REPLICATION Chapter 7

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

Models of Replication THREE HYPOTHESES FOR DNA REPLICATION 1) Semiconservative model: Daughter DNA molecules contain one parental strand and one newly-replicated strand 2) Conservative model: Parent strands transfer information to an intermediate (?), then the intermediate gets copied. The parent helix is conserved, the daughter helix is completely new 3) Dispersive model: Parent helix is broken into fragments, dispersed, copied then assembled into two new helices. New and old DNA are completely dispersed 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 1) Semiconservative model: Daughter DNA molecules contain one parental strand and one newly-replicated strand 2) Conservative model: Parent strands transfer information to an intermediate (?), then the intermediate gets copied. The parent helix is conserved, the daughter helix is completely new 3) Dispersive model: Parent helix is broken into fragments, dispersed, copied then assembled into two new helices. New and old DNA are completely dispersed

PREDICTED DENSITIES OF NEWLY REPLICATED DNA MOLECULES ACCORDING TO THE THREE HYPOTHESES ABOUT DNA REPLICATION Density Gradient Centrifugation can be used to address the question. WHAT RESULTS ARE PREDICTED?

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 While each nucleotide added to a growing DNA chain lacks an -OH group at its 2' position, it retains its 3' -OH. This hydroxyl group is used to attack the alpha phosphate group of an incoming nucleoside triphosphate. In the attack, the 3' -OH replaces the beta and gamma phosphates that are ejected from the complex as a pyrophosphate molecule. The result is the formation of the phosphodiester bond between the growing daughter strand and the next nucleotide. The 3' -OH of the newly added nucleotide is now exposed on the end of the growing chain and can attack the next nucleotide in the same way. The figure above presents a simplified schematic of a growing polynucleotide chain. The lines represent the ribose sugar with one 3' -OH branching from it. Each p represents a phosphate group. This figure illustrates a number of key points of DNA replication. First, we see that the parent strand is oriented in the 3' to 5' direction. Second, each new nucleotide added to the growing daughter strand is complementary to the nucleotide on the parent strand that is across from it and a bond forms between them. Finally, we see how the 3' -OH group displaces the two outermost phosphate groups of an incoming nucleotide in order to add it to the growing chain. The Driving Force of the Addition Reaction Each incoming nucleotide supplies the energy for its addition in the high-energy bond between the beta and gamma phosphates that are ejected upon addition. It is not the release of the pyrophosphate that drives the reaction, but rather the subsequent hydrolysis that takes place. A much larger amount of energy is released when the two are split by inorganic pyrophosphatase to yield two phosphates. Total G~ -7 kcal/mol., offsetting the loss in entropy when the nucleotide is added (~0.5 kcal/mol).

DNA Synthesis 3’-OH nucleophilic attack on alpha phosphate of incoming dNTP 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!) removal and splitting of pyrophosphate by inorganic pyrophosphatase 2 phosphates

Chain Elongation in the 5’  3’ direction 2. Reaction Template Growing strand: 5' 3' strand Pol I 3. Reaction requirements a. DNA template b. 3 - OH primer; can be RNA or DNA c. dNTP’s and Mg2+ Primer Incoming dNTP - Nucleophilic attack of primer strand 3’ OH on the phosphate of the dNTP

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 Relaxation of closed circular replicating DNA by topoisomerases. The problem: As Replication of closed circular DNA proceeds, an overwound region (in the unreplicated portion) is formed as a result of unwinding on the other side of the molecule.

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.

Gyrase--A Type II Topoisomerase Introduces negative supercoils Cuts both strands Section located away from actual cut is then passed through cut site.

Helicase Operates in replication fork Separates strands to allow DNA Pol to function on single strands. Involves breaking H-bonds and hydrophobic interactions Requires ATP 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

Initiation of Replication Replicaion 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 Binding to Ori C by DnaA protein. Is a 50 kD protein that recognizes and interacts with specific repeated sequences (9-mers) in the oriC region. This recognition sequence is a set of 9 nt, repeated 4 times. Several DnaA proteins come together to form a complex w/~150 bp DNA This formation denatures a region of the DNA (characterized by repeating 13 mers [AT rich region]) adjacent to the initial complex. Forms an open complex. it is now open and accessible to the other replications proteins to form the replication fork. Replication proceeds bidirectionally by extension of RNA primers

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 Uses ATP to separate the DNA strands At least 4 helicases have been identified in E. coli.: rep helicase helicase II Helicase III DnaB protein How was DnaB identified as the helicase necessary for replication? NOTE: Mutation in such an essential gene would be lethal, undetectable. Solution? Conditional mutants—i.e., mutation expressed only under certain conditions, ex. Elevated T. Jab isolated mutants that ceased DNA synthesis @40ºC. One had a mutation in the DnaB gene.

Liebowitz Experiment What would you expect if the substrates are separated by electrophoresis after treatment with a helicase? Helicase Assay Does DnaB Encode Rplication Helicase? Used single stranded DNA encoded by M-13 (a phage) Made a labeled 1.06 fragment complementary to the phage DNA Tested DnaB protein, DnaG protein, an single stranded DNA binding proteins

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 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’ Maintain strand separation once helicase separates strands Not only separate and protect ssDNA, also stimulates binding by DNA pol 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 Primer Proofreading High Fidelity DNA Replication Expected error rate=1 mistake/103 nt Error rate= 1 mistake/109 nucleotides Afforded by complementary base pairing and proof-reading capability of DNA polymerase

Overview of replication Overview of replication. We will start with a discussion of the primary polymerizing enzymes (pol I and pol II)

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 A very talented enzyme. First DNA pol discovered. Several activities, but occurs as a monomer despite multiple functions. Various functions in separate domains. Proteolysis yields 2 chains Larger Chain (Klenow Fragment) 68 kd C-terminal 2/3rds---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 NOTE: can be active at a nick in the strand as long as there is a 5’ phosphate functions in multiple processes that require only short lengths of DNA synthesis - has a major role in DNA repair (Cairns- deLucia mutant was UV-sensitive) - its role in DNA replication is to remove primers and fill in the gaps left behind - for this it needs the nick-translation activity

Nick translation Requires 5’-3’ activity of DNA pol I Steps At a nick (free 3’ OH) in the DNA the DNA pol I binds and digests nucleotides in a 5’-3’ direction The DNA polymerase activity synthesizes a new DNA strand A nick remains as the DNA pol I dissociates from the ds DNA. The nick is closed via DNA ligase Requires 5’-3’ activity of DNA pol I Steps At a nick (free 3’ OH) in the DNA the DNA pol I binds and digests nucleotides in a 5’-3’ direction The DNA polymerase activity synthesizes a new DNA strand A nick remains as the DNA pol I dissociates from the ds DNA. The nick is closed via DNA ligase Uses: removal of RNA primers DNA repair Great for DNA labeling with radioactive dNTPs Source: Lehninger pg. 940