2. The Problem
DNA is maintained in a compressed,
supercoiled state.
BUT, basis of replication is the formation
of strands based on specific bases pairing
with their complementary bases.
Before DNA can be replicated it must be
made accessible, i.e., it must be unwound
7. Replication as a process
Double-stranded DNA unwinds.
The junction of the unwound
molecules is a replication fork.
A new strand is formed by pairing
complementary bases with the
old strand.
Two molecules are made.
Each has one new and one old
DNA strand.
9. Features of DNA Replication
DNA replication is semiconservative
Each strand of template DNA is being copied.
DNA replication is semidiscontinuous
The leading strand copies continuously
The lagging strand copies in segments (Okazaki
fragments) which must be joined
DNA replication is bidirectional
Bidirectional replication involves two replication
forks, which move in opposite directions
10. DNA Replication-Prokaryotes
DNA replication is semiconservative.
the helix must be unwound.
Most naturally occurring DNA is slightly
negatively supercoiled.
Torsional strain must be released
Replication induces positive supercoiling
Torsional strain must be released,
again.
SOLUTION: Topoisomerases
12. Topoisomerase Type I
Precedes replicating DNA
Mechanism
Makes a cut in one strand, passes other
strand through it. Seals gap.
Result: induces positive supercoiling as
strands are separated, allowing
replication machinery to proceed.
13. Helicase
Operates in replication
fork
Separates strands to
allow DNA Pol to
function on single
strands.
Translocate along single
strain in 5’->3’ or 3’->
5’ direction by
hydrolyzing ATP
14. Gyrase--A Type II Topoisomerase
Introduces negative supercoils
Cuts both strands
Section located away from actual cut is
then passed through cut site.
15. Initiation of Replication
Replication initiated at specific sites:
Origin of Replication (ori)
Two Types of initiation:
De novo –Synthesis initiated with RNA
primers. Most common.
Covalent extension—synthesis of new strand
as an extension of an old strand (“Rolling
Circle”)
16. De novo Initiation
Binding to Ori
C by DnaA
protein
Opens
Strands
Replication
proceeds
bidirectionally
17. Unwinding the DNA by Helicase
(DnaB protein)
Uses ATP to separate the DNA strands
At least 4 helicases have been identified in
E. coli.
How was DnaB identified as the helicase
necessary for replication?
NOTE: Mutation in such an essential gene
would be lethal.
Solution?
Conditional mutants
18. Liebowitz Experiment
What would you
expect if the
substrates are
separated by
electrophoresis after
treatment with a
helicase?
19. Liebowitz Assay--Results
What do these
results indicate?
ALTHOUGH PRIMASE
(DnaG) AND SINGLE-
STRAND BINDING
PROTEIN (SSB) BOTH
STIMULATE DNA
HELICASE (DnaB),
NEITHER HAVE
HELICASE ACTIVITY
OF THEIR OWN
20. Single Stranded DNA Binding
Proteins (SSB)
Maintain strand separation once helicase
separates strands
Not only separate and protect ssDNA, also
stimulates binding by DNA pol (too much
SSB inhibits DNA synthesis)
Strand growth proceeds 5’>>3’
21. Replication: The Overview
Requirements:
Deoxyribonucleotides
DNA template
DNA Polymerase
5 DNA pols in E. coli
5 DNA pols in mammals
Primer
Proofreading
22. A total of 5 different DNAPs have been reported
in E. coli
DNAP I: functions in repair and replication
DNAP II: functions in DNA repair (proven in 1999)
DNAP III: principal DNA replication enzyme
DNAP IV: functions in DNA repair (discovered in 1999)
DNAP V: functions in DNA repair (discovered in 1999)
To date, a total of 14 different DNA polymerases
have been reported in eukaryotes
The DNA Polymerase Family
23.
24. DNA pol I
First DNA pol discovered.
Proteolysis yields 2 chains
Larger Chain (Klenow Fragment) 68 kd
C-terminal 2/3rd. 5’>>3’ polymerizing
activity
N-terminal 1/3rd. 3’>>5’ exonuclease
activity
Smaller chain: 5’>>3 exonucleolytic
activity
nt removal 5’>>3’
Can remove >1 nt
Can remove deoxyribos or ribos
25. DNA pol I
First DNA pol discovered.
Proteolysis yields 2 chains
Larger Chain (Klenow Fragment) 68 kd
C-terminal 2/3rd. 5’>>3’ polymerizing
activity
N-terminal 1/3rd. 3’>>5’ exonuclease
activity
Smaller chain: 5’>>3 exonucleolytic
activity
nt removal 5’>>3’
Can remove >1 nt
Can remove deoxyribos or ribos
27. Requires 5’-3’ activity of DNA
pol I
Steps
1. At a nick (free 3’ OH) in the DNA
the DNA pol I binds and digests
nucleotides in a 5’-3’ direction
2. The DNA polymerase activity
synthesizes a new DNA strand
3. A nick remains as the DNA pol I
dissociates from the ds DNA.
4. The nick is closed via DNA ligase
Nick Translation
Source: Lehninger pg. 940
28. 5'-exonuclease activity, working together with
the polymerase, accomplishes "nick translation"
This activity is critical in primer removal
Nick Translation 2
29. DNA Polymerase I is great, but….
In 1969 John Cairns and Paula deLucia
-isolated a mutant bacterial strain with only 1%
DNAP I activity (polA)
- mutant was super sensitive to UV radiation
- but otherwise the mutant was fine i.e. it could
divide, so obviously it can replicate its
DNA
Conclusion:
DNA pol I is NOT the principal replication
enzyme in E. coli
30. - DNAP I is too slow (600 dNTPs added/minute)
- DNAP I is only moderately processive
(processivity refers to the number of dNTPs
added to a growing DNA chain before the
enzyme dissociates from the template)
Conclusion:
There must be additional DNA polymerases.
Biochemists purified them from the polA
mutant
Other clues….
31. The major replicative polymerase in E. coli
~ 1,000 dNTPs added/sec
It’s highly processive: >500,000 dNTPs
added before dissociating
Accuracy:
1 error in 107 dNTPs added,
with proofreading final error rate of 1 in
1010 overall.
DNA Polymerase III
32. The 10 subunits of E. coli DNA polymerase III
Subunit Function
a
e
q
t
b
g
d
d’
c
y
5’ to 3’ polymerizing activity
3’ to 5’ exonuclease activity
a and e assembly (scaffold)
Assembly of holoenzyme on DNA
Sliding clamp = processivity factor
Clamp-loading complex
Clamp-loading complex
Clamp-loading complex
Clamp-loading complex
Clamp-loading complex
Core
enzyme
HoloenzymeDNA Polymerase III Holoenzyme (Replicase)
33. Activities of DNA Pol III
~900 kd
Synthesizes both leading and lagging
strand
Can only extend from a primer (either
RNA or DNA), not initiate
5’>>3’ polymerizing activity
3’>>5’ exonuclease activity
NO 5’>>3’ exonuclease activity
34. Subsequent
hydrolysis of
PPi drives the
reaction
forward
Nucleotides are added at the 3'-end of the strand
The 5’ to 3’ DNA polymerizing activity
35. Leading and Lagging Strands
REMEMBER: DNA polymerases require a
primer.
Most living things use an RNA primer
Leading strand (continuous): primer made
by RNA polymerase
Lagging strand (discontinuous): Primer
made by Primase
Priming occurs near replication fork, need to
unwind helix. SOLUTION: Helicase
Primosome= Primase + Helicase
36. The Replisome
DNA pol III extends on
both the leading and
lagging strand
Growth stops when Pol
III encounters an RNA
primer (no 5’>>3’
exonuclease activity)
Pol I then extends the
chain while removing
the primer (5’>>3’)
Stops when nick is
sealed by ligase
37. Ligase
Uses NAD+ or ATP for
coupled reaction
3-step reaction:
AMP is transferred to
Lysine residue on enzyme
AMP transferred to open
5’ phosphate via
temporary pyrophosphate
(i.e., activation of the
phosphate in the nick)
AMP released,
phosphodiester linkage
made
NADNMN + AMP
ATP ADP + PPi
38. DNA Replication Model
1. Relaxation of supercoiled
DNA.
2. Denaturation and untwisting
of the double helix.
3. Stabilization of the ssDNA in
the replication fork by SSBs.
4. Initiation of new DNA
strands.
5. Elongation of the new DNA
strands.
6. Joining of the Okazaki
fragments on the lagging
strand.
39. Termination of
Replication
Occurs @ specific site opposite ori c
~350 kb
Flanked by 6 nearly identical non-palindromic*,
23 bp terminator (ter) sites
* Significance?
Tus Protein-arrests
replication fork
motion
40. FIDELITY OF REPLICATION
Expect 1/103-4, get 1/108-10.
Factors
3’5’ exonuclease activity in DNA pols
Use of “tagged” primers to initiate
synthesis
Battery of repair enzymes
Cells maintain balanced levels of dNTPs
41. Why Okazaki Frags?
Or, why not 3’5’ synthesis?
Possibly due to problems with proofreading.
PROBLEM:
Imagine a misincorporation with a 3’5’
polymerase
How is it removed?
How is the chain extended?
Is there a problem after removing a mismatch?
42. Covalent Extension Methods
Often called “Rolling
circle”
Common in
bacteriophages
NOTE: de novo
initiation of circular
DNA results in theta
structures,
sometimes callled
“theta replication”
43. Rolling Circle I
Few rounds of theta-
replication
Nick outer strand
Extend 3’ end of outer
strand, displacing
original
Synthesis of
complementary strand
using displaced strand
as template
Concatamers cut by
RE’s, sealed
Result several copies of
circular dsDNA
44. Rolling Circle I
“Template “rolls”, extrudes leading strand
Okazaki frags made on leading strand as it
emerges.
46. Rolling Circle II
EX ΦX174
Circular ssDNA chromosome
Copy + strand using E. coli
replication proteins to make
ds circle (theta replication)
Protein A (phage) cuts +
strand
Rolling circle replication
Protein A cuts at unit length
and circularizes (ligates)
released ss chromosome
Replication continues
47. Reverse Transcription
DNA replication in retroviruses
RNA Dependent DNA polymerase
Process:
Retroviral RNA acts as template
Primer—Segment of host cell t-RNA
Result: DNA RNA hybrid
RNA strand degraded by RNAse H
DNA strand serves as template.
Also catalyzed by RT
Result:dsDNA
New DNA integrates into host genome
49. Eukaryotic DNA Replication
Much larger genomes with slower
polymerase
Solution
Multiple initiation sites
More molecules of polymerase
EX: DNA pola present in ~2-5 X105 copies/cell
Histones an issue
Still many questions
50. Completing the Ends of Non-circular
DNA
THE PROBLEM?
Solutions
Phage T-7
Eukaryotes
51. Phage Solution to Problem
Phage DNA is linear
Ends have repetitive complementary sequences
After removal of 5’end RNA primer, are left with
a 3’ overhang
Overhangs form H-bonds with complementary
overhangs, gaps filled in ligated. RESULT:
concatamers
RE cuts concatamer into unit length genomes
with 5’ overhangs
DNA pol extends 3’ ends, resulting in complete
unit length genomes.
52. Eukaryotes
Solution: Telomeres
At ends of chromosomesare non-coding
regions, >1000 tandem repeats of GC rich
sequence.
Telomeric DNA synthesized and
maintained by Telomerase
Adds tandem repeats of TTGGG
Is a ribonucleoprotein, uses internal
ribonucleotide sequences as a template