Biochemistry Sixth Edition Chapter 28 DNA Replication, Repair, and Recombination Part II: DNA...

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Transcript of Biochemistry Sixth Edition Chapter 28 DNA Replication, Repair, and Recombination Part II: DNA...

BiochemistrySixth Edition

Chapter 28DNA Replication, Repair, and

Recombination

Part II: DNA replication

Copyright © 2007 by W. H. Freeman and Company

Berg • Tymoczko • Stryer

Semi-conservative model of DNA replication

Parental strand serves as “template”

1

2

3

Problems (because of the double helix):

1. Antiparallel strands (opposite strands)

2. Double strands

3. Supercoiling, unwinding

E. coli DNA polymerase I(Klenow fragment)

Polymerase unit

3’->5’ exonuclease unit(proofreading/correction)

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??)

Active site(2 metal ions)

Holds DNA

Two bound metal ions participate in polymerase

1. Activates 3’-OH2. Stabilizes (-) charge

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

Fails to form H bonds butcan still direct addition of T 1. Shape complementarity

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

Polymerases undergo conformational changes 3. Shape selectivity

Where does this primer come from?

(5 nt)

Removed by hydrolysis

~1000 nt

Primer

Remove and fill in(DNA polymerase I)

DNA ligase

Replication fork

thermodynamically uphill rxn

nucleophilic attackbut no leaving group

Mechanism of DNA ligase

Bacterial helicase (PcrA)

ssDNA binding

ATP binding and hydrolysis

Separation of strands requires helicase and ATP

1. Both A1 and B1 bind DNA2. ATP closure, A1 releases DNA3. ATP hydrolysis open, B1 releases4. Move in 3’ 5’ direction

Conserved residues among helicaseATP-induced conformational change

Helicase: large class(5’->3’, RNA, oligomers)

Hexameric helicase (euk) ATPase (AAA family)

DNA replication must be rapid!

Ex. E. coli:4.6x106 bp, replicate in <40 mins 2000bp/sec

Differences in eukaryotic:1. Multiple origins2. Additional enzyme for telomeres

Polymerases: catalytic potency, fidelity and

processivityprocessivity (catalysis without releasing substrates)

(processive vs. distributive enzymes!)

2 subunit of DNApolymerase III

Keep polymeraseassociated with DNA Sliding DNA clamp

35 Å

How does DNA get in this? clamp loader (requires ATP)

(DnaB)

(single-strand-binding)

Topoisomerase II(add – supercoils)

3‘5’ proofreading

Interacts with SSB

DNA polymerase “holoenzyme”

Add 1000 nt before releasing & new loop

DNA pol I•Remove primers•Fill in

Where does replication begin??

In E. coli: a unique site “origin of replication” is called oriC locus

DnaAoriC:Preparation for replication

Bind to each others’ ATPase domains;

Breaking apart whenATP hydrolyzed

DnaAoriC:Preparation for replication

DnaB (hexameric helicase) + DnaC (helicase loader)SSB

“Prepriming complex”

DnaG (primase)

1

2

DNA pol III holoenzyme +Prepriming complex

ATP hydrolysis within DnaA

Breakup of DnaA(preventing addition round of replication!)

3

Eukaryotic replication, why more complex?

1. Size of DNA (6 billion bp)2. Number of chromosome (23 vs. 1)3. Linear vs. circular

Eukaryotic replication, why more complex?

1. Size of DNA2. Number of chromosome

30,000 origins!But no defined sequence

ORCs (origin of replication complexes) = prepriming complexReplicon: replication unit (how many does E. coli have?)

ORCsoriginsPreparation 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

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

!!!Topo I or 2?

Bi-directional DNA synthesis

CyclinsCyclin-dependent protein kinase (CDK)

Cell cycle

Eukaryotic replication, why more complex?

1. Size of DNA (6 billion bp)2. Number of chromosome (23 vs. 1)3. Linear vs. circular

!!!

Chromosome shorteningafter each round ofDNA replication

Chromosome ends = telomeres

(AGGGTT)n

Telomere-bindingprotein

Loop forprotection

Telomeres are replicated by telomerase

Telomerase has:1. RNA template2. Reverse transcriptase

High levels of telomerasein dividing cells

tumor and aging

Summary: DNA replication

1. Mode

2. Enzymology

3. Polymerization steps

4. Eukaryotes vs. prokaryotes

5. Telomeres