TEST BANK for Genetics From Genes to Genomes 7th Edition

TEST BANK for Genetics From Genes to Genomes 7th Edition

Test Bank for Genetics From Genes to Genomes 7th Edition Goldberg All Chapters Student name:__________ 1)

Why did Mendel perform reciprocal crosses?

A) To obtain enough plants to perform the experiments that Mendel wanted.

B) To test a hypothesis that stated the ovum carries all the information for progeny.

C) To be able to breed plants year round.

D) To determine whether the inheritance of a trait depends on which parent carries the trait.

2) What is the difference between cross- and selffertilization?

A) In cross-fertilization, the pollen from one plant is used to fertilize the egg from the same plant.

B) In cross-fertilization, the pollen from one plant is used to fertilize the egg of another plant.

C) In self-fertilization, the pollen from one plant is used to fertilize the egg from another plant.

D) In crossfertilization, insects are used to pollinate the plants, whereas in selffertilization, the investigator pollinates the plants.

3) What is the outcome of crossing two pure-breeding plants with antagonistic characters of traits?

A) Only one of the characteristics will be seen in the progeny.

B) Both characteristics will be seen in the progeny.

C) Both characteristics will be seen in the progeny in

Version 1

a 3:1 ratio.

D) Only one characteristic will be seen, and it will be that of the female.

1

A) alleles of genes assort into gametes grouped according to how they were inherited originally.

B) dominant alleles for one gene must assort into the same gamete as the dominant alleles for another gene.

C) alleles of genes on different chromosomes assort randomly into different gametes.

D) dominant alleles for one gene must assort into the same gamete as the recessive alleles for another gene.

5) An S1S2 ×S1S2 mating is performed. If the phenotypic ratio of the progeny is 3:1, then

A) one allele is dominant and the other is recessive.

B) neither allele is dominant.

C) the S1 allele is dominant to the S2 allele.

D) the S2 allele is dominant to the S1 allele.

E) the relationship between the alleles cannot be determined.

6) Which of the following probabilities is correct (according to Mendel's law of independent assortment) regarding a mating of an Ss RR individual to an individual who is Ss Rr? (A – indicates the second allele is either dominant or recessive.)

A) Homozygous recessive: 10% B) Heterozygous both alleles: 50%

Version 1

C) ss R– : 15.5% D) S– RR: 37.5%

2

What does the pattern of inheritance in this pedigree indicate about the rare

7) A) The disease allele is dominant.

B) The disease allele is recessive.

C) There is no indication that the disease allele is either dominant or recessive.

D) The disease allele is not inherited but arises only by a new mutation in affected individuals.

8) The mutant alleles of the CF gene that result in cystic fibrosis are recessive to normal alleles because

A) the protein produced by the normal allele in heterozygotes is sufficient for normal cellular function.

B) the CF mutations that cause cystic fibrosis always result in no protein being produced.

C) CF mutations result in a protein that has normal function only if normal CFTR protein also exists in the cell.

D) dominant alleles that cause a fatal disorder, such as cystic fibrosis, cannot be inherited.

9) The reason that the HD allele that causes Huntington disease is dominant to HD+ alleles is that

A) the normal HD allele does not normally produce a protein but the mutant HD allele does.

Version 1

B) the mutant HD allele suppresses protein production from the

3

normal HD allele.

C) the HD mutation results in a protein that can damage nerve cells even in the presence of the normal protein.

D) the protein produced from the mutant HD allele is nonfunctional.

10) If an individual is heterozygous for only 7 of his gene pairs (he is homozygous for all of his other genes), how many

different gamete types can he produce?

A) 49 B) 100

C) 128 D) 1024 E) 131,072

11) In some genetically engineered corn plants, a Bt gene was inserted into a chromosome. The Bt gene specifies a protein called Bt that is lethal to certain flying insect pests that eat the corn plants. If the corn plant is heterozygous for the Bt gene (one homolog has the introduced Bt gene and the other does not), what proportion of the sperm would carry the Bt gene? Is the presence of the Bt gene (a mutation) dominant or recessive to its absence (the wild type)?

A) all pollen; dominant B) 1/2; dominant

12) Suppose that in plants, smooth seeds (S) is dominant to wrinkled seeds (s), and tall plants (T) is dominant to short plants (t). An F1 plant from a mating between homozygous plants that were tall/smooth and short/wrinkled was crossed to Version 1

C) 1/3; recessive D) 1/4; dominant E) 1/8; recessive

the short/wrinkled parent.

What proportion of the progeny are expected (according to the Mendel’s 4

law of independent assortment) to be homozygous for short

A) 1/2 B) 1/4

and wrinkled alleles?

C) 1/8 D) 1/16 E) 0

13) Sickle-cell disease is a recessive trait in humans. In a cross between a father who has sickle-cell disease and a mother who is heterozygous for the sickle-cell allele, what is the probability that all of their first three children will be unaffected?

A) 1/4 B) 1/2

C) none D) 1/8 E) 1/16

14) Starting with the parental cross AA × aa, what proportion of the F2 offspring is expected to be homozygous?

A) 1/4 B) 1/2 C) 3/4

D) All are homozygotes.

E) None are homozygotes.

15) Starting with the parental cross AA bb × aa BB, what proportion of the F2 offspring is expected to be homozygous at least one of the two genes?

A) 1/4

Version 1

B) 1/2 C) 3/4

5

D) All are homozygotes.

E) None are homozygotes.

16) In the testcross Aa Bb × aa bb, what proportion of individuals are expected (according to Mendel's law of independent assortment) to be homozygous for both genes in the F1generation?

A) 1/4 B) 1/2 C) 3/4

D) All are homozygotes.

E) None are homozygotes.

17) Among the crosses shown below, which will produce a 1:1 phenotypic ratio according to Mendel's law of independent assortment?

A) AA BB × aabb B) Aa Bb × AaBb

C) Aa Bb × aabb D) AaBB × aaBB E) AA bb × aaBB

18) 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) 1/4 B) 1/2 C) 3/4

Version 1

D) All of these choices are correct.

E) None of these choices are correct.

6

19) 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 for alleles of the B gene?

A) 1/4 B) 1/2 C) 3/4

D) All of these choices are correct.

E) None of these choices are correct.

20) 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.

21) 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.

22) The alleles present in an individual make up the individual's C) dominance.

A) recombinant types.

B) recessiveness.

Version 1

7

D) phenotype.

E) genotype.

Version 1

8

23)

The first offspring from the parents are called C) F 2.

D) a testcross.

E) P 2.

A) P. B) F 1.

24) What type of cross is performed to determine the genotype of an individual with the dominant character of a trait?

D) A genotyping A) A testcross B) A dihybrid cross C) A monohybrid cross

cross E) A controlled cross

25) If the parents of a family already have two boys, what is the probability that the next two offspring will both be girls?

A) 1 B) 1/2

26) Suppose that in plants, smooth seeds (S) is dominant to wrinkled seeds (s) and tall plants (T) is dominant to short plants (t). A dihybrid tall plant with smooth seeds was crossed to a short plant with wrinkled seeds. According to Mendel’s law of independent assortment, what proportion of the

Version 1

C) 1/3 D) 1/4 E) 1/8

progeny is expected to be tall and smooth?

9

A) 1/2 B) 1/4

C) 1/8 D) 1/16 E) 0

27) A rare recessive characteristic in a pedigree is indicated by which pattern of inheritance?

A) Vertical B) Horizontal C) Diagonal

28) The dominant Huntington disease allele causes severe neural/brain damage at approximately age 40. A female whose mother has Huntington disease wants to have a child with a male whose parents are normal. It is not known if the female has the disease. Keeping in mind that the disease allele A) 25% B) 50%

D) Both vertical and horizontal E) Pure-breeding

is rare in the population, what is the probability that their firstborn will inherit the allele that causes Huntington disease?

C) 75% D) 100% E) 0%

29) Starting with the parental cross AA × aa, what proportion of the F2 offspring is expected to be heterozygous?

A) 1/4 B) 1/2 C) 3/4

Version 1

D) All are heterozygotes.

E) None are heterozygotes.

10

30) Starting with the parental cross AA BB × aa bb, what proportion of the F2 offspring is expected to be heterozygous A) 1/4 B) 1/2 C) 3/4

for both gene pairs?

D) All are heterozygotes.

E) None are heterozygotes.

31) What proportion of the F1 offspring resulting from the cross Aa Bb × aa bb is expected to be heterozygous for both gene pairs?

A) 1/4 B) 1/2 C) 3/4

D) All are heterozygotes.

E) None are heterozygotes.

32) Among the crosses shown below, which will produce offspring with a 1:1:1:1 genotypic ratio?

A) AA BB × aa bb B) AaBb × AaBb

C) Aa Bb × aabb D) Aa BB × aaBB E) AA bb × aa BB

33) What is the term for crosses between parents that are heterozygous at a single locus?

D) Dihybrid A) Testcrosses B) Cross fertilize C) Monohybrid crosses

Version 1

11

crosses E) Reciprocal crosses

Version 1

12

34)

A particular form of a single gene is known as C) a reciprocal.

D) anallele.

E) a recessive.

A) a parental.

B) a dihybrid.

35) A phenotype reflecting a new combination of alleles occurring during gamete formation is called A) a recombinant type.

B) an independent assortment.

C) heterozygous.

D) homozygous.

E) a multihybrid cross.

36) How was the approach taken by Mendel similar to the approaches taken by modern scientific inquiry?

A) Mendel repeated his experiments.

B) Mendel examined both continuous and discrete traits.

C) Mendel used the same technical methods that are

37) Pea shape is controlled by a gene that specifies an enzyme known as Sbe1 (for Starch-branching enzyme 1).

Two alleles of Sbe I exist, where is one allele is dominant

used today.

D) Mendel's experiments challenged no hypotheses that were favored at the time.

and the other is recessive.

The recessive allele most like specifies C) a different type

A) an Sbe1 enzyme with reduced function.

B) an Sbe1 enzyme with a new function.

Version 1

13

of enzyme.

D) an Sbe1 enzyme with enhanced catalytic activity.

Version 1

14

38) The normal allele of the pea color gene specifies the enzyme Sgr, which functions in a pathway to break down chlorophyll during pea maturation, resulting in yellow mature peas. A second allele of the Sgr gene produces no enzyme A) dominant B) recessive

39) Mendel’s law of independent assortment dictates that an Aa Bb dihybrid would make equal numbers of four A) A; a; B; b B) AA; BB; aa; bb C) A B; A b; a B; a b

40) According to Mendel’s law of equal segregation, an Aa monohybrid makes two types of gametes with equal A) AA and aa.

B) A and a.

and is _____ to the normal allele.

C) wild-type D) functioning

gamete types. What are these four gamete types?

D) AA BB; AA bb; Aa Bb; aa BB

frequency. These two gamete types are: C) Aa and aa.

D) AA and Aa.

41) For each of the following pedigree symbols, select the correct meaning.

Version 1

15

41.1) A) Unaffected male B) Unaffected female C) Mating

D) Affected male E) Affected female

41.2) A) Unaffected male B) Unaffected female C) Mating

D) Affected male E) Affected female

41.3) A) Unaffected male B) Unaffected female C) Mating

D) Affected male E) Affected female

41.4) A) Unaffected male B) Unaffected female C) Mating

Version 1

D) Affected male E) Affected female

16

42) A pedigree for a common human characteristic (not a disease) controlled by a single gene is shown. Shaded symbols indicate individuals exhibiting the characteristic.

Version 1

17

42.1) Identify the most likely mode of inheritance of the trait.

A) Dominant B) Recessive C) Either dominant or recessive

D) Cannot be determined

42.2) If individuals 4 and 7 have a child, what is the probability that the child will exhibit the characteristic?

A) 1/4 B) 1/2

C) 1/6 D) 2/3 E) 0

43) Below is a pedigree of a rare human genetic disease.

The filled in symbols indicate affected individuals. Assume that the disease is caused by a mutant allele of gene A.

Version 1

18

43.1)

Based on this pedigree, what is the most likely

A) Dominant B) Recessive C) Either dominant or recessive

43.2) 1?

D) aa E) Cannot be determined

What is/are the possible genotype(s) of person

A) AA B) Aa C) Either AA or Aa

Version 1

D) aa E) Cannot be determined

What is/are the possible genotype(s) of person

A) AA B) Aa C) Either AA or Aa

43.4) 3?

D) Cannot be determined

What is/are the possible genotype(s) of person

A) AA B) Aa C) Either AA or Aa

43.3) 2?

mode of inheritance?

D) aa E) Cannot be determined

19

43.5) 4?

What is/are the possible genotype(s) of person

A) AA B) Aa C) Either AA or Aa

D) aa E) Cannot be determined

43.6) If individuals 1 and 4 have a child together, what is the probability that the child will exhibit the disease?

A) 0% B) 25%

C) 50% D) 75% E) 100%

43.7) If individuals 2 and 3 have a child together, what is the probability that the child will exhibit the disease?

A) 0% B) 25%

C) 50% D) 75% E) 100%

44) In corn, having ligules (L) is dominant to liguleless (l), and green leaves (G) is dominant to white leaves (g).

Version 1

20

44.1) If a testcross is performed with a dihybrid plant with ligules and green leaves, what proportion of the progeny would be green and liguleless?

A) 1/16 B) 1/8 C) 1/4

D) 1/2 E) Cannot be determined

44.2) If a pure-breeding liguleless plant with green leaves is crossed to pure-breeding plant with ligules and white leaves, predict the proportion of F2 progeny with the genotype Ll gg.

A) 1/16 B) 1/8 C) 1/4

D) 1/2 E) Cannot be determined

44.3) If a pure-breeding plant that is liguleless and has green leaves is crossed to a pure-breeding plant with white leaves and ligules, predict the genotypes and phenotypes of the F1.

A) LL GG, green and ligules B) Ll GG, green and ligules C) Ll Gg, green and ligules

Version 1

D) ll gg, white and liguleless E) Ll gg, green and liguleless

21

Answer Key Test name: Chapter 01 Test Bank 1) D 2) B 3) A 4) C 5) A 6) D 7) B 8) A 9) C 10) C 11) B 12) B 13) D 14) B 15) C 16) A 17) D 18) B 19) E Version 1

22

20) C 21) B 22) E 23) B 24) A 25) D 26) B 27) B 28) A 29) B 30) A 31) A 32) C 33) C 34) D 35) A 36) A 37) A 38) B 39) C 40) B Version 1

23

41) Section Break 41.1) A 41.2) C 41.3) B 41.4) D 42) Section Break 42.1) B 42.2) C 43) Section Break 43.1) A 43.2) D 43.3) D 43.4) D 43.5) B 43.6) C 43.7) A 44) Section Break 44.1) C 44.2) B 44.3) C

Version 1

24

Student name:__________ 1) ABO blood type demonstrates which of the following inheritance patterns?

A) incomplete dominance B) complete dominance C) pleiotropy

2) Two alleles of gene C control hair color in horses: C¹and C². Horses homozygous for allele C¹ are red, heterozygotes are yellow, and C² homozygotes are cream.

A) pleiotropy B) complete dominance C) incomplete dominance

D) codominance E) three genes are involved

What type of allele interaction is described?

D) recessive lethality E) codominance

3) If a trait is controlled by two different genes, a possible interaction between alleles of these genes can be called A) epistasis.

B) epigenetics.

C) dominance.

D) codominance.

E) incomplete dominance.

4) Females from a pure-breeding curly-winged strain are mated with males from a pure-breeding straight-winged (wild-type) strain. The F1 mate with each other to produce an F2 generation that consists of 160 flies with curly wings and 80 with straight wings. What can you infer from this observation?

Version 1

1

A) Curly wings is a recessive trait.

B) The dominant curly wing allele is also a recessive lethal.

C) Wing shape is controlled by two codominant alleles.

D) Two interacting genes determine wing shape.

E) All of the hybrid F 1 flies had straight wings.

5) What can explain the phenomenon where in different individuals, a particular genotype might give rise to different phenotypes?

A) pleiotropy B) codominance C) incomplete dominance

6) Two alleles of gene C control hair color in horses: C¹and C². Horses homozygous for allele C¹ are red, heterozygotes are yellow, and C² homozygotes are cream. In the offspring of matings between heterozygotes, what A) 2 yellow: 1 red B) 3 red: 1 cream C) 1 red: 2 yellow: 1 cream

D) complete dominance E) penetrance and expressivity

phenotypic ratio is expected?

D) all red E) 9 red: 3 yellow: 4 cream

7) If a trait is controlled by two codominant alleles of one gene, what phenotypic ratio is expected in the offspring of a mating of two heterozygotes?

C) 1:2:1 A) 2:1 B) 3:1 Version 1

2

D) 1:1 E) 4:1

Version 1

3

8) Which genotypic ratio indicates a recessive lethal allele when two heterozygotes are mated?

A) 2:1 B) 3:1

C) 1:2:1 D) 1:1 E) 4:1

9) A particular flower can be blue, red, or white. A purebreeding red-flowered plant is crossed with a pure-breeding white-flowered one. The F 1 are then crossed to produce an F 2 generation. Which of the following phenotypic ratios in the F 2 indicate that flower color in these two strains is controlled by two genes?

A) 2:1 B) 3:1

C) 1:2:1 D) 9:3:4

10) If a gene for a trait is monomorphic in a population and two random individuals mate, what would be the most likely phenotypic ratio for that trait in the offspring?

A) 2:1 B) More information is needed

C) 1 D) 3:1

11) What genetic phenomenon might a 2:1 phenotypic ratio indicate?

A) additivity

Version 1

4

B) codominance C) complete dominance D) recessive epistasis

E) recessive lethality

12) A particular flower can be purple, blue, red, or white.

Plants from two different pure-breeding strains of whiteflowered plants are crossed and the F 1 are then crossed to produce an F 2 generation. What might a 9:7 phenotypic ratio

in the F 2 indicate about the genes that control flower color in these plants?

A) two genes with reciprocal recessive epistasis B) one gene with two codominant alleles C) two genes and dominant epistasis

D) two genes and recessive lethality E) two genes and additive interactions

13) A particular flower can be purple, blue, red, or white.

A pure-breeding purple-flowerd plant is crossed with a purebreeding white-flowered plant and the F 1 are then crossed to produce an F 2 generation. What might a 9:3:4 phenotype ratio in the F 2 indicate?

A) reciprocal recessive epistasis B) codominance C) dominant epistasis

D) recessive epistasis E) additivity

14) Which ratio in the F 2 of a cross between two purebreeding strains would indicate that a phenotype is controlled by more than one gene?

A) 3:1 B) 2:1

Version 1

C) 1:2:1 D) 9:3:3:1

5

15) A disease is caused by homozygosity for the g allele (G is the corresponding wild-type allele). However, the penetrance of the disease is 75%. Two individuals known to be heterozygotes have a child. What is the probability that the A) 1/4 B) 3/4

child exhibits the disease?

C) 1/8 D) 3/16 E) 9/16

16) Some complex traits are determined by many interacting genes, each of which may have several alleles. If such a trait is measured in a population, what phenotypic pattern is expected?

A) All individuals will have the same phenotype.

B) Two types of individuals will exist, and most will have the dominant phenotype.

C) Different individuals will each have one of a few

discrete phenotypes.

D) The population will have continuous variation in phenotypic expression.

17) In some flowers, a purple pigment is synthesized from a red precursor pigment. In the absence of all pigment, flowers are white. Pure-breeding plants with red flowers were crossed to a pure-breeding plants with white flowers. All of the F1 plants had white flowers. The F1 plants were crossed to each other, and the F2 consisted of 166 plants: 123 with white flowers, 32 with purple flowers, and 11 with red flowers.

How is flower color is determined in these plants?

A) One gene with two alleles exists and heterozygotes have a different phenotype than either homozygote.

B) The dominant allele of one gene masks the effect of a second gene.

Version 1

C) Recessive alleles of one gene mask the effect of a second gene.

D) A dominant allele of each gene is necessary for purple 6

flowers.

E) A dominant allele of either of two genes is

sufficient for purple flowers.

18) A particular flower can be purple, blue, red, or white.

A pure-breeding purple-flowered plant is crossed with a purebreeding white-flowered plant and the F1 are then crossed to produce an F2 generation. Which phenotypic ratio in the F2

may indicate that flower color in these plants is controlled by two genes that interact additively?

A) 3:1 B) 1:2:1

19) In dogs, Gene B specifies a protein required for eumelanin (dark pigment) deposition. Protein B (specified by the B allele) deposits eumelanin densely so that the dog’s hair is black. Protein b (specified by the b allele) deposits eumelanin less densely, producing brown hair (chocolate).

Allele B is dominant to allele b. Gene D specifies another protein that is also required for pigment deposition. The recessive allele d1specifies a protein that functions less efficiently than that specified by the dominant allele D. Less A) Bb d1d1 B) BB d1d1

C) 9:3:3:1 D) 9:3:4 E) 15:1

pigment is deposited in d1d1homozygous dogs than in dogs with a D allele, and so the color dictated by gene B is lighter.

Which of the following genotypes would a lightchocolate dog have?

C) bb d1d1 D) bb Dd1

20) In rats, the P gene allele for pigmentation (P) is dominant to the allele for albinism (p). The B gene allele for black pigmentation (B) is dominant to the allele for cream pigmentation (b). The pp homozygous recessive genotype is epistatic to any allele combination at gene B.

Version 1

7

20.1) Predict the genotypes and phenotypes of the F1 progeny of a cross between a pure-breeding black rat and an albino that is also homozygous for cream.

A) PP BB, black B) Pp Bb, black

C) Pp Bb, albino D) pp Bb, albino

20.2) Predict the phenotypic ratio of the F2 progeny of a parental cross between a pure-breeding black rat and an albino that is also homozygous for cream.

A) 1 black: 2 cream: 1 albino B) 9 black: 3 cream: 4 albino C) 9 black: 7 albino

D) 12 black: 3 cream: 1 albino E) 15 black: 1 albino

21) In the common daisy, genes A and B control flower color. Both genes have a dominant allele (A or B) and a recessive allele (a or b). At least one copy of each dominant allele is required for flowers to be colorful instead of white.

21.1) Predict the genotypes and phenotypes of the F1 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB. B) aa BB, white A) AA bb, white Version 1

8

C) Aa Bb, colorful D) Aa Bb, white E) aa bb, colorful

Version 1

9

21.2) Predict the phenotypic ratio of the F2 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB.

A) 3 colorful : 1 white B) 9 colorful : 7 white C) 9 white : 7 colorful

D) 15 white : 1 colorful E) 15 colorful : 1 white

21.3) The inheritance pattern of daisy flower color provides an example of what type of gene interaction?

A) additivity B) recessive epistasis C) reciprocal recessive epistasis

D) dominant epistasis E) redundancy

22) Achondroplasia is a form of dwarfism in humans. It is caused by a mutant allele of the fibroblast growth factor receptor 3 gene ( FGFR3) that produces an overactive protein. Having one copy of the mutant allele results in dwarfism. Two copies of the mutant allele results in death before birth. The mutant FGFR3 allele is completely penetrant.

22.1) What can you infer about the inheritance of the FGFR alleles?

A) The mutant FGFR3 allele is pleiotropic.

B) Some achondroplastic dwarfs are heterozygous Version 1

for the mutant allele and some are homozygous.

10

C) The wild-type and mutant FGFR3 alleles are codominant.

D) The mutant FGFR3 allele shows incomplete

penetrance.

22.2) If two people with achondroplasia have a child together, what is the probability that their child will also

have achondroplasia?

A) 0 B) 1/2

C) 2/3 D) 3/4 E) 1

23) In primroses, the dominant allele of gene K is necessary to synthesize blue flower pigment. Blue pigment synthesis is inhibited by a dominant allele of gene D. In other words, plants with the genotype K− D− will not produce pigment (and their flowers will be white) because of the presence of the D allele.

23.1) If two dihybrid plants ( Kk Dd) are crossed, what is the ratio of blue to white offspring in the progeny?

A) 3 blue: 1 white B) 7 blue: 9 white C) 4 blue: 12 white

D) 3 blue: 13 white E) 1 blue: 15 white

23.2) What type of interaction between alleles of different genes determines flower color in primroses?

Version 1

11

A) reciprocal recessive epistasis B) recessive epistasis C) redundancy

D) dominant epistasis E) complete dominance

24) Two pure-breeding phlox plants were crossed, one with dark-blue flowers and the other with pink flowers. The F1 generation all had dark-blue flowers. When the F1 were crossed with each other, the F2 generation consisted of 176 plants: 101 with dark-blue flowers, 33 with light-blue flowers, 30 with red flowers, and 10 with pink flowers.

24.1)

How is flower color controlled in phlox?

A) one gene with four alleles that form a dominance series B) one gene with two codominant alleles C) two genes whose alleles interact additively

24.2) What progeny types would result from crossing pure-breeding plants with light-blue flowers to A) all would be dark blue B) all would be red C) 1 light blue : 1 red

25) In dogs, the dominant allele E (of gene E) specifies dark pigment; the recessive e allele specifies a light pigment (cream). Genes A and K determine how the phenotype Version 1

D) two genes with dominant epistasis

pure-breeding plants with red flowers?

D) 9 dark blue : 3 light blue : 3 red : 1 pink

associated with the E allele is expressed. The dominance series for A 12

gene alleles is AY (fawn) > aw (gray) > at (tan belly). The dominance series for the gene K alleles is Kb (solid color) > kbr (brindled) > ky (gene A markings expressed normally). Note that of all the gene K alleles, only the ky allele allows the phenotypes associated with the gene A genotypes to be

expressed normally. Also, the ee genotype is epistatic to all alleles of genes A and K.

25.1) What is the coat color of a dog with the genotype Ee kyky AYaw? A) cream B) fawn

C) gray D) tan belly

25.2) Which of the following genotypes could a gray dog have?

A) ee kyky awaw B) Ee kyky AYaw

C) EE kyky awat D) Ee Kbkyawaw

25.3) Using the at-home genetic test ‘39Fetch’ you discover that your dog is ee. What color is your dog’s coat?

A) cream B) fawn

C) gray D) tan belly

Answer Key Test name: Chapter 02 Test Bank Version 1

13

Version 1

14

1) [B, D] 2) C 3) A 4) B 5) E 6) C 7) C 8) A 9) D 10) C 11) E 12) A 13) D 14) D 15) D 16) D 17) B 18) C 19) C 20) Section Break 20.1) B Version 1

15

20.2) B 21) Section Break 21.1) C 21.2) B 21.3) C 22) Section Break 22.1) A 22.2) C 23) Section Break 23.1) D 23.2) D 24) Section Break 24.1) C 24.2) A 25) Section Break 25.1) B 25.2) C 25.3) A

Version 1

16

Student name:__________ 1) In Drosophila virilis, somatic cell nuclei contain 12 chromosomes while sperm nuclei contain only 6 chromosomes. What does n equal for this species?

A) 3 B) 6 C) 12

D) 24 E) 6 or 12, depending on cell type

2) The stage of mitosis when chromosomes condense to form rod-shaped structures visible under the microscope is called A) interphase.

B) prophase.

C) metaphase.

D) anaphase.

E) telophase.

3) The stage of mitosis when sister chromatids separate from each other and migrate to opposite poles of a cell is called A) interphase.

B) prophase.

C) metaphase.

D) anaphase.

E) telophase.

4) During which of the following stages of the cell cycle would a chromosome consist of only a single chromatid?

D) mitotic A) G 1 B) G 2 C) mitotic prophase

Version 1

1

metaphase E) All of the choices are correct.

Version 1

2

5) Which of the following statements applies to homologous nonsister chromatids?

A) Their centromeres are attached during meiosis II. B) They are genetically identical.

C) They segregate from each other at anaphase of mitosis.

D) They contain the same genes in the same order

but may have different alleles of some genes.

E) More than one statement applies to homologous chromatids.

6) Drosophila melanogaster has four pairs of chromosomes (including a pair of sex chromosomes). Sperm from this species are formed by a meiotic process in which homologous chromosomes pair and segregate but do not undergo crossing-over. How many genetically different kinds of sperm could be produced by a Drosophila melanogaster male (XY)?

A) 4 B) 8

C) 16 D) 64 E) 256

7) Crossing-over between homologous chromosomes occurs at which of the following stages of meiosis?

A) S phase B) prophase I

Version 1

C) metaphase I D) anaphase I E) prophase II

3

8)

A chromosome with two arms of similar length is A) acrocentric.

B) homologous.

referred to as C) telocentric.

D) metazoan.

E) metacentric.

9) At which of the following stages of meiosis would homologous chromosomes be paired?

A) only prophase of meiosis I B) both prophase of meiosis I and prophase of meiosis II C) both prophase of meiosis I and metaphase of

meiosis II D) only metaphase of meiosis I E) only metaphase of meiosis II

10) Which of the following events occurs during mitosis but not during meiosis?

A) Sister chromatids segregate.

B) Homologous chromosomes pair.

C) Crossing-over occurs between homologous chromosomes.

D) Chromosomes align on the metaphase plate.

E) None of the choices is correct.

11) The mitotic stage during which chromosomes begin attaching to spindle fibers and moving randomly and

reversibly to the centrosomes is

A) prophase.

B) prometaphase.

Version 1

C) metaphase.

D) anaphase.

E) telophase.

4

12) The chromosomal structure to which spindle fibers attach during the mitotic divisions is the D) metaphase A) chromatid.

B) centrosome.

C) kinetochore.

plate.

E) centromere.

13) Microtubules that originate at opposite centrosomes and interdigitate near the cell's equator without attaching to chromosomes are the A) kinetochore microtubules.

B) polar microtubules.

C) astral microtubules.

D) interdigitating microtubules.

14) Mitosis results in _______ chromosome number, whereas meiosis results in _______ chromosome number.

A) a doubling of; no change in B) no change in; no change in C) a reduction by half in; no change in

15)

Cells in the G

0

D) no change in; a doubling of E) no change in; a reduction by half in

stage

A) have two chromatids per chromosome.

B) are replicating their chromosomes.

C) are about to enter the mitotic phase of the cell

D) are in an extended G 1 phase and no longer dividing.

E) are dead.

cycle.

Version 1

5

16) Cells in the G 2 stage of the cell cycle have ______ as cells of the same species in the G 1 stage.

A) twice as many crossovers B) twice as many chromatids C) half as many chromatids

D) the same number of chromatids E) half as many chromosomes

17) Which of the following is not a property of homologous chromosomes?

B) They exchange regions by crossing-over during meiosis.

C) They carry alleles for the same genes at the same chromosomal position.

D) Their centromeres are attached to each other during G 2of the cell cycle.

E) They segregate to opposite poles at anaphase of meiosis I.

18) A variable trait in corn is the presence or absence of knobs at particular sites on some chromosomes. The allele present on each chromosome determines whether that chromosome has a knob. Suppose that one member of each of two pairs of homologs in a corn plant has a knob. If this plant

is crossed with a knobless plant, what percentage of the offspring is expected to have only knobless chromosomes?

A) They pair physically during prophase of meiosis I.

A) 100% B) 75%

C) 50% D) 25% E) 0%

19) What aspect of chromosome behavior most clearly accounts for Mendel's law of segregation?

A) movement of

Version 1

6

sister chromatids to opposite poles at anaphase II of meiosis B) movement of homologous chromosomes to opposite poles at anaphase I of meiosis C) crossing-over between homologous chromosomes during prophase I of meiosis D) replication of chromosomes prior to meiosis

E) independent alignment of different homologous pairs on the metaphase I spindle

20) What aspect of chromosome behavior most clearly accounts for Mendel's law of independent assortment?

A) movement of sister chromatids to opposite poles at anaphase II of meiosis B) movement of homologous chromosomes to opposite poles at anaphase I of meiosis C) crossing-over between nonhomologous chromosomes during prophase I of meiosis

D) replication of chromosomes prior to meiosis E) independent alignment of different homologous pairs on the metaphase I spindle

21) Which aspect(s) of chromosome behavior is/are primarily responsible for the tremendous amount of genetic

variability associated with sexual reproduction?

A) segregation of sister chromatids at anaphase II of meiosis B) segregation of homologous chromosomes at anaphase I of meiosis C) crossing-over between homologous chromosomes during prophase I of meiosis D) independent alignment of different homologous pairs on the metaphase I spindle

E) both crossingover between homologous chromosomes during prophase I of meiosis and independent alignment of different homologous pairs on the metaphase I spindle

22) What is the correct order of these mitotic

Version 1

7

events?

A Chromosomes align on the metaphase plate of the cell.

B Kinetochores begin attaching to spindle fibers.

C Nuclear membrane reforms andchromosomes decondense.

opposite sides of nucleus.

E Sister chromatids separate andmove to opposite poles.

D Chromosomes condense andcentrosomes migrate to A) BDACE B) DABEC

23) What is the correct order of these meiotic events?

A Segregation of homologous chromosomes to opposite poles.

B Segregation of sister chromatids to opposite poles.

C Alignment of homologous pairs on the metaphase plate of the cell.

A) CDEAB B) DCEBA

C) DBAEC D) ABDCE E) EDBAC

D Pairing and synapsis of homologous chromosomes.

E Condensation of chromosomes in a diploid nucleus.

C) EDCBA D) EDCAB E) DCABE

24) A diploid cell with three pairs of chromosomes has the genotype Aa Bb Cc, where each gene is on a different chromosome. If this cell were to undergo meiotic division, how many genetically different types of gametes could be produced?

A) 1 B) 3

Version 1

C) 6 D) 8 E) 9

8

25) A cell with three pairs of chromosomes has the genotype Aa Bb Cc, with each gene on a different chromosome. If this cell were to undergo mitotic division,

A) 1 B) 3

26)

how many genetically different types of daughter cells could be produced?

C) 6 D) 8 E) 9

Which of the following statements is FALSE?

A) Sister chromatids of the same chromosome are identical.

B) Sister chromatids can be nonhomologous chromosomes.

C) Homologous chromosomes are not necessarily identical.

D) An example of a nonhomologous chromosome pair is paternal chromosome 1 and paternal chromosome 2 in a human.

27) At which phase of the cell cycle are sister chromatids generated?

A) G0 B) G1

C) G2 D) S E) M

28) Which term describes chromosomes whose centromeres are located closer to one end of the chromosome?

B) acrocentric A) metacentric

Version 1

9

C) holocentric D) anacentric

Version 1

10

29) Which of the following is a diploid cell destined for a specialized role in the production of gametes?

A) somatic cell B) germ-line cell

C) spermatid D) polar body E) ootid

30) Structures that appear along the synaptonemal complex during pachytene where genetic material is exchanged between nonsister chromatids are the ______.

A) chiasmosomes B) telomeres C) centromeres

D) recombination nodules E) kinetochores

31) The domestic dog has 39 pairs of homologous chromosomes. A dog’s somatic cells have ______ chromosomes and gametes have ______ chromosomes.

A) 78; 78 B) 78; 39

C) 39; 15 D) 156; 78 E) 39; 39

32) Cytokinesis in plant cells occurs by means of a cleavage furrow. ⊚

Version 1

true

⊚

false

11

33) DNA replication occurs during interphase in preparation for mitosis. ⊚ ⊚

true false

34) Mitosis and cytoplasmic division result in the formation of two genetically identical cells. ⊚ ⊚

35)

true false

Germ-line cells are haploid but gametes are diploid. ⊚ ⊚

true false

36) At the end of meiosis I, each daughter cell contains the sister chromatids from only one chromosome of each chromosomal pair. ⊚ ⊚

true false

37) The period between meiosis I and II is termed interkinesis.

Version 1

12

⊚

38)

⊚

false

Crossing-over occurs during prophase I of meiosis. ⊚ ⊚

39)

true

true false

Crossing-over allows the reassortment of linked genes. ⊚ ⊚

true false

40) The region where the homologous chromosomal pairs align is the metaphase plate. ⊚ ⊚

true false

41) The orientation of each chromosomal pair on the spindle axis is random. ⊚ ⊚

42)

true false

Crossing-over occurs during meiosis I and meiosis II.

Version 1

13

⊚

true

⊚

false

43) Chromosomal replication occurs between meiosis I and meiosis II. ⊚ ⊚

true false

44) In animal cells, cell division is accomplished by the formation of a cleavage furrow. ⊚ ⊚

45)

true false

Daughter cells produced in meiosis are identical. ⊚ ⊚

true false

46) Chromosomal replication occurs prior to both mitosis and meiosis. ⊚ ⊚

true false

Answer Key Version 1

14

Test name: Chapter 03 Test Bank 1) B 2) B 3) D 4) A 5) D 6) C 7) B 8) E 9) C 10) E 11) B 12) C 13) B 14) E 15) D 16) B 17) D 18) D 19) B

Version 1

15

20) E 21) E 22) C 23) D 24) D 25) A 26) B 27) D 28) B 29) B 30) D 31) B 32) FALSE 33) TRUE 34) TRUE 35) FALSE 36) TRUE 37) TRUE 38) TRUE 39) TRUE 40) TRUE Version 1

16

41) TRUE 42) FALSE 43) FALSE 44) TRUE 45) FALSE 46) TRUE

Version 1

17

Student name:__________ 1) Nondisjunction can occur at either the first or second division of meiosis. XXY individuals could arise from nondisjunction at the ______ meiotic division in the ______.

A) first; mother B) second; mother

C) second; father D) first; father

2) Which male and female organs are derived from the same bipotential sex organ precursors?

A) uterus and prostate B) clitoris and penis C) cervix and seminal vesicles

D) fallopian tubes and vas deferens

3) If an XY individual inherits nonfunctional alleles of the AMFR gene from both parents, A) the individual does not develop any internal sex organs.

B) the individual develops female internal sex organs.

C) the individual develops female external sex

E) the individual develops male external sex organs.

F) the individual does not develop any external sex organs.

organs.

D) the individual develops male internal sex organs.

4) Which types of chromosomes provide the basis for sex determination in most mammals?

C) autosomes A) haploid set of chromosomes B) diploid set of chromosomes Version 1

1

D) sex chromosomes E) homologous chromosomes

Version 1

2

5) Nondisjunction can occur at either the first or second division of meiosis. XYY individuals would most likely arise from nondisjunction at the ______ meiotic division in the ______.

A) first; mother B) second; mother C) first; father

D) second; father E) More than one of the choices could give rise to XYY individuals.

6) Premeiotic germ cells that divide mitotically in females are A) primary oocytes.

B) secondary oocytes.

7)

C) ootids.

D) oogonia.

E) ova.

The cells that undergo meiosis in males are A) spermatocytes.

B) spermatogonia.

C) spermatids.

D) oocytes.

E) sperm.

8) Normal fruit flies have red eyes. Flies with mutations that block the function of the X-linked white gene have white eyes. The wild-type allele is completely dominant to mutant alleles. A cross between a white-eyed female fruit fly and a red-eyed male would generate which of the following types of progeny?

Version 1

3

A) red-eyed females and white-eyed males B) white-eyed females and red-eyed males C) all red-eyed females and a 1:1 mixture of whiteeyed and red-eyed males D) all red-eyed males and a 1:1 mixture of whiteeyed and red-eyed females

E) The result cannot be predicted because it depends on whether the female is homozygous or heterozygous.

9) Red-green color blindness is controlled by an X-linked gene in humans. The allele that causes color blindness is recessive to the allele for normal vision. A man and woman both with normal vision had color-blind fathers. If this man and woman have a child, what is the probability that the child will be color blind?

A) 1/2 B) 1/4

10) Hemophilia is caused by an X-linked recessive mutation in humans. If a woman whose paternal uncle (father's brother) was a hemophiliac marries a man whose brother is also a hemophiliac, what is the probability that their first child will have hemophilia? (Assume that no other cases A) 0 B) 1/4

C) 1/8 D) 1/3 E) 0

of hemophilia and no hidden carriers exist in the pedigree.)

C) 1/2 D) 1/8 E) 1

11) Normal fruit flies have red eyes. Flies with mutations that block the

Version 1

4

function of the X-linked white gene have white eyes. The wild-type allele is completely dominant to mutant alleles.

White-eyed female fruit flies are mated to red-eyed males.

One of the hundreds of female progeny is white-eyed. What is

the likely karyotype of this white-eyed daughter?

C) XO D) XXX E) XYY

A) XX B) XXY

12) Normal fruit flies have red eyes. Flies with mutations that block the function of the X-linked white gene have white eyes. The wild-type allele is completely dominant to mutant alleles. White-eyed females are mated to red-eyed males. One of the hundreds of female progeny is white-eyed.

This white-eyed daughter likely arose from nondisjunction of

the sex chromosomes during _________ in the __________.

D) meiosis II; A) meiosis I; mother B) meiosis II; mother C) meiosis I; father

father E) meiosis I or II; mother

13) Males with one copy of an X-linked gene are said to be _________ for that gene.

A) homozygous B) heteroallelic

C) heterozygous D) hemizygous E) deficient

14) In chickens, females have two different sex chromosomes (Z and W), while the males have two Z chromosomes. A ZVersion 1

5

linked gene controls the pattern of the feathers, with the dominant B allele causing the barred pattern and the b allele causing nonbarred feathers. Which of the following crosses would produce all daughters of one type (barred or nonbarred) A) barred females × nonbarred males B) nonbarred females × barred males C) nonbarred females × nonbarred males

15) Suppose you discover a new species of worm that exists in two forms—slimy and nonslimy. You find that mating slimy females to nonslimy males produces offspring consisting of slimy males and nonslimy females, whereas mating nonslimy females with slimy males produces offspring of both sexes that are all slimy. You would conclude that the _________ allele is dominant and that ________ are the

and all sons of the other type?

D) barred females × barred males E) More than one ofthe choices is correct.

heterogametic sex (the sex with two different sex chromosomes) in this species of worm.

D) nonslimy; A) slimy; females B) slimy; males C) nonslimy; females

16) Suppose you discover a mouse that has spiky fur instead of the usual soft fur. You notice that this characteristic seems to be present only in males. To investigate this pattern, you cross a spiky-fur male with a soft-fur female, and find that all the F 1 progeny of both sexes have soft fur. You then interbreed the F 1 and observe that all the F 2 females have A) an X-linked recessive.

B) Y-linked.

C) autosomal recessive with sex-influenced expression.

Version 1

males

soft fur, but 1/4 of the F 2 males have spiky fur. You conclude that the spiky allele is

D) an X-linked dominant.

E) an autosomal dominant with sexinfluenced expression.

6

17) In fruit flies, brown eyes can be caused by recessive mutant alleles of any one of three genes: pn (prune), bw (brown), or ry (rosy). The pn gene is X-linked, bw is on the second chromosome, and ry is on the third chromosome. One wild-type allele of each of the three genes must be present for eyes to be red (wild type). Suppose that two brown-eyed flies are crossed and their progeny consist of brown-eyed sons and red-eyed daughters. Which mutation is responsible for the brown eyes in the parental female?

A) pn B) bw C) ry

D) either bw or ry E) any of the three —pn, bw, or ry

18) Color vision depends on dominant alleles of three different genes: the R gene and the G gene, both on the X chromosome, and the B gene, which is autosomal. Recessive mutant alleles of any one of these three genes can cause color blindness in homozygotes. Suppose a color-blind man marries a color-blind woman and all their offspring have normal vision. What is the genotype of the woman?

A) RR GG bb B) RR gg BB C) rr GG bb

19)

D) RR gg BB or RR GG bb E) rr GG bb or rr gg BB

Which sex chromosomes are present in only one sex?

B) X and W A) X and Z

Version 1

7

C) Y and Z D) Y and W E) X and Y

Version 1

8

20) In the following pedigree, the indicated trait is a rare disease. It is most likely caused by which type of allele?

A) autosomal recessive B) autosomal dominant C) X-linked recessive

D) X-linked dominant E) Y-linked

21) In the following pedigree, the indicated trait is a rare disease. It is most likely is caused by which type of allele?

A) autosomal

Version 1

9

recessive B) autosomal dominant C) X-linked recessive

D) X-linked dominant E) Y-linked

22) In the following pedigree, the indicated trait is a rare disease. It is most likely caused by which type of allele?

A) autosomal recessive B) autosomal dominant C) X-linked recessive

D) X-linked dominant E) Y-linked

23) In the following pedigree, the indicated trait is a rare disease. It is most likely caused by which type of allele?

Version 1

10

A) autosomal recessive B) autosomal dominant C) X-linked recessive

D) X-linked dominant E) Y-linked

24) In the following pedigree, the indicated trait is a rare disease. It is caused by what type of allele?

A) autosomal recessive B) autosomal dominant C) X-linked recessive

D) X-linked dominant E) Y-linked

25) In fruit flies, an X-linked dominant mutant allele of the Notch gene (N) causes Notched wings in heterozygous females but is lethal in hemizygous or homozygous condition.

What ratio of offspring would be observed in a cross of a Notched-wing female with a normal male?

D) 1/2 normal A) 1/3 Notched-wing females, 1/3 normal females, 1/3 normal males B) 1/4 Notched-wing females, 1/4 normal females, 1/4 Notched-wing males, 1/4 normal males C) 1/2 Notched-wing females, 1/2 normal males

Version 1

11

females, 1/2 Notched-wing males E) 2/3 Notched-wing females, 1/3 normal males

Version 1

12

26) In humans, XO individuals are females with Turner syndrome and XXY individuals are males with Klinefelter syndrome. Which of the following events cannot give rise to a Klinefelter male?

A) nondisjunction at meiosis I in the mother B) nondisjunction at meiosis II in the mother C) nondisjunction at meiosis I in the father D) nondisjunction at meiosis II in the father

27) In humans, XO individuals are females with Turner syndrome and XXY individuals are males with Klinefelter syndrome. Red-green color blindness is caused by an Xlinked recessive allele. Suppose a color-blind man and a woman with normal vision and no family history of color A) nondisjunction at meiosis I in the mother B) nondisjunction at meiosis II in the mother C) nondisjunction at meiosis I in the father D) nondisjunction at meiosis II in the father

E) All of thechoices could give rise to a Klinefelter male.

blindness had a daughter who was color blind and had Turner syndrome.

Which event could have given rise to this offspring?

E) nondisjunction ateither meiosis I or meiosis IIin the mother

28) Individuals with an XXY karyotype are ______ in humans and ______ in fruit flies.

A) male; male B) male; female C) female; male

D) female; female E) male; intersexual

29) Version 1

In animal 13

gametogenesis, a single primary spermatocyte generates ______ sperm, while a single primary oocyte generates ______ egg(s).

C) 4; 2 D) 4; 1 E) 4; 4

A) 1; 4 B) 1; 1

30) In which of the following cases will a Barr body be seen?

A) only XX B) XY C) XO

D) only XXY E) both XX and XXY

31) You examine cells with a microscope and detect two Barr bodies in each cell. What is the most likely genotype of the cells?

A) XX B) XY

C) XO D) XXY E) XXX

32) You engineer an XY mouse in which the SRY gene is deleted. What do you expect to observe in this mouse?

X chromosome.

A) The mouse will develop male anatomic features.

B) The mouse will develop female anatomic features.

C) The mouse’s Y chromosome will convert into an

Version 1

14

D) The mouse’s X chromosome will convert into a Y chromosome.

E) The mouse will exhibit both male and female morphological features.

Version 1

15

33) In human oogenesis, how many chromosomes does a primary oocyte carry?

A) 22 B) 23

C) 44 D) 46 E) 92

34) In human oogenesis, how many chromosomes does an oogonium carry?

A) 22 B) 23

C) 44 D) 46 E) 92

35) In human oogenesis, how many chromosomes does a secondary oocyte carry?

A) 22 B) 23

C) 44 D) 46 E) 92

36) In human oogenesis, how many chromosomes does an ovum carry?

C) 44 A) 22 B) 23 Version 1

16

D) 46 E) 92

Version 1

17

37) Suppose you discovered a new mutation in mice that causes a curved spine. You noticed that this mutant phenotype is present only in females. When curved-spine females are crossed with normal males, the progeny are always recovered in a 1:1:1 ratio of curved-spine females, normal females, and normal males. Explain the genetic basis for this ratio.

A) hemizygous or homozygous lethal autosomal allele B) hemizygous or homozygous lethal sex-linked

D) recessive autosomal allele E) dominant sexlinked allele

allele C) recessive sex-linked allele

38) Individuals with Turner syndrome have which of the following sex chromosome complements?

A) XO B) XX

C) XYY D) XXY E) XXX

39) Which of the following is a diploid cell destined for a specialized role in the production of gametes?

A) somatic cell B) germ-line cell

Version 1

C) spermatid D) polar body E) ootid

18

40) An XY individual with a loss-of-function mutation in the SRY gene is expected to bemorphologically female. ⊚ ⊚

41)

true false

Germ-line cells are haploid but gametes are diploid. ⊚ ⊚

Version 1

true false

19

Answer Key Test name: Chapter 04 Test Bank 1) [A, B, D] 2) [B] 3) [B, D, E] 4) D 5) D 6) D 7) A 8) A 9) B 10) A 11) B 12) E 13) D 14) A 15) A 16) C 17) A 18) A 19) D Version 1

20

20) C 21) D 22) B 23) A 24) E 25) A 26) D 27) E 28) B 29) D 30) E 31) E 32) B 33) D 34) D 35) B 36) B 37) B 38) A 39) B 40) TRUE Version 1

21

41) FALSE

Version 1

22

Student name:__________ 1) The measured distance between genes D and E in a two-point testcross is 50 map units. Which describe the relationship between genes D and E? A) D and E are on the same chromosome, at least 50 map units apart.

B) D and E are on the same chromosome, exactly 50 map units apart.

C) D and E are on different homologous

chromosomes.

D) D and E are on the same chromosome, less than 50 map units apart

2) If the recombination frequency between two genes is close to 50%, what could be true about the location of the two genes?

A) they are on nonhomologous chromosomes B) they are very close together on the same chromosome C) they are far apart on the same chromosome

3)

D) none of the choices could be true

Which process(es) can generate recombinant gametes?

A) segregation of alleles in a homozygote B) crossing-over between two linked heterozygous

loci

heterozygous loci D) crossing-over between two linked homozygous loci

C) independent assortment of two unlinked

4)

In tetrad analysis, NPD asci can result from C) single A) independent assortment of unlinked genes.

B) single crossovers between linked genes.

Version 1

1

crossovers between a gene and the centromere.

D) double crossovers between linked genes.

Version 1

2

5) Individuals heterozygous for the RB alleles can develop tumors as a result of

+

and RB

−

−

A) a somatic mutation in the RB allele that leads to homozygosity for RB +.

B) the fact that RB − is dominant to RB +.

C) a mitotic crossover that leads to homozygosity for + RB in some cells and RB − in other cells.

D) a somatic mutation in the RB + allele that leads to homozygosity for RB −.

6) Which type of tetrad contains two recombinant and two parental spores?

A) PD B) NPD C) T D) ordered tetrads

E) None of these types contain two recombinant and two parental spores.

7) The R and S genes are linked and 10 map units apart.

In the cross R s / r S ×r s / r s what fraction of the progeny will be R S / r s?

A) 5% B) 10%

C) 25% D) 40% E) 45%

8) If the map distance between genes A and B is 20 map units and the map distance between genes B and C is 35 map units, what is the map distance between genes A and C?

Version 1

3

A) 15 map units B) 55 map units C) More information is needed to distinguish between 15 and 55 map units.

D) Gene C must be located on a different nonhomologous chromosome.

9) Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the crossL M/l m ×l m/l m would be L m/l m?

A) 10% B) 25%

C) 50% D) 75% E) 100%

10) 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

11) The zipper-like connection between paired homologs in early prophase I is known as a

A) spindle fiber.

B) synaptic junction.

Version 1

C) synaptonemal complex.

D) chiasma.

4

E) None of the choices is correct.

Version 1

5

12) The Rand Sgenes are linked and 10 map units apart. In the cross R s / r S × r s / r s what percentage of the progeny will be R s / r s?

A) 5% B) 10%

C) 25% D) 40% E) 45%

13) In a dihybrid testcross (Hh Ii × hh ii), four types of offspring are produced: H I, H i, h I, and h i. What will be the degrees of freedom in a chi-square test for goodness of fit for the null hypothesis that H and I are unlinked?

A) 1 B) 2

14) Crossing-over takes place in bivalents (tetrads) consisting of ______ chromatids, and one crossover involves A) 2; 2 B) 2; 4

15) 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+ / y car), a single chiasma is observed somewhere along the X

Version 1

C) 3 D) 4 E) 0

_______ nonsister chromatids.

C) 4; 2 D) 4; 4 E) 8; 4

chromosome in 90% of the oocytes examined. No X chromosomes with multiple chiasmata are

6

observed. What fraction of the male progeny from such a

A) 5% B) 10%

female would be recombinant for y and car?

C) 45% D) 55% E) 90%

16) The Q gene is 10 map units from the R gene which is 40 map units from the S gene: 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.

17) The map of a chromosome interval is: A——10 m.u.——B——40 m.u.——C From the cross A b c / a B C × a b c / a b c, how many double crossover chromosomes would be expected among 1000 progeny?

A) 5 B) 10

C) 20 D) 40 E) 80

18) The cross L p q / l P Q × l p q / l p q is carried out. If the L gene is in the middle, between genes P Version 1

7

and Q, what would be the genotypes of the double crossover

A) L P Q and l p q B) L p Q and l P q C) l p Q and L P q

gametes in this cross?

D) L p q and l P Q E) cannot be determined

19) Suppose a three-point testcross was conducted involving genes X, Y, and Z. If the most abundant classes of progeny are X Y z and x y Z and the rarest classes are x Y Z and X y z, which gene is in the middle?

A) X B) Y

C) Z D) cannot be determined

20) In Drosophila, the genes b, c, and sp are linked and arranged as shown below: b——30 m.u.——c——20 m.u.——sp This region exhibits 90% interference. Among 1000 progeny of a three-point cross involving b, c, and sp, how many double crossover chromosomes would be recovered?

A) 3 B) 6

C) 54 D) 60 E) 600

21) In tetrad analysis, which result would indicate that two genes are linked?

A) NPD = T.

Version 1

8

B) PD = T. C) PD = NPD.

D) PD > NPD. E) PD > T.

22) When analyzing octads (ordered tetrads), seconddivision (MII) segregations result from

A) single crossovers between linked genes.

B) double crossovers between linked genes.

C) single crossovers between a gene and a centromere.

D) independent assortment of unlinked genes.

E) nondisjunction of homologs.

23) Tetrad analysis shows that crossing-over occurs at the four-strand stage (i.e., after replication) because, when two genes are linked, A) NPD > T. B) T > NPD.

C) T > PD. D) PD > NPD. E) PD > T.

24) Sturtevant's detailed mapping studies of the X chromosome of Drosophila supported what genetic principle?

A) That genes are arranged in a linear order on the chromosomes.

B) That genes arecarried on chromosomes.

C) That sex determination is controlled by the X and Y chromosomes.

D) That segregation of an allelic gene pair is

Version 1

accompanied by disjunction of homologous chromosomes.

E) That different pairs of chromosomes assort independently.

9

25) Suppose an individual is heterozygous for alternate alleles of gene A (Aa). Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells that are homozygous for a? (Choose the most precise answer.)

A) The crossover would have to occur between the A locus and the centromere and involve two homologous (nonsister) chromatids.

B) The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (nonsister) chromatids.

C) The crossover would have to occur between the A locus and the end of the chromosome and involve two

nonhomologous chromosomes.

D) The crossover would have to occur between the A locus and the centromere and involve two sister chromatids.

26) If an individual is heterozygous at two loci (A b / a B) that are located on the same chromosome arm with A closer to the centromere than B, under what conditions would a crossover in a somatic cell generate a twin spot?

A) The crossover would have to occur between the A locus and the B locus and involve two homologous, nonsister chromatids.

B) The crossover would have to occur between the B locus and the end of the chromosome and involve two homologous, nonsister chromatids.

C) The crossover would have to occur between the A locus and the centromere and involve two homologous, nonsister chromatids.

D) A double crossover would have to occur, with one

Version 1

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, nonsister chromatids.

E) No crossover in a somatic cell could generate a twin spot.

10

27) A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap) and gray body (gy). The F1 mice were wild-type for both traits. When the F1 were interbred, the F2 were distributed as follows: Females

all wild type

200

Males

wild type

91

Version 1

apricot

11

gray

9

apricot, gray

89

Which of the following statements is correct?

11

A) ap and gy are unlinked.

B) ap and gy are linked on an autosome and 10 map units apart.

C) ap and gy are linked on an autosome and 20 map units apart.

D) ap and gy are X-linked and 10 map units apart.

E) ap and gy are X-linked and 20 map units apart.

28) Suppose the map for a particular human chromosome interval is: a——1 m.u.——b——1 m.u.——c——1 m.u.——d——1 m.u.——e——1 m.u.——f In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a–f interval?

A) 0% B) 1% C) 2.5%

D) 5% E) cannot be determined

29) Mitotic recombination occurs between homologous chromosomes. In which of the following would you not expect to encountermitotic recombination?

A) E. coli B) tobacco plants

30) In some fungi, such as the bread mold Neurospora crassa, the arrangement of spores in the ascus directly reflects the order in which they were produced during meiosis. What

Version 1

C) humans D) Drosophila melanogaster

is the collection of spores produced by meiosis in Neurospora crassa called?

12

A) unordered tetrad B) unordered octad

C) ordered tetrad or ordered octad D) ordered pentad

31) Twin spotting provides evidence of what genetic event?

A) meiotic recombination B) mitotic recombination C) linkage

D) mutation E) biological evolution

32) Another name for a chromosome is a _______, because it contains alleles that are often inherited together.

A) linkage group B) crossing-over group

33) The diploid garden pea plant has 14 chromosomes.

The haploid fungus Neurospora crassa has 7 chromosomes.

Neither organism has separate male and female individuals.

A) the garden pea has 14 linkage groups, and Neurospora has 7.

B) the garden pea has 7 linkage groups, and Neurospora has 7.

C) the garden pea has 8 linkage groups, and

C) geneticrecombinant D) bivalent

Therefore, the number of linkage groups in these two organisms is Neurospora has 8.

D) the garden pea has 15 linkage groups, and Neurospora has 8.

34)

Version 1

Suppose that

13

progeny classes in a dihybrid testcross ( Aa Bb × aa bb) do not appear in a 1:1:1:1 ratio. What does this suggest?

A) Alleles at the two loci, A and B, assort independently and the two genes are present on nonhomologous chromosomes.

B) Alleles at the two loci, A and B, do not assort independently and the two genes are present on nonhomologous chromosomes.

C) Alleles at the two loci, A and B, do not assort independently and the two genes are linked and present on the

35) In a region of a chromosome, three genes are arranged in the order A--- B--- C, where the physical distance (in base pairs) between genes A and B is the same as the physical distance between genes B and C. However, a genetic map of this region shows that genes A and B are much farther apart from each other than are genes B and C. A) There is a higher probability of double crossovers occurring between genes A and B compared to between genes B and C. B) There is a higher probability of single crossovers occurring between genes A and B compared to between genes B and C. C) There is a higher probability of single crossovers

36) A two-point testcross was performed. Based on the frequencies of progeny in each class, a chi-square test for goodness of fit allows you to accept the null hypothesis that the two genes are unlinked. Which ONE of the following is A) There are 3 degrees of freedom.

B) The expected frequencies of progeny in each class

Version 1

same chromosome.

D) Alleles at the two loci, A and B, assort independently and the two genes are present on the same chromosome.

Which one of the following is a correct statement to explain these observations?

occurring between genes B and C compared to between genes A and B. D) The genetic map is much more accurate than the physical map.

sufficient to allow you to accept the null hypothesis?

differs greatly from the observed frequencies.

C) The probability

14

is less than 1/100 that the expected and observed frequencies occurred by chance.

D) The probability is greater than 1/10 that the expected and observed frequencies occurred by chance.

E) The expected frequencies for all four progeny classes are the same.

37) In Drosophila, the genes y, f, and v are all X-linked.

Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are crossed to y f v / y f v males.

The F2 are distributed as follows: yfv

3,210

y f v+

72

+

yf v

1,024

y f+v+

678

+

y fv

690

y+f v+

1,044

+ +

yfv

60

y+f+v+

3,222 10,000

Version 1

15

37.1) Which of the following linkage maps correctly shows the order and distance between the y, f, and v A) f——35 m.u.——y——15 m.u.——v B) f——22 m.u.——y——15 m.u.——v C) y——35 m.u.——f——22 m.u.——v D) y——22 m.u.——v——15 m.u.——f

genes?

E) y——15 m.u.

——v——22 m.u.

——f

37.2) What is the coefficient of coincidence in this region?

C) 0.4 D) 0.6 E) 0.8

A) 0 B) 0.2

38) Females heterozygous for the recessive mutations px, sp, and cn on the second chromosome and normal alleles are mated to a male homozygous for all three mutations. The offspring are as follows: px sp cn

1,461

px sp cn+

3,497

+

px sp cn

1

px sp+cn+

11

+

px sp cn

9

px+sp cn+

0

+

+

px sp cn

px+sp+cn+

1,539 10,000

3,482

38.1) What is the genotype of the females that gave rise to these progeny?

Version 1

16

A) px+ sp cn / px sp+ cn+ B) px+ sp cn+ / px sp+ cn C) px+ sp+ cn+ / px sp cn

38.2)

D) px sp cn+ / px+ sp+ cn E) insufficient data

Which of the three genes is in the middle?

A) px B) sp

C) cn D) insufficient data

38.3) Which of the following linkage maps correctly shows the order and approximate distance between the px, sp, and cn genes?

A) sp——0.21 m.u.——px——30 m.u.——cn B) sp——30 m.u.——px——0.21 m.u.——cn C) sp——0.30 m.u.——px——30 m.u.——cn D) px——0.21 m.u.——sp——30 m.u.——cn

E) px——30 m.u.

——sp——0.21 m.u.——cn

38.4) What is the coefficient of coincidence in this region?

A) 0 B) 0.16

Version 1

C) 0.33 D) 0.50 E) 0.66

17

39) In peas, tall (T) is dominant to short (t), red flowers (R) is dominant to white flowers (r), and wide leaves (W) is dominant to narrow leaves (w). A tall plant that has red flowers and wide leaves is crossed to a short plant that has tall, red, wide

381

tall, white, wide

122

short, red, wide

118

short, white, wide

379 Total

Version 1

white flowers and narrow leaves. The resulting progeny are shown in the table.

1000

18

39.1)

What is the genotype of the tall plant that has

red flowers and wide leaves?

C) T R W / T R W D) T R W / T r w E) T r W / t R W

A) T R W / t r w B) T R W / t r W

39.2) This cross is not useful to determine if one of the genes is linked to the others. Which gene?

D) This cross shows that all three genes are linked.

A) gene T B) gene R C) gene W

39.3) If two or more of the genes are linked, what is the map distance that separates them?

D) 50 m.u.

E) None of the genes is linked to another.

A) 4 m.u.

B) 12 m.u.

C) 24 m.u.

40) A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table.

Parent genotypes:

Cc Dd × cc dd

Progeny genotypes:

Cc Dd

222

Cc dd

280

Version 1

cc Dd

280

cc dd

218

19

1000

40.1) The chi-square value is the sum for all progeny classes of (Observed-Expected)2 / Expected. Using the chi-square test for goodness of fit, calculate the chisquare value to test the null hypothesis that genes C and D are unlinked. What is the chi-square value?

A) 0 B) 0.0576 C) 10.8

D) 14.4 E) cannot be determined

40.2) Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

A) 1 B) 2

Version 1

C) 3 D) 4

20

40.3) Given this data, use Table 5.2 to find the most accurate range within which the p value falls.

A) 0.001 < p < 0.01 B) 0.01 < p < 0.05

C) 0.05 < p < 0.10 D) 0.10 < p < 0.50

40.4) What is a reasonable conclusion based on the chi-square analysis?

A) A high probability exists that the deviation from the expected results is due to chance.

B) One can say with a high degree of confidence that genes C and D are linked.

C) The null hypothesis is probably true.

D) Genes C and D are most likely unlinked.

40.5) If only 100 progeny had been counted and the same proportions of progeny genotypes observed, how would the p value and the conclusion drawn about linkage change?

A) The p value would increase, and the likelihood of linkage decreases.

B) The p value would decrease, and the likelihood of linkage increases.

C) Neither the p value nor the likelihood of linkage

41) In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive alleles of X-linked genes. The wild-type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous

Version 1

would change.

D) The p value would decrease, and the likelihood of linkage decreases.

for sn and ct+ is crossed to

21

a sn+ct male. The F1 flies are interbred. The F2 males are

distributed as follows:

sn ct

13

sn+ ct

39

sn ct+

36

sn+ ct+

12

41.1)

What is the map distance between sn and ct?

C) 25 m.u.

D) 50 m.u.

E) 75 m.u.

A) 12 m.u.

B) 13 m.u.

41.2) Of the four genotypic classes of offspring, which arose from a parental gamete produced by the F1 females?

+

A) sn ct B) sn ct

C) sn + ct D) sn ct +

+

42) In Drosophila, singed bristles (sn) and carnation eyes (car) are mutant phenotypes caused by recessive alleles of Xlinked genes. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred. The F2 are distributed as follows: sn car

55

sn car+

45

sn+ car

45

sn+ car+

55

Version 1

Total 200

22

42.1) If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

A) Genes sn and car are linked.

B) Genes sn and car are unlinked.

C) Genes sn and car are located close together on the same chromosome.

D) Crossing-over sometimes occurs between sn and

car.

E) The recombination frequency between sn and car is 100%.

42.2) What is theχ2 value for a chi-square test for goodness of fit of the null hypothesis?

A) 0.5 B) 1.0

Version 1

C) 2.0 D) 0.4 E) 20

23

42.3)

What is the p value from this test? (Choose the

A) p > 0.5 B) 0.1 < p < 0.5

most accurate answer.)

C) p < 0.1 D) p < 0.05 E) p < 0.01

42.4) What does the data analysis allow you to conclude about linkage between sn and car?

A) A high probability exists that the deviations from the expected number of F 2 in each genotype class are due to chance.

B) The p value is high meaning that the data is significant.

C) Good evidence exists that cn and car are linked.

D) The data do not allow rejection of the null hypothesis.

43) In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

43.1) Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the A) H R / h r B) H r / h r

Version 1

probable genotype of the woman?

C) h r / h R D) H r / h R E) H R / H r

24

43.2) A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy.

If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the

baby will have normal vision and normal blood clotting?

A) 0 B) 0.03

C) 0.485 D) 0.47 E) 0.015

44) In Drosophila, the autosomal recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). Flies who are homozygous recessive at both pr and cn have orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colors: wild-type

8

brown

241

bright-red

239

orange

12

44.1)

Total 500

Which phenotypes are parental?

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.

44.2) What is the genotype of the wild-type mother of these progeny?

B) pr+ cn / pr+ cn A) pr cn / pr+ cn+

Version 1

25

C) pr+ cn / pr cn+ D) pr cn+ / pr cn+ E) pr cn / pr cn

Version 1

26

44.3) The mother of these progeny resulted from a cross between two flies from true-breeding lines. What +

+

+

+

A) pr cn / pr cn and pr cn / pr cn B) pr+ cn+ / pr+ cn+ and pr cn / pr cn C) pr+ cn+ / pr cn and pr cn / pr cn D) pr+ cn / pr cn and pr cn+ / pr cn

are the genotypes of these two lines?

E) More than one of these could be true.

44.4) 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.

45) Consider a pair of homologous chromosomes heterozygous for three genes (e.g. A B C / a b c) 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.

45.1) A crossover that would result in genetic recombination (e.g., A b c or a B C) could involve A) 1 & 2 B) 1 & 3 C) 1 & 4

Version 1

which pair of chromatids?

D) 2 & 3 E) 2 & 4 F) 3 & 4

27

45.2) Assume a double crossover occurs in this pair of chromosomes that results in chromatids of the genotypes A b C and a B c. If the first crossover (the one between A and B) involves chromatids 1 ( A B C) & 4 ( a b c), which chromatids could be involved in A) 1 & 2 B) 1 & 3 C) 1 & 4

the second crossover?

D) 2 & 3 E) 2 & 4 F) 3 & 4

46) A haploid yeast strain has mating type a and genotype his4 TRP1. A different haploid yeast strain has mating type α and genotype HIS4 trp1. The two strains mate and produce diploid offspring with genotype his4 TRP1 / HIS4 trp1. The diploid cells undergo meiosis to produce a variety of tetrad types. In the following questions, the genotypes of the spores in a tetrad are shown. For each tetrad, tell what type it is: parental (PD), non-parental (NPD), or tetratype (T).

46.1)

his4 TRP1; his4 trp1; HIS4 trp1; HIS4 TRP1

A) PD B) NPD

46.2)

C) T D) cannot be determined

his4 TRP1; his4 TRP1; HIS4 trp1; HIS4 trp1 A) PD

Version 1

28

B) NPD C) T

46.3)

his4 trp1; his4 trp1; HIS4 TRP1; HIS4 TRP

A) PD B) NPD

46.4)

C) T D) cannot be determined

HIS4 trp1; HIS4 TRP1; HIS4 TRP1; his4 trp1

A) PD B) NPD

46.5)

D) cannot be determined

C) T D) none of the above

HIS4 trp1; HIS4 trp1; HIS4 trp1; his4 TRP1

A) PD B) NPD

C) T D) none of the above

47) In Neurospora, the white-spore allele (ws) results in white spores rather than the wild-type black spores. Diploid ws / ws+ cells undergo meiosis Version 1

29

followed by one round of mitosis to form ordered tetrads.

Indicate if the pattern of spores seen in the tetrads are

47.1)

consistent with firstdivision or second-division segregation.

ws; ws; ws; ws; ws+; ws+; ws+; ws+

A) first-division segregation pattern B) second-division segregation pattern

47.2)

ws; ws; ws+; ws+; ws; ws; ws+; ws+

A) first-division segregation pattern B) second-division segregation pattern

47.3)

ws; ws; ws+; ws+; ws+; ws+; ws; ws

A) first-division segregation pattern B) second-division segregation pattern

47.4)

ws+; ws+; ws+; ws+; ws; ws; ws; ws B) second-

A) first-division segregation pattern

Version 1

30

division segregation pattern

47.5)

ws+; ws+; ws; ws; ws+; ws+; ws; ws

A) first-division segregation pattern B) second-division segregation pattern

48) Two genes are considered linked when more F 2 progeny with recombinant genotypes than parental genotypes ⊚ ⊚

appear in the offspring of a dihybrid testcross.

true false

49) Chiasmata are structures that show where recombination occurred between sister chromatids. ⊚ ⊚

true false

50) Chiasmata can be seen through a light microscope and are sites of recombination. ⊚ ⊚

Version 1

true false

31

51) A linkage group includes all of the genes on a chromosome, including genes that are so far apart from each other on a chromosome that they assort independently during meiosis. ⊚ ⊚

true false

52) The hypothesis that predicts no linkage between genes is the null hypothesis in certain chi-square tests. ⊚ ⊚

true false

53) If the p value corresponding to a given χ2 value and number of degrees of freedom is lower than 0.05, then the null hypothesis is rejected and it can be said with some confidence that the data are not a good fit to the null hypothesis. ⊚ ⊚

54)

true false

Genes that are not syntenic are unlinked. ⊚ ⊚

Version 1

true false

32

55) The genetic distance from one end of a linkage group and the other may exceed 50 m.u. because the distances between many gene pairs are added together to make the map. ⊚ ⊚

true false

56) Sectors with different phenotypes from one another in an otherwise uniform yeast colony may be evidence of mitotic recombination. ⊚ ⊚

true false

57) Large sectors suggest that mitotic recombination occurred late in the growth of a yeast colony. ⊚ ⊚

true false

58) A coefficient of coincidence of 0.5 in a region of three genes means that half as many double crossovers were observed as would have been expected if crossovers in the two intervals were independent. ⊚

Version 1

true

⊚

false

33

Version 1

34

Answer Key Test name: Chapter 05 Test Bank 1) [A, C] 2) [A, C] 3) [B, C] 4) [A, D] 5) [C, D] 6) [C] 7) A 8) C 9) B 10) B 11) C 12) E 13) C 14) C 15) C 16) B 17) D 18) A 19) B Version 1

35

20) B 21) D 22) C 23) B 24) A 25) A 26) C 27) D 28) D 29) A 30) C 31) B 32) A 33) B 34) C 35) B 36) D 37) Section Break 37.1) E 37.2) C 38) Section Break Version 1

36

38.1) D 38.2) A 38.3) A 38.4) D 39) Section Break 39.1) B 39.2) C 39.3) C 40) Section Break 40.1) D 40.2) C 40.3) A 40.4) B 40.5) A 41) Section Break 41.1) C 41.2) [C, D] 42) Section Break 42.1) B 42.2) C 42.3) A Version 1

37

42.4) [A, D] 43) Section Break 43.1) D 43.2) E 44) Section Break 44.1) B 44.2) C 44.3) A 44.4) C 45) Section Break 45.1) [B, C, D, E] 45.2) [C] 46) Section Break 46.1) C 46.2) A 46.3) B 46.4) D 46.5) D 47) Section Break 47.1) A 47.2) B Version 1

38

47.3) B 47.4) A 47.5) B 48) FALSE 49) FALSE 50) TRUE 51) TRUE 52) TRUE 53) TRUE 54) TRUE 55) TRUE 56) TRUE 57) FALSE 58) TRUE

Version 1

39

Student name:__________ 1)

Which of the following is not true of DNA? A) It is acidic.

B) It contains deoxyribose.

C) It is found in cell nuclei.

2)

The molecule of heredity is usually A) RNA. B) DNA. C) protein.

3)

C) nucleotides.

D) nucleosides.

E) polymers.

DNA is localized mainly in the A) cell membrane.

B) endoplasmic reticulum.

5)

D) carbohydrate.

E) None of the choicesis correct.

The four subunits that make up DNA are called A) phosphodiesters.

B) proteins.

4)

D) It contains phosphate.

E) It contains amino acids.

C) vacuoles.

D) chromosomes.

E) Golgi.

The polarity of DNA synthesis is A) 5'→3'.

Version 1

1

B) 3'→5'.

C) 5'→2'.

D) 2'→5'.

6) In the Hershey and Chase experiment designed to determine the molecule of heredity, what was radiolabeled with 35S? A) protein B) DNA

7)

C) RNA D) rRNA

The ratio of _____ in DNA is 1:1.

A) guanine to adenine B) adenine to thymine C) cytosine to adenine

D) uracil to cytosine

8) X-ray data showed that the spacing between repeating units along the axis of the DNA helix is A) 2.0 angstroms.

B) 3.4 angstroms.

C) 20 angstroms.

D) 34 angstroms.

9) X-ray data showed that the DNA helix undergoes one complete turn every A) 2.0 angstroms.

B) 3.4 angstroms.

Version 1

C) 20 angstroms.

D) 34 angstroms.

2

10) ________ bonds are responsible for the chemical affinity between A and T (or G and C) nucleotides.

A) Ionic B) Covalent

C) Hydrogen D) Hydrophobic

11) Which one of the following is not an element found in DNA?

A) oxygen B) sulfur

C) nitrogen D) phosphorous E) hydrogen

12) _______-form DNA spirals to the right and is the major form of naturally occurring DNA molecules.

A) A B) B

13)

The nucleotide that is present in RNA but not DNA is A) thymine.

B) uracil.

14)

C) D D) Y E) Z

C) adenine.

D) cytosine.

E) guanine.

DNA replication occurs through a _________ process.

A) conservative

Version 1

3

B) semiconservative

C) dispersive D) transformative

15) During DNA replication, an incoming nucleotide is enzymatically joined to the preceding nucleotide by a(n) __________ bond.

A) hydrogen B) ionic

C) phosphodiester D) electrostatic

16) During early interphase, the structure of a eukaryotic DNA chromosome is A) a single continuous linear double helix.

B) a double helix undergoing semiconservative replication.

C) a double helix undergoing conservative

replication.

D) singlestranded.

17) During the S phase of interphase, the state of a eukaryotic chromosome can be described as A) a single continuous linear double helix.

B) a double helix replicating semiconservatively.

C) a double helix replicating conservatively.

D) a triple helix replicating semiconservatively.

18) The step of DNA replication in which the replication proteins open up the double helix and prepare for

Version 1

E) two linear single-stranded molecules.

complementary base pairing is called

4

A) initiation.

B) elongation.

C) termination.

D) translation.

E) translocation.

19) The step of DNA replication in which the proteins connect the correct sequence of nucleotides into a continuous new strand is called A) initiation.

B) elongation.

C) termination.

D) translation.

E) translocation.

20) The step of DNA replication in which two replication forks moving in opposite directions may meet is called A) initiation.

B) elongation.

C) termination.

D) translation.

E) translocation.

21) The group of enzymes able to relax supercoils in DNA is called A) primases.

B) helicases.

C) topoisomerases.

D) telomeres.

E) ligases.

22) How many replication forks depart from an origin of replication?

A) one Version 1

5

B) two C) three

D) four

23) The protein that progressively unwinds DNA ahead of each replication fork is called A) primase.

B) helicase.

C) topoisomerase.

D) telomerase.

E) ligase.

24) In eukaryotic cells, replication proceeds from ____ origin(s) of replication.

A) zero B) one

25) Eukaryotic chromosomes have evolved ______ at the ends of linear chromosomes to ensure the replication of the A) methylases B) capping proteins C) ligases

C) two D) many

two ends of the chromosomes.

D) telomeres E) single-stranded binding proteins

26) Which of the following is not involved in ensuring the accuracy of replication of a cell's genetic information?

A) redundancy of double-stranded DNA B) repair enzymes C) precision of replication machinery

Version 1

D) DNA polymerase proofreading mechanism E) restriction

6

endonucleases

Version 1

7

27) The process of _________ is extremely important in generating genetic diversity.

A) translation B) transcription

C) recombination D) transformation

28) Repair of heteroduplex regions formed during recombination can result in a 1:3 or 3:1 segregation of parental alleles, which indicates that ______ has occurred.

A) allelic exchange B) gene conversion C) crossing over

29)

D) recombination E) DNA replication

Recombination occurs during meiotic A) anaphase.

B) interphase.

C) prophase.

D) metaphase.

30) Recombination involves the breakage and reunion of DNA molecules from A) homologous nonsister chromatids.

B) homologous sister chromatids.

C) heterologous nonsister chromatids.

31)

The nicking of DNA that initiates recombination

Version 1

D) heterologous sister chromatids.

during mitosis may be due 8

to all of the following EXCEPT A) instructions from the normal cell cycle program.

D) physical damage.

E) radiation.

B) X-rays.

C) chemical damage.

32) What radiolabeled substance did Hershey and Chase use to label the protein component of the bacteriophage in their study to determine whether protein or DNA was A) nitrogen B) carbon

necessary for phage production?

C) sulfur D) phosphorus E) iodine

33) If 35% of the bases in a region of the mouse genome are cytosine, what percentage in that region are adenine?

A) 15% B) 20% C) 30%

D) 35% E) cannot be determined

34) Two haploid strains of S. cerevisae (yeast) are crossed.

One has the genotype ABC and the other abc. Five sets of the resultant tetrads are noted below. In which set did a gene conversion event occur?

A) abc, aBc, AbC, aBC B) abc, abc, ABC, ABC C) aBc, aBc, AbC, AbC

Version 1

D) abC, abc, ABc, ABC E) Abc, Abc, aBC, aBC

9

35) Without __________ regions of the DNA during recombination, gene conversion could not occur.

A) homoduplex B) heteroduplex

C) homotriplex D) heterotriplex

36) Hershey and Chase relied on ________ to physically separate the infected bacterial cells from the phage ghosts.

A) radioactivity B) gel filtration

37) Meselson and Stahl relied on equilibrium density gradient centrifugation in a _______ solution to resolve the A) radiolabeled phosphate B) calcium chloride C) radiolabeled nitrogen

38)

DNA containing 14N from the DNA containing 15N. D) sodium acetate E) cesium chloride

The full chemical name of DNA is

A) derived nucleic acid.

B) dideoxyribonucleic acid.

C) denatured ribonucleic acid.

39)

C) ion exchange D) centrifugation

Which term describes the ability of a substance to

Version 1

D) deoxyribonucleic acid.

change the genetic 10

characteristics of an organism?

A) transcription B) translation

C) transformation D) transposition

40) Which term is used to describe the overall 5′-to-3′ direction of linked nucleotides in a DNA chain?

A) complementarity B) sequence

C) polarity D) primacy

41) Which protein is needed to lay down a segment of RNA complementary to the DNA before replication can begin?

A) RNA polymerase II B) primase

C) helicase D) topoisomerase E) ligase

42) How many origins of replication does the E. coli chromosome contain?

A) none B) 1

43)

C) 2 D) 3 E) many

A virus that infects bacterial cells is known as a

Version 1

11

A) retrovirus B) reovirus

44)

Which type of viruscauses polio and AIDS?

A) retrovirus B) reovirus

45)

C) bacteriophage D) bacteriovirus E) parvovirus

C) bacteriophage D) adenovirus E) parvovirus

Which protein relaxes supercoils by nicking the DNA?

A) topoisomerase B) helicase

C) primase D) ligase E) telomerase

46) Given the following strand of DNA: 5′ CATAGCCTTA 3′ Which of the following sequences is the complementary DNA strand?

47)

A) 5′ CATAGCCTTA 3′ B) 3′ GTATCGGAAT 5′ C) 5′ GTATCGGAAT 3′

D) 3′ GUAUCGGAAU 5′ E) 3′ CATAGCCTTA 5′

Which of the following is an example of a protein that

can read the sequence of

Version 1

12

DNA without opening the double helix?

D) transcription A) RNA polymerase B) DNA polymerase C) helicase

factor E) primase

48) Which of the following properties of RNA is most directly responsible for its ability to form complex shapes?

A) the presence of uracil instead of thymine B) the fact that RNA is primarily double-stranded C) the relative instability of RNA compared to DNA D) the fact that RNA is primarily single-stranded

E) the substitution of deoxyribose in DNA with ribose in RNA

49) Suppose a region of DNA is 100 bp long. How many unique sequences could it potentially represent?

A) 100 B) 4100

4

C) 400 D) 4,000 E) 10,000

50) Which of the following contains single-stranded DNA as its genetic material?

A) humans B) E. coli

Version 1

C) HIV D) фX174

13

51)

Transformation in bacteria results from the uptake of

A) RNA B) DNA

which molecule?

C) lipid D) protein E) polysaccharide

52) Z-form DNA is a ______ helix with ______ backbone.

A) right-handed; a smooth B) left-handed; a smooth C) right-handed; an irregular

D) left-handed; an irregular

53) Exposure of which of the following regions of a molecule of DNA permits the reading of the DNA sequence without opening the double helix?

A) phosphate groups B) phosphodiester bonds

54)

C) major groove D) minor groove

With respect to DNA structure, polarity means that

A) DNA nucleotides can be either positively or negatively charged.

B) the phosphate backbone of DNA is negatively charged.

C) DNA nucleotides can be charged or uncharged.

Version 1

D) DNA strands have a 5′ direction and a 3′ direction.

14

55) Avery and colleagues purified a white substance as the transforming principle. Treating this substance with enzymes that destroyed RNA, protein, or polysaccharides did not affect its ability to transform cells, but an enzyme that degrades A) their transforming principle was not entirely pure.

DNA destroyed the transforming ability. These results suggested that E) both DNA and proteins are required for transformation.

B) the transforming principle appeared to be DNA. C) the test for transformation was not accurate.

D) a single gene could transform bacterial cells.

56) If you repeated the Hershey and Chase experiment designed to determine the molecule of heredity using a virus with RNA as its genetic material, you would expect that A) 35S-labeled protein would be transferred to infected cells.

B) 32P-labeled DNA would be transferred to infected cells.

C) 32P-labeled RNA would be transferred to infected cells.

D) 32P-labeled RNA would be transferred to phage ghosts.

E) 35S-labeled RNA would be transferred to infected cells.

57) Suppose the experiment of Meselson and Stahl was performed on a sample of 8 cells, each containing one copy of its circular double-stranded DNA genome, and that had been growing on normal 14N medium. You then grew the cells for 3 generations in medium containing 15N. The outcome would be A) 8 cells with single-stranded DNA molecules with N, and 24 cells with single-stranded DNA molecules with 15 N. 14

Version 1

B) 16 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, 15

and 48 cells with double-stranded DNA molecules with 15N. C) 8 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, and 24 cells with doublestranded DNA molecules with 15N. D) 8 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, and 32 cells with double-

stranded DNA molecules with 15N. E) 65 cells with single-stranded DNA molecules with 15N.

58) Crossing-over that occurs only between two specific DNA target sites that are usually less than 200 base pairs long is called A) mitotic recombination.

B) integrative recombination.

C) site-specific recombination.

59) A single DNA molecule contains two specific target sites separated by an intervening sequence. If site-specific recombination between these sites causes the intervening sequence to be released as a separate circular DNA molecule A) opposite directions causing excision.

B) the same direction causing excision.

C) opposite directions causing integration.

D) the same direction causing integration.

60) Suppose that in the future, scientists identify bacteria on Mars and these organisms lack a site-specific recombination system. Which components would allow for

D) homologous recombination.

E) non-random recombination.

containing one target site, then the sites were oriented in E) the same direction causing inversion.

site-specific recombination to occur in these bacteria?

C) an FRT target

A) FRT target sites and FLP recombinase B) FRT target sites and Cas9 enzyme

Version 1

16

site and a loxP site D) a synthetic FRT homologous chromosome

Version 1

17

61) The normal functions of Spo11 and Cas9 enzymes are different because A) Spo11 produces double-strand breaks during meiosis, whereas Cas9 functions during mitosis.

B) Spo11 produces site-specific double-strand breaks during meiosis, whereas Cas9 breaks DNA at random genomic locations.

C) Spo11 produces double-strand breaks in many genomic locations during meiosis, whereas Cas9 is targeted to specific sequences by a CRISPR RNA. D) Spo11 produces double-strand breaks at

sequences targeted by RNAs expressed during meiosis, whereas Cas9 has the ability to recognize DNA sequences without an RNA. E) Spo11 is unique to yeast, whereas Cas9 occurs in most higher eukaryotes.

62) What is the correct order of these events in recombination that is resolved by the crossover pathway?

1. Double-strand break formation 2. Strand invasion 3. Resection 4. Resolution of the double Holliday junction 5. Branch migration 6. Formation of a double Holliday junction

A) 6, 1, 2, 5, 3, 4 B) 1, 2, 5, 3, 6, 4

C) 6, 4, 1, 2, 5, 3 D) 1, 3, 2, 6, 5, 4 E) 1, 2, 3, 5, 6, 4

63) During meiotic prophase in a eukaryotic cell, Spo11 initiates recombination by causing Version 1

18

a double-strand break in DNA between Gene A and Gene B on one sister chromatid. Which of these events will occur

prior to formation of a Holliday junction?

A) A heteroduplex forms due to pairing of Gene A and Gene B. B) Resolvase cuts all four chromatids.

C) Spo11 causes a double-strand break in a nonsister chromatid.

D) Strand invasion causes one strand of the uncut

chromatid to form a D loop.

E) Branch migration lengthens the heteroduplex region to include Gene A and Gene B.

64) A chromosome of genotype C D recombines with a homolog of genotype c d during meiosis when Spo11 produces a double-strand break between genes C and D. If anticrossover helicase disentangles the invading strand, the likely outcome would be A) two C D and two c d gametes.

B) two C d and two c D gametes.

C) one C D, one C d, one c D, and one c d gamete.

65) Two strains of S. cerevisae (yeast) are crossed. One has the genotype A B and the other a b. Which statement correctly describes the tetrads that can be produced by one reciprocal exchange between the A and B genes without A) Recombination occurs at the four-strand stage to produce a tetrad with A B, A B, a B, and a b chromosomes.

B) Recombination occurs at the four-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes.

C) Recombination occurs at the two-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes.

Version 1

D) four C D gametes.

E) four c d gametes.

gene conversion in the dihybrid?

D) Recombination occurs at the two-strand stage to produce a tetrad with A b, A b, a B, and a B chromosomes.

19

66) The antiparallel polarity of the two DNA strands within the double helix affects DNA replication in that A) it takes twice as long to replicate two strands sequentially from 5′ to 3′.

B) circular bacterial DNA does not have polarity, so a single strand can be copied twice.

C) DNA polymerases on each strand synthesize DNA in the 5'-to-3' direction, and one strand is synthesized

discontinuously.

D) each portion of the double helix must be unwound twice so that both strands are replicated continuously.

67) If a chemical inhibitor of DNA ligase was added to bacterial cells after DNA replication initiation, the outcome would be A) elongation of both strands would not occur without ligase to join new nucleotides together.

B) elongation would occur, but Okazaki fragments would not be joined together.

C) elongation of the lagging strand, but not the leading strand would occur.

D) normal replication would proceed as DNA ligase

68) Homologous recombination occurs in a heterozygote in which alleles D and d differ by a single base pair. The D allele has a G at one position, whereas the d allele has a C at the same position. If branch migration causes heteroduplex

is not required for this process.

E) too many supercoils would build up in the DNA, resulting in a halting of elongation.

formation across this position, what is the expected outcome? base pair and will ensure

A) Both D and d alleles will remain unchanged because G can base pair with C. B) Mismatch repair will identify an abnormal G⋮G base pair and may convert the D allele to a d allele.

C) Mismatch repair will identify an abnormal C⋮G Version 1

20

that the cell has two copies of each allele.

D) Mismatches will cause the Holliday junction to be unstable and to resolve by the noncrossover pathway.

Version 1

21

69) Meselson and Weigle grew one strain of bacteriophage lambda with alleles A and B in heavy isotopes and another strain with alleles a and b in light isotopes. Both loci are located near the end of the viral A) They visualized the broken chromosomes under a microscope and concluded that DNA was exchanged between light and heavy chromosomes during recombination.

B) They found a gradient of all densities of recombinant chromosomes and concluded that DNA was not exchanged between light and heavy chromosomes during recombination.

C) They found that some A b genetic recombinants were almost as dense as the A B parent chromosome and concluded that DNA was exchanged between light and heavy chromosomes during recombination.

chromosome. Which statement best describes their results and conclusions?

D) They found that some A b genetic recombinants were almost as dense as the A B parent chromosome and concluded that DNA was not exchanged between light and heavy chromosomes during recombination.

70) If a cell had a single nonfunctional DNA repair enzyme, how would that cell be affected?

A) The cell could not possibly be viable without that DNA repair enzyme.

B) Mutations during DNA replication would not be repaired, so replication and therefore division would not be possible.

C) The cell’s other DNA repair enzymes would most

likely compensate and cell viability would be maintained.

D) DNA polymerase would rereplicate any damaged regions of DNA.

71) How does the antiparallel polarity of DNA strands contribute most to the integrity of genetic information?

A) A DNA double helix cannot be broken or Version 1

degraded.

B) The 22

antiparallel polarity of the strands means that replication occurs more quickly, decreasing the likelihood of DNA damage.

C) Hydrogen bonding between strands is very stable, so DNA bases cannot be damaged by environmental agents.

D) Both antiparallel strands can be replicated continuously, which is more accurate than discontinuous

replication.

E) Each antiparallel strand can serve as a template to produce the other strand in the event of DNA damage.

72) Site-specific recombination systems, such as FLP recombinase/FRT sites, can function only in a single species. ⊚ ⊚

true false

73) Both Spo11 and Cas9 are enzymes that create doublestrand breaks in DNA. ⊚ ⊚

Version 1

true false

23

Answer Key Test name: Chapter 06 Test Bank 1) E 2) B 3) C 4) D 5) A 6) A 7) B 8) B 9) D 10) C 11) B 12) B 13) B 14) B 15) C 16) A 17) B 18) A 19) B Version 1

24

20) C 21) C 22) B 23) B 24) D 25) D 26) E 27) C 28) B 29) C 30) A 31) A 32) C 33) A 34) A 35) B 36) D 37) E 38) D 39) C 40) C Version 1

25

41) B 42) B 43) C 44) A 45) A 46) B 47) D 48) D 49) B 50) D 51) B 52) C 53) C 54) D 55) B 56) C 57) B 58) C 59) B 60) A 61) C Version 1

26

62) D 63) D 64) A 65) B 66) C 67) B 68) B 69) C 70) C 71) E 72) FALSE 73) TRUE

Version 1

27

Student name:__________ 1) In a segment of DNA, an A base is changed to a C and the error is copied during replication. Under which circumstances is that change considered a mutation? cell.

A) The change alters the phenotype of the organism’s offspring.

B) The change results in production of a novel protein.

C) The change is in a somatic cell, such as a skin

D) The change is in a gene, but has no effect on a gene product or phenotype.

2) Replacing a thymine nucleotide with a guanine is an example of a A) translocation.

B) transition.

C) transversion.

D) forward mutation.

E) reversion or reverse mutation.

3) Replacing an adenine nucleotide with a guanine is an example of a A) translocation.

B) transition.

C) transversion.

D) forward mutation.

E) reversion or reverse mutation.

4) Assume that a wild-type sequence is 5' AGCCTAC 3'.

Which sequence could be produced by a transversion?

A) 5' AGTCTAC 3'

Version 1

B) 5' AGCCGCCGCCGCCTAC 1

3'

E) 5' AGCCTGC C) 5' AGCCCAC 3' D) 5' ATCCTAC 3'

3'

5) Assume that the mutation rate for a given gene is 5 × −6 10 mutations per generation. For that gene, how many mutations would be expected if 10 million sperm are examined?

A) 5 × 10 B) 5 C) 50

6)

−6

D) 500 E) None of these choices are correct.

Which type of mutation is least likely to revert?

A) deletion B) transition C) transversion

D) insertion E) All are equallylikely

7) Fifty-million sperm were examined for mutations in a specific gene and 100 mutations were found. What was the mutation rate for this gene?

A) 5 × 10−6 B) 50 × 10−6

Version 1

C) 2 × 10−6 D) 2 × 10−5 E) 5 × 10−5

2

8) What does the removal of a purine base from the deoxyribose-phosphate backbone of DNA result in?

A) an apurinic site B) a modified base C) a double-strand break

9)

D) a mutation E) a deletion

Thymine dimers are formed by the action of A) X-rays.

B) active oxygen species.

C) a mutagen such as ethylmethane sulfonate (EMS).

D) depurination.

E) UV light.

10) Exposure to UV light from the sun or tanning beds results in A) depurination.

B) deamination.

C) alkylation.

D) thymine dimerization.

E) oxidation.

11) The heritable disorder fragile X syndrome, a major cause of intellectual disability, is caused by

A) production of enzymes that break the phosphate backbone.

B) UV light.

C) X-rays.

D) the presence of an extra X chromosome in the

Version 1

sperm or egg.

E) the expansion of a region of trinucleotide repeats.

3

12) Fragile X syndrome is caused by a mutant allele of an X-linked gene. If a man has one pre-mutation allele for fragile X syndrome, what is the probability that he will pass the premutation allele to his son?

A) 100% B) 75%

C) 50% D) 25% E) 0%

13) The base analog 5-bromouracil undergoes tautomerization. One tautomer behaves like thymine, and the other behaves like cytosine. If 5-bromouracil is incorporated in the DNA, what is the minimum number of cycle(s) of DNA replication that must occur to mutate the codon for proline (CCC) into the codon for alanine (GCC)?

A) one cycle B) two cycles C) three cycles

14) The base analog 5-bromouracil undergoes tautomerization. One tautomer behaves like thymine, and the other behaves like cytosine. If 5-bromouracil is incorporated in the DNA, what is the minimum number of cycle(s) of DNA A) one cycle B) two cycles C) three cycles

Version 1

D) 5-bromouracil cannot induce this mutation

replication that must occur to mutate the codon for proline (CCC) into the codon for serine (TCC)?

D) 5-bromouracil cannot induce this mutation

4

15)

Intercalating agents, such as acridine orange, are A) promote transitions.

B) remove amine groups.

C) attach to purines causing distortions.

D) add ethyl or methyl groups.

mutagens because they E) fit between stacked bases and disrupt replication.

16) Alkylating agents, such as ethylmethane sulfate (EMS), are mutagens because they A) promote deletions and insertions.

B) remove amine groups.

C) add hydroxyl groups to bases.

D) add alkyl groups (such as ethyl groups) to bases.

17) In a test tube reaction, a particular DNA polymerase has an error rate of 1 mistake in every 106 bases copied. Why is the mutation rate in a cell that uses this DNA polymerase A) The polymerase is more careful in replicating regions where genes exist.

B) Repair mechanisms correct errors made by the polymerase.

C) Not all mutations can be detected easily.

18)

E) fit between stacked bases and disrupt replication.

much lower than 1 in 106 base pairs?

D) The DNA polymerase has no proofreading function.

E) Mutations do not occur if mutagens are not present.

The first step in base excision repair is nucleotide within DNA.

A) removal of a double-stranded fragment of damaged DNA. B) nicking the backbone of one DNA strand.

C) recognizing and excising the incorrect base from a Version 1

5

D) removing extraneous groups such as methyl or oxygen added by mutagens.

E) replacing an A-T base pair with a C-G base pair.

Version 1

6

19) The genetic condition xeroderma pigmentosum, which can lead to skin cancer, results from the inability to A) repair regions with UV-induced thymine dimers.

B) process the amino acid phenylalanine.

C) produce functional hemoglobin.

D) correct transitions.

E) correct breaks in the X chromosome.

20) A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide. How is this repair system able to determine which mismatched base is incorrect?

A) The incorrect base results in a bulge on only the new strand.

B) The incorrect base is in the parental strand which is modified with methyl groups.

C) Some bases of the parental strand are methylated, while none of the new strand are immediately after DNA replication.

D) A methyl group is attached to the incorrect base in the newly synthesized DNA strand.

E) DNA polymerase is bound to the new strand at the location of the incorrect base.

21) In bacteria, what is the consequence of a mutation that inactivates the enzyme responsible for methylating the A in

the DNA sequence 5′ GATC?

A) Replication of the lagging strand cannot be completed.

B) Thymine dimers cannot be repaired.

C) Parental and new DNA strands cannot be distinguished during mismatch repair.

D) Expression of certain metabolic genes is decreased.

E) The mutation rate of all genes will be lower.

Version 1

7

22)

In the Ames test for mutagenicity,

A) mutagens cause reverse mutations, converting auxotrophs to prototrophs that survive on minimal medium.

B) colonies on minimal medium result from forward mutations that convert prototrophs to auxotrophs.

C) cells are treated with mutagens, and only cells with no mutations survive on minimal medium.

D) cells are treated with excess amino acids, and

cells with mutations are killed.

E) rat liver enzymes are present to protect cells from mutation.

23) How is the Ames test for mutagenicity useful for identifying potential carcinogens?

A) Bacteria do not get cancer so they can survive lethal carcinogens.

B) Substances that mutate bacterial DNA are likely to mutate human DNA. C) Bacteria thrive on substances that could cause cancer in humans.

D) The same genes that cause cancer in humans can

be mutated in bacteria.

E) Liver enzymes alter the bacteria so they will behave like human cells.

24) In bacteria, utilization of the SOS system for DNA repair may result in mutations. These mutations arise because sloppy (error-prone) DNA polymerases repair the DNA. Under what circumstances will a bacterial cell utilize the SOS system?

A) The normal high-fidelity DNA polymerases are not present in the cell.

B) The large amount of DNA damage overwhelms the capacity of the normal repair systems.

C) Under stressed conditions, even when there is no Version 1

DNA damage, bacteria will utilize the SOS system to increase the frequency of mutations.

D) When a bacterial cell is dividing at 8

a much faster than normal rate, the SOS system allows DNA replication to occur at a much faster than normal rate.

25) Luria and Delbruck grew many liquid cultures of bacteria, then spread a small sample of each culture on nutrient agar infused with bacteriophage in separate petri plates. They assayed the number of bacterial colonies that formed on each plate.

25.1) When Luria and Delbruck assayed bacterial growth on the petri plates, what did they find?

A) All plates had some bacterial colonies; some plates had more bacterial colonies than others.

B) Some plates had no bacterial colonies; others had various numbers of colonies.

C) All plates had a similar number of bacterial colonies.

D) Some plates had no bacterial colonies; all of the

other plates had a similar number of bacterial colonies.

E) No bacterial colonies formed on any of the plates.

25.2) In the Luria-Delbruck experiment, what did the appearance of a bacterial colony on a petri plate indicate?

A) The bacterial colony was derived from one initial cell that was resistant to infection by the bacteriophage.

B) The bacterial colony was made up of many cells that became resistant to bacteriophage infection independently.

C) A colony formed when a bacterial cell was not

Version 1

resistant to bacteriophage infection.

D) The bacterial cells that made up a colony were resistant to antibiotic.

9

25.3) What did the results of the Luria-Delbruck fluctuation experiment indicate?

A) All bacteria are naturally resistant to infection by phage.

B) Some individuals in any bacterial population are resistant to infection by phage.

C) Exposure to phage induces some bacteria to mutate and become resistant to phage.

D) Resistance to phage in bacteria is caused by

25.4) A researcher planned to duplicate the LuriaDelbruck fluctuation experiment. This researcher inoculated twenty samples of liquid nutrient media with bacteria from the same colony and let them grow overnight. The next morning the researcher noticed that all but one of the flasks had come open and were ruined. Wishing to get some information from their efforts, the researcher spread samples of bacteria from the one intact liquid culture on twenty petri plates that A) All plates had some bacterial colonies; some plates had more bacterial colonies than others.

B) Some plates had no bacterial colonies; others had varied numbers of colonies.

C) All plates had a similar number of bacterial colonies.

D) Some plates had no bacterial colonies; all of the

random spontaneous mutation.

E) Some phage mutated to produce large plaques with sharp edges.

had nutrient agar infused with bacteriophage. What results do you expect the researcher obtained from this experiment?

other plates had a similar number of bacterial colonies.

E) No bacterial colonies formed on any of the plates.

Answer Key Test name: Chapter Version 1

10

07 Test Bank 1) [A, B, C, D] 2) C 3) B 4) D 5) C 6) A 7) C 8) A 9) E 10) D 11) E 12) E 13) D 14) B 15) E 16) D 17) B 18) C 19) A

Version 1

11

20) C 21) C 22) A 23) B 24) B 25) Section Break 25.1) B 25.2) A 25.3) D 25.4) C

Version 1

12

Student name:__________ 1) Which statement(s) about the rII− strain of T4 bacteriophage that Benzer studied is (are) true?

A) It grows in E. coli K(λ), but not in E. coli B. B) It grows in E. coli B, but not in E. coli K(λ).

C) It produces larger plaques than wild-type.

D) It produces smaller plaques than wildtype.

2) Which are important aspects of a complementation test done by coinfection of bacterial cells with phage T4?

A) counting the number of plaques that are produced after infection of E. coli K(λ) cells B) using sufficient amounts of two phage strains; each with different mutations C) infecting bacteria with two phage strains that have

recessive mutations D) recovering phage from the plaques after growth and lysis

3) What level(s) of protein structure could be affected by a nucleotide substitution that results in an amino acid change in a polypeptide?

A) primary structure B) secondary structure C) tertiary structure

D) quaternary structure E) There will be no structural changes.

4) What is true about the mutations within a complementation group? gene.

A) They produce the same mutant phenotype.

B) They complement each other and are in the same Version 1

C) They do not complement each other 1

and are in the same gene.

D) They complement each other and are in different genes.

5)

E) They do not complement each other and are in different genes.

A plaque is

A) a colony of bacteria growing on a plate.

B) a clump of bacterial cells that contain phage within them.

C) a region on a plate where bacteria that have survived phage infection are living.

6) One strain of rII− phage has a deletion in the rII region, another has a point mutation in the rII region. When E. coli K(λ) cells are infected with either rII− phage (not both) no plaques form. When E. coli K(λ) cells are infected with both rII− phage simultaneously plaques do form. Why A) Recombination between the two phage genomes resulted in a wild-type phage.

B) The two phage have mutations in different rII genes and their genomes complement each other.

C) A protein made by the deletion phage can repair the DNA of the phage with the point mutation.

D) an area on a plate that has live phageresistant bacteria.

E) an area on a plate that has phage and dead bacteria.

do plaques form when E. coli K(λ) cells are infected with both types of rII− phage simultaneously?

D) Bacteria can survive infection with one phage, but they cannot survive infection with two phages.

7) What is the correct order of events that occur during one round of infection by bacteriophage T4?

1) The bacterial cell is

Version 1

2

lysed.

2) Phage proteins and DNA are synthesized, and bacterial DNA is degraded.

3) New phages are assembled.

4) The phage head proteins enter the bacterial cell.

A) 4, 2, 3, 1 B) 4, 5, 2, 1, 3

5) The phage injects DNA into the bacterial cell.

C) 5, 1, 2, 3 D) 5, 2, 3, 1 E) 4, 5, 3, 1

8) How many progeny phage are released when a single E. coli cell is lysed by phage T4?

A) between 1 and 10 B) between 10 and 100 C) between 100 and 1,000

9) A researcher is studying coat color in mice. Wild-type fur in mice is brown. Three pure-breeding strains of mice with white fur have been isolated. The strains are called milky, blanc, and weiss. White fur is a recessivecharacteristic in each strain. These mice are crossed to each other in pairs and the progeny phenotypes are recorded. milky × blanc = all white progeny A) All three strains have mutations in the same gene.

B) All three strains have mutations in different genes.

C) The milky and blanc strains have mutations in the same gene; weiss has a mutation in a different gene.

D) The milky and weiss strains have mutations in the same gene; blanc has a mutation in a different gene.

Version 1

D) about 10,000 E) about 100,000

milky × weiss = all brown progeny blanc × weiss = all brown progeny What conclusion is consistent with these results?

E) The weiss and blanc strains have mutations in the same gene; milky has a mutation in a different gene.

3

10)

Which statement about biochemical pathways is

A) All enzymes in the pathway catalyze the same reaction.

B) If an enzyme in a pathway is absent, adding excessive amounts of its substrate will restore the normal phenotype.

C) If an enzyme in a pathway is absent, adding excessive amounts of its product will restore the normal phenotype.

D) If the enzyme that catalyzes the final step in a

11) In the human genetic disorder alkaptonuria, urine turns black because of the presence of homogentisic acid in A) the presence of large amounts of homogentisic acid in the diet.

B) failure of individuals with alkaptonuria to manufacture enzymes involved in the synthesis of homogentisic acid.

C) failure of wild-type individuals to manufacture enzymes involved in the synthesis of homogentisic acid.

D) failure of the kidneys to remove homogentisic

correct? pathway is absent, all the other enzymes will be inactivated.

E) If the first enzyme in a pathway is absent, adding the final product will not restore the normal phenotype.

individuals with the trait.

This is due to acid from the urine.

E) failure of individuals with alkaptonuria to manufacture enzymes involved in the breakdown of homogentisic acid.

12) Consider the pathway for the synthesis of the amino acid arginine in Neurospora:

Mutant strains of Neurospora may carry one

Version 1

4

or more mutations. Neurospora mutant strain a is grown on minimal media plus supplements as shown. Growth is shown by (+) and no growth is shown by (o). Growing strain “a” cells accumulate citrulline.

Which statement is true about strain a?

A) It has one mutation in ARG-H. B) It has two mutations, one in ARG-F and one in ARG-H. C) It has two mutations, one in ARG-E and one in ARG-H.

D) It has two mutations, one in ARG-E and one in ARG-F. E) It has mutations in ARG-E, ARG-F, and ARG-H.

13) Consider the pathway for the synthesis of the amino acid arginine in Neurospora:

Mutant strains of Neurosporamay carry one or more mutations. Neurosporamutants strain a and strain b are grown on minimal media plus supplements as shown. Growth is shown by + and no growth by o.

A) Strain a has only one mutation and it is in ARGE. B) Strain b has only one mutation and it is in ARG-

What can you conclude from these data?

has a mutation in ARG-E. D) Strain b has mutations in ARG-E, ARG-F, and ARG-H.

H. C) Strain a has a mutation in ARG-F and strain b

14)

Which statement about amino acids is false?

Version 1

5

A) Every amino acid contains a carboxyl group.

B) The side chain or R group is the same in each amino acid.

C) In a polypeptide, amino acids are joined by

15)

peptide bonds.

D) One end of a polypeptide, the N terminus, contains a free amino group.

Which statement about amino acids is false?

A) A polypeptide is made up of several amino acids attached by peptide bonds.

B) Amino acids are linked by peptide bonds that form between two amino groups.

C) The C terminus of a polypeptide chain contains a

free carboxylic acid group.

D) Each type of amino acid has a unique R group.

16) Which condition does not involve a defect in an enzyme pathway?

A) Alkaptonuria B) Albinism C) Sickle-cell disease

D) Phenylketonuria (PKU)

17) Which interaction is not involved in maintaining tertiary structure in protein molecules?

A) covalent bonds B) hydrogen bonds C) hydrophobic interactions D) electrostatic interactions

Version 1

E) All of the choices may be involved in maintaining tertiary structure in proteins.

6

18)

Sickle-cell disease is due to A) the insertion of an amino acid.

B) the deletion of an amino acid.

C) substitution of an amino acid.

D) failure to synthesize a hemoglobin molecule.

19)

E) deletion of the β-globin gene.

Which statement about sickle-cell disease is false?

A) Individuals who are heterozygous for the sicklecell allele cannot make hemoglobin.

B) The sickle-cell hemoglobin molecule contains an amino acid substitution.

C) The hemoglobin molecules of an individual with

sickle-cell disease clump together.

D) The red blood cells of an individual with sickle-cell disease distort and elongate.

20) In a polypeptide, what level of structure refers to a localized region that takes on a particular geometry?

A) primary structure B) secondary structure C) tertiary structure

D) quaternary structure E) both tertiary and quaternary structures

21) Interaction between two distinct polypeptide chains is what type of structure in a protein?

A) primary structure B) secondary structures C) tertiary structure

Version 1

D) quaternary structure E) both primary and secondary structure

7

22)

The photoreceptor protein rhodopsin

A) is found in cone cells and is sensitive to weak light at many wavelengths.

B) is found in rod cells and is sensitive to weak light at many wavelengths.

C) is found in cone cells and is responsible for blue color vision.

D) is found in rod cells and is responsible for red color vision.

E) is missing in individuals who exhibit red-green color blindness.

23) Examination of the rhodopsin gene family provides evidence for gene evolution by A) duplication and divergence.

B) accumulation of random mutations.

C) convergent evolution.

D) spontaneous generation.

24) Red-green color blindness is more common in males than females because A) the red pigment gene is on the X chromosome, the green is on an autosome.

B) the green pigment gene is on the X chromosome, the red is on an autosome.

C) the rhodopsin gene is on the X chromosome.

D) both the red and the green pigment genes are on

25) Assume that a series of compounds has been discovered in Neurospora. Compounds A–F appear to be intermediates in a biochemical pathway. Conversion of one intermediate to the next is controlled by enzymes that are Version 1

the X chromosome.

E) both the red and the green pigment genes are on an autosome.

encoded by genes. Several mutations in these genes have been identified and Neurospora strains 1–4 8

each contain a single mutation. Strains 1–4 are grown on minimal media supplemented with one of the compounds A– F. The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth).

Which biochemical pathway fits the data presented?

Media Supplement

Version 1

9

Strain

A

B

C

D

E

F

1

o

o

o

+

+

+

2

o

o

o

o

+

+

3

o

o

o

o

+

o

4

o

o

+

+

+

+

A) A → B → C →D→E→F B) A → B → C →F→D→E C) F → B → C → D→A→E D) A → B → C →D→F→E E) A → B → F → E→C→D

26) Shown below are the maps of a series of rII− deletion strains (1–5). The deleted region is indicated as (......) and the intact region as ______.

1 ___________(...........)_______________ 2 _________________(...........)_________ 3 (.....................)_____________________ 4 ________________________(................)

5 _____(..........)______________________ rII− phage strains A-E have point mutations in the rII region.

E.coli K(λ) cells are coinfected with one phage that has a deletion and one phage that has a point mutation. The presence of wild-type progeny phage is assessed by the presence (+) or absence (o) of plaques.

1

2

3

4

5

A

+

o

+

+

+

B

o

+

+

+

+

Version 1

C

+

+

+

o

+

D

+

+

o

+

+

E

+

+

o

+

o

10

26.1) Indicate the order of the point mutations in the rII region.

A) CADBE B) DEBAC

26.2)

The test described here is a recombination test.

⊚ ⊚

true false

Version 1

C) BADCE D) ABDEC E) CEADB

11

1

2

3

4

5

A

+

+

o

+

o

B

+

o

+

+

+

C

+

+

o

+

+

D

o

+

+

+

+

E

+

+

+

o

o

Version 1

12

27.1)

Indicate the order of the point mutations in the

A) CADBE B) DEBAC

rII region.

C) BADCE D) ABDEC E) CEADB

27.2) The test described here depends on recombination. ⊚ ⊚

true false

28) A researcher is studying the rII locus of phage T4. − FourrII strains are obtained: A, B, C, and D. In the first experiment, E. coli strain K(λ) is coinfected with two rII− strains simultaneously and the results are recorded.

Infection with A and B phage = lysis occurs Infection with A and C phage = lysis occurs Infection with B and C phage = no lysis occurs Infection with B and D phage = no lysis occurs Infection with C and D phage = no lysis occurs In a second experiment, coinfections are performed first in E. coli strain B, then the progeny phage are used to infect E. coli strain K(λ).

Progeny of A and B phage = plaques form Progeny of B and C phage = plaques form Progeny of C and D phage = plaques form Progeny of B and D phage = no plaques from

Version 1

13

A) Strains A and B carry mutations in the same gene.

B) Strains B and D both carry the same mutation.

C) Strains B, C, and D carry mutations in the same gene.

D) Strain A carries a mutation in a different gene than strains B, C, and D.

28.2)

Experiment 1 depends on recombination.

⊚ ⊚

true false

28.3)

Experiment 2 depends on recombination.

⊚ ⊚

true false

E) Strains B and C both carry the same mutation.

F) A, B, C, and D carry mutations in the same gene.

29) Fruit flies normally have red eyes. Eight different true-breeding strains of fly with white eyes have been identified (A-H). In each strain, the white eye trait is due to an autosomal recessive allele.

It is possible that all eight strains have mutations in the same gene.

Alternatively, some or all of the strains may have mutations in different Version 1

14

genes. To determine how many genes are involved in eye color in these flies, all possible pairwise crosses are performed. The offspring phenotypes resulting from each cross are observed. (+ = wild-type; − = mutant)

29.1) Which strains’ mutations represent a complementation group?

A) C, D, F B) A, B

C) E D) A, B, E, G, H E) A, B, E, H

29.2) Based on these crosses, how many different eye color genes are mutant in all of these strains collectively?

A) 1 B) 2

Version 1

C) 3 D) 4 E) 8

15

Answer Key Test name: Chapter 08 Test Bank 1) [B, C] 2) [B, C] 3) [A, B, C, D] 4) C 5) E 6) B 7) D 8) C 9) C 10) C 11) E 12) B 13) C 14) B 15) B 16) C 17) E 18) C 19) A Version 1

16

20) B 21) D 22) B 23) A 24) D 25) D 26) Section Break 26.1) B 26.2) FALSE 27) Section Break 27.1) A 27.2) FALSE 28) Section Break 28.1) [B, C, D] 28.2) FALSE 28.3) TRUE 29) Section Break 29.1) E 29.2) C

Version 1

17

Student name:__________

A) Protein→RNA→DNA. B) DNA→RNA→Protein.

C) RNA→DNA→Protein.

2)

1) The Central Dogma describes the flow of genetic information as D) DNA→Protein→RNA.

The site of protein synthesis is the C) centrioles.

D) ribosome.

A) nucleus.

B) mitochondria.

3)

A codon is a three-base sequence of

A) mRNA that codes for an amino acid.

B) rRNA that codes for an amino acid.

C) tRNA that codes for an amino acid.

D) mRNA or tRNA that codes for an amino acid.

4) Which of these statements does NOT describe a characteristic of the genetic code?

A) The number of codons and amino acids is the same.

codons.

D) The code is degenerate.

B) The code is used by nearly every living organism.

C) Some amino acids are specified by multiple

5)

Version 1

Charles Yanofsky

1

helped decipher the genetic code working with the _____ biosynthesis genes in _______.

A) leucine; Salmonella enteritidis B) phenylalanine; Klebsiella pneumoniae C) tryptophan; Escherichia coli

D) glycine; Serratia marcescens

6) The following DNA sequence is part of one exon and contains the beginning of a gene’s open reading frame 5' ATGCCTGAATCAGCTTTA 3'. How many amino acids does this sequence encode?

A) 5 B) 6

7) In an in vitro translation experiment, how many different amino acids could be encoded by the synthetic A) 0 B) 1

8)

C) 7 D) 18

mRNA sequence 5' UGCUGCUGC 3'?

C) 2 D) 3 E) 6

The universal genetic code includes A) 3 start codons and 1 stop codon.

B) 2 start codons and 2 stop codons.

C) 1 start codon and 3 stop codons.

Version 1

D) 0 start codons and 4 stop codons.

2

9)

A ribosome begins translating mRNA at the initiation

codon which establishes

A) synthesis platform.

B) code degeneration.

C) peptide transition.

D) reading frame.

Version 1

3

10) Sydney Brenner determined that there are three ________ codons that do not encode amino acids.

A) nonsense B) missense

11)

C) sense D) antisense

Nonsense codons are D) start codons.

A) codons that codefor multiple amino acids.

B) codons that do notcode for an amino acid.

C) codons that can be read forward or backward.

12)

Which of these is NOT a step in transcription?

A) translation B) initiation

C) termination D) elongation

13) During maturation of a eukaryotic mRNA, sequences that are spliced out are

A) extremes.

B) exons.

Version 1

C) inclusions.

D) introns.

E) spliceosomes.

4

14)

The sequences found in a mature eukaryotic mRNAs A) extremes.

B) exons.

are C) inclusions.

D) introns.

E) spliceosomes.

15) The retroviruses, including HIV, are unique because they use A) replication.

B) reverse transcription.

C) splicing.

D) reverse translation.

E) posttranslational modification.

16) A complex of proteins and small nuclear RNAs that form discrete particles to remove introns is a A) holoenzyme.

B) spliceosome.

C) nucleosome.

D) ribosome.

E) chromosome.

17) RNA molecules that act as enzymes to catalyze specific biochemical reactions are called A) RNases.

B) splice acceptors.

C) ribozymes.

D) tRNAs. E) restriction enzymes.

18) Experiments that are done outside a living cell are described by the Latin term Version 1

5

A) in vino.

B) in vito.

C) in vivo.

D) in vitro.

E) de vito.

19) Experiments that are done inside a living cell are described by the Latin term A) in vino.

B) in vito.

C) in vivo.

D) in vitro.

20) Transcription occurs ________ and translation occurs __________ of eukaryotic cells.

A) in the nucleus; in the cytoplasm B) in the cytoplasm; in the nucleus C) in the nucleus; at the cell centrosomes

D) in the cytoplasm; at the cell centrosomes

21) Transcription occurs ________ and translation occurs __________ of prokaryotic cells.

A) at the centrosomes; in the cytoplasm B) in the cytoplasm; in the nucleus C) in the cytoplasm; at the centrosomes D) in the cytoplasm; in the cytoplasm

E) in the nucleus; in the cytoplasm

22) Select the statement that describes the stability of DNA and RNA under typical cellular conditions.

A) Double-

Version 1

6

stranded DNA in the nucleus and single-stranded RNA in the cytoplasm are equally stable.

B) Double-stranded DNA in the nucleus is more stable than single-stranded RNA in the cytoplasm.

C) Single-stranded RNA in the cytoplasm is more stable than double-stranded DNA in the nucleus.

D) Both RNA in the cytoplasm and doublestranded DNA in the nucleus can never be degraded within a cell.

23) In the processing of eukaryotic transcripts into mRNA, ______ cap and ______ tail are usually added to the transcript.

A) an acetylated guanine; a poly-C B) a poly-G; a methyl group C) a poly-T; an acetyl group

D) a methylated guanine; a poly-A E) a methylated adenine; a poly-G

24) RNA polymerase binds to ______ before beginning transcription.

D) a replication A) a promoter B) an operator C) a structural gene

25)

origin E) the ShineDalgarno sequence

A prokaryotic large ribosomal subunit is composed of A) DNA and protein.

B) introns and protein.

C) rRNA and protein.

Version 1

D) only proteins folded together.

E) only rRNAs folded together.

7

26)

Aminoacyl-tRNA synthetase is the enzyme that

A) folds tRNA molecules into their proper configuration.

B) causes tRNA molecules to bind to the aminoacyl site of a ribosome.

C) produces tRNA by reading DNA molecules.

D) adds the appropriate amino acid to an uncharged tRNA.

27) Which enzyme forms peptide bonds between adjacent amino acids at the ribosome?

A) RNA polymerase B) phosphotransferase

C) peptidyl transferase D) ribonuclease

28) An anticodon is a physical component of _______ molecule.

A) a tRNA B) an mRNA

C) an rRNA D) a DNA

29) Which of these processes are coupled in prokaryotes but NOT in eukaryotes?

A) transcription and translation B) replication and transcription C) replication and translation

Version 1

D) transcription and splicing

8

30) A mutation that is characterized by a change in the DNA base-pair sequence, but no change in the resulting A) frameshift mutation.

B) missense mutation.

31) A mutation that changes a codon sequence, and thereby changes the amino acid that should have been placed A) frameshift mutation.

B) missense mutation.

protein’s amino acid sequence, is a C) silent mutation.

D) nonsense mutation.

at that point in the polypeptide chain, is a C) silent mutation.

D) nonsense mutation.

32) A mutation that changes a codon that originally coded for an amino acid into a stop codon is a A) frameshift mutation.

B) missense mutation.

33) A mutation that occurs when a base is inserted into or deleted from a gene’s DNA base-pair sequence, completely A) frameshift mutation.

B) missense mutation.

C) silent mutation.

D) nonsense mutation.

altering the subsequent sequence of codons, is a C) silent mutation.

D) nonsense mutation.

34) Drugs like AZT, ddC, or ddI work against the AIDS virus because they

Version 1

9

A) destroy viral proteins.

B) look like hairpins to cause premature termination of viral mRNA. C) look like tRNAs and block viral translation.

D) look like nitrogenous bases and block viral

reverse transcriptase.

E) prevent the virus from releasing its genome into the cytoplasm.

35) A mutant tRNA that recognizes a nonsense codon and inserts an amino acid where protein synthesis would have otherwise ended is a D) uncharged A) nonsense tRNA. B) nonsense suppressor tRNA. C) revertant tRNA.

tRNA.

36) The ___________ codon is used as the start codon by nearly all organisms.

A) AUG methionine or formylmethionine B) UGA arginine or methylarginine C) AUG cysteine or methylcysteine

D) UGA methionine or formylmethionine

37) Mutations that completely abolish the function of a wild-type allele are A) null mutations.

B) hypomorphic mutations.

C) hypermorphic mutations.

Version 1

D) conditional mutations.

E) neomorphic mutations.

10

38) Mutations that result in the production of much less of a protein or a protein that functions less efficiently than wild type are A) null mutations.

B) hypomorphic mutations.

C) hypermorphic mutations.

D) conditional mutations.

E) neomorphic mutations.

39) A neomorphic mutation could result in an allele whose product is expressed in the normal time and place, but A) produces no gene product.

B) produces a nonfunctional gene product.

C) produces a novel protein that gives rise to a new phenotype.

D) produces mutant protein subunits that aggregate with wild-type subunits, inactivating them.

E) produces an altered protein that functions more efficiently than the wild-type protein does.

40) The appearance of a novel phenotype resulting from the substitution of a single base pair might be due to

A) a change in the amino acid sequence.

B) a change in the amount of protein expressed.

C) an alteration in a gene that codes for a nontranslated RNA. D) a change in the developmental time or location at

41) A mutation creates a dominant negative allele of a particular gene. The gene encodes a protein that forms a trimer within the cell. If one or more of the subunits has the

Version 1

which a gene is expressed.

E) All of the choices are possible consequences of a single base-pair substitution.

mutant structure, the entire trimeric protein is inactive.

In a heterozygous cell, if

11

the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be active?

A) 100% B) 87.5%

C) 50% D) 33% E) 12.5%

42) A neomorphic mutation in the Antennapedia ( Antp) gene of Drosophila causes A) kinks to form in their tails.

B) growth of antenna where legs should be.

C) total loss of color vision.

D) extra antennae on the head.

E) growth of legs where the antennae should be.

43) To prevent active proteases from damaging the secretory cells of the digestive system, some protease enzymes must be secreted in an inactive form and then activated later by proteolytic cleavage once outside of the secretory cells. What is the category of precursor proteins of this type?

A) ubiquitins B) zymogens

C) glycosides D) activins

44) Which of the following is required for initiation of eukaryotic translation?

B) 5' cap A) Shine-Dalgarno sequence Version 1

12

C) release factors D) N-formylmethionine E) introns

Version 1

13

45) The DNA sequence 5′ ATGGGGGACACC 3′ encodes the amino acids Met Gly Asp Thr. A mutation changes the sequence to 5′ ATGGGTGACACC 3′, but the same amino A) frameshift mutation B) missense mutation

46) The DNA sequence 5′ ATGGGGGACACC 3′ encodes the amino acids Met Gly Asp Thr. A mutation changes the sequence to 5′ ATGGGGAACACC 3′, which encodes the A) frameshift mutation B) missense mutation

acids are encoded. What type of mutation has occurred?

C) silent mutation D) nonsense mutation

amino acids Met Gly Asn Thr. What type of mutation has occurred?

C) silent mutation D) nonsense mutation

47) A DNA sequence 5′ ATGGGGGACACC 3′ encodes the amino acids Met Gly Asp Thr. A mutation changes the sequence to 5′ ATG AGGGGACACC 3′, which encodes the amino acids Met Arg Gly His. What type of mutation has occurred?

A) frameshift mutation B) missense mutation

C) silent mutation D) nonsense mutation

48) A mutant allele results in a truncated form of a signaling protein. This mutant protein can still

Version 1

14

bind its receptor but it does not activate the receptor and the mutant protein inhibits wild-type signaling proteins from binding the receptor. What type of allele is this mutant allele?

D) antimorphic A) hypomorphic allele B) hypermorphic allele C) neomorphic allele

49) Many cancers have a mutation that changes a valine to a glutamic acid in the RAF kinase protein which normally phosphorylates proteins that promote cell division. This mutant RAF protein is a constitutively active kinase. What A) hypomorphic allele B) hypermorphic allele

50) A large deletion removes the promoter and first three exons of a gene. None of the gene’s mRNA or protein is A) hypomorphic allele B) hypermorphic allele

allele

type of allele is this mutant allele?

C) null allele D) antimorphic allele

detected. What type of allele is this mutant allele?

C) null allele D) antimorphic allele

51) Which of the following mutant alleles is most likely to be dominant to the wild-type allele? activity A) a silent mutation B) a premature stop codon that results in a partial loss of protein function C) a missense mutation that increases enzyme

Version 1

D) a promoter mutation that reduces gene transcription

15

52) An anticodon of sequence 5′ GUA 3′ will recognize the codon sequence

A) 5′ GTA 3′.

B) 5′ UAC 3′.

C) 5′ CAU 3′.

Version 1

D) 5′ AUG 3′.

E) None of the codon sequences are correct.

16

53) An anticodon of sequence 5′ GUA 3′ will recognize the codon sequence

A) 5′ GTA 3′.

B) 5′ UAU 3′.

C) 5′ CAU 3′.

Version 1

D) 5′ AUG 3′.

E) None of the codon sequences are correct.

17

54) A tRNA with the anticodon sequence 5′ IUA 3′ will not normally exist because

Version 1

18

A) 5′ UAA 3′ is a stop codon.

B) I is not a base present in tRNAs. C) it will recognize codons for different amino acids.

D) 5′ UAU 3′ and 5′ UAC 3′ both specify tyrosine.

55) A tRNA with the anticodon sequence 5′ IUC 3′ will not normally exist

Version 1

19

because

Version 1

20

A) 5′ GAU 3′ and 5′ GAC 3′ specify aspartic acid, but 5′ GAA 3′ specifies glutamic acid.

B) 5′ GAU 3′ and 5′ GAC 3′ specify different amino acids.

C) I is not a base present in tRNAs. D) 5′ GAU 3′ and 5′ GAC 3′ both specify aspartic acid.

56) When the anticodon of a tRNA and the codon of an mRNA interact, the two sequences are A) identical.

B) antiparallel.

57)

The different tRNAs are produced by

A) different aminoacyl-tRNA synthetases that produce tRNAs from ribonucleotides in the cytoplasm.

B) RNA editing of transcripts of a single tRNA gene.

C) alternative splicing of a single tRNA gene.

58)

D) transcription of different tRNA genes in the genome.

The wobble base of a tRNA is A) the 5′ base of the anticodon.

B) the 3′ base of the anticodon.

C) the 5′ base of the codon.

59) can

C) parallel.

D) not related.

D) any base in the anticodon.

Proteins that bind to the enhancers of eukaryotic genes A) stimulate

Version 1

21

translation initiation.

B) promote the binding of sigma factor to RNA polymerase.

C) stabilize the interaction between the promoter and

basal factors.

D) facilitate formation of the 30S subunit ribosomal subunit.

60) In eukaryotes, recognition of the promoter region by RNA polymerase requires the clearing of __________ by __________ A) introns; release factors.

B) histones; coactivator proteins.

C) ribosomes; exons.

D) untranslated regions; spliceosomes.

61) In prokaryotes, mutations affecting the ribosome binding site would most likely A) lower the affinity of mRNA for the small ribosomal subunit.

B) result in premature initiation of translation.

C) produce longer-than-normal protein.

D) shift the reading frame.

62) In eukaryotes, a mutation in a splice-donor site of a gene would result in which of the following?

A) RNA polymerase would be unable to bind to the promoter sequence.

B) The gene would not be transcribed.

C) The reading frame would be disrupted.

Version 1

D) Proteins bound at the enhancer would be unable to remove histones.

22

Version 1

23

Answer Key Test name: Chapter 09 Test Bank 1) B 2) D 3) A 4) A 5) C 6) B 7) D 8) C 9) D 10) A 11) B 12) A 13) D 14) B 15) B 16) B 17) C 18) D 19) C Version 1

24

20) A 21) D 22) B 23) D 24) A 25) C 26) D 27) C 28) A 29) A 30) C 31) B 32) D 33) A 34) D 35) B 36) A 37) A 38) B 39) C 40) E Version 1

25

41) E 42) E 43) B 44) B 45) C 46) B 47) A 48) D 49) B 50) C 51) C 52) B 53) B 54) A 55) A 56) B 57) D 58) A 59) C 60) B 61) A Version 1

26

62) C

Version 1

27

Student name:__________

A) integrates into the host chromosome B) contains a selectable marker C) contains an origin of replication D) has several unique recognition sites for restriction

1) Which is a characteristic of a plasmid used as a cloning vector? enzymes E) can hold an insert of up to 300 kb

2) What is one difference between using restriction endonucleases and mechanical shearing of DNA?

A) Restriction endonucleases cut DNA at known specific sites while shearing occurs at random sites.

B) Restriction endonucleases cut DNA at random sites while shearing occurs at known specific sites.

C) DNA cut with restriction enzymes can be used for

cloning and DNA broken by shearing cannot.

D) Restriction enzymes produce larger fragments on average than shearing.

3) What is the origin and physiological function of restriction endonucleases?

A) Bacteriophage produce them to gain resistance to antibiotics.

B) Restriction endonucleases are required for DNA replication in yeast.

C) Bacteria produce them to protect against viral

invasion.

D) Scientists invented them and synthesize them in labs.

4) Why isn’t the bacterial chromosome digested by the restriction enzymes the bacteria produce?

Version 1

1

A) The sequence recognized and cleaved by a restriction enzyme is not found in the bacterial genome.

B) Bacterial cells have high concentrations of ligase which immediately repairs the any sites that are cleaved.

C) Methyl groups added to the bases of restriction enzyme recognition sites within the bacterial chromosome block cleavage.

D) Bacterial DNA is modified with phosphate groups

that prevent the function of restriction enzymes.

E) Restriction enzymes are found in the cytoplasm and DNA is only in the nucleus.

5) What is the purpose of ethidium bromide in DNA electrophoresis? by size A) To tag DNA fragments so they can be viewed under UV light B) To introduce mutations into the DNA C) To fragment the DNA before separating the pieces

D) To join fragments of DNA together

6) The recognition site for a restriction enzyme is 5′ ACCGG^T 3′. The ^ symbol indicates the site of cleavage.

After digestion of DNA with this restriction enzyme, how will the ends of the resulting DNA fragments look?

A) They will have sticky, single-stranded ends with 3′ overhangs.

B) They will have sticky, single-stranded ends with 5′ overhangs.

C) They will have blunt ends.

D) They will have sticky, double-stranded ends.

E) The result cannot be predicted using only the information given.

7)

Version 1

The recognition site 2

for a restriction enzyme is 5′ ACC^GGT 3′. The ^ symbol indicates the site of cleavage. After digestion of DNA with this restriction enzyme, how will the ends of the resulting DNA fragments look?

A) They will have sticky, single-stranded ends with 5′ overhangs.

B) They will have sticky, single-stranded ends with 3′ overhangs.

C) They will have blunt ends.

D) They will have sticky, double-stranded ends.

E) The result cannot be predicted using only this information.

8) In gel electrophoresis, DNA fragments move at different rates because they have different A) electrical charge densities.

B) sizes.

C) base sequences.

D) amino acid compositions.

E) overall electrical charges.

9) Imagine that DNA is composed of equal proportions of six different types of nucleotides, instead of the normal four. The six nucleotides are A, C, G, T, X, and Y. In this new type of DNA molecule, A pairs with T, C pairs with G, and X pairs with Y. What would be the average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with a four-base recognition sequence?

A) 125 bp B) 256 bp

Version 1

C) 425 bp D) 1056 bp E) 1296 bp

3

10) Which is an example of a recombinant DNA molecule?

A) a single-stranded RNA hybridized to a singlestranded DNA B) a human genomic DNA fragment ligated into a bacterial plasmid vector C) a bacteriophage chromosome in a bacterial cell

D) a bacterial plasmid cut with a restriction enzyme and then resealed using ligase

11) What is the function of the amp r gene in a plasmid vector?

A) It provides several unique restriction enzyme recognition sites for opening up the plasmid.

B) It allows a researcher to select for bacterial cells with a plasmid.

C) It contains an origin for replication of the bacterial chromosome.

D) It allows a researcher to distinguish plasmids with

12) Which vector would be best to use if you want to construct a genomic library with the largest fragments A) YAC B) BAC

13) is to

an insert from those without an insert.

E) It is required for packing plasmid DNA into bacteriophage heads.

possible?

C) Plasmid D) Virus

In Sanger sequencing, the role of the DNA polymerase

A) synthesize new Version 1

4

DNA strand that is complementary to the template strand.

B) proofread base pairs and correct errors.

C) synthesize a primer in the 5'→3' direction.

D) attach DNA fragments together with covalent bonds

14) In Sanger sequencing, what causes DNA synthesis to terminate at a specific base?

A) Restriction endonucleases cleave DNA after particular bases.

B) Fluorescent molecules produce high-energy light that inactivates DNA polymerase.

C) Incorporation of a deoxynucleotide that lacks a base causes DNA polymerase to stall.

D) Incorporation of a dideoxynucleotide that lacks a 3′ hydroxyl.

E) Incorporation of a ribonucleotide that lacks a 5' phosphate.

15) Why is it necessary to obtain overlapping sequences when sequencing a genome for the first time?

A) To assemble the whole genome-sequence virtually from random fragments B) To quantify the amount of DNA in the genome C) To make sure all the clones in a library are used

D) To allow for sequencing errors to be corrected

16) Which method of fragmenting DNA would not be useful for sequencing a genome for the first time?

A) Complete digestion with a restriction enzyme B) Partial digestion with a restriction enzyme C) Forcing DNA through a thin needle

Version 1

D) Exposing DNA to ultrasound energy

5

17) Paired-end sequencing provides a method for assembling genome sequences of organisms that contain a large number of long repeat sequences, such as transposable elements, in their genomes. Which one of the following A) Sequencing of the entire insert DNA fragment of BAC clones from a BAC clone library.

B) Sequencing of the entire insert DNA of clones from a library made using ~ 2 kb DNA fragments.

C) Sequencing of both ends of DNA inserts of BAC

18) The mouse genome consists of 2700 Mb. A genomic library is constructed using BAC vectors and overlapping fragments of the mouse genome. The average size of the

approaches is utilized for paired-end sequencing?

clones from a BAC clone library.

D) Sequencing of only one end of DNA inserts of BAC clones from a BAC clone library.

mouse DNA inserts is 200 kb.

18.1) How many clones would make up one genomic equivalent?

A) 135 B) 270

C) 13,500 D) 270,000

18.2) A mouse genomic library is constructed so that the number of clones equals one genomic equivalent. Is it likely that this library contains the entire mouse genome?

A) No; by chance some parts of the genome will be represented more than once and others not at all.

B) Yes; that is a sufficient number of clones for every part of the genome to be represented one time.

C) No; the restriction enzyme used to fragment the

Version 1

genome will delete some DNA on the ends of clones.

6

18.3) What is the minimum number of clones that should be in this mouse genomic library to have a 95% chance that it contains the entire genome?

A) 2,700 B) 13,500

C) 54,000 D) 135,000

19) You performed a Sanger sequencing reaction and obtained the following read. In this figure, A = green, C = purple, G = black, T = red. The height of the peaks is unimportant. The smallest molecule is at the left of the trace.

19.1) What is the sequence of the DNA synthesized in the sequencing reaction?

A) 5' TTTGCTTTGTGAGCGGATAACAA 3' B) 3' TTTGCTTTGTGAGCGGATAACAA 5' C) 5' AAACGAAACACTCGCCTATTGTT 3'

Version 1

D) 3' AAACGAAACACT CGCCTATTGTT 5'

7

19.2) What is the sequence of the template DNA used for this sequencing reaction?

A) 5' TTTGCTTTGTGAGCGGATAACAA 3' B) 3' TTTGCTTTGTGAGCGGATAACAA 5' C) 5' AAACGAAACACTCGCCTATTGTT 3'

D) 3' AAACGAAACACT CGCCTATTGTT 5'

20) In frogs, green skin is determined by gene B. The wild-type sequence of the 5′ end of the RNA-like strand of the entire first exon is shown below. The gene encodes an enzyme that functions to convert a brown compound into a green pigment in the skin.

5′ ACTCAAGCACAGGTCGCATAAATGTTCCTGTTATTTG G 3′

Version 1

8

20.1) Baby frog #1 was born with brown skin instead of the normal green. DNA was collected from the baby frog and the DNA sequence of part of gene B was determined using the primer 5′ ACTCAAGCACAGGTCG 3′. What is the entire sequence of the smallest DNA fragment produced in the sequencing reaction? (See the diagram for sequencing data.)

Version 1

9

A) C B) 5′ ACTCAAGCACAGGTCGC C) 5′ AGGTCGG

D) 5′ CATAAATGTCCT GTTATTTGG E) 5′ CGACCTG

20.2) Why was another nucleotide not added to the smallest fragment?

A) DNA polymerase became inactive after adding the final nucleotide.

B) The final nucleotide added does not have a 3′ hydroxyl group.

C) The final nucleotide added does not have a 5′ phosphate group.

D) The DNA strand reached a length where it cannot

Version 1

grow any longer.

E) DNA polymerase reached the end of the template.

10

Answer Key Test name: Chapter 10 Test Bank 1) [B, C, D] 2) A 3) C 4) C 5) A 6) A 7) C 8) B 9) E 10) B 11) B 12) A 13) A 14) D 15) A 16) A 17) C 18) Section Break 18.1) C Version 1

11

18.2) A 18.3) C 19) Section Break 19.1) A 19.2) D 20) Section Break 20.1) B 20.2) B

Version 1

12

Student name:__________ 1) What information cannot be determined from the base-pair sequence of a cDNA clone?

A) exon sequences B) sequence of the promoter C) similarity to previously identified sequences D) amino acid sequence of the encoded polypeptides

E) sequence of the spliced mRNA transcript

2) Which enzyme copies an mRNA sequence into DNA during cDNA library construction?

A) RNA polymerase B) DNA polymerase C) reverse transcriptase

D) restriction endonuclease

3) Reverse transcriptase is an enzyme encoded in the genomes of A) red blood cells.

B) phage λ.

C) retroviruses.

D) yeast.

E) E. coli.

4) How do the genes in the β-globin locus differ from each other?

A) They use different template strands for transcription.

B) They are expressed at different times in development.

Version 1

C) They evolved from different ancestral genes.

D) They are dispersed on different 1

chromosomes.

E) Some are expressed in all cells; others are

5)

expressed only in red blood cell precursors.

Pseudogenes are evidence of A) gene duplications.

B) integrated viruses, or proviruses.

C) natural selection for functional genes.

D) de novo genes.

E) proteins that are produced but have no function.

6) Use the BLAST tool at https://blast.ncbi.nlm.nih.gov/Blast.cgi and click “Protein blast” to find out the identity of the protein with the following amino acid sequence: CCKHPEAKRM PCAEDYLSVV LNQLCVLHEK TPVSDRVTKC CTESLVNRRP

A) myosin B) actin C) albumin

D) DNA polymerase E) tubulin

7) A BLAST search can be used for which of these applications?

A) to compare a nucleotide or amino acid sequence to databases from a variety of species B) to display the reading frame of a nucleotide sequence C) to determine whether a gene has an essential

Version 1

function in an organism D) to analyze microarray data E) to visualize the locations of all the genes on a single chromosome

2

8) What is the advantage of having different forms of hemoglobin?

A) Multiple forms allows for speciation.

B) The multiple forms all have the same function so that one can be mutated without affecting the individual.

C) There is no advantage because the forms arose by random chance.

D) The different forms are adapted to most

efficiently carry oxygen in different stages of development.

E) The different forms are adapted to most efficiently carry oxygen in different climates.

9) How are open reading frames or exons predicted from genomic sequence?

A) If the open reading frame is at least 10 sense codons long, it is considered an exon.

B) If the sequence is absent from mRNA, it is considered an exon.

C) If there is a splice donor and acceptor site at the ends of the open reading frame, it is considered an exon.

D) If an open reading frame is longer than 21 sense

codons, it is predicted to be an exon.

E) If an enhancer and promoter sequence are present, the downstream sequence is considered an exon.

10) Approximately what proportion of the human genome is composed of the exome?

A) < 0.1% B) 2%

Version 1

C) 10% D) 20% E) 45%

3

11) Which of the following is a way that protein function can be modified AFTER translation?

A) protein cleavage B) transcription regulation C) gene duplication

D) exon shuffling E) gene rearrangement

12) Which of the following processes can generate orthologous genes over evolutionary time?

A) protein cleavage B) transcription regulation C) gene duplication

D) chromosomal inversion E) mutation that creates a new ATG

13) Which of the following is a potential mechanism through which proteins can acquire novel domain architectures over evolutionary time?

A) protein cleavage B) exon shuffling C) mutation in a pseudogene

D) duplication and divergence E) de novo mutation

14) Which disease results from deletion of the β-globin gene cluster LCR?

Version 1

4

A) sickle-cell disease B) mild α-thalassemia C) severe α-thalassemia

D) mild βthalassemia E) severe βthalassemia

15) What do you predict would be the effect of an inversion ofthe β-globin gene cluster with respect to the LCR?

A) No globin genes will be expressed and the fetus will die in utero.

B) ε-globin will be expressed first during development.

C) The order of globin gene expression will be

reversed; the effect on the fetus cannot be predicted.

D) The individual will develop sickle-cell disease in adult life.

16) What is one way that genes can be predicted in genome sequences?

A) identifying the sequences most conserved between different species B) identifying the sequences least conserved between different species C) identifying the regions most conserved between individuals of the same species

17)

D) identifying regions that have undergone the most rapid evolution E) identifying sequences that have been duplicated

What is a reference sequence?

A) A single, complete, annotated version of a species’ genomic DNA sequence that can be accessed for Version 1

bioinformatic studies.

B) A constantly updated collection of DNA 5

sequences from all species that give the latest information for bioinformatic studies.

C) A program to compare genome sequences.

D) A program that identifies open reading frames.

18) Which experiment would best determine which genes are expressed in a Drosophila embryo?

A) analysis of a Drosophila genomic DNA library B) analysis of a cDNA library from adult male Drosophila C) analysis of a cDNA library from Drosophila embryos

D) identifying all the open reading frames in the Drosophila genome E) identifying all the promoters in the Drosophila genome

19) Which piece of evidence is most likely to indicate a gene within a genome?

A) A unique sequence within a genome is 80% similar in human, mouse, and zebrafish.

B) A sequence is unique to a nematode genome.

C) An identical open reading frame is found in two species of fruit fly.

D) A sequence is repeated multiple times in the

human genome.

E) A sequence is 50% identical between human and chimp but does not include an open reading frame.

20) A student analyzes a genomic DNA sequence of 100 nucleotides in order to determine if it is part of a gene. When she translates the sequence beginning at the first nucleotide, she finds that the tenth codon is TAA (stop codon). What should she do next?

A) Tell her professor that the sequence does not

Version 1

represent an exon.

B) Tell her

6

professor that the sequence encodes a peptide with nine amino acids.

C) Translate the sequence in the other five reading frames.

21)

D) Use the genomic DNA to make a cDNA of the sequence.

A gene desert is

A) a unique sequence of DNA in a genome.

B) a region of a genome containing very few genes.

C) a gene that is transcribed only in dry climates.

D) a region of a genome containing many open reading frames but no promoter.

E) a region of a genome in which all genes are transcribed at low levels.

22) If four genes are in the same region of a eukaryotic chromosome, then A) the four promoters must be oriented such that two genes are transcribed using the Watson strand as a template and two genes use the Crick strand.

B) the four promoters must be oriented such that all four genes are transcribed using the same strand as the template.

C) the four promoters may be oriented such that some of the genes are transcribed using the Watson strand as

23)

template, and others use the Crick strand.

D) the genes can be transcribed from different strands only if their promoters are more than 1 kilobase apart.

E) the genes must be in a gene family.

Which of these examples represents a gene family?

A) Many genes in a genome that encode myosin motor proteins B) All of the genes that are required for gamete formation in humans

Version 1

C) Orthogolous genes encoding a specific kinase in humans, chimps, and mice D) Many

7

pseudogenes present in a single genome E) All of the genes in an organism that are expressed

in a single tissue

24) The arms of human chromosome 2 are syntenic with two different chimpanzee chromosomes. What mechanism

could have caused this pattern of synteny?

A) an inversion in the human chromosome B) an inversion in one of the chimpanzee chromosomes C) gene duplication in the human chromosome D) exon shuffling in the chimpanzee chromosomes

25) A single eukaryotic gene can produce two different forms of a protein, one that is membrane bound and one that is secreted. Which of these mechanisms allows a single gene A) alternative splicing B) gene divergence C) gene duplication

26) Each human α-globin locus contains two α-globin genes, α1 and α2, and most individuals have two copies of the locus. Which of the genotypes would result in the most severe A) α1 α1 / α2 α2 B) α1 – / α2 –

E) a translocation involving two chromosomes

to produce two different proteins from one type of primary transcript?

D) exon shuffling E) gene rearrangement

mutant phenotype? (A"–" represents a deletion allele.)

C) α1 α1 / – – D) α1 – / α2 α2 E) α1 – / – –

Answer Key Version 1

8

Test name: Chapter 11 Test Bank 1) B 2) C 3) C 4) B 5) A 6) C 7) A 8) D 9) D 10) B 11) A 12) C 13) B 14) E 15) C 16) A 17) A 18) C 19) A

Version 1

9

20) C 21) B 22) C 23) A 24) E 25) A 26) E

Version 1

10

Student name:__________

A) Individuals with Huntington disease have more trinucleotide repeats in the coding region of the HD gene than normal individuals.

B) A single base near the neurofibromatosis gene can be a G or a T; phenotype is not affected.

C) A single base change in the gene for β globin changes an amino acid and results in sickle-cell disease.

1) Which is an example of a SNP? D) The most common cystic fibrosis allele has a deletion of three base pairs in the coding region of the CFTR gene.

2) Which are typically used for positional cloning of a disease gene in humans?

A) analysis of several human pedigrees in which the disease is segregating B) genotyping of diseased and healthy individuals at millions of anonymous polymorphisms using microarrays

C) sequencing candidate genes D) a multigenerational selective breeding program

3) What mechanisms contribute to the formation of deletions and insertions?

A) Repair of double-strand breaks in DNA B) Repair of thymine dimers in DNA C) Errors in DNA replication

D) Unequal crossing-over due to mispairing of homologs

4) Which are limitations of preimplantation embryo diagnosis?

A) The technique is technologically complex and expensive.

Version 1

B) The test increases the rate of identical twin births.

1

C) Removal of one cell from an eight-cell embryo results in permanent damage to the embryo.

D) Only certain well-defined genetic diseases can be

tested.

5) How can databases of variants be used to help pinpoint a disease-causing mutation?

A) Some amino acids are conserved in genes across species; mutations in conserved amino acids are not likely to cause disease.

B) Polymorphisms found in individuals that do not have the syndrome can sometimes be eliminated from consideration.

C) Common polymorphisms are unlikely to cause a

rare disease and they can be eliminated from consideration.

D) Mutations that occur between genes are most likely to cause a genetic disease.

6) If an individual is a compound heterozygote at the CFTR locus, the gene that causes cystic fibrosis, what can be inferred?

A) If the individual has children, all the children will have cystic fibrosis.

B) At least one of the individual’s parents had cystic fibrosis.

C) The individual is a carrier for cystic fibrosis, but

does not have the disease.

D) A drug that effectively treats one allele may not treat the other.

7) Most polymorphisms do not result in a phenotypic difference because they are typically

A) nonsense mutations.

Version 1

B) either missense or neutral mutations.

2

C) either silent mutations or are in non-coding regions.

D) either missense mutations or are in promoter

8) true?

regions.

Which statement about SNPs in the human genome is

A) Most SNPs have an effect on phenotype.

B) Any two human genome copies will have on average 3 million single nucleotide polymorphisms.

C) SNPs refer only to deletions or insertions, not

base substitutions.

D) Most SNPs are located in the introns of genes, and thus effect phenotype.

9) The most common mutant allele of the PAH gene, which is responsible for the metabolic disorder PKU, has a SNP in the splice donor site of one intron. What is the simplest way to detect this allele?

A) PCR using genomic DNA as template and two primers, one on either side of the SNP, followed by gel electrophoresis B) PCR using genomic DNA as template and two primers, one complementary to the region containing the SNP, followed by gel electrophoresis C) PCR using two primers, one on either side of the

SNP, followed by sequencing the PCR products D) Direct exome sequencing using a microarray E) Whole-genome sequencing

10) Fragile X syndrome is caused by expansion of a trinucleotide repeat in the 5′ UTR of the FMR-1 gene. What

is the simplest way to detect this expansion?

A) PCR using genomic DNA as template and two Version 1

primers that are complementary to repeats 3

on either side of the expansion, followed by gel electrophoresis B) PCR using genomic DNA as template and two primers, one complementary to the repeat region and the other complementary to unique sequence, followed by gel electrophoresis C) PCR using two primers that are complementary to unique sequences on either side of the repeat region, followed

11)

by either electrophoresis or sequencing of the PCR products.

D) Direct exome sequencing using a microarray E) Whole-genome sequencing

Which does a successful PCR require?

A) at least 100 starting DNA template molecules B) some sequence information about the region to be amplified C) a cloned cDNA of the region to be amplified D) a double-stranded DNA template of at least 100

kb to amplify E) an undamaged DNA template with intact chromosomes

12) If a double-stranded DNA molecule is amplified by PCR, how many double-stranded DNA molecules will exist after FIVE cycles?

A) 6 B) 8

C) 10 D) 32 E) 64

13) Allele-specific oligonucleotides (ASOs) for the normal and disease alleles of a gene are in one section of a microarray. The disease is a recessive trait. If a probe made from an individual’s genomic DNA hybridizes with both ASOs, what can be inferred about the individual?

Version 1

4

A) the individual has two normal alleles B) the individual has the disease C) Each of the individual's children has a 50%

14)

chance of being a carrier.

D) one of the individual’s parents has the disease

Positional cloning depends on knowing what? gene A) the function of a gene B) the expression pattern of a gene C) the map location of markers that are linked to a

D) the sequence of a gene

15) Which is the most common type of DNA polymorphism?

D) copy number A) single nucleotide polymorphism B) deletion/insertion polymorphism C) simple sequence repeat

16)

variant

Simple sequence repeats are

A) found only in coding regions of genes.

B) sequences of 1 to 10 base pairs repeated a variable number of times in tandem.

C) sequences that are 10 base pairs long or longer

and found in several locations in the genome.

D) members of a gene family.

17) Simple sequence repeat loci are highly variable because of what process?

Version 1

5

light A) repair of double strand breaks that result from exposure to X-rays B) exposure to DNA damaging chemicals found in food and water C) the formation of thymine dimers by ultraviolet

D) slipped mispairing during DNA replication

18) How many bases do deletion–insertion polymorphisms (DIPs) most frequently involve?

A) 1–2 B) multiples of 3

C) 10–100 D) more than 100

19) The classical form of the metabolic disease phenylketonuria is caused by a mutation in the gene encoding the enzyme phenylalanine hydroxylase (PAH), which converts phenylalanine to tyrosine. A variant form of phenylketonuria is caused by a mutation in a separate gene that encodes a different enzyme involved in the synthesis of a cofactor needed for PAH to function. Which of the following phenomena is illustrated by these two forms of phenylketonuria?

A) Allelic heterogeneity B) Compound heterozygosity C) Locus heterogeneity

D) Anonymous polymorphism

20) Examine this pedigree that shows segregation of Huntington disease and the DNA Version 1

6

marker G8. What was the most likely genotype of individual V-3 at the DNA marker?

A) BC B) AB C) CC

D) AC E) All of the possible genotypes are equally likely.

21) If the 13 CODIS SSR loci are analyzed, why does each person (except identical twins) have a unique DNA profile (unique combination of SSR alleles)?

A) Each SSR locus has many alleles, each with a different number of repeats.

B) SSRs have a very low mutation rate in individuals.

C) SSR inheritance follows Mendel's law of

segregation.

D) Mitotic recombination results in new combinations of SSRs in different parts of an individual’s body.

22) Which is a challenge of using pedigrees for positional cloning?

A) Some matings may not be informative.

Version 1

B) Recombination occurs during meiosis.

7

C) Many individuals in a pedigree are unaffected.

D) Pedigrees are not based on DNA sequences.

23) Which is part of a DNA sequencing technique that is useful for high-throughput sequencing, but is not part of Sanger DNA sequencing?

A) Using dideoxynucleotides as chain terminators B) Separating DNA by size using gel electrophoresis C) Hybridization between the template and a primer D) Removal of a chemical group that blocks the 3′

end of the new DNA strand E) Labeling deoxynucleotides with fluorescent tags

24) What technique can help identify a disease gene quickly by narrowing the focus?

A) Compare the genome sequences of different species to identify amino acids that are conserved in various encoded proteins.

B) Filter the results to include only common polymorphisms.

C) Determine genotypes of parents and children at all SSR loci.

D) Use microarrays to identify silent mutations.

25) Which polymorphism is most likely to affect an individual’s phenotype?

A) A SNP in an intron of a protein-coding gene B) An SSR within a telomeric sequence C) A DIP in an intragenic region between genes Version 1

D) A SNP in the start codon of a proteincoding gene

8

26) What is the name of the process for removing fluid that surrounds a fetus to obtain fetal cells for genetic testing?

A) amniocentesis B) polymerase chain reaction C) preimplantation genetic diagnosis

27) You are studying a disease that is known to be caused by a single nucleotide change in a single gene, although the effect this change ultimately has on the protein's structure and function is unknown. You have DNA samples from multiple patients that you suspect of having this disease. What is the A) Preimplantation genetic diagnosis B) PCR amplification followed by Sanger DNA sequencing C) PCR amplification followed by gel electrophoresis

D) in vitro fertilization

most efficient way to test the samples for the relevant mutation?

D) Highthroughput exome sequencing E) Highthroughput genome sequencing

28) In the United States, most newborns undergo a screening test for up to 60 genetic disorders. A few drops of blood is taken from each baby and the levels of various blood components are determined. The screening identifies babies who might have a genetic disorder, but is not diagnostic.

Further testing is required if the screen shows a blood component is out of the normal range. Which of the following technologies could potentially be a low-cost replacement for the current newborn screening and would test most babies directly for the presence of many genetic disorders?

A) PCR to Version 1

9

amplify the PKU gene B) exome sequencing and comparison with known mutations in databases C) DNA profiling using CODIS SSRs

D) preimplantation genetic diagnosis

29) Identifying candidates for a disease gene by positional cloning depends on distinguishing DNA markers that are close to the disease gene from markers that are far away.

When the disease gene is close to a marker, there will be

A) a high number of mutations in the intervening DNA. B) a low rate of recombination between the disease gene and the marker.

C) many alleles at the marker locus.

D) a high chance of locus heterogeneity.

30) Herceptin® is a drug that is given to treat certain breast cancers. However, it is most effective on tumors that are overexpressing the gene HER2. Which of the following strategies would be best for determining whether Herceptin® would be effective in a given patient?

A) Sequence DNA from the patient’s sperm or eggs to determine whether the patient’s germ cells contain gain-offunction mutations in HER2, and treat with Herceptin® only if they do.

B) Determine the patient’s HER2 sequence and compare it to a database of known HER2 mutations to determine whether the overexpression allele is present; if the overexpression allele is present, give Herceptin® .

C) Determine the patient’s HER2 sequence and compare it to a database of known HER2 mutations to

Version 1

confirm that the patient has a mutation in the gene; give Herceptin® if any mutation is found.

D) Perform microarray analysis to determine the location of the patient's HER2 gene and give Herceptin® if the HER2 gene is not in the normal location.

10

31) How can microarrays differentiate between a wildtype allele and a disease allele that differ at only one nucleotide?

A) The DNA sequences of both alleles are determined.

B) Oligonucleotides hybridize with the two alleles differently.

C) Polymerase chain reaction primers amplify DNA

from one allele, but not the other.

D) Microarrays are not capable of detecting a difference between those alleles.

32) Linkage between a gene of interest and a DNA marker is being studied. A mating results in 6 parental and 3 recombinant offspring.

32.1) What is the maximum Lod score for this mating?

A) 0.22 B) 1.1

32.2) The results of this mating are sufficient to conclude with confidence that the gene of interest and ⊚ ⊚

Version 1

C) 6.3 D) 11

the DNA marker are linked.

true false

11

32.3) What additional data would increase your confidence that the gene of interest and the DNA marker are linked?

A) Analysis of a mating from a different pedigree that produced 6 parental and 2 recombinant offspring.

B) Analysis of another mating from the same pedigree that produced 4 parental and 4 recombinant offspring.

C) Analysis of a mating from a different pedigree that produced 4 parental and 4 recombinant offspring.

D) Analysis of another mating from the same pedigree that produced 6 parental and 2 recombinant offspring.

33) Linkage between a gene of interest and a DNA marker is being studied. A mating results in 19 parental and 1 recombinant offspring.

33.1) What is the maximum Lod score for linkage at this locus?

A) 0.22 B) 1.1

33.2) The results of this mating are sufficient to conclude with confidence that the gene of interest and ⊚ ⊚

Version 1

C) 6.3 D) 11

the DNA marker are linked.

true false

12

34) In 2008, Margaret Binkele was murdered in her Georgia home. The number of repeats at three CODIS simple sequence repeat (SSR) loci was determined using crime scene samples and blood from three suspects. The results are Crime scene Suspect 1 sample

Suspect 2

Suspect 3

SSR locus 1

10, 10

8, 10

6, 7

10, 10

SSR locus 2

11, 11

11, 11

9, 15

10, 12

SSR locus 3

6, 9

6, 7

6, 9

6, 9

Version 1

summarized in the table.

Genotypes at three SSR loci

13

34.1) According to the data, the crime scene sample matches the sample from which suspect?

C) Suspect 3 D) None of the suspects is a match

A) Suspect 1 B) Suspect 2

34.2) In 2008, Margaret Binkele was murdered in her Georgia home. The number of repeats at three CODIS simple sequence repeat (SSR) loci was determined using crime scene samples and blood from three suspects. The results are summarized in the table.

Genotypes at three SSR loci

Crime scene Suspect 1 sample

Suspect 2

Suspect 3

SSR locus 1

10, 10

8, 10

6, 7

10, 10

SSR locus 2

11, 11

11, 11

9, 15

10, 12

SSR locus 3

6, 9

6, 7

6, 9

6, 9

Version 1

A) A receipt that showed the man used his credit card in Texas on the night of the murder B) The genotype at additional SSR loci that show a match at seven and a mismatch at three loci.

C) Discovery of the man’s fingerprints on the 14

murder weapon D) A match between his blood and the crime scene

Version 1

sample at more than three SSR loci

15

Answer Key Test name: Chapter 12 Test Bank 1) [B, C] 2) [A, B, C] 3) [A, C, D] 4) [A, D] 5) [B, C] 6) D 7) C 8) B 9) C 10) C 11) B 12) D 13) C 14) C 15) A 16) B 17) D 18) A 19) C Version 1

16

20) D 21) A 22) A 23) D 24) A 25) D 26) A 27) B 28) B 29) B 30) B 31) B 32) Section Break 32.1) A 32.2) FALSE 32.3) [A, D] 33) Section Break 33.1) C 33.2) TRUE 34) Section Break 34.1) D Version 1

17

34.2) [C, D]

Version 1

18

Student name:__________ 1) The enzyme that some eukaryotic organisms utilize for RNA-dependent DNA synthesis at the ends of linear chromosomes is called

A) DNApolymerase.

B) telomerase.

C) DNA ligase.

D) replicase.

2) When a cell loses telomerase activity, what is the predicted effect?

A) Telomeres will increase slightly with each cell division.

B) After 30–50 divisions, the cell will show signs of senescence and then die.

C) An immune stem cell will increase white blood cell production.

D) A normal somatic cell will gain the ability to divide indefinitely.

E) The cell will die immediately.

3) The components of a chromosome include one long DNA molecule and A) phospholipids.

B) proteins.

4)

C) carbohydrates.

D) steroids.

E) chromophores.

By weight, chromatin consists roughly of A) 1/3 DNA, 1/3 histones, and 1/3 nonhistones.

B) 1/3 DNAand2/3 acidic proteins.

C) 1/2 DNA, 1/4 histones, and 1/4 basic proteins.

Version 1

D) 1/4 DNA, 1/4 RNA, 1/4 histones, and 1/4 nonhistones.

1

5) Histones and DNA have a strong attraction for each other because A) DNA is positively charged and histones are negatively charged.

B) both DNA and proteins are hydrophobic.

C) DNA is negatively charged and histones are

positively charged.

D) like substances share common charges.

6) Select the statement that correctly describes the roles of histone and nonhistone proteins found in chromatin?

A) Only histones have a structural role in chromatin.

B) Only nonhistone proteins have a structural role in chromatin.

C) Histone proteins function primarily in chromosome segregation.

D) Both histone and nonhistone proteins have a

7)

structural role in chromatin.

E) Nonhistone proteins function only in replication and chromosome segregation.

The first level of compaction of DNA consists of

A) DNA winding around histones to form nucleosomes.

B) tight coiling of DNA with nucleosomes into higher order structures.

C) compaction into metaphase chromosomes.

Version 1

D) histone, DNA, and nonhistone covalent bonding.

2

A) H4.

B) H3.

C) H2.

D) H1.

9) In DNA, the 300 Å fiber is formed into structural loops. This function appears to be associated with A) histone H4.

B) nucleosomes.

C) histone H1.

D) certain nonhistone proteins.

E) remodeling complexes.

10) In what stage of mitosis are the highly-compacted chromosomes used for FISH? A) prophase B) metaphase C) anaphase

D) telophase E) never achieves this level of compaction

11) The total compaction of DNA as seen in metaphase chromosomes is approximately ______-fold.

A) 10,000 B) 50,000

C) 100,000 D) 40

12) A technique for visualizing mitotic chromosomes involves staining of chromosomes using Giemsa stain. At high resolution we see _______ of bands in a normal human diploid karyotype.

Version 1

3

A) hundreds B) thousands

13)

During DNA replication in a typical human cell,

A) only one origin of replication exists for each chromosome.

B) multiple origins of replication function consecutively.

C) multiple origins of replication function simultaneously.

14)

C) millions D) billions

D) the origins of replication are at the centromeres.

E) the origins of replication are at the telomeres.

In mammalian cells, replication proceeds A) unidirectionally B) bidirectionally.

C) unidirectionally from many origins.

15) Yeast research with autonomously replicating sequences (ARS), along with digestion of chromatin with A) DNA is not protected inside a nucleosome.

B) ARSs are really plasmids.

C) ARSs bond irreversibly to replication enzymes.

16) In a typical human cell, DNA polymerase synthesizes DNA at a rate of approximately 50 nucleotides per second.

Human chromosome 1 contains about 250 million base pairs.

Version 1

D) bidirectionally from a single origin.

DNase I, have led scientists to determine that D) origins of replication are accessible regions of DNA devoid of nucleosomes.

If a single origin of

4

replication were used, how long would it take the cell to copy A) less than 24 hours B) 1 to 2 days C) 1 to 2 weeks

17) Telomeres have at least two roles in chromosome integrity. Telomeres are composed of repeat sequences identical to the sequence of RNA in telomerase used for A) allow the chromosome to shorten each generation to speed up replication.

B) interact with shelterin complexes that protect chromosome ends from degradation.

C) permit the hairpin turn at the end to be cleaved.

this chromosome?

D) 50 to 60 days E) 110 to 120 days

extending chromosome ends. Second, they D) carry genes needed for DNA replication.

18) The two chromatids of each replicated chromosome must separate from one another and segregate during A) mitosis.

B) meiosis I. C) meiosis II.

19)

D) both mitosis and meiosis I. E) both mitosis and meiosis II.

In DNA, most satellite sequences are found in A) chromosome arms.

B) telomeres.

Version 1

C) centromeres.

D) spaces around the dark bands.

5

20) One of the functions of centromere is to form kinetochore for proper chromosome segregation. The other function is to A) hold sister chromatids together.

B) ensure that proper chromosome arm length is maintained.

C) allow easy karyotyping.

21)

Cohesins are multisubunit protein complexes that

A) hold sister chromatids together.

B) ensure that proper chromosome arm length is maintained.

C) allow easy karyotyping.

22)

D) allow replication of chromosome ends.

During mitosis, kinetochores assemble during A) telophase.

B) anaphase.

23)

D) allow replication of chromosome ends.

C) metaphase.

D) prophase.

In yeast chromosomes, centromeres A) help distinguish one chromosome from another.

B) are closely related in sequence.

C) are only 10–15 bp long.

Version 1

D) contain satellite DNA.

6

A) one spindle fiber.

B) one spindle fiber on each side.

C) multiple spindle fibers.

D) multiple repeating structural subunits.

25) The single-celled yeast, Saccharomyces cerevisiae, was used to construct the first eukaryotic artificial chromosome. This was a significant achievement because

A) yeast centromeres are identical to human centromeres.

B) it defined chromosomal regions that serve as centromeres and origins of replication in yeast.

C) it revealed the rate of DNA replication in yeast.

D) it demonstrated that some yeast genes are not

26) Which sequence of elements represents a complete yeast artificial chromosome (YAC) that is most likely to A) telomere — ARS — centromere B) telomere — ARS — centromere — 100,000 bp insert C) telomere — centromere — 100,000 bp insert — telomere D) telomere — ARS — centromere — 11,000 bp

essential.

E) it defined the chromosomal region with the TRP + gene for tryptophan synthesis in yeast.

replicate and segregate correctly? insert — telomere — suitable selectable markers E) telomere — ARS — centromere — 100,000 bp insert — telomere

27) Sites of transcription and therefore most of the genes along the length of the chromosome appear to be found in C) constitutive A) heterochromatin.

B) euchromatin.

Version 1

7

heterochromatin.

D) centromeric regions.

Version 1

8

28) When a chromosomal rearrangement, such as an inversion, places a known gene into or next to a A) will turn on.

B) may be amplified.

heterochromatic region, the gene's transcription C) may cease.

D) may be inverted.

29) Position-effect variegation of red and white eye color in Drosophila produces

A) one red eye and one white eyes.

B) two red or two white eyes depending on which gene is dominant.

C) eyes alternating red and white eye facets.

D) eyes with red and white patches of varying sizes and positions.

30) In which of the following cases will a Barr body be seen?

A) only XX B) XY C) XO

31) During development, an XX human female embryo will condense one of the X chromosomes in each cell into a

D) only XXY E) both XX and XXY

Barr body. This usually occurs C) at the 16-cell

A) at the 500–1000 cell stage, about two weeks after zygote formation.

B) by the end of the first trimester.

Version 1

stage.

D) in the X that has the most recessive

9

alleles.

Version 1

10

32)

FISH analysis is useful for determining the A) order of DNAfragments in a BAC. B) pattern of expression of a cloned gene.

C) chromosomal location of a gene.

D) map order of two closely linked genes.

33) You treat a sample of chromatin with DNase and run the digested fragments out on an electrophoretic gel. You find mostly long DNA fragments. What can you conclude about this sample of chromatin?

A) The sample contains mostly euchromatin.

B) The sample contains relatively few histone proteins.

C) The sample contains mostly heterochromatin.

D) The DNA is from yeast cells.

E) The sample likely represents actively transcribed genes.

34) You are studying a histone complex that contains an unusually high level of modification with acetyl groups. What can you conclude about the DNA bound to this histone complex?

A) It contains actively transcribed genes.

B) It is most likely heterochromatin.

C) It is most likely telomeric.

D) It is most likely derived from a condensed Barr body.

35) Which of the following predictions Version 1

11

could you make about a cell in which the HP1 protein was

A) Methylation of DNA would increase.

B) All of the cell’s chromosomes would be completely condensed into heterochromatin.

C) Histone methyltransferase would bind to

disabled? chromatin more easily.

D) Heterochromatin would not be formed.

36) Which of the following is a predicted consequence of kinetochores not achieving the correct level of tension from the mitotic spindle?

A) Sister chromatids will no longer be held together at their centromeres.

B) All chromosomes will migrate to one pole instead of to two opposite poles.

C) Mitosis will not progress beyond metaphase.

D) No prediction can be made.

37) When constructing a YAC, why must a large DNA insert be included?

A) The inserted DNA base pairs with DNA in the genome.

B) YACs that are too small do not segregate properly during mitosis.

C) The inserted DNA functions as telomeres.

D) The inserted DNA contains origins of replication

38)

needed for maintenance of the YAC. E) The inserted DNA allows YACs to be used in both eukaryotic and prokaryotic cells.

What is the best definition for the FISH technique?

Version 1

12

A) A method to fluorescently label different genes on metaphase chromosomes.

B) A method to fluorescently label methylated histones associated with chromosomes.

C) A method to fluorescently label heterochromatin.

D) A method to fluorescently label euchromatin.

39) What is true about new nucleosome formation during DNA replication?

A) Nucleosomes will be composed of different combinations of H3/H4 tetramers and H2A/H2B dimers that are either newly synthesized or were from previous histone octamers.

B) Nucleosomes will be composed of different combinations of H3/H2A tetramers and H4/H2B dimers that are either newly synthesized or were from previous histone octamers.

C) One DNA strand will have newly synthesized octamers and the other will have previously synthesized

40)

octamers.

D) Nucleosomes will be composed of different combinations of H2A/H2B/H3/H4 tetramers that are either newly synthesized or were from previous histone octamers.

What are telomeres?

A) short, repetitive DNA sequences found at the ends of linear eukaryotic chromosomes B) long, repetitive DNA sequences found at the ends of linear eukaryotic chromosomes C) short, repetitive DNA sequences found at the ends

Version 1

of linear prokaryotic chromosomes D) long, unique DNA sequences found at the ends of linear eukaryotic chromosomes

13

41)

Which statement correctly describes the similarity of

centromeres in different

A) Both yeast and plant centromeres have two 10 bp to 15 bp conserved elements.

B) Yeast centromeres have fewer satellite DNA repeats than do human centromeres.

C) Both human and plant centromeres are composed of repeated satellite DNA sequences.

D) Yeast centromeres have fewer satellite DNA repeats than do plant centromeres.

Version 1

14

42) The technique of preparing mitotic chromosomes involves a staining procedure using Giemsa stain. In this procedure, chromosomes are first ______ and then exposed to A) heated; hundreds of reproducible bands B) heated; thousands of reproducible bands C) fragmented; hundreds of reproducible bands

stain, producing ______ in a standard human karyotype.

D) fragmented; hundreds of variable bands

43) The creation of yeast artificial chromosomes preceded the production of the first yeast A) synthetic chromosome that includes sequences from the human genome.

B) synthetic chromosome that lacks about 50,000 bp from the normal yeast third chromosome.

C) essential chromosome that combines the 1,000

essential yeast genes into one chromosome.

D) circular chromosome that does not require telomere sequences.

44) Suppose a cell lacks functional shugoshin. What is the predicted effect on cohesin?

A) Cohesin will be cleaved prior to metaphase in mitosis.

B) Cohesin will not be cleaved during mitosis or meiosis.

C) Cohesin at the centromeres of sister chromatids will be cleaved during meiosis I.

D) Cohesin on the arms of sister chromatids will be cleaved during meiosis II. E) Cohesin will be cleaved normally during mitosis and meiosis I.

45) Version 1

The position of a 15

gene located within a light G band on the short arm of chromosome 5 could be designated as A) 5p13-pter.

B) 5p13-qter.

C) p513.

D) 5q13-qter.

E) q513.

46) Acetylated histone tail lysines are associated with chromatin that is A) open.

B) closed.

C) transcriptionally inactive.

D) newly replicated.

E) about to replicate.

47) Methylated histone tail amino acids are associated with chromatin that is A) either open or closed.

B) open only.

C) closed only.

D) newly replicated.

E) about to replicate.

48) Probes used in fluorescence in situ hybridization (FISH) and spectral karyotyping (SKY) techniques contain A) DNA only.

B) fluorescent tag(s) only.

C) DNA attached to fluorescent tag(s).

Version 1

D) proteins.

E) proteins attached to fluorescent tag(s).

16

49) When eukaryotic chromosomes are gently treated with detergents to remove histones and some nonhistone proteins and then observed with an electron microscope, the chromosomal DNA appears as _____________ formed by A) helixes; cohesins B) loops; condensins

50) A barrier element between a gene and heterochromatin would prevent that gene from being transcriptionally silenced ⊚ ⊚

protein complexes, called _________________.

C) fibers; histones D) beads; scaffold

by heterochromatin spreading.

true false

51) Kinetochores include motor proteins that exert pulling forces on spindle microtubules. ⊚ ⊚

Version 1

true false

17

Answer Key Test name: Chapter 13 Test Bank 1) B 2) B 3) B 4) A 5) C 6) D 7) A 8) D 9) D 10) B 11) A 12) A 13) C 14) B 15) D 16) D 17) B 18) E 19) C Version 1

18

20) A 21) A 22) D 23) B 24) C 25) B 26) E 27) B 28) C 29) D 30) E 31) A 32) C 33) C 34) A 35) D 36) C 37) B 38) A 39) A 40) A Version 1

19

41) C 42) A 43) B 44) C 45) A 46) A 47) A 48) C 49) B 50) TRUE 51) TRUE

Version 1

20

Student name:__________

A) silent point mutations B) translocations C) duplications

2)

1) FISH analysis is likely to detect which type of change in DNA? D) frameshift point mutations E) deletions

Which process can cause duplications?

A) repair of one double-strand break by homologous recombination B) repair of two double-strand breaks that occur in different places on different sister chromatids C) repair of two double-strand breaks that occur in the same chromatid by nonhomologous end-joining

D) misalignment of homologous chromosomes at repetitive sequences followed by crossing-over E) a mutagen that introduces point mutations

3) Which are possible results of an intragenic inversion (an inversion contained within a gene)?

A) The order of genes along the chromosome may be different than normal.

B) A normal protein may be produced.

C) Some of the gene’s DNA sequences will be adjacent to DNA sequences to which they are not normally

adjacent.

D) All of the gene’s A, C, G, and T bases remain in the same order as normal.

4) When a crossover occurs within the inversion loop of a pericentric inversion, each recombinant chromatid will have A) a dicentric bridge.

B) a duplication of some genes in the inverted region.

Version 1

C) one copy of each gene, with some genes in a different order 1

than normal.

D) a deletion of some genes in the inverted region.

E) one copy of each gene in the normal order.

5) Which gamete genotypes are expected to contribute to viable offspring when an individual is heterozygous for a paracentric inversion and crossing-over occurs as shown in the diagram? (The ● symbol represents a centromere.)

A) e b c D E B) ● A B C D E C) ● A B C d e D) A B C ● d a

E) e b c ● D E F) ● a d c b e G) A B C ● D E H) a d ● c b e I) ● A B C d a ●

6) Which gamete genotypes are expected to contribute to viable offspring when an Version 1

2

individual is heterozygous for a pericentric inversion and crossing-over occurs as shown in the diagram? (The ● symbol represents a centromere.)

Version 1

3

A) ● A B C D E B) ● A B C d a ● C) ● a d c b e D) e b c D E

E) ● A B C d e F) A B C ● D E G) A B C ● d a H) a d ● c b e I) e b c ● D E

7) In plants, genes E, F, and G are on one chromosome arm. Alternate dominant and recessive alleles of all three genes determine visible traits: E = normal leaf number, e = extra leaves, F = normal stems, f = furry stems, G = normal flower size, g = giant flowers. A plant that is heterozygous for an inversion involving all three genes is crossed to a plant with extra leaves, furry stems, and giant flowers. The arrangement of alleles is shown in the diagram. If you look at thousands of offspring of this cross, what phenotypes do you expect to see?

A) Most, but not all, offspring will either be normal for all three traits or have extra leaves, furry stems, and gigantic flowers.

B) Half of the offspring will be normal for all three traits and half will have extra leaves, furry stems, and giant flowers.

C) Due to crossovers between E and F in parent 1, a small number of offspring will have only extra leaves, or only furry stems and giant flowers.

D) Due to double crossovers between E and F and F and G, a very small number of offspring will have only

Version 1

furry stems, or only extra leaves and giant flowers.

E) Due to crossovers between F and G, a small number of offspring will have only extra leaves and furry stems, or only giant flowers.

4

8)

Which can result in Down syndrome?

A) nondisjunction resulting in an extra chromosome

21 B) deletions of a segment of chromosome 21 C) having the triploid number of chromosomes

D) a reciprocal translocation between acrocentric chromosomes 14 and 21

9) What is true about reciprocal translocation heterozygotes and inversions?

A) They are both used to make Balancer chromosomes.

B) The amount of DNA in the genome remains the same.

C) The genes at the boundaries may be disrupted,

while those in the middle are most likely unaffected.

D) Up to 50% of gametes may be unbalanced, resulting in semisterility.

10) Which is a mechanism by which transposable element mobilization may be controlled by the cell?

A) synthesizing an enzyme that removes transposable elements from the genome B) blocking transcription of transposable element genes C) blocking translation of transposable element transcripts

D) increasing the mutation rate of transposable element DNA sequences E) alternative splicing of transposable element transcripts

11) On the island of Madeira, two populations of house mice have accumulated different Robertsonian translocations.

Which are true about these two populations of mice?

Version 1

5

A) They have fewer chromosomes than mice in other areas of the world.

B) Mice from different populations are unable to mate with each other.

C) They are becoming extinct because of semisterility.

D) They are isolated from each other geographically.

E) If mice from different populations mate, their offspring are sterile.

12) What might result from movement of transposable elements in a species' genome?

A) The level of expression of one or more genes could be changed.

B) A gene could be moved from one chromosome to another.

C) The rate of point mutation could increase.

D) A segment of DNA could be duplicated or deleted.

13) In what way can defective transposable elements alter genomes even if they cannot mobilize?

A) Genes located between two transposable elements could be moved to a nonhomologous chromosome.

B) Crossing-over between transposable elements on the same chromosome can result in a deletion or an inversion.

C) Crossing-over between transposable elements on nonhomologous chromosomes can result in a reciprocal

translocation.

D) Proteins produced by the defective transposable elements introduce point mutations in other areas of the genome.

14) What are the four major classes of chromosomal rearrangements?

A) inversions, Version 1

6

duplications, translocations, deletions B) duplications, reciprocal translocations, nonreciprocal translocations, inversions C) deletions, inversions, duplications, point mutations

D) translocations, pericentric inversions, paracentric inversions, deletions

15) The type of mutation that results in a loss of material from the genome is called what?

A) inversion B) duplication

16)

C) deletion D) translocation

What type of mutation adds material to the genome?

A) inversion B) duplication

C) deletion D) translocation

17) Which type of mutation is most likely to result in lethality?

A) inversion B) duplication

C) deletion D) translocation

18) What is the movement of part of one chromosome to another chromosome called?

Version 1

7

A) inversion B) duplication

C) deletion D) translocation

19) What is a segment of DNA that can use transposase to move from one place in the genome to another called?

A) regulatory region B) duplication C) translocation

D) DNA transposon E) retrotransposon

20) An acentric fragment is one result of crossing-over between a normal chromosome and a chromosome that has undergone what kind of mutation?

A) paracentric inversion B) duplication C) reciprocal translocation

D) pericentric inversion

21) Karyotypes generally remain constant within a species because A) rearrangements occur frequently.

B) changes in chromosome number occur infrequently.

C) genetic instabilities produced by genomic changes

Version 1

areusually at a selective disadvantage.

D) genetic imbalances are often at a selective advantage.

8

22)

Despite selection against chromosomal variations,

A) related species almost always have the same karyotype.

B) related species almost always have different karyotypes.

C) closely related species diverge by many

chromosomal rearrangements.

D) distantly related species diverge by only a few chromosomal rearrangements.

23) An individual with one normal homolog and one homolog with a deletion is called what?

A) a deletion heterozygote B) a deletion homozygote C) dosage compensation

D) haploinsufficient

24) In a deletion heterozygote, the normal chromosome will form what structure during prophase of meiosis I?

A) inversion loop B) cruciform

25)

C) slipped mispairing D) deletion loop

Which event could result in an inversion?

°

A) A 360 rotation of a chromosomal region following two double-strand breaks in a chromosome's DNA. B) A crossover between two DNA sequences in different places on the same chromosome and that are

Version 1

inverted repeats of each other.

C) A crossover between two repeated DNA sequences that are in different places on the 9

same chromosome and are oriented in the same direction.

D) A crossover between two repetitive sequences on

26)

different chromosomes.

Inversions may be difficult to detect because they

A) never visibly change chromosome banding patterns.

B) increase recombination in heterozygotes.

C) do not always cause an abnormal phenotype.

D) normally are removed immediately in natural populations.

27) Robertsonian translocations result from which of the following?

A) Breaks occur at or near the centromeres of two acrocentric chromosomes followed by the reciprocal exchange of broken parts.

B) Two homologous chromosomes break and exchange parts.

C) Unequal crossing-over occurs during meiosis.

D) Two small chromosomes fuse end-toend resulting in one chromosome with two centromeres.

28) Which of the following chromosomal rearrangements usually results in normal meiosis?

A) translocation heterozygote B) translocationhomozygote C) paracentric inversion heterozygote

29)

Which segregation pattern in a translocation

Version 1

D) pericentric inversion heterozygote

heterozygote is likely to 10

result in a normal zygote after fertilization?

A) alternate B) adjacent-1

C) adjacent-2 D) nondisjunction

30) An individual with which type of chromosomal rearrangement is expected to have the lowest fertility?

A) chromosomal duplication B) pericentric inversion C) translocation heterozygote

31)

D) translocation homozygote E) paracentric inversion

What is true in translocation heterozygotes?

A) Genes that are normally located on different homologs appear to be linked.

B) Fertility is reduced because crossing-over prevents the production of balanced gametes.

C) The translocation chromosome is used as a Balancer because it cannot cross over with a normal

chromosome.

D) Crossing-over between normal and translocation chromosomes results in an inversion.

32) Which is a characteristic of retrotransposons, but not of DNA transposons?

D) may not have a A) include a gene that encodes reverse transcriptase B) may be present in a genome from one to millions of times C) found only in humans Version 1

11

function in their host E) have short DNA sequences that are inverted repeats of each other on the ends

Version 1

12

33) One similarity between DNA transposons and retrotransposons is that

A) during transposition, both go through an RNA form that is copied back into DNA. B) both can encode an enzyme required for mobilization.

C) transposition results in movement of the original

transposon to a different place in the genome.

D) they both have poly-A segments at one end.

34) What might be the outcome of repeated duplications of one or a few genes in a species' genome?

A) tetraploidy B) semisterility C) the presence of gene families D) repetitive DNA that is of no use to the cell

E) fusion of segments of two chromosomes into one

35) Which is a mechanism by which chromosomal rearrangements result in proteins with altered amino acid sequences? genes.

A) Chromosome sets from two different species are combined in one individual.

B) Translocations may fuse two different open reading frames.

C) Deletions may remove regulatory regions of

Version 1

D) Inversions may move a gene from euchromatin to heterochromatin.

13

36)

Deletions are most useful for gene mapping

A) when you have a mutant allele of a gene that is recessive to normal alleles and amorphic.

B) when you have a mutant allele of a gene that is dominant to wild-type alleles.

C) if you have a mutant allele of a gene that results in

no detectable mutant phenotype when homozygous.

D) if you have a gain-of-function mutant allele of a gene.

37) Inversions are most likely to affect an organism’s phenotype

A) if one or both of the inversion's breakpoints lies within the transcribed region of a gene.

B) if the inversion's breakpoints are on either side of a gene and its regulatory regions and the entire gene is reversed in the genome.

C) if the inversion is pericentric.

D) if the inversion is paracentric.

E) if the inversion is small.

38) What mechanism used for mobilization of transposable elements can lead to deletions in genomes?

A) Transposase sometimes deletes genomic DNA while removing DNA transposons from the genome.

B) Alignment of two transposable elements is required for mobilization and misalignment can result in deletions.

C) RNA polymerase transcribes transposable elements; when it makes mistakes, DNA nucleotides are

39)

Duplications are most likely to have phenotypic

Version 1

sometimes deleted.

D) Piwi RNAs inhibit movement of transposons by deleting DNA that includes and surrounds the inverted repeats.

consequences if 14

A) a gene within the duplication is haploinsufficient.

B) an additional dose of a gene within the duplication does not affect tissue physiology.

C) the expression of a gene near one duplication

breakpoint is altered.

D) an allele of a gene within the duplication is amorphic.

40) Which of the following features is shared by mobile transposable elements in the human genome?

A) RNA intermediates B) functional transposase genes

C) mutant LTRs D) mutant pol genes

41) Match the description with the appropriate term. a. inversion b. duplication c. deletion d. translocation e. transposable element

41.1) A segment of DNA that moves from place to place in the genome

41.2) Addition of one or more copies of a segment of DNA to the genome

Version 1

15

41.3)

A change in the orientation of a DNA sequence

41.4)

Loss of a segment of DNA from the genome

41.5) Fusion of a segment of a chromosome with part of a nonhomologous chromosome

42) Match the description with the appropriate term a. retrotransposon b. DNA transposon c. transposable element d. transposase e. reverse transcriptase

42.1) Broad term for any DNA segment that moves about in the genome

42.2) Moves in the genome through an RNA intermediate

42.3) DNA is removed from the genome and moved to a new position.

Version 1

16

42.4)

Version 1

An enzyme that catalyzes the movement of

DNA transposons

17

Answer Key Test name: Chapter 14 Test Bank 1) [B, C, E] 2) [B, D] 3) [B, C] 4) [B, D] 5) [B, F] 6) [F, H] 7) [A, D] 8) [A, D] 9) [B, C, D] 10) [B, C, E] 11) [A, D, E] 12) [A, B, D] 13) [B, C] 14) A 15) C 16) B 17) C 18) D 19) D Version 1

18

20) A 21) C 22) B 23) A 24) D 25) B 26) C 27) A 28) B 29) A 30) C 31) A 32) A 33) B 34) C 35) B 36) A 37) A 38) A 39) D 40) B Version 1

19

41) Section Break 41.1) e 41.2) b 41.3) a 41.4) c 41.5) d 42) Section Break 42.1) c 42.2) a 42.3) b 42.4) d

Version 1

20

Student name:__________

A) inversion in chromosome 21 B) nondisjunction resulting in an extra chromosome

1) Which can result in Down syndrome?

D) having a triploid number of chromosomes

21 C) deletion of a segment of chromosome 21

2) Triticale is an allopolyploid hybrid between wheat and rye. Some strains of Triticale show agricultural promise because A) they combine the high yields of wheat with the ability to adapt to unfavorable environments like rye.

B) the grain is high in protein, especially the amino acid lysine.

C) they are sterile because the wheat and rye

3) During evolution, the additional copies of genes resulting from whole genome duplication events in new A) They are always lost.

B) They may mutate beyond recognition.

C) They may acquire new functions.

chromosomes cannot pair during meiosis.

D) they combine desirable traits from wheat and rye in one crop plant.

species exhibit the following feature(s).

D) They never undergo any change.

4) How would you determine if a present-day species evolved from an ancestral species by whole genome duplication?

A) Present-day species will always be tetraploid.

B) Genome sequence of present-day species reveals

Version 1

blocks of homologous gene sequences on two different chromosomes.

1

C) Genome sequence of present-day species shows that within blocks of homologous gene sequences, the order of genes is mostly conserved.

D) Genome sequence of present-day species shows

5)

that within blocks of homologous gene sequences, the order of genes is never conserved.

Which is not an example of aneuploidy?

A) monosomy B) tetraploidy

C) trisomy D) nullisomy

6) What is a reason that aneuploidy in sex chromosomes is generally better tolerated than aneuploidy in autosomal chromosomes in humans?

A) In somatic cells, most of the genes on only one X chromosome are transcriptionally active.

B) Autosomal aneuploidy leads to heart defects and death in utero.

C) Y chromosome duplication results in only minor changes in testosterone levels.

D) Sex chromosome aneuploids may occur as the result of fertilization, but extra sex chromosomes are removed

from the developing embryo during subsequent mitosis.

E) Any change in autosome number results in increased susceptibility to infection.

7) Which sex chromosome aneuploidy is not usually seen in live births?

A) XO B) XXY Version 1

C) YO D) XXX

2

8) Turner syndrome (XO) is a sex chromosome aneuploidy. Of the effects listed below, which one is not

usually seen in this syndrome?

D) unusually long

A) unusually short stature B) infertility C) skeletal abnormalities

limbs

9) In Drosophila, a gynandromorph, which is composed of equal parts of male and female tissue, results from A) an XX female losing one X chromosome during the first mitotic division after fertilization.

B) an egg carrying an X chromosome fertilized by a Y-carrying sperm.

C) a normal egg fertilized by both an X-carrying

10)

Which is not an example of euploidy?

A) Seedless watermelons have three copies of each chromosome.

B) A normal human has two copies of each chromosome.

C) A human with Down syndrome has three copies

11)

sperm and a Y-carrying sperm.

D) the fusion of a female embryo with a male embryo.

of chromosome 21.

D) Commercially grown strawberry plants have eight copies of each chromosome.

Triploid organisms usually result from D) a fertilization A) the union of monoploid and diploid gametes.

B) nondisjunction during mitosis.

C) propagation of fused cell lines.

Version 1

3

event that involves three monoploid gametes.

E) normal fertilization of gametes produced by triploid parents.

Version 1

4

12) During mitosis, if all of the chromosomes in a diploid cell fail to separate and instead segregate into one daughter cell, the result will be what?

A) one monoploid daughter cell and one triploid daughter cell B) one 2 x − 1 daughter cell and one 2 x + 1 daughter cell

C) two 2 x daughter cells D) only one tetraploid daughter cell

13) Hybrids in which the chromosome sets come from two distinct, though related, species are known as

A) autopolyploids.

B) allopolyploids.

C) aneuploids.

D) bivalents.

14) Why are organisms with an odd number of chromosome sets usually sterile?

A) Almost all gametes will be unbalanced.

B) Chromosomes will fail to segregate independently during meiosis I C) Chromosomes will fail to segregate independently during meiosis II

15)

D) Because an odd number of chromosomal sets is present, meiosis will not occur at all.

Monoploids are useful in plant breeding because

Version 1

5

A) they can be used to select for plants with desirable recessive traits.

B) they will not express any undesirable recessive traits.

C) they are one step in the path to creating desirable

16) Wild oysters are diploid and have 20 chromosomes in their somatic cells. Scientists have generated tetraploid oysters using colchicine to prevent separation of sister chromatids. Crossing a diploid oyster with a tetraploid oyster resulted in triploid oysters that are commercially

16.1)

C) 20 D) 40

What is n in wild oysters?

C) 20 D) 40

A) 5 B) 10

16.3) A) 5 B) 10

Version 1

advantageous because they are larger and do not have a reproductive phase that usually interrupts the harvesting season.

What is x in wild oysters?

A) 5 B) 10

16.2)

polyploid plants.

D) they are typically resistant to commercial herbicides.

What is x in tetraploid oysters?

C) 20 D) 40

6

16.4)

What is n in tetraploid oysters?

A) 5 B) 10

C) 20 D) 40

16.5) What is the probability that a gamete produced by a triploid oyster will be balanced?

A) 1/1024 B) 1/512

Version 1

C) 1/20 D) very close to 1

7

Answer Key Test name: Chapter 15 Test Bank 1) [B] 2) [A, B, D] 3) [B, C] 4) [B, C] 5) B 6) A 7) C 8) D 9) A 10) C 11) A 12) D 13) B 14) A 15) A 16) Section Break 16.1) B 16.2) B 16.3) B Version 1

8

16.4) C 16.5) B

Version 1

9

Student name:__________ 1) A bacterium is found that is resistant to the antibiotic gentamicin. The bacterium was isolated in a hospital where patients were routinely given gentamicin for a variety of infections. What was the pressure that selected for this resistant population?

A) presence of gentamicin in the environment B) high mutation rate for the bacterium C) growth situation for the bacterium

D) patients that did not receive the antibiotic

2) What is a typical characteristic of bacterial chromosomes?

A) On average 5% of the genome encodes proteins.

B) It has telomeres.

C) It is linear.

D) On average a gene occurs once in every 1000 bp.

E) It is singlestranded DNA.

3) If the Tn10 transposon is used to mutagenize a bacterium for the purpose of creating a mutation in the his operon, what would be the most appropriate selection media?

A) rich media that contains all of the amino acids supplemented with tetracycline B) minimal media supplemented only with histidine C) minimal media supplemented with tetracycline

4)

A strain of E. coli is trp− his− lac−. Which medium

Version 1

and histidine D) minimal media supplemented with tetracycline alone

would this bacterium grow 1

on? sugar A) minimal medium with lactose as the sugar, supplemented with histidine and tryptophan B) minimal medium with glucose as the sugar, supplemented with histidine and tryptophan C) minimal medium with glucose and lactose as the

D) minimal medium with glucose as the sugar

5) The episome in an Hfr strain is inserted near the trp (tryptophan operon) locus. This Hfr strain is grown with an F− Trp− strain. From this mating, we isolate an F+ Trp+ strain that can readily impart the Trp+ phenotype to F− Trp− strains, but no other traits are ever transferred. (When the original Hfr strain is mated with F− strains, traits other than Trp+ can be transferred, although at lower frequency than Trp+.) What most likely has happened?

A) An F′ trp + plasmid has been generated.

B) Mutation occurred.

C) Transduction occurred.

D) Transformation occurred.

6) Viruses are isolated from wild-type E. coli cells that have been infected with wild-type bacteriophage λ. These viruses are used to infect a Gal− strain of E. coli. A few bacterial colonies that can grow on galactose are obtained, while no bacteria that can grow on galactose are obtained from cells that were not infected. What has happened?

+

A) A λ gal phage was generated.

B) Reversion mutations have occurred.

C) The bacteria underwent conjugation with a Gal+

D) The bacteria underwent transformation with a wild-type E. coli strain.

strain.

Version 1

2

7) Some plasmids can replicate in several distinct species of bacteria. Frequently these plasmids have transposons carrying several different antibiotic resistance genes. In a hospital that has an outbreak of several species of bacteria each carrying resistance to three drugs (streptomycin, gentamicin, and penicillin), what is the best way to determine whether or not this resistance is due to a single shared plasmid with all three resistance genes?

A) Demonstrate that all the resistant bacteria have a plasmid.

B) Isolate plasmids from all resistant bacterial species and demonstrate that drug-sensitive cells from these same species become resistant to all three drugs when this plasmid is transformed. Confirm that the same plasmids are present in all species by sequencing the plasmid DNA. C) Sequence the genomes of the resistant species and demonstrate mutations in the same homologous genes.

8)

D) Isolate plasmids from one resistant species and transform a drug-sensitive strain of the other bacterial species to see if resistance occurs in the transformants.

Penicillin stops bacterial cell wall formation by

A) inhibition of a transpeptidase.

B) inhibiting synthesis of NAM. C) inhibiting synthesis of NAG

D) inhibiting transport of NAG and NAM past the permeable membrane.

9) What would happen if a transposon were to integrate into the replication origin of the bacterial genome?

A) the transposon would be replicated with the

Version 1

genome B) the transposon 3

would "jump" immediately to another location C) the transposon would inactivate the replication origin, so no DNA replication would be possible

D) the transposon would express its genes constitutively

10) What characteristics make bacteria like E. coli attractive as model organisms?

A) short generation time, simple genome relative to that of humans, easily mutagenized, haploid, gene control mechanisms identical to eukaryotes B) short generation time, simple genome relative to that of humans, easily mutagenized, haploid C) short generation time, genome as complex as that

of humans, easily mutagenized, haploid D) short generation time, simple genome relative to that of humans, easily mutagenized, diploid

11) Two F+ E. coli strains are co-cultured in the same flask. One is Strr Thr+ and the other is Genr Thr−. The culture is then plated on minimal media that is supplemented with gentamycin only (not threonine) and a few colonies grow.

What sort of genetic exchange is most likely occurring?

A) transformation B) conjugation

Version 1

C) transduction D) electroporation

4

12) Which technique would most likely be used for mapping genes that are separated by 2 ×105 base pairs?

Version 1

5

A) specialized transduction B) generalized transduction

C) conjugation D) transformation

13) In an Hfr strain, the F episome is integrated between genes A and B. When this Hfr mates with an F− strain, gene B is always transferred first to the F− cell. What is the last bacterial gene that could possibly (at least in theory) be transferred to the F− from the host chromosome?

A) agene that is 90 minutes from gene A B) agene that is at 30 minutes from gene A

C) gene A D) agene that is at 60 minutes from gene A

14) If an organism is isolated that has no nuclear membrane and the DNA is condensed into a body in the cytoplasm, this organism would most likely be classified as

A) a prokaryote.

B) a eukaryote.

C) a pathogen.

D) either a eukaryote or prokaryote.

15) What would you anticipate would be true about the metabolism of a bacterial species isolated from water samples, one at the surface and another from a hot vent on the ocean floor?

A) The species from the water surface would be more likely to be able to conduct photosynthesis.

B) The species from the hot vent would be more Version 1

likely to be able to conduct photosynthesis.

C) The species from the hot vent would be 6

more likely to be able to fix nitrogen D) The species from the hot vent would be more

16)

likely to be able to degrade oil.

What is a property of a pathogenic bacterium?

A) It may produce a protein that interferes with basic cellular functions.

B) All strains of the same species are also pathogenic.

C) It probably conducts photosynthesis.

D) It is resistant to all antibiotics.

17) Which gene would you predict would be most likely to be part of a core genome and not the pangenome?

A) gene encoding DNA polymerase B) gene conferring antibiotic resistance C) gene conferring metal resistance

18)

D) gene encoding an enzyme required for lactose utilization

What is a difference between a Tn and IS element?

A) Tn elements can carry antibiotic resistance genes.

D) IS elements integrate into plasmids and genomes.

B) IS elements carry a transposase gene.

C) Tn elements integrate into plasmids and genomes.

19) Metagenomics may be useful because

Version 1

7

A) it could be used to detect or predict human diseases.

B) it is a technique for sequencing a bacterial genome rapidly.

20)

C) metagenomes do not change rapidly.

D) metagenomes are not responsive to environmental changes.

Horizontal gene transfer

A) can accelerate evolution of a species B) can inhibit evolution of a species.

C) has no role in evolution.

D) has a role in evolution only if the gene transfer is to a different species.

21) A strain of penicillin-resistant N. gonorrhoeae that does not have a plasmid is isolated. A lysate made from the bacterium cannot degrade penicillin in an in vitro assay.

When a culture of this strain is grown in a solution that contains radioactive penicillin, the amount of penicillin found within cells is less than the amount found within cells from a nonresistant strain. The DNA sequence of the porin gene from the resistant strain does not have any mutations. What is the most likely mechanism of resistance in this strain?

A) A mutation in the mtr gene has occurred, resulting in fewer efflux pumps.

B) A penA nonsense mutation has occurred.

C) A penA missense mutation has occurred.

D) The bacterial genome carries thepenr gene.

22) Researchers often work with temperature-

Version 1

8

sensitive mutations ______, because null alleles of these genes create mutants that cannot grow under any conditions.

A) that confer antibiotic resistance B) in essential genes C) that affect colony morphology

23) To produce a mutation in a specific E. coli gene, an engineered fragment of DNA in which sequences from each A) is transformed into cells and replicates independently of the genomic DNA because it has a centromere.

B) is transformed into cells and maintained as a plasmid.

C) is introduced into cells and transposes into the

24) Three genes ( X, Y, and Z) are in the same region of a bacterial chromosome. To determine the order of these genes, you infect X − Y − Z − cells with a lysate from wild-type cells infected with a generalized transducing phage.

When X + cells are selected, 70% of cells are also Y + and 5% of cells are Z +. When Y + cells are selected, 68% of A) Z is in the middle, but closer to X. B) Y is in the middle, but closer to Z. C) X is in the middle, but closer to Y.

25) by

D) that affect metabolism

end of that gene flank an antibiotic resistance gene bacterial gene.

D) is introduced into cells and undergoes homologous recombination with the bacterial gene.

cells are also X + and 1% are Z +. What is the relative arrangement of the three genes?

D) X is in the middle, but closer to Z.

Complementation studies can be performed in bacteria A) gene targeting.

Version 1

9

B) forming merodiploids with one copy of a bacterial gene on an F′ plasmid.

C) forming merodiploids with two copies of a bacterial gene on F′ plasmids.

D) forming diploids with two copies of the bacterial chromosome.

26) One way N. gonorrhoeae became resistant to penicillin was by acquiring a plasmid that carries the pen r penicillin resistance gene, which encodes a protein that A) allows cells to synthesize penicillin.

B) cleaves penicillin to an inactive form.

C) pumps penicillin out of the cell.

D) attaches penicillin to the cell wall.

27) In transduction, the DNA is being moved between the two cells by

A) a third bacterial cell.

B) a phage.

28)

C) transposons.

D) membrane bound transport proteins.

A bacteriophage is a

A) bacterium that infects a virus.

B) virus that infects a bacterium.

C) particle made up of viral nucleic acid surrounded by bacterial protein.

Version 1

D) particle made up of bacterial nucleic acid surrounded by bacterial protein.

10

29)

When a bacteriophage carrying bacterial DNA infects

a new bacterium,

A) the recipient is usually killed.

B) the recipient keeps the transferred DNA in storage but does not replicate it.

C) it transfers bacterial DNA from the donor bacterium to the recipient bacterium.

D) only viral DNA is transferred to the recipient bacterium.

Version 1

11

30) Plasmids are created exclusively in research laboratories and do not occur naturally. ⊚ ⊚

true false

31) Bacteria that live in extreme environments, such as thermal vents on the ocean floor, have evolved genes that allow them to use the resources within that environment. ⊚ ⊚

true false

32) All the genes in a species’ core genome are located on the bacterial chromosome, whereas the genes of the ⊚ ⊚

true false

33) Mapping genes by conjugation can be performed by interrupted-mating experiments in which conjugation is ⊚ ⊚

pangenome are carried on plasmids.

physically disrupted at 1minute intervals.

true false

34) Scientists can identify the gene mutated Version 1

12

in a bacterial auxotroph by transforming the cells with a genomic library in which fragments of the auxotroph’s ⊚ ⊚

genome have been cloned into plasmids.

true false

35) Multidrug resistance is not a major health concern because new antibiotics are easily and frequently identified. ⊚ ⊚

Version 1

true false

13

Answer Key Test name: Chapter 16 Test Bank 1) A 2) D 3) C 4) B 5) A 6) A 7) B 8) A 9) C 10) B 11) A 12) C 13) C 14) A 15) A 16) A 17) A 18) A 19) A Version 1

14

20) A 21) A 22) B 23) D 24) C 25) B 26) B 27) B 28) B 29) C 30) FALSE 31) TRUE 32) FALSE 33) TRUE 34) FALSE 35) FALSE

Version 1

15

Student name:__________

A) the cytoplasm B) the nucleus C) the space between the inner and outer mitochondrial membranes

1) In plant cells, DNA molecules are found inside which structures?

D) the mitochondrial matrix E) the chloroplast stroma

2) Which is an accurate description of organelle genomes?

A) Like bacteria, mitochondria use N-formyl methionine and tRNA fmet in translation.

B) mtDNA and cpDNA are usually organized into nucleosomes by histones.

C) Like bacterial genes, mitochondrial and chloroplast genes do not have introns.

D) Translation in mitochondria and chloroplasts is

often inhibited by bacterial antibiotics (e.g., chloramphenicol).

E) Like bacterial genomes, mitochondrial and chloroplast genomes in all species consist of one circular DNA molecule.

3) Inhibitors of bacterial translation, such as chloramphenicol and erythromycin, usually A) inhibit mitochondrial protein synthesis.

B) have no effect on mitochondrial protein synthesis.

C) inhibit eukaryotic cytoplasmic protein synthesis.

D) inhibit chloroplast protein synthesis.

E) have no effect on chloroplast protein synthesis.

4) Where are the proteins needed for photosynthesis produced?

Version 1

1

A) in the nucleus B) in the cytoplasm C) in the chloroplasts

5)

D) in the mitochondria

What doesthe codon 5′ UGA specify?

A) a stop signalin human mtDNA B) formyl-methionine in mtDNA C) a stop signalin the universal genetic code

6) LHON is a mitochondrial disease caused by weak hypomorphic mutations that affect the mitochondrial electron transport chain. Among the first symptoms of the disease is blindness. People with LHON are usually homoplasmic because retinas must be homoplasmic mutant for a person to have disease symptoms. Which is true about a female who is A) She may have children who are blind due to LHON. B) Some cells in her body are probably homoplasmic.

C) She is likely to develop the disease in other parts

D) the amino acid tryptophan in human mtDNA

heteroplasmic for the mutation that causes LHON?

of her body.

D) All of her children will be homoplasmic for the wildtype mitochondria.

7) LHON is a rare mitochondrial disease caused by weak hypomorphic mutations that affect the mitochondrial electron transport chain. The first symptoms of the disease include blindness. People with LHON are usually homoplasmic because retinas must be homoplasmic mutant for a person to

have disease symptoms.

Which is true about a male who is heteroplasmic for the mutation that causes LHON?

A) He is likely to develop the disease in other parts of his body.

B) All of his children will most likely be homoplasmic for wild-type

Version 1

2

mitochondria.

C) He may have children who are blind due to LHON.

D) Some cells in his body are probably homoplasmic.

8) Extranuclear genes show a biparental inheritance pattern in yeast since

A) progeny from diploid vegetative growth can have the phenotype of the dominant phenotype.

B) progeny from diploid vegetative growth can have the phenotype of the α parent.

C) progeny from diploid vegetative growth can have

the phenotype of only the a parent.

D) progeny from diploid vegetative growth can have the phenotype of either the a or the α parent.

9) In four o'clock plants, reciprocal crosses are performed with a green plant and a white plant. How will the

offspring of the reciprocal crosses look?

A) The offspring of both crosses will be green.

B) The offspring will be white if the egg came from the white plant and green if the egg came from the green plant.

C) The offspring will be white if the sperm came

from the white plant and green if the sperm came from the green plant.

D) The offspring of both crosses will be variegated.

10) Where are the proteins that carry out photosynthesis encoded?

A) All are encoded in chloroplast DNA. B) Some are encoded in chloroplast DNA and others in nuclear DNA. C) All are encoded in nuclear DNA. Version 1

D) Some are encoded in nuclear DNA and others in mitochondrial DNA.

3

11) What does DNA sequencing suggest about the relationship between the nuclear and organellar genomes?

A) The same subset of mitochondrial genes has moved to the nuclear genome in all species.

B) Mitochondrial genes that are found in the nuclear genome do not have introns suggesting that movement occurred via an RNA intermediate.

C) The mitochondrial and nuclear genomes are entirely independent of each other.

D) Some cells have two types of mitochondria that

have different genome sequences.

E) Some proteins found in mitochondria are imported from the cytoplasm.

12) Variegated four o'clock leaves have white patches among the green areas due to A) the presence of cells in which most of the mitochondria have a mutation that blocks electron transport.

B) the presence of cells in which most of the chloroplasts have a mutation that prevents synthesis of chlorophyll.

C) an incompletely penetrant mutation in a nuclear

gene that prevents production of chlorophyll.

D) a recessive mutation in a nuclear gene that results in a defect in ribosomes.

13) Which is a true statement about mitochondrial genomes?

A) In most cases, the mitochondrial genome is transmitted, largely intact, from one parent to offspring.

B) Mitochondrial inheritance shows Mendel's principle of segregation.

C) Mitochondrial inheritance shows Mendel's

Version 1

principle of independent assortment.

D) All of the choices are correct.

4

14) For what purpose is a researcher likely to shoot plant cells with a gene gun?

A) to kill the cell B) to introduce foreign DNA into the chloroplast genome C) to introduce a mutation in the nuclear genome

15)

Why is a variegated plant considered heteroplasmic?

A) The plant has mutant and wild-type chloroplast DNAs distributed amongst the cells in different ratios.

B) The plant has only two types of cells, one with only mutant chloroplast DNAs and one with only wild-type chloroplast DNAs. C) All of the plant’s cells have mutant and wild-type

16)

chloroplast DNAs. D) All of the plant’s cells have mitochondria and chloroplasts.

The codon5′ UGA specifies

A) a stop signal in the universal genetic code and in human mtDNA. B) the amino acid tryptophan in the universal genetic code and in human mtDNA. C) formyl-methioninein mtDNA.

17)

D) to add a transgene to the mitochondrial genome

D) a stop signal in the universal genetic code and the amino acid tryptophan in human mtDNA.

Which has been associated with aging in humans?

A) an increase in the number of mitochondria Version 1

B) no change in the sequence of the 5

mitochondrial genome C) adecrease in mitochondrial function

D) an increase in mitochondrial function

18) Which of the following has been associated with Alzheimer's Disease? cells A) loss of mitochondria B) 5kb and 7.4kb deletions in mtDNA of heart cells C) mutations in cytochrome oxidase c genes in brain

D) no change in mitochondrial function.

19) Which step in producing ATP is correctly paired with the location where it occurs?

A) Krebs cycle—mitochondrial matrix B) formation of an electrochemical gradient— cytoplasm C) glucose is broken down to produce pyruvate— space between inner and outer mitochondrial membranes

D) electron transport chain—outer mitochondrial membrane E) breakdown of pyruvate to form NADH and FADH 2—cytoplasm

20) What would be the most likely result if a scientist removes mitochondria from a human cell and attempts to grow them in culture?

A) The mitochondria originated from a prokaryotic cell, so the mitochondria will grow independently in culture.

B) Mitochondria require the products of nuclear genes to function, so the mitochondria will die soon after being removed from the cell.

Version 1

C) The mitochondria will continue to produce ATP indefinitely only if a source of pyruvate is provided.

D) The 6

mitochondria will grow more quickly in the presence of antibiotics that kill bacteria and reduce competition for

resources with bacterial cells.

21) What did comparisons of the sequence of the mitochondrial genome in trypanosomes with the sequence of mature mitochondrial RNAs demonstrate?

A) The RNA sequences are similar to the genome sequence, but have insertions and deletions of uracil residues.

B) The RNA sequences align perfectly with the genomic DNA from which it is transcribed.

C) All of the proteins necessary for translation are encoded in the mitochondrial genome.

D) Translation is initiated in mitochondria only at 5′ AUG codons.

E) Some mature RNAs do not have start and stop codons.

22) Due to past gene transfer between the organelles and the nucleus,

A) mitochondrial genomes in all species have the same genes.

B) a chloroplast from one plant species can be transferred to any other plant species.

C) mitochondria and chloroplasts have genes that allow them to live independently.

D) some proteins must be imported into organelles

for their function.

E) organelle genomes experience lower rates of mutation than they would if gene transfer did not occur.

23) Which statement describes evidence in support of the endosymbiont theory?

Version 1

7

A) The sequence of mitochondrial rRNA genes are most similar to the rRNA genes in purple bacteria.

B) Mitochondrial and chloroplast DNA is organized into nucleosomes.

C) Mitochondria and chloroplasts have genes that allow them to live independently.

D) Mitochondria, chloroplasts, and cytoplasmic translation are inhibited by antibiotics such as chloramphenicol.

24) What are characteristics of the pedigrees of families with mitochondrial diseases?

A) All children of affected mothers are usually affected regardless of their sex.

B) Only the male children of affected mothers are affected, female children are not affected.

C) Only female children of affected mothers are also

25)

Oocyte nuclear transfer can be used to

A) allow women with mitochondrial diseases to have normal offspring.

B) allow women with any genetic disease to have normal offspring.

C) allow parents to control all of the phenotypic

26)

affected, male children are not affected.

D) All children of affected males are usually affected.

characteristics of their children.

D) create embryos for stem cell therapy.

Which is true about mitochondria?

A) The mitochondrial genome is circular in all species.

B) Mitochondria are found in all eukaryotic cells,

Version 1

except those with chloroplasts.

C) Mitochondrial genomes vary in length

8

from 6 kb to 2400 kb.

D) Translation in mitochondria utilizes the universal

genetic code.

27) This sequence is the 5′end of a mRNA: 5′ UUCGACCAUUAACGGUUGAAGGUAG 3′

27.1) If this mRNA is translated in the cytoplasm, what amino acid sequence is encoded?

A) Phe Asp His B) Met Asn Gly Trp C) Met Asn Gly

D) Translation will not beginwithin the sequence shown.

27.2) If this mRNA is translated in human mitochondria, what amino acid sequence is encoded?

Version 1

9

A) Phe Asp His B) Met Asn Gly Trp

28) Researchers combined two yeast strains of opposite mating type; the a strain was resistant to chloramphenicol and the α strain was sensitive to chloramphenicol. The diploid progeny were allowed to undergo several cell divisions, and

C) Met Asn Gly D) translation will not begin

then the cells were replica plated on glycerol medium with and without chloramphenicol.

28.1) What results would you expect if mitochondria are inherited biparentally?

A) Some, but not all, yeast colonies found on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

B) All of the yeast colonies on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

C) Many yeast colonies will grow on the plate without chloramphenicol and no colonies will grow on

the plate with chloramphenicol.

D) No colonies will grow on either plate.

28.2) What results would you expect if mitochondria are inherited uniparentally from the a strain?

A) Some, but not all, yeast colonies found on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

B) All of the yeast colonies on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

C) Many yeast colonies will grow on the plate without chloramphenicol and no colonies will grow on Version 1

the plate with chloramphenicol.

D) No colonies will grow on either plate.

10

28.3) What results would you expect if mitochondria are inherited uniparentally from the α strain?

A) Some, but not all, yeast colonies found on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

B) All of the yeast colonies on the plate without chloramphenicol will also grow on the plate with chloramphenicol.

C) Many yeast colonies will grow on the plate without chloramphenicol and no colonies will grow on

Version 1

the plate with chloramphenicol.

D) No colonies will grow on either plate.

11

Answer Key Test name: Chapter 17 Test Bank 1) [B, D, E] 2) [A, D] 3) [A, D] 4) [B, C] 5) [C, D] 6) [A, B] 7) [B, D] 8) D 9) B 10) B 11) B 12) B 13) A 14) B 15) A 16) D 17) C 18) C 19) A Version 1

12

20) B 21) A 22) D 23) A 24) A 25) A 26) C 27) Section Break 27.1) D 27.2) B 28) Section Break 28.1) A 28.2) B 28.3) C

Version 1

13

Student name:__________

A) at transcription initiation.

B) during RNA splicing.

C) during nuclear export of mRNA.

1) A crucial step in the regulation of most bacterial genes occurs D) when nonsense suppressors translate mRNAs.

2) What would be the phenotype of a null mutation in the gene encodingLac repressor?

A) constitutive expression of the lac operon B) inducible expression of the lac operon C) permanently repressed expression of the lac

D) cannot predict what would happen to expression of the lac operon

operon

3) Transcription and translation can be coupled in bacteria but not eukaryotes because

A) no nuclear membrane exists in prokaryotes.

B) no nuclear membrane exists in eukaryotes.

C) the bacterial DNA is in a single chromosome.

D) eukaryotic chromosomes are found in nucleoids.

4) How do negative regulators such as the Lac repressor prevent RNA polymerase from initiating transcription?

A) by blocking the ribosome binding site B) by forming a loop in the operator that restricts the passage of the polymerase C) by physically blocking the DNA binding site of Version 1

RNA polymerase D) by binding to the polymerase to inhibit its catalytic activity

1

5) How does a positive regulator affect transcription by RNA polymerase in prokaryotes?

A) by allowing passage of the polymerase through an operator B) by interacting with RNA polymerase to increase the frequency of transcription initiation C) by causing the helix to unwind in the operator

allowing easier initiation D) by making the transcription start site more exposed to the polymerase

6) Proteins whose conformations change when they are bound to an effector molecule are called

A) enzymes.

B) allosteric proteins.

C) regulatory proteins.

D) activator proteins.

E) inhibitory proteins.

7) Catabolic pathways that break down complex substances into more usable units are usually regulated in response to the A) end products of the pathway.

B) levels of the molecule that is to be broken down.

C) other metabolites that are limiting.

D) levels of enzymes in the pathway.

8) Anabolic pathways involved in the synthesis of essential molecules are usually regulated in response to A) the end

Version 1

2

product of the pathway.

B) a substrate of the pathway.

C) other metabolites that are limiting.

D) the levels of enzymes in the pathway.

9) Which statement accurately describes global gene regulation in bacteria?

A) Sigma factors are not involved in gene regulation, only in attachment of the polymerase to the promoter.

B) Alternative sigma factors that recognize different promoters are a mechanism of global gene regulation.

C) E. coli cells devote more energy to the production

of ribosomes during stress so that global gene regulation can occur.

D) All promoters are recognized by all sigma factors.

10) How is glucose involved in catabolite repression of the lac operon?

A) Glucose causes cAMP levels to increase, which leads to increased CRP binding and the lac operon is activated even when lactose is present.

B) Glucose causes cAMP levels to decrease, which leads to decreased CRP binding and the lac operon is repressed even when lactose is present.

C) Glucose is also a substrate for β-galactosidase and

thus competes with lactose for this enzyme.

D) Glucose activates transcription of the lacI gene to increase the amount of Lac repressor in the cell.

11) The function of a helix-turn-helix motif in a transcription factor is to DNA. A) recruit RNA polymerase.

B) form dimers with other transcription factors.

C) bind a specific sequence in the major groove of

Version 1

D) unwind the double helix.

3

12) How does tryptophan, the end product of the trp operon, function in the regulation of the operon?

A) Trp binds to and inhibits the repressor, thus allowing transcription of the operon.

B) Trp binds to and activates the repressor, which then binds to DNA to allow transcription of the operon.

C) Trp binds directly to DNA and inhibits transcription of the operon.

13) What term describes regulation of the trp operon that can occur even in trpR− mutants, thus suggesting that A) modulation B) derepression

D) Trp binds to and changes the conformation of the repressor, which can then bind DNA and block transcription of the operon.

the mechanism is repressor-independent?

C) attenuation D) amplification

14) In E. coli,the heat-shock response, which alters the proteins produced by the cell at different temperatures, is mediated by

A) degradation of chaperone proteins at high temperatures.

B) denaturing of DNA in the promoters in the genes of heat-sensitive proteins.

C) synthesis of alternative sigma factors at high temperatures to activate transcription of heat-shock genes.

D) increasing the promoter affinity of already existing polymerase sigma factors at high temperatures.

15) As a general principle of gene regulation through operons, regulatory genes encode

Version 1

4

A) trans-acting proteins that interact with cis-acting DNA elements.

B) cis-acting proteins that interact with cis-acting DNA elements.

C) cis-acting proteins that interact with trans-acting

DNA elements.

D) trans-acting proteins that interact with trans-acting DNA elements.

16) In the trp operon, attenuation occurs through translation of two Trp codons in the leader sequence. What would happen if these two codons were mutated to stop codons?

A) Region 1 will bind to region 3 in the absence of Trp. B) The antiterminator loop will form in the presence or absence of Trp. C) Region 3 will bind to region 4 in and transcription

will proceed in the presence or absence of Trp. D) Region 3 will bind to region 1 in the presence of Trp.

17) Initiation of transcription by RNA polymerase involves the binding of which of the following subunits to the core enzyme?

A) delta B) sigma

C) gamma D) alpha E) zeta

18) The transition from transcriptional initiation to elongation involves A) binding of sigma factor.

Version 1

B) release of sigma factor.

5

C) release of RNA polymerase from DNA. D) binding of RNA polymerase to DNA.

19)

E) binding of rho factor.

Which molecules are examples of effector molecules?

A) Allolactose and Trp B) Lac repressor and Trp repressor C) The promoter and operator

D) lacY and lacZ E) Lactose and glucose

20) A single DNA unit that enables the simultaneous regulation of more than one coding region in response to environmental changes is called a(n) C) regulator.

D) inducer.

E) operon.

A) promoter.

B) operator.

21) Alterations to DNA sites such as promoters and operators can act A) only in cis.

B) only in trans.

C) either in trans or in cis.

22)

D) These terms apply only to posttranscriptional regulation.

The scientists who proposed the operon theory are B) Watson and A) Monod and Jacob.

Version 1

Crick.

6

C) Hardy and Weinberg.

D) Darwin and Mendel.

E) Hershey and Chase.

23) Which of the following is an example of a reporter gene?

A) The linkers that are attached to RNA molecules to perform RNA-Seq B) The lacZ coding sequence under the control of a prokaryotic promoter C) The DNase I enzyme used to digest DNA for a

footprint assay D) A human protein expressed under the control of a prokaryotic promoter

24) The sigma factor that mediates a global heat-shock response in E. coli is A) sigma 70.

B) sigma 32.

25) a(n)

C) sigma 34.

D) sigma 72.

E) sigma 36.

In the regulation of the trp operon, tryptophan acts as

A) repressor.

B) attenuator.

26) Fifteen different strains of bacteria are auxotrophic for maltose and their mutations do not complement each other.

The mutations in these bacteria map to the same DNA region

Version 1

C) activator.

D) effector.

E) operator.

using transformation of random fragments of genomic DNA. Only two

7

different enzymes participate in the maltose biosynthetic pathway, and the genes for both are within a single operon.

What does this data suggest?

Version 1

8

A) There are fifteen enzymes in maltose metabolism.

B) All the strains have mutations that prevent transcription of the operon.

C) Fifteen genes encode maltose biosynthesis

enzymes, but only two genes are crucial.

D) Maltose is the preferred sugar source for the mutant bacteria.

27) Why is it advantageous for transcription factors to be multimeric and for their binding sites to be clustered?

A) The strength of the DNA-protein interactions is increased.

B) The DNA sequence specificity of each DNAprotein interaction is increased.

C) Each DNA-protein complex is better able to bind to RNA polymerase.

28)

D) Competition for binding between different transcription factors for the same regulatory region is decreased.

Small RNAs can regulate gene expression by

A) binding target RNAs and inhibiting translation.

B) binding the rho gene and preventing termination.

C) binding target promoter regions and preventing

transcription.

D) binding target operator regions and preventing transcription.

29) What is the relationship between the promoters for antisense RNA and the promoters of their target genes?

A) The promoters are oriented so that they transcribe opposite strands of the same segment of DNA. B) The promoters are found in different Version 1

chromosomal regions but are controlled by the same sigma factors.

C) The sense and 9

antisense RNAs use the same promoter, both RNAs are produced from a single transcript that is processed.

D) The sense and antisense RNAs are transcribed

using the same promoter that is bidirectional.

30) What is an advantage of using the lac regulatory region to control the expression of exogenous proteins in bacterial cells?

A) Expression can be controlled by the addition of an inducer to the media.

B) This regulatory region functions in both bacterial and mammalian cells.

C) By using this regulatory region, eukaryotic genomic DNA can be expressed in bacterial cells.

D) Host regulatory mechanisms are not effective when this regulatory region is cloned into a plasmid.

31) Why is the analysis of transcriptomes useful for genetics research?

A) It can reveal many aspects of gene regulation.

B) It is easy to do without computer programs.

C) It can help discover nontranscribed DNA regions.

D) It directly identifies transcription factor binding sites.

32) How was E. coli was used to aid in the cloning of the bioluminescence genes from Vibrio fischeri?

A) Gene fragments from V. fischeri were transformed into E. coli and bioluminescent E. coli were identified.

B) Gene fragments from V. fischeri were cloned into plasmids, transformed into E. coli, and then retransformed

Version 1

back into V. fischeri.

C) Gene fragments from E. coli were used to disrupt the bioluminescence genes of

10

V. fischeri by homologous recombination.

D) Transposons from E. coli were used to mutagenize the bioluminescence genes of V. fischeri.

33) V. fischeri senses the density of V. fischeri cells in the environment by

A) secretion of a molecule that can then be internalized and bound to a receptor in the cytoplasm.

B) cell surface receptors that interact with other cell surface receptors on nearby cells.

C) a cell surface receptor that binds to a molecule

secreted by cells in the environment.

D) a cytoplasmic receptor that binds to the LuxC protein.

34) What is one possible advantage to targeting quorumsensing mechanisms for antibiotic development?

A) There may be no selective advantage to individual cells that are resistant to quorum-sensing inhibitors.

B) There is a selective advantage to those cells lacking a quorum-sensing system.

C) Quorum-sensing mechanisms are found in all

pathogenic bacteria.

D) Quorumsensing mechanisms are found only in human pathogens.

35) A mutation in a ribosome binding site in the lac operon would likely result in C) complete loss A) reduction in the amount of LacZ, LacY, or LacA protein.

B) increased production of the LacZ, LacY, and LacA proteins.

Version 1

11

of transcription of the lac operon.

D) a lac mRNA that is shorter than wild-type.

Version 1

12

36) Cells of which bacterial genotype would be able to utilize lactose as an energy source?

D) I + o + Z−Y + A) ISo + Z + Y + B) I + o + Z + Y −/ F′ ( I S o + Z − Y +) C) I + o c Z + Y − /F′ ( I + o + Z − Y +)

37) A fragment of DNA with a radioactive label is incubated with DNase in the presence and absence of a protein that acts as a positive regulator of transcription. The samples are then run on a gel and visualized with autoradiography. Which statement describes a possible result A) The banding pattern on the gel is the same with and without the transcriptional regulator, therefore the protein binds the DNA fragment.

B) Some bands on the gel are absent in the sample with the transcriptional regulator, therefore the protein does not bind the DNA fragment.

C) Some bands on the gel are absent in the sample with the transcriptional regulator, therefore the protein binds

and conclusion from such an experiment?

the DNA fragment.

D) Some bands on the gel are absent in the sample without the transcriptional regulator, therefore the protein does not bind the DNA fragment.

38) An enzyme functions in a biosynthetic pathway in bacteria. You predict that the product of the reaction may

regulate the levels of the protein by

A) binding an aptamer in the RNA leader sequence that alters the conformation to block the ribosome binding site.

B) binding an aptamer in the RNA leader sequence that alters the conformation to expose the ribosome binding site.

C) binding the promoter region to prevent transcription.

D) binding an aptamer in the RNA leader sequence that causes an antiterminator to form.

Version 1

13

39) You have identified a 50 nucleotide deletion that is not within an open reading frame in an otherwise wild-type prokaryotic genome. You identify five mRNAs that are not translated in these mutant bacteria; the five mRNAs are transcribed from five different genes located far from the deletion. Select the most likely explanation for the effect of this deletion.

A) The deletion is within an antisense RNA that regulates translation of all five genes.

B) The deletion prevents expression of a small RNA that regulates translation of all five genes.

C) The deletion removes an aptamer in an RNA

leader sequence.

D) The deletion removes an effector molecule that regulates transcription.

40) When RNA-Seq is performed before and after heat shock of bacterial cells, computer analysis would be expected to show A) very few changes in the cDNAs sequenced because the response to heat shock occurs primarily at the protein level.

B) after heat shock, an increase in the average length of sequence reads of cDNAs that encode molecular chaperone proteins.

C) after heat shock, an increase in the number of sequence reads of cDNAs that encode molecular chaperone

proteins.

D) after heat shock, a decrease in the number of sequence reads of cDNAs that encode molecular chaperone proteins.

Answer Key Test name: Chapter 18 Test Bank Version 1

14

Version 1

15

1) A 2) A 3) A 4) C 5) B 6) B 7) B 8) A 9) B 10) B 11) C 12) D 13) C 14) C 15) A 16) C 17) B 18) B 19) A 20) E 21) A Version 1

16

22) A 23) B 24) B 25) D 26) B 27) A 28) A 29) A 30) A 31) A 32) A 33) A 34) A 35) A 36) C 37) C 38) A 39) B 40) C

Version 1

17

Student name:__________

A) They can be more than 10 kilobases from the genes they regulate.

B) They contain TATA boxes.

C) They may bind to more than one transcription factor at the same time.

D) They can increase gene transcription levels above

1) Which is true of enhancer DNA sequences? the basal level.

E) They retain function if their nucleotide sequence is moved or inverted.

2) In Drosophila females, Sxl protein directly alters the splicing pattern of the transcripts from which gene(s)?

A) yp1 B) fru

C) Sxl D) tra E) Dsx

3) Which is the best method to demonstrate the function of a putative enhancer?

A) Compare DNA sequences from two or more species and identify conserved regions located outside of coding regions.

B) Introduce two separate pieces of DNA into a eukaryotic cell; one is a fragment of DNA with the putative enhancer and the other is a plasmid that contains a reporter gene.

C) Clone a region of DNA with the putative enhancer into a reporter construct, introduce the construct into eukaryotic cells, and measure expression of the reporter protein.

D) Clone the putative enhancer into a reporter

Version 1

construct and measure the level of reporter gene expression in a prokaryotic cell.

E) Crosslink genomic DNA and associated proteins, fragment the DNA, use an antibody to isolate DNA sequences bound to a particular transcription factor, and sequence the DNA.

1

4) The enhancer shown in the diagram is able to regulate transcription of which genes?

A) Gene A B) Gene B

C) Gene C D) Gene D

5) Transcription in prokaryotes and eukaryotes is similar in that A) transcriptional machinery controls the structure of chromatin.

B) the mRNA produced can undergo alternative splicing.

C) both are regulated by promoters that can be located far away from the gene.

D) both occur within the cell nucleus.

E) both are regulated by binding of proteins to DNA near the gene being transcribed.

6) Which of these types of gene regulation occurs earliest in the process of gene expression?

A) alternative splicing B) protein modification C) export of mRNA to the cytoplasm

D) translation initiation

7) A protein with a zinc-finger domain most likely has what type of activity?

Version 1

2

A) modifying proteins with phosphate groups B) DNA binding C) mRNA splicing

D) initiation of translation

8) DNA sequences that are binding sites for transcription factors are called A) cross-reacting.

B) cis-acting elements.

C) trans-acting factors.

D) origins of transcription.

E) transcription factors.

9) What is the name of the cis-acting DNA sequence at which the transcriptional initiation complex forms?

A) promoter B) terminator C) enhancer

D) activator E) transcription factor

10) What is the term for a cis-acting DNA sequence that may function at a distance from the gene they are regulating?

A) promoter B) terminator C) enhancer

Version 1

D) activator E) transcription factor

3

11)

What is a trans-acting factor that increases

transcription above the

A) repressor B) activator C) promoter D) enhancer E) basal factor

Version 1

4

12) A putative regulatory sequence is fused to an “enhancerless” GFP gene in a reporter construct, and that construct is then used to generate a transgenic organism. If the putative regulatory sequence contains an enhancer that binds with an eye-specific combination of activators, how will the transgenic organism look?

A) The entire body will glow green at all times.

B) Only the eyes will glow green at all times.

C) Only the eyes will glow green when exposed to a particular wavelength of light.

D) No parts of the transgenic organism will glow green.

13) Where do mutations that alter the amount of protein synthesized from a reporter construct occur most often?

A) the reporter gene B) cis-acting DNA elements C) X-gal coding sequence

14)

D) RNA polymerase II gene E) None of the choicesis correct.

A trans-acting factor functions by

A) binding to enhancers or promoters leading to increased or decreased transcription.

B) regulating the expression of the gene from which it was transcribed.

C) transporting mRNAs from the nucleus to the cytoplasm.

Version 1

D) binding to and increasing the activity of DNA polymerase.

E) binding only to promoters and increasing the binding of RNA polymerase.

5

15) Trans-acting DNA-binding proteins that influence transcription are generally referred to as D) transcription A) DNA polymerases.

B) tumor suppressors.

C) enhancers.

factors.

E) promoters.

16) What do the promoters of nearly all eukaryotic genes contain?

A) a binding site for an activator B) a TATA box C) an ATG

D) a translation start site E) a binding site for a repressor

17) What is a transcription factor that associates with an enhancer and causes an increase in initiation of transcription?

A) initiator B) activator

C) repressor D) enhancer E) demethylase

18) What is the primary function of basal transcription factors?

A) repress transcription initiation of specific genes B) assist RNA polymerase binding to the promoter C) increase transcription by binding enhancer sequences

Version 1

D) bind to activators or repressors and modify histone tails E) add poly-A to the 3′ end of transcripts

6

19) Which is a type of DNA-binding domain found in transcription factors?

A) TAF B) zinc finger

20) Two identical Jun polypeptides associate with each other via their leucine zippers to form an active transcription A) monomer B) oligomer

21)

C) major groove D) leucine zipper E) corepressor

factor. What is the transcription factor called?

C) heterodimer D) homodimer E) diplomer

The leucine zipper motif functions to

A) mediate the physical association of two polypeptides.

B) anchor transcriptional activator proteins to enhancer sequences.

C) release leucines from misfolded transcription factors.

D) integrate leucines and isoleucines into newly translated transcriptional activators.

E) separate DNA strands for transcription.

22) What is the term for transcription factors that decrease transcriptional activity?

A) deregulators B) depleters

Version 1

C) regulators D) repressors E) TBPs

7

23) The action of which small RNA inhibits the movement of transposable elements?

C) piRNA D) tRNA

A) miRNA B) siRNA

24) Recessive mutations in the Sxl gene in Drosophila are lethal in homozygous XX females because

A) without Sxl protein the expression of genes on both X chromosomes is increased.

B) increased Sxl protein signals male-specific development.

C) these mutations are the result of a loss of both X chromosomes.

D) Sxl is required for transcription of malespecific genes.

E) recessive mutations are always lethal in homozygotes.

25) A) interfere with the association between enhancers and specific promoters.

B) remodel chromatin structure.

26)

Insulators act to

C) bind with repressors.

D) bind with activators.

Enhancers can be identified by

A) sequence comparison with promoter sequences that are already known.

Version 1

B) introducing two separate pieces of DNA into a eukaryotic

8

cell. One is a fragment of DNA with the putative enhancer and the other is a plasmid that contains a reporter gene.

C) cloning the putative enhancer into a plasmid and measuring the level of reporter gene expression in a prokaryotic cell.

D) cloning a region of DNA with the putative enhancer in a reporter plasmid and demonstrating an increase

in gene expression in specific cell types when the construct is introduced as a transgene into a eukaryotic organism.

27) The ChIP technique can be used to identify putative DNA binding sites for various transcription factors because it

A) identifies DNA sequences bound to specific proteins.

B) identifies all sites bound by proteins in the nucleus.

C) identifies the size of different protein-DNA

complexes.

D) identifies proteins that can interact with mRNA.

28) How can one primary transcript result in several polypeptides with different amino acid sequences?

A) alternative splicing of exons B) phosphorylation of proteins after translation C) incorporation of an amino acid at a stop codon

D) addition of a poly-A tail to the mRNA

29) Why is the number of X chromosomes crucial for sex determination in Drosophila?

A) The Sxl promoter is responsive to the concentration of four transcription factors produced from

Version 1

genes on the X chromosomes.

B) The Sxl

9

promoter is responsive to the concentration of four transcription factors produced from genes on the Y chromosome.

C) The X chromosome produces a single transcription factor that results in a male if at a low concentration and a female in a high concentration.

D) The Sxl promoter is repressed by increased concentrations of transcription factors that are produced from genes on the X chromosome.

30) Which statement is true about sex determination in Drosophila?

A) Alternative splicing of Sxl transcripts in females results in an active protein that directs alternative splicing of tra transcripts in females.

B) Sxl protein is produced early in males and it directs splicing of its own RNA to produce more Sxl in males.

C) Four autosomal genes produce transcription factors in males and the transcription factors activate male-

specific gene expression.

D) Tra and Tra2 act together in males to alternatively splice dsx transcripts so that Dsx-M is produced.

31) Interactions between an enhancer and a repressor initiate a series of events that result in a change in gene expression. In what order can the events occur? (Some steps may not be used. Put the remaining steps in chronological order from first to last.)

1. Transcription of the gene of interest is reduced.

2. A corepressor is recruited.

3. Chromatin around the gene of interest expands.

4. Histone tails may be methylated or deacetylated.

5. Chromatin around the gene of interest becomes more compact.

6. Proteins attached to the promoter and enhancer associate with each other and DNA bends.

A) 2, 4, 5, 1

Version 1

10

B) 2, 6, 5, 1 C) 6, 4, 3, 1

D) 2, 6, 4, 5, 1 E) 4, 5, 1, 2

32) Which could be accomplished using ribosome profiling?

A) identifying enhancers that activate genes in muscle tissue B) identifying all of the genes regulated by a particular transcription factor C) quantifying the level at which the mRNAs in a

cell are being translated D) calculating the location and number of genes in a genome

33) Gene B is usually expressed only in skin cells. To learn about the mechanism by which expression of the gene B is regulated, you make clones that contain a GFP reporter and various parts of the upstream and downstream intergenic regions of genomic DNA that normally surround gene B (black lines) as shown in the figure below. The resulting clones were introduced into frog skin cells growing in the lab and levels of GFP expression was monitored by measuring green fluorescence. The table below shows relative GFP expression, with more +s Version 1

11

equal to more expression. (one + = basal expression).

33.1) Which area is likely to contain a cis-acting regulatory element?

D) D E) E F) F

A) A B) B C) C

33.2)

Which area contains the gene B promoter?

D) D E) E F) F

A) A B) B C) C

33.3)

Version 1

Which area contains an enhancer?

12

A) A B) B C) C

D) D E) E F) F

33.4) Which area is likely to bind to proteins that activate gene B? A) A B) B C) C

Version 1

D) D E) E F) F

13

Answer Key Test name: Chapter 19 Test Bank 1) [A, C, D, E] 2) [C, D] 3) [C] 4) [A, B, C] 5) E 6) A 7) B 8) B 9) A 10) C 11) B 12) C 13) B 14) A 15) D 16) B 17) B 18) B 19) B Version 1

14

20) D 21) A 22) D 23) C 24) A 25) A 26) D 27) A 28) A 29) A 30) A 31) A 32) C 33) Section Break 33.1) [B, D] 33.2) D 33.3) B 33.4) B

Version 1

15

Student name:__________ 1) X-chromosome inactivation in mammalian females appears to be an epigenetic phenomenon, although the molecular nature of the “cellular memory” transmitted during cell division is unknown. In addition to being bound by Xist lncRNA, what else distinguishes the inactive X chromosome?

A) change in the base-pair sequence B) methylation of CpGs C) acetylation of histones

D) demethylation of DNA E) methylation of histones

2) Which of the following exhibit “unprogrammed” transgenerational epigenetic inheritance?

A) Silenced epialleles in plants B) Genomic imprinting C) Memory of transposable element (TE) invasion in Drosophila

D) Xchromosome inactivation in mammalian females E. Hox gene silencing in Drosophila

3) What describes a situation in which an allele’s expression depends on the parent from which it was inherited?

A) genomic imprinting B) chimerism C) mosaicism

Version 1

D) homodimerization E) autosomal linkage

1

A) enhancer trap.

B) replication origin.

C) Mendelian trait.

D) mutant chromosome.

E) epigenetic phenomenon.

5) What is the relationship between methylation and genomic imprinting?

A) Differential methylation of DNA in males and females results in only one of the parental alleles being transcriptionally active in the offspring.

B) DNA methylation silences the alleles inherited from both parents, so no alleles are transcriptionally active.

C) Reversible methylation of DNA inherited from the father results in the paternal alleles being expressed only

Version 1

late in development.

D) Reversible methylation of DNA inherited from the mother results in the maternal alleles being expressed only late in development.

2

6) This pedigree shows the segregation of a rare disease in a family. What is the most likely mode of inheritance of this disease? (Assume complete penetrance.)

A) autosomal recessive B) autosomal dominant C) X-linked recessive D) X-linked dominant E) maternal inheritance

F) maternal imprinting G) paternal imprinting

7) This pedigree shows the segregation of a rare disease in a family. What is the most likely mode of inheritance of this disease? (Assume complete penetrance.)

Version 1

3

A) autosomal recessive B) autosomal dominant C) X-linked recessive D) X-linked dominant E) maternal inheritance

F) maternal imprinting G) paternal imprinting

8) This pedigree shows the inheritance pattern of a disease caused by a paternally imprinted allele in a family.

Genetic testing of individuals I-1 and II-5 indicates they are carriers of the disease allele. No one else in the family agreed to be tested. If first cousins III-4 and III-5 have a child together (IV-1) what is the chance the child will have the disease? (Assume complete penetrance.)

A) 0 B) 1/9

C) 1/4 D) 1/2 E) 1/3

9) Epialleles are common in plants and once Version 1

4

established, a silenced epiallele can be transmitted to progeny for hundreds of generations. What epigenetic mechanism in plants silences epialleles?

A) Histone modifications B) DNA methylation C) Change in the base-pair sequence

10) In plants, heterochromatin is re-formed following cell division through modifications of DNA and histones at the replication fork during DNA replication. Identify the correct order of events that result in a feedback loop that copies DNA A) H3K9me marks on nucleosomes→binding of a histone methyltransferase (HMT) →binding of a DNA methyltransferase and DNA methylation→feedback loop.

B) Binding of a histone methyltransferase (HMT) → H3K9me marks on nucleosomes→binding of a DNA methyltransferase and DNA methylation→feedback loop.

C) H3K9me marks on nucleosomes→binding of a DNA methyltransferase and DNA methylation →binding of a

D) Binding after DNA replication of a sequence-dependent DNAbinding protein

methylation and histone modification marks.

histone methyltransferase (HMT) → feedback loop.

D) Binding of a DNA methyltransferase and DNA methylation→ binding of a histone methyltransferase (HMT).

11) Hox genes encode transcription factors that function as master regulators. Maintenance of Hox gene silencing in Drosophila occurs in part through an epigenetic mechanism.

Which of the following is sufficient for Hox gene silencing in Drosophila?

A) Methylation of a CpG island.

B) A histone methyltransferase that binds directly to a site near the Hox gene promoter.

C) A histone deacetylase that acts as a corepressor.

Version 1

D) A cis-acting DNA sequence that binds a repressor and a corepressor with histone methyltransferase activity.

5

12) In Drosophila, hybrid dysgenesis (the sterility of hybrid progeny—that is, progeny where one parent has P elements and the other has none) depends on the sex of the parent that carries the P element transposon. This phenomenon involves piRNAs that prevent mobilization of the P element. In the following figure, two crosses are shown —the cross on the left has P elements in the male parent but not in the female parent; the cross on the right has P elements in the female parent but not in the male parent.

12.1) Which parent in these crosses produces gametes that contain P element piRNAs? A) Male with P elements B) Female with P elements C) Male with no P elements

12.2)

D) Female with no P elements

Which cross will give rise to sterile progeny?

A) The cross where the male parent has P elements and the female parent has no P elements B) The cross where the female parent has P

elements and the male parent has no P elements

13) The following figure shows two cross schemes. The difference between the two schemes is that the female parent in

Version 1

6

the P generation in the scheme on the left (Experiment) was fed a diet rich in methyl donors; the P generation female parent in the scheme on the right (Control) was fed a normal diet.

Version 1

7

13.1) Based on how DNA methylation influences the expression of the AVY allele, which pattern of inheritance is exhibited in the figure?

A) Mendelian inheritance B) intergenerational epigenetic inheritance C) transgenerational epigenetic inheritance

D) organellebased maternal inheritance

13.2) Why are the F1 AVYa mice from the cross on the left darker on average than those from the cross on the right?

VY

A) The A allele present in the F 1 mice receives the methyl groups from methylated a allele contributed by the female parent that was fed a diet rich in methyl donors.

B) Methylation of the AVY allele of the AVYa embryo occurs in the female eating the methyl-enriched diet.

C) The methylated state of the AVY allele is stable and this methylated state is passed without any changes

Version 1

through successive generations.

D) Feeding the female parents diet rich in methyl donors alters the a allele contributed by the female parent to an AVY allele which is methylated.

8

Answer Key Test name: Chapter 20 Test Bank 1) [B, E] 2) [A, C] 3) A 4) E 5) A 6) F 7) C 8) C 9) B 10) C 11) D 12) Section Break 12.1) B 12.2) A 13) Section Break 13.1) B 13.2) B

Version 1

9

Student name:__________ 1)

To create transgenic mice, DNA is injected into A) a pronucleus of a zygote.

B) the cytoplasm of a zygote.

C) the nucleus after the pronuclei have fused.

D) a pronucleus of one cell from an embryo.

2) What is the purpose of the P element in the creation of transgenic Drosophila? cells.

A) Efficient transfer of DNA into Drosophila chromosomes.

B) Transfer of DNA into a specific site in the host chromosome.

C) Transfer of DNA only into the nuclei of somatic

3)

D) Transfer of DNA only into primitive eye cells.

Recombinant Ti plasmids allow for cells.

A) the transfer of a transgene from the plasmid in Agrobacterium to a plant cell.

B) the transfer of plant DNA to Agrobacterium.

C) the transfer of DNA between two Agrobacterium

D) the transfer of the plasmid from Agrobacterium to E. coli.

4) Transgenic organisms are used to clarify which gene may be responsible for a mutant phenotype because

A) individuals with a mutant phenotype can carry mutations in more than one gene.

Version 1

B) the genes that cause diseases are always dominant.

1

C) the genes that cause diseases are always recessive.

D) strains used to create transgenic organisms carry

no mutations, and all their genes are wild-type.

5) How can transgenic organisms be used to identify transcriptional enhancers?

A) A reporter gene can be constructed so that its expressionis controlled by an enhancer that may potentially exist within a fragment of genomic DNA. B) A reporter gene can be expressed in every tissue.

C) A gene that is normally expressed in the organism can be cloned under the control of different regulatory elements so that it is expressed in all tissues.

D) A transgene can be integrated in a site that disrupts a gene to determine whether that gene is essential for the cell or organism.

6) Although bacteria can be used to produce proteins for therapeutic uses, what can be a disadvantage to using a prokaryotic system?

A) Proteins may not be posttranslationally processed correctly.

B) Transcripts from eukaryotic open reading frames are degraded quickly in prokaryotic cells.

C) Some prokaryotic promoters do not work well to transcribe eukaryotic open reading frames.

D) The structures that eukaryotic mRNAs fold into are frequently used as rho termination signals.

7) What is a possible disadvantage of using genetically modified crops?

A) The transgene may make its way into wild plants.

Version 1

B) Transgenic 2

strains are typically less robust than the usual strains used for farming.

C) The ability of transgenic plants to use nutrients is reduced so more fertilizer has to be applied.

D) The ability of transgenic strains to use water is

poor so these strains are not drought tolerant.

8) Transgenic animals made by pronuclear injection can be limited for studies of the links between genes and disease because genes.

A) the transgene allele must be dominant to the normal, endogenous allele, and some diseases are caused by recessive mutations.

B) the transgene allele must be recessive to the normal, endogenous allele, and some diseases are caused by dominant mutations.

C) all diseases are caused by mutations in multiple

D) diseases are not caused by specific alleles of a gene, but by posttranslational modifications of proteins.

9) A trait unique to embryonic stem cells that makes them useful for generating knockout animals is that they A) are totipotent.

B) can be grown in culture.

C) have been isolated from mice.

10)

D) are extremely well characterized at the molecular level.

Why are conditional knockout animals desirable?

A) genes that are essential for development can be deleted in the adult or only in specific tissues.

B) animal cell lines with deletions can be created.

C) they are easier to create than standard knockout

Version 1

animals.

D) the mutations they create are more stable than those created by standard knockout

3

techniques.

Version 1

4

11) Knockin mice can be created using the Cre/loxP system. In this procedure, the loxP sites are used to

A) delete the neomycin-resistance marker gene.

B) recombine the mutation into the genome.

C) express Cre protein.

12)

D) flip the orientation of genomic DNA after the mutation is inserted.

CRISPR sequences occur naturally as A) an antiviral immune system in bacteria.

B) an antibacterial immune system in eukaryotes.

C) an antiviral immune system in eukaryotes.

D) a gene editing system in bacteria.

13) Currently the two main vectors for delivery of therapeutic genes for human gene therapy are

A) AAV and retroviruses.

B) AAV and Agrobacterium.

C) Retroviruses and Agrobacterium.

D) AAV and P elements.

14) What is one problem associated with using retroviruses as human gene therapy vectors?

A) Their random integration into the host chromosome may result in detrimental mutations.

Version 1

B) They need to integrate through an RNA intermediate.

5

C) The level of proteins produced from them is usually very low.

D) The level of proteins produced from them is

usually very high.

15) If integration of DNA provided by pronuclear injection occurs after three cell divisions, the resulting mouse will A) not contain the transgene in any cells.

B) be a mosaic of cells with some cells containing the transgene and others not.

C) contain the transgene in only germ cells.

D) contain the transgene in only somatic cells.

16) A researcher has a laboratory strain of flies that have loss-of-function mutations in both gene X and white. These flies are used as hosts for P element transformation: Embryos of the double mutant strain are injected with two plasmids, one containing a wild-type copy of the white gene ( w +) and a wild-type copy of gene X, side-by-sidewithin P element inverted repeats, and the other plasmid containing only the P element transposase gene. Transgenic offspring of the injected flies will A) have both plasmids integrated randomly into the genome.

B) have white eyes and express transposase.

C) have red eyes and a wild-type copy of gene X integrated into a random location in the genome.

17) After exposure to Agrobacterium tumefaciens bacteria containing both a recombinant T-DNA vector and a helper plasmid, new plants can be grown from single cells in

Version 1

D) have red eyes and a wild-type copy of gene X in place of the mutant copy in their genome.

the presence of herbicide.

Without the herbicide,

6

A) neither transformed nor nontransformed cells would survive.

B) only nontransformed plants would grow.

C) only cells with the helper plasmid would grow.

18) A scientist wants to determine whether a mutant phenotype is due to the loss of gene C or gene D, which are both deleted in a mutant fly with abnormal eyes. Mutant flies with transgenic gene C have wild-type eyes, whereas mutant A) Either gene C or gene D is required for normal eye development.

B) Both gene C and gene D are required for normal eye development.

C) Gene C is required for normal eye development.

D) both transformed and nontransformed plants would grow.

flies with transgenic gene D have abnormal eyes.

What can be concluded from these results?

D) Gene D is required for normal eye development.

19) The coding sequence of the green fluorescent protein gene ( GFP) is placed under the control of an enhancer active in neurons. When this transgene is introduced into flies, GFP will be expressed in A) all neurons or a subset of neurons.

B) all fly cells that have the transgene.

C) all cells in the early embryo and then GFP will be

silenced.

D) all cells in the late embryo after neurons form.

20) Pharming refers to the use of transgenic animals and plants to produce protein drugs, such as A) the production of the blood factor antithrombin III in goat milk.

Version 1

B) the production of factor VIII protein, the blood-clotting factor, by

7

bacteria.

C) GM plants that have enhanced nutritional value.

D) a transgenic monkey model for Huntington

disease.

21) The FDA-approved GM Atlantic salmon have a growth hormone transgene that is expressed year-round,

which is advantageous because

A) the GM salmon are more nutritious than normal salmon.

B) the GM salmon achieve their full weight faster than normal salmon.

C) the GM salmon produce more offspring than

normal salmon.

D) the GM salmon are more resistant to infection than normal salmon.

22) Fragile X syndrome is a human disease caused by a trinucleotide repeat expansion that results in loss-of-function of the FMR1 gene on the X chromosome. An animal model with most features of this syndrome could be created by

A) adding a transgene containing a mutant copy of the FMR1 gene to a mouse or primate genome.

B) adding a transgene containing a wild-type copy of the FMR1 gene to a mouse or primate genome.

C) knocking out both copies of the FMR1 gene from

a mouse or primate genome.

D) knocking out one copy of the FMR1 gene from a mouse or primate genome.

23) To create a knockout mouse, after introducing to ES cells a DNA construct in which a specific gene is mutagenized by the insertion of a drug resistance marker, the cells A) that have incorporated the transgene by

Version 1

homologous recombination grow in the presence of the

8

drug and these are injected into a host blastocyst.

B) that lack the transgene grow in the presence of the drug and these are injected into a host blastocyst.

C) that have deleted the transgene grow in the presence of the drug and the nuclei are injected into a host blastocyst.

D) that lack the transgene grow in the presence of the drug and the nuclei are used for nuclear transfer.

24) A scientist might create a conditional knockout of a mouse gene when A) a homozygous loss-of-function mutation in the gene causes embryonic lethality.

B) a homozygous loss-of-function mutation in the gene produces a wild-type phenotype.

C) a homozygous loss-of-function mutation in the

25) To generate a knockin mouse, a scientist introduces to ES cells a construct with a mutant exon and within an adjacent intron, loxP sites flanking a neomycin resistance A) Homologous recombination between the mutant DNA and the corresponding gene in the mouse genome.

B) Recombination between the loxP sites.

C) Transplantation of cells into host blastocyst.

gene affects only ears.

D) a homozygous loss-of-function mutation in the gene affects only coat color.

gene. What event will occur next?

D) Crossing of heterozygous mice to a Cre-expressing strain.

26) Using the CRISPR/Cas9 system, a geneticallyengineered sgRNA complementary to a target site in the genome binds to Cas9 endonuclease. A mutation can occur at this site when A) Cas9 cleaves the DNA and nonhomologous endjoining results in a small deletion.

Version 1

B) Cas9 cleaves the DNA and homologous recombination with the 9

homologous chromosome is used to repair the break.

C) Cas9 cleaves the DNA to remove the complementary sequence.

D) Cas9 cleaves the DNA and nonhomologous end-

joining correctly repairs the break.

27) The disease Leber congenital amaurosis (LCA), caused by loss-of-function mutation of a gene called RPE65, has been treated with gene therapy by injecting recombinant AAV vectors containing a normal copy of the RPE65 gene

into their retinal epithelial cells. Patients receiving this therapy

A) may acquire detrimental mutations due to integration of the virus.

B) may need to repeat the treatment as the viral DNA is degraded over time.

28) Huntington disease is caused by expansion of the trinucleotide repeat region of the HD gene that results in the production of a huntingtin protein with an expanded number of glutamines. An animal model with most features of this A) adding a transgene containing a disease-causing mutant allele of the HD gene to a mouse or primate genome.

B) adding a transgene containing a wild-type copy of the HD gene to a mouse or primate genome.

C) knocking out both copies of the HD gene from a

29) Scientists are using CRISPR/Cas9 to ameliorate sickle-cell disease symptoms in human patients. Which of the A) embryonic stem cells

Version 1

C) will pass the mutation to their offspring.

D) now have the normal RPE65 gene in all cells.

syndrome could be created by

mouse or primate genome.

D) knocking out one copy of the HD gene from a mouse or primate genome.

following is required for this method of gene repair?

B) aT-DNA plasmid vector

10

C) a viral vector D) a single guide RNA

Version 1

11

Answer Key Test name: Chapter 21 Test Bank 1) A 2) A 3) A 4) A 5) A 6) A 7) A 8) A 9) A 10) A 11) A 12) A 13) A 14) A 15) B 16) C 17) D 18) C 19) A Version 1

12

20) A 21) B 22) C 23) A 24) A 25) A 26) A 27) B 28) A 29) D

Version 1

13

Student name:__________ 1) Which of the following is the least important characteristic that geneticists look for in a model organism?

A) ease of cultivation B) rapid reproduction

C) small size D) small genome

2) Which statement best describes the limitations of using humans in genetic studies of developmental biology?

A) The variation in human development between individuals is too large for genetic studies.

B) It is impractical and unethical to conduct experimental manipulation or targeted mutagenesis of humans to examine the role of genes in development.

C) It is impractical and unethical to examine and describe abnormalities of human development, such as birth

defects.

D) Genetic studies of human development are not necessary, because most problems in human development are due to environmental influences and not genetics.

3) The observation that ______ provides evidence that all living forms are related.

A) some organisms use mosaic determinism, whereas others use regulative determinism B) only histone genes are conserved C) many genes and genetic pathways are conserved

D) sex determination occurs by different mechanisms in many species

4) The Aniridia gene in humans is involved in eye formation. Although eye development is very different in fruit flies, this gene is highly conserved. What is the Drosophila homolog of Aniridia?

Version 1

1

C) no-eye D) ommatidia

A) eyeless B) Pax-6

5) A temperature-sensitive (ts) allele is often due to a ______ mutation that affects protein folding or activity.

C) missense D) silent

A) null B) nonsense

6) What is the most significant advantage of using RNAi to study development?

C) RNA is easy to A) Geneticists have only been able to identify a subset of the genes involved in development.

B) Genetic studies of a single gene can be completed without creating new mutant organisms.

isolate.

D) RNAi causes phenotypes not typical of in vivo development.

7) A) depletes the mRNAs of specific genes.

B) uses dsRNA to bind to transcription factors.

C) can be used to create functional knockin mutants.

RNAi

D) is not useful for the study of development.

8) A researcher performs RNAi against a gene that is required for viability in C. elegans embryos, and 70% of the embryos die during development. Some of the embryos may have survived because

Version 1

2

A) RNAi did not eliminate the mRNA (and protein) of the gene completely.

B) the dsRNA was used to translate additional proteins to compensate for the mRNA that was degraded.

C) RNAi eliminated the protein product of the gene.

D) of issues related to fertilization.

9) The presence of a homeodomain in a protein suggests what about its function?

A) It has kinase activity.

B) It is a membrane bound receptor.

C) It is a secreted protein.

D) It is a transcription factor.

10) How would you best follow the timing of expression of a given protein during development in a live animal?

A) in situ hybridization B) RT-PCR C) using a transgenic organism that expresses a GFP fusion protein

11)

D) microarray analysis E) using antibodies against the protein of interest

A homeodomain is a region within a protein that A) acts as a receptor.

B) has kinase activity.

C) binds to specific base-pair sequences.

Version 1

D) binds to transcription factors.

3

12)

Maternal effect mutations can be identified when

A) homozygous mutant females exhibit an abnormal phenotype, whereas mutant males appear wild-type.

B) homozygous mutant females can develop normally, but their offspring exhibit a mutant phenotype.

C) heterozygous mutant females can develop

normally, but their offspring exhibit a mutant phenotype.

D) any aspect of embryonic development is abnormal.

13) In Drosophila, two maternal transcripts that are distributed evenly throughout the oocyte prior to fertilization are A) caudal and knirps.

B) bicoid and nanos.

C) caudal and hunchback.

D) sevenless and even-skipped.

14) Which of the following is a zygotic gap gene in Drosophila?

A) knirps ( kni) B) even-skipped ( eve)

C) caudal( cad) D) hedgehog ( hh)

15) Which of the choices gives the correct sequence of early embryonic development in Drosophila? blastoderm, cellular A) zygote, syncytialblastoderm, multinucleate syncytium, cellular blastoderm B) zygote, cellularblastoderm, syncytial blastoderm, multinucleate syncytium.

C) zygote,multinucleate syncytium, syncytial Version 1

4

blastoderm.

D) syncytial blastoderm, multinucleate syncytium, zygote, cellular blastoderm

Version 1

5

16) Nurse cells surround Drosophila oocytes and produce maternal mRNA and proteins that are deposited into the egg. Which of the following products would be produced by nurse cells?

A) bicoid mRNA B) Antp mRNA

C) Caudal protein D) Hedgehog protein

17) A molecule whose concentration determines the developmental fate of a cell is called A) monomorphic.

B) a maternal effect gene.

18)

C) a signaling molecule.

D) a morphogen.

In primary mutant screens,

A) mutant flies are identified by morphology, and then the mutant genes are isolated and studied.

B) mutant flies are created by RNAi using random RNA sequences and then the genes affected identified.

C) mutant flies are created by gain-of-function

expression of random genes.

D) two mutant flies are mated to determine if there is an effect on morphology.

19) What is a major advantage of using a modifier screen as opposed to a primary screen?

A) Modifier screens can be used to identify Version 1

pleiotropic genes.

B) Modifier 6

screens are easier to perform.

C) Modifier screens use a random selection of flies based on their morphology and therefore should identify all genes involved in development.

D) Modifier screens are performed with chemically mutagenized flies so all genes that affect development should

be identified more rapidly than screening for spontaneous mutants.

20) How do genetic mosaics help determine the focus of action of a gene?

A) Mosaics are composed of cells with different genotypes, so the phenotype of the cells that lack a gene product can be compared to that of neighboring cells that have the gene product.

B) Mosaics are created by mating flies that are homozygous recessive for two different genes and the morphology of the double mutant progeny is studied.

C) Mosaics are created by mating two flies that carry homozygous gain-of-function mutations in different genes

and the effects of the mutations on the morphology studied.

D) Mosaics are created by the Cre-loxP system which allows for the deletion of specific genes at controlled stages of development.

21) How are temperature-sensitive mutants used to determine when genes act during development?

A) Embryos can be shifted to the restrictive temperature to inactivate the mutant gene, and the effect of inactivation on subsequent stages can be observed.

B) Embryos can be shifted to the permissive temperature to inactivate the mutant gene, and the effect of inactivation on subsequent stages observed.

C) Embryos can be incubated at the restrictive temperature for several hours. If the embryo develops normally, then the gene product is required during that period

Version 1

of development.

D) Embryos can be raised at the permissive temperature and if the embryo does not die, then the gene product is not required for development.

7

22) Gene order can be inferred from gene expression patterns because

A) an upstream gene can affect downstream gene expression.

B) mRNAs for proteins in a pathway are produced at the same time, but then are translated only when the proteins are needed.

C) proteins in a pathway are produced at the same time, but the proteins that are not needed are degraded.

D) mRNAs for proteins in a pathway are produced at

the same time, and those that are not needed are destabilized by miRNAs. E) gene expression does not depend on upstream genes.

23) How were mutants used to order the sevenless–Ras pathway?

A) Cells that expressed a constitutively active Ras mutant protein did not require Sevenless activation to become R7.

B) Cells that expressed a constitutively active Sevenless mutant protein did not require Ras activation to become R7.

C) Cells that expressed an inactive Ras mutant

protein did not require Sevenless activation to become R7.

D) Cells that expressed a constitutively inactiveSevenless mutant protein did not require Ras activation to become R7.

24) What is the order of action of these classes of genes during Drosophila development?

A) maternal effect, gap, pair rule, segment polarity B) maternal effect, gap, segment polarity, pair rule C) maternal effect, pair rule, segment polarity, gap

Version 1

D) maternal effect, segment polarity, gap, pair rule

8

25)

Expression of homeotic genes genes.

A) assigns a unique identity to each segment.

B) defines anterior and posterior ends of the fly.

C) controls the expression of the segment polarity

26)

D) subdivides the embryo into a precise number of segments.

Which statement describes a primary mutant screen?

A) A researcher uses CRISPR to create mutations in mouse homologs of three genes that function in heart development in the fruit fly.

B) A scientist uses RNAi to simultaneously knock down the expression of two genes in C. elegans to determine whether they function redundantly.

C) A scientist examines a large number of mutagenized zebrafish embryos for abnormalities in blood

27) The pattern of embryonic expression for a specific gene shown by RNA in situ hybridization and a GFP fusion ⊚ ⊚

vessel formation.

D) A researcher treats a large number of Arabidopsis plants with different chemicals to determine which chemicals inhibit plant growth.

protein will always be identical.

true false

Answer Key Test name: Chapter 22 Test Bank Version 1

9

Version 1

10

1) D 2) B 3) C 4) A 5) C 6) B 7) A 8) A 9) D 10) C 11) C 12) B 13) C 14) A 15) C 16) A 17) D 18) A 19) A 20) A 21) A Version 1

11

22) A 23) A 24) A 25) A 26) C 27) FALSE

Version 1

12

Student name:__________ 1) T cells have a protein on their cell surface called PD1. PD-1 interacts with a protein called PDL-1. Where is PDL1 present?

A) In normal cells in the body B) Only in cancer cells C) In some cancer cells

D) Only in normal cells in the body E) In T cells

2) How are T cells activated to target cancer cells for cancer treatment? cells.

A) By treatment with monoclonal antibodies that block PD-1.

B) By increasing expression of PD-1 on T cells.

C) By treatment with monoclonal antibodies that block PDL-1.

D) By increasing expression of PDL-1 on cancer

E) By treatment with monoclonal antibodies that block PD-1 and PDL-1.

3) Sequencing of cancer cell genomes has led to the identification of mutations in a number of genes. Which of the following would be considered a “driver” mutation in a cancer cell?

A) A loss-of-function mutation in a tumor-suppressor gene.

B) A loss-of-function mutation in a proto-oncogene.

C) A gain-of-function mutation in a proto-oncogene.

D) A gain-of-function mutation in a tumor-

4)

The enzymatic activity of CDKs is regulated by

Version 1

suppressor gene.

E) A mutation in the gene that codes for the β-polypeptide of adult hemoglobin.

forming a complex with

1

which proteins?

A) p53 B) p21

C) growth factors D) cyclins

5) Which statement best describes the action of polypeptide growth factors?

A) They bind to intracellular receptors and trigger intracellular signal transduction pathways.

B) They bind to cell surface receptors and trigger intracellular signal transduction pathways.

C) They bind to cell surface receptors that

phosphorylate cyclins.

D) They bind an intracellular receptor forming a complex that then acts as a transcription factor.

6) The retinoblastoma (Rb) protein controls progression into S phase by regulating ___________ activity.

A) cyclin B) p53

7)

The activity of Rb is controlled by A) its phosphorylation state.

B) cyclin A. C) E2F.

8)

C) CDK4 D) E2F E) CDC28

Which of the following is not likely to occur if p53 is

Version 1

D) its poly-A tail.

E) its level of methylation.

inactivated?

2

A) The appearance of homogenously staining regions on chromosomes B) Increased propensity to arrest in G1 C) Inhibition of the G 1 to S checkpoint D) An increase in gene amplification in affected cells

9)

E) Generation of chromosomes that lack telomeres and centromeres

A cellular process that results in cell death is A) apoptosis.

B) contact inhibition.

C) posttranslational control.

D) metastasis.

10) Which characteristic is typical of a cancer cell but not of a normal cell? tissue.

A) Cell division is inhibited when the cell contacts neighboring cells.

B) The cell cycle lacks S phase.

C) The cell has the ability to invade surrounding

D) Cell death occurs after a limited number of divisions.

11) The enzyme telomerase is essential for unlimited cell division because

A) it is essential for DNA replication initiation at origins of replication.

B) a lack of telomerase will result in chromosome shortening and cell death.

C) it is essential for kinetochore attachment to the Version 1

spindles.

D) a lack of telomerase results in decreased p53 activity.

3

12) A karyotype of chromosomes from a cancer cell would be most likely to exhibit

A) polyploidy.

B) a single missing chromosome.

C) a single extra chromosome.

D) a normal set of chromosomes because only point mutations have occurred.

13) The ability of a cell to move from the tissue to which it belongs to other parts of the body is known as

A) metastasis.

B) contact inhibition.

14)

The current model of cancer development isthat

A) cancer develops from a single cell that evolves through one to twomutations.

B) cancer develops from a single cell that evolves through multiple mutations.

C) cancer develops from multiple cells that evolve

15)

C) immune surveillance.

D) angiogenesis.

through multiple mutations.

D) cancer develops from multiple cells that evolve through one to two mutations.

In its active form, the Ras protein is associated with B) ATP. A) DNA.

Version 1

4

C) GTP. D) RNA. E) GDP.

Version 1

5

16) What is one piece of evidence that supports the hypothesis that cancers are clonal?

A) A tumor grows in a single anatomical site.

B) All cells from the same cancer have identical mutations.

C) All cells from the same tumor have identical karyotypes.

D) All cancer cells from a heterozygous woman have

the same X chromosome inactivated.

E) Individuals of all ages, including children, can develop cancer.

17) What is the term for mutant alleles that are dominant to normal alleles at the cellular level and act as cancer drivers?

A) Polymerases B) Oncogenes

C) Activators D) Tumorsuppressor genes

18) How can a mutation in a tumor-suppressor gene behave as a recessive allele at the cellular level, but appear as a dominant allele in pedigree analysis?

A) Some functional protein can still be produced in a cell heterozygous for a loss-of-function allele; however, an additional mutation can inactivate the second allele, eliminating all functional protein allowing cancer to occur in many family members.

B) A single functional allele is sufficient to regulate the cell cycle, but not sufficient to promote apoptosis,

Version 1

allowing cancer to develop in every generation.

C) A cell with half the amount of the functional tumorsuppressor protein can form benign tumors that will occur in every

6

generation in the pedigree.

D) Some functional protein can still be produced in a cell heterozygous for a loss-of-function allele; but loss-offunction tumor-suppressor alleles are so common in the

population that many individuals are homozygous.

19) The human papillomavirus (HPV) carries a gene that functions as an oncogene by inactivating the p53 protein. The fact that the loss of p53 function is oncogenic suggests that A) p53 normally functions to prevent uncontrolled cell division.

B) p53 is a proto-oncogene.

C) the HPV protein activates transcription of p53.

D) p53 functions at origins of replication in DNA.

20) The BRCA2 protein functions in double-strand break repair. If a woman is heterozygous for a loss-of-function allele of BRCA2, she

A) will develop breast cancer because two functional copies of BRCA2 are needed in every cell.

B) will not exhibit breast cancer because BRCA1 can compensate for the loss of BRCA2.

C) will develop breast cancer after several mutations occur, even if the mutations occur in different clonal

populations of cells.

D) may develop breast cancer because mutation of the wild-type BRCA2 allele could destroy an important DNA repair mechanism.

21) What is the relationship between genomic instability and proliferation?

A) No relationship exists between genomic

Version 1

instability and proliferation rate.

7

B) Genomic instability increases the probability of developing cells with a high rate of proliferation.

C) Genomic instability reduces the probability of developing cells with a high rate of proliferation.

D) Cells that proliferate rapidly have the most stable genomes.

22) Drugs such as Gleevec® and Herceptin® represent a modern approach to cancer treatment because they

A) target specific gene alterations that occur in some cancers.

B) target metabolic pathways that are common to all cells.

C) enhance apoptotic pathways in all cells.

D) target proteins that are present in both normal and cancer cells, but that cancer cells rely on more than normal cells do.

23) What is one reason why tumor-suppressor genes make poor druggable targets?

A) Cancer-causing mutations in tumor-suppressor genes result in loss of function.

B) Mutant alleles of tumor-suppressor genes generally result in a greater activity of the protein, which makes inhibiting the proteins more difficult.

C) Most drugs cannot distinguish the normal and

mutant forms of the tumor suppressor proteins.

D) Drugs that can target-tumor suppressor proteins are difficult to deliver to cancer cells.

24) What is one current limitation of whole-genome sequencing for personalized cancer treatment?

A) The human genome is too large to be sequenced for individual patients.

Version 1

B) Drugs may not exist to target the mutant genes that are identified.

8

C) Whole-genome sequencing reveals only mutations in coding sequences, and regulatory mutations may also be important.

D) The genomes of cells with abnormal karyotypes,

such as cancer cells, cannot be sequenced reliably.

25) Which of the following scenarios exemplifies cancer therapy targeted toward a specific mutation?

A) A patient with leukemia receives chemotherapeutic drug that inhibits DNA synthesis.

B) A benign tumor is surgically removed from a patient’s kidney.

C) A patient with melanoma receives a drug that targets a signal transduction pathway protein.

26) Rous sarcoma virus was identified when a scientist discovered that a virus passing between chickens was causing their tumors. The genome of this retrovirus includes the src gene, which encodes a tyrosine kinase that functions in A) cyclin.

B) proto-oncogene.

C) reverse transcriptase.

D) A patient with two mutant tumorsuppressor alleles receives radiation and a chemotherapeutic drug that blocks mitosis.

signaling pathways leading to proliferation. The src gene is a D) tumorsuppressor gene.

27) A man inherited a chromosome with a deletion that removes a tumor-suppressor gene. The protein encoded by this gene normally functions in DNA damage repair. Which

series of events would be most likely to lead to cancer?

A) Loss-of-heterozygosity of the tumor-suppressor gene followed by driver mutations in two different protooncogenes B) A loss-of-function mutation in a proto-oncogene

followed by a gain-of function mutation in a tumor-suppressor gene C) Loss-ofheterozygosity of the

Version 1

9

tumor-suppressor gene followed by passenger mutations in several genes of unknown function D) A gain-of-function mutation in a proto-oncogene followed by a chromosomal duplication of a region including

a tumor-suppressor gene that functions the G 1-to-S checkpoint

28) When autocrine stimulation is occurring, cancer cells divide in response to growth signals produced by the

surrounding noncancerous cells.

⊚ ⊚

true false

29) Genomic instability in cancer cells includes both nucleotide mutations and chromosomal aberrations. ⊚ ⊚

true false

30) If cancer does not develop within one year following exposure to a mutagen, then the mutagen did not cause DNA damage. ⊚ ⊚

true false

31) If a female is heterozygous for a mutant allele of a tumor-suppressor gene, her children will be predisposed to ⊚

Version 1

true

develop cancer only if their father is also a carrier. ⊚

false

10

Version 1

11

Answer Key Test name: Chapter 23 Test Bank 1) [A, C] 2) [A, C, E] 3) [A, C] 4) D 5) B 6) D 7) A 8) B 9) A 10) C 11) B 12) A 13) A 14) B 15) C 16) D 17) B 18) A 19) A Version 1

12

20) D 21) B 22) A 23) A 24) B 25) C 26) B 27) A 28) FALSE 29) TRUE 30) FALSE 31) FALSE

Version 1

13

Student name:__________ 1)

What does biological ancestry describe?

A) the ancestors from whom mitochondria were inherited B) interaction between different generations in a population C) the ancestors from whom specific alleles of

nuclear genes were inherited D) family trees or pedigrees

2) The sum total of all alleles carried in all members of a population is called the A) gene pool.

B) genome.

3) In forensics, a DNA sample is analyzed to determine which alleles of 13 SSR loci are present. How is the probability of that specific combination of SSR alleles A) The frequency of each genotype is known from determining which allele is the rarest for each locus and then determining if the individual sample contains those alleles.

B) The total number of alleles for all loci in the population is calculated.

C) The Hardy-Weinberg Law is used to predict the

C) ploidy.

D) polygenic sum.

E) polymorphism.

existing in the population calculated? genotype frequency at each locus.

D) The total number of alleles in the population is multiplied by 2.

4) The proportion of individuals in a population that are of a particular phenotype is the A) phenotype frequency.

Version 1

B) genotype frequency.

1

C) allele frequency.

D) None of thechoices is correct.

5) The proportion of individuals in a population that are of a particular genotype is the A) phenotype frequency.

B) genotype frequency.

C) allele frequency.

D) None of thechoices is correct.

6) Which of the following is one of the assumptions of the Hardy-Weinberg law?

A) The population is small.

B) Non-random mating occurs within the population.

C) Mutations appear in the gene pool at high frequency.

D) Migration occurs into or out of the population.

E) The ability of all genotypes to survive and reproduce is the same.

7) Why is the Hardy-Weinberg equation more accurate in predicting genotypic frequencies in the short run than in the long run?

A) The equation does not take epistasis into consideration.

B) The Hardy-Weinberg equation is based on Mendel's second law of inheritance.

C) The equation can be applied only to traits

Version 1

determined by single genes.

D) Populations do not conform to the assumptions of the HardyWeinberg law.

2

8)

What is the most common reason that new mutations

are lost from a population? fitness.

A) Most mutations are caused by exposure to environmental mutagens that result in many mutations in the same individual, leading to death.

B) Most mutations are silent.

C) All new mutations result in lower reproductive

D) Populations are of a finite size and new mutations can be lost by chance.

9) Why are many human recessive disease alleles maintained in populations despite continuing selection against them?

A) Heterozygotes have a higher fitness than either homozygote.

B) Mutations convert recessive alleles to dominant alleles.

C) Even when the allele frequency is low, many

gametes contain the recessive allele.

D) Allele frequency remains constant from generation to generation.

10) What evidence supports the hypothesis that humans originated in Africa?

A) H. sapiens genomes contain some alleles from Neanderthal genomes.

B) One Y chromosome lineage that originated 1,000 years ago is found in 8% of modern-day Asian men.

C) Alleles of nuclear genes are less variable in African populations than in other populations.

D) Mitochondrial DNA sequence variation is higher in African populations than in other populations.

11) Sequencing the genomes of modern humans, nonhominin

Version 1

3

primates, Neanderthals, and Denisovans has allowed

A) gene flow between our ancestors and populations of other hominins did occur.

B) Homo sapiens emerged from Africa after the Neanderthal and Denisovan lineages had died out.

C) Neanderthals and Denisovans interbred with one another but not with our ancestors.

D) modern humans are the direct descendants of

investigators to conclude that Neanderthals.

E) modern humans do not share alleles with the Neanderthal and Denisovan lineages.

12) What is the term for the loss of genetic variation due to an event that decreases the size of the entire population?

A) gene flow B) genetic drift C) founder effect

D) population bottleneck E) selection

13) Why did the alleles that conferred insecticide resistance in mosquitoes become less frequent in the population when insecticide application was discontinued in the 1960s in Thailand?

A) The mosquitoes experienced a bottleneck and by chance the surviving insects had a low proportion of resistance alleles.

B) The resistance alleles were linked to a body color variation that led to resistant mosquitoes experiencing a higher rate of predation.

C) A mosquito population with a low proportion of resistant alleles migrated to Thailand and interbred with the

Version 1

original population.

D) In the absence of pesticide, mosquitos homozygous for the resistance alleles had lower fitness than the other genotypes.

4

14) In a population of 400 salamanders, some individuals have long toes and others have short toes. This phenotypic difference is due to a single gene with dominant and recessive alleles, T and t, where T confers long toes. If the toe length gene is in Hardy-Weinberg equilibrium, and 256 A) More information is needed to determine the answer.

15) Brachydactyly is a condition caused by an allele dominant to wild-type in humans, in which the fingers and toes are shortened. In a population of 10,000 people, 1,600 are

individuals have short toes, how many individuals are heterozygous at the toe length locus?

B) 32 C) 128 D) 144

BB, 4,800 are Bb, and 3,600 are bb.

15.1) What percentage of people in this population have short fingers and toes?

A) 0% B) 16%

15.2)

C) 36% D) 64% E) 84%

What is the frequency of the b allele?

A) 0.6 B) 0.4

C) 0.36 D) 0.48 E) 0.16

15.3) This population is in Hardy-Weinberg equilibrium. ⊚ Version 1

true

⊚

false 5

16) In 2013, a man was arrested for transporting 400 pounds of marijuana across state lines. His DNA was tested, entered into the CODIS database, and found to match the crime scene sample in a murder case (see Table).

Crime scene sample

Suspect 1

SSR locus 1

10, 10

10, 10

SSR locus 2

11, 11

11, 11

SSR locu s3

6, 9

6, 9

16.1) In a group of 100,000 people, about how many would be expected to have the same genotype as the crime scene sample? For simplicity, assume all SSR alleles are present in the population at a frequency of 0.2.

A) 0.000256 B) 6 C) 13

D) 51 E) None of the choices is correct.

16.2) What additional evidence would increase your confidence that the man arrested for transporting

marijuana committed the murder?

A) a match between his blood and the crime scene

sample at more than three SSR loci

Version 1

6

B) a receipt that showed the man used his credit card in Texas on the night of the murder C) the genotype at additional SSR loci that show a match at seven and a mismatch at three loci

D) discovery of the man’s fingerprints on the murder weapon

17) A large population of fish live in a lake. The color of their scales is determined by two different alleles of the S gene, S1 and S2. Homozygous S1S1 fish have yellow scales, S2S2 homozygotes have blue scales, and heterozygotes are green. Scientists catch 100 fish at random and record their color. Among those 100 fish are 30 yellow, 50 blue, and 20 green ones.

After the original sample is analyzed, an asteroid landed in the lake and killed all the fish that happened to be in one location. When 100 fish were sampled again, 15 were yellow, 75 were blue, and 10 were green.

17.1) What is the frequency of the S 1 allele in the population before the asteroid hit?

A) 0.15 B) 0.40

C) 0.50 D) 0.60 E) 0.80

17.2) The fish population is in Hardy-Weinberg equilibrium before the asteroid hit. ⊚ ⊚

Version 1

true false

7

17.3) What is the frequency of the S 1 allele in the fish population after the asteroid hit?

A) 0.15 B) 0.20

C) 0.25 D) 0.40 E) 0.80

17.4) By what mechanism did allele frequencies change due to the asteroid?

A) Allele frequencies did not change because of the asteroid.

B) The asteroid caused mutations in the fish population.

C) The asteroid changed the environment, so fish with allele S 2 had increased fitness.

17.5) Hardy-Weinberg conditions were established in the lake after the asteroid hit. The fish population mated and produced the next generation. Scientists sampled 100 fish that were all offspring of the postasteroid population. How many yellow fish will be A) 4 B) 8

D) The asteroid killed fish because of their location, so the allele frequency changed by chance.

among the 100 offspring sampled?

C) 15 D) 30 E) 32

Answer Key Test name: Chapter 24 Test Bank Version 1

8

Version 1

9

1) D 2) A 3) C 4) A 5) B 6) E 7) D 8) D 9) A 10) D 11) A 12) D 13) D 14) C 15) Section Break 15.1) D 15.2) A 15.3) TRUE 16) Section Break 16.1) C 16.2) [A, D] Version 1

10

17) Section Break 17.1) B 17.2) FALSE 17.3) B 17.4) D 17.5) A

Version 1

11

Student name:__________

A) are affected by both genetic and environmental factors.

B) usually do not result in continuous phenotypes.

C) are not usually affected by environmental

2)

1) Most complex quantitative traits conditions.

D) usually result in one or two phenotypic values such as red or white eye color in Drosophila.

The total phenotype variance ( VP) is D) independent of

A) the sum of genetic variance ( VG) and environmental variance ( VE).

B) the difference between VG and VE. C) independent of VG.

3)

E) always constant.

Broad-sense heritability is defined as A) VG. B) VP.

4)

VE.

C) VE. D) VG/VP. E) VP/VG.

The genetic relatedness of two siblings is A) 2.0.

B) 1.0.

C) 0.5.

D) 0.25.

E) 0.75.

5) Which of the following is not true about monozygotic twins?

Version 1

1

A) They share all alleles at all loci.

B) They have a genetic relatedness of 0.5.

C) They have agenetic relatedness of 1.0.

D) They come from the fertilization of a single egg

6)

by a single sperm cell.

E) They are the result of a split of the zygote after fertilization.

The response to selection, R, is equal to A) the heritability ( h2) of a trait.

B) the strength ( S) of selection.

C) the difference between h2 and S.

D) S/h2.

E) h2S.

7) Multiple interacting genes and the environment contribute to a complex trait, which can result in A) discontinuous distribution of discrete phenotypes.

B) continuous variation of the trait in a population.

C) strict dominance of a single allele so that one

8)

phenotype is most common.

D) continuous variation of the trait in a single individual.

Complex traits often show

A) a distribution with a large proportion of individuals at one extreme.

B) a distribution with a large proportion of individuals at the two extremes.

C) a normal distribution with an equal number of

Version 1

points above and below the mean.

D) a normal distribution with an equal number of points at each value within the range

2

9)

Twins from two individual zygotes (dizygotic twins) A) are as related genetically as monozygotic twins.

B) share no genetic similarities.

C) are 100% genetically identical.

D) are as related genetically as non-twin siblings.

10) In an experiment to distinguish genetic versus environmental effects on a trait, the test individuals A) should be as genetically related as possible.

B) should have variable alleles at as many loci as possible.

11)

C) must have the same parents.

D) must be grown in a natural environment.

Introgressions are used to D) rough map A) fine map QTLs. B) establish linkage disequilibrium.

C) establish linkage equilibrium.

12)

QTLs.

By definition, linkage disequilibrium means that

A) alleles at separate loci occur together more than predicted by random chance.

B) alleles at separate loci segregate randomly.

C) two allelic variants are at least 60 m.u. apart.

D) two allelic variants are 50 m.u. apart.

13) A SNP that can be either A or T is associated Version 1

3

with a human disease. In the Case population, there were 2000 A alleles and 1500 T alleles. In the Control population, there were 2500 A alleles and 3000 T alleles. The allelic odds ratio indicates that the presence of an A confers a ______ A) 1.33 B) 1

times greater risk for the disease than does the T allele.

C) 0.64 D) 1.6

14) A SNP that can exist as a C or G is associated with a human disease. The genotypic odds ratio of the homozygote CC relative to the homozygote GG is 2.2, which indicates that A) both CC and GG individuals are equally likely to get the disease because 2.2 is not a significant value.

B) heterozygotes are 2.2 times as likely to get the disease as GG homozygotes.

C) CC homozygotes are 2.2 times as likely to get the disease as GG homozygotes.

D) CC homozygotes are 2.2 times less likely to get the disease as GG homozygotes.

15) If a trait is heritable, then individuals with a genetic relatedness of 0.5 are expected to A) exhibit more phenotypic similarity than two random individuals in the population.

B) be as similar in phenotype as a pair of random individuals in the population.

C) have identical phenotypes because they are twins.

D) have an average phenotypic value that is close to the mean of the population.

16) A farmer performs truncation selection to try to increase the weight of peaches in his orchard. Which statement correctly describes a possible outcome of this selection?

Version 1

4

A) If peach weight is not heritable, the average weight in the next generation will increase.

B) If peach weight is heritable, the average weight in the next generation will increase.

C) The average peach weight will decrease due to

17) In a GWAS experiment, 500,000 SNPs across the genome are genotyped in thousands of Case and Control A) A chi-square test for goodness of fit will test the null hypothesis that the distribution of each SNP allele is the same in Cases and Controls.

B) A chi-square test for goodness of fit can be used because this is a small number of SNPs. C) A chi-square test for independence should be used to test the null hypothesis that the distribution of each SNP

this inbreeding.

D) If peach weight is heritable, the average weight in the next generation will decrease.

genomes. Which statistical analysis should be applied? allele is the same in Cases and Controls.

D) A chi-square test for independence should be used to test the null hypothesis that each SNP is linked to a QTL.

18) When the results of a GWAS experiment using 500,000 SNPs are plotted in a Manhattan plot, significantly associated SNPs are A) data points with a ‒log ( p value) less than 0.05.

B) data points with a ‒log ( p value) greater than 7.

C) data points with a ‒log ( p value) greater than 7

that fall within a known coding sequence.

D) any data points above the line of correlation.

19) During truncation selection, the realized heritability of a trait depends on the selection differential, which is the difference between the average trait value of the selected parents and the average trait value of the entire parental population.

Version 1

5

⊚

true

20) To test for statistical significance of the association of a SNP with a trait, the null hypothesis in a chi-square test for independence is that the one of the SNP alleles is more ⊚ ⊚

Version 1

⊚

false

common in the Cases than in the Controls.

true false

6

Answer Key Test name: Chapter 25 Test Bank 1) A 2) A 3) D 4) C 5) B 6) E 7) B 8) C 9) D 10) A 11) A 12) A 13) D 14) C 15) A 16) B 17) C 18) B 19) TRUE Version 1

7

20) FALSE

Version 1

8