Name: Genetics From Genes to Genomes 2nd Canadion Edition By Leland Hartwell - Test Bank - Eazyquizes
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Genetics From Genes to Genomes 2nd Canadion Edition By Leland Hartwell – Test Bank

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Chapter 02

Mendelian Genetics

Multiple Choice Questions

The first generation of offspring from the parents is called
A.P_.
B. F_1.
C. F_2.
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

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

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

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

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

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

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

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

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

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

Phenylketonuria (PKU) is caused by _____?
A.Recessive allele
B. Dominant allele
C. Polygenic
D. Multifactorial
E. Monohybrid allele

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

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

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

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%

In a monohybrid cross AA´aa, 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

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

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

In the dihybrid cross AaBb´aabb, 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

______ 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

Assuming independent 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

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

In a dihybrid cross AAbb´aaBB, 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

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

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

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.

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

In a dihybrid cross for which the parental cross is AABB´aabb, 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.

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

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

Which of the crosses listed below will give a 1:1:1:1 genotypic ratio in the F₁ generation? Assume independent assortment.
A.AABB´aabb
B. AaBb´AaBb
C. AaBb´aabb
D. AaBB´aaBB
E. AAbb´aaBB

For the cross AaBb´aabb, 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.

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

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

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

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

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. 1/64
D. 1/128
E. 1/256

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.AABB´AABB
B. AABB´aabb
C. aabb´aabb
D. aaBB´aabb
E. AAbb´aabb

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

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

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

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

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

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

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

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

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

Which of the following ratios show codominance?
A.2:1
B. 3:1
C. 1:2:1
D. 1:1
E. 4:1

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

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

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

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

Chapter 04

Linkage, Recombination, and the Mapping of Genes on Chromosomes

Multiple Choice Questions

The R/r and S/s genes are linked and 10 map units apart. In the cross Rs/rS´rs/rs what fraction of the progeny will be RS/rs?
A.5%
B. 10%
C. 25%
D. 40%
E. 45%

Blooms: Apply
Difficulty: Medium
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

The T/t and S/s genes are linked and 10 map units apart. In the cross Ts/tS´ts/ts what fraction of the progeny will be Ts/ts?
A.5%
B. 10%
C. 25%
D. 40%
E. 45%

If the map distance between genes A and B is 10 map units and the map distance between genes B and C is 25 map units, what is the map distance between genes A and C?
A.15 map units
B. 35 map units
C. either 15 map units or 35 map units, depending on the order of the genes
D. The map distance between A and C cannot be predicted from these data.
E. 5 map units

In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn⁺ and ct⁺) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct⁺ is crossed to a sn⁺ct male. The F₁ flies are interbred. The F₂ males are distributed as follows:

sn ct	13
sn ct⁺	36
sn⁺ ct	39
sn⁺ ct⁺	12

What is the map distance between sn and ct?
A. 12 m.u.
B. 13 m.u.
C. 25 m.u.
D. 50 m.u.
E. 75 m.u.

Blooms: Analyze
Difficulty: Medium
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive, X-linked alleles. The wild-type alleles (sn⁺ and car⁺) are responsible for straight bristles and red eyes, respectively. A sncar female is mated to a sn⁺car⁺ male and the F₁ progeny are interbred. The F₂ males are distributed as follows. What is the value of c² for a test of the hypothesis that the sn and car genes are unlinked?

sn car	55
sn car⁺	45
sn⁺ car	45
sn⁺ car⁺	55

0.5
B. 1.0
C. 2.0
D. 0.4
E. 20

Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross LM/lm ´ lm/lm would be Lm/lm?
A.10%
B. 25%
C. 50%
D. 75%
E. 100%

The pairwise map distances for four linked genes are as follows: A-B = 22 m.u., B-C = 7 m.u., C-D = 9 m.u., B-D = 2 m.u., A-D = 20 m.u., A-C = 29 m.u. What is the order of these four genes?
A.ABCD
B. ADBC
C. ABDC
D. BADC
E. CADB

Blooms: Analyze
Difficulty: Medium
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

The zipper-like connection between paired homologs in early prophase is known as a
A.spindle fiber.
B. synaptic junction.
C. synaptonemal complex.
D. chiasma.
E. none of the choices are correct.

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis

The measured distance between genes D and E in a two-point test cross is 50 map units. What does this mean in physical terms?
A.D and E are on different pairs of chromosomes.
B. D and E are linked and exactly 50 map units apart.
C. D and E are linked and at least 50 map units apart.
D. either D and E are on different pairs of chromosomes or D and E are linked and exactly 50 map units apart.
E. either D and E are on different pairs of chromosomes or D and E are linked and at least 50 map units apart.

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis

In humans, the genes for red-green colour blindness (R = normal, r = colour-blind) and haemophilia A (H = normal, h = haemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are colour-blind but have normal blood clotting and two have haemophilia but normal colour vision. What is the probable genotype of the woman?
A.HR/hr
B. Hr/hr
C. hr/hR
D. Hr/hR
E. HR/Hr

Blooms: Analyze
Difficulty: Easy
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis

In humans, the genes for red-green colour blindness (R = normal, r = colour-blind) and haemophilia A (H = normal, h = haemophilia) are both X-linked and only 3 map units apart. A woman has a mother who is colour-blind and her father has haemophilia A. Her parents were otherwise normal. The woman is pregnant with a boy and wants to know the probability that he will have normal vision and normal blood clotting. What is the probability?
A..03
B. .15
C. .485
D. .47
E. .015

Blooms: Evaluate
Difficulty: Easy
Learning Objective: 04-01 Discuss the physical basis for genetic linkage.
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Topic: 04-02 Some genes on the same chromosome do not assort independently—instead, they are linked

Recombination frequencies near 50 percent suggest that
A.two genes are on different chromosomes.
B. two genes lie very close together on the same chromosome.
C. two genes are on the same chromosome but lie very far apart.
D. two genes are on different chromosomes, or two genes are on the same chromosome but lie very far apart.
E. none of the choices are correct.

Which of the following processes can generate recombinant gametes?
A.Segregation of alleles in a heterozygote.
B. Crossing over between two linked heterozygous loci.
C. Independent assortment of two unlinked heterozygous loci.
D. Crossing over between two linked heterozygous loci and independent assortment of two unlinked heterozygous loci.
E. Segregation of alleles in a heterozygote, crossing over between two linked heterozygous loci and independent assortment of two unlinked heterozygous loci.

Crossing over takes place in paired bivalents consisting of ______ chromatids, and involves _______ of the chromatids.
A.2; 2
B. 2; 4
C. 4; 2
D. 4; 4
E. 8; 4

Blooms: Remember
Difficulty: Medium
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y⁺car⁺/ycar), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny from such a female would be recombinant for y and car?
A.5%
B. 10%
C. 45%
D. 55%
E. 90%

Blooms: Apply
Difficulty: Medium
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

The map of a chromosome interval is Q-10 m.u.-R-40 m.u.-S. Which interval would likely show the higher ratio of double to single chiasmata?
A.Q–R
B. R–S
C. The ratios would be the same in the two intervals.
D. Two chiasmata never occur in the same interval.

Blooms: Apply
Difficulty: Medium
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

Which of the following genes is located on chromosome 8 in humans?
A.DMD
B. HTT
C. HFE
D. ANK1
E. RB1

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-06 Construct a simple genetic map based upon data from testcrosses.
Topic: 04-16 Multiple-factor crosses help establish linkage groups

Which of the following genes is responsible for Fragile X syndrome?
A.ATM
B. HFE
C. FMR1
D. MEN1
E. BRCA 1

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-06 Construct a simple genetic map based upon data from testcrosses.
Topic: 04-16 Multiple-factor crosses help establish linkage groups

ATM gene is responsible for which of the following diseases?
A.Spherocytosis type 1
B. Ataxia-telangiectasia
C. Retinoblastoma
D. Marfan syndrome
E. Huntington Disease

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-06 Construct a simple genetic map based upon data from testcrosses.
Topic: 04-16 Multiple-factor crosses help establish linkage groups

Haemochromatosis is caused by a problem in gene _____ which is located on chromosome _____.
A.RB1, 13
B. HFE, 6
C. HTT, 4
D. FBN1, 15
E. PKD1, 16

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-06 Construct a simple genetic map based upon data from testcrosses.
Topic: 04-16 Multiple-factor crosses help establish linkage groups

The map of a chromosome interval is A-10 m.u.-B-40 m.u. – C. From the cross Abc/aBC´abc/abc, how many double crossovers would be expected out of 1000 progeny?
A.5
B. 10
C. 20
D. 40
E. 80

Blooms: Apply
Difficulty: Medium
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

In an individual with genotype Lpq/lPQ the L gene is in the middle. What would be the genotypes of the gametes produced by a double crossover surrounding the L locus?
A.LPQ and lpq
B. LpQ and lPq
C. lpQ and LPq
D. Lpq and lPQ
E. cannot be determined

Blooms: Analyze
Difficulty: Medium
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

Suppose a three-point testcross was conducted involving the genes X, Y, and Z. If the most abundant classes are XYz and xyZ and the rarest classes are xYZ and Xyz, which gene is in the middle?
A.X
B. Y
C. Z
D. cannot be determined
E. either Z or Y

Blooms: Evaluate
Difficulty: Medium
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-10 Recombination frequencies between two genes never exceed 50 percent

In Drosophila, the genes b, c, and sp are linked and in the order given, the distances being b-c = 30 m.u. and c-sp = 20 m.u. These intervals exhibit 90% interference. How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?
A.3
B. 6
C. 54
D. 60
E. 600

Consider a pair of homologous chromosomes heterozygous for three genes (e.g. ABC/abc) during prophase I of meiosis. Let the sister chromatids of one homolog be numbered 1 and 2; and the sister chromatids of the other homolog be numbered 3 and 4. A crossover that would result in genetic recombination (e.g., Abc or aBC) could involve which chromatids?
A.1 & 2 or 3 & 4
B. 1 & 3 or 2 & 4
C. 1 & 4 or 2 & 3
D. 1 & 3 or 1 & 4 or 2 & 3 or 2 & 4
E. any two of the four chromatids

Sturtevant’s detailed mapping studies of the X chromosome of Drosophila established what genetic principle?
A.That genes are arranged in a linear order on the chromosomes.
B. That genes are carried on chromosomes.
C. That sex determination is controlled by the X and Y chromosomes.
D. That segregation of an allelic gene pair is accompanied by disjunction of homologous chromosomes.
E. That different pairs of chromosomes assort independently.

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-08 Why recombination?

Suppose an individual is heterozygous for a pair of alleles (e.g., A/a). Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells homozygous for a? (Pick the most precise answer.)
A.The crossover would have to occur between the A locus and the centromere and involve two homologous (non-sister) chromatids.
B. The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (non-sister) chromatids.
C. The crossover would have to occur on the same chromosome arm as the A locus and involve two homologous (non-sister) chromatids.
D. The crossover would have to occur on the same chromosome as the A locus and involve two homologous (non-sister) chromatids.
E. The crossover would have to occur between the A locus and the centromere and involve two sister chromatids (not homologous) chromatids.

Blooms: Apply
Difficulty: Easy
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-05 Recombination: A Result of Cross-Over During Meiosis

Suppose an individual is heterozygous for two linked pairs of alleles on the same chromosome arm, Ab/aB such that the A locus is closer to the centromere than the B locus. Under what conditions would a crossover in a somatic cell generate a twin spot, i.e., two adjacent clones of cells, one clone homozygous for a and the other clone homozygous for b?
A.The crossover would have to occur between the A locus and the centromere and involve two homologous (non-sister) chromatids.
B. The crossover would have to occur between the A locus and the B locus and involve two homologous (non-sister) chromatids.
C. The crossover would have to occur between the B locus and the end of the chromosome locus and involve two homologous (non-sister) chromatids.
D. A double crossover would have to occur, with one crossover between the A locus and the centromere and a second crossover between the A and B loci and both crossovers would have to involve two homologous (non-sister) chromatids.
E. No crossover in a somatic cell could generate a twin spot.

Individuals heterozygous for the RB⁺ and RB^– alleles can develop tumours as a result of
A.a mitotic crossover that leads to homozygosity for both RB⁺ and RB^–.
B. a somatic mutation in the RB⁺ allele that leads to homozygosity for RB^–.
C. a somatic mutation in the RB^– allele that leads to homozygosity for RB⁺.
D. the fact that RB^– is dominant to RB⁺.
E. either a mitotic crossover that leads to homozygosity for both RB^– or a somatic mutation in the RB⁺ allele that leads to homozygosity for RB^–.

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-06 Construct a simple genetic map based upon data from testcrosses.
Topic: 04-13 Three-point crosses provide faster and more accurate mapping

What happens physically during the process of crossing over?
A.Two homologous chromatids break and rejoin at random sites within each chromatid.
B. The genetic information on one chromatid is replaced by copying genetic information from a homologous chromatid without there being any physical exchange between the chromosomes.
C. Two homologous chromatids break and rejoin at precisely the same site along the chromosome so that there is no loss or gain of material on either product.
D. A segment of one chromatid is cut out and then pasted into the other chromatid.
E. It is not known what occurs during crossing over.

Some of the larger human chromosomes typically contain multiple chiasmata during meiotic prophase. If you were to carefully study the distribution of these chiasmata, what would you find?
A.Chiasmata are randomly distributed along chromosomes.
B. All chromosome pairs have the same number of chiasmata.
C. A single chromosome pair always has the same number of chiasmata in every meiotic cell.
D. Chiasmata are spaced along a chromosome arm more regularly than would be expected by chance.
E. Chiasmata are spaced more irregularly along a chromosome arm than would be expected by chance.

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-07 Chismata mark the sites of recombination

In fruit flies, the recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). The double mutant prcn combination has orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colours:

wild-type	8
brown	241
bright-red	239
orange	12
	500

Which classes are the parental types?
A. wild-type and orange
B. brown and bright-red
C. wild-type and brown
D. bright-red and orange
E. There is no way to determine this.

Blooms: Apply
Difficulty: Medium
Learning Objective: 04-01 Discuss the physical basis for genetic linkage.
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Topic: 04-02 Some genes on the same chromosome do not assort independently—instead, they are linked

In fruit flies, the recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). The double mutant prcn combination has orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colours:

wild-type	8
brown	241
bright-red	239
orange	12
	500

What is the genotype of the mother of these progeny?
A. prcn/pr⁺cn⁺
B. pr⁺cn/pr⁺cn
C. pr⁺cn/prcn⁺
D. prcn⁺/prcn⁺
E. prcn/prcn

Blooms: Analyze
Difficulty: Medium
Learning Objective: 04-01 Discuss the physical basis for genetic linkage.
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Topic: 04-02 Some genes on the same chromosome do not assort independently—instead, they are linked

In fruit flies, the recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). The double mutant prcn combination has orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colours:

wild-type	8
brown	241
bright-red	239
orange	12
	500

The mother of these progeny resulted from a cross between two flies from true breeding lines. What are the genotypes of these two lines?
A. prprcn⁺cn⁺ and pr⁺pr⁺cncn
B. pr⁺pr⁺cn⁺cn⁺ and prprcncn
C. pr⁺prcn⁺cn and prprcncn
D. pr⁺prcncn and prprcn⁺cn
E. Either prprcn⁺cn⁺ and pr⁺pr⁺cncn or pr⁺pr⁺cn⁺cn⁺ and prprcncn could be true.

In fruit flies, the recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). The double mutant prcn combination has orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colours:

wild-type	8
brown	241
bright-red	239
orange	12
	500

What is the map distance between the pr and cn genes?
A. 20 m.u.
B. 2 m.u.
C. 4 m.u.
D. 46 m.u.
E. 8 m.u.

Which of the following best describes a linkage map that contains a high density of genetic markers?
A.recombinant
B. saturated
C. three-point
D. random
E. sporadic

Which of the following was useful in the identification of the gene that causes cystic fibrosis (CF)?
A.observation of independent assortment of PON and CF
B. sequencing of the complete human genome
C. identification of another trait that is often inherited with CF
D. comparison of karyotypes of affected and unaffected individuals
E. chorionic villus sampling

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-04 Describe when during the cell cycle a crossover occurs, the mechanism, and its potential outcomes.
Topic: 04-06 Reciprocal exchanges between homologues are the physical basis of recombination

Observation of which of the following led to the discovery of mitotic recombination?
A.twin spots
B. chromosome knobs
C. linkage mapping
D. interference
E. chiasmata

Blooms: Remember
Difficulty: Easy
Learning Objective: 04-05 Demonstrate how genetic distance is estimated between two or three linked loci.
Topic: 04-12 Comparisons of two-point crosses establish relative gene positions

Syntenic genes are
A.always recombinant
B. always unlinked
C. assort independently
D. always linked
E. located on the same chromosome

Blooms: Understand
Difficulty: Easy
Learning Objective: 04-01 Discuss the physical basis for genetic linkage.
Learning Objective: 04-02 Explain why Mendel’s law of independent assortment does not apply to linked genes.
Topic: 04-02 Some genes on the same chromosome do not assort independently—instead, they are linked

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Genetics From Genes to Genomes 2nd Canadion Edition By Leland Hartwell – Test Bank

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