Study Questions Set 4 Flashcards

1
Q

List proteins that are involved in DNA replication in E. coli. Explain the role of each of these proteins in replication.

A

• DnaA: initiates replication only of DNA is negatively supercoiled
• DnaB (helicase): moves along and separates dsDNA; requires DnaC
• DnaC: escorts DnaB to DnaA, forming the pre-priming complex
• Single-strand binding proteins (SSB): small proteins that combine into tetramers, which binds to ssDNA to prevent dsDNA reformation
• Primase (DnaG) (RNA polymerase): 1 primer for the leading strand, hundreds for lagging. Functions with the helicase to form the next priming site: primosome
• RNase H: removes most of the RNA primers
• DNA polymerase I (DNAP I): also removes rNTPs and replaces them with dNTPs
o 2 subuints: smaller which removes the rNTPs, & the bigger which adds the dNTPs
• DNA polymerase III (DNAP III): catalyzes synthesis of DNA from RNA primers
o : active site for nucleotide addition in 5’-3’ direction
o : 3’-5’ exonuclease (proof reads)
o : unclear function
• DNA ligase: links the fragments on the lagging strand via a phosphodiester bond
• Topoisomerase: enzymes that regulate supercoiling
o Topo II has the major role in replication – introduces negative supercoils necessary for initiation by DnaA
 After helicase creates positive supercoils, it turns them to negative

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2
Q

What is meant by replication being bidirectional? Semiconservative? Continuous and discontinuous?

A
  • Bidirectional: means there are 2 replication forks, one replication bubble, and a leading & lagging strand
  • Semiconservative: means that one strand from the double helix ends up in each of the daughter cells
  • Continuous: replication on the leading strand – 5’-3’, no breaks in the DNA formed
  • Discontinuous: replication on the lagging strand – 5’-3’ but since the DNAP III is moving in the 3’-5’ direction, it creates many small okazaki fragments of DNA which need to be linked together via DNA ligase
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3
Q

Contrast the role of DNApol I and III during prokaryotic replication

A

III polymerizes new DNA on the RNA primer using the template strand. I will remove the RNA primers and replace them with dNTPs, and also fix any errors made by III

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4
Q

Which subunit of DNApol III provides processivity? which protein loads this subunit on the DNA?

A

The B-subunit functions as a clamp, but associates with the core protein at the 3’ end of the growing strand, thereby increasing the processivity of the strand. The gamma subunit will load the beta subunit onto the DNA with the use of ATP

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5
Q

how can the discontinuous synthesis of the lagging strand keep up with the continuous synthesis of the leading strand?

A

DNApol III allows for the leading and lagging strand to be replicated at the same pace at the same time
trombone model has the lagging strand stay linear in the region that DNApol III binds to

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6
Q

why is decatenation required in circular DNAs?

A
  • Decatenation: the separation of catenanes
  • Catenanes: nearly complete daughter helices; equivalent to sister chromosomes
  • It is necessary so that the catenanes can segregate into the two daughter cells
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7
Q

why do eukaryotes need telomeres but prokaryotes do not?

A

• Since the shortening of chromosome ends results in the loss of genetic information, telomeres could be a safeguard to regulate the cell’s life time. As well, prokaryotes have circular DNA and not linear, so there is ‘end’ to them that could cause the DNA replication proteins to mistake the end of the chromosome as a break

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8
Q

What would be the components necessary to make DNA in vitro by using DNA polymerase I.

A
  • Primase

- Pol a

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9
Q

What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA ligase?

A
  • DNA ligase is what links all of the DNA fragments together on the lagging strand. If it was temperature sensitive, any time the cell was outside the working range, the lagging strand would not come together and thus not produce a replicate.
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10
Q

Compare and contrast major euk and pro DNApols

A

Prokaryotic:
- DNA polymerase I (removes RNA primers and fills in necessary nucleotides; not highly processive; exonuclease activity)
- Prokaryotic: DNA polymerase III : no primase function, only catalyzes synthesis of DNA from RNA primers. Can proof read itself
Eukaryotic:
9 different DNA polymerases
- Pol alpha: primase function, no proof reading ability
- Pol delta: takes over polA on lagging strand, proof reads Pol A, fills gaps on lagging strand
- Pol epsilon: takes over polA on leading strand & continues synthesis

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11
Q

What is telomerase and why is it important?

A
  • Telomerase: enzyme that repairs/replicates the ends of chromosomes
  • It extends the telomere by one repeat unit on the 3’ end, as the 3’ end is what degrades due to replication not being able to replicate the last 100 units or so
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12
Q

What is the major difference between bacterial and eukaryotic replication that allows a eukaryotic cell to replicate its DNA in a reasonable amount of time?

A
  • The major difference is that prokaryotes only have one point of origin on their DNA so replication can only start in one section and go all the way around. In eukaryotes, the DNA has multiple points of origin & so the replication can be happening at multiple locations on the DNA strand at once.
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13
Q

Describe the events that occur at an origin of replication during initiation of replication in E. coli?

A
  • Origin recognition complex (ORC) is a protein complex in which it’s binding identifies an origin of replication
    o Cdc6 binds, CDT1 and MCM then bind, Cdc6 and CDT1 then released
    o End up with the prereplication complex – cell now can enter S phase
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14
Q

What are cis- elements? What are trans- factors?

A

• Cis elements: sites or sequences on DNA
o DnaA box, OriC, TATA box etc.
• Trans factors: the proteins which recognize the cis elements
o DnaA, DnaB, DnaC

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15
Q

17 Is the following statement true or false: Regardless of whether a gene is expressed in a given cell type, it will replicate at the same, characteristic time during S phase. Explain your reasoning.

A

• False. Euchromatin is replicated first, followed by heterochromatin (facilitative then constituative)

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16
Q

Thinking question: Is making of RNA primers (by primase), which have to be subsequently removed and replaced with dNTPs (by DNA polymerase I) actually wasteful and energetically inefficient process? (Hint: think about fidelity of primer-making vs. proofreading capability of DNA pol I, in other words what is the only purpose of RNA primers – is the accuracy of this process very important at this point?)

A
  • RNA primers need not be accurate since they will be excised and replaced by DNA polymerase; more efficient than using DNA primers which would need to be exact
  • Since a 3’ OH group is needed, these primers are necessary to achieve replication and since DNA pol I has proofreading capability, this process is not overly wasteful
17
Q
  1. What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA polymerase I?
A

• DNAP I corrects any errors from DNAP III and removes the remaining rNTPs and replaces them with dNTPs. If it had a temperature sensitive mutation, then outside it’s working range the DNA would have remaining RNA fragments left over & possible mistakes from DNAP III, which would likely lead to non-functioning proteins, meaning the cell may possible not survive

18
Q

You preformed Cot analysis using genomic DNA samples obtained from a 2 year-old child and a 76 year-old individual. Results of this analysis show that one of the most rapidly reassociating classes of DNA is substantially reduced in the older individual with respect to the 2 year-old. How can you explain this finding? (hint: think about termination of linear DNA replication.)

A
  • This would be a result of the telomere degradation that occurs in linear chromosomes after each DNA replication. The telomeres consist of primarily repeating units, which account for the rapidly reassociating class. Therefore this makes sense that the older individual would have less of the rapid reassociating class, as those are the telomeres which they no longer have many of.