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Molecular Biology > Molecular Biology Lecture 8 - DNA Replication - Telomere Problem > Flashcards

Molecular Biology Lecture 8 - DNA Replication - Telomere Problem Flashcards

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1
Q

Eukaryotic initiation of DNA replication

A
  1. Multiple replicators (initiation sites)
    • every ~ 30 kb
    • not every replicator is always used
  2. One round of DNA replication per cell cycle
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2
Q

Chromosonal DNA replication takes place only during the S-phase of the cell cycle.

A

True

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3
Q

The initiation of cells replication in eukaryotic cells involve two event that occur at distinct time in the cell cycle.

A
  1. Replicator selection
    - The process of identifying the sequence that will initiate replication & occurs in G1 phase.
  2. Origin of activation
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4
Q

Eukaryotics use a pre-replicative complex (pre-RC)

A
  1. ORC (origin recognition complex) binds
    • probably remains bound throughout the cell cycle
  2. ORC recruits Cdc6 and Cdt1 (helicase loading proteins)
  3. Helicases are assembled (MCM2-7)
  4. Assembled in G1 of cell cycle
    • DNA replication is not initiated
    • loaded replicators waiting for signal
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5
Q

Activation of pre-RCs by protein kinases

A
  1. Two kinases
    • CdK (cyclin-dependent kinase)
    • Dbk (Dbf4-dependent kinase)
  2. Cell cycle regulated
    • active in late G1/S transition
  3. Phosphorylate Cdc6 and Cdt1 and other proteins: inactivates Cdc6 and Cdt1
  4. Activate helicases and recruit replication machinery
  5. Activation of pre-RC
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6
Q

Effect of CDk activity on the pre-RC formation

A

High CDk activity is required for existing pre-RC complexes to initiate DNA replication. These same levels of CDk activity completely inhibit formation of new PRe-RC & vice versa.

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7
Q

Cell cycle regulation Cdk activity and DNA replication

A

Figure 8-33

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8
Q

Finishing DNA replication reveals Two problems

A
  1. Untangling the linked chromosomes (catenanes)

2. End problem with linear chromosomes - constant shortening

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9
Q

Untangling replicated DNA molecules in circular DNA is done by the enzyme-______?

A

topoisomerase II

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10
Q

End replication problem

A

The requirement of RNA primers to initiate all new DNA synthesis means that a complete template of the lagging strand template cannot be made even if the end of the last RNA primer anneals to its base.

  1. Chromosomes shorten with each round of replication
  2. Problem w/i and b/w generations
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11
Q

One End replication problem______

A

Protein primer

  1. No base pairing needed - starts at very end.
  2. Protein remains linked to 5’ end. Uses Amino acid to provide an -OH that replaces the 3’-OH normally provided by an RNA primer
  3. Certain bacteria and bacterial and animal viruses
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12
Q

End replication problem______ A more common solution

A

Telomeric DNA and telomerase

Telomerase consists of two components

  1. Protein
    • RNA-dependent DNA polymerase (RT).
    • TERT - telomerase reverse transcriptase
    • probably has helicase activity
  2. RNA
    • RNA template for TERT
    • TER -telomerase RNA
    • 150-1300 bases
    • template sequence: 3’AAUCCCAAUC 5’telomere repeat: 5’ TTAGGG… 3’
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13
Q

Telomerse

A

Uses its RNA component to anneal to the 3’ end of the ssDNA region of the telomere. Telomerase the uses its reverse transciptase activity to synthesize DNA to the end of the RNA template. Telomerase the displaces the RNA from the the DNA product and rebinds at the end of the telomere and repeats the process

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14
Q

Telomerase extends 3’ overhang

A

..

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15
Q

What controls telomere length

A
  1. Species-specific mechanisms
    • telomere length varies greatly between species
    • e.g. yeast - 300-600
      mouse - 25 kb
      human - 10 kb
  2. telomere-binding proteins
    • bind telomere sequence (DS DNA)
    • inhibit telomerase additively - inhibition increases with telomere length
    • feedback mechanism
  3. telomerase activity
    • cells w/o telomerase shorten telomeres with cell division
    • cell type, cell age
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16
Q

Telomerase-binding proteins and telomere length

A

Negative feedback

- binding proteins inhibit telomerase
- maintain telomere at a certain length
17
Q

cell senescence

A
  1. Cell senescence
    • somatic cells eventually stop dividing
      • loss of telomerase activity
    • Hayflick limit
      • cells in culture divide a certain number of times
      • can be extend by adding telomerase
  2. Cancer
    • 90% of human cancers have telomerase activity
    • mice with transgenic telomerase gene have increased rates of cancer (2-3 fold)
  3. Aging
    • Hypothesis: loss of telomerase activity leads to cell senescence and cell death (apoptosis), which leads to aging
18
Q

Mutations and DNA Repair

A

Mutations in genes are usually deleterious, but are evolutionarily necessary

- usually reduce viability and reproductive fitness
- rarely increase viability and fitness
- without mutations there would be no evolution
19
Q

Sources of mutations

A
  1. Replication errors
  2. DNA damage (mutagens)
  3. Transposable elements
20
Q

Tautomeric shifts

A

Thymine (enol) mimics cytosine (amino).
Cytosine imino mimics thymine (keto).

Guanine (enol) mimics adenine (amino).
Adenine (imino) mimics guanine (keto).

21
Q

Tautomeric shifts - changes in hydrogen bonding

A

Draw & label

22
Q

Mutagens increase mutation frequency

A

True

23
Q

Base modifications - chemical mutagens

A 
Types of modifications
methylation (alkylation)
deamination 
oxidation
base analogs

Change hydrogen bonding of bases
- mimic tautomeric shifts

24
Q

Alkylation - ethylmethane sulfonate (EMS)

A

EMS essentially causes a tautomeric shift.

25
Q

Deamination - nitrous acid (HNO2)

A

Deamination of cytosine spontaneously or induced by nitrous acid essentially causes a tautomeric shift.

26
Q

Deamination - nitrous acid (HNO2)

A

Deamination of adenine spontaneously or induced by nitrous acid essentially causes a tautomeric shift.

27
Q

Deamination of 5-methylcytocine

A

5-mC is a common base in mammalian DNA (methylation).

Deamination of 5-mC produces thymine

28
Q

Base analog - 5-bromouracil

A

Different from mutagens that change bases.
Substitute for normal nucleotides in DNA synthesis.
Shifts to enol form more readily than thkymine does.

29
Q

Production of ROS by the mitochondria

A

Reduction of O2 produces ROS during electron transport
O2 + e- → O2- superoxide radical
O2- + e- + 2 H+ → H2O2 hydrogen peroxide
H2O2 +e- + H+ → H2O + .OH hydroxyl radical
.OH + e- + H+ → H2O

30
Q

Mutagens - X-rays and γ-rays

A
  1. High energy
    • can break phosphodiester backbone
    • can produce DS breaks
      Ionizing radiation
      • generates loss or gain of electrons
      • can produce free radicals (e.g. ROS)
31
Q

Mutagens - UV light

A
  1. Produces pyrimidine dimers (usually thymine dimers)

2. Prevents normal base pairing - blocks DNA replication and transcription.

32
Q

DNA repair - E. coli

A
  1. Direct reversal of DNA damage (pyrimidine dimers)
  2. Mismatch repair
  3. Excision repair
    • base excision repair
    • nucleotide excision repair
    • transcription-coupled nucleotide excision repair
  4. Recombinational repair (double-strand break repair)
  5. Translesion polymerase

Molecular Biology (5 decks)

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