Lecture 4: DNA Synthesis at Replication Fork, Finishing DNA Synthesis

Question 1

What is the replisome?

Answer 1

The combination of all of the proteins that function at the replication fork such as helicase, DNA polymerase III holozyme, primase, etc

Question 2

What is the origin of replication known as?

Answer 2

The specific sites at which DNA unwinding and initiation of replication occur. There is usually one OoR per chromosomes in prokaryotes, whereas eukaryotes can have upto 1000 OoR per chromosomes.

Question 3

What is the definition of a replicon?

Answer 3

All of the DNA that is replicated from one particular organ of replication. This means that eukaryotic can be dividied into multiple replicons, whereas, prokaryotic DNA has just one replicon.

Question 4

What are the two components that control the initiation of replication per replicon model?

Answer 4

Replicator and Initiator

Question 5

What is the replicator in a replicon model? Explain its function in initiating DNA synthesis.

Answer 5

The replicator are cis-acting DNA sequences which help to direct the initiation of DNA replication. A fraction of the replicator is the origin of replication. The replicator includes a binding site for the initiator protein that helps the assembly of the replication initiation machinery and also contains a stretch of AT-rich DNA that unwinds readily but not spontaneously.

Question 6

What is the initiator in the replicon model and what is its function in initiating DNA synthesis?

Answer 6

This is a protein which specifically helps to recognize a DNA element in the replicator and activates the initiation of the replication. It helps to select the sites that will become the origin of replication. These proteins are regulated by ATP binding and hydrolysis.

This is the only sequence-specific protein DNA binding protein that is involved in the initiation of replication.

Question 7

What are three examples of structure of replicators and initiator protein?

Answer 7

1. E.coli: Single replicator - oriC
2. Eukaryotic virus SV40: Origin
3. S. cerevisiae: Replicator
   (Slide 5- Lecture 4)

Question 8

Explain the structure of the E.coli single replicator known as oriC.

Answer 8

• Contains 9-mer (repeated 5 times) and 13-mer.
  motifs (repeated 3 times)
  - 9-mer motifs serves as a binding site for initiator.
  protein
  - 13-mer motif is the initial site of ssDNA formation
  during initiation

Question 9

Explain the structure of the eukaryotic virus SV40 origin.

Answer 9

• Contains P and EP sites
  
  • Four pentamer binding sites (P) for the initiator
    protein called large T antigen
  • A 20-bp early palindrome (EP) that is the site of DNA
    unwinding

Question 10

Explain the structure of the S.cerevisiae replicator.

Answer 10

• Contains three elements: A, B1, and B2
  
  • A and B1 elements bind to the initiator ORC (Origin
    recognition complex)
  • B2 element facilities the binding of the DNA helicase
    to the origin

Question 11

What are the different functions of initiator proteins during the initiation of replication?

Answer 11

1. They bind to the replicator DNA, often using a specific binding site
2. They interact with additional factors and recruit these factors for replication
3. They distort or unwind a region of DNA adjacent to their binding site providing a ssDNA template for replication

Question 12

What is the initiator protein for E.coli? How many binding domains does this protein have?

Answer 12

The protein is DnaA and it has 2 DNA binding domains. This protein is able to perform all three functions of initiator proteins.

The regulation of E.coli replication is linked to the DnaA initiator protein.

Question 13

How do protein-protein and protein-DNA interactions help direct the initiation process in E.coli for replication? Explain the step-by-step process.

Answer 13

1. Multiple DnaA.ATP proteins bind to the repeated 9-mer sequences within the oriC
2. This leads to strand separation in the 13-mer repeats
3. A complex between DNA helicase (DnaB) and the DNA helicase loader (DnaC) associates with the DnaA-bound origin
4. The DNA helicase loader helps to catalyze the opening of the DNA helicase protein ring and the replacement of the ring around the ssDNA at the origin
5. The DNA helices each recruit a primase that synthesizes an RNA primer on each ssDNA template. This primer causes the helicase loader to release from the helicase and this helps to activate the DNA helicase.
6. The newly synthesized primers and the helicase are recognized by the clamp loader components of the DNA Pol III holoenzyme
7. The sliding clamps are assembled onto each of the RNA primers and the synthesis of the leading-strand is initiated by one of the three core DNA Pol III enzymes on the holoenzyme
8. Once the DNA helicase has moved 1000 bp, a second RNA primer is synthesized on each lagging-strand template and a sliding clamp is loaded
9. This results in a primer:template junction which is recognized by a second DNA Pol III core enzyme and this initiates the synthesis of the lagging strand
10. Now, the synthesis of both the leading and lagging strand is initiated at each replication fork
11. The third core DNA Pol III enzyme is also involved in the synthesis of the lagging strand
12. Each replication fork will continue to the end of the template or until it meets another replication fork moving in the opposite direction

(Slides 7 and 8 - Lecture 4)

Question 14

How many bases are eukaryotic genome replicators typically separated by?

Answer 14

~30 kb (1 kb = 1000 bases)

Question 15

Why must each base pair in each chromosome be replicated only once in a eukaryotic cell?

Answer 15

Addition of even one or two more copies of critical regulator genes can lead to catastrophic defects in gene expression, cell division, or the response to environemental signals. This can be a challenge for eukaryotic chromosomes because they have multiple origins of replication.

Question 16

Is a replicator able to initiate replication again if it has already been replicated?

Answer 16

No replicator can initiate after it has been replicated. Therefore, cheater a replicator is activated to cause its own replication or replicated by a replication fork derived from a adjacent replicator, it must be inactivated until the next round of cell division.

Question 17

What is the importance of Type II Topoimerases in the finishing of replication?

Answer 17

Once replication has occurred for circular DNA, the two circular DNA molecules become concatenated (linked together). Type II Topoimerase comes in to separate these molecules. This type of topoimerase is also important for the segregation of large linear molecules.

Question 18

What is the replication end problem?

Answer 18

The replication end problem occurs as the lagging-strand replication machinery reaches the end of the chromosome and primase no longer has enough space to synthesize a new RNA primer. When this occurs, there is incomplete replication and short ssDNA region at the 3’ end of the lagging strand final product DNA. When this final product DNA is replicated in the next round, one of the two products will be shortened and will lack the region that was not fully copied in the previous round of replication. Therefore, in each round of DNA replication, the 5’ end if DNA loses about 50-100 nt, leading to the shortening of one of the two daughter chromosomes and the loss of genes near the end of the chromosome. If this were to continue, this would disrupt the complete propagation of the genetic material from generation to generation and slowly the genes would be lost.

Question 19

What are some solutions to the end replication problem?

Answer 19

One solution is to use a protein instead of an RNA as the primer for the last Okazaki fragment at each end of the chromosome. This way, the priming protein bind to the lagging strand template and uses an amino acid to provide an OH that replaces the 3’-OH normally provided by the RNA primer. By priming the last strand, the protein used to prime would become covalently linked to the 5’ end of the chromosome. This method is known as protein priming.

Question 20

How are eukaryotic cells able to solve the end replication problem?

Answer 20

They use telomeres and telomerase. We know that telomeres are the TG-rich DNA sequences which appear the ends of the chromosome. Telomerase is a specialized DNA polymerase that solves the replication problem by extending the 3’ end of the chromosome. This polymerase consists of multiple protein subunits and an RNA component known as ribonucleoprotein. In this case, an exogenous DNA template Is not required in order to direct the addition of new dNTPs onto the 3’ end of the DNA. Instead, the RNA component of the telomerase serves as a template for adding the telomeric sequence to the 3’ terminus at the end of the chromosome. The extension of the 3’ end provides additional template for the lagging-strand machinery.

Question 21

What is the RNA component of telomerase called?

Answer 21

Telomerase RNA (TER)

Question 22

What are the stepped involved in the solving the end replication problem using telomeres and telomerase?

Answer 22

1. Telomerase uses it RNA component (TER) to anneal to the 3’ end of the ssDNA region of the telomere
2. The enzyme uses its reverse transcription activity to synthesize DNA to the end of the RNA template
3. Telomerase then displaces the RNA from the DNA product and rebinds at the end of the telomere and repeats the process

(Slide 14 - Lecture 4)

Question 23

What hypothesis is derived from the finding of elevated telomerase activity in cancer cells?

Answer 23

Telomeres may represent a method to limit the growth capacity of cells that have lost normal growth control.