REPLICATION

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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.
REPLICATION

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1. REPLICATION Chapter 7

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. Models of 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 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

4. 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 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

5. Density Gradient Centrifugation can be used to address the question. WHAT RESULTS ARE PREDICTED?Density Gradient Centrifugation can be used to address the question. WHAT RESULTS ARE PREDICTED?

6. Meselson and StahlMeselson and Stahl

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. 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.

9. 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). 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).

10. DNA Synthesis 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!) 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. 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 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

12. Semi-discontinuous Replication All known DNA pols work in a 5?>>3? direction Solution? Okazaki fragments

13. Okazaki Experiment Predictions: By using short pulses of label, one should be able to label and catch the predicted short pieces before they are stitched together. 2. If one knocks out the enzyme responsible for stitching (ligase) they should be able to accumulate these small fragments. Predictions: By using short pulses of label, one should be able to label and catch the predicted short pieces before they are stitched together. 2. If one knocks out the enzyme responsible for stitching (ligase) they should be able to accumulate these small fragments.

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 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.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.

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. 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. 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 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 ATPOperates 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. Introduces negative supercoils Cuts both strands Section located away from actual cut is then passed through cut site.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?) 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?) 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 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 ? DNA replication begins at a specific site. Example: oriC site from E. coli. ? 245 bp out of 4,000,000 bp ? contains a tandem array of three 13-mers; GATCTNTTNTTTT ?GATC common motif in oriC ?AT bp are common to facilitate duplex unwindingBinding 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 ? DNA replication begins at a specific site. Example: oriC site from E. coli. ? 245 bp out of 4,000,000 bp ? contains a tandem array of three 13-mers; GATCTNTTNTTTT ?GATC common motif in oriC ?AT bp are common to facilitate duplex unwinding

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 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 @40?C. One had a mutation in the DnaB gene. 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 @40?C. One had a mutation in the DnaB gene.

24. Liebowitz Experiment 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 proteinsHelicase 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

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 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

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? Maintain strand separation once helicase separates strands Not only separate and protect ssDNA, also stimulates binding by DNA pol 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?

27. 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 PoI has 3 catalytic sides: 5?->3? polymerase activity 3?->5? nucleases activity 5?->3? nuclease activity Purpose of 3? nuclease activity is to remove incorrect matches from the growing chain. PoI can not elongate improperly mismatched bp. 5? exonuclease acts on the dsDNA removes nucleotides from 5? end important for primers removalHigh 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 PoI has 3 catalytic sides: 5?->3? polymerase activity 3?->5? nucleases activity 5?->3? nuclease activity Purpose of 3? nuclease activity is to remove incorrect matches from the growing chain. PoI can not elongate improperly mismatched bp. 5? exonuclease acts on the dsDNA removes nucleotides from 5? end important for primers removal

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

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 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 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

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 PoI has 3 catalytic sides: 5?->3? polymerase activity 3?->5? nucleases activity 5?->3? nuclease activity Purpose of 3? nuclease activity is to remove incorrect matches from the growing chain. PoI can not elongate improperly mismatched bp. 5? exonuclease acts on the dsDNA removes nucleotides from 5? end important for primers removal Activity depends on the presence of Mg2+ (RE: Taq Pol) Why the exonuclease activities? The 3'-5' exonuclease activity serves a proofreading function It removes incorrectly matched bases, so that the polymerase can try again PoI has 3 catalytic sides: 5?->3? polymerase activity 3?->5? nucleases activity 5?->3? nuclease activity Purpose of 3? nuclease activity is to remove incorrect matches from the growing chain. PoI can not elongate improperly mismatched bp. 5? exonuclease acts on the dsDNA removes nucleotides from 5? end important for primers removal Activity depends on the presence of Mg2+ (RE: Taq Pol) Why the exonuclease activities? The 3'-5' exonuclease activity serves a proofreading function It removes incorrectly matched bases, so that the polymerase can try again

33. 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 Uses: removal of RNA primers DNA repair Great for DNA labeling with radioactive dNTPs 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

34. 5'-exonuclease activity, working together with the polymerase, accomplishes "nick translation" 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: DNA pol 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 - 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

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 The "real" replicative polymerase in E. coli ~ 1,000 dNTPs added/sec It?s highly processive: >500,000 dNTPs added before dissociating It?s accurate: makes 1 error in 107 dNTPs added, with proofreading, this gives a final error rate of 1 in 1010 overall. The "real" replicative polymerase in E. coli ~ 1,000 dNTPs added/sec It?s highly processive: >500,000 dNTPs added before dissociating It?s accurate: makes 1 error in 107 dNTPs added, with proofreading, this gives a final error rate of 1 in 1010 overall.

38. DNA Polymerase III Holoenzyme (Replicase) ? 5? to 3? polymerizing activity ? 3? to 5? exonuclease activity q a and e assembly (scaffold) ? Assembly of holoenzyme on DNA ? Sliding clamp = processivity factor ? Clamp-loading complex d Clamp-loading complex d? Clamp-loading complex c Clamp-loading complex y Clamp-loading complex ? 5? to 3? polymerizing activity ? 3? to 5? exonuclease activity q a and e assembly (scaffold) ? Assembly of holoenzyme on DNA ? Sliding clamp = processivity factor ? Clamp-loading complex d Clamp-loading complex d? Clamp-loading complex c Clamp-loading complex y Clamp-loading complex

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 ~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 ~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

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 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 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 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 The replisome? All of the enzymes that function at the replication fork. DNA pol holoenzyme +~20 accessory enzymes and proteins. Two catalytic cores of DNA Pol III The entire replisome moves along the parental ds helix Actually the replisome is probably stationary and the DNA is pulled through the replisome Replication proceeding from left to right The Replication Apparatus Prepriming proteins + primosomes + replisomes 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 The replisome? All of the enzymes that function at the replication fork. DNA pol holoenzyme +~20 accessory enzymes and proteins. Two catalytic cores of DNA Pol III The entire replisome moves along the parental ds helix Actually the replisome is probably stationary and the DNA is pulled through the replisome Replication proceeding from left to right The Replication Apparatus Prepriming proteins + primosomes + replisomes

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 (i.e., activation of the phosphate in the nick) AMP released, phosphodiester linkage made NAD?NMN + AMP ATP ?ADP + PPi How is energy obtained for condensation of nt to primer or growing chain? Uses hydrolysis of 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 WHAT HAPPENS TO THE PPi? How is energy obtained for condensation of nt to primer or growing chain? Uses hydrolysis of 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 WHAT HAPPENS TO THE PPi?

44. DNA Replication Model Relaxation of supercoiled DNA. Denaturation and untwisting of the double helix. Stabilization of the ssDNA in the replication fork by SSBs. Initiation of new DNA strands. Elongation of the new DNA strands. Joining of the Okazaki fragments on the lagging strand. Relaxation of supercoiled DNA. Catalyzed by topoisomerases activity is required for the replication fork to move Topoisomerase I ? single stranded break to provide axis of rotation Topoisomerase II (DNA gyrase) ? double stranded breaks to remove supercoils Denaturation and untwisting of the double helix. Catalyzed by DNA helicases. By repeated ATP hydrolysis move along the ssDNA and untwist any dsDNA encountered. Two helicases at the replication fork moving in opposite direction on the leading and lagging strands. Stabilization of the ssDNA in the replication fork. Single-strand DNA binding proteins (SSB proteins) stabilize the ssDNA so doesn?t reform double helix or fold on itself (ssb gene) Binding is cooperative ? binding of the 1st stimulates the binding of additional SSB More than 200 of these proteins bind to each replication fork Initiation of new DNA strands. RNA primers are laid as template Leading strand is synthesized in the same direction as the direction of fork movement Lagging strand is synthesized in the opposite direction and requires multiple RNA primers to generate the Okazaki fragments Primase + helicase = primosome Elongation of the new DNA strands. New DNA is added to the RNA primer by Pol III Continuous synthesis on the leading strand 5??3? Pol III dissociates on the lagging strand as it the previous RNA primer to make the Okazaki fragments Joining of the Okazaki fragments on the lagging strand. Pol I 5??3? exonuclease activity removes RNA primer while Pol I 5??3? polymerase activity fills in DNA ligase seals the gap by catalyzing the phosphodiester bond b/w 3?OH and the 5? phosphate groups on either side of the gap thus sealing the gap Relaxation of supercoiled DNA. Catalyzed by topoisomerases activity is required for the replication fork to move Topoisomerase I ? single stranded break to provide axis of rotation Topoisomerase II (DNA gyrase) ? double stranded breaks to remove supercoils Denaturation and untwisting of the double helix. Catalyzed by DNA helicases. By repeated ATP hydrolysis move along the ssDNA and untwist any dsDNA encountered. Two helicases at the replication fork moving in opposite direction on the leading and lagging strands. Stabilization of the ssDNA in the replication fork. Single-strand DNA binding proteins (SSB proteins) stabilize the ssDNA so doesn?t reform double helix or fold on itself (ssb gene) Binding is cooperative ? binding of the 1st stimulates the binding of additional SSB More than 200 of these proteins bind to each replication fork Initiation of new DNA strands. RNA primers are laid as template Leading strand is synthesized in the same direction as the direction of fork movement Lagging strand is synthesized in the opposite direction and requires multiple RNA primers to generate the Okazaki fragments Primase + helicase = primosome Elongation of the new DNA strands. New DNA is added to the RNA primer by Pol III Continuous synthesis on the leading strand 5??3? Pol III dissociates on the lagging strand as it the previous RNA primer to make the Okazaki fragments Joining of the Okazaki fragments on the lagging strand. Pol I 5??3? exonuclease activity removes RNA primer while Pol I 5??3? polymerase activity fills in DNA ligase seals the gap by catalyzing the phosphodiester bond b/w 3?OH and the 5? phosphate groups on either side of the gap thus sealing the gap

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? Occurs @ specific site opposite ori c ~350 kb Flanked by 6 nearly identical non-palindromic*, 23 bp terminator (ter) sites * Significance? Ter sites are polar, providing directionality. Allow replication forks to enter the terminus, but not leave it. Tus ( Terminator Utilization Substance) Protein ---arrests replication fork motion. Is a 309 aa monomer. Tus Binds to terminator sites, probably interacts with helicase ?stops replication fork. NOTE: Mutants that lack rep terminus still re[plicate DNA and replication stops. Termination system very highly conserved in prokaryotes Final step: unlinking circular DNA, probably by a topoisomerase Occurs @ specific site opposite ori c ~350 kb Flanked by 6 nearly identical non-palindromic*, 23 bp terminator (ter) sites * Significance? Ter sites are polar, providing directionality. Allow replication forks to enter the terminus, but not leave it. Tus ( Terminator Utilization Substance) Protein ---arrests replication fork motion. Is a 309 aa monomer. Tus Binds to terminator sites, probably interacts with helicase ?stops replication fork. NOTE: Mutants that lack rep terminus still re[plicate DNA and replication stops. Termination system very highly conserved in prokaryotes Final step: unlinking circular DNA, probably by a topoisomerase

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 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 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? 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? Or, why not 3??5? synthesis? Possibly due to problems with proofreading. PROBLEM: imagine a misincorporation with a 3??5? polymerase How is the chain extended? How is it removed? How is the chain extended? Is there a problem after removing a mismatch? If have a 5? monophosphate, can?t extend the chain. ? would have to evolve an energizing system to reactivate the edited product Or, why not 3??5? synthesis? Possibly due to problems with proofreading. PROBLEM: imagine a misincorporation with a 3??5? polymerase How is the chain extended? How is it removed? How is the chain extended? Is there a problem after removing a mismatch? If have a 5? monophosphate, can?t extend the chain. ? would have to evolve an energizing system to reactivate the edited product

48. 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?

49. 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 After initial replication of the circular P22 DNA by theta-replication, the DNA is replicated by rolling circle replication. Rolling circle replication generates a long concatemer of linear, double-stranded DNA that can be packaged into phage heads Rolling Circle? Common among bacteriophages Circular DNA serves as template for production of a concatamer. After initial replication of the circular P22 DNA by theta-replication, the DNA is replicated by rolling circle replication. Rolling circle replication generates a long concatemer of linear, double-stranded DNA that can be packaged into phage heads Rolling Circle? Common among bacteriophages Circular DNA serves as template for production of a concatamer.

50. Rolling Circle I ?Template ?rolls?, extrudes leading strand Okazaki frags made on leading strand as it emerges. ?Rolling Circle? Common among bacteriophages Circular DNA serves as template for production of a concatamer (a DNA base sequence tandemly repeated many times. Ex in computer programming)?Rolling Circle? Common among bacteriophages Circular DNA serves as template for production of a concatamer (a DNA base sequence tandemly repeated many times. Ex in computer programming)

51. Rolling Circle I

52. Rolling Circle II EX FX174 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 The A protein is cis-acting fX174 RF (ds) DNA is a template for synthesizing single-stranded viral circles. The A protein remains attached to the same genome through indefinite revolutions, each time nicking the origin on the viral (+) strand and transferring to the new 5' end. Circular ssDNA chromosome?the + strand Copy + strand using E. coli replication proteins to make ds circle (Replicative form) Protein A (phage) cuts + strand Rolling circle replication Protein A cuts at unit length and circularizes (ligates) released ss chromosome Replication continues The A protein is cis-acting fX174 RF (ds) DNA is a template for synthesizing single-stranded viral circles. The A protein remains attached to the same genome through indefinite revolutions, each time nicking the origin on the viral (+) strand and transferring to the new 5' end. Circular ssDNA chromosome?the + strand Copy + strand using E. coli replication proteins to make ds circle (Replicative form) Protein A (phage) cuts + strand Rolling circle replication Protein A cuts at unit length and circularizes (ligates) released ss chromosome Replication continues

53. 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 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 NOTES ON RT: Lacks proofreading ability Has RNAse and polymerase functions Target of AZT Useful tool in lab 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 NOTES ON RT: Lacks proofreading ability Has RNAse and polymerase functions Target of AZT Useful tool in lab

54. cDNA Library Made from mRNA Steps 1st strand RNAse H 2nd strand Tailing Insertion Transform COMPARE with genomic library. mRNA w/ poly-A tail RT synthesizes DNA from RNA template RT Second strand synthesis Treat with RNAse H Use DNA pol I for Insert into vector after tailing with terminal deoxinucleotidyl transferase (TDT or terminal transferase) Transform into bacterium Now use PCR to fish out a specific transcript. COMPARE with genomic library. mRNA w/ poly-A tail RT synthesizes DNA from RNA template RT Second strand synthesis Treat with RNAse H Use DNA pol I for Insert into vector after tailing with terminal deoxinucleotidyl transferase (TDT or terminal transferase) Transform into bacterium Now use PCR to fish out a specific transcript.

55. Eukaryotic DNA Replication Much larger genomes with slower polymerase Solution Multiple initiation sites More molecules of polymerase EX: DNA pol? present in ~2-5 X105 copies/cell Histones an issue Still many questions

56. Completing the Ends of Non-circular DNA THE PROBLEM? Solutions Phage T-7 Eukaryotes THE PROBLEM? STATE OR FIGURE OUT THE PROBLEM. THE PROBLEM? STATE OR FIGURE OUT THE PROBLEM.

57. 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. 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. 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.

58. 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 Ends of chromosomes would shorten at each round of replication unlessEnds of chromosomes would shorten at each round of replication unless

59. Telomeres Elongation Translocation Elongation New primer synthesis Dna Replication Primer removal Repeat Elongation Translocation Elongation New primer synthesis Dna Replication Primer removal Repeat Correlation between telomerase and cancer? Normal somatic cells---no telomerase activity Cancer cells?in some types, telomerase is reactivated ? telomerase inhibitor may be a candidate for treatmentElongation Translocation Elongation New primer synthesis Dna Replication Primer removal Repeat Correlation between telomerase and cancer? Normal somatic cells---no telomerase activity Cancer cells?in some types, telomerase is reactivated ? telomerase inhibitor may be a candidate for treatment


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