Unit III Replication, Repair, Recombination
I. Replication - general (whole chromosome level).
A. Circular E. Coli chromosome remains so during replication, forms θ (theta) structure.
B. 1. Replication starts at a specific place on the “circle” each time.
2. Evidence: in rapidly growing culture all genes are present at between 1 and 2 times the
frequency of the “tre” gene. Those at 2 X “tre” are near the origin.
C 1. Proceeds in both directions from the origin
2. Evidence: “label” DNA synthesis with moderately radioactive dTTP. Then switch to medium
with highly radioactive dTTP. Autoradiograph (lay a piece of X-ray or photographic film over layer of
cells. Radiation exposes; causes spots on film) Fig 30-4
3. Results: Darkly exposed regions occur on both sides of the lightly exposed one.
II. Enzymes of replication
A. (DNA) ligase: joins 3' OH to 5' mono PO4, using energy of ATP --> AMP + PPi. The two strands
joined must each be base paired to a “template” in the region of joining: no gap, adjacent. (ligase
“seals a nick” in double stranded DNA) Fig 30-20
B. Topoisomerase: changes the supercoiling of DNA. Fig 29-17
1. Supercoiling is more or less twist (or bps per turn) than the DNA would have as a result of
interactions within it and between it and the solution (i.e., differences from its “natural twist”).
2. Type I relaxes supercoiling, does not need ATP (uses mechanical energy of supercoil). Type
IA, present in all cells, relaxes negative supercoiling; type IB relaxes positive or negative.
3. A clue to the mechanism of type IA was the observation that it catenates (interlinks) and
decatanates ssDNA circles. Fig 29-25
4. Type IA’s make a single-strand break by attack of one of its tyrosyl side chain hydroxyls on the
5' PO4 of the DNA. (The tyr side chain and the 5' PO4 become covalently linked.) Then, the other ss
DNA is “passed through” the break and the 5' PO4 is transferred back to the 3'OH to which it was
originally linked. (Unwinds one twist)
5. The mechanism begins with the tight binding of one of the strands in a groove in domain I.
This places one of its PO4 groups in proximity to the active site tyr, represented by the phe side chain
F328 of domain III in Fig 29-26.
6. Formation of the tyr-5’ PO4 linkage occurs along with a conformational change that moves
domain III (which holds the 5’ PO4) away from domain I (which holds the ssDNA segment with the 3’
OH ). This creates a gap in the cut DNA strand and in the enzyme that the other DNA strand can pass
through. The 5’ PO4 is passed back to the 3’ OH and the enzyme releases the product. Fig 29-27
C. Gyrase: a type II topoisomerase
1. Gyrase introduces negative supertwists using ATP --> ADP. It also catenates and
decatanates dsDNA circles.
2. Negative supercoiling is required:
a. to keep the E. Coli chromosome in compact form.
b. to initiate replication at the origin.
c. in order for replication to proceed.
d. to initiate transcription.
3. Type II’s make a double-strand break and pass a ds DNA through; this unwinds two twists.
4. Both strands’ 5'PO4's are linked to tyr OH’s as a covalent intermediate during “pass through”.
5. When the first DNA segment (G: “gate”) binds, the gyrase conformation changes with the A’
and B’ subunits holding it. Fig 29-32
6. ATP binding and the binding of a second DNA segment (T: “transported”) induce the
conformation to change again, with the tyr side chains attacking the G segment. This allows the T
segment to pass through.
7. After T-segment release it can revert to the G-bound form.
D. DNA polymerase I
1. Polymerizes in the 5' --> 3' direction. This is the same rxn as RNA Pol: the 3’ OH on the
existing chain attacks the alpha phosphate of dNTP and ester links to it, releasing pyrophosphate. Fig
30-1 Which dNTP is used is determined by the template strand. After about 20 nucleotides are added,
Pol I dissociates (moderately processive). It polymerizes at the rate of about 600 residues per second
per Pol I molecule.
2. Proofreads: “checks” each newly added nucleotide for “correct fit” (base pairing), hydrolyzes
the bond just formed if not correct. This is 3' --> 5' exonuclease.
3. Also corrects errors “in front” of new polymer. This is a 5' --> 3' exonuclease activity (see
“Okazaki” below).
4. Each activity is on a different region of DNA Pol.I. Pol I has a deep cleft which partially
surrounds the DNA.
5. The combination of its polymerizing and its 5’ --> 3’ exonuclease activities has an effect called
“nick translation”: the nick “moves” along the DNA strand. Fig 30-12
E. DNA Polymerases II and III.
1. Discovered in a mutant strain of E.Coli with defective DNA Pol I. This strain grows (replicates)
normally but is very sensitive to DNA damage.
2. Both have the same requirements and enzymatic activities as DNA Pol I, except that they lack
the 5’ --> 3' exonuclease.
3. Pol III has a “sliding clamp” which holds the DNA in a “closed circle”. Pol III polymerizes at
about 9000 residues per second per molecule. Fig 30-13
F. Primase
1. DNA Pols can’t initiate polymerization on a template, linking residues 1 and 2 of a polymer,
but RNA polymerases do.
2. A special RNA Pol, “primase”, forms a short RNA oligomer of about 11 residues (resulting in
an ~11 bp DNA-RNA hybrid). Fig 30-7 Notice the error in this figure: the DNA is polymerized from the
3’ OH, the head of the green arrow; there should be no gap between it and the DNA (the Okazaki
fragment).
3. This is the primer (a base-paired oligomer or polymer with a free 3' OH) that is needed by
the DNA Pols as the starting point for polymerization.
G. Helicase: an enzyme that separates the strands of dsDNA. (Helicase does not unwind DNA.)
1. DnaB is an E Coli hexamer of identical subunits that surrounds one strand of the DNA (the
“lagging strand”, see below) and moves along it using ATP hydrolysis so as to separate the parental
strands and advance the replication fork, much like the T7 helicase/primase. Fig 30-15
2. Rep helicase binds to the other DNA strand (the “leading strand”) and separates the strands
by an “active, rolling mechanism”. It doubles the rate of E Coli replication, but is not required. Fig 30-17
a. When either of its identical subunits is bound to ssDNA, the “2nd” subunit can bind to ATP
and dsDNA.
b. The “2nd” subunit separates the dsDNA and remains bound to the same strand as the “1st”
subunit.
c. The “2nd” subunit releases ADP and Pi, while the “1st” subunit releases the ssDNA. This
returns the complex to its starting form, from which the process can repeat.
III. Replication - Specific events
A. Semi-discontinuous replication
1. Dilemma: all polymerases work in the 5' --> 3' direction, but both parental strands serve as
template and only one of them opens in the 3' --> 5' direction. (Can’t polymerize 5' --> 3' on a 5' --> 3'
template, strands of duplex must be antiparallel)
2. Solution: the “leading” strand is synthesized continuously, complimentary to the template
strand that is opening 3' --> 5'. But synthesis of the “lagging” strand is done by making short (10002000 residues) fragments (Okazaki fragments), which are then linked together. The leading strand is
made simultaneous with movement of replication fork, but 1-2000 bases of the template for the
lagging strand must be exposed before an Okazaki fragment can be formed in the opposite direction
to movement of replication fork. Fig 30-5
3. Evidence for Okazaki fragments:
a. When E Coli are labeled for 30 seconds with [ 3H] thymidine, the radioactivity is in small
1000-2000 residue DNA fragments.
b. If the cells are transferred to “cold” medium after the 30 second [ 3H] pulse, the radioactivity
is in larger DNA fragments the longer the cells are allowed to grow.
c. Electron micrographs show ssDNA regions on opposite strands at replication forks on
opposite ends of replication eyes. Fig 30-6
B. Initiation of Replication: Fig 30-29
1. The origin (Ori C) has a specific sequence (TTATCCACA) repeated five several times, the
“DnaA boxes”.
2. DnaA protein binds to these segments only if DNA is negatively supercoiled, forming a DNA
loop. DnaA causes the dsDNA to melt at an AT rich sequence repeated 3 times at Ori C’s “left”
boundary.
3. DnaB-DnaC hexamers then bind to the ends of the melted region. After DnaC is released,
ATP is hydrolysed and five more DnaA subunits bind. Then, DnaB helicase begins to advance the
replication forks.
4. As helicase moves and separates the strands, single strand binding (ssb) protein binds each
of the ssDNAs to keep the 2 templates exposed.
C. 1. The opening of the duplex and the advancement of the replication forks “sends positive
supertwists” up beyond the replication forks. These are released by action of gyrase, which keeps the
chromosome negatively supercoiled.
2. The priming of leading strand synthesis on each parental strand is done by a complex of
primase and RNA Polymerase. This only needs to be done once, at the origin. Primase primes the
synthesis of each Okazaki fragment on the lagging strand at each fork, so this is done 2000-4000
times in completing the replication of the chromosome.
D 1. A dimer of the very large DNA Pol III (~900,000 MW vs 103,000 for DNA pol I monomer) then
begins to polymerize DNA on both templates (that for the lagging strand “looping back” through one
subunit of the dimer while the other subunit advances on the leading strand). Fig 30-28
2. Each Okazaki fragment must be “primed”. When Pol III reaches the previous Okazaki
fragment, it releases the lagging strand template and reinitiates on a new primer polymerized by
primase after 1000-2000 residues of the lagging strand template have been exposed.
3. Pol I binds to the nick at the junction of each Okazaki fragment and the growing lagging
strand, removes the primer using its 5' --> 3' exonuclease activity, and replaces it with DNA using its
polymerase activity (nick translation).
4. Ligase then “seals the nick”, joining each Okazaki fragment to the growing lagging strand.
This is repeated 1000-2000 times on each of the lagging strands, linking the Okazaki fragments to
form half of the total DNA on each daughter strand
(Section E, which follows, was skipped in 2006)
E. Termination of Replication Fig 30-35
1. There are seven nearly identical terminator sequences (Ter A, B, C, D, E, F, and G) of about
23 bp around the point “directly across the circle” from the origin (“the half-way point” around the
circular chromosome). There are three on one side of this point and four on the other.
2. When each replication fork reaches the Ter sites preceding this point, it can move through,
but each of the Ter sites beyond half-way is oriented to stop the fork. As a result, whichever fork
reaches this region first must wait there for the other.
3. Tus protein binds Ter sequences and apparently interacts with the DnaB helicase to stop
movement of the fork. When Tus is fused using recombinant-DNA technology to other sequencespecific DNA-binding proteins, it stops forks at the other proteins site.
4. Tus binds to more than two thirds of the of the phosphates in a 13 bp segment of Ter, and the
side chains of the residues of a β sheet that connects its two domains makes bonds with atoms of
bases in the major groove. It partially unwinds the end of the Ter DNA segment that allows forks to
move through. Fig 30-36
5. It is unclear how this system is beneficial, and it is not essential for survival, but it is widely
used in bacteria.
IV. DNA Polymerase Mechanisms
A. DNA Pol Structure Fig 30-9
1. Upon the binding of a properly base-paired dNTP, the O, O1, and O2 helices close over the
bound dNTP. This is required before the catalytic process can occur.
2. The involvement of this conformational change was shown in studies on crystals of a
fragment of DNA Pol I from Thermus Aquaticus (Taq) that contains the polymerase activity (Klenow
Fragment).
a. When the crystals were incubated in solutions of ddCTP (2’, 3’ – dideoxy CTP), ddCTP
was linked to the 3’ end of a bound 11bp dsDNA that had a GGAAA-5’ ssDNA overhang. The enzyme
was active in the crystals. Another ddCTP was bp’d to the second G on the template. The enzyme
was in the “closed” conformation it takes on when a dNTP is bp’d at the active site.
b. When these crystals were then soaked in a solution without ddCTP, the ddCTP dissociated
from the active site and the enzyme reverted to the open conformation.
B. Catalytic Mechanism Fig 30-10
1. All known DNA polymerases have 2 metal ions, usually both are Mg2+, bound to two aspartate
side chains (R = -CH2CO2-) at the active site.
a. One of the Mg2+ makes ionic bonds to all three phosphates of the bound dNTP.
b. The other Mg2+ is bound to the α-phosphate and the O of the 3’ OH, where it enhances the
nucleophillic attack of this O on the P of the α-phosphate.
C. The “Clamp Loader” for DNA Pol III
1. Each time Pol III reaches the primer of the previous Okazaki fragment on the lagging strand, it
releases the lagging strand. The Pol III core that has the enzymatic activities also dissociates from the
sliding clamp, which consists of two β subunits. See Fig 30-28
2. The sliding clamp “circle” that remains at the site of the dissociation is opened up to a “C”
shape by the δ (delta) subunit of Pol III holoenzyme, so as to release the DNA. The δ subunit has a “β
interaction element”, which consists of an alpha helical segment with leu and phe side chains. The
binding of this portion to the clamp stabilizes the open form. Fig 30-32
3. Meanwhile, the clamp loader has fitted another clamp onto the lagging strand where the
primer for the next Okazaki fragment is synthesized. The core then binds to the clamp and begins
synthesis of the next Okazaki fragment. The core has a very high affinity for the sliding clamp in
association with primed ssDNA, and this enables it to displace the clamp loader.
4. The clamp loader consists of 5 subunits (γτ2δδ’; one gamma, two taus, one delta, and one
delta prime). (γ is produced from the same gene as tau, and is a truncated or shortened form of it
consisting of its N-terminal domain. Since it is this domain that is involved in the clamp loader, it is
often symbolized as γ3δδ’.) Fig 30-34
5. The clamp loader forms a circle of about the same diameter as the clamp. When the three
gammas have each bound one ATP (which they must do in sequence, as the conformation changes
upon ATP binding to uncover the next ATP binding site, finally exposing the δ subunit’s β-binding
element), the loader can bind a closed clamp consisting of two β subunits. This “springs” the clamp
open, allowing it to “load” onto the DNA. Once DNA binds, ATP is hydrolysed and the clamp opening
reverses.
V. Eukaryote Replication
1. Much of the process of replication is similar in eukaryotes to that in prokaryotes, above. One
of the major differences is the much greater length of eukaryote chromosomes. Also, the
polymerization rate is much slower. To compensate, eukaryote chromosomes have numerous
replication origins rather than a single one as in prokaryotes. Fig 30-42
2. The ends of the linear eukaryote chromosomes cannot be replicated as above. Even if an
RNA primer was initiated opposite residue #1, that would still leave the 11 residue primer as
unreplicated DNA. The telomeres or chromosome ends have several thousand repeats of a short
sequence, TTGGGG in vertebrates, TTAGGG in tetrahymena.
3. Ribonucleoprotiens (RNA-protein complexes) called telomerases synthesize these ends. The
one in tetrahymena contains RNA with a sequence complimentary to the telomere sequence. This
RNA serves repeatedly as the template for telomere synthesis. Fig 30-47
4. Specific proteins bind to telomeres and “cap” them to prevent them from linking to the ends of
other chromosomes. Somatic cells lack telomerases, so their repeated sequences get shorter the
more times they divide. The shorter the repeat, the more likely a capping protein will not be bound, so
that chromosomes will be fused, and as a result, broken during mitosis. Eventually, this is the end of
that cell line and this is the basis of the limit on the number of divisions a somatic cell can undergo.
5. Evidence of the relationship between shortened telomeres and cell mortality.
a. Production of telomeres by telomerase allows tetrahymena to divide indefinitely
(immortality). When tetrahymena telomerase is mutationally impaired, it can no longer divide
indefinitely.
b. The cells of people with progeria, a disease of rapid premature aging, have short telomeres
and reduced ability to undergo repeated cell divisions.
VI. General Recombination - exchange of segments of DNA between 2 duplexes in which there is
homology (same sequence) between exchanged segments
A. Holliday model steps: Fig 30-64
1. Alignment of alleles (alleles: alternate forms of a gene at corresponding locations on 2
chromosomes)
2. Nicking (phosphodiester bond hydrolysis) by endonuclease of one strand on each duplex
3. “Invasion” of the other duplex by single stranded DNAs, and base pairing (sequence
homology), resulting in hybrid duplexes.
4. Nick sealing --> covalent hybrids
5. Branch migration, the continued exchange of strands; the cross-over point “moves”
6. Finally, 2 nicks, and 2 ligations resolve the crossover to produce the recombinants
B. Experimental evidence: Fig 30-67
1. Treat “figure 8" plasmid dimers with restriction endonuclease (RE).
a. Plasmids are “mini chromosomes” of bacteria that contain a replication origin and a few
genes which often confer resistance to antibiotics.
b. REs recognize a specific sequence and nick both strands of dsDNA at the sequence,
which occurs only ONCE on this plasmid. In cutting two separate, identical plasmids, it acts at the
same relative location on each.
2. Results: X (chi) forms with 2 pairs of arms - within a pair the arms are always equal in
length.
3. Conclusion: crossover is at a point of homology. (Since the RE cuts at the same place on
each plasmid, the ends of the X arms represent the same place on each plasmid. Going an equal
distance from the ends on each of the equal-length arms of a pair to the cross-over point, one is at the
“same place” on each plasmid, where each has the same sequence)
VII. Enzymes of Recomination
A. RecA:
1. A helical filament of RecA forms around a single stranded DNA (ssDNA) or dsDNA that has
gap in one strand creating a ssDNA region. Fig 30-68, 69
2. A filament of RecA on ssDNA can then bind to dsDNA and promote invasion of it if it has a
strand complimentary to the ssDNA.
3. RecA partially separates the strands of the dsDNA and rotates the residues so that the bases
are accessible for base pairing to the ssDNA. Fig 30-70
4. The ssDNA base pairs with the complimentary strand of dsDNA, displacing the homologous
strand
5. Rotation of the RecA helical filament is driven by ATP hydrolysis on RecA. This promotes
strand exchange by unwinding dsDNA and “spooling in” the ssDNA. Fig 30-72
6. The exchange of strands must begin from the 3’ end of the strand in the dsDNA that is
complimentary to the ssDNA, not the from the 5’ end or in the midst of a ds segment. This indicates
the directionality of the filament rotation and strand exchange process.
7. One of the strands must have a free end.
B. RecBCD: Fig 30-73
1. Consists of 3 proteins which form a complex that has helicase and nuclease activities.
2. Binds to end of linear dsDNA and moves along it using ATP hydrolysis, separating the two
strands and “chewing” both strands into little fragments using its nuclease activity.
3. When it reaches a specific sequence (5’GCTGGTGG3’) called “chi”, it stops degrading the
strand with the 3’ end and accelerates the degrading of the other. Chi occurs at an average of about
every 5 kbp (about 800 chi per E Coli chromosome).
4. As RecBCD continues to move along and do this, it frees a ssDNA, to which RecA can bind.
C. RecA-like proteins of Eukaryotes
1. Yeast RAD51 is 30% homologous to recA. It forms helical filaments on DNA like RecA does.
2. RAD51 functions in ATP-dependent DNA repair and replication in the same way as RecA.
3. Chickens, mice and humans have homologs of RAD51. Apparently, RecA-like filaments
function in recombination in all organisms.
D. RuvABC in Branch Migration and Holliday Junction Resolution
1. RuvA forms a homotetramer that binds Holliday junctions. It has four grooves in its surface
that are lined with positively charged residues that bind the DNA. Fig 30-74
2. The “pins” or projections at the center of the tetramer have negatively charged residues that
repel the ssDNA segments and facilitate strand exchange in branch migration.
3. In the presence of DNA, RuvB forms a homohexamer with a central tunnel that binds the
DNA. Fig 30-75, 76
4. One face of the hexamer has a β hairpin on each subunit that binds RuvA. RuvB binds DNA
weakly in the absence of RuvA, but strongly in its presence.
5. One RuvB hexamer binds to each of the dsDNAs extending from two opposite sides of RuvA.
Rotation of each hexamer screws the dsDNAs in toward the center of RuvA, from which they emerge
newly-partnered as the perpendicular dsDNAs.
6. RuvC is a homodimeric nuclease that, by binding to center of the face of RuvA, may be able
to catalyze the production of the two nicks required to resolve the crossover. There are two problems
with this that are as yet unresoved: (1): RuvA tetramers probably associate face-to-face most of the
time, denying RuvC access to the junction. (2): The conformation of RuvC observed in its crystal
structure doesn’t fit the RuvA-junction complex so as to bind the DNA very well.
VIII. Transposition: Recombination Without Homology. Segments of prokaryote and eukaryote
chromosomes move to random locations distant from their previous sites, at low freqency.
A. Insertion Sequences (ISs) are the simplest mobile genetic elements. These transposons have
no replication origin so their existence separated from main chromosome is not perpetuated. Instead,
they move from place to place within the main chromosome.
1. They can be inserted almost anywhere, without homology.
2. They encode (contain a gene for) transposase, the protein enzyme which catalyses the
insertion.
3. Insertion results in duplication of a short sequence on the main chromosome at the point of
insertion.
4. Insertion sequences (IS) contain “inverted repeats” at their ends, which are binding sites for
transposase. Fig 30-80
5. Transposase makes staggered cuts, the IS is joined to end of each ssDNA “overhang”, and
then the gaps are filled in by DNA Pol I and ligase seals the nicks. Fig 30-81
B. General recombination between 2 insertion sequences: the results depend on their orientations
1. Two ISs inverted on the same chromosome ---> inversion of genes between (this is
physiologically important in determining flagellar protein of salmonella) Fig 30-91
2. Two ISs in the same orientation on the same chromosome ---> deletion of genes between
them (possibly producing a complex transposon, which is like an IS, but contains additional genes)
3. Two in the same orientation on homologous chromosomes (say, in a eukaryote, in which the
transposons are near misaligned alleles) ---> gene duplication (and deletion)
C. Cut and Paste Transposition of Tn5
1. Tn5 has 4 genes, one for Tn5 transposase, and three antibiotic resistance genes which code
for enzymes that degrade antibiotics.
2. It is bounded on each end by “outside end” (OEs) sequences of 19bp, the binding sites for
Tn5 transposase. Fig 30-84
3. When bound to the OEs, the Tn5 transposases bind to each other (dimerize) and become
catalytically active.
4. Each transposase activates a water molecule to hydrolyze the bond on the end of its OE,
leaving the OE with a free 3’ OH on its end.
5. It then activates the 3’OH to attack the other strand, forming a hairpin on each end of Tn5 and
excising it from the chromosome. The hairpin is then hydrolyzed, yielding a blunt-ended dsDNA.
6. The Tn5 dimer binds elsewhere on the main chromosome or on a plasmid. The 3’ OHs on the
ends of Tn5 attack phosphates on opposite strands 9 bp apart, inserting Tn5. This leaves a 9 base
ssDNA gap on each end, which is filled in by DNA Pol I, duplicating this sequence, and ligase seals
the nicks. Fig 30-85
7. The dsDNA breaks left where Tn5 was excised must be repaired.
D. Transposons in Humans
1. About 3% of the human genome (genome = total DNA or chromosomal content) consists of
sequence derived from DNA-based transposons like those above (ISs, Tn5), but most are not active in
transposition.
2. About 20% of the human genome consists of one type of transposon that transposes by way
of an RNA intermediate, a retrotransposon.
3. The involvement of an RNA intermediate was shown by modifying the sequence content of
Ty1, the most common transposon of budding yeast.
a. A yeast intron was inserted in the midst of it.
b. A galactose-sensitive promoter was placed in front of it.
4. Results:
a. The transposition rate varied with the galactose concentration, paralleling the transcription
rate.
b. The transposed elements (DNA segments inserted elsewhere in the genome) did not
contain the intron (indicating it had been spliced out while the transposon was in the form of the RNA
intermediate).
5. Ty1 is related to, and probably a descendent of a retrovirus, which is an RNA virus that has a
dsDNA form of its chromosome inserted into the host’s chromosome at one point in its life cycle.
a. A retrovirus genome is flanked by LTRs (long terminal repeats: direct repeats of the same
sequence in the same orientation) of 250 to 600 bp on its ends. Ty1 is flanked by a LTRs. Fig 30-96
b. The retroviral genome (including that of HIV, the AIDS virus) contains 3 genes:
i. gag, which codes for the proteins of the viral core that associate with the progeny
chromosomes.
ii. pol, which codes for reverse transcriptase (RT), the enzyme that produces a dsDNA
“copy” of its ssRNA chromosome, and integrase, the enzyme that inserts this dsDNA into the host
chromosome.
ii. env, which codes for the proteins on the outer surface of the viral progeny. The env
proteins encase the core during the process of progeny release from the host cell.
c. Ty1 has two genes, which correspond to gag and pol, but lack an env type of gene.
Consequently, the “daughters” of Ty1 cannot be released from the host cell, instead remaining in it
until their dsDNA form is inserted into the host chromosome.
6. Mechanism of retrotransposition: Fig 30-97
a. The reverse transcriptase (RT) has an endonuclease activity that nicks the chromosomal
DNA at a point complimentary to the 3’ end of the LTR.
b. The retrotransposon RNA bps here, with the chromosomal DNA providing a primer for RT,
which synthesizes a DNA complimentary to the RNA (the reverse of transcription).
c. RT catalyses hydrolysis of the RNA bp’d to the DNA, and then synthesizes a DNA strand
complimentary to the DNA.
d. The dsDNA form is integrated at this point in the chromosome
IX. Other Mobile Genetic Elements.
A. The F factor plasmid: contains genes for proteins involved in sexual conjugation (a “pilus” forms
a connection between 2 cells, allowing DNA to be transferred). Fig 31-3
1. An F+ (F plus) cell contains the F plasmid, can transfer it to an F - (F minus) cell (which
doesn’t). Fig 31-4
a. A nick is made in one strand of the F plasmid.
b. The 5’ end of the nicked strand moves into the F - cell, while DNA polymerase works from
the 3’ OH of the nicked strand (“rolling circle” replication). Fig 30-26
c. After transfer of a unit length single strand, replication on this template in the recipient
(formerly F-) cell produces a duplex F plasmid. (Both cells now F +).
2. Alternatively, the F plasmid may be integrated into the main chromosome, forming an “Hfr”
chromosome (so called because these cells undergo recombination at high frequency). In conjugation,
Hfr cells pass the entire Hfr chromosome to the F — cell, by the same mechanism as F+. (But mating is
usually interrupted before transfer is completed, so that only part of the F factor genes and part of the
main chromosome genes are transferred.) Fig 31-5
3. The F plasmid may “loop out” of the Hfr chromosome, reforming the F plasmid and main
chromosome. This may occur at the wrong site on rare occasions, forming F’ (F prime) (which
contains some of the main chromosome genes). F’ may be transferred in conjugation.
B. The λ phage chromosome may also incorporate into the E Coli main chromosome, where it lies
dormant (no transcription) and is inherited for many generations.
1. When it “loops out” it forms λ and main chromosome, except on rare occasions in which it
brings with it genes from the main chromosome adjacent to the attachment site. These are
transducing phage. (Other phage transduce at high frequency)
2. How does the λ phage chromosome “loop in” (integrate) and “loop out” (excise) from the main
chromosome?
a. Site-specific recombination occurs between the POP’ (attP-attachment) site on λ’s and the
BOB’ (att B) site on the main chromosome. Fig 33-40, 41
b. O is a 15 bp segment of homology and P, P’, B and B’ are the segments adjacent (33-34).
c. Integrase is a λ coded enzyme that recognizes O and has “nicking” and “resealing”
activities.
d. Acting with integration host factor (IHF), an E Coli protein, integrase catalyses integration
by way of a 6 base pair overlap in the O region. (IHF recognizes att sequences)
e. Another λ protein, excisionase, which recognizes POB is required (along w/integrase and
IHF) for excision. (Guess whether λ int and λ xis are early, middle or late genes!)
C. R factor plasmids carry genes which confer resistance to antibiotics (genes code for proteins
which can inactivate antibiotics) (R and F plasmids contain origin of replication)
1. Some contain RTF (resistance transfer factor) genes, which enable transfer of the plasmid
during conjugation (like F). They may even transfer to a different species resulting in acquisition of
resistance (new strain).
2. Simple R plasmids without RTF may recombine (integrate) with plasmids with RTF to form
complex R factor plasmids with multiple resistances (to many antibiotics).
X. DNA Repair
A. Repair of alkylated bases or phosphates
1. Methyl and other alkyls are removed from O6 of G by a DNA methyl transferase, which
accepts the alkyl group on one of its cys residues. This inactivates the protein.
2. The same protein accepts methyl groups from DNA phosphates, but on a different cys
residue. This is also irreversible. When methylated at this residue (cys 69) the protein binds to a
transcription activating sequence preceding its operon, which also contains the coding sequences for
other repair enzymes.
B. Repair of thymine or other pyrimidine dimers, which form when DNA is hit by UV light. These
dimers block replication and transcription. Fig 30-52
1. Nucleotide excision repair (NER)
a. A fragment of about twelve nucleotides containing the dimer is cut out by a specific
nuclease (“UvrABC”) which detects the dimer. Fig 30-55
b. DNA Pol I “fills the gap” by polymerizing
c. Ligase seals the nick
d. Defective NER in humans causes Xeroderma Pigmentosum
2. Light repair - photolyase uses the energy of a photon to break the dimer bonds
a. Photolyase binds to the dimer.
b. The photon is absorbed by a bound folate or flavin, which transfers the energy to a bound
FADH .
c. This activates FADH- to transfer an electron to the T dimer, splitting it. (The T - transfers the
electron back to FADH·.)
C. Why does DNA contain T instead of U? U is also diketo; it forms the same H bonds for base
pairing as T.
1. Addition of CH3 to make Ts is “expensive” (ATPs) so why do it?
2. C spontaneously deaminates, forming U.
3. If U were a normal component of DNA, then U formed from C could not be recognized as
aberrant, would not be repaired.
D. Repair of U from C. Fig 30-56
1. Uracil - DNA glycosidase removes base (U)
2. This “apyrimidinic” site is recognized by an endonuclease which makes a nick on the 5' side.
3. DNA Pol I uses 5’ --> 3' exonuclease, polymerization.
4. ligase seals the nick.
5. Human Uracil–DNA Glycosylase (UDG): Fig 30-57
a. UDG binding distorts the DNA, “pinching it”.
b. The U is given a “push” by a leu side chain that takes its place.
c. A pocket on the enzyme has a high affinity for U and “pulls” on the U, but has a low affinity
for other bases, does not pull on them.
d. This causes Us that are weakly bound to Gs to “flip out” of the double helix for hydrolysis.
e. UDG remains bound to AP sites to prevent them from undergoing cytotoxic reactions until
AP endonuclease binds and displaces it.
E. Mismatch Repair Fig 30-58
1. The incidence of mismatches in DNA, such as A with C or T with G, is lower than the rate at
which DNA Pols create them. There is a system for fixing them.
2. Defects in the human mismatch repair system cause a high incidence of cancer.
3. E Coli Mismatch repair
a. A homodimer of MutS binds the mismatched/unpaired bases.
b. A homodimer of MutL binds to this complex.
c. Each MutS-MutL pair “moves” in the opposite on the dsDNA without releasing the other
pair: a loop of DNA is pulled up between the pairs.
d. When a GATC sequence that has only one of its As methylated (hemimethylated) is
reached (which may be more than 1000 bp from the mismatch), MutH binds and nicks the strand that
has the unmethylated A. This is the daughter strand; all As in all GATCs are methylated over time,
but newly-synthesized strands aren’t yet.
e. A helicase moves from the nick back past the mismatch and an exonuclease degrades the
daughter strand.
f. DNA Pol III fills in the gap and ligase seals the nick.
XI. The SOS Response to Severe DNA damage Fig 30-59
1. When rec A is bound to ss DNA it has the effect of a specific protease. It stimulates the selfhydrolysis of a specific bond in a specific protein, lex A.
2. Lex A is a protein which binds to specific operator sequences on the E Coli chromosome, “in
front” of the rec A, SSB, uvr ABC (excises T dimers) genes and other genes of repair and
recombination (a total of 43 genes).
3. When lex A is bound to these sites it represses the transcription of these genes and the
production of the corresponding proteins. When rec A “cleaves” lex A, lex A dissociates and the
genes for these repair proteins are transcribed and the proteins are produced.
4. When DNA is badly damaged the amount of single stranded DNA (ssDNA) increases, as
does the likelihood of a ssDNA-rec A complex forming and lex A being hydrolysed.
XII. Double-Strand Break Repair: nonhomologous end-joining (NHEJ) Fig 30-61, 62
1. Ku protein, a heterodimer of homologous subunits, binds to the end of a double-strand break
(DSB). It has a band or strap that forms a closed circular ring that fits around the dsDNA about 10 bp
from the end.
2. The Ku-DNA complex dimerizes so as to align the DNAs and bring them together end-to-end.
3. The DNAs may be blunt-ended or have 1-4 bp complimentary ssDNAs.
4. Polymerases may fill in gaps, and/or nucleases may trim ends (which would create deletion
mutations).
5. A specialized ligase (ligase IV) complexed with a specific accessory protein (Xrcc4) ligates the
strands.
XIII. Detection of Carcinogens (Mutagens) (Ames test) Fig 30-63
1. The test uses a his- mutant of Salmonella (a mutant that requires histidine in medium in order
to grow). Another mutation will sometimes revert the mutant back to wild type (his +).
2. Attempt to grow colonies from 109 his- cells on an agar plate; this produces about 30 colonies.
3. A small filter paper disk is soaked in the test substance and placed on the center of the agar
plate. In the presence of powerful mutagens, 1000’s of colonies are produced.
4. Some mutagens are metabolites of exposure compounds, so mammalian liver homogenate
may be included in the agar.
IV. Recombinational Repair
A. Lesions at Replication Forks
1. A damaged site (lesion) which exists during passage of the replication fork (or which is
produced immediately after passage while still ss DNA) will prevent replication on this template,
leaving a ssDNA gap. Eventually, progress of the replication fork ceases, with the Pol III cores
releasing the templates (fork collapse).
2. This site can’t be repaired by nucleotide excision repair (NER) because that requires a
template across from the lesion. Fig 30-77
3. Regression of the replication fork back past the lesion, which involves the reassociation of the
parental strands and the association of the daughter strands, may be mediated by Rec A or by RecG,
an ATP driven helicase that promotes branch migration.
4. On the associated daughter strands, replication on the ssDNA extension (that was created by
the progress of the replication fork) produces a segment that can bp with the damaged region after the
branch migration is reversed.
5. Replication can resume and NER can fix the lesion.
B. Single-Strand Breaks at Replication Forks
1. The replication fork collapses beyond the break. Fig 30-78
2. Rec BCD binds to the end of the dsDNA and degrades it back to the next chi sequence,
where it produces the 3’-ended ssDNA for RecA to bind.
3. Rec A will coat the ss region, find the complimentary strand on the other daughter
chromosome (which is ds DNA after its replication), and promote strand exchange.
4. Branch migration back beyond the end of RecBCD action produces a produces a standard
Holliday junction.
5. Once the junction is resolved, replication can resume.
C. Double-Strand Break Repair: Homologous End-Joining (HEJ)
1. This process avoids the mutations that may be necessary in NHEJ. Fig 30-79
2. The dsDNA ends are trimmed so as to produce 3’-ended ssDNA overhangs, as in RecBCD
action.
3. RecA, or its homolog, RAD51 in eukaryotes, binds to the ssDNAs and promotes strand
invasion and strand exchange with each ssDNA. This produces two Holliday junctions.
4. Branch migration of the junctions provides templates for the synthesis of the DNA that was
trimmed and ligation results in 2 intact dsDNAs with two Holliday junctions that can be resolved.
Humans with defects in either of two genes that code for proteins that interact with RAD51 in
recombination have an 80% chance of developing cancer.