breeding varieties are NEVER heterozygous for the trait of interest
C. true-breeding varieties are more vigorous
D. true-breeding varieties are easier to cross-pollinate
E. true-breeding varieties are NEVER homozygous for the trait of interest

38. In people with sickle-cell disease the red blood cells break down, clump, and clog the blood vessels. The blood vessels and the broken cells accumulate in the spleen. Among other things this leads to physical weakness, heart failure, pain, and brain damage. Such a suite of symptoms can be explained by _____.
A. the polygenic nature of sickle-cell disease
B. the pleiotropic effects of the sickle-cell allele
C. an epistatic interaction between the sickle-cell allele and a proteolytic enzyme gene
D. a bacterial infection interacting with the sickle-cell allele
E. side effects of the drugs used to cure sickle-cell disease

39. In a certain plant, the alleles A, B, and C are completely dominant to the alleles a, b, and c. A plant with the genotype AABbcc will have the same phenotype as the plant with the genotype _____.
A. AAbbcc
B. aabbcc
C. AaBBcc
D. AABBCc
E. none of the above

40. Cystic fibrosis can be inherited even if neither parent has the disease. This is because the disease _____.
A. requires certain environmental conditions to be expressed
B. occurs only in polyploid individuals
C. is caused by a recessive allele
D. is caused by a dominant allele
E. occurs only in individuals who are already weak from other causes

41. Individuals have different phenotypes because the underlying genotypes contain instructions for different sets of _____.
A. chromosomes
B. polypeptides
C. ribosomes
D. chippendales
E. nucleotides

42. Assume tall (T) is dominant to dwarf (t). If a homozygous dominant individual is crossed with a homozygous dwarf, the offspring will ______.
A. all be intermediate in height
B. all be tall
C. be 1/2 tall and 1/2 dwarf
D. be 3/4 tall and 1/4 dwarf
E. all be short

43. Phenylketonuria is an autosomal recessive disorder. Using "P" and "p" to represent the alleles, what is the genotype of a phenylketonuric person?
A. PP
B. Pp
C. pp
D. Pp or pp
E. all of the above

44. Homologous pairs of chromosomes frequently _____.
A. carry different genes for different traits
B. differ in length
C. contain different alleles
D. are not both present in diploid somatic cells
E. are paired up in the G2 phase

45. It is far more common to find human genetic disease caused by __________ alleles than by __________ alleles because __________.
A. dominant ... recessive ... dominant alleles have an evolutionary advantage over recessive alleles
B. dominant ... recessive ... dominant alleles became dominant because they aid the survival of the organism carrying them
C. recessive ... dominant ... harmful recessive alleles can survive in the heterozygote without any selection pressure against them
D. recessive ... dominant ... even when homozygous, recessive alleles usually do not cause as much damage as dominant alleles
E. none of the above

46. Assume that a diploid cell is heterozygous at a certain locus. The cell will contain __________ during the __________ of the cell cycle.
A. two different types of alleles at the locus ... G1, S, and G2 phases
B. four complete chromosome sets ... G1 and G2 phases
C. two identical alleles at the locus ... G1 and S phases
D. four different types of alleles at the locus ... telophase and G2 phases
E. one allele ... prophase

47. Pea plants are tall if they have the genotype TT or Tt, and they are short if they have genotype tt. A tall plant is mated with a short plant. Half the offspring are tall, and half are short. This allows us to conclude that the tall plant _____.
A. was homozygous
B. was heterozygous
C. had two tall parents
D. all of the above
E. none of the above

48. The results of a testcross (that is, the offspring of this cross) were 50% phenotypically dominant and 50% phenotypically recessive. The genotype of the phenotypically dominant parent in this cross was _____.
A. homozygous dominant
B. homozygous recessive
C. heterozygous
D. homozygous dominant and homozygous recessive
E. the answer cannot be determined from the information

49. The genetic disease cystic fibrosis is caused by a defective allele that _____.
A. encodes a dysfunctional enzyme that fails to break down brain lipids
B. causes hemoglobin molecules to malfunction
C. encodes a defective chloride-channel membrane transport protein
D. encodes a neurotoxin
E. encodes an enzyme that breaks down muscle fibers

50. What is indicated when a single-character testcross yields offspring in a 1:1 phenotypic ratio?
A. The parent with the dominant phenotype was homozygous.
B. The parent with the dominant phenotype was heterozygous.
C. Epistasis has occurred.
D. Alleles are co-dominant.
E. Both parents are heterozygous.

51. If a homozygous dominant is crossed with a homozygous recessive for a given trait, the offspring will be __________.
A. all of the dominant phenotype
B. 1/4 of the recessive phenotype
C. all homozygous dominant
D. all homozygous recessive
E. present in a 9:3:3:1 ratio

52. Michelle and Keith are apparently normal, but their daughter was born with alkaptonuria, an inherited metabolic disorder. If alkaptonuria is like most other human hereditary disorders, the probability of their next child being born with alkaptonuria is _____.
A. 0
B. 1/4
C. 1/2
D. 2/3
E. 3/4

53. Suppose we have a pea plant with purple flowers, determined by the dominant allele P. How might you determine whether the plant is homozygous (PP) or heterozygous (Pp)?
A. Perform a testcross: Cross the plant with a known heterozygote, Pp.
B. Cross the plant with a true-breeding purple plant.
C. Examine the plant's chromosomes with a microscope.
D. Perform a testcross: Cross the plant with a white one, which must be homozygous recessive, or pp.
E. Attempt to cross the plant with a pink-flowered snapdragon.

54. In a cross between two heterozygotes (Aa), the next generation will be _____.
A. in the ratio 1:3 heterozygotes to homozygotes
B. all heterozygotes
C. in the ratio 1:1 homozygotes to heterozygotes
D. all homozygotes
E. in the ratio 1:3 homozygotes to heterozygotes

55. Fetal cells may be removed along with fluid from the womb by a process known as _____.
A. karyotyping
B. uterine tapping
C. amniocentesis
D. sonogramming
E. chorionic villus sampling

56. If the two traits that Mendel looked at in his dihybrid cross of smooth yellow peas with wrinkled green peas had been controlled by genes that were located on the same chromosome, then the F2 generation _____.
A. would have contained four phenotypes in a 9:3:3:1 ratio
B. would have contained only individuals that were heterozygous at both loci
C. would have deviated from the 9:3:3:1 phenotypic ratio that is predicted by the law of independent assortment
D. would have contained no individuals that were heterozygous at both loci
E. none of the above

57. In Mendel's monohybrid cross of purple-flowered and white-flowered peas, all members of the F1 generation were of the __________ phenotype because their genotype was __________ at the flower-color locus.
A. white-flowered ... homozygous recessive
B. white-flowered ... heterozygous
C. purple-flowered ... homozygous recessive
D. purple-flowered ... homozygous dominant
E. purple-flowered ... heterozygous

58. Huntington's disease is an example of a genetic disorder caused by _____.
A. a late-acting lethal dominant allele
B. a nonlethal dominant allele
C. a late-acting recessive allele
D. homozygous recessive alleles
E. multiple alleles

59. Seeds from a single sexually reproducing plant are harvested and later planted under identical conditions. What results should be expected?
A. The seedlings should be identical to each other but not to the parental plant.
B. The seedlings should be identical to each other and to the parental plant.
C. The seedlings should differ from each other, depending on the environmental conditions under which each is planted.
D. The seedlings should differ from each other, depending on their genetic constitution.
E. The seedlings should differ from each other and be identical to the parental plant.

60. Tay-Sachs disease runs in Rebecca's family. On a family pedigree, she saw a shaded circle. This represented _____.
A. a male with Tay-Sachs
B. a female carrier
C. a male carrier
D. a female with Tay-Sachs
E. a male of unknown genotype

61. Several inherited disorders are much more common in close-knit religious communities, such as the Amish, than in the general population. This is at least partly due to the fact that _____.
A. people in such communities are more likely to marry relatives
B. shared environmental conditions such as diet can increase mutation rate
C. modern medical care is not widely available in such communities
D. community members care for each other, and such disorders are contagious
E. individuals in such communities are more likely to have large families

62. John and Jane are planning a family, but since each has a brother who has sickle-cell disease, they are concerned that their children may develop sickle-cell disease. Neither John, Jane, nor their respective parents have the disease. They consult a genetic counselor who tells them that _____.
A. there is no chance that any of their children will have sickle-cell disease
B. all of their children will have sickle-cell disease
C. one of four of their children can be expected to have sickle-cell disease
D. it's possible that none of their children will have the disease, but blood tests on them both will be required to make sure
E. it is impossible to tell whether their children will have the disease

63. An AABb individual is mated with another AABb individual. The possible number of genetically different kinds of offspring is _____.
A. 1
B. 2
C. 3
D. 4
E. 9









64. If each parent can produce 100 genetically distinct gametes, how many genetically distinct offspring can two parents produce?
A. 10,000
B. 200
C. 1,000
D. 100
E. 1,000,000

65. Flower color in snapdragons is an example of incomplete dominance. If a red-flowered plant is crossed with a white-flowered plant, the F1 generation has pink flowers. If a pink-flowered plant is crossed with a pink-flowered plant, the progeny plants will be __________.
A. 100% pink
B. 100% red
C. 25% red, 50% pink, and 25% white
D. 50% pink and 50% red
E. 25% white and 75% red

66. A couple, both members of which are descended from eastern European (Ashkenazi) Jews, visit a genetics counselor before trying to have children. In view of their ethnic background, the counselor recommends that they be tested to see if they are carriers of _____.
A. cystic fibrosis
B. hemophilia
C. sickle-cell disease
D. hypercholesterolemia
E. Tay-Sachs

67. Pea flowers may be purple (W) or white (w). Pea seeds may be round (R) or wrinkled (r). What proportion of the offspring from the cross WwRr x WwRr will have white flowers and wrinkled seeds?
A. 1/16
B. 3/16
C. 8/16
D. 9/16
E. all of them

68. Heart disease, diabetes, cancer, alcoholism, and many mental illnesses can best be described as _____.
A. simple Mendelian disorders
B. infectious diseases caused by microorganisms
C. multifactorial disorders with a possible polygenic component
D. all symptoms of Huntington's disease.
E. the results of a bad lifestyle

69. R = can roll tongue; r = can't roll tongue. A couple who both have the ability to roll their tongues have a son who is also a tongue-roller. The son is curious about whether he is homozygous or heterozygous for the tongue-rolling trait. How could he find out? (p. 252)
A. Have children with a woman who is also a tongue-roller. If their children are tongue-rollers, then he is a homozygous dominant.
B. Have children with a woman who is also a tongue-roller. If all their children are non-rollers, then he is a heterozygote.
C. Have children with a woman who is a non-roller. If any of their children is a non-roller, then he is a heterozygote.
D. Have children with a woman who is a non-roller. If any of their children is a tongue-roller, then he is a homozygous recessive.
E. He will have to wait to have grandchildren before the question can be answered.

70. A = big apples; R = red apples; a = small apples; r = yellow apples. You have one tree that produces big yellow apples and another tree that produces small red apples. When the two are crossed, you find that half the offspring trees produce big red apples and half produce big yellow apples. What are the genotypes of the parents?
A. AArr and aaRr
B. Aarr and aaRr
C. AARr and AArr
D. AaRr and AaRr
E. AaRr and aarr





71. When two average-height parents give birth to a child exhibiting achondroplastic dwarfism, it is most likely due to a new mutation. This is because _____.
A. the frequency of achondroplasia is unknown
B. achondroplasia is a relatively rare disorder
C. achondroplasia is caused by an allele that is always expressed, so the parents couldn't have the allele
D. such mutations are statistically predictable
E. none of the above

72. While on a field trip in the jungle you find a new species of mouse. You catch a pair and take them back to the lab. In mice, black coat color, B, is dominant to brown b, yet the female mouse gives birth to a large litter in which nine of the offspring are albino. Which of the following statements best explains the results of this cross?
A. A new mutation has occurred in the mice.
B. This is an example of polygenic inheritance.
C. There is probably an epistatic interaction involving another gene that influences coat color.
D. The coat color alleles are co-dominant.
E. The coat color alleles are incompletely dominant.

73. A man who has type B blood and a woman who has type A blood could have children of which of the following phenotypes?
A. A or B only
B. AB only
C. AB or O
D. A, B, or O
E. A, B, AB, or O

74. Which of the following matings can NOT produce a child with blood type O? The letters refer to blood types (phenotypes).
A. A x A
B. A x B
C. O x AB
D. O x O
E. all of the above

75. A single genetic locus that controls more than one trait is said to be _____.
A. polygenic
B. photogenic
C. pleiotropic
D. autotrophic
E. somatic

76. If one parent is blood type AB and the other is type O, what fraction of their offspring will be blood type A?
A. 1/4
B. 1/2
C. 3/4
D. all of the above
E. none of the above

77. Cystic fibrosis, which is usually lethal before the age of reproduction, is a homozygous recessive trait. Why do cases continue to arise, even though people with the disease rarely reproduce?
A. because new mutations continually introduce this harmful condition into the population
B. because the harmful allele "hides" inside heterozygous individuals and 1/4 of the offspring of two heterozygotes should be afflicted
C. because mosquitoes can transfer the disease from person to person
D. because people continue to make inappropriate lifestyle choices
E. none of the above

78. A red bull is crossed with a white cow and all of the offspring are roan, a shade between red and white. This is an example of genes that are _____.
A. nonhomologous
B. epistatic
C. co-dominant
D. incompletely dominant
E. completely dominant



79. An organism homozygous for red color was crossed with an organism homozygous for blue color. The offspring of this first fertilization were green. Two of the green organisms were crossed. What is (are) the expected phenotypes in this second group of offspring?
A. red only
B. red or blue
C. green only
D. red, blue, or green
E. red or green

80. Tom's brother suffers from phenylketonuria, a recessive disorder. The brothers' parents do not have PKU. What are the chances that Tom, who is normal for this trait, is a carrier of PKU?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 4/3

81. In human hypercholesterolemia, the ability of liver cells to remove cholesterol from the blood is controlled at one locus. People who are HH have normal cholesterol, those having Hh have heart attacks by age 30, and people who are hh have heart attacks by age 2. If an HH person mates with an Hh person, will that person have any severely afflicted (hh) offspring?
A. no
B. yes, 1/4 of their offspring
C. yes, 1/2 of their offspring
D. yes, 3/4 of their offspring
E. yes, all of their offspring

82. Sometimes, one gene pair will interact to control the expression of a second gene pair (for example, coat color in Labrador retrievers). This gene interaction is called __________.
A. incomplete dominance
B. epistasis
C. gene regulation
D. second pair control
E. complete dominance

83. A woman with type O blood is expecting a child. Her husband is type A. Both the woman's father and her husband's father had type B blood. What is the probability that the child will have type O blood?
A. 100%
B. 75%
C. 50%
D. 25%
E. 0%