Lecture 3 DNA Replication 1 Mechanism

Question 1

Watson and Crick 1953

Answer 1

Molecular structure of nucleic acid paper published in 1953 Nature recognised specific base pairing

Question 2

Semi conservative model supported

Answer 2

Parent/ template strand codes a daughter strand
So semi conservative - each new helix contains one original+one new strand

Question 3

Meselson-stahl experiment 1958

Answer 3

Confirmed semiconservative theory

DNA extracted
Bacteria grown in 15N then transferred to 14N. Cacl aq and centrifuge used to establish visible DNA layers

Initial heavy DNA 15N layer low in tube

First gen daughter molecules hybrid DNA
(15N and 14N strand in each) layer mid tube

Second gen daughter molecules (2 helices are 15N and 14N and two are light DNA - all 14N) this results in a later mid tube and a layer near the surface in the tube.

Question 4

Detection of fork movement using radioisotopes

Answer 4

1972 D.M. Prescott and P.L. Kuempel
used 3H-thymidine labeling experiments to show that replication of the E. coli chromosome involved two replication forks growing in opposite directions (bidirectional replication)rather than a single, unidirectional replication fork.

Question 5

Discontinuous DNA replication

Answer 5

Experiments by Kiwako Sakabe and Reiji Okazaki (1967) provide evidence DNA replication is a discontinuous process

Properties of first DNA polymerase show continuous synthesis only possible in 5’3’ direction (leading strand) to form the 3’5’ strand sections of 5’3’ would need to be made and connected in a delayed fashion - lagging strand

Question 6

Okazaki fragments

Answer 6

Pulse labelling of E. Coli chromosomes found evidence for short segments making up 4’5’ strand. Confirmed using temp. sensitive mutants of the polynucleotide ligase an enzyme that links short DNA strands together. These short strands are now known as Okazaki fragments and are 1000-2000 BP long

Question 7

Polymerisation of nucleic acid chains

Answer 7

Arthur Konenburg 1956 E. Coli DNA polymerase 1 - enzyme that synthesises DNA discovered.

First example of a template being used in biosynthetic reaction. All DNA polymerases add nucleotides at 3’ end and require:
1)a template strand
2) a primer strand (DNA or RNA) synthesised on template to provide a 3’OH group that can be extended
3) dNTPs deoxynucleoside triphosphates as building blocks for DNA

Question 8

DNA replication primed by RNA

Answer 8

RNA synth doesn’t require primer so DNA synth is usually primed by a small segment of RNA. RNA primer synthesised by DNA primase. In E. Coli this is DnaG which synthesises 10-12 nucleotide long oligoribonucleotides (RNA) using DNA as template.

Question 9

DNA polymerase corrects errors

Answer 9

When a mispair occurs DNA polymerase repositions mispaired bases to exonuclease site. Using 3’5’ exonuclease activity (hydrolysis) to remove wrongly inserted nucleotides aka proofreading. The 3’ terminus is then repositioned in the insertion site and polymerase continues to insert correct nucleotides.
This process leaves only 1 mistake every 5x10^7 bp

Question 10

Elongation of DNA chain

Answer 10

DNA polymerase 1:
Insertion site - where nucleotide addition occurs
Post Insertion dite- where newly formed base pair migrates after it is added

Question 11

Priming leading and lagging strands

Answer 11

Leading strand primed once at origin of replication. Lagging strand primed separately for each Okazaki fragment. Hence leading strand synthesized continuously and lagging discontinuously. Known as semi discontinuous DNA replication.

Question 12

Primer removal

Answer 12

Another 5’3’ exonuclease removes RNA primers. Synthesises new DNA strand as RNA primer is removed - sometimes called nick translation as it starts at a nick.

DNA polymerase 1 removes RNA extends DNA chain and nicks are sealed by DNA ligase

Question 13

DNA ligase closes nicks in DNA backbone between Okazaki fragments

Answer 13

DNA ligase joins (ligates) DNA sections by catalysing formation of phosphodiester bond. E. Coli DNA ligase (LigA) uses nicotinamide adenine dinucleotide (NAD) as cofactor in the reaction.

Any gaps of one or more nucleotide cannot be repaired. The enzyme needs a 5’ phosphate (5’P) and 3’ hydroxyl (3’OH)

Question 14

Separation of parental strands

Answer 14

Strands unwound by DNA helicase. In E. Coli this replicative helicase is called DnaB. It is a 3’5’ DNA helicase, unwinding parental DNA as it moves. Consists of 6 subunits ( a homohexamer) forming a ring that confers high processivity ( rarely falls off). DnaB tracks along single strand - lagging strand template to unwind parental duplex ahead of replication fork

Question 15

DnaG primase binds DnaB helicasr

Answer 15

The tethering puts it in correct place on template for RNA primer synthesis. Activity of both are stimulated in this interaction

Question 16

DNA Poly 1 is not replicative poly in E. Coli

Answer 16

It’s DNA Poly 3