iGenetics A Molecular Approach 3rd Edition By Peter J. Russell – Test Bank

 

 

To Purchase this Complete Test Bank with Answers Click the link Below

 

https://tbzuiqe.com/product/igenetics-a-molecular-approach-3rd-edition-by-peter-j-russell-test-bank/

 

If face any problem or Further information contact us At tbzuiqe@gmail.com

Sample Test

iGenetics: A Molecular Approach, 3e (Russell/Bose)

Chapter 3   DNA Replication

 

MATCHING

 

Please select the best match for each term.

 

1.    A) A complex of helicase and primase on the template DNA

2.    B) A DNA sequence that contains the specific region where replication begins

3.    C) The distance between the origin and the termination of replication where two replication forks fuse

4.    D) A complex of proteins and RNA that replicates the ends of eukaryotic chromosomes

5.    E) A complex of key replication proteins at the replication fork in coli and bacteriophage DNA

 

1) Primosome

Skill:  Factual recall

 

2) Replicon

Skill:  Factual recall

 

3) Replisome

Skill: Factual recall

 

4) Telomerase

Skill: Factual recall

 

5) Replicator

Skill: Factual recall

 

Answers: 1) A 2) C 3) E 4) D 5) B

 

 

MULTIPLE CHOICE

 

6) During replication, the direction of synthesis of new DNA from the leading and lagging strands is

1.    A) 5′ to 3′ only.

2.    B) 3′ to 5′ only.

3.    C) from left to right only.

4.    D) both 5′ to 3′ and 3′ to 5′.

5.    E) different, depending on whether the cell is prokaryotic or eukaryotic.

Answer:  A

Skill:  Factual recall

 

7) DNA replication is always

1.    A) discontinuous.

2.    B) bidirectional.

3.    C) conservative.

4.    D) semiconservative.

5.    E) dispersive.

Answer:  D

Skill:  Factual recall

8) In E. coli, replication begins at which chromosome site?

1.    A) The replication fork

2.    B) ter

3.    C) oriC

4.    D) TBP

5.    E) All of these

Answer:  C

Skill:  Factual recall

 

9) Which of the following did Kornberg use to detect DNA synthesis?

1.    A) Radioactively labeled coli cells

2.    B) Fluorescently labeled coli cells

3.    C) Radioactively labeled deoxynucleotide triphosphates

4.    D) Fluorescently labeled deoxynucleotide triphosphates

5.    E) Nonlabeled deoxynucleotide triphosphates

Answer:  C

Skill:  Factual recall

 

10) How many replication forks are produced when DNA denatures at an origin?

1.    A) 0

2.    B) 1

3.    C) 2

4.    D) 3

5.    E) The number varies

Answer:  C

Skill:  Factual recall

 

11) What is a replication bubble?

1.    A) A complex of replication enzymes on the DNA template strand

2.    B) A DNA sequence that initiates replication

3.    C) A tangle of denatured DNA strands near the replication fork

4.    D) A locally denatured segment of DNA where replication originates

5.    E) A localized site in the nucleus where chromosomes are replicating

Answer:  D

Skill:  Factual recall

 

12) Which of the following are necessary for DNA replication in vitro?

1.    A) RNA, helicase, primers, DNA polymerase

2.    B) Okazaki fragments, helicase, DNA polymerase

3.    C) Template DNA, DNA polymerase, four dNTPs, primers, magnesium ions

4.    D) Template DNA, four dNTPs, magnesium ions

5.    E) DNA can’t replicate in vitro

Answer:  C

Skill:  Factual recall

13) In Meselson and Stahl’s experiment, what kind of DNA molecules would be found after four replication cycles?

1.    A) Only heavy DNA (15N-15N)

2.    B) Only intermediate DNA (15N-14N)

3.    C) Only light DNA (14N-14N)

4.    D) Both heavy (15N-15N) and light DNA (14N-14N)

5.    E) Both heavy (15N-15N) and intermediate DNA (15N-14N)

Answer:  D

Skill:  Conceptual understanding

 

14) The two most basic steps of DNA replication are

1.    A) primase causes primer to bind the template and ligase copies the template.

2.    B) helicase unwinds the template and DNA polymerase binds the template.

3.    C) leading strand is copied first and lagging strand is copied second.

4.    D) the new strand is denatured and a template is synthesized.

5.    E) the template is denatured and a new strand is synthesized.

Answer:  E

Skill:  Conceptual understanding

 

15) Where does the initiator protein bind DNA at the start of replication?

1.    A) At a replication fork

2.    B) At an origin of replication

3.    C) At any AT-rich region

4.    D) At a promoter region

5.    E) At a start codon

Answer:  B

Skill:  Factual recall

 

 

16) As helicase unwinds the DNA molecule, what keeps the strands apart?

1.    A) DNA polymerase

2.    B) Reverse transcriptase

3.    C) Replication fork

4.    D) Single-strand binding proteins

5.    E) Okazaki fragments

Answer:  D

Skill:  Factual recall

 

17) After removal of the RNA primers and replacement with DNA nucleotides, the single-stranded nick adjacent to Okazaki fragments is filled in through a reaction that involves which enzyme?

1.    A) DNA primase

2.    B) SSB protein

3.    C) RNA polymerase

4.    D) DNA ligase

5.    E) DNA helicase

Answer:  D

Skill:  Factual recall

18) Which enzyme elongates the new DNA strand starting at an RNA primer?

1.    A) DNA polymerase I

2.    B) DNA polymerase III

3.    C) RNA primase

4.    D) DNA ligase

5.    E) RNA polymerase

Answer:  B

Skill:  Factual recall

 

19) After a region of DNA has been replicated, ________ removes the RNA primers.

1.    A) DNA polymerase I

2.    B) DNA polymerase III

3.    C) DNA helicase

4.    D) RNA primase

5.    E) DNA ligase

Answer:  A

Skill:  Factual recall

 

20) Which enzyme replaces RNA primers with DNA after elongation?

1.    A) DNA polymerase I

2.    B) DNA polymerase III

3.    C) RNA polymerase

4.    D) RNA primase

5.    E) DNA ligase

Answer:  A

Skill:  Factual recall

 

 

21) Which kind of enzyme prevents DNA from tangling up by introducing negative supercoils as the replication fork migrates during replication?

1.    A) Helicase

2.    B) Ligase

3.    C) DNA polymerase I

4.    D) DNA polymerase III

5.    E) Topoisomerase

Answer:  E

Skill:  Factual recall

 

22) The base-pairing error rate remains low during replication because

1.    A) DNA repair mechanisms can fix the mispaired bases.

2.    B) bases that are mispaired can excise themselves.

3.    C) UV light radiation corrects any base mispairs.

4.    D) mispaired bases cause a cell to die before replication is complete.

5.    E) None of these; base-pairing errors are not possible

Answer:  A

Skill:  Factual recall

23) The enzymatic activity of a telomerase is best described as a

1.    A) polymerase.

2.    B) ligase.

3.    C) topoisomerase.

4.    D) reverse transcriptase.

5.    E) exonuclease.

Answer:  D

Skill:  Factual recall

 

24) Rolling circle replication of DNA is characterized by the absence of

1.    A) the DNA

2.    B) a nick in the DNA

3.    C) the RNA primers.

4.    D) the replication bubble.

5.    E) the Okazaki

Answer:  D

Skill:  Conceptual understanding

 

25) Which enzyme activity is associated with the proofreading mechanism of DNA polymerase I?

1.    A) 5′-to-3′ exonuclease activity

2.    B) 3′-to-5′ exonuclease activity

3.    C) 5′-to-3′ polymerase activity

4.    D) Both A and B

5.    E) All of these

Answer:  B

Skill:  Factual recall

 

 

TRUE/FALSE

 

26) A new nucleotide is added to a growing strand of DNA at the 3′ end.

Answer:  TRUE

Skill:  Factual recall

 

27) Only the leading strand of a DNA molecule serves as a template during replication.

Answer:  FALSE

Explanation:  Both the leading and lagging strands serve as templates.

Skill:  Factual recall

 

28) At the growing end of a DNA chain, DNA polymerase catalyzes the formation of a disulfide bond between the 3′-OH group of the deoxyribose on the last nucleotide and the 5′-phosphate of the dNTP precursor.

Answer:  FALSE

Explanation:  DNA polymerase catalyzes the formation of a phosphodiester bond between nucleotides.

Skill:  Factual recall

 

29) DNA primase is an RNA polymerase.

Answer:  TRUE

Explanation:  DNA primase catalyzes the reaction to synthesize a short RNA primer molecule.

Skill:  Factual recall

30) Okazaki fragments are made from the lagging strand of the DNA double helix.

Answer:  TRUE

Skill:  Factual recall

 

31) DNA polymerase III is very inaccurate at matching bases during replication, with errors in one out of every 100 base pairs.

Answer:  FALSE

Explanation:  DNA polymerase III is very accurate, causing errors in only one out of every 1,000,000 base pairs.

Skill:  Factual recall

 

32) In eukaryotic cells, histone proteins are actively synthesized during the S phase of the cell cycle.

Answer:  TRUE

Skill:  Factual recall

 

33) In eukaryotes, DNA replication begins at a single origin of replication on each chromosome.

Answer:  FALSE

Explanation:  Replication begins at multiple origins of replication on each chromosome.

Skill:  Factual recall

 

 

34) Topoisomerase and SSB proteins are important components of the replication process in prokaryotes, but they are not found in eukaryotes.

Answer:  FALSE

Explanation:  These proteins also play key roles in eukaryotic DNA replication.

Skill:  Factual recall

 

35) Mg2+ ions are required for optimal DNA polymerase activity.

Answer:  TRUE

Explanation:  DNA polymerases often require magnesium ions as cofactors.

Skill:  Factual recall

 

SHORT ANSWER

 

36) What are the key replication enzymes at the replisome, and how is DNA replication on both leading and lagging strands made efficient through the conformation of the DNA at the replisome in prokaryotes?

Answer:  The key replication enzymes at the replisome are helicase, primase, and DNA polymerase III. To make replication more efficient, the lagging-strand DNA is folded so that its DNA polymerase III is complexed with the DNA polymerase III on the leading strand (forming the DNA Pol III holoenzyme). The folding of the lagging-strand template also makes production of sequential Okazaki fragments more efficient by bringing the 3′ end of each completed Okazaki fragments near the site where the next Okazaki fragment will start.

Skill:  Conceptual understanding

 

37) Why is DNA replication referred to as semiconservative?

Answer:  The progeny double helices consist of one parental DNA strand and one newly synthesized strand.

Skill:  Conceptual understanding

38) Why is an AT-rich sequence characteristic of DNA replicators in all organisms?

Answer:  AT-rich regions of DNA are relatively easy to denature to single strands. AT base pairs are held together by only two hydrogen bonds, while GC pairs are held together by three.

Skill:  Conceptual understanding

 

39) What are the differences in replication between leading and lagging strands in terms of continuity and directionality in relation to the replication fork?

Answer:  The leading strand is copied continuously from the 3′ end toward the replication fork, while the lagging strand is copied in fragments away from the replication fork.

Skill:  Factual recall

 

40) What experiment demonstrated that the 5′-to-3′ exonuclease activity of DNA polymerase I was essential for cell viability?

Answer:  In E. coli, the DNA Pol I polAex1 mutant strain survives at 37˚C but dies at 42˚C. This was shown to be because the mutant DNA Pol I enzyme had normal activity at 37˚C but a defective 5′-to-3′ activity at 42˚C. This demonstrated that the 5′-to-3′ exonuclease activity of DNA Pol I was essential for cell survival.

Skill:  Factual recall

 

41) A cross is made between yeast cells with different alleles for a set of linked genes: pr+q × p+rq+. The resulting tetrads show a 3:1 ratio for r+ to r instead of the expected 2:2 ratio. Can you explain how this could have occurred?

Answer:  Gene conversion by a mismatch repair mechanism could have caused this deviation from expected ratios. During pairing of homologous chromosomes in meiosis, recombination between the two inner chromatids occurred, resulting in heteroduplex (mismatched) DNA strands. Both mismatches were repaired by excision and DNA synthesis to match the parent DNA with the r+ allele.

Skill:  Analytical reasoning

 

42) Describe the method devised by Arthur Kornberg which first successfully achieved DNA synthesis in vitro, including its components and their uses.

Answer:  Kornberg mixed together DNA fragments, all four dNTPs (DNA precursors), and an E. coli lysate to achieve DNA synthesis in vitro. The DNA fragments acted as a template for the synthesis of new DNA. The dNTPs were precursors of the new DNA strand, and the cell lysate contained the enzyme necessary to catalyze DNA synthesis (DNA Pol I).

Skill:  Factual recall

 

43) The diploid set of chromosomes in Drosophila embryos replicates six times faster than the single E. coli chromosome, even though there is about 100 times more DNA in Drosophila than in E. coli and the rate of movement of the replication fork in Drosophila is much slower. How is this so?

Answer:  Eukaryotic chromosomes duplicate rapidly because DNA replication initiates at many origins of replication throughout the genome. In E. coli, there is only one replicon, while in eukaryotes there are multiple, smaller replicons.

Skill:  Application of knowledge

44) What are some key differences in replication between E. coli DNA and λ phage DNA?

Answer:  Both start replication as a circular molecule, and in E. coli, the parental DNA strands remain in a circular form throughout the replication cycle. A replication bubble opens to form two replication forks, and replication proceeds bidirectionally. Lambda phage DNA is replicated by a rolling circle mechanism, in which a nick is made in one of the two strands of the circle, and the 5′ end of the cut DNA strand is rolled out as a free “tongue”of increasing length as replication proceeds. The parental DNA circle is replicated continuously, while the linear displaced strand is replicated discontinuously. As long as the circle continues to roll, concatamers of phage DNA can be produced. This is later cut up into linear chromosomes and packaged into new phage heads.

Skill:  Factual recall

 

45) How do the DNA polymerase repair mechanisms work?

Answer:  Both DNA Pol I and DNA Pol III have 3′-to-5′ exonuclease activity and can remove nucleotides from the end of a DNA chain as part of an error correction mechanism. If an incorrect base is inserted by DNA polymerase, and the error is recognized immediately, the exonuclease activity excises the erroneous nucleotide from the new strand. After excision, the DNA polymerase resumes motion in the forward direction and inserts the correct nucleotide.

Skill:  Factual recall

 

46) How did Meselson and Stahl rule out the conservative model of DNA replication using equilibrium density gradient centrifugation?

Answer:  First, they grew E. coli cells in a medium containing only high-density nitrogen (15N). Then they allowed the cells to undergo successive rounds of replication in the presence of normal density nitrogen (14N). They used equilibrium density gradient centrifugation after each round of replication to separate the DNA produced by density. If the conservative model was correct, they would have found no intermediate-density DNA after a round of replication. However, after one round of replication, they found that the entire amount of DNA had a density exactly intermediate between 14N and 15N. Subsequent rounds of the experiment showed that semiconservative replication was the correct model.

Skill:  Conceptual understanding

 

47) How were the sequences that compose the replication origins in yeast discovered?

Answer:  Yeast cells were grown in heavy isotope media to produce denser DNA and then shifted to media with normal, light isotopes. After a few minutes, DNA from these cells was extracted and cut into small pieces. The lighter DNA fragments (which should contain the origins of replication) were collected, fluorescently labeled, and used to hybridize a microarray of yeast sequences. Determination of the sequences to which this light, fluorescently labeled DNA hybridized led to the identification of a number of yeast replication origins.

Skill:  Factual recall

 

48) How will DNA replication be affected if DNA polymerase I has a mutation that inactivates 5′-to-3′ exonuclease activity?

Answer:  The 5′-to-3′ exonuclease activity of DNA polymerase I removes the RNA primers from the ends of Okazaki fragments. If this activity is missing, the primers may not be removed from the growing DNA strands.

Skill:  Conceptual understanding

49) When the RNA primers are removed from the 5′ ends of eukaryotic chromosomes after replication, DNA polymerase is unable to fill in the gaps. What prevents the chromosomes from getting shorter and shorter with each round of replication?

Answer:  The enzyme telomerase maintains chromosome lengths by adding telomere repeats to the chromosome ends. Telomerase contains an RNA component that is complementary to the telomere repeat unit of the chromosome. After binding to the overhanging repeat, the telomerase RNA is used as a template to synthesize new chromosomal telomere repeats.

Skill:  Factual recall

 

50) Why do tumor cells in mammals have telomerase activity?

Answer:  Tumor cells are “immortal” cells, and the enzyme telomerase is required for long-term cell viability. It has been demonstrated that mutations in genes such as TLC1 and EST1 decrease cell longevity by continuous shortening of telomere length.

Skill:  Conceptual understanding