Deck 20: DNA Replication, repair, and Recombination
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Deck 20: DNA Replication, repair, and Recombination
1
Which DNA polymerase is most responsible for chain elongation in E.coli?
A)DNA polymerase I.
B)DNA polymerase II.
C)DNA polymerase III.
D)DNA polymerase IV.
A)DNA polymerase I.
B)DNA polymerase II.
C)DNA polymerase III.
D)DNA polymerase IV.
DNA polymerase III.
2
The protein machine that carries out DNA replication is called ________.
A)DNA polymerase
B)the replication fork
C)the Klenow fragment
D)the replisome
A)DNA polymerase
B)the replication fork
C)the Klenow fragment
D)the replisome
the replisome
3
Approximately ________ base pairs are added per second during DNA in vivo replication in E.coli.
A)10
B)100
C)1000
D)1,000,000
A)10
B)100
C)1000
D)1,000,000
1000
4
The epsilon subunit of DNA polymerase III is responsible for its ________ activity.
A)5' → 3' polymerase
B)sliding clamp
C)3' → 5' exonuclease (proof reading)
D)ribosome assembly
A)5' → 3' polymerase
B)sliding clamp
C)3' → 5' exonuclease (proof reading)
D)ribosome assembly
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5
DNA that has been liberally labeled with radioactive 14C is used as the template for replication.Replication is carried out in a medium containing only unlabeled nucleotides.After two rounds of replication,what percent of DNA molecules are still labeled?
A)25%
B)50%
C)75%
D)100%
A)25%
B)50%
C)75%
D)100%
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6
How many replisomes are required for one complete round of DNA replication in E.coli?
A)One
B)Two
C)Four
D)More than 4
A)One
B)Two
C)Four
D)More than 4
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7
The beta subunits of E.coli DNA polymerase III are responsible for its ________.
A)5'→ 3' polymerase
B)sliding clamp
C)3' → 5' exonuclease and proof-reading
D)ribosome assembly
A)5'→ 3' polymerase
B)sliding clamp
C)3' → 5' exonuclease and proof-reading
D)ribosome assembly
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8
The correct assembly of proteins needed for DNA replication at the correct site on DNA is the ________ stage.
A)termination
B)disassembly
C)elongation
D)initiation
A)termination
B)disassembly
C)elongation
D)initiation
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9
During the elongation stage
A)replication proteins assemble at the DNA replication site.
B)DNA is replicated semiconservatively.
C)the protein machine disassembles.
D)replication ends.
A)replication proteins assemble at the DNA replication site.
B)DNA is replicated semiconservatively.
C)the protein machine disassembles.
D)replication ends.
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10
Enzymes that remain bound to their nascent chains through many polymerization steps are said to be ________.
A)constitutive
B)distributive
C)conjugated
D)processive
A)constitutive
B)distributive
C)conjugated
D)processive
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11
The 3' → 5' exonuclease activity of E.coli DNA polymerase III accounts for the ________ of polymerization.
A)low error rate
B)high speed
C)directionality
D)All of the above
A)low error rate
B)high speed
C)directionality
D)All of the above
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12
The processivity of E.coli DNA polymerase III holoenzyme accounts for all of the following except
A)the relatively small number of enzymes needed to replicate the entire chromosome.
B)the directionality of polymerization.
C)the rapid rate of polymerization.
D)the ability of a single holoenzyme to add many nucleotides to a growing DNA chain.
A)the relatively small number of enzymes needed to replicate the entire chromosome.
B)the directionality of polymerization.
C)the rapid rate of polymerization.
D)the ability of a single holoenzyme to add many nucleotides to a growing DNA chain.
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13
At the normal rate of replication,about how many minutes does it take for the entire E.coli chromosome (4.6 × 103 kbp)to be duplicated?
A)5
B)20
C)40
D)80
A)5
B)20
C)40
D)80
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14
Meselson and Stahl's experiments involved growing E.coli on a medium containing 15N (heavy nitrogen)which increased the density of the bacterial DNA.Cells were then grown on a medium containing only 14N.DNA samples were obtained for analysis after one,two and three rounds of replication (generations)and centrifuged on a density gradient.The double-helical DNA forms bands in the gradient depending on its density.How many bands did Meselson and Stahl see after two generation of cells were grown on the 14N medium?
A)One with DNA containing both 15N and 14N.
B)Two,one with DNA containing only 15N and one with DNA containing only 14N.
C)Two,one with DNA containing both 15N and 14N and one with DNA containing only 14N.
D)Three,one with DNA containing only 15N,one with DNA containing both 15N and 14N and one with DNA containing only 14N.
A)One with DNA containing both 15N and 14N.
B)Two,one with DNA containing only 15N and one with DNA containing only 14N.
C)Two,one with DNA containing both 15N and 14N and one with DNA containing only 14N.
D)Three,one with DNA containing only 15N,one with DNA containing both 15N and 14N and one with DNA containing only 14N.
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15
E.coli DNA polymerase catalyzes polymerization in the ________ direction(s).
A)5' → 3'
B)3' → 5'
C)2' → 5'
D)both 5' → 3' and 3' → 5'
E)All of the above
A)5' → 3'
B)3' → 5'
C)2' → 5'
D)both 5' → 3' and 3' → 5'
E)All of the above
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16
DNA polymerase III requires a(n)________ for synthesis of DNA to occur.
A)DNA template
B)RNA primer
C)free 3'-OH end of a DNA strand
D)All of the above
A)DNA template
B)RNA primer
C)free 3'-OH end of a DNA strand
D)All of the above
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17
Both strands of double-stranded DNA are copied so that each daughter molecule consists of one parental strand and one newly synthesized strand.This is called ________ replication.
A)bidirectional
B)conservative
C)semiconservative
D)duplicative
A)bidirectional
B)conservative
C)semiconservative
D)duplicative
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18
It takes about the same amount of time for eukaryotic chromosomes to be replicated in vivo as it does for E.coli.Why is this true?
A)Eukaryotes have multiple origins of replication;E.coli has one unique origin.
B)Most chromosomes are approximately the same size.
C)Eukaryotic replication occurs at a faster rate,enabling replication of the larger chromosomes in about the same amount of time.
D)The replication of DNA in eukaryotes is faster due to a quicker,more accurate repair of mismatches.
A)Eukaryotes have multiple origins of replication;E.coli has one unique origin.
B)Most chromosomes are approximately the same size.
C)Eukaryotic replication occurs at a faster rate,enabling replication of the larger chromosomes in about the same amount of time.
D)The replication of DNA in eukaryotes is faster due to a quicker,more accurate repair of mismatches.
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19
The replication of DNA is ________ because ________.
A)conservative;one strand of parental DNA is retained in each daughter DNA
B)conservative;each daughter molecule has two new strands copied from the parental DNA template
C)semiconservative;one strand of parental DNA is retained in each daughter DNA
D)semiconservative;each daughter molecule has two new strands copied from the parental DNA template
A)conservative;one strand of parental DNA is retained in each daughter DNA
B)conservative;each daughter molecule has two new strands copied from the parental DNA template
C)semiconservative;one strand of parental DNA is retained in each daughter DNA
D)semiconservative;each daughter molecule has two new strands copied from the parental DNA template
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20
The incorporation of new nucleotides into a growing strand of DNA occurs during ________.
A)the elongation stage
B)nucleation
C)initiation
D)assembly of the protein machine
A)the elongation stage
B)nucleation
C)initiation
D)assembly of the protein machine
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21
Nick translation is a process for
A)filling in and sealing the gap in DNA after removal of the RNA primer.
B)synthesis of the protein for primase from mRNA.
C)removal of the RNA primer by DNA polymerase I.
D)removal of Okazaki fragments.
A)filling in and sealing the gap in DNA after removal of the RNA primer.
B)synthesis of the protein for primase from mRNA.
C)removal of the RNA primer by DNA polymerase I.
D)removal of Okazaki fragments.
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22
Why does E.coli need both DNA polymerase III and DNA polymerase I?
A)DNA polymerase III lacks the 5' → 3' exonuclease activity needed to remove RNA primers.
B)Each polymerase is specific for only one strand of DNA.DNA polymerase III acts only on the leading strand,and DNA polymerase I acts only on the lagging strand.
C)DNA polymerase I is needed to join the Okazaki fragments.
D)The DNA replication is bidirectional;one polymerase is used for each direction.
E)Only DNA polymerase I has proofreading ability.
A)DNA polymerase III lacks the 5' → 3' exonuclease activity needed to remove RNA primers.
B)Each polymerase is specific for only one strand of DNA.DNA polymerase III acts only on the leading strand,and DNA polymerase I acts only on the lagging strand.
C)DNA polymerase I is needed to join the Okazaki fragments.
D)The DNA replication is bidirectional;one polymerase is used for each direction.
E)Only DNA polymerase I has proofreading ability.
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23
Which proteins are responsible for the unwinding of the double-stranded DNA during replication?
A)Ligases.
B)Helicases.
C)Topoisomerases.
D)Primases.
E)Lyases.
A)Ligases.
B)Helicases.
C)Topoisomerases.
D)Primases.
E)Lyases.
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24
Which subunit of the replisome leads at the replication fork?
A)β-clamp.
B)SSB.
C)Primosome.
D)Polymerase core complex.
A)β-clamp.
B)SSB.
C)Primosome.
D)Polymerase core complex.
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25
What is a reasonable length for an RNA primer in E.coli?
A)About 20 nucleotides.
B)50 nucleotides.
C)100 nucleotides.
D)1000 nucleotides.
A)About 20 nucleotides.
B)50 nucleotides.
C)100 nucleotides.
D)1000 nucleotides.
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26
Fragments on the lagging strand are joined by the enzyme ________.
A)DNA polymerase I
B)DNA synthase
C)DNA primosome
D)DNA ligase
A)DNA polymerase I
B)DNA synthase
C)DNA primosome
D)DNA ligase
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27
The RNA primer at the beginning of each Okazaki fragment is removed by
A)DNA polymerase I.
B)DNA polymerase II.
C)DNA polymerase III.
D)3' → 5' exonuclease subunit of DNA polymerase III.
A)DNA polymerase I.
B)DNA polymerase II.
C)DNA polymerase III.
D)3' → 5' exonuclease subunit of DNA polymerase III.
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28
The primosome is composed of polypeptides including primase which is
A)DNA polymerase III.
B)RNA dependent DNA polymerase.
C)DNA dependent RNA polymerase.
D)Parental DNA.
A)DNA polymerase III.
B)RNA dependent DNA polymerase.
C)DNA dependent RNA polymerase.
D)Parental DNA.
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29
RNA primers are synthesized about once every ________.
A)millisecond
B)second
C)minute
D)40 minutes
A)millisecond
B)second
C)minute
D)40 minutes
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30
Okazaki fragments are ________.
A)the smallest subunits of DNA polymerase III
B)short stretches of DNA formed on the lagging strand
C)short RNA primers needed for initiation of polymerization
D)fragments of DNA polymerase I that lack 5' → 3' exonuclease activity
A)the smallest subunits of DNA polymerase III
B)short stretches of DNA formed on the lagging strand
C)short RNA primers needed for initiation of polymerization
D)fragments of DNA polymerase I that lack 5' → 3' exonuclease activity
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31
What is the function of single-strand binding protein (SSB)(helix stabilizing protein)?
A)It stabilizes nicks on the lagging strand during replication until DNA ligase seals the nick.
B)It recognizes thymine dimers in DNA,which make the DNA single-stranded,and serves as a signal for the activation of repair proteins.
C)It stabilizes many forms of RNA which are single-stranded.
D)During replication it binds to single-stranded regions of the lagging strand and prevents it from folding back on itself into double-helical regions.
A)It stabilizes nicks on the lagging strand during replication until DNA ligase seals the nick.
B)It recognizes thymine dimers in DNA,which make the DNA single-stranded,and serves as a signal for the activation of repair proteins.
C)It stabilizes many forms of RNA which are single-stranded.
D)During replication it binds to single-stranded regions of the lagging strand and prevents it from folding back on itself into double-helical regions.
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32
The best enzyme to use in the Sanger method for DNA sequence determination is
A)DNA polymerase I.
B)Klenow fragment of DNA polymerase I.
C)DNA polymerase from a high temperature bacterium.
D)SSB protein.
A)DNA polymerase I.
B)Klenow fragment of DNA polymerase I.
C)DNA polymerase from a high temperature bacterium.
D)SSB protein.
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33
Which is not part of the replisome?
A)Helicase DnaB.
B)The primosome.
C)Gyrase.
D)Single-stranded binding protein.
A)Helicase DnaB.
B)The primosome.
C)Gyrase.
D)Single-stranded binding protein.
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34
Polymerization of the lagging strand is catalyzed by ________.
A)a second core complex on the same polymerase holoenzyme that polymerizes the leading strand
B)a completely separate replisome from that which polymerizes the leading strand
C)DNA polymerase I holoenzyme
D)the same polymerase holoenzyme that polymerized the leading strand,but with opposite directionality
A)a second core complex on the same polymerase holoenzyme that polymerizes the leading strand
B)a completely separate replisome from that which polymerizes the leading strand
C)DNA polymerase I holoenzyme
D)the same polymerase holoenzyme that polymerized the leading strand,but with opposite directionality
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35
What is a Klenow fragment?
A)A fragment of DNA polymerase I that lacks 5' → 3' exonuclease activity.
B)A topoisomerase responsible for unwinding DNA at the replication fork.
C)A portion of the replisome responsible for RNA primer synthesis.
D)Short stretches of DNA formed during polymerization of the lagging strand.
A)A fragment of DNA polymerase I that lacks 5' → 3' exonuclease activity.
B)A topoisomerase responsible for unwinding DNA at the replication fork.
C)A portion of the replisome responsible for RNA primer synthesis.
D)Short stretches of DNA formed during polymerization of the lagging strand.
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36
In E.coli,________ helps regulate the frequency of DNA replication.
A)DnaA
B)DnaB
C)Tus
D)RecA
A)DnaA
B)DnaB
C)Tus
D)RecA
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37
Which is not an activity for DNA polymerase I?
A)5' → 3' polymerase activity.
B)Joining of Okazaki fragments.
C)5' → 3' exonuclease activity.
D)Nick translation.
E)3' → 5' proofreading activity.
A)5' → 3' polymerase activity.
B)Joining of Okazaki fragments.
C)5' → 3' exonuclease activity.
D)Nick translation.
E)3' → 5' proofreading activity.
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38
In E.coli the helicase ________ is one of the subunits of the ________ in the replisome.
A)gyrase;primosome
B)gyrase;β-clamp
C)DnaB;primosome
D)DnaB;β-clamp
A)gyrase;primosome
B)gyrase;β-clamp
C)DnaB;primosome
D)DnaB;β-clamp
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39
The lagging strand loops back through the replisome so that
A)SSB tetramers can open them.
B)it catches up with the leading strand.
C)both leading and laggings strands are made in the same direction.
D)gyrase can open the helix.
A)SSB tetramers can open them.
B)it catches up with the leading strand.
C)both leading and laggings strands are made in the same direction.
D)gyrase can open the helix.
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40
What is the function of the primosome?
A)Removal of RNA from the lagging strand.
B)Synthesis of short RNA fragments.
C)Assembly of DNA polymerase III holoenzyme.
D)Detection of the origin of replication.
A)Removal of RNA from the lagging strand.
B)Synthesis of short RNA fragments.
C)Assembly of DNA polymerase III holoenzyme.
D)Detection of the origin of replication.
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41
Which is not a common type of DNA damage (lesion)?
A)Pyrimidine dimers.
B)Cross-linking of DNA strands.
C)Alkylation of bases.
D)Deamination of thymines.
A)Pyrimidine dimers.
B)Cross-linking of DNA strands.
C)Alkylation of bases.
D)Deamination of thymines.
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42
Why is only one copy of DNA made per cell division cycle in eukaryotes despite the presence of multiple origins of replication?
A)There is only one origin of replication per chromosome.
B)The rate of elongation proceeds at a rate that allows only one copy to be made by the time the cell divides.
C)The activity of an S-phase protein kinase controls the formation of the pre-replication complex.
D)Primase is activated by phosphorylation only at the onset of S-phase by S-phase protein kinase.
A)There is only one origin of replication per chromosome.
B)The rate of elongation proceeds at a rate that allows only one copy to be made by the time the cell divides.
C)The activity of an S-phase protein kinase controls the formation of the pre-replication complex.
D)Primase is activated by phosphorylation only at the onset of S-phase by S-phase protein kinase.
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43
In eukaryotes,after completion of S-phase and G2-phase,SPK (S-phase protein kinase)is broken down.This allows
A)binding of ORC (origin recognition complex).
B)prevention of new DNA formation.
C)the beginning of mitosis.
D)termination of replication.
A)binding of ORC (origin recognition complex).
B)prevention of new DNA formation.
C)the beginning of mitosis.
D)termination of replication.
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44
The sequence of proteins used in bacterial recombination is:
A)RuvC → RuvB → RecA → RecBCD
B)RecBCD → RecA → RuvAB → RuvC
C)RecA → RecBCD → RuvAB → RuvC
D)RuvAB → RuvC → RecBCD → RecA
A)RuvC → RuvB → RecA → RecBCD
B)RecBCD → RecA → RuvAB → RuvC
C)RecA → RecBCD → RuvAB → RuvC
D)RuvAB → RuvC → RecBCD → RecA
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45
The general excision repair pathway for DNA repair has the following order:
A)endonuclease cuts → helicase or endonuclease removes → DNA polymerase → DNA ligase
B)DNA ligase → DNA polymerase → helicase or endonuclease removes → endonuclease cuts
C)endonuclease cuts → DNA polymerase repairs → helicase opens up → DNA ligase
D)endonuclease cuts → DNA ligase removes → endonuclease removes → DNA ligase
A)endonuclease cuts → helicase or endonuclease removes → DNA polymerase → DNA ligase
B)DNA ligase → DNA polymerase → helicase or endonuclease removes → endonuclease cuts
C)endonuclease cuts → DNA polymerase repairs → helicase opens up → DNA ligase
D)endonuclease cuts → DNA ligase removes → endonuclease removes → DNA ligase
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46
In packaging the eukaryotic DNA into nucleosomes,it is thought that the requirement for ________ and their synthesis is a cause of the slower rate of movement of the replication fork.
A)Okazaki fragments
B)larger DNA molecules
C)increase number of accessory proteins
D)histone proteins
E)lagging strands
A)Okazaki fragments
B)larger DNA molecules
C)increase number of accessory proteins
D)histone proteins
E)lagging strands
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47
Deaminated bases must be removed from DNA because
A)they distort the molecule.
B)they form thymine dimers.
C)they pair incorrectly during replication.
D)the mRNA formed may be irregular.
A)they distort the molecule.
B)they form thymine dimers.
C)they pair incorrectly during replication.
D)the mRNA formed may be irregular.
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48
A major difference between prokaryotic and eukaryotic DNA replication mechanisms is
A)no Okazaki fragments are made in eukaryotes.
B)both leading and lagging strands are continuous in eukaryotes.
C)no primer synthesis is needed in eukaryotes.
D)the replication fork moves slower in eukaryotes.
A)no Okazaki fragments are made in eukaryotes.
B)both leading and lagging strands are continuous in eukaryotes.
C)no primer synthesis is needed in eukaryotes.
D)the replication fork moves slower in eukaryotes.
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49
In DNA repair,DNA glycosylases recognize deaminated ________ within the DNA polymer and remove them.
A)nucleoside
B)nucleotide
C)thymine dimers
D)bases
A)nucleoside
B)nucleotide
C)thymine dimers
D)bases
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50
How are the BRCA proteins and their genes implicated in breast cancer?
A)The BRCA proteins damage DNA by generating double-stranded breaks which can lead to cancer.
B)Cells become more susceptible to damage if the genes for these proteins are damaged.
C)BRCA proteins inhibit excision repair mechanisms.
D)BRCA proteins initiate recombination between non-homologous DNA sequences.
A)The BRCA proteins damage DNA by generating double-stranded breaks which can lead to cancer.
B)Cells become more susceptible to damage if the genes for these proteins are damaged.
C)BRCA proteins inhibit excision repair mechanisms.
D)BRCA proteins initiate recombination between non-homologous DNA sequences.
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51
In the Holliday model describing recombination
A)homologous sequences line up.
B)strands of DNA are nicked.
C)DNA strands associate with the homologous strand ("strand invasion").
D)strands rotate then are cleaved at the crossover point.
E)All of the above.
A)homologous sequences line up.
B)strands of DNA are nicked.
C)DNA strands associate with the homologous strand ("strand invasion").
D)strands rotate then are cleaved at the crossover point.
E)All of the above.
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52
Why is DNA damage in skin cells from exposure to excessive UV light not completely reversed by photoreactivation?
A)DNA photolyase in humans is activated only under conditions of heat shock.
B)Humans lack a photoreactivation repair mechanism.
C)The presence of histones greatly slows photoreactivation,thus humans depends primarily on other forms of DNA repair.
D)Photoreactivation is specific for the removal of thymine dimers.Thymine dimers do not form in human DNA.
A)DNA photolyase in humans is activated only under conditions of heat shock.
B)Humans lack a photoreactivation repair mechanism.
C)The presence of histones greatly slows photoreactivation,thus humans depends primarily on other forms of DNA repair.
D)Photoreactivation is specific for the removal of thymine dimers.Thymine dimers do not form in human DNA.
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53
Functions of DNA polymerases and accessory proteins in eukaryotic DNA replication include
A)PCNA protein,a sliding clamp unit for DNA.
B)β- DNA polymerase,for repair of damaged DNA.
C)RPA protein,and equivalent of SSB proteins in bacteria.
D)RPC protein,used for mitochondrial DNA synthesis.
E)All of the above.
A)PCNA protein,a sliding clamp unit for DNA.
B)β- DNA polymerase,for repair of damaged DNA.
C)RPA protein,and equivalent of SSB proteins in bacteria.
D)RPC protein,used for mitochondrial DNA synthesis.
E)All of the above.
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54
The reaction catalyzed by DNA photolyase is an example of what type of repair?
A)Recombination.
B)Excision.
C)Mismatch.
D)Direct.
A)Recombination.
B)Excision.
C)Mismatch.
D)Direct.
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55
Which molecules have repair mechanisms in most cells?
A)DNA.
B)RNA.
C)Proteins.
D)DNA and proteins.
E)DNA,RNA and proteins.
A)DNA.
B)RNA.
C)Proteins.
D)DNA and proteins.
E)DNA,RNA and proteins.
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56
In photoreactivation,DNA damage caused by UV light is reversed by DNA photolyase which
A)is activated by UV light to form a complex.
B)removes nicks in the DNA backbone.
C)forms a complex with thymine dimers and the complex absorbs visible light which reverses the dimerization.
D)binds to the adenine bases opposite the thymine dimer.
E)All of the above.
A)is activated by UV light to form a complex.
B)removes nicks in the DNA backbone.
C)forms a complex with thymine dimers and the complex absorbs visible light which reverses the dimerization.
D)binds to the adenine bases opposite the thymine dimer.
E)All of the above.
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57
Transposons are mobile genetic elements that jump from chromosome to chromosome by
A)homologous recombination.
B)non-homologous recombination.
C)site specific recombination.
D)mutation.
A)homologous recombination.
B)non-homologous recombination.
C)site specific recombination.
D)mutation.
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58
What distinguishes direct repair from other DNA repair mechanisms?
A)Direct repair does not require the breaking of any bonds in the DNA phosphodiester backbone.
B)Direct repair proteins do not require activation by light or other chemicals.
C)Direct repair systems activate the proofreading activity of DNA polymerase.
D)Direct repair systems remove and replace single defective nucleotides.
A)Direct repair does not require the breaking of any bonds in the DNA phosphodiester backbone.
B)Direct repair proteins do not require activation by light or other chemicals.
C)Direct repair systems activate the proofreading activity of DNA polymerase.
D)Direct repair systems remove and replace single defective nucleotides.
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59
What is the DNA sequence (5' → 3')determined in this gel? Assume the samples were applied near the top of the gel. 
A)AGGCTGGACGA
B)AGGGGCTAGCA
C)AGGCTGACGGA
D)AGGCAGTCGGA

A)AGGCTGGACGA
B)AGGGGCTAGCA
C)AGGCTGACGGA
D)AGGCAGTCGGA
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60
During recombination in E.coli,________ recognize(s)regions of sequence similarity and promote(s)strand invasion,while ________ bind(s)to a Holliday junction promoting strand migration.
A)RuvA and RuvB;RecBCD
B)RecBCD;RecA
C)RecA;RecBCD
D)RecA;RuvA and RuvB
A)RuvA and RuvB;RecBCD
B)RecBCD;RecA
C)RecA;RecBCD
D)RecA;RuvA and RuvB
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61
DNA polymerase I possesses 5' → 3' exonuclease activity.
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62
In a sequencing gel,all fragment in one of the four lanes terminates in the same type of nucleotide base.
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63
A short primer is added to the DNA polymerase reaction mix for Sanger DNA sequencing that has a 5' free hydroxyl to which new nucleotides can be added.
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64
DNA polymerase I which is inhibited by dideoxynucleoside triphosphates is used in the Sanger method for DNA sequencing.
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65
The chromosome of the fruit fly (D.melanogaster)is about twice as large as the E.coli chromosome.
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66
All DNA ligase enzymes use ATP as a cosubstrate.
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67
Proteins bind to both origin sites and termination sites on DNA during DNA replication.
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68
Prokaryotic DNA replication occurs in two steps.First,ATP provides a phosphate to the growing DNA chain.This is followed by addition of a nucleoside.
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69
The β subunits of E.coli DNA polymerase form a sliding clamp that surrounds the DNA strands at the replication fork.
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70
The purpose of SSB is to prevent single-stranded DNA from folding back on itself to form double-helical regions.
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71
The polymerization of DNA is essentially irreversible due to the hydrolysis of pyrophosphate.
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72
In the Sanger method for DNA sequencing,the gels are read from bottom to top,since the smallest fragments travel farthest in the sequencing gel.
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73
The Klenow fragment can be used for DNA sequencing if proteins such as SSB protein are added.
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74
Automated DNA sequencing methods that use amplification clusters of DNA fragments attached to glass slides and a unique DNA polymerase are much faster than the Sanger method.
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75
On average about twenty errors are made during each replication of the human genome.
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76
Sanger DNA sequencing requires inhibition of DNA polymerization by a specific dideoxynucleoside every time it is copied in the replication process.
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77
Eukaryotic cells contain at least five DNA polymerases all of which are responsible for nuclear DNA replication.
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78
Since the rate of fork movement in eukaryotes is much slower than that of bacteria,chromosomes require more than an hour to replicate.
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79
In E.coli replication begins at the origin of replication and proceeds in one direction until the entire circular DNA molecule has been copied.
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80
DNA polymerase III is the largest DNA polymerase in E.coli.
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