Chapter 25: DNA Metabolism

Question 1

nucleases, DNases

Answer 1

• enzymes that degrade DNA
• two types
  
  • Exonucleases
    
    • degrade nucleic acids from one end of the molecule
    • operate only in 5’→3’ or the 3’→5’ direction
    • removing nucleotides only from the 5’ or the 3’ end, respectively
  • Endonucleases
    
    • can begin to degrade at specific internal sites in a nucleic acid strand or molecule
• A few degrade only singlestranded DNA
• a few cleave only at specific nucleotide sequences

Question 2

DNA polymerase

Answer 2

• DNA polymerase from E. coli
• encoded by the polA gene
• fundamental reaction is a phosphoryl group transfer
• nucleophile is the 3’-hydroxyl group of the nucleotide at the 3’ end
• Nucleophilic attack occurs at the α phosphorus of the incoming deoxynucleoside 5’-triphosphate
• Inorganic pyrophosphate is released in the reaction
• has two Mg²⁺ ions at the active site has and three Asp residues, two of which are highly conserved
  
  • One Mg²⁺ ion helps deprotonate the 3’-hydroxyl group, rendering it a more effective nucleophile
  • the othe Mg²⁺ ion binds to the incoming dNTP and facilitates departure of the pyrophosphate
  • Both ions stabilize the structure of the pentacovalent transition state
• reaction proceeds with a minimal change in free energy
• noncovalent basestacking and base-pairing provide stabilization
• formation of products is facilitated by the 19 kJ/mol generated in the subsequent hydrolysis of the pyrophosphate product by the enzyme pyrophosphatase
• PDF pg 1045, Figure 25-5

Question 3

primer

Answer 3

• a strand segment (complementary to the template)
• has a free 3’-hydroxyl group to which a nucleotide can be added
• the free 3’ end of the primer is called the primer terminus
• Many are oligonucleotides of RNA
• synthesized by primase
• removed and replaced by DNA, functions of DNA polymerase I

Question 4

DNA polymerase active site

Answer 4

• has 2 parts: insertion site and postinsertion site
• incoming nucleotide is initially positioned in the insertion site and the phosphodiester bond is formed
• then the polymerase slides forward on the DNA and the new base pair is positioned in the postinsertion site.

Question 5

processivity

Answer 5

• average number of nucleotides added before a polymerase dissociates

Question 6

discrimination between correct and incorrect nucleotides relies on

Answer 6

• the hydrogen bonds that specify the correct pairing between complementary bases
• The standard A═T and G☰C base pairs have very similar geometries, and an active site sized to fit one will generally accommodate the other
• An incorrect nucleotide (G═T) may be able to hydrogen-bond with a base in the template, but it generally will not fit into the active site and will be rejected before the phosphodiester bond is formed
• PDF pg 1046, figure 25-6

Question 7

Proofreading

Answer 7

• DNA polymerases insert one incorrect nucleotide for every 10⁴ to 10⁵ correct ones
• Mistakes occur because a base is briefly in an unusual tautomeric form, allowing it to hydrogen bond with an incorrect partner
• 3’→5’ exonuclease activity
  
  • double-checks each nucleotide after it is added
  • permits enzyme to remove a newly added nucleotide
  • highly specific for mismatched base pairs
• process
  
  • polymerase added wrong nucleotide
  • translocation of polymerase to position where next nucleotide is to be added is inhibited
  • kinetic pause provides opportunity for correction
  • 3’→5’ exonuclease activity removes mispaired nucleotide
  • polymerase begins again
• not the reverse of polymerization reaction because pyrophosphate is not involved
• improves accuracy by 10² to 10³-fold
• In the monomeric DNA polymerase I, the polymerizing and proofreading activities have separate active sites
  
  • exonuclease activity behind the polymerase activity
• When base selection and proofreading are combined, DNA polymerase leaves behind one net error for every 10⁶ to 10⁸ bases added
• PDF pg. 1047, figure 25-7

Question 8

DNA polymerase I

Answer 8

• not suited for replication, has slow rate at which it adds nucleotides
• relatively low processivity
• performs a host of cleanup functions during replication, recombination, and repair
• has three domains, catalyzing its DNA polymerase, 5’→3’ exonuclease (in front), and 3’→5’ exonuclease activities.
• 5’→3’ exonuclease activity
  
  • located in a structural domain that can be separated from the enzyme by mild protease treatment
  • when removed the remaining fragment (Mr 68,000), the large fragment or Klenow fragment, retains the polymerization and proofreading activities
  • can replace a segment of DNA (or RNA) paired to the template strand
  • Most other DNA polymerases lack a 59S39 exonuclease activity
• removes RNA primers and replaces them w/DNA in E. coli
• PDF pg. 1047, table 25-1

Question 9

DNA polymerase II

Answer 9

• involved in DNA repair
• PDF pg. 1047, table 25-1

Question 10

DNA polymerase III

Answer 10

• principal replication enzyme in E. coli
• more complex than DNA polymerase I
• polymerization and proofreading activities reside in its α and ε (epsilon) subunits respectively
• θ subunit associates with α and ε to form a core polymerase, which can polymerize DNA with limited processivity
• core polymerases can be linked
  
  • can be linked by another set of subunits, a clamp-loading complex, or γ complex, consisting of five subunits of four different types, τ2γδδ’
    
    • linked through the τ (tau) subunits
    • forms a core polymerase
    • can polymerize DNA w/limited processivity
    • AAA1 ATPase
  • Two additional subunits, χ (chi) and ψ (psi), are bound to the clamp-loading complex.
    
    • assembly of 13 protein subunits (nine different types) is called DNA polymerase III*
    • can polymerize DNA, but with a much lower processivity
    • increase in processivity is provided by addition of β subunits, four of which complete the DNA polymerase III holoenzyme.
    • β subunits associate in pairs to form donut-shaped structures that encircle the DNA and act like clamps
    • β sliding clamp prevents dissociation of DNA polymerase III from DNA, dramatically increasing processivity
• PDF pg. 1047, table 25-1
• structure PDF pg. 1049, figure 25-9

Question 11

DNA replicase system or replisome

Answer 11

• a large protein complex that carries out DNA replication, starting at the replication origin
• contains several enzymatic activities, such as helicase, primase and DNA polymerase and creates a replication fork to duplicate both the leading and lagging strand

Question 12

helicases

Answer 12

• enzymes that move along the DNA and separate the strands
• uses chemical energy from ATP
• Strand separation creates topological stress in the helical DNA structure which is relieved by the action of topoisomerases
• separated strands are stabilized by DNA-binding proteins

Question 13

Nick Translation

Answer 13

• catalyzed by DNA polymerase I
• The 5’→3’ exonuclease domain in front of the enzyme degrades DNA strand ahead as it moves along the strand
• it synthesizes a new strand behind
• a break or nick in the DNA is moved along with the enzyme
• has a role in DNA repair and in the removal of RNA primers during replication
• strand of nucleic acid removed (DNA or RNA) is shown in purple, the replacement strand in red.
• DNA synthesis begins at a nick (a broken phosphodiester bond, leaving a free 39 hydroxyl and a free 59 phosphate)
• A nick remains where DNA polymerase I eventually dissociates in the DNA backbone in the form of a broken phosphodiester bond
• These nicks are sealed by DNA ligases
• PDF pg. 1048, Figure 25-8

Question 14

Replication of the E. coli Chromosome

3 stages of replication:

initiation, elongation, and termination

Answer 14

E. coli replication origin, oriC

• 245 bp
• DNA sequence elements that are highly conserved
• highly enriched in GATC sequences
• DNA is methylated by the Dam methylase
  
  • methylates the N⁶ position of adenine within the palindromic sequence (5’)GATC
• R sites
  
  • five repeats of a 9 bp sequence
  • binding sites for the key initiator protein DnaA
  • binds ATP- or ADP-bound DnaA
• DUE (DNA unwinding element)
  
  • region rich in A═T base pairs
• I sites
  
  • three additional DnaA-binding sites
  • binds ATP-bound DnaA
• binding sites for IHF (integration host factor) and FIS (factor for inversion stimulation), required components of certain recombination reactions
• HU (a histonelike bacterial protein), does not have a specific binding site
• PDF pg. 1050, Figure 25-10

Enzymes and Proteins

• PDF pg. 1050, Table 25-3
• participate in initiation phase of replication
• open the DNA helix at the origin and establish a prepriming complex
• DnaA protein
  
  • crucial component in the initiation
  • member of the AAA+ ATPase family
  • form oligomers and hydrolyze ATP relatively slowly which mediates interconversion of the protein between two states
  • ATP-bound form is active
  • ADP-bound form is inactive
  • cycling between the forms is between 20-40 minutes
  • DnaA has a higher affinity for the R sites
• DnaC (AAA+ ATPase)
• Hda (AAA+ ATPase)

Process

• Eight ATP-bound DnaA form a helical complex encompassing the R and I sites in oriC and bind to the R sites
• HU, IHF, and FIS also bind, facilitating DNA bending
• tight right-handed wrapping of DNA around this complex creates a positive supercoil and the strain denatures the DUE region
• A hexamer of DnaC, each subunit bound to ATP, forms a tight complex with the hexameric, ring-shaped DnaB helicase
• DnaCDnaB interaction opens the DnaB ring, aided by DnaA
• Two DnaB are loaded in the DUE, one onto each ssDNA
• ATP bound to DnaC is hydrolyzed, releasing the DnaC and leaving the DnaB bound to the DNA
• DnaB migrates along ssDNA in the 5’→3’ direction, unwinding the DNA
• DnaB helicases loaded onto the two DNA strands travel in opposite directions, creating two replication forks
• molecules of single-stranded DNA–binding protein (SSB) bind to and stabilize the separated strands
• DNA gyrase (DNA topoisomerase II) relieves topological stress induced ahead of the fork by the unwinding reaction
• DNA polymerase III holoenzyme and β subunits binds to ssDNA
• Hda binds to β subunits and interacts with DnaA stimulating hydrolysis of its bound ATP and DnaA disassembles

Regulation

• only phase of DNA replication that is known to be regulated
• replication occurs only once in each cell cycle
• timing of replication initiation is affected by DNA methylation and interactions with the bacterial plasma membrane

AFTER REPLICATION

• DNA is hemimethylated: the parent strands have methylated oriC sequences but the newly synthesized strands do not
• hemimethylated oriC sequences are sequestered by The plasma membrane and binding to SeqA
• oriC is released from the plasma membrane
• SeqA dissociates
• DNA must methylated by Dam methylase before it can again bind DnaA and initiate a new round of replication

Question 15

Replication of the E. coli Chromosome

3 stages of replication:

initiation, elongation, and termination

Answer 15

primosome

• replication complex
• DnaB helicase is bound in front of DNA polymerase III
  
  • unwinds DNA at the replication fork (Fig. 25–13a)
  • travels along the lagging strand template in the 5’→3’ direction
• DnaG primase occasionally associates with DnaB helicase and synthesizes a short RNA primer (for the lagging strand)
• Both strands are produced by a single asymmetric DNA polymerase III dimer; this is accomplished by looping the DNA of the lagging strand (Fig. 25–13a)
  
  • one core polymerase synthesizes the leading strand continuously
  • the other cycles from one Okazaki fragment to the next on the looped lagging strand

Leading strand synthesis

• primase (DnaG) synthesizes a short (10 to 60 nucleotide) RNA primer at the replication origin
• first primer laid down primes the leading strand
• DnaG interacts with DnaB helicase to carry out this reaction
• primer synthesized in the direction opposite to which DnaB helicase is moving
  
  • DnaB helicase moves along the the lagging strand
  • primase primes leading strand in the opposite direction.
• Deoxyribonucleotides are added to the primer by the DNA polymerase III in the primosome
• Leading strand synthesis then proceeds continuously, keeping pace with the unwinding of DNA at the replication fork

Lagging strand synthesis

• accomplished in short Okazaki fragments
• RNA primer is synthesized by primase
• DNA polymerase III dimer
  
  • loops the DNA of the lagging strand
  • binds to the RNA primer and adds deoxyribonucleotides
• DnaB helicase continues to unwinds DNA at the replication fork (Fig. 25–13a) ahead of DNA polymerase III
• DnaG primase associates with DnaB helicase and synthesizes a short RNA primer for the next Okazaki fragment (Fig. 25–13b)
• when synthesis of an okazaki fragment is complete
  
  • replication halts
  • core polymerase of DNA polymerase III dissociate from β sliding clamp and completed Okazaki fragment
  • DNA polymerase I removes RNA primer w/its 5’→3’ exonuclease activity and replaces it with DNA
  • The remaining nick is sealed by DNA ligase
    
    • catalyzes formation of a phosphodiester bond between a 3’ hydroxyl at the end of one DNA strand and a 5’ phosphate at the end of the other
    • phosphate must be activated by adenylylation
• clamp-loading complex of DNA polymerase III (Fig. 25–14c)
  
  • binds 3 ATP and to a dimeric β sliding clamp
  • Binding imparts strain on the β clamp opening the ring at one subunit interface
  • newly primed lagging strand is slipped into the ring
  • Hydrolysis of bound ATP closes β clamp around the DNA
• synthesis of new fragment begins
• PDF pg. 1053, Fig. 25–13
• PDF pg. 1054, Table 25-4

Question 16

Replication of the E. coli Chromosome

3 stages of replication:

initiation, elongation, and termination

Answer 16

• the two replication forks meet at a Ter region
  
  • a terminus region
  • contains mulltiple copies of a 20 bp sequence
  • arranged on chromosome to create a trap that a replication fork can enter but not leave
• binding sites for the protein Tus (terminus utilization substance)
• prevents overreplication by one fork if the other is delayed or halted (DNA damage, etc.)
• Tus-Ter complex stops a replication fork from only one direction
• Only one Tus-Ter complex functions per replication cycle—the complex first encountered by either replication fork
• the other fork halts when it meets the first (arrested) fork
• final few hundred base pairs of DNA between these large protein complexes are replicated (unknown mechanism), completing two topologically interlinked (catenated) circular chromosomes aka catenanes
  
  • The circles are not covalently linked, but because they are interwound and each is covalently closed, they cannot be separated
• topoisomerase IV (a type II topoisomerase) separates the catenanes and they segregate into daughter cells
• PDF PG 1055, figure 25-17

Question 17

Replication in Eukaryote

Regulation

Answer 17

• eukaryotic replication is regulated and coordinated with the cell cycle
• ensures that all cellular DNA is replicated once per cell cycle
• cyclins and the cyclin-dependent kinases
  
  • proteins that regulate replication
  • rapidly destroyed by ubiquitin-dependent proteolysis at the end of the M phase (mitosis)
• Replication requires S-phase cyclin-CDK complexes (cyclin E–CDK2 complex) and CDC7DBF4.
  
  • activate replication by binding to and phosphorylating proteins in the prereplicative complexes (pre-RCs) on replication initiation sites
  • pre-RCs form at the end of the M phase (mitosis) in fast growing cells
  • pre-RCs form at the end of the G1 in slow growing cells
  • pre-RCs renders the cell competent for replication (licensing)
• Other cyclins and CDKs inhibit formation of more pre-RC complexes once replication has been initiated
• PDF pg. 1056

Question 18

Replication in Eukaryote

Initiation

Answer 18

• Origins of replication more well-characterized in lower than higher eukaryotes
• In vertebrates, a variety of APT-rich sequences may be used for replication initiation
• Site may vary from one division to the next
• Yeast (Saccharomyces cerevisiae)
  
  • has defined replication origins called autonomously replicating sequences (ARS), or replicators
  • span, 150 bp
  • 400 replicators are distributed among the 16 chromosomes of the haploid

Process

• key event is loading of the helicase
  
  • ring-shaped MCM2–7 helicase
  • heterohexameric complex of minichromosome maintenance (MCM) proteins
• MCM2–7 helicase is loaded onto the DNA by ORC (origin recognition complex)
  
  • 6 protein complex
  • has five AAA1 ATPase domains
  • analogous to DnaA
  • CDC6 and CDT1 are required to load the helicase
• PDF pg 1057

Question 19

Replication in Eukaryote

Elongation

Answer 19

• rate of movement of the replication fork in eukaryotes (50 nucleotides/s) is only one-twentieth that observed in E. coli
• proceeds bidirectionally from many origins, spaced 30 to 300 kbp apart
• DNA polymerase α
  
  • multisubunit enzyme
  • similar structure and properties in all eukaryotic cells
  • one subunit has a primase activity
  • largest subunit (Mr ≈180,000) contains the polymerization activity.
  • has no proofreading 3’→5’ exonuclease activity
    
    • unsuitable for high-fidelity DNA replication
  • synthesises only short primers (either RNA or DNA)
• DNA polymerase δ
  
  • extends primers
  • stimulated by proliferating cell nuclear antigen (PCNA; Mr 29,000)
    
    • found in large amounts in the nuclei of proliferating cells
    • PCNA structure similar/analogous to β subunit of E. coli DNA polymerase III
• DNA polymerase ε
  
  • used in DNA repair and removal of primers
  • analogous to that of the bacterial DNA polymerase I
• RPA (replication protein A)
  
  • single-stranded DNA–binding protein
  • equivalent in function to the E. coli SSB protein
• RFC (replication factor C)
  
  • a clamp loader for PCNA
  • facilitates the assembly of active replication complexes
  • subunits have significant sequence similarity to subunits of bacterial clamp-loading (γ) complex
• PDF pg 1057

Question 20

Replication in Eukaryote

Termination

Answer 20

• involves the synthesis of telomeres
• PDF pg. 1057

Question 21

Ames test

Answer 21

• carcinogens, based on their mutagenicity
• A strain of Salmonella typhimurium w/a mutation inactivating the histidine biosynthetic pathway is plated on a histidine free medium
• the few cells grow carry spontaneous backmutations that permit the histidine biosynthetic pathway to operate
• Three identical nutrient plates are inoculated with an equal number of cells, and a filter paper disk w/progressively lower concentrations of a mutagen is placed in the middle
• mutagen increases the rate of back-mutation and # of colonies
• clear areas around the filter indicate where the concentration of mutagen is so high that it kills the cells
• As the mutagen diffuses away from the filter paper, it is diluted to sublethal concentrations that promote back-mutation
• Mutagens are compared on the basis of their effect on mutation rate
• Because many compounds undergo a variety of chemical transformations after entering cells, compounds are sometimes tested for mutagenicity after first incubating them with a liver extract.
  
  • Some substances have been found to be mutagenic only after this treatment.
• PDF pg. 1059

Question 22

Mismatch Repair

Answer 22

• mismatches are corrected to reflect the information in the old (template) strand
• repair system discriminates between the methylated template and new unmethylated strand
• efficiently repairs mismatches up to 1,000 bp

Process in Bacteria

• Immediately after passsage of the replication fork, there is a short period (few seconds or minutes) where template strand is methylated by Dam methylase but the new one isn’t
  
  • Dam methylase
    
    • methylates DNA at the N⁶ position of all adenines within (5’)GATC
    • GATC is a palindrome present in opposite orientations on the two strands
• unmethylated GATC permits the new strand to be distinguished from the template strand
• MutL protein forms a complex with MutS protein and binds to all mismatched base pairs (except C–C)
• MutH protein binds to MutL-MutS complex and to GATC sequences
  
  • has site-specific endonuclease activity
  • inactive until complex encounters a hemimethylated GATC sequence
• DNA on both sides of mismatch is threaded through the MutL-MutS complex creating a DNA loop
• Simultaneous movement of both legs of the loop through the complex is equivalent to the complex moving in both directions at once along the DNA
• MutH catalyzes cleavage of unmethylated strand on the 5’ side of the G in GATC flagging it for repair
• When mismatch is on the 5’ side of cleavage site
  
  • unmethylated strand is unwound and degraded from 3’→5’, from cleavage site to mismatch, replaced with new DNA
  • requires DNA helicase II (UvrD helicase), SSB, exonuclease I or exonuclease X (both degrade strands of DNA from 3’→5’), DNA polymerase III, DNA ligase
• When mismatch is on the 3’ side of cleavage site
  
  • similar process
  • requires exonuclease VII (degrades ssDNA in 5’→3’ or 3’→5’ direction) or RecJ nuclease (degrades ssDNA in the 5’→3’ direction)
• DNA helicase II, SSB, and one of four different exonucleases removes a segment of the new strand between the MutH cleavage site and a point just beyond the mismatch
• resulting gap is filled in by DNA polymerase III
• nick is sealed by DNA ligase

Process in Eukaryotes

• eukaryotic cells have several proteins structurally and functionally analogous to the bacterial MutS and MutL (but not MutH) proteins
• MutS homologs in most eukaryotes, from yeast to humans, are MSH2 (MutS homolog), MSH3, and MSH6
• Heterodimers of MSH2 and MSH6 generally bind to single base-pair mismatches
• longer mismatches (2 to 6 bp) are bound by a heterodimer of MSH2 and MSH3, or by both types in tandem
• Homologs of MutL, predominantly a heterodimer of MLH1 and PMS1 (post-meiotic segregation), bind to and stabilize the MSH complexes
• PDF pg. 1060

Question 23

Base-Excision Repair

Answer 23

• DNA glycosylases
  
  • recognize DNA lesions (such as the products of cytosine and adenine deamination) and remove affected bases by cleaving the N-glycosyl bond
  • cleavage creates an apurinic or apyrimidinic site in the DNA, aka AP site or abasic site
  • specific for one type of lesion
  • exmples
    
    • Uracil DNA glycosylases
      
      • remove uracil from DNA that results from spontaneous deamination of cytosine
      • doesn’t remove uracil residues from RNA or thymine residues from DNA
    • UNG
      
      • most abundant human uracil glycosylase
      • eliminates occasional U residue inserted in place of a T during replication
    • hSMUG1
      
      • removes any U residues that occur in single-stranded DNA during replication or transcription
    • TDG and MBD4
      
      • remove either U or T residues paired with G, generated by deamination of cytosine or 5-methylcytosine, respectively
• the deoxyribose 5’-phosphate left behind is removed and replaced with a new nucleotide
  
  • AP endonucleases cut the DNA strand containing the AP site
    
    • position of incision relative to the AP site (5’ or 3’ to the site) varies with the type of AP endonuclease
  • segment of DNA including the AP site is then removed
  • DNA polymerase I replaces the DNA
  • DNA ligase seals the remaining nick
• PDF pg. 1061, 1062

Question 24

Nucleotide-Excision Repair

Answer 24

• repairs DNA lesions that cause large distortions in the helical structure
• the sole repair pathway for pyrimidine dimers in humans
• excinuclease, a multisubunit enzyme, hydrolyzes two phosphodiester bonds, one on either side of the distortion
  
  • In bacteria
    
    • hydrolyzes 5th phosphodiester bond on 3’ side and 8th phosphodiester bond on the 5’ side generating a fragment of 12 to 13 nucleotides (depending if lesion involves 1 or 2 bases.
  • In E. coli
    
    • ABC excinuclease has three subunits, UvrA (Mr 104,000), UvrB (Mr 78,000), and UvrC (Mr 68,000)
    • catalyze two specific endonucleolytic cleavages
    • A complex of UvrA and UvrB proteins (A2B) scans the DNA and binds to the site of a lesion
    • UvrA dimer dissociates, leaving a tight UvrB-DNA complex
    • UvrC binds to UvrB
    • UvrB makes an incision at 5th phosphodiester bond on 3’ side of the lesion
    • UvrC-mediated incision at 8th phosphodiester bond on 5’ side
  • In human
    
    • hydrolyzes 6th phosphodiester bond on 3’ side and 22nd phosphodiester bond on the 5’ side producing a fragment of 27 to 29 nucleotides
    • mechanism is quite similar to bacterial enzyme
      
      • 16 polypeptides with no similarity to the E. coli excinuclease subunits are required
• excised oligonucleotides are released
• resulting gap is filled—by DNA polymerase I in E. coli and DNA polymerase ´ in humans
• DNA ligase seals the nick
• PDF pg. 1063

Question 25

Direct Repair

Answer 25

• damage is repaired without removing a base or nucleotide
• Pyrimidine dimers result from a UV-induced reaction, and photolyases use energy derived from absorbed light to reverse the damage
• one by DNA photolyases
  
  • contain 2 cofactors that serve as light-absorbing agents, or chromophores
  • One of the chromophores is always FADH2
  • In E. coli and yeast, the other chromophore is a folate.
  • not present in the cells of placental mammals
• reaction mechanism entails generation of free radicals
• PDF pg. 1336

Question 26

error-prone translesion DNA (TLS) / sos response

Answer 26

• in certain types of lesions, such as double-strand breaks, double-strand cross-links, or lesions in a single-stranded DNA, the complementary strand is itself damaged or is absent
• Double-strand breaks and lesions in ssDNA arise when a replication fork encounters an unrepaired DNA lesion or from ionizing radiation and oxidative reactions
• At a stalled bacterial replication fork, there are two avenues for repair In the absence of a second strand
  
  • genetic info must come from a separate, homologous chromosome, thus involving homologous genetic recombination
  • error-prone translesion DNA

error-prone translesion DNA (TLS) / SOS response

• desperation strategy
• DNA repair becomes significantly less accurate
• part of a cellular stress response to extensive DNA damage known as the SOS response
• In bacteria
  
  • UvrA and UvrB are present at higher levels
  • UmuD protein is cleaved to a shorter form (UmuD’)
  • UmuD’ complexes with UmuC and RecA to create a specialized DNA polymerase, DNA polymerase V (UmuD’₂UmuC RecA), that can replicate past DNA lesions
    
    • umuC & umuD genes
      
      • fully induced only late in the SOS response
      • not activated for translesion synthesis initiated by UmuD cleavage unless levels of DNA damage are high and all replication forks are blocked
  • mutations resulting from DNA polymerase V kill some cells and create deleterious mutations in others
  • DNA polymerase IV, is also induced by SOS response
    
    • Replication is also highly error-prone
    • product of the dinB gene
• In eukaryotes
  
  • Mammals have many low-fidelity DNA polymerases of the TLS polymerase family
  • most have specialized functions in DNA repair
  • DNA polymerase η (eta) promotes translesion synthesis primarily across cyclobutane T–T dimers
    
    • Few mutations result because it preferentially inserts two A residues across from the linked T residues
    • frequency of mutation is minimized by the very short lengths (often one nucleotide) of DNA synthesized
  • DNA polymerases β, ι (iota), and λ, have specialized roles in eukaryotic base-excision repair
    
    • all have 5’-deoxyribose phosphate lyase activity in addition to polymerase activity
    • After base removal by a glycosylase and backbone cleavage by an AP endonuclease, these polymerases remove the abasic site (a 5’-deoxyribose phosphate) and fill in the very short gap

Question 27

Three Types of Genetic recombination events

Answer 27

• Homologous genetic recombination
  
  • involves genetic exchanges between any two DNA molecules that share an extended region of nearly identical sequence.
  • actual sequence of bases is irrelevant, as long as it is similar in the two DNAs
• site-specific recombination
  
  • exchanges occur only at a particular DNA sequence
• DNA transposition
  
  • involves a short segment of DNA with capacity to move from one location in a chromosome to another

Question 28

recombinational DNA repair

Answer 28

• In bacteria, homologous genetic recombination is primarily a DNA repair process
• directed at the reconstruction of replication forks that have stalled or collapsed at the site of DNA damage or during conjugation

Recombination Process

• replication fork encounters a damaged site under repair
• one arm of the replication fork becomes disconnected by a double-strand break and the fork collapses
• end of that break is processed by degrading the 5’ ending strand
  
  • PDF pg. 1071
  • RecBCD nuclease/helicase binds to linear DNA at a free (broken) end
  • RecB and RecD
    
    • helicases
    • molecular motors that propel the complex along the DNA
    • requires ATP
    • RecB moves 3’→5’ along one strand
      
      • degrades both strands
      • cleaves 3’-ending strand more often than 5’-ending strand.
    • RecD moves 5’→3’ along the other
  • RecC
    
    • binds to chi site ((5’)GCTGGTGG) on the 3’-ending strand, preventing its further degradation and is looped out creating an extension
  • enzyme continues to unwind and degrade the 5’-ending strand
  • RecA
    
    • its active form is an ordered, helical filament of up to several thousand subunits that assemble
    • forms on single-stranded DNA, such as that produced RecBCD enzyme
    • SSB are normally present and impedes RecA from binding on first few subunits to DNA
    • RecBCD loads RecA loader, facilitating the nucleation of RecA filament on ssDNA coated with SSB
    • filaments assemble/disassemble in 5’→3’
    • promotes central steps of homologous recombination, including DNA strand invasion
  • enhancement declines as the distance from
• the 3’ end invades intact duplex DNA connected to the other arm of the fork and pairs with its complementary sequence
• This creates a branched DNA structure (a point where three DNA segments come together)
• branch can be moved (branch migration) promoted by RuvAB to create an X-like crossover structure known as a Holliday intermediate
• Once a Holliday intermediate has been created, it can be cleaved by a specialized nuclease called RuvC
• nicks are sealed with DNA ligase
• PDF pg. 1070

Replication Fork Reassembles

• After recombination steps are completed the replication fork reassembles in a process called origin independent restart of replication
• Four proteins (PriA, PriB, PriC, and DnaT) act with DnaC to load the DnaB helicase onto the reconstructed replication fork
• DnaG primase synthesizes an RNA primer
• DNA polymerase reassembles on DnaB to restart DNA synthesis
• requires DNA polymerase II (role unknown) this polymerase II activity gives way to DNA polymerase III

Question 29

homologous genetic recombination

Answer 29

• In eukaryotes
• has several roles in replication and cell division, including the repair of stalled replication forks
• occurs with the highest frequency during meiosis
• contributes to the repair of several types of DNA damage
• provides a transient physical link between chromatids promoting orderly segregation of chromosomes at the first meiotic cell division
• enhances genetic diversity in a population

Process

• PDF pg. 1074
• Step 1: begins with replication of DNA in diploid germ-line cell
  
  • textbook example has 4 homologous chromosomes
• Step 2: Chromosomes duplicate
  
  • creates sister chromatids
  • held together by cohesins at their centromeres
• Step 3: homologous chromosomes pair up
  
  • synapsis begins (very close pairing association) at the centromeres
  • three-part synaptonemal complex develops between homologous chromosomes holding them 2gether forming tetrads
  • each paired but not fused homologous pair has 4 chromatids
  • recombination, crossover occurs
    
    • not a random process; “hot spots” have been identified
    • frequency
      
      • roughly proportional to the distance between two points
      • allows determination of the relative positions of and distances between different genes
    • double-stranded DNA in each inner chromatid unwinds, is cut & repaired → segments of nonsister chromatids are exchanged 
    • two models: Holliday and double-strand-break model
    • double strand break model
      
      • synapsis breaks down
      • DNA brakes closed up by joining each broken end to segments of non-sister chromatid (paternal chromatid is joined to a piece of maternal chromatid)
      • These points of crossing over become visible as chiasmata after synaptonemal complex disassembles & homologs move slightly apart from each other
  • centromeres of paired chromosomes move apart
  • the two homologs remain attached at each chiasma
• Step 4: First meiotic division
  
  • homologous pairs of chromosomes align along the metaphase plate
  • orientation of each pair w/respect to others is random → Mendel’s Law of Independent Assortment
  • homologous pairs disjoin & move to opposite poles
    
    • If improper spindle fiber attachment occurs, a cellular kinase senses lack of tension, spindle attachments are removed and cell tries again
  • sister chromatids remain attached and travel together
• Step 5: nuclear envelope reforms and 2 haploid cells
• Step 6: Second meiotic division
  
  • chromosomes line up on metaphase plate
  • sister chromatids facing opposite poles
  • sister chromatids separate & pulled to opposite poles
  • each now a chromosome
• Step 7: Meisosis II results in 4 unique haploid gametes
  
  • chromosome # has doubled

meiosis I - Prophase I

• has 5 stages: leptotene, zygotene, pachytene, dyplotene, diakinesis
• after 5 stages, nuclear membrane breaks down & spindle forms
• Leptotene
  
  • “thin thread”
  • chromosomes condense & become visible
  • homologous chromosomes come closer to each other
  • Double-strand breaks are introduced and processed
• zygotene
  
  • “paired thread”
  • chromosomes continue to condense
  • homologous chromosomes pair up and begin synapsis (very close pairing association)
  • bivalent or tetrad: each paired but not fused homologous pair has 4 chromatids
• pachytene
  
  • “thick thread”
  • pairing is complete
  • condensation continues
  • recombination, crossover
  • three-part synaptonemal complex develops between homologous chromosomes holding them 2gether
  • DNA synthesis happens during homologous recombination/cross-over
  • requires double-stranded DNA in each inner chromatid to unwind, be cut & repaired → segments of nonsister chromatids are exchanged
  • two models: Holliday and double-strand-break model
• diplotene
  
  • broken end to segments of non-sister chromatid (paternal chromatid is joined to a piece of maternal chromatid)
  • These points of crossing over become visible as chiasmata after synaptonemal complex disassembles & homologs move slightly apart from each other
• diakenisis
  
  • “moving apart”
  • centromeres of paired chromosomes move apart
  • the two homologs remain attached at each chiasma

Question 30

homologous genetic recombination

Double Strand Breaks

Answer 30

• PDF pg. 1075
• step 1
  
  • homologous chromosomes are aligned
  • a double-strand break in a DNA molecule is created
  • exposed ends are processed by an exonuclease
  • Strands with 3’ ends are degraded less than those with 5/ ends, producing 3’ single-strand extensions with a free 3’-hydroxyl group at the broken end
  • similar to “sticky ends”
  • Each of the 39 ends can then act as a primer for DNA replication
• step 2
  
  • and exposed 3’ ends invade the intact duplex DNA of the homolog
  • it pairs with its complement in the intact homolog
  • The other strand of the duplex is displace
• step 3
  
  • invading 3’ end is extended by DNA polymerase & branch migration
  • this generates a DNA molecule with two crossovers in the form of branched structures called Holliday intermediates
• step 4
  
  • exposed non-invading 3’ end aligns w/displaced strand from intact duplex DNA homolog
  • exposed non-invading 3’ is extended by DNA polymerase (using displaced strand as a template)
  • this replaces DNA missing from the original double strand break
• step 5
  
  • Holliday intermediate resolvases cleave the Holliday intermediate
    
    • RuvC-like nuclease
  • This generates two recombination products
  • If cleaved one way, the DNA flanking the region containing the hybrid DNA is not recombined
    
    • Product set 1 (Figure 25-35a, pg. 1075)
    • Picture cleavage of Holliday horizontally down the middle between the two homologs
  • if cleaved the other way, the flanking DNA is recombined
    
    • Product set 2 (Figure 25-35a, pg. 1075)
    • Picture the recombination of all homologs staying the same as the Holliday
    • Only the ends are recombined
      
      • The right ends of the top two chromatids combine with the left ends of the bottom two chromatids
      • The left ends of the top two chromatids combine with the right ends of the bottom two chromatids
      • middle portions stay the same

Question 31

Site-Specific Recombination

Answer 31

• Results in Precise DNA Rearrangements
• limited to specific sequences
• occur in virtually every cell
• recombinase
  
  • acts on the recombination site
    
    • short (20 to 200 bp), unique DNA sequence
    • partially asymmetric (nonpalindromic)
    • align in the same orientation during the recombinase reaction
    • If the two sites are on the same DNA molecule
      
      • reaction either inverts or deletes the intervening DNA
      • determined by if recombination sites have the opposite or same orientation
      • PDF pg. 1079, Figure 25-38.a
    • If the sites are on different DNAs
      
      • intermolecular recombination
      • if one or both DNAs are circular, the result is an insertion
      • PDF pg. 1079, Figure 25-38.b
  • acts as a site-specific endonuclease and ligase
• One or more auxiliary proteins regulate the timing or outcome of the reaction
• two general classes: Rely on either Tyr or Ser residues in the active site
• In both types of system, the exchange is always reciprocal and precise, regenerating the recombination sites

Tyr Reaction

• PDF pg. 1078, Figure 25-37
• step 1
  
  • Tyr residues act as nucleophiles at the active site
  • reaction is carried out within a tetramer of identical subunits
  • Recombinase subunits bind to a specific sequence on recombination site
• step 2
  
  • One strand in each DNA is cleaved at particular points in the sequence
  • nucleophile is the —OH group of an active-site Tyr residue
  • the product is a covalent phosphotyrosine link between protein and DNA
• step 3
  
  • After isomerization , the cleaved strands join to new partners, producing a Holliday intermediate.
• step 4
  
  • a second reaction similar to the first occurs
  • The original sequence of the recombination site is regenerated after recombining the DNA flanking the site

Ser Reaction

• PDF pg. 1078
• both strands of each recombination site are cut concurrently and rejoined to new partners without the Holliday intermediate

On a circular bacterial chromosome

• PDF pg 1079, Fig. 25–39
• sometimes generates deleterious byproducts
• resolution of a Holliday intermediate at a replication fork by a nuclease (RuvC) followed by completion of replication can produce either
  
  • two monomeric chromosomes
  • or a contiguous dimeric chromosome
    
    • covalently linked chromosomes cannot be segregated to daughter cells at cell division
    • XerCD system, converts the dimeric chromosomes to monomeric chromosomes so that cell division can proceed

Question 32

Transposon

Answer 32

• segments of DNA
• found in virtually all cells
• transposition
  
  • move, or “jump,” from one place on a chromosome to another on the same or a different chromosome
  • target is random
• DNA sequence homology is not required
• tightly regulated cuz insertion in essential gene can kill cell
• molecular parasites, adapted to replicate passively within the chromosomes
• bacteria have 2 classes
  
  • 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 one or more genes in addition to those needed for transposition
    • extra genes might confer additional capabilities likes resistance to antibodies
  • vary in structure
    
    • have short repeated sequences at each end that serve as binding sites for the transposase
• In Eukaryotes
  
  • structurally similar to bacterial transposons
  • some use similar transposition mechanisms
  • others seem to involve an RNA intermediate

Transposition Pathway

• Direct or simple tranposition
  
  • PDF pg. 1080, Fig. 25–40
  • cuts on each side of the transposon excise it
  • transposon moves to a new location
  • the double-strand break in the donor DNA that must be repaired
  • At the target site, a staggered cut is made
  • transposon is inserted
  • a short sequence at the target site (5 to 10 bp) is duplicated to form an additional short repeated sequence that flanks each end of the inserted transposon
  • These duplicated segments result from the cutting mechanism used to insert a transposon into the DNA at a new location
  • sequences are generally only a few base pairs long
• replicative transposition
  
  • the entire transposon is replicated, leaving a copy behind at the donor location
  • cointegrate is an intermediate in this process, consisting of the donor region covalently linked to DNA at the target site
  • Two complete copies of the transposon are present in the cointegrate, both having the same relative orientation in the DNA
  • In some well-characterized transposons, the cointegrate intermediate is converted to products by site-specific recombination, in which specialized recombinases promote the required deletion

Question 33

antibody diversity

Answer 33

• PDF pg. 1336, Figure 25-42
• A human (like other mammals) is capable of producing millions of different immunoglobulins (antibodies) with distinct binding specificities, even though the human genome contains only ,29,000 genes
• Recombination allows an organism to produce an extraordinary diversity of antibodies
• the generation of complete immunoglobulin genes from separate gene segments is one example
• The genes for these polypeptides are divided into segments, and the genome contains clusters with multiple versions of each segment
• The joining of one version of each gene segment creates a complete gene