Final Genetics Flashcards


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1

Which of the following mutations could lead to constitutive expression of the genes of the lac operon?

A. A super repressor mutation

B. A mutation in the operator sequence

C. A mutation in the lac-Y gene

D. A mutation in the lac-Z gene

B

2

Which of the following best describes the biological role of the lac operon?

A. It ensures that bacterial cells produce lactose only when no other food sources are available.

B. It prevents other sugars from being metabolized until all available lactose has been used.

C. It ensures that a cell dedicates resources to the production of enzymes involved in lactose metabolism only when lactose is available in the environment.

D. It ensures that a cell produces enzymes involved in lactose metabolism in a constitutive manner.

C

3

The placement of the operator sequence between the promotor and the structural genes is critical to the proper function of the lac operon.

A. True

B. False

A

4

Which of the following occurs as a result of an abundance of tryptophan in E. coli?

A. The 5 trp genes (TrpA – TrpE) are not transcribed

B. The 5 trp genes (TrpA – TrpE) are transcribed, but not translated.

C. The leader sequence is not transcribed.

D. Stalling of the ribosome at trp codons in the leader sequence

A

5

Which of the following features of the trp operon is likely least essential to the process of attenuation?

A. The order of the structural genes, E, D, C, B, A

B. Trp codons near the beginning of the leader sequence

C. The ability of sequences within the leader mRNA to pair with one another

D. Transcription and translation of the leader sequence occur simultaneously.

A

6

Attenuator systems such as the one described for regulation of tryptophan synthesis would be just as likely to occur in eukaryotes as in prokaryotes.

A. True

B. False

B

7

Which of the following best describes the mechanism by which steroid hormones control gene expression?

A. Steroid hormones enter a cell, bind directly to hormone response elements (HREs), and enhance transcription

B.Chaperone proteins transport steroid hormones into the cell and guide them to their target genes.

C. Steroid hormones transport mRNA from the nucleus into the cytoplasm, where it is translated into protein.

D. Steroid hormones that enter the cell activate receptors. These hormone-receptor complexes then bind HREs and influence gene expression

D

8

Which of the following best describes the role of chaperone proteins in the regulation of gene expression by steroid hormones?

A. Chaperone proteins enter the cell and bind receptor molecules.

B. Chaperone proteins maintain functionality of the receptor.

C. Chaperone proteins activate receptor proteins.

D. Chaperone proteins directly enhance transcription

B

9

The reason some cells respond to the presence of a steroid hormone while others do not is that _______.

A. the receptors necessary for regulation differ among cells of various types

B. only certain cells contain the gene that is targeted by a given steroid hormone

C. the specific HRE is present only in certain cells

D. chaperone proteins block the hormone response elements (HREs) in some cells

A

10

Which statements about the modification of chromatin structure in eukaryotes are true?

Select that apply

A. Deacetylation of histone tails in chromatin loosens the association between nucleosomes and DNA.

B. Methylation of histone tails in chromatin can promote condensation of the chromatin.

C. Acetylation of histone tails is a reversible process.

D. DNA is not transcribed when chromatin is packaged tightly in a condensed form.

E. Acetylation of histone tails in chromatin allows access to DNA for transcription.

F. Some forms of chromatin modification can be passed on to future generations of cells.

B, C, D, E, & F

11

The diagram below shows two stretches of DNA in the genome of an imaginary eukaryotic cell. The top stretch of DNA includes the fantasin gene, along with its promoter and one of its enhancers. The bottom stretch of DNA includes the imaginin gene, its promoter, and one of its enhancers. The slash marks (//) indicate that more than 1,000 nucleotides separate the promoter and enhancer of each gene.

Which statements about the regulation of transcription initiation in these genes are true? Select that Apply

A. Control elements A, B, and C are proximal control elements for the fantasin gene.

B. The fantasin gene and the imaginin gene have identical enhancers.C.

C. The imaginin gene will be transcribed at a high level when repressors specific for the imaginin gene are present in the cell.

D. The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell.

E. Both the fantasin gene and the imaginin gene will be transcribed at high levels whenever general transcription factors are present in the cell.

F. Control elements C, D, and E are distal control elements for the imaginin gene.

G. Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell.

D, E, & G.

12

The enzymes involved in lactose metabolism are

A. needed when lactose is present.

B. needed when lactose is absent.

A

13

The enzymes involved in tryptophan metabolism are

A. needed when tryptophan is present.

B. needed when tryptophan is absent.

B

14

The crp gene, which encodes the CAP protein.

A. Absence of a functional crp would compromise the positive control exhibited by CAP.

B. Presence of a functional crp would compromise the positive control exhibited by CAP.

A

15

The CAP-binding site within the promoter.

A. With a CAP binding site there would be a reduction in the inducibility of the lac operon.

B. Without a CAP binding site there would be a reduction in the inducibility of the lac operon.

B

16

Is the binding of a transcription factor to its DNA recognition sequence necessary and sufficient for an initiation of transcription at a regulated gene?

A. Transcription factors are proteins that are necessary for the initiation of transcription. However, they are not sufficient for the initiation of transcription.

B. Transcription factors are proteins that are not necessary for the initiation of transcription.

C. Transcription factors are proteins that are necessary for the initiation of transcription. They are sufficient for the initiation of transcription.

A

17

What else plays a role in this process?

A. Transcription factors contain at least two functional domains: one binds to the DNA sequences of promoters and/or enhancers, while the other interacts with RNA polymerase or other transcription factors.

B. Transcription factor that binds to the DNA sequences of promoters and/or enhancers.

C. Transcription factor that interacts with RNA polymerase.

D.All transcription factors bind to other transcription factors without themselves binding to DNA.

A

18

Which of the following describes the relationship between the amount of methylation and the expression of luciferase?

A. As methylation increases, luciferase expression increases.

B. As methylation increases, luciferase expression decreases.

C. There is no relationship between the amount of methylation and the expression of luciferase.

B

19

Which produces a larger decrease in luciferase expression?

A. methylation within the transcription unit

B. methylation outside the transcription unit

A

20

At very high levels of methylation (593 CpGs), which of the following is true?

A. Methylation within the transcription unit has the strongest effect on luciferase expression.

B. Methylation outside the transcription unit has the strongest effect on luciferase expression.

C. Methylation has the same effect regardless of region.

C

21

Which term would be applied to a regulatory condition that occurs when protein is associated with a particular section of DNA and greatly reduces transcription?

A. stimulation

B. activation

C. positive control

D. negative control

E. induction

D

22

Which term refers to a contiguous genetic complex that is under coordinated control?

A. attenuation

B. operon

C. allosteric

D. prototroph

E. lysogen

B

23

Which term most appropriately refers to a regulatory protein in prokaryotes?

A. RNA processing

B. helicase activation

C. gyrase action

D. translation

E. DNA binding protein

E

24

In the lac operon, the product of structural gene lacZ is capable of ________.

A. splitting the β-linkage of lactose

B. forming lactose from two glucose molecules

C. forming ATP from pyruvate

D. nonautonomous replication

E. replacing hexokinase in the early steps of glycolysis

A

25

Which of the following terms best characterizes catabolite repression associated with the lac operon in E. coli?

A. constitutive

B. repressible system

C. positive control

D. inducible system

E. negative control

C

26

When referring to attenuation in regulation of the trp operon, it would be safe to say that when there are high levels of tryptophan available to the organism, ________.

A. tryptophan is inactivating the repressor protein

B. the trp operon is being transcribed at relatively high levels

C. ribosomes are stalling during translation of the attenuator region

D. translational termination is likely

E. transcriptional termination is likely

E

27

Genetic regulation in eukaryotes can take place at a variety of levels from transcriptional to posttranslational. At what level is genetic regulation considered most likely in prokaryotes?

A. polyadenylation of the 3' end of the mRNAs

B. exon processing

C. capping

D. transcriptional

E. intron processing

D

28

Regarding eukaryotic and prokaryotic genetic regulation, what process seems to be the most similar between the two?

A. transcriptional regulation

B. 5'-capping regulation

C. intron/exon shuffling

D. RNA splicing regulation

E. poly-A tail addition

A

29

DNA methylation may be a significant mode of genetic regulation in eukaryotes. Methylation refers to ________.

A. altering translational activity, especially of highly methylated tRNAs

B. altering RNA polymerase activity by methylation

C. addition of methyl groups to the cytosine of CG doublets

D. changes in DNA-DNA hydrogen binding

E. alteration of DNA polymerase activity by addition of methyl groups to glycine residues

C

30

Which of the following clusters of terms applies when addressing enhancers as elements associated with eukaryotic genetic regulation?

A. cis-acting, variable position, fixed orientation

B. cis-acting, variable orientation, variable position

C. cis-acting, fixed position, fixed orientation

D. trans- and cis-acting, variable position

E. trans-acting, fixed position, fixed orientation

B

31

What advantages do regulatory systems provide to bacteria?

Select the two correct answers.

A. Regulatory systems enable transcription of different genes in different environments.

B. Regulatory systems enable bacteria to function normally in the absence of nutrient medium.

C. Regulatory systems provide an efficient response to protect bacteria from harmful environmental factors.

D. Regulatory systems allow the necessary mutation of bacterial genes to enable them to adapt in different environments.

A & C

32

Why are genes related to common functions found together in operons?

A. Having genes with related functions together in operons provides for inhibition of undesirable processes.

B. Having genes with related functions together in operons provides for coordinated responses.

C. Having genes with related functions together in operons provides for the correct function of the operon even if some genes are mutated.

B

33

A bacterial operon is responsible for production of the biosynthetic enzymes needed to make the theoretical amino acid tisophane (tis). The operon is regulated by a separate gene, R, deletion of which causes the loss of enzyme synthesis. In the wild-type condition, when tis is present, no enzymes are made; in the absence of tis, the enzymes are made. Mutations in the operator gene (O−) result in repression regardless of the presence of tis.

Is the operon under positive or negative control?

A. positive

B. negative

A

34

If the frequency of the M allele in the human MN blood group system is 0.65 in a population at equilibrium, then the frequency of the N allele must be 0.04.

A. True

B. False

B

35

If a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the hetero-zygote carriers for this disease? A. 0.043

B. 0.022

C. 0.956

D. 0.0005

A

36

In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds?

A. Disruptive

B. Stabilizing

C. Directional

D. There would be no selection in this population.

A

37

________ is a group of individuals belonging to the same species that live in a defined geographic area and actually or potentially interbreed.

A. Pool

B. Pod

C. Cohort

D. Population

E. Hybrid

D

38

Which term is given to the total genetic information carried by all members of a population?

A. gene pool

B. breeding unit

C. genome

D. chromosome complement

E. race

A

39

A number of mechanisms operate to maintain genetic diversity in a population. Why is such diversity favored?

A. Homozygosity is an evolutionary advantage.

B. Genetic diversity helps populations avoid diploidy.

C. Genetic diversity may better adapt a population to inevitable changes in the environment.

D. Diversity leads to inbreeding advantages.

E. Greater genetic diversity increases the chances of haploidy.

C

40

In a population of 100 individuals, 49 percent are of the NN blood type. What percentage is expected to be MN assuming Hardy-Weinberg equilibrium conditions?

A. 9 percent

B. 51 percent

C. 21 percent

D. 42 percent

E. There is insufficient information to answer this question.

D

41

Albinism is an autosomal recessive trait in humans. Assume that there are 100 albinos (aa) in a population of 1 million. How many individuals would be expected to be homozygous normal (AA) under equilibrium conditions?

A. 19,800

B. 100

C. 980,100

D.10,000

E. 999,900

C

42

In chickens, having feathers on legs is dominant to having featherless legs. In a population of 200 chickens, 128 have featherless legs. How many chickens are heterozygous for the “feathered leg” allele? Assume the population is in Hardy-Weinberg equilibrium.

A. 32

B.36

C. 64

D. 72

E. Not enough information is provided.

C

43

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the allele frequency of the recessive allele?

A. 0.2

B. 0.4

C. 0.64

D. 0.8

E. Not enough information is provided.

E

44

Applying Hardy-Weinberg equilibrium, how many genotypes are predicted for a gene that has three alleles?

A. 3

B. 4

C. 6

D. 9

C

45

When does allele frequency equilibrium occur?

Select that apply

A. when Δ p I = 0

B. when p I is greater than p C

C. when p C = 0

D. when m(p C - p I) = 0

E. when m = 0

A, D, & E

46

In small isolated populations, gene frequencies can fluctuate considerably. The term that applies to this circumstance is ________.

A. allelic separation

B. stabilizing selection

C. natural selection

D. genetic isolation

E. genetic drift

E

47

Red-green color blindness is an X-linked recessive disorder. In a population (n = 1000) that contains an equal number of males and females, the frequency of the normal allele, X 1, is 0.90 and the frequency of the mutant allele that causes color blindness, X 2, is 0.10.

How many females in this population are carriers of the red-green colorblindness trait? Assume that the population is in Hardy-Weinberg equilibrium.

A. 5

B. 10

C. 90

D. 180

E. More information is needed to answer this question.

C

48

What is the basis for homology among chromosomes?

A) autoradiographic pattern

B) position of the centromere

C) overall length

D) type and location of genes

E) banding patterns

D

49

Which mode of inheritance produces heterozygotes with phenotypes that differ from either homozygote but typically more closely resembles one homozygous phenotype than the other?

A) complete dominance

B) incomplete dominance

C) codominance

D) epistasis

E) incomplete penetrance

B

50

Crosses in which F1 plants heterozygous for a given allele are crossed to generate a 3:1 phenotypic ratio in the F2 generation are known as ________.

A) replicate crosses

B) reciprocal crosses

C) test crosses

D) monohybrid crosses

E) dihybrid crosses

D

51

A cell in the G0 stage _____.

A) has withdrawn from the cell cycle

B) is in the process of dividing

C) is replicating its DNA

D) has just completed cell division

E) is preparing for cell division

A

52

Palomino horses have a golden yellow coat, chestnut horses have a brown coat, and cremello horses have a coat that is almost white. A series of crosses between the three different types of horses produced the following offspring:

Cross Offspring

Palomino x palomino 13 palomino, 6 chestnut, 5 cremello

Chestnut x chestnut 16 chestnut

Cremello x cremello 13 cremello

Palomino x chestnut 8 palomino, 9 chestnut

Palmino x cremello 11 palomino, 11 cremello

Chestnut x cremello 23 palomino

a) dominance

b) Recessive

c) codominance

d) incomplete dominance

e) sex-linked

D

53

The Chi-square test involves a statistical comparison between measured (observed) and predicted (expected) values. One generally determines degrees of freedom as ________.

A) the number of categories being compared

B) one less than the number of classes being compared

C) one more than the number of classes being compared

D) ten minus the sum of the two categories

E) the sum of the two categories

B

54

In a healthy male, how many sperm cells would be expected to form from (a) 400 primary spermatocytes? (b) 400 secondary spermatocytes?

A) (a) 800; (b) 800

B) (a) 1600; (b) 1600

C) (a) 1600; (b) 800

D) (a) 400; (b) 400

E) (a) 100; (b) 800

C

55

In a trihybrid cross of a tall, purple-flowered pea plant with round seeds (TtPpRr) with a tall, whiteflowered pea plant with wrinkled seeds (Ttpprr), what is the probability: a short, white-flowered plant with wrinkled seeds will be produced?

A) 0.054

B) 0.0625

C) 0.0754

D) 0.0863

B

56

A Mendelian test cross is used to determine whether:

A) an allele is dominant or recessive.

B) flowers are purple or white.

C) the genotype or phenotype is more important.

D) an individual is homozygous or heterozygous.

E) segregation or independent assortment is occurring.

D

57

Organized by centrioles, what structures are important in the movement of chromosomes during cell division?

A) Cell Walls

B) Chloroplasts

C) Spindle Fibers

D) Mitochondria

E) Centromeres

C

58

Which answer gives the stages of mitosis in proper chronological order?

A) Anaphase, prophase, metaphase, telophase

B) Metaphase, anaphase, telophase, prophase

C) Prophase, metaphase, anaphase, telophase

D) Interphase, prophase, metaphase, anaphase, telophase

C

59

Describe Mendel's conclusions about how traits are passed from generation to generation.

A) Mendel hypothesized that traits in peas are controlled by specific unit factors. He suggested that unit factors occur in pairs and that unit factors separate from each other during gamete formation.

B) Mendel hypothesized that traits in peas are controlled by DNA sequences he called genes. He suggested that genes occur in pairs and stay together during gamete formation.

C) Mendel hypothesized that traits in peas are controlled by specific unit factors. He suggested that unit factors occur in pairs and stay together during gamete formation.

D) Mendel hypothesized that traits in peas are controlled by DNA sequences he called genes. He suggested that genes occur in pairs and members of each gene pair separate from each other during gamete formation.

A

60

What is the probability that, in an organism with a haploid number of 13, a sperm will be formed that contains all 13 chromosomes whose centromeres were derived from maternal homologs?

A) 9.77 X 10-4

B) 1.22 X10-4

C) 2.41 X 10 -4

D) 5.67 X 10-4

E) 7.97 X 10-4

B

61

A recessive allele in tigers causes the white tiger. If two normally pigmented tigers are mated and produce a white offspring, what percentage of their remaining offspring would be expected to have normal pigmentation?

A) 25%

B) 50%

C) 75%

D) 0%

E) 100%

C

62

Assume that a black guinea pig crossed with an albino guinea pig produced 5 black offspring. When the albino was crossed with a second black guinea pig, 4 black and 3 albino offspring were produced. What genetic explanation would apply to these data?

A) albino = dominant; black = incompletely dominant

B) albino and black = codominant

C) albino = recessive; black = recessive

D) albino = recessive; black = dominant

E) None of the answers listed are correct.

D

63

What is the outcome of synapsis, a significant event in meiosis?

A) Monad movement to opposite poles

B) Side-by-side alignment of homologous chromosomes

C) Dyad formation

D) Side-by-side alignment of nonhomologous chromosomes

E) Chiasma segregation

B

64

In an organism with 52 chromosomes, how many bivalents would be expected to form during meiosis?

A) 208

B) 104

C) 13

D) 26

E) 52

D

65

Polydactyly is expressed when an individual has extra fingers and/or toes. Assume that a man with six fingers on each hand and six toes on each foot marries a woman with a normal number of digits. Having extra digits is caused by a dominant allele. The couple has a son with normal hands and feet, but the couple's second child has extra digits. What is the probability that their next child will have polydactyly?

A) 1/32

B) 1/8

C) 7/16

D) 3/4

E) 1/2

E

66

Which mode of inheritance results in both alleles being detected equally in the heterozygous phenotype?

A) complete dominance

B) incomplete dominance

C) codominance

D) epistasis

E) incomplete penetrance

C

67

Considering the Mendelian traits round versus wrinkled and yellow versus green, consider the crosses below and determine the genotypes of the parental plants by analyzing the phenotypes of their offspring. Parental Plants Offspring Wrinkled, yellow X round, yellow 6/16 (wrinkled, yellow) 2/16 (wrinkled, green) 6/16 (round, yellow) 2/16 (round, green)

A) wwGg x WWgg

B) wwGg x WWGg

C) wwGg x Wwgg

D) wwGg x WwGg

E) wwGg x wwgg

D

68

A cell in G1of interphase has 24 chromosomes. How many DNA molecules will be found per cell when this original cell progresses to metaphase 1 of meiosis?

A) 12

B) 24

C) 48

C

69

30. (3 Pts) The fundamental Mendelian process that involves the separation of contrasting genetic elements at the same locus would be called ________.

A) segregation

B) independent assortment

C) continuous variation

D) discontinuous variation

E) dominance or recessiveness

A

70

Nondisjunction in the first meiotic division in a human male could result at fertilization in _____.

A. the XYY condition

B. Turner syndrome

C. Klinefelter syndrome

D. XXX syndrome

E. either B or C

E

71

The maternal-effect mutation bicoid (bcd) is recessive. In the absence of the bicoid protein product, embryogenesis is not completed. Consider a cross between a female heterozygous for bicoid alleles (bcd + /bcd - ) and a male homozygous for the mutation (bcd - /bcd - ). Which statement best explains how it is possible for a male homozygous for the mutation to exist?

A) This is a hypothetical situation, because it is not possible for a male homozygous for the mutation (bcd - /bcd - ) to exist.

B) The bicoid mutation (bcd - ) results in a dominant-negative mutation which suppresses the function of the normal bicoid allele (bcd + ).

C) The required bicoid protein product was provided by the mother whose genotype had to have been heterozygous (bcd + /bcd - ) to produce a male homozygous for the mutation (bcd - /bcd - ).

D) Since bicoid (bcd) is a maternal-effect mutation, it’s only involved in the embryogenesis of female embryos.

C

72

Species I is diploid (2n=8) with chromosomes AABBCCDD; related species II is diploid (2n=8) with chromosomes MMNNOOPP. What type of chromosome mutation does the individual organism has with the following sets of chromosomes? AABBCCDDMMNNOOPP

A) Autotriploidy of species I

B) Autotriploidy of species II

C) Allotriploidy of species I and II

D) Allotetraploidy of species I and II

E) Autotetraploidy of species I

D

73

In foxes, two alleles of a single gene, F and f, may result in lethality (ff), platinum coat (Ff), or silver coat (FF). What ratio is obtained when platinum foxes are interbred?

A) 1/2 platinum; 1/2 silver

B) 1/3 platinum; 2/3 silver

C) 3/4 platinum; 1/4 silver

D) 2/3 platinum; 1/3 silver

E) 1/4 platinum; 3/4 silver

D

74

While working in a genetics lab over the summer, you isolated a true-breeding strain of wingless Drosophila. After sharing your results with your mentor, you learn that six other true-breeding strains of Drosophila with the same mutant phenotype have been isolated independently in your lab. Your mentor asks you to determine if the mutants belong to the same complementation group. What is true about flies that belong to the same complementation group?

A) They all have the same mutation in the same wing- development gene. All flies in a complementation group have identical DNA sequences for this gene.

B) They all have some mutation in the same wing-development gene. Each strain may have a different mutation, but the same gene is mutated in all strains in a complementation group.

C) They all have some mutation in some wing-development gene. Each strain may have a different mutation in a different gene, but all strains within a complementation group have the same phenotype.

D) None of the above is correct

B

75

As this illustration shows, four o’clock plants can have green leaves, white leaves, or variegated leaves (leaves with areas of white and areas of green). The white regions of the leaves lack functioning chloroplasts.

The transmission of leaf color in four o'clock plants demonstrates organelle inheritance. In this problem, you will explore this inheritance pattern.

Which of the following crosses between four o’clock plants could produce progeny plants with green leaves? Select all that apply.

i) white (female) x variegated (male)

ii) green (female) x white (male)

iii) variegated (female) x variegated (male)

iv) white (female) x green (male)

v) variegated (female) x white (male)

A) I, ii, and iii

B) ii, iv and v

C) ii, iii and v

D) i, iii and iv

C

76

Regarding familial Down syndrome, which of the following statements is most likely true?

a) The affected individuals (with Down) have 47 chromosomes.

b) The carriers have 46 chromsomes.

c) The affected individuals have 46 chromosomes.

d) The carriers have nondisjunction during meiosis.

C

77

Which statement best describes human sex determination?

a) Females are heterogametic.

b) Individuals with a Y chromosome are male.

c) Individuals with two X chromosomes are female.

d) Individuals with at least twice as many X chromosomes as Y chromosomes are female.

e) Individuals with one X chromosome are male.

B

78

Which of the following best describes the Lyon hypothesis?

a) Barr bodies are inactivated X chromosomes.

b) The paternal X chromosome of a female is inactivated in an early embryonic stage.

c) The maternal X chromosome of a female is inactivated in an early embryonic stage.

d) One of the two female X chromosomes is randomly inactivated in an early embryonic stage.

e). Each cell of a female randomly inactivates one X chromosome shortly after mitosis

D

79

How does nondisjunction in human female gametes give rise to Klinefelter and Turner syndrome offspring following fertilization by a normal male gamete?

A) Nondisjunction results in ova that carry either two X chromosomes or none. The former results in Turner syndrome when fertilized by a Y-containing sperm, and the latter results in Klinfelter syndrome when fertilized by an X-containing sperm.

B) Nondisjunction of complete sets of chromosomes leads to ova that are either diploid, or completely devoid of chromosomes. A diploid ovum results in Klinefelter syndrome when fertilized by a Y-containing sperm, and results in Turner syndrome when fertilized by an X-containing sperm

C) Nondisjunction results in ova that carry either two X chromosomes or none. An ovum with no X chromosomes results in Klinefelter syndrome when fertilized by a Y-containing sperm, and in Turner syndrome when fertilized by an X-containing sperm.

D) Nondisjunction results in ova that carry either two X chromosomes or none. An ovum with two X chromosomes results in Klinefelter syndrome when fertilized by a Y-containing sperm, and in Turner syndrome when fertilized by an X-containing sperm

E) Nondisjunction results in ova that carry either two X chromosomes or none. The former results in Klinefelter syndrome when fertilized by a Y-containing sperm, and the latter results in Turner syndrome when fertilized by an X-containing sperm.

E

80

In Drosophila, the sex of a fly with the karyotype XX:2A is ________. A) male

B) female

C) Metafemale

D) Metamale

B

81

Which genetic region of the human Y chromosome is shown to have homology to the X chromosome?

A) PARs = pseudoautosomal regions,

B) NRY = nonrecombining region of the Y,

C) MSY= Male –specific region

D) SRY = sex-determining

A

82

All forms of extranuclear inheritance involve _____.

a) DNA carried by organelles other than the nucleus

b) endosymbiotic cells carried in the cytoplasm

c) gene products carried by the ooplasm

d) transmission from the maternal parent to all offspring

e) transmission of information via the cytoplasm

E

83

A chromosome has the following segments where “*” represents the centromere. A B C D E * F G

What types of chromosome mutations are required to change this chromosome into the following chromosome (A B C D E E D C* F G) a) Duplication

b) Duplication and pericentric inversion

c) Paracentric inversion and duplication

d) Translocation and duplication

e) Translocation

C

84

The XYY condition in humans is caused by nondisjunction in _____. A. males in meiosis I

B. males in meiosis II

C. males in meiosis I or II

D. females in meiosis I

E. females in meiosis II

B

85

An individual with Klinefelter syndrome generally has ___________ Barr body.

A) 0

B) 1

C) 2

D) 3

B

86

The genes encoding the red- and green-color-detecting proteins of the human eye are located next to one another on the X chromosome and probably evolved from a common ancestral pigment gene. The two proteins demonstrate 76 percent homology in their amino acid sequences. A normalvisioned woman (with both genes present on each of her two X chromosomes) has a red-color-blind son who was shown to have one copy of the green-detecting gene and no copies of the red-detecting gene. Which of the following is the most plausible explanation for these observations?

a) The son received an X chromosome lacking the red-detecting gene from his father.

b) An X chromosome lacking the red-detecting gene was produced by a crossover event between the red-detecting allele and the green-detecting genes of the mother’s X chromosomes

c) The ovum that was fertilized to produce the son contained the mother’s two green-detecting genes, while the red-detecting genes ended up in the polar bodies.

d) The ovum that was fertilized to produce the son contained one of the mother’s green-detecting genes, while the other green-detecting gene and the two copies of the red-detecting genes ended up in the polar bodies.

e) The son received a dominant red-detecting gene on the Y chromosome

B

87

A boy with Klinefelter syndrome (47,XXY) is born to a mother who is phenotypically normal and a father who has the X- linked skin condition called anhidrotic ectodermal dysplasia. The mother’s skin is completely normal with no signs of the skin abnormality. In contrast, her son has patches of normal skin and patches of abnormal skin.

A) The father must have contributed the abnormal X-linked gene.

B) Both parents must have contributed the abnormal X-linked gene.

C) The mother must have contributed the abnormal X-linked gene.

D) Cannot be determined

A

88

With which of the following would hemizygosity most likely be associated?

A) codominance

B) incomplete dominance

C) trihybrid crosses

D) X-linked inheritance

E) sex-limited inheritance

D

89

Duchenne muscular dystrophy is caused by a recessive X-linked allele. A man with this disorder _____.

a) could have inherited it from either parent

b) must have inherited it from his mother

c) must have inherited it from both parents

d) can pass it along to all of his children

e) can pass it along to only his sons

B

90

What essential criteria must be met in order to execute a successful mapping cross? i) One parent should be heterozygous and the other homozygous recessive for all the genes being mapped. ii) Recombination must occur in the sex that displays genetic heterogeneity. iii) The genotypes of the offspring must be readily determined from their phenotypes. iv) One parent should be homozygous dominant and the other homozygous recessive for all the genes being mapped.

A) i), III), and iv)

B) i), ii), and III)

C) ii), III), and iv)

D) i), ii), and iv)

B

91

The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany, ebony female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes for the single crossovers?

a) mahogany and ebony

b) ebony and wilid-type

c) wild type and mahogany-ebony

d) mahogany and mahogany-ebony

C

92

The discontinuous aspect of replication of DNA in vivo is caused by ________.

A) polymerase slippage

B) trinucleotide repeats

C) the 5' to 3' polarity restriction

D) topoisomerases cutting the DNA in a random fashion

E) sister-chromatid exchanges

C

93

The phenomenon in which one crossover decreases the likelihood of crossovers in nearby regions is called ________.

A) reciprocal genetic exchange

B) negative interference

C) mitotic recombination

D) positive interference

E) chiasma

D

94

In an analysis of the nucleotide composition of double-stranded DNA to see which bases are equivalent in concentration, which of the following would be true?

A) A = C

B) A = G and C = T

C) A + C = G + T

D) A + T = G + C

E) A = G and C = T and A + C = G + T are both true.

C

95

DNA polymerase III adds nucleotides ________.

A) to the 3' end of the RNA primer

B) to the 5' end of the RNA primer

C) in the place of the primer RNA after it is removed

D) to both ends of the RNA primer

E) to internal sites in the DNA template

A

96

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with ________.

A) sex-linked inheritance with 30% crossing over

B) linkage with approximately 33 map units between the two gene loci

C) independent assortment

D) 100% recombination

E) linkage with 50% crossing over

B

97

In E. coli, the “proofreading” of DNA replication is mediated by:

a. Adenine methylation and mismatch repair system.

b. Nucleotide excision repair system

c. 5’-3’ exonuclease activity of DNA polymerase I.

d. 3’-5’ exonuclease activity of DNA polymerases.

D

98

Given the structure below, please answer the following questions. Is the accompanying figure DNA or RNA? ________ Is the arrow closest to the 5' or 3' end? ________ Spleen diesterase is an enzyme that breaks the covalent bond that connects the phosphate to the 5' carbon. Assume that the dinucleotide is digested with spleen diesterase. To which base and to which carbon on the sugar is the phosphate now attached, A or T?

A) DNA; 3' end; T, 5'

B) RNA; 3' end; A, 3'

C) DNA; 5' end; A, 5'

D) DNA; 3' end; A, 3'

E) RNA; 3' end; T, 3'

D

99

For single crossover, the frequency of recombinant gametes is half the frequency of crossing over because

a) A test cross between a homozygote and heterozygote produces ½ heterozygous and ½ homozygous progeny

b) The frequency of recombination is always 50%

c) Each crossover takes place between only two of the four chromatids of a homologous pair

d) Crossovers occur in about 50% of meiosis

C

100

The term normally applied when two genes fail to assort independently is ________.

a) epistasis

b) nondisjunction

c) segregation

d) linkage

e) dominance and/or recessiveness

D

101

DNA polymerase I is thought to add nucleotides ________. A) to the 5' end of the primer

B) to the 3' end of the primer

C) in the place of the primer RNA after it is removed

D) on single-stranded templates without need for an RNA primer

E) in a 5' to 5' direction

C

102

Structures located at the ends of eukaryotic chromosomes are called ________.

A) centromeres

B) telomerases

C) recessive inversions

D) telomeres

E) permissive mutations

D

103

In a three-point mapping experiment, how many genotypic classes are expected?

a) Two

b) Four

c) Six

d) Eight

D

104

Which of the following contains the three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes?

A) removal of exons, insertion of introns, capping

B) 5'-poly(A) tail addition, insertion of introns, capping

C) heteroduplex formation, base modification, capping

D) 5'-capping, 3'-poly(A) tail addition, splicing

E) 3'-capping, 5'-poly(A) tail addition, splicing

D

105

In the Meselson–Stahl experiment, if DNA is replicated conservatively, after two generations of replications there would be _____.

a) one "hybrid" band

b) one "hybrid" band and one new, light band

c) one "hybrid" band and one old, heavy band

d) one old, heavy band, one new, light band, and one "hybrid" band

e) one old, heavy band and one new, light band

E

106

If 10 percent of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine?

A) 15 percent

B) 30 percent

C) 35 percent

D) 40 percent

E) 70 percent

D

107

Describe the role of 15N in the Meselson-Stahl experiment.

i) A comparison of the density of DNA samples at various times in the experiment allows an estimation of amounts of old and new DNA .

ii) Labeling the pool of nitrogenous bases of the DNA of E. coli with the heavy isotope 15N labels allows old DNA to be followed.

iii) Labeling the pool of nitrogenous bases of the RNA with 15N allows for the detection of newly synthesized DNA.

iv) Growing the cells for many generations in medium containing 15N and transferring to 14N medium allows for the detection of new DNA.

A) i), ii), and iv)

B) ii), iii), and iv)

C) i), ii), and iii)

D) i), iii), and iv)

A

108

All Old Order Amish families from Lancaster County with Ellis-van Creveld syndrome can trace their genealogies to Mr. and Mrs. Samuel King, who immigrated to Lancaster County in 1744. This is an example of ________.

A) founder effect

B) mutation

C) gene flow

D) Hardy-Weinberg equilibrium

E) bottleneck effect

A

109

Which aspect of population genetics refers to chance fluctuations of allele frequencies resulting from sampling error and is more prominent in small, isolated populations?

A) founder effect

B) mutation

C) gene flow

D) gene drift

E) bottleneck effect

D

110

Which aspect of population genetics can lead to changes in allele frequencies in a new mixed population following migration?

A) natural selection

B) mutation

C) gene flow

D) founder effect

E) bottleneck effect

A

111

Regulatory regions of a eukaryotic gene all contain which of the following sequences, which regulate transcription of genes located on the same chromosome as the sequences?

A) zinc fingers

B) cis-acting regulatory sequences

C) homeodomains

D) trans-acting regulatory sequences

E) leucine zippers

B

112

You have identified a mutation in a gene that also seems to decrease transcription of another gene 2000 bp away from the mutation site. What regulatory sequence, which may be found within another gene, has likely been mutated in this instance?

A) core promoter

B) proximal elements

C) enhancer sequence

D) homeodomain motif

E) upstream activator sequence

C