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Biochemistry Sixth Edition Chapter 28 DNA Replication, Repair, and Recombination Part II: DNA replication Copyright © 2007 by W. H. Freeman and Company.

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Presentation on theme: "Biochemistry Sixth Edition Chapter 28 DNA Replication, Repair, and Recombination Part II: DNA replication Copyright © 2007 by W. H. Freeman and Company."— Presentation transcript:

1 Biochemistry Sixth Edition Chapter 28 DNA Replication, Repair, and Recombination Part II: DNA replication Copyright © 2007 by W. H. Freeman and Company Berg Tymoczko Stryer

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3 Semi-conservative model of DNA replication Parental strand serves as “template”

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8 1 2 3

9 Problems (because of the double helix): 1.Antiparallel strands (opposite strands) 2.Double strands 3.Supercoiling, unwinding

10 E. coli DNA polymerase I (Klenow fragment) Polymerase unit 3’->5’ exonuclease unit (proofreading/correction)

11 DNA polymerases catalyze the formation of polynucleotide chains 1.Template-directed enzyme (base pair-dependent) 2.Catalyzes nucleophilic attack by 3’-OH on the “  ” phosphate 3.Requires a primer with free 3’-OH (RNA polymerase??)

12 Active site (2 metal ions) Holds DNA

13 Two bound metal ions participate in polymerase 1.Activates 3’-OH 2.Stabilizes (-) charge

14 Is hydrogen bond-base pairing enough for adding the right nucleotide?

15 Fails to form H bonds but can still direct addition of T  1. Shape complementarity

16 Structural study Residues of the enzyme form H bonds with minor groove side of the base pairs in the active site  2. Minor groove interactions major minor ruler

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18 Polymerases undergo conformational changes  3. Shape selectivity

19 Where does this primer come from? (5 nt) Removed by hydrolysis

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21 ~1000 nt Primer Remove and fill in (DNA polymerase I) DNA ligase Replication fork

22 thermodynamically uphill rxn nucleophilic attack but no leaving group

23 Mechanism of DNA ligase

24 Bacterial helicase (PcrA) ssDNA binding ATP binding and hydrolysis

25 Separation of strands requires helicase and ATP 1.Both A1 and B1 bind DNA 2.ATP  closure, A1 releases DNA 3.ATP hydrolysis  open, B1 releases 4.Move in 3’  5’ direction

26 Conserved residues among helicase  ATP-induced conformational change Helicase: large class (5’->3’, RNA, oligomers) Hexameric helicase (euk)  ATPase (AAA family)

27 DNA replication must be rapid! Ex. E. coli: 4.6x10 6 bp, replicate in <40 mins  2000bp/sec Differences in eukaryotic: 1.Multiple origins 2.Additional enzyme for telomeres Polymerases: catalytic potency, fidelity and processivity processivity (catalysis without releasing substrates) (processive vs. distributive enzymes!)

28  2 subunit of DNA polymerase III Keep polymerase associated with DNA  Sliding DNA clamp 35 Å How does DNA get in this?  clamp loader (requires ATP)

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30 (DnaB) (single-strand-binding) Topoisomerase II (add – supercoils)

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32 3‘  5’ proofreading Interacts with SSB DNA polymerase “holoenzyme”

33 Add 1000 nt before releasing & new loop DNA pol I Remove primers Fill in

34 Where does replication begin?? In E. coli: a unique site “origin of replication” is called oriC locus

35 DnaA  oriC: Preparation for replication Bind to each others’ ATPase domains; Breaking apart when ATP hydrolyzed

36 DnaA  oriC: Preparation for replication DnaB (hexameric helicase) + DnaC (helicase loader) SSB “Prepriming complex” DnaG (primase) 1 2

37 DNA pol III holoenzyme + Prepriming complex ATP hydrolysis within DnaA Breakup of DnaA (preventing addition round of replication!) 3

38 Eukaryotic replication, why more complex? 1.Size of DNA (6 billion bp) 2.Number of chromosome (23 vs. 1) 3.Linear vs. circular

39 Eukaryotic replication, why more complex? 1.Size of DNA 2.Number of chromosome 30,000 origins! But no defined sequence ORCs (origin of replication complexes) = prepriming complex Replicon: replication unit (how many does E. coli have?)

40 ORCs  origins Preparation for replication Cdc6, Cdt1  MCM2-7 (licensing factors  formation of initiation complex) Replication protein A (=SSB) Two distinct polymerases: Pol  (initiator) Pol  (replicative) 1 2 Eukaryotic DNA replication Polymerase switching 3

41 Pol  (initiator)  Primase + polymerase (20nt) Replication factor C (RFC): displaces pol  recruits PCNA (  2 of pol III) Replication until replicons meet (Primers, ligase) 3 4 Eukaryotic DNA replication 5

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43 !!! Topo I or 2?

44 Bi-directional DNA synthesis

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47 Cyclins Cyclin-dependent protein kinase (CDK) Cell cycle

48 Eukaryotic replication, why more complex? 1.Size of DNA (6 billion bp) 2.Number of chromosome (23 vs. 1) 3.Linear vs. circular

49 !!! Chromosome shortening after each round of DNA replication

50 Chromosome ends = telomeres (AGGGTT) n Telomere-binding protein Loop for protection

51 Telomeres are replicated by telomerase Telomerase has: 1.RNA template 2.Reverse transcriptase

52 High levels of telomerase in dividing cells tumor and aging

53 Summary: DNA replication 1.Mode 2.Enzymology 3.Polymerization steps 4.Eukaryotes vs. prokaryotes 5.Telomeres


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