Chapter 25 Exam 2 Flashcards

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1
Q

How are bacterial genes named?

A

Genes typically named using three italicized lowercase letters reflecting a function
examples: dna, uvr, and rec

capital letters added to abbreviation reflect order of discovery, not enzymatic order
examples: dnaA, dnaB, and dnaQ

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2
Q

How are bacterial proteins named?

A

Proteins often named after their genes using nonitalicized, roman type with the first letter capitalized
examples: DnaA (encoded by dnaA) and RecA (encoded by recA)

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3
Q

How are eukaryotic genes named?

A

*no single convention exists for all eukaryotic systems
*in Saccharomyces cerevisiae, gene names are three italicized uppercase letters followed by an italicized number
example: COX1

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4
Q

How are eukaryotic proteins named?

A

Protein naming is complex and variable

in yeast, some proteins have long common names
example: cytochrome oxidase

other yeast proteins have the same name as the gene, with one uppercase and two lowercase letters in roman type, followed by a number and the letter “p”
example: Rad51p (encoded by RAD51)

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5
Q

How is DNA replication semiconservative?

A

semiconservative replication = each DNA strand serves as a template for the synthesis of a new strand
produces two new DNA molecules, each with one new strand and one old strand
established by Meselson and Stahl in 1957

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6
Q

What is a replication fork?

A

replication forks = dynamic points where parent DNA is being unwound and separated strands replicated

both DNA strands are replicated simultaneously

both ends of the bacterial chromosome have active replication forks (bidirectional replication)

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7
Q

What is a denaturation map and what does it show?

A

denaturation mapping = selective denaturing of sequences unusually rich in A=T base pairs to provide landmarks along the DNA molecule
generates a reproducible pattern of single-strand bubbles

origin = location where replication loops are initiated
*It shows that replication loops always initiate at an origin

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8
Q

How does DNA synthesis proceed?

A

DNA Synthesis Proceeds in a 5′→3′ Direction and is Semi-discontinuous
*new DNA strands are always synthesized in the 5′ → 3′ direction
the free 3′-OH serves as the point of elongation

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9
Q

What are Okazaki fragments?

A

Okazaki fragments = short DNA fragments that are synthesized in the replication of one of the new DNA strands
~150 to 200 nucleotides long in eukaryotes
~1,000 to 2,000 nucleotides long in bacteria
spliced together by DNA ligase

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10
Q

What is a leading strand?

A

leading strand = 5′ → 3′ synthesis proceeds in the same direction that the replication fork moves
synthesized continuously

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11
Q

What is a lagging strand?

A

lagging strand = 5′ → 3′ synthesis proceeds in the opposite direction that the replication fork moves
synthesized discontinuously through the synthesis of Okazaki fragments
*new DNA strands are always synthesized in the 5′→3′ direction
*the template is read in the 3′→5′ direction

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12
Q

What are nucleases?

A

nucleases = enzymes that degrade nucleic acids
DNases = specifically degrade DNA

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13
Q

What are exonucleases?

A

exonucleases = degrade nucleic acids from one end of the molecule
many operate in only one direction (either 5′→3′ or 3′→5′)

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14
Q

What are endonucleases?

A

endonucleases = degrade nucleic acids at specific internal sites in the molecule

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15
Q

What is DNA polymerase?

A

DNA polymerase = complex enzymes that can synthesize DNA
-many have additional activities
-E. coli cells have at least 5 DNA polymerases, including DNA polymerase I

*the general reaction for DNA polymerization is:

(dNMP)n + dNTP → (dNMP)n+1 + PPi

where dNMP and dNTP are a deoxynucleoside 5′-monophosphate and 5′-triphosphate, respectively

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16
Q

What is the DNA polymerase reaction?

A

the reaction is a phosphoryl group transfer

the 3′-OH of the nucleotide at the 3′ end of the strand (the nucleophile) attacks the α phosphorus of the incoming dNTP

one Mg2+ ion helps deprotonate the 3′-OH group
makes it a more effective nucleophile

Other Mg2+ ion binds the incoming dNTP and facilitates departure of PPi

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17
Q

What does DNA polymerase require?

A

DNA Polymerases Require a Template and a Primer

polymerization is guided by a template DNA strand according to Watson and Crick base-pairing rules

primer = a strand segment with a free 3′-OH group to which a nucleotide can be added
must be complementary to the template
many are RNA oligonucleotides

primer terminus = the free 3′ end of the primer

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18
Q
A
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19
Q

What is an insertion site?

A

= the portion of the active site where the incoming nucleotide binds
One of the active sites of DNA polymerase

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20
Q

What is the postinsertion site?

A

the portion of the active site where the new base pair resides when the polymerase slides forward
One of the active sites of DNA polymerase

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21
Q

What is processivity in regards to DNA polymerases?

A

processivity = the average number of nucleotides added before a polymerase dissociates

DNA polymerases vary greatly in processivity.
ranges from a few nucleotides to many thousand

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22
Q

How accurate is replication?

A

errors in E. coli occur every 1 in 109 to 1 in 1010 nucleotides added
*equates to an error only once per 1,000 to 10,000 replications
*the active site of DNA polymerase excludes base pairs with the incorrect geometry
*DNA polymerases insert one incorrect nucleotide for every 104 to 105 correct nucleotides
-may occur because a base is in a tautomeric form

Contribution of Base-Pair Geometry to the Fidelity of DNA Replication

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23
Q

How does error correction happen in DNA replication?

A

Error Correction by DNA Polymerase I
translocation of the enzyme is inhibited when an incorrect nucleotide is added

many DNA polymerases have intrinsic 3′→5′ exonuclease proofreading activity
permits the enzyme to remove a newly added nucleotide

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24
Q

What is proofreading?

A

proofreading = DNA polymerase activity that involves replacement of the incorrect nucleotide
requires the expenditure of three high-energy bonds

proofreading improves the accuracy 102 - to 103-fold

one error in every 106 to 108 bases remains after base selection and proofreading
the higher measured accuracy of E. coli replication comes from mismatch repair

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25
Q

What is DNA polymerase I in E.coli and what is its function?

A

polA, proofreading, has 3-5 prime and 5-3 prime exonucleases (1 subunit)

DNA polymerase I is abundant, but insufficient for replication of the E. coli chromosome
rate (600 nucleotides/min) is slower than observed for replication fork movement
low processivity
the primary function of DNA polymerase I is cleanup during replication, recombination, and repair

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26
Q

What is the function of DNA polymerase II (E.coli)?

A

polB, proofreading (3-5 prime), 7 subunits
Involved in DNA repair

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27
Q

What is DNA polymerase III in E.coli and what is its function?

A

polC (dnaE) proofreading (3-5 prime), 9 subunits
The principal replication enzyme in E.coli

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28
Q

What is DNA polymerases IV and V in E.coli and what are their functions?

A

dinB (1 subunit), umuC (3 subunits) no proofreading
Involved in an unusual form of DNA repair

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29
Q

How does polA show exonuclease activity differently?

A

the 5′→3′ domain is in front of the enzyme and performs nick translation
nick translation = a break or nick in the DNA is moved along with the enzyme
important in:
DNA repair
the removal of RNA primers during replication
mild protease treatment separates this domain from the remainder of the enzyme (the large fragment or Klenow fragment) retains the polymerization and 3′ → 5′ proofreading activities.

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30
Q

What is nick translation?

A

nick translation = a break or nick in the DNA is moved along with the enzyme
important in:
DNA repair
the removal of RNA primers during replication
(When the 5′→3′ exonuclease domain is removed, the remaining fragment (𝑀r 68,000), the large fragment or Klenow fragment, retains the polymerization and proofreading activities.)

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31
Q

What are the subunits of Polymerase III?

A

9 subunits
Theta or θ subunit associates with α and ε subunits to form a core polymerase
can polymerize DNA but with limited processivity
up to three core polymerases are linked by a clamp-loading complex (τ3δδ′)

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32
Q

How does β Subunits convert DNA polymerase III* to DNA Polymerase III Holoenzyme?

A

DNA polymerase III* = the entire assembly of 16 protein subunits (eight different types)
β subunits = provide an increase in processivity by preventing dissociation of DNA polymerase III from DNA
one dimer of β subunits associates with each active core subassembly
addition of β subunits converts DNA polymerase III* to DNA polymerase III holoenzyme

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33
Q

What is DNA replicase system?

A

DNA replicase system (replisome) = the entire complex of enzymes and proteins required for replication in E. coli
consists of 20+ different enzymes and proteins
*helicases, topoisomerases, DNA-binding proteins, primases, DNA poly I, RNase H1, DNA ligases

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34
Q

What is are helicases?

A

helicases = enzymes that move along the DNA and separate the strands
requires chemical energy from ATP
enzyme

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35
Q

What are topoisomerases?

A

topoisomerases = relieve topological stress created by strand separation
enzyme

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36
Q

What are DNA-binding proteins?

A

DNA-binding proteins = stabilize separated strands
enzyme

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37
Q

What are primases?

A

enzymes
primases = synthesize short segments of RNA to serve as primers

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38
Q

What is DNA polymerase I?

A

enzyme
DNA polymerase I = removes and replaces RNA primers

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39
Q

what are RNase H1?

A

Enzyme
RNase H1 = a specialized nuclease that degrades RNA in RNA-DNA hybrids
can remove RNA primers

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40
Q

What are DNA ligases?

A

Enzyme
DNA ligases = seals nicks in the DNA backbone following removal and replacement of an RNA primer

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41
Q

What are the three stages of the E.coli chromosome replication?

A

DNA synthesis can be divided into three stages:
initiation
elongation
termination

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42
Q

What is initiation of the E.coli chromosome?

A

*the E. coli replication origin (oriC) consists of:
five repeats of a 9 bp sequence (R sites) that serve as binding sites for the key initiator protein, DnaA
*the DNA unwinding element (DUE) = a region rich in A=T base pairs
*three additional DnaA-binding sites (I sites)
*binding sites for the proteins IHF (integration host factor) and FIS (factor for inversion stimulation)

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43
Q

What are some of the proteins required for initiation at the E.coli origin? (10 proteins)

A

DnaA protein- recognizes oriC sequence; opens duplex at specific sites in origin
DnaB protein (helicase)- unwinds DNA
DnaC protein- required for DnaB binding at origin
HU- histonelike protein; DNA-binding protein; stimulates initiation
FIS and IHF- DNA binding proteins that stimulate initiation
Primase (DnaG protein)- synthesizes RNA primers
Single-stranded DNA binding-protein (SSB)- binds single-stranded DNA
DNA gyrase (DNA top II)- Relives torsional strain generated by DNA unwinding
DNA methylase- methylates 5’ GATC sequence at oriC

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44
Q

What is the role of Dna A proteins?

A

DnaA Proteins Bind at R and I Sites in oriC
eight ATP-bound DnaA molecules bind to the R (5) and I (3) sites in oriC

strand separation in the A=T-rich DUE occurs due to:
strain due to positive supercoiling
binding of DnaA to the DUE region
*DnaA has a higher affinity for R sites than I sites, and it binds R sites equally well in its ATP- or ADP-bound form. The I sites, which bind only the ATP-bound DnaA, allow discrimination between the active and inactive forms of DnaA.

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45
Q

How does DnaB helicase get loaded?

A

DnaC protein = an AAA+ ATPase that loads DnaB protein onto the separated DNA strands
-two ring-shaped DnaB hexamers are loaded in the DUE, one onto each DNA strand

DnaB protein (helicase) = migrates along ssDNA in the 5′→3′ direction and unwinds DNA

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46
Q

What is the role of SSB and DNA gyrase in initiation?

A

single-stranded DNA-binding protein (SSB) stabilizes separated strands

DNA gyrase (DNA topoisomerase II) relieves topological stress ahead of the replication forks

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47
Q

Why does initiation only occur once in each cell cycle?

A

Hda protein = an AAA+ ATPase that binds to the β subunits of DNA polymerase III and stimulates hydrolysis of the ATP bound to DnaA
-causes disassembly of the DnaA complex
homologous to DnaA

release of ADP by DnaA and rebinding of ATP occurs on a time scale of 20 to 40 minutes

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48
Q

How is replication regulated by methylation?

A

Dam (DNA adenine methylation) methylase = methylates the N6 position of adenine within the palindromic sequence (5′)GATC of oriC DNA

hemimethylated oriC sequences are sequestered by:
interaction with the plasma membrane
binding of the protein SeqA

after SeqA dissociation, oriC sequences are released from the membrane and Dam methylase fully methylates DNA (both strands) to allow new DnaA to bind

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49
Q

How is the leading strand elongated?

A

the more straightforward of the two strands

primase (DnaG) synthesizes a short (10 to 60 nucleotide) RNA primer at the replication origin
requires interaction with DnaB helicase
Primase and DnaB move in opposite directions

DNA polymerase III adds nucleotides to the 3′ of the primer
linked to the DnaB tethered to the opposite DNA strand

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50
Q

How is the lagging strand elongated?

A

accomplished in short Okazaki fragments

as in leading strand synthesis, primase synthesizes an RNA primer and DNA Pol III adds nucleotides to the 3′ end

DNA of the lagging strand loops around so one asymmetric DNA Pol III dimer complex can synthesize both strands

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51
Q

How are the leading and lagging strands synthesized by DNA Poly III?

A

DNA polymerase III uses one set of its core subunits to synthesize the leading strand continuously
the other two sets of core subunits cycle from one Okazaki fragment to the next on the looped lagging strand

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52
Q

What is the first step of the leading and lagging strand sythesis?

A

DnaB helicase travels along the lagging strand template in the 5′→3′ direction and unwinds the DNA

DnaG primase occasionally associates with DnaB helicase and synthesizes a short RNA primer

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53
Q

What is the second step of leading and lagging strand synthesis?

A

a new β sliding clamp is positioned at the primer by the clamp-loading complex of DNA polymerase III

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54
Q

What is the third step in the synthesis of the leading and lagging strand?

A

replication halts after synthesis of an Okazaki fragment is completed and DNA polymerase III core subunits dissociate from the β sliding clamp

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55
Q

What is the fourth step of the synthesis of the leading and lagging strand?

A

DNA polymerase III core subunits associate with a new β sliding clamp
initiates synthesis of a new Okazaki fragment

56
Q

How does the DNA polymerase III clamp loader work?

A

clamp-loading complex = and AAA+ ATPase that binds ATP and the new β sliding clamp
-consists of parts of the three τ (tau) subunits along with the δ and δ′ subunits
-ATP binding opens the ring at one subunit interface
-ATP hydrolysis closes the clamp around the DNA

57
Q

What is the 5th and final step of lagging strand synthesis?

A

DNA polymerase I or RNase H1 = removes the RNA primer and replaces it with DNA

DNA ligase = seals the remaining nick

58
Q

How does DNA ligase work?

A

DNA ligase = catalyzes the formation of a phosphodiester bond between a 3′-OH and a 5′ phosphate
the phosphate must be activated by adenylylation using ATP (virus and eukaryotes) or NAD+ (bacteria)

59
Q

How is E.coli chromosome replication terminated?

A

-replication forks meet at a region with multiple copies of a 20-bp sequence called Ter
-Ter sequences are found near each other, but in opposite directions
–creates a site that replication forks cannot leave
-Ter is also a binding site for the protein Tus (terminus utilization sequence)
-causes a replication fork to stop
only one Tus-Ter complex functions per replication cycle

60
Q

How are Catenanes separated?

A

Separation of Catenanes by Topoisomerase IV
catenanes = two topologically interlinked circular chromosomes

separation of the catenated circles in E. coli requires topoisomerase IV (a type II topoisomerase)

61
Q

What are replicators? (eukaryotic)

A

replicators = yeast (S. cerevisiae) replication origins
also called autonomously replicating sequences (ARSs)
yeast have ~400 replicators
each spans ∼150 bp and contains several essential, conserved sequences

human chromosomes have ~30,000 to 50,000 replication origins

62
Q

How does replication ensure that cellular DNA is replicated only once per cycle? (EU)

A

regulation involves cyclins and the cyclin-dependent kinases (CDKs)
cyclins undergo ubiquitin-dependent proteolysis at the end of the M phase (mitosis)

63
Q

What are Pre-replicative complexes? (EU)

A

Pre-replicative complexes (pre-RCs) = established on replication initiation sites in the absence of cyclins
their formation (sometimes called licensing) renders the cell competent for replication

64
Q

What is the ORC (initiation of replication in eukaryotes)?

A

ORC (origin recognition complex) = a six protein complex that recognizes and finds the origin
functions like DnaA

65
Q

How is replication of eukaryotes initiationed?

A

initiation involves the loading of a replicative helicase composed of minichromosome maintenance (MCM) proteins (MCM2 to MCM7)

the MCM2–7 helicase translocates 3′→5′ along the leading strand template
similar to DnaB helicase

66
Q

How does the assembly of a pre-replicative complex at a eukaryotic replication origin happen?

A

1) the initiation site (origin) is bound by ORC and CDC6
2) ORC, CDC6, and CDTI facilitate loading of two MCM2-7 helicase complexes, which form the pre-RC
3)The pre-RC is subsequently activated by addition of CDC45 and the GINS proteins, triggering DNA denaturation
4)Replisome components are added

67
Q

What is the rate of eukaryotic replication?

A

the replication fork moves ~50 nucleotides/second
~1/20th the rate seen in E. coli

eukaryotes compensate for the slower rate by having many origins

68
Q

What is DNA polymerase ε (EU)?

A

DNA polymerase ε = synthesizes the leading strand
highly processive
has 3′→5′ proofreading exonuclease activity

69
Q

What is DNA polymerase δ (EU)?

A

DNA polymerase δ = synthesizes the lagging strand
has 3′→5′ proofreading exonuclease activity

70
Q

What is DNA polymerase α (EU)?

A

a DNA polymerase/primase that synthesizes RNA primers and also extends them to initiate synthesis of each Okazaki fragment
does not have 3′→5′ proofreading exonuclease activity

71
Q

What is the function of PCNA? eu

A

proliferating cell nuclear antigen (PCNA) = a protein that forms a circular clamp to enhance the processivity of two polymerases
analogous function to the β subunit

72
Q

What is the function of RPA? eu

A

RPA (replication protein A) = single-stranded DNA-binding protein
equivalent in function to the E. coli SSB protein

73
Q

What is the function of RFC? eu

A

RFC (replication factor C) = a clamp loader for PCNA that facilitates the assembly of active replication complexes
significant sequence similarity to the subunits of the bacterial clamp-loading (γ) complex

74
Q

How does eukaryotic replication terminate?

A

termination occurs when replication forks operating from nearby origins converge

75
Q

What does viral DNA polymerases provide a target for?

A

Antiviral Therapy
acyclovir = inhibits the DNA polymerase of the herpes simplex virus
converted to acyclo-GTP

acyclo-GTP competitively inhibits viral DNA polymerases and terminates replication

76
Q

What is a mutation?

A

mutation = a permanent change in the nucleotide sequence
substitution mutation = replacement of one base pair with another
insertion mutation = the addition of 1+ base pairs
deletion mutation = the deletion of 1+ base pairs

77
Q

What is a silent mutation?

A

a mutation that affects nonessential DNA or has a negligible effect on gene function

78
Q

What is the ames test?

A

the Ames test = measures the potential of a given chemical compound to promote certain easily detected mutations in a strain of S. typhimurium that cannot synthesize histidine
–strains that grow contain mutations that restore the ability to synthesize histidine
(However, of the compounds known to be carcinogenic from extensive animal trials, more than 90% are also found to be mutagenic in the Ames test. Because of this strong correlation between mutagenesis and carcinogenesis, the Ames test for bacterial mutagens is still widely used as a rapid and inexpensive screen for potential human carcinogens.)

79
Q

What is mismatch repair?

A

mismatches are corrected to reflect template strand information
distinguished from the newly synthesized strand by the presence of methyl group tags on the template DNA

in E. coli, Dam methylase methylates DNA at the N6 position of adenines within (5′)GATC sequences

80
Q
A
81
Q

What is the model for the early steps of methly-directed mismatch repair?

A

MutS recognizes the mismatch

MutH recognizes the sequence (5′)GATC and has site-specific endonuclease activity

MutL binds to both MutS and MutH

82
Q

Where is the mismatch repair cleavage site?

A

Mismatch Repair on the 5′ Side and 3′ Side of the Cleavage Site
the unmethylated strand is unwound and degraded in the 3′→5′ direction from the cleavage site through the mismatch

this segment is replaced with new DNA by:
SSB
exonuclease I, exonuclease X, or exonuclease VII
DNA polymerase III
DNA ligase

3’ side repair is similar to mismatches on the 5′ side

the exonuclease is either exonuclease VII or RecJ nuclease

83
Q

What is the eukaryotic mismatch repair systems?

A

several proteins are structurally and functionally analogous to the bacterial MutS and MutL proteins (But no MutH)

many details of eukaryotic mismatch repair are unknown

identification of newly synthesized DNA strands does not involve GATC sequences

84
Q

What is Base-Excision repair?

A

DNA glycosylases = recognize common DNA lesions and remove the affected base by cleaving the N-glycosyl bond in the process of base-excision repair
generally specific for one lesion type

AP site or abasic site = an apurinic or apyrimidinic site in the DNA resulting from removal of a base by DNA glycosylase

85
Q

What are DNA Glycosylases?

A

uracil DNA glycosylases = specifically remove from DNA the uracil that results from spontaneous deamination of cytosine
does not remove uracil residues from RNA or thymine residues from DNA
deamination is 100-fold faster in ssDNA
*other DNA glycosylases recognize:
-formamidopyrimidine (from purine oxidation)
-8-hydroxyguanine (from purine oxidation)
-hypoxanthine (from adenine deamination)
-alkylated bases
-pyrimidine dimers

86
Q

How are AP sites repaired in bacteria?

A

the deoxyribose 5′-phosphate left behind is removed and replaced with a new nucleotide

AP endonucleases = cut the DNA strand containing the AP site

DNA polymerase I replaces the DNA

DNA ligase seals the remaining nick

87
Q

What is Nucleotide-Excision repair?

A

nucleotide-excision system = repairs DNA lesions that cause large distortions in the DNA helical structure

excinuclease = a multisubunit enzyme that hydrolyzes two phosphodiester bonds, one on either side of the distortion

DNA polymerase I (E. coli) or DNA polymerase ε (humans) fills the gap

DNA ligase seals the nick

(In E. coli and other bacteria, the enzyme system hydrolyzes the 5th phosphodiester bond on the 3’ side and the 8th phosphodiester bond on the 5’side to generate a fragment of 12 to 13 nucleotides (depending on if lesion is one or two bases).
In humans and other eukaryotes, the enzyme system hydrolyzes the 6th phosphodiester bond on the 3’ side and the 22nd phosphodiester bond on the 5’ side, producing a fragment of 27 to 29 nucleotides.)

88
Q

What is the ABC Excinuclease?

A

has three components:
–UvrA = a dimeric ATPase that scan DNA and binds to the lesion site
–UvrB = binds UvrA and makes an incision at the fifth phosphodiester bond on the 3′-side of the lesion after UvrA dissociates from the lesion
–UvrC = binds UvrB and makes an incision at the eighth phosphodiester bond on the 5′-side of the lesion after UvrB does

removes a fragment of 12 to 13 nucleotides

89
Q

What are eukaryotic Excinuclease?

A

similar mechanism to that of the bacterial enzyme

requires 16+ polypeptides with no similarity to the E. coli excinuclease subunits

hydrolyzes the sixth phosphodiester bond on the 3′ side and the twenty-second phosphodiester bond on the 5′ side
produces a fragment of 27 to 29 nucleotides

90
Q

What is direct repair?

A

pyrimidine dimers result from a UV-induced reaction
DNA photolyases = promote direct photoreactivation of cyclobutane pyrimidine dimers
use energy from absorbed light
generally contain two cofactors: FADH2 and folate
DNA photolyases are not found in placental mammals (including humans)

91
Q

How do nucleotides with alkylation damage form?

A

O6-methylguanine = a modified nucleotide that forms in the presence of alkylating agents
common and highly mutagenic lesion
tends to pair with thymine rather than cytosine

92
Q

How do nucleotides with alkylation damage repair happen?

A

O6-methylguanine-DNA methyltransferase = catalyzes transfer of the methyl group of O6-methylguanine to one of its own Cys residues
–a single methyl transfer event permanently methylates the protein, inactivating it
-consumption of an entire protein to correct a single damaged base illustrates the priority given to maintaining the integrity of cellular DNA

93
Q

What can lead to error-prone translesion DNA synthesis?

A

The Interaction of Replication Forks with DNA Damage Can Lead to Error-Prone Translesion DNA Synthesis
double-strand breaks and lesions in ssDNA arise from:
a replication fork encountering an unrepaired DNA lesion
ionizing radiation
oxidative reactions

94
Q

What are two ways to repair a stalled bacterial replication fork?

A

two avenues:
homologous genetic recombination
error-prone translesion DNA synthesis (TLS)

95
Q

What is Error-prone TLS in bacteria?

A

part of the SOS response (a cellular stress response to extensive DNA damage)

other SOS proteins include UvrA, UvrB + UmuC, and UmuD
Umu = unmutable (because lack of umu genes eliminates error-prone repair)

cleaved UmuD′ and UmuC bind with RecA to create DNA Pol V (a specialized polymerase)
the polymerase can process past the damaged area

96
Q

What is the purpose of error-prone TLS?

A

error-prone TLS serves as a desperation strategy

the resulting mutations kill some cells and create deleterious mutations in others

at least a few mutant daughter cells survive

the resultant mutations contribute to evolution

97
Q

What should we know about DNA polymerase IV?

A

induced during the SOS response

replication by DNA polymerase IV is highly error-prone

lack a proofreading exonuclease and have a more open active site than other DNA polymerases
accommodates damaged template nucleotides

98
Q

How do TLS polymerases differ in mammals?

A

mammals have many low-fidelity TLS polymerases
most also have specialized functions in DNA repair
typically limited to short lengths to minimize mutagenic potential

DNA polymerase η = promotes translesion synthesis primarily across cyclobutane T–T dimers

DNA polymerases β, ι, and λ have specialized roles in eukaryotic base-excision repair

99
Q

What is homologous genetic recombination?

A

= involves genetic exchanges between any two DNAs that share an extended region of nearly identical sequence.
It is largely a pathway to repair double-strand breaks in DNA.

100
Q

What is site-specific recombination?

A

involves genetic exchanges only at a particular DNA sequence

101
Q

What is DNA transposition?

A

involves a short segment of DNA with the capacity to move from one location in a chromosome to another (“jumping genes”)

102
Q

What is nonhomologous end joining?

A

nonhomologous end joining (NHEJ) = an alternative process for double-strand break repair
does not entail recombination

103
Q

What is recombinational DNA repair?

A

recombinational DNA repair = homologous genetic recombination as a DNA repair process
typically directed at the reconstruction of stalled or collapsed replication forks

104
Q

How are collapsed replication forks reconstructed?

A

the replication fork collapses when it encounters a damaged site in a template strand
*the 5′-ending strand is degraded
*the 3′ single-stranded extension is bound by a recombinase that pairs with a complementary sequence in the intact DNA duplex
*creates a branched DNA structure (a point where three DNAs come together)
*branch migration = process that moves the DNA branch
–creates a Holliday intermediate (an X-like crossover structure)
a special class of nucleases cleaves The Holliday intermediate
ligation restores the replication fork

105
Q

What are the RecBCD Helicase/Nuclease?

A

RecBCD nuclease/helicase = promotes DNA end-processing in E. coli
coupled to ATP hydrolysis

RecB (moves 3′→5′) and RecD (moves 5′→3′) are helicase motors

RecC binds tightly to the chi sequence ((5′)GCTGGTGG)
decreases degradation of the 3′-terminal strand
increases degradation of the 5′-terminal strand
creates ssDNA with a 3′ end for use in recombination

1,009 chi sequences scattered throughout the E. coli genome enhance the frequency of recombination about 5- to 10-fold within 1,000 bp of each chi site.

106
Q

What are RecA protein Filaments?

A

RecA = the bacterial recombinase
active form is an ordered, helical filament of thousands of subunits that assemble cooperatively on ssDNA
promotes central steps of homologous recombination

RecBCD acts directly as a RecA loader

filaments assemble and disassemble predominantly in a 5′→3′ direction

107
Q

How are Holliday intermediate (+) resolved?

A

bacterial RuvC protein = specialized nuclease that cleaves the Holliday intermediate

108
Q

How is reassembly of the replication fork after recombination?

A

origin-independent restart of replication
Different combinations of four proteins (PriA, PriB, PriC, and DnaT) act with DnaC in several pathways to load DnaB helicase onto the reconstructed replication fork.

The DnaG primase then synthesizes an RNA primer, and DNA polymerase III reassembles on DnaB to restart DNA synthesis.

109
Q

What is the origin-independent restart of replication?

A

the process of reassembling the replication fork after recombination
tightly intertwined with replication

110
Q

What is the replication restart primosomes?

A

complexes that include the proteins required for origin-independent restart of replication

111
Q

What is meiosis?

A

the process by which diploid germ-line cells with two sets of chromosomes divide to produce haploid gametes
involves a high frequency of recombination

homologous chromosomes are segregated into separate daughter cells at the end of meiosis I

sister chromatids are separated during meiosis II
(requires eukaryotic homologous recombination)

112
Q

How does recombination in meiosis take place?

A

The DNA strand invasion in eukaryotes is catalyzed by RecA-like recombinases called Rad51 and Dmc1. Loading of Rad51 onto DNA is promoted by Rad51 loading protein BRCA2 (analogous to the bacterial RecF, RecO, and RecR proteins).

113
Q

What is recombination during Prophase 1?

A

also referred to as crossing over
occurs at “hot spots” on chromosomes

chiasmata = points where two pairs of sister chromatids are linked due to crossing over

this process aligns sister chromosomes for proper segregation and increases genetic diversity

114
Q

What is a model of homologous recombination in meiosis?

A

double-strand break repair model = 3′ ends are used to initiate the genetic exchange

key features of the model:
homologous chromosomes align
a double-strand break occurs in one chromosome and an exonuclease leaves a free 3′-OH
3′ ends “invade” the intact duplex DNA of the homolog
cleavage of the crossovers creates recombinant products

115
Q

What is the functions of homologous recombination in eukaryotes?

A

homologous recombination:
contributes to the repair of several types of DNA damage
provides a transient physical link between chromatids that promotes the orderly segregation of chromosomes at the first meiotic cell division
enhances genetic diversity in a population

116
Q

How are Some Double-Strand Breaks are Repaired by Nonhomologous End Joining?

A

nonhomologous end joining (NHEJ) = process by which broken chromosome ends are processed and ligated back together
occurs when recombinational DNA repair is not feasible because replication is not occurring and sister chromatids are not present
mutagenic process
does not randomly join ends

117
Q

How does Nonhomologous end joining happen?

A

the Ku70-Ku80 complex binds the DNA ends, followed by a complex including DNA-PKcs (kinase) and Artemis (nuclease)

remaining steps:
the broken DNA ends are synapsed
Artemis (after phosphorylation by DNA-PKcs) removes single-strand extensions or hairpins
DNA ends are separated by a helicase
strands from the two different ends are annealed

118
Q

What is the process of NHEJ?

A

(DNA Ligase) Small DNA gaps are filled by a DNA polymerase, Pol m or Pol l .

the nicks are sealed by a protein complex consisting of XRCC4 (x-ray cross complementation group), XLF (XRCC4-like factor), and DNA ligase IV.

119
Q

What is Site-specific recombination?

A

Site-Specific Recombination Results in Precise DNA Rearrangements

site-specific recombination = involves genetic exchanges only at specific sequences
uses recombinase enzymes with Tyr or Ser residues in the active site

120
Q

What is the process of Site-specific recombination(with Tyr)?

A

Site-Specific Recombination with Tyr Recombinase Enzymes
the recombinase is covalently linked to the DNA at the cleavage site

cleaved DNAs are rejoined to form a Holliday intermediate

exchange is reciprocal and precise

We can view a recombinase as a site-specific endonuclease and ligase in one package.

121
Q

What is the process of Site-specific recombination (Ser)?

A

both strands of each recombination site are cut concurrently and rejoined to new partners

no Holliday intermediate is formed

exchange is reciprocal and precise

122
Q

*What are potential outcomes of site-specific recombination?

A

inversion occurs if two sites are on the same DNA molecule and the recombination sites have the opposite orientation

deletion occurs if two sites are on the same DNA molecule and the recombination sites have the same orientation

intermolecular recombination occurs if recombination sites are on different DNAs

insertion occurs if one or both DNAs are circular

123
Q

What are effects of site-specific recombination?

A

outcome depends on the location and orientation of the recombination sites in a dsDNA molecule

124
Q

What is the XerCD system?

A

resolution of a Holliday intermediate can generate a contiguous dimeric chromosome

XerCD = a specialized site-specific recombinase in E. coli
converts the dimer to monomers
allows chromosome segregation and cell division to proceed

125
Q

What are transposons?

A

= transposable elements that “jump” from one place on a chromosome (the donor site) to another on the same or a different chromosome (the target site)
found in virtually all cells

126
Q

What is transposition?

A

movement of transposons
determined more or less randomly
tightly regulated and infrequent

127
Q

What are bacterial classes of transposons?

A

insertion sequences (simple transposons) = contain only the sequences required for transposition and the genes for the proteins (transposases) that promote the process

complex transposons = contain 1+ genes in addition to those needed for transposition
example: antibiotic-resistance genes

128
Q

How are transposon’s inserted?

A

short repeated sequences at each end of the transposon serve as binding sites for the transposase
become duplicated following transposon insertion

129
Q

What is direct transposition?

A

the transposon is excised via cuts on each side and moves to a new location

a staggered cut is made at the target site and the transposon is inserted into the break

DNA replication fills in the gaps to duplicate the target-site sequence

This leaves a double-strand break in the donor DNA that must be repaired.

130
Q

What is replicative transposition?

A

the entire transposon is replicated, leaving a copy behind at the donor location

cointegrate = the donor region covalently linked to DNA at the target site
serves as an intermediate in the process
contains two complete copies of the transposon

131
Q

What are two general pathways for transposition?

A

direct transposition = “cut and paste”

replicative transposition = “copy and paste”

132
Q

How are IgM Genes assembled by recombination?

A

millions of antibodies (immunoglobulins) can be created from the separate gene segment sets
this is despite the human genome containing only ~20,000 genes

recombination occurs as bone marrow stem cells mature into B lymphocytes
each B lymphocyte produces one kind of antibody

133
Q

What is the structure of IgM?

A

immunoglobulins consist of:
two heavy polypeptide chains
two light polypeptide chains (can be in the kappa or lambda family)

each chain has a variable region and a constant region

134
Q

How does recombination of IgM genes happen?

A

during formation of a mature B lymphocyte, one V segment (first 95 aa) and one J segment (remaining 12 aa) are brought together

40 x 5 = 200 possible V–J combinations
additional variation at the V-J junction increases this to 500

135
Q

Recombination of IgM genes?

A

Fig depicts the organization of the DNA encoding the kappa light chains of human IgG and shows how a mature kappa light chain is generated.
The V (variable) segment encodes the first 95 amino acid residues of the variable region, the J (joining) segment encodes the remaining 12 residues of the variable region, and the C segment encodes the constant region.

136
Q

What is the mechanism of recombination of IgM genes?

A

recombination signal sequences (RSS) = found between the V and J segments

RSS bind to proteins RAG1 and RAG2.
RAG = products of the recombination activating gene

RAG proteins catalyze formation of a double-strand break between segments to be joined