replication

DNA REPLICATION

1

Nature of the Genetic Material• Property 1 - it must contain, in a stable

form; information encoding the organism’s structure, function, development and reproduction

• Property 2 - it must replicate accurately so progeny cells have the same genetic makeup

• Property 3 - it must be capable of some variation (mutation) to permit evolution 2

Replication of DNA and Chromosomes

• Speed of DNA replication: 3,000 nucleotides/min in human 30,000 nucleotides/min in E.coli

• Accuracy of DNA replication: Very precise (1 error/1,000,000,000

nt)

3

Meselson and Stahl (1958)

4

5

Bi-directional replication in E. coli

6

Only One Replication Origin in E. coli

7

Multiple Origins in Eukaryotes• Eukaryotes replicate their DNA only in S-

phase• Eukaryotes have larger chromosomes• Replication speed 2,600 npm. • Largest Drosophila chromosome is 6.5 x 107

nucl., but it can replicate in 3-4 min. From a single origin, bidirectional replication would take 8.5 days. ==> The chromosome must have some 7,000 origins of replication.

8

A replicating Drosophila chromosome

9

Origins initiate replication at different times.

10

Replication as a Process

1. Double-stranded DNA unwinds.

2. The junction of the unwound

molecules is a replication fork.

3. A new strand is formed by pairing complementary bases with theold strand.

4. Two molecules are made.

Each has one new and one old

DNA strand.

11

Continuous synthesis

Discontinuous synthesis

DNA Replication

12

• The two strands are antiparallel, that is they run in opposite directions.

• Therefore, one parent strand - the one running 3' to 5' and called the leading strand - can be copied directly down its entire length.

13

• DNA polymerase enzymes are only able to join the phosphate group at the 5' carbon of a new nucleotide to the hydroxyl (OH) group of the 3' carbon of a nucleotide already in the chain

14

• As a result, DNA can only be synthesized in a 5' to 3' direction while copying a parent strand running in a 3' to 5' direction.

• Remember as mentioned above, each DNA strand has two ends.

15

• The 5' end of the DNA is the one with the terminal phosphate group on the 5' carbon of the deoxyribose; the 3' end is the one with a terminal hydroxyl (OH) group on the deoxyribose of the 3' carbon of the deoxyribose.

16

P

P

P

P

P

P

P

PP P

CH2

CH2

CH2

OH

OH

O

O

OBase

Base

Base

CH2

CH2

CH2

OH

O

O

OBase

Base

Base

5' end of strand

3' end of strand3'

5'

3'

H20+

Synthesis reaction

DNA SYNTHEIS REACTION

products

17

• In addition, DNA polymerase enzymes cannot begin a new DNA chain from scratch. They can only attach new nucleotides onto 3' OH group of a nucleotide in a preexisting strand

18

• However, the other parent strand - the one running 5' to 3' and called the lagging strand - must be copied discontinuously in short fragments (Okazaki fragments) of around 100-1000 nucleotides each as the DNA unwinds.

19

• Therefore, to start the synthesis of the leading strand and each DNA fragment of the lagging strand, an RNA polymerase complex called a primosome or primase is required.

20

• The primase, which is capable of joining RNA nucleotides without requiring a preexisting strand of nucleic acid, first adds several comlementary RNA nucleotides opposite the DNA nucleotides on the parent strand. This forms what is called an RNA primer.

21

Conclusions

1. There is a covalent linkage between ribonucleotides and deoxyribonucleotides in the newly synthesised DNA.

2. RNA fragments (10 to 20 nucleotides) are located at the 5’ end of the nascent fragments and are required for priming de novo DNA synthesis.

How is the DNA primed?

22

1. Evidence that RNA primers are required for replication.

2. The replicative enzyme, DNA polymerase III

3. Additional features of the replication process.

4. The central dogma in biology: DNA RNA Protein.

5. Transcription.

23

DNA Must Be “Primed” Before DNA Polymerase Can Replicate It

DNA polymerase cannot initiate polymerisation de novo.

Okazaki and colleagues provided evidence for short stretches of RNA linked to nascent chains of DNA during replication.

1. Sugino et al., (1972) isolated Okazaki fragments after pulsing with 3H-U (incorporates into RNA and not DNA) and found it associated with newly replicated DNA.

2. Initiation of DNA replication, but not continuation, was shown to be sensitive to rifampicin (an antibiotic that inhibits RNA polymerases).

3. In follow up experiments Sugino et al., (1973) isolated Okazaki fragments after a short pulse (3H-dT) by banding on a CsCl gradient.

4. Treatment of the Okazaki fragments with alkali (hydrolyses RNA but not DNA) or ribonuclease resulted in a small shift in density (size).

24

The subunits of E. coli DNA polymerase III

Subunit Function

’

5’ to 3’ polymerizing activity3’ to 5’ exonuclease activity and assembly (scaffold)Assembly of holoenzyme on DNASliding clamp = processivity factorClamp-loading complexClamp-loading complexClamp-loading complexClamp-loading complexClamp-loading complex

CoreEnzymedimer

Ho

loen

zym

e

25

Enzymes in DNA replication

Helicase unwinds parental double helix

Binding proteinsstabilise separatestrands

DNA polymerase IIIbinds nucleotides to form new strands

Ligase joins Okazaki fragments and seals other nicks in sugar-phosphate backbone

Primase adds short primer to template strand

DNA polymerase I (Exonuclease) removes RNA primer and inserts the correct bases 26

• Binding proteins prevent single strands from rewinding.

• Helicase protein binds to DNA sequences called origins and unwinds DNA strands.

5’ 3’

5’

3’

• Primase protein makes a short segment of RNA complementary to the DNA, a primer.

3’ 5’

5’ 3’

Replication

27

Overall directionof replication

5’ 3’

5’

3’

5’

3’

3’ 5’

DNA polymerase enzyme adds DNA nucleotides to the RNA primer.

Replication

28

DNA polymerase enzyme adds DNA nucleotides to the RNA primer.

5’

5’

Overall directionof replication

5’

3’

5’

3’

3’

3’

DNA polymerase proofreads bases added and replaces incorrect nucleotides.

Replication

29

5’

5’ 3’

5’

3’

3’

5’

3’Overall directionof replication

Leading strand synthesis continues in a 5’ to 3’ direction.

Replication

30

3’ 5’ 5’

5’ 3’

5’

3’

3’

5’

3’Overall directionof replication

Okazaki fragment

Discontinuous synthesis produces 5’ to 3’ DNA segments called Okazaki fragments.

Replication

31

5’

5’ 3’

5’

3’

3’

5’

3’

3’

5’ 5’ 3’

Leading strand synthesis continues in a 5’ to 3’ direction.

Discontinuous synthesis produces 5’ to 3’ DNA segments called Okazaki fragments.

Replication

32

5’

5’

3’ 3’

5’

3’

5’ 3’

5’

3’

3’

5’

Exonuclease activity of DNA polymerase I removes RNA primers.

Polymerase activity of DNA polymerase I fills the gaps

Replication

33

.

Ligase forms bonds between sugar-phosphate backbone.

3’

5’

3’

5’ 3’

5’

3’

3’

5’

Replication

34

Topoisomerase nicks DNA to relieve tension from unwinding

23

1

4

56

7

Pol III synthesises leading strand

Helicase opens helix

Primase synthesises RNA primer

Pol III elongates primer; produces Okazaki fragment

Pol I excises RNA primer; fills gap

DNA ligase links Okazaki fragments to form continuous strand

DNA REPLICATION

35

Simultaneous Replication Occurs via Looping of the

Lagging Strand

•Helicase unwinds helix

•SSBPs prevent closure

•DNA gyrase reduces tension

•Association of core polymerase with template

•DNA synthesis

•Not shown: pol I, ligase36

Replication Termination of the Bacterial Chromosome

ori

ter

Origin

5’3’

3’5’

BIDIRECTIONAL REPLICATION

37

Procaryotic (Bacterial)Chromosome Replication

ori

ter

Bidirectional Replication Produces a Theta Intermediate

Replication Forks

38

Replication Termination of the Bacterial Chromosome

Termination: meeting of two replication forksand the completion of daughter chromosomes

Region 180o from ori contains replication fork traps:

ori

Ter sites

Chromosome

39

Replication Termination of the Bacterial Chromosome

TerATerB

One set of Ter sites arrest DNA forks progressing in the clockwise direction, asecond set arrests forks in the counterclockwise direction:

Chromosome

40

Replication Termination of the Bacterial Chromosome

Ter sites are binding sites for the Tus protein

Tus: 35.8 kDDNA binding at TerMonomer

TusDNA

Ter

Replication forkarrested in polar

manner

Tus may inhibit replication fork progressionby directly contacting DnaB helicase, inhibiting DNA unwinding 41

Summary

DNA replication proteins:DNA PolIIIDNA PolIDNA Ligase

Primase (DnaG)Helicase (DnaB)SSB

Replication terminationReplication fork traps opposite oriCTer sitesTus protein

42

Summary of Lecture 1

1. DNA replication is semi-conservative (Meselson-Stahl, 1958).

2. Replication requires a DNA polymerase, a template, a primer and the 4 nucleotides and proceeds in a 5’ to 3’ direction (Kornberg, 1957).

3. Replication of the Escherichia coli genome (a single circular DNA) starts at a specific site (ori) and is bi-directional (Cairns, 1963).

4. Replication is continuous on leading strand and discontinuous on lagging strand and requires RNA primers (Okazaki’s, 1968).

5. Lagging strand synthesis involves Okazaki fragments.

6. DNA polymerase III is the replicative enzyme of E. coli (Cairns and deLucia, 1969).

43

Match the work with the correct definition or description

1. Helicase2. SSB 3. Topoisomerase

A. keeps DNA wound to the correct tension ( i.e not overwound or underwound)

B. catalyze the unwinding of the DNA double helixC. stabilizes single stranded DNA

44

Introns are:

A. Coding regions of DNA

B. Noncoding regions of DNA

C. a type of proteins

D. a type of tRNA

45

Transcription is directly involved in which of the following steps in the flow of genetic

information?

A. DNA to RNA

B. RNA to DNA

C. DNA to DNA

D. RNA to protein

46

Okazaki fragments are formed on the _______ and in the ______ direction.

A. Leading; 3’-5’

B. Lagging; 3’-5’

C. Leading; 5’-3’

D. Lagging; 5’-3’

47

1. Purine base2. Pyrmidine base3. Would this be a strand found in:

a. DNA b. RNA

48