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Cell & Molec FINAL > DNA replication > Flashcards

DNA replication Flashcards

1
Q

DNA structure

A
  • Strands run in opposite polarity
  • New nucleotides added in the 5’ –> 3’ direction (added onto the 3’ end)
  • Complementary base pairing (held together by hydrogen bonds)
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2
Q

Requirements for DNA replication

A
  • template DNA to be copied
  • raw materials (nucleotides)
  • enzymes
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3
Q

DNA replication must be . . .

A
  • accurate
  • fast
    (bacteria is 1000 bp/sec)
    (humans are 33 bp/sec)
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4
Q

Who is credited for discovering DNA replication?

A

Watson and Crick publish DNA has a double helix in 1953

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5
Q

DNA replication potential models

A
  • Conservative
  • Semi- conservative
  • Dispersive
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6
Q

Conservative potential model

A

Both strands serve as template (parental and new)

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7
Q

Semi conservative potential model

A

One strand serves as template

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8
Q

Dispersive

A

Parental DNA is “Dispersed” randomly

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9
Q

When did Matt Meselson and Franklin Stahl meet?

A

Met in summer 1954

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10
Q

Meselson & Stahl Experiment

A
  • they grew E. coli in a medium containing 15N- incorporated into DNA bases, which made DNA “Heavy”
  • grew long enough so both strands are labeled
  • Parent DNA is Heavy:Heavy (H:H)
  • Transferred the DNA to medium containing 14N and grew several generations
  • “Light” nucleotides are incorporated into newly made DNA
  • Pattern reveals model of replication
  • Labeled DNA analyzed on CESIUM CHLORIDE gradients
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11
Q

Conservative

A

L:L and H:H

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12
Q

Semi-conservative

A

H:L and H:L

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13
Q

Dispersive

A

H:L and H:L

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14
Q

What does DNA migrate to in the experiment?

A

its buoyant density: heavy labeled DNA runs further
- H:H DNA runs near bottom of tube
- H:L DNA (15 N & 14N) runs in between
- L:L DNA runs near top of tube

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15
Q

When were the different models ruled out and which model won?

A

Conservative was ruled out first because bands did not match up. Dispersive was ruled out after second round because two different bands. Therefore, Semi-conservative model won!

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16
Q

DNA replication is bi-directional

A

DNA synthesis initiates at an origin of replication and proceeds outwards in opposite directions

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17
Q

Prokaryotic DNA replication

A
  • Millions of base pairs
  • 1000 bp/sec
  • ONE origin of replication
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18
Q

Eukaryotic DNA replication

A
  • 100s of millions of base pairs (billions
  • 33 bp/sec
  • MULTIPLE origins of replication (solves slow rate of elongation)
19
Q

Replicon model

A

Proposed in E. coli
- Contains:
~ Origin of replication (replicator)
~ Initiator (DnaA)
~ Replication forks

20
Q

Origin of replication (replicator)

A

sequence where DNA replication starts

21
Q

Initiator (DnaA)

A

protein(s) that bind to the origin and initiate replication

22
Q

Replication forks

A

move away from the origin in a bidirectional fashion

23
Q

DNA replication in eukaryotes

A
  • multiple origins of replication
  • opens multiple replication bubbles that fuse together to complete replication
24
Q

First step in DNA Replication

A

1.) Find the origins of replication
- Short stretches of DNA having a specific nucleotide sequence
- DnaA (the initiator) binds to OriC (the origin)
- Looping causes stress to form in nearby AT-rich sequence= short stretch unwinds
- Unwound section provides access for DNA replication enzymes

25
Q

Second step in DNA Replication

A

2.) Two DNA helicase attach and unwind the DNA molecule by breaking hydrogen bonds between nucleotides
- Move away from each other and separate the two parental strands, making them available as template strands
- Creates replication forks: Y-shaped region where DNA is unwound

26
Q

Third step in DNA Replication

A

3.) Untwisting of the double helix causes strain ahead of the replication fork
- Topisomerase keeps the tension down

27
Q

Fourth step in DNA Replication

A

4.) Single-Stranded Binding Proteins (SSB) bind to unpaired DNA strands, keeping them from repairing
- keeps unraveled DNA single-stranded/stabilizes DNA

28
Q

Fifth step in DNA Replication

A

5.) Unwound sections of parental DNA are now able to serve as templates for synthesis of new complementary DNA strands

29
Q

One problem with DNA replication (5th step)

A

Enzymes that synthesize DNA cannot initiate synthesis of a polynucleotide
- Can only add nucleotides to the end of an existing chain that is base-paired with the template strand

30
Q

Solution to problem with DNA replication (5th step)

A

PRIMASE
- can synthesize a short stretch of RNA called the RNA primer
- new DNA strand can start from the 3’ end of the RNA primer

31
Q

Sixth step of DNA Replication

A

6.) DNA polymerase III catalyzes the synthesis of new DNA
- Adds complementary (to the parental DNA strand) nucleotides to the 3’ -OH end of the RNA primer
- Only adds in the 5’ –> 3’ direction!
- each nucleotide added comes from triphosphate nucleotide –> when added, releases two phosphates as pyrophosphate (broken down by pyrophosphatases)

32
Q

Seventh step of DNA Replication

A

7.) DNA added to both strands in different ways depending on if leading strand (continuous) or lagging strand (not continuous)
- helicases continue to unwind parental DNA
- LEADING strand: synthesizes continuously
- LAGGING strand: once enough of the parental DNA is unwound, primase lays down another RNA primer (DNA polymerase III extends unit it hits the previous primer)

33
Q

Leading strand

A

Template strand with 5’ end at replication fork
- Continuous synthesis in the 5’ –> 3’ direction
- Grows towards the replication fork
- Only ONE primer needed and DNA polymerase III remains attached to this strand

34
Q

Lagging strand

A

Template strand with 3’ end at replication fork
- Discontinuous synthesis in the 5’ –> 3’ direction
- Grows away from the replication fork
- MULTIPLE primers needed to be laid down by primase
- DNA polymerase III extends until it hits the previous primer = creates Okazaki fragments

35
Q

Which strand creates Okazaki fragments?

A

Lagging strand

36
Q

Eighth step of DNA Replication

A

8.) RNA primers are removed and Okazaki fragments are ligated

37
Q

DNA polymerase 1

A

removes the RNA primer and fills gap with DNA nucleotides

38
Q

DNA ligase

A

covalently links the Okazaki fragments together

39
Q

When does termination occur?

A

When replication bubbles merge

40
Q

Issues with the 5’ end of lagging strand

A
  • RNA primer removed from lagging strand, but DNA polymerase III cannot add to it because can only add to 3’ end
  • Therefore, next round of replication: shorter leading strand and shortened/staggered lagging strand
41
Q

Do prokaryotes have an issue with 5’ end of lagging strand?

A

No issue due to circular chromosome

42
Q

What is the solution for the issue with the 5’ end of the lagging strand in Eukaryotes?

A

Telomeres!
- Multiple repetitions of one short nucleotide sequence (no genes!)
- (ex. humans is TTAGGG)

43
Q

Issues with proofreading

A
  • DNA polymerase III works from 5’ to 3’
  • It proofreads (3’ to 5’) so each nucleotide against its template as soon as it is added to the growing strand (opposite direction)

Cell & Molec FINAL (8 decks)

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