1

Lecture 24: DNA replication

Figure 16.7a, c (c) Space-filling model

C

T

A

A

T

CG

GC

A

C G

AT

AT

A T

TA

C

TA0.34 nm

3.4 nm

G

1 nm

G

T

C

T

A

A

T

CG

GC

A

C G

AT

AT

A T

TA

C

TA0.34 nm

3.4 nm

G

1 nm

G

T

Lecture Outline

• Overview of DNA replication• DNA has polarity (5’P, 3’OH)

• Rules of Replication• Details of the replication machine

– Proofreading– Mutation– Special problems at the ends of

chromosomes

Review

• Watson-Crick model of DNAstructure

• Base composition ratios– A=T; G=C

• Radioactive labeling experimentsshow that– DNA is the genetic material– Replication is semi-conservative

C

T

A

A

T

CG

GC

A

C G

AT

AT

A T

TA

C

TA0.34 nm

3.4 nm

G

1 nm

G

T

Chromosomes

• Bacterial chromosomes tend to becircular, eukaryotic chromosomes linear

• DNA plus associated proteins =chromatin– Euchromatin vs heterochromatin

2

DNA StructureN H O CH3

N

N

O

N

N

N

N H

Sugar

Sugar

Adenine (A) Thymine (T)

N

N

N

NSugar

O H N

H

NH

N OH

H

N

Sugar

Guanine (G) Cytosine (C)Figure 16.8

H

C

T

A

A

T

CG

GC

A

C G

AT

AT

A T

TA

C

TA0.34 nm

3.4 nm

G

1 nm

G

T

5’ Carbon has PO4

3’ Carbon has OH

• In DNA replication– The parent molecule unwinds, and two new

daughter strands are built based on base-pairing rules

(a) The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.

(b) The first step in replication is separation of the two DNA strands.

(c) Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand.

(d) The nucleotides are connected to form the sugar-phosphate backbones of the new strands. Each “daughter” DNA molecule consists of one parental strand and one new strand.

A

CT

AG

A

CT

A

G

A

CT

A

G

A

CT

A

G

T

GA

TC

T

GA

TC

A

CT

AG

AC

T

A

G

T

GA

T

C

TGA

T

C

T

GA

T

C

T

G

A

T

C

Figure 16.9 a–d

3

Replication overview

• Must maintain integrity of the DNA sequencethrough successive rounds of replication– (think of the game “telephone”)

• Need to:– unwind DNA, add an RNA primer, find an

appropriate base, add it to the growing DNAfragment, proofread, remove the initial primer, fillin the gap with DNA, ligate fragments together

• All of this is fast, about 100 bp/secondanimation from DNAi Figure 16.13

New strand Template strand5′ end 3′ end

Sugar A TBase

C

G

G

C

A

C

T

PP

P

OH

P P

5′ end

Pyrophosphate

Phosphate

Elongating a New DNA Strand

• DNApolymerases,addnucleotidesto the 3′ endof a growingstrand

Nucleosidetriphosphate

P

A

OHP

T

P

C

P

C

P

A

P

T

P

G

P

C

OH P

T

P

G

P

G

P

A

P

A

P

T

P

A

3’

3’ 5’

5’

Template

Primer witha free 3’OH

P

T

OH

Rules of Replication1. Semi-conservative2. Primer is required3. Template is required4. Elongation occurs 5’ -> 3’5. Semi-discontinuous6. Bidirectional

4

Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.

DNA synthesisNeeds a template

Must add a base toan existing 3’ OH

This end has a 5’ phosphate

This end has a 3’ OH

• Circular bacterial chromosomes have oneorigin

• Eukaryotic chromosomes have hundreds oreven thousands of replication origins

Origin of replication

Bubble

Parental (template) strand

Daughter (new) strand

Replication fork

Two daughter DNA molecules

0.25 µm

Close-up of a replication fork

Parental DNA

DNA pol Ill elongatesDNA strands only in the5′ 3′ direction.

1

Okazakifragments

DNA pol III

Templatestrand

Lagging strand3

2

Templatestrand DNA ligase

Overall direction of replication

One new strand, the leading strand,can elongate continuously 5′ 3′ as the replication fork progresses.

2

The other new strand, thelagging strand must grow in an overall3′ 5′ direction by addition of shortsegments, Okazaki fragments, that grow5′ 3′ (numbered here in the orderthey were made).

3

DNA ligase joins Okazakifragments by forming a bond betweentheir free ends. This results in a continuous strand.

4

Figure 16.14

3′5′

5′3′

3′5′

21

Leading strand

1

• Synthesis of leading and laggingstrands during DNA replication Model for the “replication machine,” or replisome

5

Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.

Model for the “replication machine,” or replisome Detail of the lagging strand

Overall direction of replication

3′

3′

3′

3′5′

3′5′

3′5′

3′5′

3′5′

3′5′

3′5′

3′ 5′

5′

1

1

21

12

5′

5′

12

3′5′

Templatestrand

RNA primer

Okazakifragment

Figure 16.15

Primase joins RNA nucleotides into a primer.

1

DNA pol III adds DNA nucleotides to the primer, forming an Okazaki fragment.

2

After reaching the next RNA primer (not shown), DNA pol III falls off.

3

After the second fragment is primed. DNA pol III adds DNAnucleotides until it reaches the first primer and falls off.

4

DNA pol 1 replaces the RNA with DNA, adding to the 3′ end of fragment 2.

5

DNA ligase forms a bond between the newest DNAand the adjacent DNA of fragment 1.

6 The lagging strand in this region is nowcomplete.

7

First look at leading strand:• Helicase unwinds a bit of

DNA• Primase adds an RNA primer• Polymerase adds

complementary nucleotidesto 3’ OH and slides down themolecule, continuing toextend the sugar-phosphatechain.

• (E. coli has about 40,000 turnsin the circular chromosome--each of them must be unwoundto separate the DNA strands)

Now look at lagging strand• Can’t form continuous molecule,

since synthesis only happens 5’to 3’. SO, need many primersand lots of short fragments

• Primase adds an RNA primer andPolymerase adds nucleotidesuntil it reaches the previousprimer

• Another polymerase removesRNA primer on adjacent standand adds dNTPs

• Ligase connects the fragments

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Table 16.1 Bacterial DNA replication proteins andtheir functions Try this at home:

• Draw a replication bubble, label thebases on the template, primer, andnewly synthesized DNA using theshorthand DNA notation.

• Convince yourself that synthesis mustgo 5’ to 3’ and that replication must besemi-discontinuous.