dna replication – process lecture 17 1. forms of dna helices 2 dna replication template dna...

DNA Replication – Process

Lecture 17

1

Forms of DNA Helices

2

DNA Replication

Template DNA (parent DNA)

Replication of parent DNA can theoretically generate 2 possible conformations of daughter

DNA

S Phase S Phase

M PhaseM Phase


3

DNA ReplicationDesign an experiment to differentiate these two

possibilities

DNA is unique among biomolecules due to very high concentration of Phosphorus

𝑃→ 𝑆❑32 +𝛾❑

32

Grow cells in P32 enhanced

media

32P labeled DNA

Single cell cycle in unenhanced

media


4

DNA Replication

How do bacteria go about replicating DNA?

Linearize DNA for replication?

Keep DNA circular?

The experiment:Grow cells in 3H-thymidine

enhanced media


5

DNA Replication

Where does replication start and is it uni or bidirectional?

Heavy staining on both sides of the replication eye

indicates that replication is almost always bidirectional

in bacteria.

Bacterial replication also starts at the same spot indicating a single origin or replication


6

Dealing with DNA Superstructure

What enzyme is capable of modifying

DNA topology?Topoisomerase (DNA

Gyrase)

Parent DNA must be unwound at the replication fork

For this to occur at biological rates (1000 nt per second for E. coli), genomic DNA must twist at a rate of 100 rps.

E. coli’s genome is circular, so this is not possible!


7

Semi-discontinuous Replication

DNA’s 2 strands are replicated at the same time

DNA Polymerase synthesize DNA in the

5’ 3’ direction

So how does the other strand elongate?

5’5’ 3’

The ‘Lagging Strand’ is synthesized as small fragments (1000 – 2000 nuclotides in bacteria or 100-200 nt in eukaryotes) called Okazaki fragments.

Okazaki fragments are combined by a DNA Ligase


8

DNA Replication

How will a dNTP be added to the existing chain?

How might the reaction be activated?


9

Requirement for priming

This model of DNA replication requires a 3’ OH for strand elongation

Nucleophile

3’ 5’

5’

Analysis of Okazaki fragments indicated that the 5’ end was always composed of RNA

160 nucleotides in length

These will have to be replaced at some point


10

Enzyme Requirement for Replication

• DNA Polymerase• Catalyze the DNA chain elongation using the ‘parent’ strand as a template

• RNA Primer Synthesis• RNA Polymerase

• 460 kDa (E. coli)• catalyzes primer synthesis for leading strand only

• Primase• 60 kDa• Responsible for priming Okazaki fragment• Works synergistically with RNA Polymerase to prime leading strand

• Topoisomerase (Gyrase)• Unwind supercoiled template DNA• Energy dependent process

• SSB• Binds to ssDNA and prevents reannealing

• Ligase• Join Okazaki fragments


11

Formation of the Replication Fork

1. 4 DnaA proteins bind to the oriC region (recognizes 9 nucleotide segments)1. Additional DnaA monomers bind forming a histone like complex of tightly wound

DNA2. Localized melting of a 13bp repeats

2. DnaB (6 subunit helicase) binds and unwinds the DNA1. Topoisomerase functions downstream to relieve stress

3. Single Strand DNA binding proteins prevent the ssDNA from reannealing


12

Formation of the Replication Fork

DnaA helical filament – 8 monomers per turn

178 Å

• dsDNA wraps around the filament

• This increase Writhing number? • Enables some untwisting?

L = W + t


13

The Big Picture (in E.coli)Topoisomerase: relieves

topological strain Helicase - unwinds dsDNA at replication fork

Primase: synthesizes RNA primers for lagging strand

DNA Polymerase III : elongate primed DNA from 5’3’

3’5’ exonuclease activity limits errors

Pol III elongates lagging strand until strained. It then releases the template and rebinds at a new primed location.

SSB: keeps ssDNA from reannealing

Pol I: closes any gaps in lagging strand and

replaces RNA primer

Ligase: seals off backbone nicks


14

DNA Polymerase

What do we need in a DNA polymerase?

• Template DNA binding site• Single Strand or Double Strand?

• Room for growing dsDNA

• dNTP binding site

• Mechanism to differentiate between the 4 possible dNTPs

• Self Correcting?

Forms of DNA HelicesE. coli DNA Polymerase I

Palm Domain • ~22 Å x 30 Å • Ideal shape to bind B-DNA • Lined with basic amino

acids

Thumb• Guides newly

formed DNA• Responsible for

processivity

Pol I• 3’ 5’ and 5’ 3’ exonuclease activity• Small N-terminal domain contains the 5’3’ exonuclease activity

• Completely independent from active site or 3’ 5 site

Fingers• Contains dNTP

binding site • Close in on

dNTP/Growing Strand once successful Watson-Crick base pairing made


16

Taq DNA Polymerase

Projected path of ssDNA template

5’

Thumb folds down over elongating dsDNA preventing it from escaping

Initial inspection of the groove between the thumb and fingers suggests the DNA would follow this cleft. This is NOT the case!Active Site

Growing Strand

Template

3 base pair closest to active site are A-Form

DNA


5’

Active Site

Growing Strand

Template

Forms of DNA HelicesTaq DNA Polymerase

Thumb folds down over elongating dsDNA preventing it from escaping

Initial inspection of the groove between the thumb and fingers suggests the DNA would follow this cleft. This is NOT the case!


18

Taq DNA Polymerase


dNTP Binding Site – allows for rapid sampling of the dNTPs

Thumb

Knuckles?


19

Taq DNA Polymerase

Growing Strand

Template

Watson-Crick Base Pair between Template and incoming dNTP at active

Mg situated for activation

of 3’OH

DNA polymerases apparently select incoming dNTP based on SHAPE of Watson-Crick base pair (Purine-Pyrimidine)

5’ end of growing strand is 2’-3’-dideoxy

CTP

Only one combination will provide most favorable pair


20

3’ 5’ Exonuclease

The structure of E. coli Pol I has been solved with DNA arranged in the 3’5’ exonuclease active site

Growing strand peels away from active site

The base pair closest to the polymerization active site is significantly weakened

Exonuclease site (Zn2+ activated mechanism)

Which bond will be cleaved?

NEED 3’-hydroxyl for elongation


21

Model for Proofreading in DNA Polymerases

• dNTPs are in rapid equilibrium at active site

• When reaction occurs, proper base pair will form a strong tight bonding pair

• When an error occurs, the base pair is not as tight, resulting in increased favorability for the growing chain to be positioned at the 3’5’ exonuclease site

• A-form DNA of DNA @ polymerization active site makes this process easier (less tightly wrapped and less stable helix)


22

5’ 3’ Exonuclease

Pol I from Thermus aquiticus lacks 3’5’ exonuclease activity

N-terminal domain contains 5’ 3’

exonuclease

Taq polymerase can translate a single nick in DNA by a mechanism that involves both the polymerase

active site and the 5’3’ exonuclease active site


23

Taq DNA Polymerase ExonucleaseWhat process in DNA replication requires a

5’ 3’ exonuclease process?

Need to remove RNA primers from lagging strand!


24

DNA Ligase

• Needed to seal the phosphodiester backbone following removal of the RNA primer by Pol I

• Energy Dependent reaction

• Hydrolysis of ATP or NADH


25

DNA LigasepdbID 2OWO


26

DNA Ligase

• AMP covalently bound to active site

pdbID 2OWO

AMP phospoamide intermediate with Lysine shown

biochemically


27

DNA Ligase Reaction

AMP phospoamide intermediate forms with Lys


28

DNA Ligase

Covalent Lys-AMP intermediate displaced by 5’ Phosphate of nicked DNA


29

DNA Ligase3’ OH attacks 5’ Phosphate displacing the AMP


30

DNA Helicase

• Helicase functions as a hexamer

• Each subunit has a babbb fold

• Forms a clamp around the DNA that slides and unwinds dsDNA

• Unwinds dsDNA by ‘active rolling’ mechanism


31

DNA Primase

• Primase must be able to oppose the direction of translocation

• Forms a non-covalent bond with DnaB (helicase)

• Synthesizes primers ~11 nt in vivo but ~60 in vitro

• Basic groove nonspecifically binds to ssDNA and directs it to the active site


32

Single Stranded DNA Binding Protein

• SSB binding to DNA prevents: • reannealing• Stem loop structures (self

complimentary sequences)

• Nuclease degradation

• Very basic proteins• Bind to ssDNA with no sequence

specificity• Binds as a tetramer• Can bind in a number of different

conformations


33

Replication Termination

There are multiple termination elements in E. coli’s genome.

Each are similar sequences long and promote binding of TUS proteins

Consensus Sequence:ANNTAGTATGTTGTAACTA

Clockwise moving replication forks pass through TerE, TerD and TerA but stop at TerC

CCW moving replication forks pass through TerG, TerF, TerB and TerC but stop at TerA

This verifies that each replication fork will finish at same place in genome

Makes specific contacts with

DnaB – termination is dependent on this interaction

dna replication – process lecture 17 1. forms of dna helices 2 dna replication template dna...

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