Chapter 9 - The replication of DNA Flashcards

1
Q

Why is DNA replication always 5’ to 3’?

A

Hypothetically 3’ to 5’ strand growth could do proceed as there would be no high-energy bond to be cleave.

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2
Q

How is DNA synthesis catalysed by DNA polymerase?

A

DNA pol is like a hand that grips the template-primer junction. The palm region contains the catalytic site and checks base-pair accuracy. The finger region closes when a correct dNTP comes in.

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3
Q

How does DNA polymerase provide proof-reading?

A

The right nucleotide has higher affinity to the DNA pol compared to the incorrect one. An incorrectly bound nucleotide is most likely yo dissociate. The next error-correcting feature is 3’ to 5’ exo proof-reading activity. Lastly, 5’-3’ exo activity is needed for primer removal.

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4
Q

What is the role of the different DNA polymerases?

A

DNA pol I and pol II are mostly involved in damage repair and proof-reading activity.

DNA pol III is the main replicase in E. coli.

In eukaryotes 
Pol Alpha - Lagging strand replication.
Pol Beta - DNA repair.
Pol Gamma - mitochondrial replication.
Pol Delta - Leading strand replication.
Pol Epsilon - replication ?
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5
Q

What are the two different error control mechanism of DNA pol and how do they differ?

A
  1. Presynthetic error control

2. Proof reading control

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6
Q

Name the two fragments of DNA pol I and their functions.

A
  1. The large cleavage product called the Klenow fragment. This fragment has polymerase and 3’-5’exonuclease activity.
  2. The second fragment possesses a 5’-3’ exo activity. This is needed to remove RNA primers.
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7
Q

What are primases?

A

Are RNA polymerase that form short RNA primers at the origin to provide this free 3’ OH end.

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8
Q

Name the part of the DNA pol III holoenzyme.

A

Sliding clamp, sliding clamp holder, Pol III core, tau proteins, and flexible linkers.

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9
Q

Briefly explain the steps in replication and which enzymes are involved.

A

Initiation
Involves recognition of an origin by a complex of proteins, called primosome. The primo some includes helices, gyrase, SSB and primases. In this stage the DNA helix is unwound to form a bubble which forms the beginning of the replication fork. Helices do the unwinding as the SSB protein stabilise the single stranded structure.

Elongation of new strands is the other name of polymerisation. The key enzyme in this process is DNA polymerases. They together constitute the replisome.

Termination is the final step. After this step he new chromosomes are formed and they separate from each other in a semiconservative manner.

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10
Q

Describe in detail initiation in bacteria.

A
  1. The bacterial replication origin has 3 x 13 bp repeats and 4x 9 bp repeats.
  2. DNA A monomers bind at the 9 bp repeats.
  3. Once these sites are recognised 20-40 DNA A monomers form large aggregates.
  4. DNA strand melts at the 13 bp repeats.
  5. DNA B and DNA C join the complex forming the replication fork.
  6. DNA B also activates the DNA G primate action.
  7. Further gyros provides the swivel that allows one strand to rotate around the other, which results unwinding or in other words generates replication fork.
  8. The protein SSB stabilises the single strand DNA as soon as it is formed.
  9. ATP hydrolysis is needed in two steps, during helicase and gyros action.
  10. Primase starts synthesis of RNA primers, 10-12 bases long.
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11
Q

What is the difference between DNA topoisomers I and II?

A

Type I are called nicking-closing enzymes and function to relax supercoiled circular molecules. They increased he linking number in the s step of 1. This enzyme function without the input of energy in the form of ATP.

Type II cuts double strands changing the linking number in the step of 2. These enzymes can either relieve supercoiling or introduce supercoiling into a DNA. They require the energy of ATP hydrolysis for their action. They make double strand breaks and can reseal it. They are also called gyrases. They require ATP for their action.

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12
Q

Describe nick resealing by topoisomerase I.

A

It has a catalytic tyrosine in the active site. Joining of tyrosine provide the high energy bond, which allows nucleophilic attack by the 3’ -OH end.

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13
Q

What is the function of RNase H?

A

Removes RNA primers which recognises DNA-RNA hybrid and cleaves the RNA bases leaving only a few.

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14
Q

Explain the action of ligase.

A

Ligase creates high energy bond by fusing AMP to the 5’-Phosphate. The 3’OH of the next Okazaki fragment attacks this bond between the two. AMP is released in the reaction.

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