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BS209: Molecular Biology > Chapter 6: DNA Replication > Flashcards

Chapter 6: DNA Replication Flashcards

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1
Q

because replication is (), each strand of the parental double helix acts as a template for the synthesis of a new daughter DNA strand

A

semi-conservative

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2
Q

3 phases of DNA synthesis

A
  1. initiation
  2. elongation
  3. termination
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3
Q

replication starts at special sites called ()

A

origins of replication

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4
Q

replication moves away from the origin in (1), forming a (2)

A
  1. both directions
  2. replication bubble
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5
Q

the DNA double helix is opened at the origin of replication and unwound to form ()

A

replication forks

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6
Q

at replication forks, () is exposed and DNA synthesis can occur

A

single-stranded DNA (ssDNA)

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7
Q

for bacteria, the origin of replication is typically a single site called ()

A

ori

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8
Q

in most eukaryotes, the origins of replication are () along the chromosome

A

multiple, not sequence-specific

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9
Q

the origin of replication is recognized by an () that opens up the double helix and recruits helicases

A

initiator protein

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10
Q

() unwind the double helix to expose ssDNA

A

DNA helicases

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11
Q

ssDNA is coated with () to prevent it from forming secondary structures or re-pairing, and to protect it from endonucleases

A

ssDNA binding proteins

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12
Q

DNA synthesis is carried out by (), but it can only add nucleotides to an existing 3’ end

A

DNA polymerase (III)

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13
Q

because DNA polymerase cannot synthesize new strands de novo, DNA synthesis needs a (1) synthesized by an RNA polymerase called (2)

A
  1. primer
  2. primase
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14
Q

DNA polymerase us recruited by the (1) to the DNA at the (2)

A
  1. sliding clamp
  2. primer 3’ terminus
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15
Q

after the RNA primer is synthesized, the (1) loads the (2) onto the DNA template strand

A
  1. clamp loader
  2. sliding clamp
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16
Q

in eukaryotes, the first few nucleotides are synthesized by (1), which forms (2)

A
  1. DNA polymerase alpha
  2. polymerase alpha - primase complex
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17
Q

DNA synthesis on both template strands occurs in a () direction

A

5’ to 3’

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18
Q

due to the 5’ to 3’ direction of DNA synthesis, synthesis is continuous on the (1), but discontinuous on the (2)

A
  1. leading strand
  2. lagging strand
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19
Q

polymerases, helicases, and primases are organized into the () at the replication fork

A

replisome

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20
Q

(dimeric) polymerases on both leading and lagging strands travel () at the replication fork

A

together behind the helicases

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21
Q

termination of DNA synthesis occurs when:

A
  1. 2 different forks meet
  2. the fork reaches the end of a linear chromosome
  3. polymerase meets the previously replicated strand
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22
Q

because RNA primers were used to start DNA synthesis, they must be () during termination

A

removed and replaced with DNA

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23
Q

in the case of discontinuous synthesis or replacement of RNA primers, () connects the adjacent strands

A

DNA ligase

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24
Q

bacterial DNA synthesis termination occurs at a specific region called ()

A

ter

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25
Q

DNA polymerases catalyze the (1) to the (2) in the newly growing strand

A
  1. addition of a new nucleotide
  2. 3’ OH of the last nucleotide
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26
Q

the structure of DNA polymerase resembles a right hand, with 3 domains called:

A
  1. thumb
  2. palm
  3. fingers
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27
Q

ssDNA is fed past the () domain of DNA polymerases

A

fingers

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28
Q

nucleotide addition is catalyzed in the active site in the () domain of DNA polymerase, which forms a cleft into which the growing dsDNA fits

A

palm

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29
Q

the () domain of DNA polymerase holds the elongating dsDNA

A

thumb

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30
Q

DNA polymerase adds nucleotides (dNTPs) to the nascent DNA strand by ()

A

nucleophilic attack

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31
Q

the kind of nucleotides added by DNA polymerase to the growing DNA strand

A

deoxyribonucelotide triphosphates (dNTPs)

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32
Q

the active site of DNA polymerase catalyzes a (1) reaction linking the 5’ phosphate of the incoming dNTP to the 3’ OH of the growing DNA to form a (2)

A
  1. phosphoryl transfer
  2. phosphodiester bond
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33
Q

nucleophilic attack by the nascent chain 3’ OH on the alpha-phosphate of the incoming dNTP releases ()

A

pyrophosphate (PPi)

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34
Q

during the phosphoryl transfer reaction, DNA polymerase uses () in the active site to promote the reaction

A

2 metal ions (Mg2+)

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35
Q

subsequent () drives the phosphoryl transfer reaction in DNA synthesis forward

A

hydrolysis of released pyrophosphate

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36
Q

initiator proteins bind to () regions on the unwound DNA strands to allow helicases to continue unwinding the DNA

A

A-T

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37
Q

why are A-T rich regions easier to unwind than G-C rich areas

A

A is bound to T by only 2 H-bonds, as compared to the 3 H-bonds connecting G and C

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38
Q

the A-T rich region is called a ()

A

DNA unwinding element

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39
Q

for bacteria and a few eukaryotic viruses, the DNA unwinding element consists of ()

A

specific DNA sequence elements

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40
Q

in most eukaryotes, the DNA unwinding element lacks defined sequence elements; instead, the initiator proteins bind at ()

A

many sites

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41
Q

initiator proteins are (), which are associated with a variety of cellular activities

A

AAA+ ATPases

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42
Q

in bacteria, the initiator protein is called ()

A

DnaA

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43
Q

in eukaryotes, the initiator protein is called the ()

A

origin recognition complex (ORC)

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44
Q

the ORC is composed of (1) subunits, labeled (2)

A
  1. 6
  2. Orc1-6
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45
Q

a special case for eukaryotes: in (), ORC binds to a specific DNA sequence

A

S. cerevisiae

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46
Q

the ORC in S. cerevisiae binds to a specific DNA sequence called the ()

A

ARS (autonomously replicating sequence)

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47
Q

the binding of ORC to DNA is likely influenced by the ff.

A
  1. chromatin structure
  2. nucleosomes
  3. sequence-specific DNA binding proteins
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48
Q

origins of replication are activated at random throughout the ()

A

S phase

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49
Q

in E. coli, the origin (called 1) has a 245 bp sequence, with seven 9bp (2) that bind DnaA

A
  1. OriC
  2. DnaA boxes
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50
Q

when bound to ATP, the AAA+ domains of DnaA multimerize into a ()

A

spiral filament

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51
Q

the () between the E. coli DNA and initiator proteins at the origin distorts the DNA, which facilitates unwinding at the adjacent AT-rich region

A

filament interaction

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52
Q

as well as generating ssDNA, DnaA also recruits bacterial () to the origin

A

bacterial DNA helicase DnaB

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53
Q

in E. coli, helicase DnaB is loaded onto the DNA by ()

A

helicase loader DnaC

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54
Q

bacterial helicase is loaded onto (1), while eukaryotic helicase is loaded onto (2)

A
  1. ssDNA
  2. dsDNA
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55
Q

the ORC recruits 2 proteins:

A
  1. Cdc6
  2. Cdt1
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56
Q

the Cdc6 and Cdt1 proteins recruited by the ORC sequentially load 2 ring-shaped () in a head-to-head orientation

A

MCM2-7 hexamers

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57
Q

the MCM2-7 pair loaded by the Cdc6 and Cdt1 proteins is then activated by accessory proteins such as ()

A

Cdc45 and Sld3

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58
Q

other proteins are loaded (including polymerase) to the activated MCM2-7 pair, forming the (1), which can be activated by (2) to unwind DNA and start replication

A
  1. full helicase complex
  2. phosphorylation
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59
Q

the full helicase complex is called the (1) for (2)

A
  1. CMG complex
  2. Cdc45, MCM, GINS
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60
Q

the eukaryotic replicative helicase CMG is a ring composed of ()

A

6 MCM subunits, plus additional factors

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61
Q

the ssDNA binding protein for bacteria is called ()

A

SSB protein

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62
Q

the ssDNA binding for eukaryotes is called ()

A

replication protein A (RPA)

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63
Q

the primer in bacterial DNA synthesis is a ()

A

short piece of RNA

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64
Q

on the lagging strand, the discontinuous synthesis of DNA results in the formation of ()

A

Okazaki fragments

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65
Q

the primer in eukaryotic DNA synthesis is a ()

A

short piece of RNA+DNA

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66
Q

the DNA polymerase in the polymerase alpha-primase complex is not replicative polymerase, so to complete (1), it must add (2) before to the primer allowing replicative polymerases to take over

A
  1. polymerase switching
  2. a short piece of DNA
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67
Q

addition of (1) after primer synthesis is done by the interaction with the sliding clamp, which is loaded at the (2) by the clamp loader

A
  1. replicative DNA polymerase
  2. template-primer junction
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68
Q

in bacteria, the sliding clamp is called

A

beta protein

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69
Q

in eukaryotes, the sliding clamp is called ()

A

PCNA

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70
Q

the high processivity of replicative DNA polymerase is due to the ()

A

sliding clamp that keeps it tethered to the DNA

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71
Q

the clamp loader (called 1 in eukaryotes) is a (2) ring structure, and some clamp-loader subunits are (3)

A
  1. replication factor C, RFC
  2. 5-subunits
  3. AAA+ ATPases
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72
Q

the cycle of sliding clamp loading to DNA is driven by (1) driven about by the binding of (2) and its subsequent hydrolysis

A
  1. conformational changes
  2. ATP
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73
Q

in the absence of ATP, the clamp loader has () for the sliding clamp

A

low affinity

74
Q

the clamp loader-sliding clamp complex has a high affinity with the ()

A

primer-template junction

75
Q

binding of the clamp loader-sliding clamp complex stimulates (), which closes the sliding clamp and releases the clamp loader

A

ATPase activity

76
Q

after the release of the clamp loader, the sliding clamp remains associated with DNA, and recruits DNA polymerase to the primer-template junction via ()

A

short polymerase binding motifs

77
Q

the clamp loader must be () in order to repeat the cycle

A

recharged with ATP

78
Q

polymerase switching in eukaryotes is needed because polymerase alpha is already at the primer 3’ end and must be replaced with replicative polymerase (1) on the lagging strand or (2) on the leading strand

A
  1. delta
  2. epsilon
79
Q

polymerases can be grouped according to the () of the rest of the protein

A

evolutionary lineage

80
Q

polymerases are more similar within groups than within organisms, suggesting ()

A

early divergence of polymerase groups

81
Q

DNA polymerases that copy RNA into DNA

A

reverse transcriptases

82
Q

reverse transcriptases are typically encoded by viruses and by DNA elements in eukaryotes called ()

A

retrotransposons

83
Q

a specialized reverse transcriptase that is needed to synthesize chromosome ends

A

telomerase

84
Q

3 quality control steps of DNA replication

A
  1. low polymerase error rate
  2. proofreading by the polymerase
  3. mismatch repair
85
Q

during addition, the polymerase recognizes the correct nucleotide due to the () when base-paired with the template strand

A

precise fit in the active site

86
Q

proofreading by the replicative polymerase is achieved by (), where an incorrect nucleotide is removed before the next nucleotide is added

A

3’ to 5’ exonuclease activity

87
Q

in the event of the addition of an incorrect nucleotide to the newly-synthesized strand, it is moved to the () and removed

A

exonuclease site

88
Q

replicative polymerases stay attached for many 1000s of nucleotides; this synthesis is ()

A

processive

89
Q

in (), Okazaki fragments are joined after synthesis; this process is different in eukaryotes and bacteria

A

Okazaki fragment maturation

90
Q

exonucleases can only cut from

A

a strand end

91
Q

in E. coli, () disassociates from the DNA when it reaches the next primer

A

DNA polymerase III

92
Q

in E. coli, () removes the RNA primer and then fills the gap with DNA; however, it leaves a nick in the DNA because it cannot attach the end of the synthesized DNA to the next DNA fragment

A

DNA polymerase I

93
Q

() seals the DNA nick in E. coli

A

DNA ligase

94
Q

DNA polymerase I in E. coli is able to remove the RNA primer due to () activity

A

5’ to 3’ exonuclease

95
Q

in eukaryotes. () continues to make DNA when it meets the RNA primer

A

DNA polymerase delta

96
Q

the flap created by DNA polymerase delta in eukaryotic Okazaki fragment maturation is cleaved by ()

A

Flap endonuclease 1 (Fen1)

97
Q

Endonucleases can cut and digest DNA strands from the ()

A

middle

98
Q

in eukaryotes, if Fen1 doesn’t cleave the DNA, a long flap forms, which is then cleaved by ()

A

Dna2

99
Q

the bacterial replisome has 2 copies of DNA polymerase III, linked by (), which associates with the clamp loader and helicase

A

tau

100
Q

leading and lagging strand synthesis in eukaryotes is coordinated at the replication fork by ()

A

Ctf4

101
Q

Ctf4 in eukaryotes couples (1), (2) and (3) at the fork

A
  1. helicase CMG
  2. polymerase epsilon
  3. polymerase alpha-primase
102
Q

there is no eukaryotic equivalent of bacterial tau protein; instead, eukaryotic DNA strands are coupled through ()

A

Ctf4-polymerase alpha-primase hub

103
Q

unwinding of DNA introduces torsional stress, resulting in (1) ahead of the fork = (2)

A
  1. overwinding
  2. positive supercoiling
104
Q

substrate of a topoisomerase and the product are chemically identical, but ()

A

topologically different

105
Q

replication of bacterial circular chromosomes results in daughter molecules that are still interlinked - ()

A

catenated

106
Q

bacterial circular chromosomes are unlinked (decatenated) by ()

A

topoisomerases (type1A or type II)

107
Q

() topoisomerases break 1 DNA strand and do not require ATP

A

type I

108
Q

type I topoisomerases have an active site () that attacks the phosphodiester bond of one strand in duplex DNA

A

tyrosine-OH

109
Q

the attacking of topoisomerase I results in a transient covalent phosphodiester linkage between ()

A

tyrosine and 5’ or 3’ end of cleaved strand

110
Q

type IA topoisomerases use the () mechanism to rebind cut strands

A

strand passage

111
Q

how to type IB topoisomerases rebind the cut strands

A

allow the free end to rotate to release supercoils

112
Q

() topoisomerases are dimeric and break both DNA strands and do require ATP

A

type II

113
Q

type II topoisomerases use the () mechanism to rebind cut strands

A

strand passage

114
Q

an unusual type II topoisomerase that introduces negative supercoils -> makes strand unwinding ahead of fork more favorable

A

bacterial DNA gyrase

115
Q

ter sites in bacterial DNA are bound by the (), which is orientation specific and allows passage in only 1 direction

A

Tus proteins

116
Q

bacterial replication forks meet in a termination zone that has ()

A

10 ter sites

117
Q

initiation of replication in bacteria is tightly regulated by factors such as (1) and (2)

A
  1. cell size
  2. nutritional state
118
Q

primary methods of regulating initiation of replication in E. coli

A
  1. regulatory inactivation of DnaA (RIDA)
  2. initiation at oriC by methylation
  3. reduction of available DnaA amount by sequestration
  4. recycling DnaA-ADP to DnaA-ATP
119
Q

after initiation, the () protein stimulates hydrolysis of DnaA-ATP to DnaA-ADP; used in RIDA

A

Hda

120
Q

main consequence of RIDA

A

DnaA-ADP dissociated from oriC and cannot reinitiate firing -> prevents overstimulation of replication initiation

121
Q

major regulatory step of preventing initiation in E. coli

A

regulatory inactivation of DnaA

122
Q

enzyme that methylates the A residue in the sequence GATC

A

DNA adenine methylase (Dam)

123
Q

action of Dam results in () of DNA, where only 1 of the 2 strands has a methyl tag

A

hemi-methylation

124
Q

protein that binds to hemi-methylated GATC sites

A

SeqA

125
Q

main consequence of SeqA binding to GATC sites

A

SeqA binding temporarily blocks binding of methylase and thus prevents complete methylation of DNA -> prevents DnaA binding for replication initiation temporarily

126
Q

DNA region in E. coli that lies close to oriC and where (about 25% of) DnaA can also bind

A

datA DNA region

127
Q

DNA regions far away from oriC; serve as cofactors to stimulate the exchange of ADP to ATP on DnaA

A

DARS regions (DnaA reactivating sequences)

128
Q

main difference between bacterial initiation and eukaryotic initiation (of replication)

A
  1. origins of bacteria can fire initiation more than once in the cell cycle
  2. replication origins in eukaryotes must only fire a maximum of only once per cell cycle
129
Q

initiation at each oriC in a bacterial cell is (1) - therefore the number of oriC in a cell is always a power of (2)

A
  1. synchronous
  2. two
130
Q

selection of origins in G1 of eukaryotic cell cycle

A

origin licensing

131
Q

in the (1), eukaryotic helicases are activated to initiate origin unwinding; once fired, origins cannot be reused until next (2)

A
  1. S phase
  2. G1 phase
132
Q

() are proteins involved in restricting initiation steps and origin firing to the appropriate timing in eukaryotes

A

cell cycle kinases

133
Q

cell cycle kinases mainly work by () target proteins

A

phosphorylating

134
Q

how do yeast cells prevent helicase loading outside of G1 phase

A

regulation is achieved by exporting proteins from the nucleus

135
Q

how do metazoa prevent helicase loading outside of G1 phase

A
  1. proteolysis of replication complex proteins after S-phase
  2. binding of Geminin to Cdt1
136
Q

how does Geminin binding to Cdt1 regulate initiation in metazoa

A

when bound to Geminin, Cdt1 cannot load MCM to origin of replication

137
Q

related factors that control whether or not a pre-replication complex will fire in the cell cycle

A
  1. origin efficiency
  2. origin timing
138
Q

probability that an origin will fire

A

origin efficiency

139
Q

timing (earlier or later) of firing compared to other origins

A

origin timing

140
Q

most origin firing in eukaryotes is thought to be largely stochastic, with evidence of () as a limiting factor

A

replication protein availability

141
Q

what is the end-replication problem

A

mechanism for lagging strand synthesis cannot replicate the very end of a linear chromosome

142
Q

consequence of the end-replication problem

A

substantial portions of chromosome ends could be lost over several rounds of replication

143
Q

some causes of chromosome portion loss due to end-replication problem

A
  1. RNA primer removal
  2. dissolution of the replication fork
144
Q

() leaves a gap which cannot be filled in at the end of linear chromosomes

A

RNA primer removal

145
Q

many eukaryotes use () to circumvent the end-replication problem

A

telomeres

146
Q

in () several copies of the linear chromosome are stuck together before replication, reducing the number of ends -> way to avoid end-replication problem

A

T4 viruses

147
Q

in (), the two DNA strands are joined at the end, making a hairpin so replication can just continue around the loop -> way to avoid end-replication problem

A

vaccinia viruses

148
Q

in (), replication is primed from a terminal protein -> way to avoid end-replication problem

A

bacteriophage phi29

149
Q

telomeres are (); their ends get shorter after each round of replication

A

simple sequence repeats

150
Q

telomeres can be elongated by adding () at the ends

A

new specific telomere sequences

151
Q

in Drosophila, () transpose onto chromosome ends to maintain length

A

transposable elements

152
Q

telomere length in most eukaryotes in maintained by (); they add telomere DNA sequences to the chromosome ends

A

telomerase

153
Q

telomerase synthesizes one strand of the telomere - the (1) strand - at the (2) end

A
  1. G-rich
  2. 3’
154
Q

if telomerase synthesizes the G-rich strand, the complementary C-rich strand is filled in by ()

A

DNA polymerase alpha

155
Q

the action of telomerase results in (), which counterbalances loss due to incomplete end-replication

A

net elongation of telomere sequence

156
Q

telomerase is a unique polymerase made of a catalytic protein (1) and an (2)

A
  1. reverse transcriptase
  2. integral bound RNA molecule
157
Q

the catalytic protein in telomerase is specifically called (); it is well-conserved in eukaryotes

A

telomerase reverse transcriptase (TERT)

158
Q

role of integral bound RNA molecule in telomerase

A
  1. provides template for synthesis of telomere repeats
  2. involved in enzyme activity
159
Q

it is not yet understood how the distribution of telomere length in a cell is regulated, but () are thought to be involved

A

telomere binding proteins

160
Q

2 factors that help establish equilibrium of telomere length

A
  1. number of telomeres elongated in a cell cycle
  2. number of repeats added in a single binding event
161
Q

mutations in telomere binding proteins can lead to ()

A

aberrant telomere lengthening

162
Q

short telomeres induces the () response

A

DNA damage

163
Q

loss of telomerase or other factors can lead to ()

A

progressive telomere shortening

164
Q

mutations in telomerase cause inherited degenerative diseases called ()

A

telomere syndromes

165
Q

telomerase is the main way that cells maintain telomeres, but some cells can used specialized recombination called ()

A

gene conversion

166
Q

gene conversion takes advantage of the () at telomere ends

A

repetitive sequences

167
Q

in gene conversion, the 3’ end of one chromosome can () and provide a template for telomere elongation by DNA polymerase

A

invade another chromosome

168
Q

chromatin consists of DNA plus specific proteins called (), which are bound to DNA

A

histone proteins

169
Q

correct recruitment of histones and maintenance of chemical modifications (e.g. methylation) form the basis for ()

A

epigenetic inheritance

170
Q

a process where the parental H3 and H4 histones are divided among the daughter cell (each get 1/2) -> allows histone modification state to be inherited by both cells

A

parental histone segregation

171
Q

after parental histone segregation, new () are deposited and modified in line with the old histones

A

H3 and H4 histones

172
Q

copying chemical “decorations” of the parent strand is important for eukaryotes because these help ()

A

dictate which genes to express in certain cells

173
Q

re-assemble histones on new daughter strands after replication fork disrupts nucleosomes

A

histone chaperone proteins

174
Q

proteins that help to disrupt the nucleosome ahead of the fork and transfer the old H3/H4 from ahead of the fork to behind the fork

A
  1. FACT (histone chaperone)
  2. ASF protein
175
Q

as daughter strands are synthesized, new () histones are loaded to make half a nucleosome

A

H2A and H2B

176
Q

new H3/H4 histones are assembled by () and then added to the initial half of replicated nucleosome (composed of H2A/B histones)

A

CAF1 protein

177
Q

because synthesis of the main histone subunits is tightly coupled to DNA synthesis, they are called (1) or (2)

A
  1. replication-dependent histones
  2. S phase histones
178
Q

newly synthesized H3 and H4 histones are acetylated at specific (1, AA) in their (2) tails

A
  1. lysine
  2. N-terminal
179
Q

deposition of new acetylated H3/4 histones on newly replicated DNA is performed by (1, 2 proteins) plus other histone chaperones

A

CAF1 and ASF1

180
Q

() binds strongly to H3/H4 and to the sliding clamp to target H3/H4 to DNA behind the fork

A

CAF1

181
Q

() binds half a nucleosome tetramer (2 H3 and 2 H4) and deposits this on DNA with the help of CAF1

A

ASF1

BS209: Molecular Biology (8 decks)

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