DNA Replication Flashcards

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1
Q

conservative replication

A

original molecule completely conserved and new molecule completely new

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2
Q

semi-conservative replication

A
  • both original strands act as a template
  • each newly formed double-helix has 1 original polynucleotide strand and 1 new polynucleotide strand
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3
Q

dispersive replication

A

the 2 resulting molecules contain both old and new double helix DNA

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4
Q

what did Meselson and Stahl’s experiment do?

A

provided evidence to reject conservative and dispersive replication

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5
Q

why did Meselson and Stahl use E. coli

A
  • in optimum conditions, they have a short generation time of ~ 50 minute
  • DNA easier to obtain as free in cytoplasm and not surrounded by a nuclear envelope
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6
Q

1st step of Meselson and Stahl experiment

A
  • 2 populations of E. coli grown on separate agar plates, one with 14N (light isotope) and one with 15N (heavy isotope) nitrogen
  • take up N for nitrogenous bases, so have 14N or 15N DNA
  • sample of DNA removed and centrifuged, with 14N settling at top and 15N at bottom of test tube
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7
Q

why was nitrogen used in Meselson and Stahl’s experiment?

A

nitrogen taken up by bacteria and used to form nitrogenous bases of their DNA

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8
Q

2nd step of Meselson and Stahl

A
  • 15N DNA washed and transferred to 14N DNA and allowed to divide once
  • centrifuged and settle in middle
  • rules out conservative as all DNA contains both 14N and 15N
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9
Q

why is the 15N DNA washed before dded to 14N agar

A

to remove 15N agar so only 14N can be used in division

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10
Q

3rd step of Meselson and Stahl

A
  • DNA allwed to divide again
  • centrifuged
  • half settles at the top and half at the middle
  • rules out dispersive as would all settle in middle if dispersive
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11
Q

4th step of Meselston and Stahl

A

with each further division, amount of 14N DNA increases and amount of mixed DNA stays the same

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12
Q

7 steps of DNA replication

A
  1. DNA helicase unwinds and unzips double helix, breaking H-bonds between strands
  2. both the 2 polynucleotides have exposed bases which act as templates
  3. free DNA nucleotides in the nucleoplasm align themselves against ech template strand
  4. H-bonds form between complementary base pairs, joining the aligned nucleotides to their exposed base pairs on the template (A-T and C-G)
  5. DNA polymerase catalyses the formation of the new polynucleotide strands in a condensation reaction forming phosphodiester bonds
  6. Enzymes rewind the DNA molecule back into a double helix
  7. As both original strands acted as a template, each newly formed DNA molecule has 1 original and 1 newly formed polynucleotide strand
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