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Molecular Biology > DNA replication, repair and recombination 1 (lecture 3) > Flashcards

DNA replication, repair and recombination 1 (lecture 3) Flashcards

1
Q

their are approximately ___ new single-nucleotide mutations in the offspring’s grem-line when compared to parental germ-line

A

70

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2
Q

Mutation rate of one nucleotide change per ______ nucleotides per generation

A

10^8

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3
Q

The human mutation rate is very to the E. coli rate of one change per ____ nucleotides per cell generation

A

10^10

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4
Q

Most DNA mutations are corrected by ____ and _____

A

proofreading and DNA repair

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5
Q

besides from proofreading and DNA repair errors can be further corrected by _______

A

post-replication repair mechanisms

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6
Q

Multicellular organisms need _______ _____ replication

A

high fidelity

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7
Q

Germ cells have to have low mutation rates to

A

Maintain species

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8
Q

Somatic cells need low mutation rates to

A

avoid uncontrolled proliferation/cancer

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9
Q

DNA duplication rates are as high as ______ nucleotides per second

A

1000

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10
Q

DNA polymerase synthesizes DNA by catalyzing what reaction

A

(DNA)n residues + dNTP (deoxyribonucleoside triphosphate aka primer strand) —-> (DNA) n+residues + P2O7^4-

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11
Q

DNA replication requires separation of

A

the two parental strands

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12
Q

DNA replication requires

A

dATP, dGTP, dCTP, and dTTP

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13
Q

DNA polymerase requires a primer with a ____ to begin

A

free 3’ -OH (hydroxyl end)

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14
Q

The newly synthesized DNA strand polymerizes in the _____ direction

A

5’ to 3’

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15
Q

DNA synthesis is catalyzed by DNA polymerase and the reaction is driven by

A

a large, favorable fee-energy change, caused by the release of pyrophosphate and its subsequent hydrolysis to two molecules of inorganic phosphate

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16
Q

The proper base-pair geometry of a correct incoming deoxyribonucleoside triphosphate causes the polymerase to ____ around the base pair, thereby initiating the

A

Tighten around the base pair, thereby initiating the nucleotide addition reaction

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17
Q

During synthesis of DNA the dissociation of pyrophosphate ______ the polymerase, allowing

A

loosens the polymerase, allowing translocation of the DNA by one nucleotide so the active site of the polymerase is ready to receive the next deoxyribonucleoside triphosphate

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18
Q

Is the replication fork asymmetric

A

yes

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19
Q

Both strands are _____ replicated

A

simultaneously

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20
Q

DNA polymerase can only synthesize DNA in the _______ direction

A

5’-to-3’

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21
Q

The ___ strand is synthesized continuously

A

leading

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22
Q

The ____ strand is synthesized in segments

A

laggins

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23
Q

Okazaki fragment are polymerized only in the _____ direction

A

5’-to-3’

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24
Q

DNA polymerase makes a mistake out of every ____ nucleotides copied

A

10^9

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25
Q

What is the first step in DNA polymerase proofreading

A

Before the nucleotide is covalently added to the growing chain. The correct nucleotide has a higher affinity for the moving polymerase than does the incorrect nucleotide, because the correct pairing is more energetically favorable. moreover, after nucleotide binding but before the nucleotide is covalently added to the growing chain the enzyme must tighten around the active site. it is slower when adding incorrect base pairs thus incorrect base pairs are harder to add and may diffuse away before the polymerase can mistakenly add them.

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26
Q

What is the second step in DNA polymerase proofreading

A

Exonucleolytic proofreading

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27
Q

When does exonucleolytic proofreading take place

A

immediately after incorrect base is added

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28
Q

3’-to-5’ exonuclease clips off unpaired residues at the ____ primer terminus

A

3’

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29
Q

If a rare tautomeric form of C (C) happens to base-pair with A and is thereby incorporated by DNA polymerase into the primer strand. Then rapid tautomeric shift of C to normal cytosine (C) destroys its base-pairing with A. The unpaired 3’- OH end of primer blocks further elongation of primer strand by DNA polymerase. How does 3’-to-5’ exonuclease fix this

A

3’-to-5’ exonuclease activity attached to DNA polymerase chews back to create a base-paired 3’-OH end on the primer strand. DNA polymerase continues the process offering nucleotides to the base-paired 3’-OH end of the primer strand

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30
Q

If there were DNA polymerases that added deoxyribonucleoside triphosphates in the 3’-5’ direction, the growing 5’ end of the chain, rather than the incoming mono nucleotide, would have to

A

provide the activating triphosphate needed for the covalent linkage.

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31
Q

Lagging strand is replicated through ______ process

A

backstitching process

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32
Q

Before the backstitching process can begin ______ synthesizes approximately 10 nucleotide long RNA primer to prime DNA synthesis

A

DNA primase

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33
Q

DNA polymerase can’t initiate ______- this would increase the mutation rate

A

de novo synthesis

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34
Q

RNA primer is erased by ______ (recognizes RNA/DNA hybrids) and replaced with DNA; ____ joins the ends

A

RNAseH, DNA ligase

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35
Q

unlike DNA polymerase, DNA primes can

A

start a new polynucleotide chain by joining two nucleoside triphosphate together

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36
Q

_____ unwinds DNA

A

DNA helicase

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37
Q

DNA helicase has _____ identical subunits that binds and hydrolyzes ____

A

6, ATP

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38
Q

DNA helicase has 6 identical subunits that ____ and ____ ATP

A

bind and hydrolyze

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39
Q

DNA helicase is capable of prying apart the helix at rates of _____ nucleotide pairs/second

A

1000

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40
Q

the binding and hydrolization of ATP by DNA helicase causes

A

conformational change that propels it like a rotary engine along single stranded DNA, passing it through a center hole

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41
Q

DNA ligase seals a broken ____ bond. DNA ligase uses a molecule of ATP to activate the ___ end at the nick

A

Phosphodiester bond, 5’

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42
Q

_____ bind tightly and cooperatively to exposed Single stranded DNA at the replication fork

A

Single-Stranded DNA binding proteins

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43
Q

Single-stranded binding proteins functions

A
  • Help stabilize unwound DNA
  • Prevent formation of hairpins
  • DNA bases remain exposed
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44
Q

cooperative binding of single-stranded DNA binding proteins _______ regions of the DNA chain

A

straighten

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45
Q

Keeps DNA polymerase on DNA when moving; releases when double stranded DNA is encountered

A

Sliding clamp

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46
Q

Proteins at the DNA replication fork

A
  • Single-stranded DNA binding proteins
  • Sliding clamp
  • clamp loader
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47
Q

function of sliding clamp

A
  • keeps DNA polymerase on DNA when moving; releases when double stranded DNA is encountered
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48
Q

hydrolyzes ATP as it loads the sliding clamp onto a primer-template junction

A

Clamp loader

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49
Q

Assembly of sliding clamp requires a

A

clamp loader

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50
Q

The sliding clamp releases when

A

double stranded DNA is encountered

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51
Q

On the leading strand the sliding clamp

A

remains associated to DNA polymerase for long stretches

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52
Q

on the lagging strand the clamp loader stays close so it can assemble a new clamp at start of

A

each new okazaki fragment

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53
Q

Removes (almost all) errors missed by proofreading by detecting distortion caused by mispairing

A

Mismatch Repair

54
Q

in mismatch repair ____ binds to mismatch

A

MutS

55
Q

in mismatch Repair ____ scans for the nick and triggers degradation of nicked strand

A

MutL

56
Q

In mismatch Repair what allows one to distinguish which strand is correct

A
  • in E. coli it depends on methylation to distinguish the new strand
  • In humans, depends on single strand breaks
    • present on lagging strand before Okazaki fragments are ligated
    • leading strand not known
57
Q

Mutations in mismatch repair gene lead to

A
  • cells accumulate mutations at high rates

- Example is hereditary nonpolyposis colon cancer (HNPCC)

58
Q

Paul Modrich won the Noble prize for work with _____ an enzyme that aded methyl groups to DNA

A

Dam methylase

59
Q

Matthew Meselson collaborated with Paul Modrich and created a virus with mismatches in its sequence. When the virus infected bacteria, the bacteria repaired the mismatch. When the viral DNA was methylated,

A

the bacteria only repaired the strand that was not methylated

60
Q

As the replication fork moves, it creates a ______ problem for the parental helix

A

winding

61
Q

For every ____ Base pairs replicated there is one turn in the parental helix

A

10

62
Q

_____ is a reversible enzyme that breaks a phosphodiester bond to change superhelicity, thereby relieving supercoiling

A

DNA topoisomerase

63
Q

Type I topoisomerase

A

catalyze the relaxation of supercoiled DNA, a thermodynamically favorable process
work by creating transient single strand break in DNA which allows the DNA on either side of the nick to rate freely relative to each other; uses the other phosphodiester bond as a swivel point

64
Q

Topoisomerases catalyze the relaxation of supercoiled DNA, a ______ process

A

thermodynamically favorable

65
Q

The nick in phosphodiester linkage by Topoisomerase is resealed rapidly and doesn’t require any additional energy since energy is stored in the ______

A

phosphotyrosine linkage

66
Q

Type I DNA topoisomerase have ____ at the active site

A

Tyrosine

67
Q

DNA topoisomerase covalently attaches to a DNA ____ thereby breaking a phosphodiester linkage in one DNA strand

A

phosphate

68
Q

Type II Topoisomerase makes a transient ______ break in the DNA

A

double-stranded

69
Q

Unlike Type I Topoisomerases Type II topoisomerases

A

hydrolyze ATP, which is needed to release and reset the enzyme after each cycle

70
Q

Why are Type II topoisomerases been effective targets for anticancer drugs

A

because Type II topoisomerases are largely confined to proliferating cells in eukaryotes (note that some these drugs inhibit it after it makes a double stranded break thus killing the cells. )

71
Q

Type ____ topoisomerase can prevent tangling problems that would arise during DNA replication

A

II

72
Q

Type II topoisomerase are activated at sites on chromosome where

A

two double-stranded helices cross each other

73
Q

Type II topoisomerase use ATP to

A
  • break one double-stranded helix reversibly to create “gate”
  • Causes second strand to pass through
  • Reseals break and dissociates
74
Q

_____ can separate “decaffeinate” 2 interlocked DNA circles

A

Type II Topoisomerase

75
Q

DNA replication origins are ____ rich regions where sequence attracts initiator proteins to pry open DNA

A

A-T

76
Q

____ is the only point of control for E. coli, so it’s highly regulated

A

initiation

77
Q

Initiation is the only point of control of control in E. coli, so it’s highly regulated and proceeds only when

A

sufficient nutrients are present

78
Q

E. coli are able to ensure only one round of DNA replication occurs for each cell division by

A
  • after initiated, the initiator protein is inactivated by hydrolysis of its bound ATP molecule, and the origin of replication experiences a “refractory period”.
  • The refractory period is caused by a delay in the methylation of newly incorporated A nucleotides in the origin
79
Q

Bacterial genomes are able to replicate in about ____ minutes

A

40

80
Q

Traveling at 50 nt/sec, it would take _____ hours to do an average eukaryotic chromosome with a single origin of replication

A

800

81
Q

Steps of initiation of DNA replication in bacteria

A
  • initiator proteins bind to specific sites in ORI, forming complex
  • complex attracts DNA helicase + helicase loader
  • Helicase is placed around a SS DNA exposed by assembly of complex
  • Helicase loader remains engaged until helicase is properly loaded
  • Helicase unwinds DNA so primes can make RNA primer on leading strand; remaining proteins assemble to create 2 replication forks with complexes moving in opposite direction with respect to ORI
82
Q

Eukaryotic DNA Replication occurs during _____ phase which lasts about ____ hrs for mammalian cells

A

DNA synthesis phase (S), about 8 hours

83
Q

Eukaryotic DNA replication is activated in ____ consisting of _____ origins

A

clusters (replication units), 20-80 origins

84
Q

different regions of each eukaryotic chromosome are replicated in a reproducible order during S phase, depending on _______. ____ is late-replicating

A

chromatin structure, Heterochromatin (timing is related to packing of DNA in chromatin)

85
Q

minimum requirements for sequence to be ORI (Yeast)

A
  • must have binding site for ORC (origin recognition complex)
  • must have an A-T rich stretch for easy unwinding
  • Must have binding site for proteins that help attract ORC
86
Q

DNA replication in eukaryotes occurs _____ in contrast to bacteria in which occurs

A

DNA replication in eukaryotes only occurs during S phase in contrast to bacteria who when growing rapidly replicated DNA continuously

87
Q

The eukaryotic McM helicase moves along the _______ template, whereas the bacterial helicase moves along the _____ template

A

leading strand, lagging-strand

88
Q

During eukaryotic DNA replication the replicated helicases are loaded onto DNA next to ORC to create the pre replicative complex (helicase, helices loading proteins, Cdc6 and Cdt1) in what phase of the cell cycle

A

G1

89
Q

Regulation of DNA replication

A
  • pre replicative complex must be formed in G1 phase
  • activated Cdks lead to:
    • dissociation of helicase loading proteins
    • activation of helicase (phosphorylated)
    • unwinding of DNA
    • loading of DNA polymerase
    • phosphorylates ORC and thus makes it inactive
  • preven assembly of new ORC until next M phase resets cycle
    • single change to form in G1 when Cdk activity is low
    • second window for pre-replicative complexes to be activated and disassembled in S phase when Cdks activity is high
90
Q

______ structures is important for mammalian ORIs

A

Chromatin

91
Q

Specific human sequences have been identified that can serve as ORIs; they are ______ of nucleotide pairs in length

A

1000s

92
Q

ORIs in humans can still function if moved to a different locals if placed where

A

chromatin is uncondensed

93
Q

____ DNA in beta-globin cluster is required for expression of the genes in the cluster

A

distant

94
Q

Replication requires not only DNA replication but synthesis and assembly of

A

new proteins

95
Q

New ______ assemble behind replication fork

A

Nucleosomes

96
Q

Eukaryotes have ____ copies of genes for each histone

A

multiple

97
Q

Histone proteins are synthesized mainly in ____ phase; amount made is ____ regulated to meet requirements

A

S phase; highly

98
Q

For efficient replication _______ are needed to destabilize DNA-histone interface

A

chromatin-remodeling proteins

99
Q

As replication fork passes through chromatin, histone octamer breaks into

A
  • H3-H4 tetramer, distributed randomly to daughter duplexes

- 2 H2A-H2B dimers which are released from the DNA

100
Q

The reassembly of histones requires ______

A

histone chaperones (NAP1 and CAF1)

101
Q

Explain reassembly of histones behind the replication fork

A
  • Freshly made H3-H4 tetramers are added to the newly synthesized DNA to fill in the “spaces” and H2A-H2B dimers (half of which are old and half new) are then added at random to complete the nucleosomes
102
Q

The length of each okazaki fragment is determined by the point at which DNA polymerase-gamma is blocked by a ______.

A

newly formed nucleosome

103
Q

The histone chaperones, along with their cargoes, are directed to newly replicated DNA through a specific interaction with the

A

eukaryotic sliding clamp called PCNA

104
Q

Patterns of histone modification can be

A

inherited

105
Q

some daughter nucleosomes contain only ____ histones or only ___ ones, but most are

A

parental, new, but most are hybrids of old and new

106
Q

Parental patterns of histone modification are spread through

A

reader-writer complexes

107
Q

patterns of histone modification may be responsible for some types of ____ inheritance

A

epigenetic

108
Q

Function of NAP-1

A

histone chaperone that loads H2A-H2B Dimer to form nucleosome

109
Q

Function of CAF-1

A

histone chaperone that loads newly synthesized H3-H4 tetramer to fill in spaces and form nucleosomes in newly replicated DNA

110
Q

______ replicates chromosome ends

A

Telomerase

111
Q

End replication problem on ____ strand : because ____

A

lagging strand, the final RNA primer synthesized on the lagging-strand template cannot be replaced by DNA because there is no 3’-OH end available for the repair polymerase

112
Q

Why do bacteria not need telomeres

A

they have circular genomes

113
Q

without Telomeres

A

DNA would be lost form the ends of all chromosomes each time a cell divides

114
Q

In humans, the telomere sequence is a repeating unit of _______, that is reacted roughly a _______ times at each chromosome

A

GGGTTA, thousand

115
Q

Telomere DNA sequences are recognized by sequence-specific DNA-binding proteins that attract an enzyme, called _____, that replenishes these sequences each time a cell divides

A

Telomerase

116
Q

Telomerase replenishes telomeres by elongating parental strand in ______ direction using RNA template on the enzyme

A

5’-to-3’

117
Q

Telomerase resembles reverse transcriptase but it

A

also carries its own RNA template that it then converts to DNA

118
Q

After extension of parental strand by telomerase, replication of lagging strand can be completed by ____, using extension as template

A

DNA polymerase

119
Q

How are the telomeres protected from being rapidly repaired if accidentally broken

A
  • A specialized nuclease chews back the 5’ end of a telomere leaving a protruding single-strand end. this protruding end- in combination with the GGGGTTA repeats in telomeres- attracts a group of proteins that form a protective chromosome cap known as shelterin.
  • When human telomeres artificially are cross-linked they form t-loops. T-loops are regulated by shelterin and provide additional protection form the ends of chromosomes
120
Q

Function of shelterin

A
  • hides telomeres from the cell’s damage detectors that continually monitor DNA
  • regulate t-loops, which provide additional protection to the ends of chromosomes
121
Q

After extension of parental strand by telomerase replication of lagging strand can be completed by DNA polymerase using extension as template. This mechanism, plus a _______ ensures 3’ end is longer, leaving a protruding SS end that loops back and tucks into the repeat (AKA t-loop)

A

5’ nuclease

122
Q

T- loops function

A

structures that protect ends and distinguishes them from broken ones that need to be repaired

123
Q

Protective chromosome cap made up of proteins

A

Shelterin

124
Q

Our somatic cells are born with full complement of ____ repeats while stem cells retain full _____ activity

A

telomere, telomerase

125
Q

each chromosome end in a given cell contains _____ number of telomere repeated depending on age

A

variable

126
Q

except for in some stem cells that retain full telomerase activity Telomere repeats are lost with each generation due to _____

A

insufficient telomerase activity

127
Q

Replicative senescence is what

A

After many generations, daughter cells will have defective chromosomes (lacking telomere function) and stop dividing; in this way the cell’s lifetime is regulated to guard against cancer

128
Q

Human fibroblasts normally divide ___ times before undergoing replicative senescence

A

60

129
Q

_____ may be responsible for aging in animals

A

replicative senescence

130
Q

Transgenic mice that lack ____ develop progressive defects in highly proliferative tissues. they also show ______ and are prone to ___

A

telomerase, premature aging, prone to cancer

131
Q

humans with Dyskeratosis congenital carry a mutant and thus develop ______ and die or _____

A
  • mutant telomerase RNA gene
  • thus develop prematurely shortened telomeres
  • die of progressive bone marrow failure
132
Q

why is it risky for an organism to control cell proliferation with telomeres

A
  • not all cells stop dividing thus gives rise to variant cells that lead to cancer

Molecular Biology (14 decks)

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