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REPLICATION
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
THREE HYPOTHESES FOR DNA REPLICATION
Models of Replication
(a) Hypothesis 1:
Semi-conservative
replication
(b) Hypothesis 2:
Conservative replication
Intermediate molecule
(c) Hypothesis 3:
Dispersive replication
MODELS OF DNA REPLICATION
PREDICTED
DENSITIES OF
NEWLY
REPLICATED
DNA
MOLECULES
ACCORDING
TO THE
THREE
HYPOTHESES
ABOUT DNA
REPLICATION
Meselson and Stahl
Conclusion: Semi-conservative replication of DNA
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.
Extending the Chain
 dNTPs are added individually
 Sequence determined by pairing with
template strand
 DNA has only one phosphate between
bases, so why use dNTPs?
Deoxyribonucleoside triphosphates are the building blocks of DNA.
However, a complete polynucleotide strand of DNA has only one phosphate
group and that through this phosphate group each nucleotide is attached to
the next. Why then is the substrate a triphosphate instead of just a
monophosphate? The answer to this question lies in the chemistry
underlying the addition of nucleotides to a growing daughter strand of
DNA.
Extending the Chain
DNA Synthesis
2 phosphates
3’-OH nucleophilic attack
on alpha phosphate of
incoming dNTP
removal and splitting of pyrophosphate
by inorganic pyrophosphatase
nucleotide gets positioned through H-
bonding with template
- 3’-OH nucleophilic attack on alpha
phosphate of incoming dNTP.
loss of entropy; not much gain in
bond-energy
reaction is driven by removal and
splitting of pyrophosphate
because of requirement for 3’-OH and
5’ dNTP substrate, DNA polymerase
can only catalyze reaction in the 5’ 3’
direction (direction of new strand!)
Chain Elongation in the 5’  3’ direction
Semi-discontinuous Replication
 All known DNA pols work in a 5’>>3’
direction
 Solution?
 Okazaki fragments
Okazaki Experiment
Continuous synthesis
Discontinuous synthesis
DNA replication is semi-discontinuous
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
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
The Problem of Overwinding
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.
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
Gyrase--A Type II Topoisomerase
 Introduces negative supercoils
 Cuts both strands
 Section located away from actual cut is
then passed through cut site.
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”)
De novo Initiation
 Binding to Ori
C by DnaA
protein
 Opens
Strands
 Replication
proceeds
bidirectionally
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
Liebowitz Experiment
What would you
expect if the
substrates are
separated by
electrophoresis after
treatment with a
helicase?
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
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’
Replication: The Overview
 Requirements:
 Deoxyribonucleotides
 DNA template
 DNA Polymerase
 5 DNA pols in E. coli
 5 DNA pols in mammals
 Prime
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
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
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
The structure of the
Klenow fragment of
DNAP I from E. coli
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
 5'-exonuclease activity, working together with
the polymerase, accomplishes "nick translation"
This activity is critical in primer removal
Nick Translation 2
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:
 DNAP I is NOT the principal replication
enzyme in E. coli
- 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….
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
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)
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
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
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
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
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
 AMP released, phosphodiester
linkage made
 NADNMN + AMP
 ATP ADP + PPi
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.
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
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
Why Okazaki Frags?
 why not 3’5’ synthesis?
 Possibly due to problems with proofreading.

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Dna replication

  • 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
  • 3. THREE HYPOTHESES FOR DNA REPLICATION Models of Replication
  • 4. (a) Hypothesis 1: Semi-conservative replication (b) Hypothesis 2: Conservative replication Intermediate molecule (c) Hypothesis 3: Dispersive replication MODELS OF DNA REPLICATION
  • 6. Meselson and Stahl Conclusion: Semi-conservative replication of DNA
  • 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.
  • 8. Extending the Chain  dNTPs are added individually  Sequence determined by pairing with template strand  DNA has only one phosphate between bases, so why use dNTPs? Deoxyribonucleoside triphosphates are the building blocks of DNA. However, a complete polynucleotide strand of DNA has only one phosphate group and that through this phosphate group each nucleotide is attached to the next. Why then is the substrate a triphosphate instead of just a monophosphate? The answer to this question lies in the chemistry underlying the addition of nucleotides to a growing daughter strand of DNA.
  • 10. DNA Synthesis 2 phosphates 3’-OH nucleophilic attack on alpha phosphate of incoming dNTP removal and splitting of pyrophosphate by inorganic pyrophosphatase nucleotide gets positioned through H- bonding with template - 3’-OH nucleophilic attack on alpha phosphate of incoming dNTP. loss of entropy; not much gain in bond-energy reaction is driven by removal and splitting of pyrophosphate because of requirement for 3’-OH and 5’ dNTP substrate, DNA polymerase can only catalyze reaction in the 5’ 3’ direction (direction of new strand!)
  • 11. Chain Elongation in the 5’  3’ direction
  • 12. Semi-discontinuous Replication  All known DNA pols work in a 5’>>3’ direction  Solution?  Okazaki fragments
  • 14. Continuous synthesis Discontinuous synthesis DNA replication is semi-discontinuous
  • 15. 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
  • 16. 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
  • 17. The Problem of Overwinding
  • 18. 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.
  • 19. 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
  • 20. Gyrase--A Type II Topoisomerase  Introduces negative supercoils  Cuts both strands  Section located away from actual cut is then passed through cut site.
  • 21. 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”)
  • 22. De novo Initiation  Binding to Ori C by DnaA protein  Opens Strands  Replication proceeds bidirectionally
  • 23. 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
  • 24. Liebowitz Experiment What would you expect if the substrates are separated by electrophoresis after treatment with a helicase?
  • 25. 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
  • 26. 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’
  • 27. Replication: The Overview  Requirements:  Deoxyribonucleotides  DNA template  DNA Polymerase  5 DNA pols in E. coli  5 DNA pols in mammals  Prime
  • 28. 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
  • 29.
  • 30. 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
  • 31. 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
  • 32. The structure of the Klenow fragment of DNAP I from E. coli
  • 33. 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
  • 34.  5'-exonuclease activity, working together with the polymerase, accomplishes "nick translation" This activity is critical in primer removal Nick Translation 2
  • 35. 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:  DNAP I is NOT the principal replication enzyme in E. coli
  • 36. - 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….
  • 37. 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
  • 38. 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)
  • 39. 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
  • 40. 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
  • 41. 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
  • 42. 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
  • 43. 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  AMP released, phosphodiester linkage made  NADNMN + AMP  ATP ADP + PPi
  • 44. 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.
  • 45. 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
  • 46. 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
  • 47. Why Okazaki Frags?  why not 3’5’ synthesis?  Possibly due to problems with proofreading.