kupdf com lehninger principles of biochemistry test bank ch 25pdf

DNA replication Page: 955 Difficulty: 3 Ans: E The proofreading function of DNA polymerase involves all of the following except: A a 3' → 5' exonuclease.. DNA replication Pages: 956-957

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Multiple Choice Questions

1 DNA replication

Page: 950 Difficulty: 2 Ans: C

The Meselson-Stahl experiment established that:

A) DNA polymerase has a crucial role in DNA synthesis

B) DNA synthesis in E coli proceeds by a conservative mechanism

C) DNA synthesis in E coli proceeds by a semiconservative mechanism

D) DNA synthesis requires dATP, dCTP, dGTP, and dTTP

E) newly synthesized DNA in E coli has a different base composition than the preexisting DNA

Page: 951 Difficulty: 2 Ans: D

When a DNA molecule is described as replicating bidirectionally, that means that it has two: A) chains

B) independently replicating segment

C) origins

D) replication forks

E) termination points

An Okazaki fragment is a:

A) fragment of DNA resulting from endonuclease action

B) fragment of RNA that is a subunit of the 30S ribosome

C) piece of DNA that is synthesized in the 3' → 5' direction

D) segment of DNA that is an intermediate in the synthesis of the lagging strand

E) segment of mRNA synthesized by RNA polymerase

Pages: 952-958 Difficulty: 2 Ans: C

Which one of the following statements about enzymes that interact with DNA is true?

A) E coli DNA polymerase I is unusual in that it possesses only a 5' → 3' exonucleolytic activity

B) Endonucleases degrade circular but not linear DNA molecules

C) Exonucleases degrade DNA at a free end

D) Many DNA polymerases have a proofreading 5' → 3' exonuclease

E) Primases synthesize a short stretch of DNA to prime further synthesis

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E coli DNA polymerase III:

A) can initiate replication without a primer

B) is efficient at nick translation

C) is the principal DNA polymerase in chromosomal DNA replication

D) represents over 90% of the DNA polymerase activity in E coli cells

E) requires a free 5'-hydroxyl group as a primer

Page: 955 Difficulty: 3 Ans: E

The proofreading function of DNA polymerase involves all of the following except:

A) a 3' → 5' exonuclease

B) base pairing

C) detection of mismatched base pairs

D) phosphodiester bond hydrolysis

E) reversal of the polymerization reaction

The 5' → 3' exonuclease activity of E coli DNA polymerase I is involved in:

A) formation of a nick at the DNA replication origin

B) formation of Okazaki fragments

C) proofreading of the replication process

D) removal of RNA primers by nick translation

E) sealing of nicks by ligase action

Prokaryotic DNA polymerase III:

A) contains a 5' → 3' proofreading activity to improve the fidelity of replication

B) does not require a primer molecule to initiate replication

C) has a β subunit that acts as a circular clamp to improve the processivity of DNA synthesis

D) synthesizes DNA in the 3' → 5' direction.

E) synthesizes only the leading strand; DNA polymerase I synthesizes the lagging strand

At replication forks in E coli:

A) DNA helicases make endonucleolytic cuts in DNA

B) DNA primers are degraded by exonucleases

C) DNA topoisomerases make endonucleolytic cuts in DNA

D) RNA primers are removed by primase

E) RNA primers are synthesized by primase

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Pages: 964-965 Difficulty: 2 Ans: D

In contrast to bacteria, eukaryotic chromosomes need multiple DNA replication origins because: A) eukaryotic chromosomes cannot usually replicate bidirectionally

B) eukaryotic genomes are not usually circular, like the bacterial chromosome is

C) the processivity of the eukaryotic DNA polymerase is much less than the bacterial enzyme D) their replication rate is much slower, and it would take too long with only a single origin per chromosome

E) they have a variety of DNA polymerases for different purposes, and need a corresponding variety

of replication origins

Page: 965 Difficulty: 2 Ans: B

The function of the eukaryotic DNA replication factor PCNA (proliferating cell nuclear antigen) is

similar to that of the β-subunit of bacterial DNA polymerase III in that it:

A) facilitates replication of telomeres

B) forms a circular sliding clamp to increase the processivity of replication

C) has a 3' → 5' proofreading activity.

D) increases the speed but not the processivity of the replication complex

E) participates in DNA repair

12 DNA repair

The Ames test is used to:

A) detect bacterial viruses

B) determine the rate of DNA replication

C) examine the potency of antibiotics

D) measure the mutagenic effects of various chemical compounds

E) quantify the damaging effects of UV light on DNA molecules

13 DNA repair

In a mammalian cell, DNA repair systems:

A) are extraordinarily efficient energetically

B) are generally absent, except in egg and sperm cells

C) can repair deletions, but not mismatches

D) can repair most types of lesions except those caused by UV light

E) normally repair more than 99% of the DNA lesions that occur

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14 DNA repair

Pages: 967-969 Difficulty: 3 Ans: A

Which of these enzymes is not directly involved in methyl-directed mismatch repair in E coli?

A) DNA glycosylase

B) DNA helicase II

C) DNA ligase

D) DNA polymerase III

E) Exonuclease I

15 DNA repair

Page: 968 Difficulty: 2 Ans: B

The role of the Dam methylase is to:

A) add a methyl group to uracil, converting it to thymine

B) modify the template strand for recognition by repair systems

C) remove a methyl group from thymine

D) remove a mismatched nucleotide from the template strand

E) replace a mismatched nucleotide with the correct one

16 DNA repair

When bacterial DNA replication introduces a mismatch in a double-stranded DNA, the methyl-directed repair system:

A) cannot distinguish the template strand from the newly replicated strand

B) changes both the template strand and the newly replicated strand

C) corrects the DNA strand that is methylated

D) corrects the mismatch by changing the newly replicated strand

E) corrects the mismatch by changing the template strand

17 DNA repair

In base-excision repair, the first enzyme to act is:

A) AP endonuclease

B) Dam methylase

C) DNA glycosylase

D) DNA ligase

E) DNA polymerase

18 DNA repair

The ABC excinuclease is essential in:

A) base-excision repair

B) methyl-directed repair

C) mismatch repair

D) nucleotide-excision repair

E) SOS repair

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19 DNA repair

The repair of cyclobutane pyrimidine dimers by bacterial DNA photolyase involves the cofactor: A) coenzyme A

B) coenzyme Q

C) FADH–

D) pyridoxal phosphate (PLP)

E) thiamine pyrophosphate (TPP)

20 DNA repair

Pages: 976-977 Difficulty: 2 Ans: E

An alternative repair system by error-prone translesion DNA synthesis can result in a high mutation rate, because:

A) alternative modified nucleotides can be incorporated more readily

B) interference from the RecA and SSB proteins hinders the normal replication accuracy

C) replication proceeds much faster than normal, resulting in many more mistakes

D) the DNA polymerases involved cannot facilitate base-pairing as well as DNA polymerase III E) the DNA polymerases involved lack exonuclease proofreading activities

21 DNA recombination

In homologous recombination in E coli, the protein that moves along a double-stranded DNA,

unwinding the strands ahead of it and degrading them, is:

A) chi

B) DNA ligase

C) RecA protein

D) RecBCD enzyme

E) RuvC protein (resolvase)

In homologous recombination in E coli, the protein that assembles into long, helical filaments that

coat a region of DNA is:

A) DNA methylase

B) DNA polymerase

C) histone

D) RecA protein

E) RecBCD enzyme

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Pages: 982-983 Difficulty: 2 Ans: A

In homologous genetic recombination, RecA protein is involved in:

A) formation of Holliday intermediates and branch migration

B) introduction of negative supercoils into the recombination products

C) nicking the two duplex DNA molecules to initiate the reaction

D) pairing a DNA strand from one duplex DNA molecule with sequences in another duplex,

regardless of complementarity

E) resolution of the Holliday intermediate

Which of the following statements is false? In vitro, the strand-exchange reaction:

A) can include formation of a Holliday intermediate

B) is accompanied by ATP hydrolysis

C) may involve transient formation of a three- or four-stranded DNA complex

D) needs RecA protein

E) requires DNA polymerase

The bacteriophage λ can lysogenize after infecting a bacterium, i.e integrate into the host bacterial chromosome by site-specific recombination, and may reside there for many generations before an

excision event regenerates the viral genome in an infective form Which one of the following is not a

component of these events?

A) Excision requires two host proteins and two virally-encoded proteins

B) Integration requires a viral-specific protein, called integrase

C) RecA protein is required to catalyze the insertional recombination event

D) The excision event relies on different sequences than the integration event

E) The virus and the host DNAs share a 15 bp “core” region of perfect homology

Short Answer Questions

Pages: 950-951 Difficulty: 2

Describe briefly how equilibrium density gradient centrifugation was used to demonstrate that DNA

replication in E coli is semiconservative

Ans: Equilibrium density gradient centrifugation separates DNA molecules of slightly different

buoyant density For example, molecules containing 15N-labeled (“heavy”) DNA are separable from identical molecules containing 14N (“light”) DNA Meselson and Stahl grew E coli for many

generations in a medium containing 15N, producing cells in which all DNA was heavy These cells were transferred to a medium containing 14N, and the buoyant density of their DNA was determined (by equilibrium density gradient centrifugation) after 1, 2, 3, etc., generations After one generation, all DNA was of a density intermediate between fully heavy and fully light, indicating that each double-stranded DNA molecule had one heavy (parental) and one light (newly synthesized) strand;

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replication was semiconservative (See Fig 25-2, p 950.)

Page: 952 Difficulty: 2

The DNA below is replicated from left to right Label the templates for leading strand and lagging strand synthesis

(5')ACTTCGGATCGTTAAGGCCGCTTTCTGT(3')

(3')TGAAGCCTAGCAATTCCGGCGAAAGACA(5')

Ans: The polarity of the strands indicates that the top strand is the template for lagging strand

synthesis, and the bottom strand is the template for leading strand synthesis (See Fig 25-4, p 952.)

All known DNA polymerases catalyze synthesis only in the 5' → 3' direction Nevertheless, during

semiconservative DNA replication in the cell, they are able to catalyze the synthesis of both daughter

chains, which would appear to require synthesis in the 3' → 5' direction Explain the process that

occurs in the cell that allows for synthesis of both daughter chains by DNA polymerase

Ans: During DNA replication, one strand is synthesized continuously and the other is synthesized by

a discontinuous mechanism The daughter chain, which appears to be growing in the 3' → 5'

direction (the “lagging strand”), is actually being synthesized by continual initiation of new chains

and their elongation in the 5' → 3' direction

Pages: 952, 960 Difficulty: 2

What is an Okazaki fragment? What enzyme(s) is (are) required for its formation in E coli?

Ans: An Okazaki fragment is an intermediate in DNA replication in E coli It is a short fragment of

newly synthesized DNA, attached to the 3' end of a short RNA primer Such fragments are produced

by the combined action of primase (part of the primosome) and DNA polymerase III during

replication of the lagging strand (See Fig 25-13, p 960.)

Diagram the reaction catalyzed by DNA polymerase that occurs between deoxyribose at the end of a

DNA chain and the 5' phosphates of a deoxyribonucleoside triphosphate Include the chemical

structure of the phosphate group, indicate the locations of the sugar and base, and show the

rearrangements of electrons that occur

Ans: See Fig 25-5, p 953

Nucleotide polymerization appears to be a thermodynamically balanced reaction (because one

phosphodiester bond is broken and one is formed) Nevertheless, the reaction proceeds efficiently both in a test tube and in the cell Explain

Ans: Base-stacking and base-pairing interactions in the polymerized DNA product stabilize it and tend

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to make the overall reaction more exergonic In the cell, pyrophosphatase may make a contribution by coupling polymerization to the highly exergonic hydrolysis of the pyrophosphate product

A suitable substrate for DNA polymerase is shown below Label the primer and template, and

indicate which end of each strand must be 3' or 5'.

To observe DNA synthesis on this substrate in vitro, what additional reaction components must be added?

Ans: The top strand (the primer) has its 5' end to the left; the bottom (template) strand has the

opposite polarity For DNA synthesis with this substrate in vitro, one would have to add DNA

polymerase, the four deoxynucleoside triphosphates, Mg2+, and a suitable buffer

Pages: 954, 958 Difficulty: 2

All known DNA polymerases can only elongate a preexisting DNA chain (i.e., require a primer), but cannot initiate a new DNA chain Nevertheless, during semiconservative DNA replication in the cell, entirely new daughter DNA chains are synthesized Explain the process that occurs in the cell that allows for the synthesis of daughter chains by DNA polymerase

Ans: In the cell, initiation of DNA chains occurs via the synthesis of an RNA primer by an RNA

polymerase type of enzyme (primase) This primer is elongated by DNA polymerase to produce the

daughter DNA chain The RNA is removed by 5' exonucleolytic hydrolysis before replication is

completed

DNA replication in E coli begins at a site in the DNA called the (a) _ At the replication

fork the (b) _ strand is synthesized continuously while the (c) _ strand is

synthesized discontinuously On the strand synthesized discontinuously, the short pieces are called (d) fragments An RNA primer for each of the fragments is synthesized by an enzyme called (e) , and this RNA primer is removed after the fragment is synthesized by the enzyme (f) _, using its (g) _ activity The nicks left behind in this process are sealed by the enzyme (h) _

Ans: (a) origin; (b) leading; (c) lagging; (d) Okazaki; (e) primase; (f) DNA pol I; (g) 5' → 3'

exonuclease; (h) DNA ligase

Briefly describe the biochemical role of the following enzymes in DNA replication in E coli:

(a) DNA helicase; (b) primase; (c) the 3' → 5' exonuclease activity of DNA polymerase; (d) DNA 1igase; (e) topoisomerases; (f) the 5' → 3' exonuclease activity of DNA polymerase I

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Ans: (a) Helicase unwinds double-stranded DNA during replication (b) Primase synthesizes short

RNA primers during lagging strand replication (c) The 3' → 5' exonuclease activity of DNA

polymerase proofreads newly synthesized DNA, removing mismatched nucleotides (d) DNA ligase seals nicks in the DNA at the boundaries between Okazaki fragments (e) Topoisomerases relieves the topological stress produced by the unwinding of double-stranded DNA at the replication fork (f)

The 5' → 3' exonuclease activity of DNA polymerase I removes RNA primers

DNA synthesis on the lagging strand in E coli is a complex process known to involve several

proteins Initiation of a new chain is catalyzed by the enzyme (a) _, and elongation is catalyzed by the enzyme (b) Synthesis is discontinuous, yielding short segments called (c) _, which are eventually joined by the enzyme (d) , which requires the cofactor (e) _

Ans: (a) primase; (b) DNA pol III; (c) Okazaki fragments; (d) DNA ligase; (e) NAD+

List two proteins or enzymes, other than DNA polymerase III, that are found at the replication fork in

E coli; describe each of their functions with no more than one sentence

Ans: The proteins are listed in Table 25-4, p 962 They include (a) DNA polymerase I, which fills

gaps and excises RNA primers; (b) primase (the DnaG protein), which synthesizes short RNA

primers; (c) DNA ligase, which seals nicks; and (d) proteins that aid in DNA unwinding and

supercoiling

In the bacterial cell, what are catenated chromosomes, when do they arise, and how does the cell resolve the problem posed by their structure?

Ans: Catenanes are topologically interlinked circular chromosomes, which are the normal end result

of DNA replication of the parental circular genome when the bidirectional replication forks meet They are unlinked by the bacterial topoisomerase IV (a type II enzyme), and thus become free to segregate into daughter cells upon cell division (See Fig 25-17, p 964.)

Why is the drug acyclovir effective against the herpes simplex virus?

Ans: Acyclovir is a guanine nucleoside with an incomplete ribose ring, which can be phosphorylated

much more efficiently by the viral thymidine kinase than the host enzyme Further conversion forms acyclo-GTP, which competitively inhibits the viral DNA polymerase more strongly than the host enzyme, and when incorporated into DNA is a chain terminator, because it lacks a 3' hydroxyl group

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40 DNA repair

Page: 966-967 Difficulty: 2

The high fidelity of DNA replication is due primarily to immediate error correction by the 3' —> 5' exonuclease (proofreading) activity of the DNA polymerase Some incorrectly paired bases escape this proofreading, and further errors can arise from challenges to the chemical integrity of the DNA List the four classes of repair mechanisms that the cell can use to help correct such errors

Ans: The four classes are listed in Table 25-5 (p 967), and consist of (1) mismatch repair, (2)

base-excision repair, (3) nucleotide-base-excision repair, and (4) direct repair

41 DNA repair

List three types of DNA damage that require repair

Ans: The defects in DNA that require repair include (a) mismatches that occur during replication; (b)

abnormal bases; (c) pyrimidine dimers produced by UV irradiation Other answers are possible

42 DNA repair

Match the damage type or repair step at the left with a related enzyme at right Only one answer will

be the most direct for each

_ binds to mismatch in DNA (e) uracil N-glycosylase

_ direct chemical reversal (i) O6-methylguanine

of pyrimidine dimer formation methyltransferase

(m) restriction endonuclease

Ans: e; b; a; c; f; d; j; i; h; m; g

43 DNA repair

Explain the role of DNA glycosylases in DNA repair

Ans: When spontaneous deamination converts cytosine in DNA to uracil, or adenine to hypoxanthine,

DNA glycosylase breaks the N-glycosidic bond to the defective base, creating an “abasic” or “AP”

site The region containing the AP site is then excised by AP endonuclease, and the resulting gap is closed by DNA polymerase I and sealed by DNA ligase (See Fig 25-23, p 972.) Other DNA glycosylases recognize other types of modified or damaged bases

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