Lecture 3: Replication Fork Initiation Unwinding

Question 1

Why must the two strands of the double helix be separated prior to DNA synthesis?

Answer 1

This is because both strands are replication at the same time (simultaneously) and this requires us to have two separate templates.

Question 2

Define the replication fork.

Answer 2

Junction between the newly separated template strands and the unreplicated duplex DNA

Question 3

Describe the replication fork in bacteria.

Answer 3

Since prokaryotes have circular chromosomes, they have a bidirectional replication fork with a single origin. Once the two bi directional forks fuse together, replication has occurred.

Question 4

What are the two type of strands in DNA synthesis? What direction does DNA synthesis occur?

Answer 4

Leading (Continious strand) and Lagging (Discontinuous strand)

Occurs in the 5’ to 3’ direction for both the leading and lagging strand

Question 5

Describe the leading strand and the lagging strand

Answer 5

Leading strand can be replicated continuously along with the DNA polymerase as they both move in the same direction of the replication fork allowing for continue replication.

Lagging strand directs the DNA polymerase to move in the opposite direction of the replication fork resulting in discontinuous replication. Lagging strand must be replicated in Osaka strands.

Question 6

What is the role of DNA polymerase III in DNA synthesis?

Answer 6

DNA pol III is a dimeric holoenzyme (multiprotein complex) and is the primary enzyme unbolted in the replication of the chromosome. The DNA pol III holoenzyme has 3 copies of core the DNA pol III enzyme and 1 copy of the 5 subunit (3 t proteins, 1 flexible linker, sliding clamp) sliding clamp loader.

Question 7

How do multiple DNA polymerases remain linked at the replication while DNA synthesizes on both the lagging and leading strand templates?

Answer 7

Due to the flexibility of the DNA and the t protein. Essentially, as the helicase unwinds the DNA at the replication fork, the leading strand is exposed and immediately one of the DNA polymerase III core enzymes begins to act on it which initates the synthesis of the continuous DNA strand. However, the lagging strand template is not immediately acted on but rather it is spooled out as ssDNA that is rapidly bound by SSBs (Single stranded DNA binding protein). Primase interacts with the DNA helicase and is activated in order to synthesize a new RNA primer on the lagging-strand template. This (the RNA:DNA hybrid) is known as the primer:template junction and this is where the sliding Clamp DNA loader will attach and a second core DNA polymerase III molecule will initiate the synthesis of the lagging strand. (Page 285 in textbook)

Question 8

What is the purpose of the third core DNA polymerase on the DNA polymerase holoenzyme?

Answer 8

It is believed that the third core DNA polymerase enzyme is used to initiate the synthesis of the following Okazaki fragment before the previous one is finished synthesizing. This is because the process of forming the primer:template junction on the lagging strand will begin happening as soon as the previous one is finished, and once the next junction is made, the sliding clamp loader will load on.

Question 9

What is required of for the initiation of a new strand of DNA?

Answer 9

RNA primer with a free 3’ OH

Question 10

What is primase?

Answer 10

DNA-dependent RNA polymerase which is used to make short RNA primers (5-10 nucleotides long) on the ssDNA template. Both the leading and lagging strand of DNA require primase.

Question 11

How does primase initiate RNA synthesis?

Answer 11

Primase prefers to initiate RNA synthesis using a ssDNA template which contains a particular trimer (In the case of E.coli, this trimer is known as GTA). Primase is activated when it interacts with DNA helicase which unwinds the DNA at the replication fork (creating a ssDNA template).

Question 12

What ensures that primase is only active at the replication fork?

Answer 12

The need for a ssDNA template and the DNA helicase associate ensures this as helicase separates the DNA into ssDNA at the replication fork and this is the only region where the requirements for activating primase are met.

Question 13

What are Okazaki fragments?

Answer 13

They are transient intermediates in DNA replication. They typically vary in length from 1000 to 2000 nucleotides in bacteria and 100-400 nucleotides in eukaryotes. The synthesis of the lagging strand can require thousands of Okazaki fragments. RNA primers are needed for the synthesis of each single Okazaki fragment and they are Covalently joined together in order to generate a contious, intact strand of new DNA after synthesis.

Question 14

What are the two nucleases required for the removal of RNA primers? Why the removal of these primers important?

Answer 14

Nucleases: RNase H and 5’ to 3’ exonuclease

Important because they need to be removed prior to the binding of two Okazaki fragments.

Question 15

Explain the role of RNase H in RNA primer removal.

Answer 15

This is an enzyme which helps to degrade RNA that is base-paired with the DNA. RNase H removes all of the RNA primer except for the ribonucleotides which are directly linked to the DNA end. This is because RNase H can only break bonds between two ribonucleotides.

Question 16

Why can RNase H not remove the ribonucleotide that is directly linked to the DNA end.

Answer 16

This is a bond between a dNTP and rNTP which needs to be removed by the 5’ to 3’ exonuclease. The RNase H enzyme can only break bonds between two ribonucleotides.

Question 17

Explain the role of the 5’ to 3’ exonuclease in the removal of RNA primer

Answer 17

This exonuclease is used to remove the final ribonucleotide. Once this is removed, a primer:template junction is formed.

Question 18

What is the purpose of the primer:template junction formed after RNA primer removal?

Answer 18

DNA polymerase III will use this in order to synthesize new DNA to help fill the gaps within the Okazaki fragments. This will leave a complete DNA molecule except for a nick in the repaired strand.

Question 19

How is the nick in the lagging DNA strand fixed?

Answer 19

DNA ligase will come and use high energy co-factors such as ATP to create a phosphodiester bond between the adjacent 5’ phosphate and 3’ OH helping to repair the nick and complete the DNA strand.

Question 20

What is the purpose of DNA helicase in DNA synthesis?

Answer 20

DNA helicase is used to unwind the double helix DNA in advance of the replication fork.

Question 21

How is DNA helicase able to unwind the double helix DNA?

Answer 21

Hexameric protein (motor proteins) wool form ring-shaped complexes which encircle one of the two ssDNA strands at the replication fork. This protein binds to and moved directionally along the ssDNA by using the energy go ATP binding and hydrolysis in order to displace any DNA strand that is annealed to the bound DNA.

Question 22

What Is the polarity of DNA helicase?

Answer 22

Either 5’ to 3’ or 3’ to 5’ direction. The direction is defined according to the strand of DNA to which the helicase is bound.

Question 23

What is the benefit of helicase in terms of processivity?

Answer 23

Helicase has a high processivity similar to DNA polymerase which means that every time helicase associates with the DNA, it is able to unwind multiple base pairs of DNA.

Question 24

What are SSBs and what is their purpose in replication?

Answer 24

SSBs are single stranded DNA-binding proteins which help to stabilize the ssDNA once it has been separated by the helicase so that it does not hybridize back and the parental strands do not anneal together again after the helicase has passed.

Question 25

What are the binding properties of SSBs?

Answer 25

SSB proteins have cooperative binding which means that binding of one SSB protein promotes the binding of another SSB protein to the adjacent ssDNA. (due to SSB-SSB interaction between the proteins and DNA-SSB interactions between DNA and the protein)

SSB proteins do not have sequence specificity and they are able to make electrostatic interactions with the phosphate backbone as well as stacking interactions with the bases of the ssDNA

Question 26

What is the purpose of Topoisomerase in DNA synthesis?

Answer 26

Topoisomerase’s help to remove supercoils which can be produced when the DNA is unwinding at the replication fork. This is because the topoisomerase is able to catalyze the changes in the topological state of DNA and helps to relax the supercoiled forms by cutting and re ligating both strands of the DNA helix.

Question 27

How does Topoisomerase help with supercoils in the DNA double helix?

Answer 27

Topoisomerase removes the supercoiling by breaking either one or both of the DNA strands and passing the same number of DNA strands as the supercoil through the break. (Slide 11 - Lecture 3)

Question 28

How does positive supercoil occur in DNA?

Answer 28

When the DNA helicase is unwinding the strands, the DNA in front of the replication fork can become positively supercoiled and accumulation of these supercoils ends up placing strain on the DNA at the replication fork which can result in issues in replication.

Question 29

Topoisomerase acts on both circular chromosome in prokaryotes and linear chromosomes in eukaryotes. (T/F)

Answer 29

True

Question 30

What are the names of the enzymes that function at the replication fork?

Answer 30

Primase
DNA helicase
SSB (Single Strand-DNA binding proteins)
Topoisomerase

Question 31

Howe many DNA polymerase exist in eukaryotes? Name the three essential ones required to duplicate the genome.

Answer 31

Over 15 different Polymerases in Eukaryotes

3 essential ones include DNA polymerase Delta, DNA polymerase epsilon, and DNA polymerase alpha / primase. These three are involved in the replication of the genome whereas most of the other DNA polymerases are used in the repair processes.

Question 32

What is DNA polymerase Alpha/ primase? What is the function of the polymerase in DNA replication?

Answer 32

This is a protein complex which consists of two subunits of DNA polymerase alpha and two subunits of primase which is used to synthesize RNA primers.

The primase synthesizes an RNA primer which helps to form the RNA primer:template junction which the DNA polymerase alpha will associate to in order to initiate DNA synthesis. This polymerase has a low processivity rate but it is rapidly replaced by the other DNA polymerases in a process known as switching.

Question 33

What is the function of DNA polymerase delta and epsilon?

Answer 33

These polymerases go on to replace the DNA polymerase alpha. It is likely that the function of these polymerases is the replication of two different DNA strands at the replication fork. They have high rates of processivity and are able to synthesize DNA of 100-10,000 nucleotides.

Question 34

What the names and functions of the primary DNA polymerases in both prokaryotes and eukaryotes?

Answer 34

Prokaryotes:
- DNA polymerase I : RNA primer removal, DNA repair
- DNA polymerase II: DNA repair
- DNA polymerase III (core enzymes): Chromosome replication (3 subunits)
- DNA polymerase III (holoenzyme) : Chromosome replication (9 subunits including core enzymes)

Eukaryotes:
- DNA polymerase Alpha: Primer synthesis during DNA replication

• DNA polymerase Delta: Lagging strand DNA synthesis, nucleotide and base excision repair
• DNA polymerase Epsilon: Leading Strand DNA synthesis: nucleotide and base excision repair

Question 35

How are DNA polymerase at the replication fork able to have such high processivity?

Answer 35

This is due to the their associate with the proteins called the sliding DNA clamps. The sliding DNA clamps are formed of multiple identical subunits and the association of the sliding DNA clamp prevents the DNA polymerase from diffusing away from the DNA. Once the DNA synthesis has occurs, the absence of a primer:template junction results in the releasing of the DNA polymerase from the sliding clamp. If these proteins are not present, the DNA polymerases are only able to synthesize at 20-100 bp which is much lower than the rate at that they able to synthesize DNA with the proteins.

Question 36

What is the benefit of the leading and lagging strand being synthesized simultaneously?

Answer 36

This helps to limit the amount of ssDNA present during the replication process which helps to prevent DNA break and mutations.