Dna replication

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

1. 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

5. PREDICTED DENSITIES OF NEWLY REPLICATED DNA MOLECULES ACCORDING TO THE THREE HYPOTHESES ABOUT 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.

9. Extending the Chain

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

13. Okazaki Experiment

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. 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

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. The structure of the Klenow fragment of DNAP I from E. coli

32. 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 Nick Translation Source: Lehninger pg. 940

33. 5'-exonuclease activity, working together with the polymerase, accomplishes "nick translation" This activity is critical in primer removal Nick Translation 2

34. 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

35. - 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….

36. 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 10 7 dNTPs added, with proofreading final error rate of 1 in 10 10 overall. DNA Polymerase III

37. 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 Holoenzyme DNA Polymerase III Holoenzyme (Replicase)

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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.

44. 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

45. FIDELITY OF REPLICATION Expect 1/10 3-4 , get 1/10 8-10 . 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

46. Why Okazaki Frags ? why not 3’ 5’ synthesis? Possibly due to problems with proofreading.