DNA replication Flashcards

1
Q

Who demonstrated that DNA replication was semi-conservative?

How did they show this?

A

Meselson & Stahl

Bacteria growing in 15N heavy medium transferred to 14N light medium - showed strands of incorporation of both 15N & 14N DNA as well as strand with only 14N DNA

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2
Q

On which sequence in E. coli does DNA replication begin?

In 5 steps, how is the replication bubble formed?

How is DnaB activated?

Why is DNA temporarily bound by single-stranded DNA binding proteins (SSP)

What are 13mers rich in?

A

OriC consensus sequence

  1. DnaA-ATP binds to 9mers
  2. Allows for interaction with 13mers due to ATP present resulting in separation between double strands
  3. DnaA recruits DnaB (inactive helicase) and DnaC (helicase loader) complex to the DNA
  4. DnaB recruits DNA primase to form RNA primer (3’OH end) for DNA polymerase
  5. When complex loaded onto DNA by DnaC & DnaC dissociates from complex, DnaB is activated

Once loaded by DnaC

Prevent re-annealing after helicase unwinds & during replication

AT

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3
Q

What is the recruitment of DNA pol III holoenzyme triggered by?

What is the DNA pol III holoenzyme made up of?

What does the sliding clamp ensure?

What does this feature give the holoenzyme?

A

Primer:template junction created by DNA primase

3 core pol III (2 on lagging, 1 on leading)
Sliding clamp loader
3 sliding clamps
Cores joined by tail proteins & linker regions to the sliding clamps & loader

Structure is bound to the replication fork as replication proceeds so remains bound to DNA as translocates

High processivity

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4
Q

What are the 9 steps of the Trombone Model of the lagging strand? (assuming 1st DNA polymerase subunit is already bound & produced a fragment)

What happens after the Okazaki fragments are made?

A
  1. RNA primer is made by association of DnaB helicase and DnaG DNA primase
  2. Clamp loader loads a sliding clamp at the Primer-Template junction and DNA primase is released (as RNA primer has been formed)
  3. 2nd DNA polymerase subunit recognises the sliding clamp (loaded onto junction after 3’OH of primer) and begins synthesising 2nd Okazaki fragment
  4. When Okazaki fragment is completed, the 1st DNA polymerase subunit (further on in the loop) dissociates
  5. Single-stranded DNA loop is further looped out from the replisome (replication fork)
  6. DnaG re-associates with DnaB at the replisome & forms a 2nd RNA primer
  7. Clamp loader loads another sliding clamp onto the primer-template junction & DNA primase released
  8. 1st DNA polymerase subunit then associates sliding clamp & proceeds to make the 3rd Okazaki fragment
  9. When 3rd Okazaki fragment completed, 2nd DNA polymerase subunit dissociates (from 1st Okazaki fragment)
    Cycle is repeated

DNA polymerase I removes the RNA primers & replaces it with DNA
2. DNA ligase fills the gaps between the fragments

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5
Q

What is the problem with DNA replication of circular chromosomes?

How is this solved?

Before DNA replication, the DNA needs to be relaxed. How is this achieved?

Why does DNA need to be relaxed?

How is the stress compensated?

A

Closed circular DNAs (ccDNA) are linked by catenanes by concatenation of DNA

Topoisomerase II nicks the double-helix DNA strands & passes one DNA circle through the other & reanneals = decatenation

Topoisomerase I negatively supercoils it by nicking one strand in double helix & reanneal to regulate helical density

To open the replication bubble to be accessed by DNA polymerase III

Positive supercoiling ahead of the complex/fork

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6
Q

What is the problem with DNA replication of linear chromosomes?

The first solution is protein priming in prokaryotes. How does this solve it?

In eukaryotes- telomeres solve this problem. What is a telomere?

How does the telomerase extend the chromosome in 5 steps?

A

Lagging daughter strand is incompletely replicated at 5’ due to the position of the Okazaki fragment & when primer removed - which produces shorter chromosomes as there is no 3’ OH for DNA polymerase to fill in the gap

Tyrosine used instead of RNA primer which provides 3’OH for DNA synthesis of 5’ end

TG rich repeated sequence found at 3’ end of eukaryotic chromosomes - recognised by telomerase to bind to 3’ end and extend

  1. Telomere recognised by telomerase with temolerase RNA partly complimentary to 3’ end of chromosome
  2. Subunit of telomerase is reverse transcribing & synthesises 3’ end of DNA complimentary to telomerase RNA using telomere sequence as template
  3. Telomerase then translocates, rebinds with complimentary RNA sequence & polymerases again until overhang filled/
  4. Extended 3’ end of DNA can now act as a template for a new Okazaki fragment
  5. Chromosome repaired but end up with telomere extension of 3’ overhang
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7
Q

What do telomerase binding proteins do?

What do they signal?

What happens in human telomeres?

What is the telomere hypothesis?

What is this associated with?

What is the problem?

A

Bind along telomere region of chromosome - increasing in numbers as the extension increases.

They inhibit telomerase binding & identify the end of the chromosome

Form a T-loop - identify & protect end of chromosome

That gradually chromosomes get shorter & short telomeres/chromosomes are linked to ageing & disease

Hayflick number - limited number of cell divisions in a life time

Telomerase activity in somatic cells is very low, but cancer cells have a very high telomerase activity

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8
Q

How is a deoxyribonucleotide integrated into a DNA strand? 4 steps

What is catalysis determined by?

What type of bonds are phosphodiester bonds?

A
  1. 3’OH on primer initiates nucleophilic attack of incoming nucleotide group on its alpha phosphate
  2. Effective nucleophilic attack between the 3’OH of the primer (or nucleotide) and the alpha phosphate of incoming nucleotide can occur with correct complementarity & as a result alignment
  3. Correct alignment/complementarity allows for hydrogen bonding between polymerase & the minor groove of DNA & allows 3’OH to make a nucleophilic attack
  4. nucleophilic attack results in release of 2 phosphates (beta & gamma) which on reaction with pyrophosphotase produces two separate inorganic phosphates which contribute to the energy requirement (ADP + Pi -> ATP) to bind the next nucleotide

Correct base complimentary to template for sufficient alignment - dictated by sequence specificity

High energy

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9
Q

What activity in DNA polymerase I is used for its nick translation?

What is nick translation?

What rejoins the extension?

A

5’ to 3’ exonuclease

Recognises nick/gap between RNA primer & Okazaki fragment, degrades RNA primer & extends fragment by 10-12bp before dissociation

DNA ligase

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10
Q

For the following questions of DNA polymerase III holoenzyme: how many subunits are there, what is the function, and any extra info?

alpha

ε

θ

τ

ß

γ

What is the primosome?

A
  1. 3- each with 5’ to 3’ polymerase activity. Each catalytic core made of 1 a, ε and θ
  2. 3- 3’ to 5’ exonuclease proofreading activity
  3. 3- 5’ to 3’ exonuclease activity nick translation
  4. 2- dimerise 2 of the core pol III enzymes
  5. 3 - sliding DNA clamps for each pol core
  6. 1- clamp loader for lagging strand. made of 7 protein complex encoded by 5 genes

Complex responsible for making RNA primers including DnaB helicase & DnaG primase, DnaC helicase loader & others

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11
Q

What is the definition of DNA processivity in DNA pol III holoenzyme?

What is the palm?

What is the thumb?

What are the fingers?

What happens to the DNA after nucleotide incorporated?

In what conformation is the template DNA and why?

A

Number of nucleotides added each time enzyme core binds to primer:template junction

active site

structural - electrostatic interactions between phosphate backbone & thumb

close in on correct nucleotide alignment to facilitate hydrogen bonds forming between minor groove of DNA & palm domain

DNA partially releases from polymerase so can rebind 1 nucleotide along

Partially bent - efficient nucleotide incorporation so base is properly exposed in palm

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12
Q

What happens in the polymerase core when a nucleotide incorrectly incorporates?

What happens to the rate of catalysis?

What does the 3’ to 5’ exonuclease activity do?

What happens to the enzyme core after base is removed?

Why is DNA replication accurate?

Why is this so important?

A

Hydrogen bonds within minor groove of DNA & palm don’t fit right due to lack of alignment/complementarity

Slows

Conformation of enzyme core changes to relocate last incorporation, base is removed & conformation reverses

Template moves back into palm for 5’ to 3’ polymerase activity to accept incoming nucleotide - catalysis rate back to normal

DNA is smaller than RNA due to lack of 2’OH - so RNA is sterically excluded from active site

10x more RNA in cell than DNA

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13
Q

What are nucleotide precursor analogues?

How do they inhibit DNA replication?

What happens when these analogues are incorporated into DNA by replication?

What is the difference between obstructing machinery via inter-strand links and intra-strand links?

A

Poor quality nucleotides/analogues

Diminish supply of correct nucleotides (defective & interfere with replication of cancer cells)

Inhibit replication - terminate chain due to incorrect position of 3’OH

Inter - forms links between opposing strands in DNA (mustard gas)
Intra - between bases in same DNA strand

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