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REPLICATION
Chapter 7
The Problem
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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
Models of Replication
THREE HYPOTHESES FOR DNA REPLICATION
MODELS OF DNA REPLICATION
(a) Hypothesis 1:
(b) Hypothesis 2:
(c) Hypothesis 3:
Semi-conservative
replication
Conservative replication
Dispersive replication
Intermediate molecule
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
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dNTPs are added individually
Sequence determined by pairing with
template strand
DNA has only one phosphate between
bases, so why use dNTPs?
Extending the Chain
DNA Synthesis
3’-OH nucleophilic attack
on alpha phosphate of
incoming dNTP
removal and splitting of pyrophosphate
by inorganic pyrophosphatase
2 phosphates
Chain Elongation in the 5’  3’ direction
Semi-discontinuous Replication
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All known DNA pols work in a 5’>>3’
direction
Solution?
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Okazaki fragments
Okazaki Experiment
Continuous synthesis
Discontinuous synthesis
DNA replication is semi-discontinuous
Features of DNA Replication
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DNA replication is semiconservative
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DNA replication is semidiscontinuous
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Each strand of template DNA is being copied.
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
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DNA replication is semiconservative.
the helix must be unwound.
Most naturally occurring DNA is slightly
negatively supercoiled.
 Torsional
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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
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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.

Gyrase--A Type II Topoisomerase
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Introduces negative supercoils
Cuts both strands
Section located away from actual cut is
then passed through cut site.
Helicase
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Operates in replication
fork
Separates strands to
allow DNA Pol to
function on single
strands.
Involves breaking Hbonds and hydrophobic
interactions
Requires ATP
Initiation of Replication
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Replicaion initiated at specific sites: Origin
of Replication (ori)
Two Types of initiation:
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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
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Binding to Ori
C by DnaA
protein
Opens
Strands
Replication
proceeds
bidirectionally
Unwinding the DNA by Helicase
(DnaB protein)
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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?
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Conditional mutants
Liebowitz Experiment
What would you
expect if the
substrates are
separated by
electrophoresis after
treatment with a
helicase?
Liebowitz Assay--Results
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What do these
results indicate?
ALTHOUGH PRIMASE
(DnaG) AND SINGLESTRAND BINDING
PROTEIN (SSB) BOTH
STIMULATE DNA
HELICASE (DnaB),
NEITHER HAVE
HELICASE ACTIVITY
OF THEIR OWN
Single Stranded DNA Binding
Proteins (SSB)
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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
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Requirements:
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Deoxyribonucleotides
DNA template
DNA Polymerase
5 DNA pols in E. coli
 5 DNA pols in mammals
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Primer
Proofreading
DNA pol I
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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
Nick translation
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
Source: Lehninger pg. 940