Genetics Canadian 2nd Edition Hartwell

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MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

1) The first generation of offspring from the parents is called

A) F2. B) P. C) backcross. D) testcross. E) F1.
Answer: E

2) Which of the following terms is not a type of mating cross?

A) reciprocal
B) dihybrid
C) monohybrid
D) dominant
E) testcross
Answer: D

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

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) 100 B) 1024 C) 128 D) 49 E) 131,072
Answer: C

5) If the parents of a family already have two boys, what is the probability that the next two offspring
will be girls?
A) 1/4 B) 1/3 C) 1/8 D) 1 E) 1/2
Answer: A

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) 1/2 B) 1/4 C) 1/3 D) all pollen E) 1/8
Genetics Canadian 2nd Edition Hartwell Test Bank
Answer: A

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) Hurler's disease
E) Tay-Sachs disease
Answer: C

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8) The gene responsible for the defective protein in cystic fibrosis is located on which of the following
chromosomes?
A) 11 B) 15 C) X D) 7 E) 4
Answer: D

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) Dominant
C) Environmental polygenic
D) Multifactorial
E) Recessive
Answer: D

10) Most single-gene diseases in humans that are not of late-onset are caused by which of the following?
A) Dominant alleles
B) Reciprocal allele
C) Vertical pattern of inheritance
D) Recessive alleles
E) Horizontal pattern of inheritance
Answer: D

11) Phenylketonuria (PKU) is caused by ________.

A) Multifactorial
B) Recessive allele
C) Dominant allele
D) Polygenic
E) Monohybrid allele
Answer: B

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) 0 B) 1/4 C) 1/2 D) 1/8 E) 1/16
Answer: B

13) A rare recessive trait in a pedigree is indicated by which pattern of inheritance?

A) vertical
B) diagonal
C) both vertical and horizontal
D) father to daughter inheritance
E) horizontal
Answer: E

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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) none
B) 1/4
C) 1/8
D) 1/2
E) 1/16 will be albino
Answer: C

15) Adominant 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) 100% C) 75% D) 50% E) 0%
Answer: A

16) Ina monohybrid cross AA × aa, what proportion of homozygotes is expected among the F 2
offspring?
A) 1/2
B) 1/4
C) 3/4
D) All are homozygotes.
E) None are homozygotes.
Answer: A

17) Anallele that expresses its phenotype even when heterozygous with a recessive allele is called
________.
A) recessive.
B) recombinant.
C) independent.
D) dominant.
E) parental.
Answer: D

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) all C) 3/4 D) 1/4 E) 1/2
Answer: A

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19) Inthe dihybrid cross AaBb × aabb, what proportion of individuals are expected to be homozygotic
for both genes in the F1 generation?
A) 1/2
B) 3/4
C) 1/4
D) All are homozygotes.
E) None are homozygotes.
Answer: C

20) ________ is a/are cross(es) between parents that differ in only one trait.
A) Self-fertilization
B) Reciprocal crosses
C) Monohybrid crosses
D) Artificial selection
E) Cross fertilization
Answer: C

21) Assumingindependent assortment, which of the crosses below will produce a 1:1 phenotypic ratio
among the F1 progeny?
A) AABB × aabb
B) AaBB × aaBB
C) AaBb × AaBb
D) AaBb × aabb
E) AAbb × aaBB
Answer: B

22) The actual alleles present in an individual make up the individual's

A) zygote.
B) allele.
C) dominant allele.
D) genotype.
E) recombinant types.
Answer: D

23) In
a dihybrid cross AAbb × aaBB, what proportion of the F2 offspring is expected to be
homozygotic for at least one gene?
A) 3/4
B) 1/2
C) 1/4
D) All are homozygotes.
E) None are homozygotes.
Answer: A

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24) A phenotype reflecting a new combination of genes occurring during gamete formation is called
A) a recombinant type.
B) heterozygous.
C) a multihybrid cross.
D) an independent assortment.
E) homozygous.
Answer: A

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) all B) 1/4 C) 1/2 D) 3/4 E) none
Answer: C

26) The diploid cell formed by the fertilization of the egg by the sperm during sexual reproduction is a
A) monohybrid.
B) gamete.
C) reciprocal.
D) dihybrid.
E) zygote.
Answer: E

27) A gamete is ________.

A) A zygote
B) Either an egg or a sperm
C) Only a sex chromosome
D) Only a sperm
E) Only an egg
Answer: B

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

29) An alternative form of a single gene is known as

A) reciprocal. B) dihybrid. C) parental. D) allele. E) recessive.
Answer: D

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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) 1/2 B) 3/4 C) all D) none E) 1/4
Answer: C

31) Which of the crosses listed below will give a 1:1:1:1 genotypic ratio in the F 1 generation? Assume
independent assortment.
A) AAbb × aaBB
B) AABB × aabb
C) AaBB × aaBB
D) AaBb × AaBb
E) AaBb × aabb
Answer: E

32) For the cross AaBb × aabb, what proportion of F 1 offspring will be heterozygous for both gene
pairs? Assume independent assortment.
A) 3/4
B) 1/2
C) 1/4
D) All are heterozygotes.
E) None are heterozygotes.
Answer: C

33) Ifa dog breeder chooses the parents for a desired next generation, the dog breeder is using a process
called ________.
A) evolution
B) mutation
C) random selection
D) natural selection
E) artificial selection
Answer: E

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

35) Which of the following was not involved in the rediscovery of Mendel's work?
A) Correns B) Morgan C) Watson D) de Vries E) Tschermak
Answer: B

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36) What does a vertical pattern of inheritance in a pedigree likely indicate?
A) mulitgenic inheritance
B) common recessive trait
C) environmental impact
D) rare dominant trait
E) rare recessive trait
Answer: D

37) Calculate the probability of either

all-dominant or all-recessive genotypes for the alleles A, B, E,
and F in the following cross: AaBbccddEeFf × AaBbCcddEeFf
A) 1/32 B) 1/16 C) 1/256 D) 1/128 E) 1/64
Answer: D

38) Insome 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) aabb × aabb
B) AABB × AABB
C) aaBB × aabb
D) AABB × aabb
E) AAbb × aabb
Answer: D

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

40) After across 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 (dwarf), dd (tall)
C) DD (dwarf), DD (tall)
D) dd (dwarf), Dd (tall)
E) DD (tall), dd (dwarf)
Answer: A

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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
Answer: A

42) Which of the following is not a phenotypic description of allele interactions affecting the expression
of traits?
A) polymorphic
B) codominance
C) incomplete dominance
D) multifactorial
E) pleiotropic
Answer: D

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

44) Which of the following diseases show pleiotropism?

A) albinism
B) muscular dystrophy
C) male pattern baldness
D) sickle cell anaemia
E) colour blindness
Answer: D

45) Adeviation from normal Mendelian ratios, which may be resolved by counting and/or controlled
crosses, is seen in which of the following terms?
A) complete dominance
B) penetrance and expressivity
C) incomplete dominance
D) codominance
E) pleiotropy
Answer: B

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

47) Which of the following ratios show codominance?

A) 4:1 B) 3:1 C) 1:2:1 D) 2:1 E) 1:1
Answer: C

48) Which of the following ratios indicates a lethal gene?

A) 1:2:1 B) 1:1 C) 2:1 D) 3:1 E) 4:1
Answer: C

49) A person who has type O blood has

A) anti-A antibodies.
B) both anti-A and -B antibodies.
C) anti-AB antibodies.
D) anti-B antibodies.
E) no surface antigens.
Answer: B

50) If two or more forms of the same gene exist, the different forms are called_______
A) alleles.
B) pleiotropic.
C) penetrance and expressivity.
D) incomplete dominance.
E) dihybrid.
Answer: A

51) The blood groups A, B, and O are different types of

A) heterozygotes.
B) alleles.
C) incomplete dominance.
D) penetrance and expressivity.
E) pleiotropy.
Answer: B

52) The blood groups A, B, and O show

A) complete dominance.
B) codominance.
C) recessiveness.
D) complete dominance, recessiveness, and codominance.
E) corecessiveness.
Answer: D

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53) Which of the following monohybrid ratios can describe incomplete dominance and codominance?
A) 1:3 B) 4:1 C) 1:2:1 D) 3:1 E) 2:1
Answer: C

54) Which of the following ratios demonstrate gene interaction?

A) 1:2:1 B) 2:1 C) 3:1 D) 1:3 E) 9:3:4
Answer: E

55) A ________ results whenever the nucleotide sequence is changed.

A) mutation B) phenotype C) character D) genotype E) trait
Answer: A

56) When the same gene is related to respiratory problems and sterility, it can be described as
A) complete dominance.
B) codominance.
C) pleiotropy.
D) incomplete dominance.
E) penetrance and expressivity.
Answer: C

57) Another name for a normal gene is

A) pleiotropy.
B) recessive.
C) wild-type.
D) codominant.
E) dominant.
Answer: C

58) The phenotypic ratio 1:2:1 may indicate

A) complete dominance.
B) epistasis.
C) codominance.
D) codominance and epistasis.
E) recessive lethal.
Answer: C

59) The phenotypic ratio 3:1 may indicate

A) incomplete dominance.
B) codominance and epistasis.
C) codominance.
D) complete dominance.
E) epistasis.
Answer: D

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60) Thephenotypic ratio 2:1 may indicate
A) recessive lethal.
B) codominance and epistasis.
C) codominance.
D) complete dominance.
E) epistasis.
Answer: A

61) Thephenotypic ratio 9:7 may indicate

A) complementary gene action.
B) codominance.
C) complete dominance.
D) recessive lethal.
E) epistasis.
Answer: A

62) Thephenotypic ratio 9:3:4 may indicate

A) codominance and epistasis.
B) complete dominance.
C) codominance.
D) epistasis.
E) recessive lethal.
Answer: D

63) Which of the following phenotypic ratios show independent assortment?

A) 7 B) 5 C) 4 D) 9 E) 13:3
Answer: E

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) conditional on other factors
B) co-dominant
C) indistinguishable from wild-type
D) continuously variable
E) lethal
Answer: C

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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) incomplete dominance
B) variable expressivity
C) incomplete penetrance
D) epistasis
E) complementation
Answer: C

66) If a mother is phenotype A and her child is phenotype B then the father's genotype is ________?
A) ii
B) IA IB
C) IA IA
D) IA i
E) Cannot be determined
Answer: B

67) Which of the following options is considered the universal donor of blood?
A
A) I i B) ii C) IA IB D) IB i E) IAIA
Answer: B

68) Which of these is not an example of a continuous trait?

A) occurrence of phenylketonuria (PKU)
B) human skin colour
C) birth weight of mice
D) age at death
E) plant height
Answer: A

69) Which of the following statements about continuous traits is not true?
A) They are relevant to medicine.
B) They are called complex traits.
C) They are also called quantitative traits.
D) They are relevant to agriculture.
E) They do not obey Mendel's laws.
Answer: E

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

71) How does penetrance differ from expressivity?

A) Penetrance is dependent on environment; expressivity is not.
B) Expressivity is dependent on environment; penetrance is not.
C) Penetrance is qualitative (presence or absence); expressivity is quantitative.
D) Penetrance involves multiple genes; expressivity involves a single gene.
E) Expressivity is qualitative (presence or absence); penetrance is quantitative.
Answer: C

72) When a gene has a more subtle and secondary effect on the phenotype, the gene is usually called
________.
A) Lethal
B) Recessive
C) Permissive
D) Modifier
E) Conditional
Answer: D

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

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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
Answer: C

75) Which of the following is not useful in a complementation test?

A) recessive alleles
B) alleles dominant to wild-type
C) sexual reproduction
D) F1 progeny
E) pure breeding lines
Answer: B

76) Iftwo 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) 0 wild-type : 1 mutant
B) 1 wild-type : 0 mutant
C) 2 wild-type : 1 mutant
D) 1 wild-type : 1 mutant
E) 1 wild-type : 2 mutant
Answer: B

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 × aaBb? Note the genotypes in the cross carefully. Assume independent assortment
of each gene.
A) all (red & tall)
B) 3 (red & tall): 1 (red & short): 3 (white & tall): 1 (white & short)
C) 9 (red & tall): 3 (red & short): 3 (white & tall): 1 (white & short)
D) 3 (red & tall): 1 (white & tall)
E) 1 (red & tall): 1 (red & short): 1 (white & tall): 1 (white & short)
Answer: B

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 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 one gene and the alleles are co-dominant.
B) The trait is controlled by one gene and both alleles are dominant.
C) The trait is controlled by two genes and the alleles are co-dominant.
D) The trait is controlled by one gene and the alleles are incompletely dominant.
E) The trait is controlled by two genes and the alleles are incompletely dominant.
Answer: D

TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false.

79) Phenotype for a given trait can be influenced by an environmental factors such as temperature.
Answer: True False

80) The mating of parents with antagonistic traits produces hybrids.

Answer: True False

81) Mendel'slaw of segregation states that two alleles for each trait unite in a specific, predictable
manner during gamete formation.
Answer: True False

82) Dihybrid crosses helped reveal the law of independent assortment.

Answer: True False

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.
Answer: True False

84) Using the product rule, one would calculate the probability of parents having six children who are
all boys as (1/2)6.
Answer: True False

85) The sum rule states that the probability of both of two mutually exclusive events occurring is the
sum of their individual probabilities.
Answer: True False

86) If you know the phenotype and the dominance relation of the alleles you can predict the genotype.
Answer: True False

87) An individual can be a heterozygote for one trait and a homozygote for another.
Answer: True False

88) A testcross is a cross between two heterozygotes.

Answer: True False

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 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.
Answer: True False

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.
Answer: True False

91) If a 4 generation family pedigree shows that the disease manifests for the first time in the 4 th
generation then it's likely that the pedigree would show consanguinity.
Answer: True False

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

Answer: True False

93) During gamete formation, different pairs of alleles on different chromosomes segregate
independently of each other.
Answer: True False

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.
Answer: True False

95) Several single-gene disorders are more common in some populations of people than in others.
Answer: True False

96) When examining a dominant trait, affected children always have at least one affected parent.
Answer: True False

97) Two affected parents can produce unaffected children in a recessive trait.
Answer: True False

98) Consanguineous mating increase the likelihood of a dominant trait.

Answer: True False

99) Incomplete dominance means that the hybrid does not resemble either pure-breeding parent.
Answer: True False

100) A lethal disorder does not include the inheritance of traits that cause death in adulthood.
Answer: True False

101) Cross-fertilization is the same as reciprocal cross.

Answer: True False

102) Traits such as human height are considered as a type of discrete traits.
Answer: True False

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103) When a sperm cell fertilizes an egg cell the result is called zygote.
Answer: True False

104) The following genotype: Gg is called heterozygote.

Answer: True False

105) Parental generation is designated as (P) and the progeny of the parental generation is designated as
F1.
Answer: True False

106) The law of segregation is a Mendelian law that states that both alleles must separate during gamete
formation.
Answer: True False

107) Multifactorial inheritance is when a phenotype arises as a result of multiple genes interacting with
each other and/or the environment.
Answer: True False

108) The flower colours white, pink, and red indicate codominant inheritance.
Answer: True False

109) A phenotype that is expressed in 87% of individuals with the same genotype shows complete
penetrance.
Answer: True False

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.
Answer: True False

111) In codominance, F1 hybrids show the traits of both parents.

Answer: True False

112) Different alleles indicate unique genes.

Answer: True False

113) Mutations are the source of new alleles.

Answer: True False

114) A wild-type allele is any allele whose frequency is closest to 100%.

Answer: True False

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.
Answer: True False

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116) A mutant allele has a rare occurrence in a population.
Answer: True False

117) Genes with more than one wild-type allele are termed polymorphic.
Answer: True False

118) The mouse agouti gene has one wild-type allele and several mutant alleles.
Answer: True False

119) The phenomenon of a single gene determining a number of distinct and seemingly unrelated
characteristics is known as pleiotropy.
Answer: True False

120) Hbß s Hbß s homozygous are resistant to Plasmodium falciparum.

Answer: True False

121) In epistasis, one gene's alleles mask the effects of another gene's alleles.
Answer: True False

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.
Answer: True False

123) Epistasis in which a dominant allele of one gene hides the effects of another gene is called recessive
epistasis.
Answer: True False

124) When an organism has two genes that perform the same function, these genes are called redundant
genes.
Answer: True False

125) In complementary gene action, dominant alleles of two or more genes are required to generate a
particular trait.
Answer: True False

126) Mutant alleles at one of two or more different genes can result in the same phenotype.
Answer: True False

127) Dominant epistasis II is also known as dominant suppression.

Answer: True False

128) To produce a particular normal phenotype, the dominant allele of two interacting genes can both be
necessary.
Answer: True False

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ESSAY. Write your answer in the space provided or on a separate sheet of paper.

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 IAIA
hh blood type. How do you interpret the evidence presented and how does it influence your decision
in this case?
Answer: 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.

130) Can a phenotype O be the father of a child who is phenotype B if the mom is phenotype A?
Answer: No

131) Calculate the probability of the production of a homozygous recessive genotype for the following
cross: AaBbccddEeFf × AaBbCcddEeFf
Answer: 1/4 × 1/4 × 1/2 × 1 × 1/4 × 1/4 = 1/512

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?
Answer: 2/3

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?
Answer: 40%

19
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.
Answer:
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

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.
Answer: 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).

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

20
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 F 2 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.
Answer:
# 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 Brown
2 l+l bb Short Brown
1 ll b+b+ Long Black
2 ll b+b Long Black
1 ll bb Long Brown

138) What does a diamond symbol in a pedigree indicate?

Answer: Sex unspecified

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?
Answer: 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.

140) List 3 criteria to recognize dominant traits?

Answer: 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).

21
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.
Answer:
# 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 fk fk gls+gls+ Forked bristles
2 fk fkgls+gls Forked bristles
1 fk fk gls gls Forked bristles and glassy eyes

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.
Answer:
Genotype Phenotype
fk+fk gls+gls Wild type

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 F 1 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.
Answer:
Genotype Phenotype
l+l b+b short, black

22
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.
Answer:
Genotype Phenotype
AaMm purple, 2 locules

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 F 1. Assume independent
assortment.
Answer:
Genotype Phenotype
LlGg Ligules/Green

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 ________.
Answer: reciprocal cross

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?
Answer: 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
role in the differences you have identified.

148) List two diseases that are caused by a dominant allele?

Answer: Hypercholesterolaemia, Huntington

23
149) List two diseases that are caused by a recessive allele?
Answer: Sickle-cell anemia, cystic fibrosis, Tay-Sachs Phenylketonuria, Thalassemia.

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 F 2.
Answer:
# 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

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.
Answer:
# 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

24
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.

Answer: 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).

153) In corn, three dominant genes are necessary for aleurone colour. 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 BbDdRr × BbDdRr
Phenotype: 27 coloured; 37 colourless
Answer:
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
25
 2 BBDdRR
4 BbDdRR
1 bbDDRR
2 bbDdRR
1 BBddRR
2 bbDdRR
1 bbddRR

154) In corn, three dominant genes are necessary for aleurone colour. 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 BbDdRR × BbDdRR
Answer: 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

26
155) In corn, three dominant genes are necessary for aleurone colour. 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 BbDdRR × BbDdrr
Answer: 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

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.
Answer:
Genotype Phenotype
PpBb Black

27
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.
Answer: 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

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.
Answer:
Genotype Phenotype
AaBb Colour

28
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.
Answer: 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

160) In poultry, if a Black Longshank male with feathered shanks is crossed with a Buff Rock female
with unfeathered shanks the F 1 are all feathered and the F2 show 90 feathered to 6 unfeathered.
Infer the genotypes of the parents.
Answer: AABB × aabb; The ratio is a 15:1 which is a dihybrid ratio; therefore the parents are
homozygous and produce a heterozygous F1.

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 GG × gg, what phenotype distribution would be expected in a
population of 400 F 2 plants?
Answer: 250 dark green (GG + 75% Gg); 150 light green (gg + 25% Gg)

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?
Answer: Blood types A, B, AB, and O are possible.

29
163) What phenotypes and genotypes would you expect from the following cross of blood-related genotypes?
IB i rh+ rh+ × IAi rh+ rh
Answer:

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

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+c × ch ch
(B) c+c+ × ch cch
(C) c+ c × chc
(D) c c × chcch
(E) c+ ch × ch cch
(F) c+ cch × chcch
(G) c c × c+cch.
Answer: (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.

30
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?

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

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.

Answer: The trait is a recessive trait. Individual #2 and individual #3 are both carriers, therefore, there
is a 1/4 chance their offspring will be homozygous for the recessive allele.

31
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+]

Answer: 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.

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 F 2 population for each hypothesis and b) what crosses
would you complete next to definitively test your two hypotheses?

Answer: 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 [aa × AA = 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 [Yyrr × yyRR = either yyRr or YyRr].

32
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 F 1 and F2 progeny of the cross AAbb × aabb
Answer: F1 = Aabb/All colourless; F2 = 1AAbb: 2Aabb: 1aabb/All colourless

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 F 1 and F2 progeny of the cross aaBB × aabb
Answer: F1 = aaBb/All colourless; F2 = 1aaBB: 2aaBb: 1aabb/All colourless

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 F 1 and F2 progeny of the cross AAbb × aaBB
Answer: F1 = AaBb coloured; 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

33
Answer Key
Testname: UNTITLED2

1) E
2) D
3) E
4) C
5) A
6) A
7) C
8) D
9) D
10) D
11) B
12) B
13) E
14) C
15) A
16) A
17) D
18) A
19) C
20) C
21) B
22) D
23) A
24) A
25) C
26) E
27) B
28) C
29) D
30) C
31) E
32) C
33) E
34) B
35) B
36) D
37) D
38) D
39) B
40) A
41) A
42) D
43) B
44) D
45) B
46) A
47) C
48) C
49) B
50) A
34
Answer Key
Testname: UNTITLED2

51) B
52) D
53) C
54) E
55) A
56) C
57) C
58) C
59) D
60) A
61) A
62) D
63) E
64) C
65) C
66) B
67) B
68) A
69) E
70) A
71) C
72) D
73) B
74) C
75) B
76) B
77) B
78) D
79) TRUE
80) TRUE
81) FALSE
82) TRUE
83) TRUE
84) TRUE
85) FALSE
86) FALSE
87) TRUE
88) FALSE
89) FALSE
90) TRUE
91) TRUE
92) FALSE
93) TRUE
94) TRUE
95) TRUE
96) TRUE
97) FALSE
98) FALSE
99) TRUE
100) FALSE
35
Answer Key
Testname: UNTITLED2

101) FALSE
102) FALSE
103) TRUE
104) TRUE
105) TRUE
106) TRUE
107) TRUE
108) FALSE
109) FALSE
110) TRUE
111) TRUE
112) FALSE
113) TRUE
114) FALSE
115) FALSE
116) TRUE
117) TRUE
118) TRUE
119) TRUE
120) TRUE
121) TRUE
122) FALSE
123) FALSE
124) TRUE
125) TRUE
126) TRUE
127) TRUE
128) TRUE
129) 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.
130) No
131) 1/4 × 1/4 × 1/2 × 1 × 1/4 × 1/4 = 1/512
132) 2/3
133) 40%

36
Answer Key
Testname: UNTITLED2

134)
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

135) 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).
136)
Genotype Phenotype
LlGg Ligules/Green
Llgg Ligules/Non-green
llGg Liguleless/Green
llgg Liguleless/Non-green

137)
# 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 Brown
2 l+l bb Short Brown
1 ll b +b + Long Black
2 ll b +b Long Black
1 ll bb Long Brown

37
Answer Key
Testname: UNTITLED2

138) Sex unspecified

139) 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.
140) 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).
141)
# 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 fk fk gls+gls+ Forked bristles
2 fk fkgls+gls Forked bristles
1 fk fk gls gls Forked bristles and glassy eyes

142)
Genotype Phenotype
fk+fk gls+gls Wild type

143)
Genotype Phenotype
l+l b +b short, black

144)
Genotype Phenotype
AaMm purple, 2 locules

145)
Genotype Phenotype
LlGg Ligules/Green

38
Answer Key
Testname: UNTITLED2

146) reciprocal cross

147) 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 d
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.
148) Hypercholesterolaemia, Huntington
149) Sickle-cell anemia, cystic fibrosis, Tay-Sachs Phenylketonuria, Thalassemia.
150)
# 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

151)
# 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

152) 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).

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153)
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

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154) 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

155) 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

156)
Genotype Phenotype
PpBb Black

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157) 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

158)
Genotype Phenotype
AaBb Colour

159) 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

160) AABB × aabb; The ratio is a 15:1 which is a dihybrid ratio; therefore the parents are homozygous and
produce a heterozygous F1.
161) 250 dark green (GG + 75% Gg); 150 light green (gg + 25% Gg)
162) Blood types A, B, AB, and O are possible.

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163)

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

164) (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.
165) Persons 1, 2, 3 are Aa. Person 4 is AA.
166) The trait is a recessive trait. Individual #2 and individual #3 are both carriers, therefore, there is a 1/4
chance their offspring will be homozygous for the recessive allele.
167) 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.
168) 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 F 2 yellow flowers will be
homozygous recessive and give rise to only orange flowers in the F3 population [aa × AA = 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 [Yyrr × yyRR = either yyRr or YyRr].
169) F1 = Aabb/All colourless; F2 = 1AAbb: 2Aabb: 1aabb/All colourless
170) F1 = aaBb/All colourless; F2 = 1aaBB: 2aaBb: 1aabb/All colourless

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Answer Key
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171) F1 = AaBb coloured; 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

Genetics Canadian 2nd Edition Hartwell Test Bank

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