HL 1 Biology

3.4 & 7.2 DNA Replication

<<Semi-Conservative Replication>>

DNA replication referred to be semi-conservative because:

  • 2 strands of double helix separate, when cell prepares to divide (Each original strands are templates to create new strands)
  • Result: 2 DNA molecules both composed of original strand + a newly synthesized strand

Process of DNA Replication:

  • Many enzymes involved = the helix has to be unwound and separated by the enzyme DNA helicase for the enzymes to gain access to the DNA
  • Nucleotides hydrogen bond to complementary bases within template
  • DNA polymerase link to phosphate of newest nucleotide to sugar of nucleotide by covalent bond

Difference in Replication between Prokaryotes & Eukaryotes

  • Origins of Replication:site in which replication starts
      • Prokaryotic cells have one origin, and eukaryoic cells have many.
  • Nucleoside Triphosphate: new unit added to the growing nucleic acid during replication
      • nucleoside triphosphate hydrogen, bonds to complementary base in DNA molecule
      • hydrolysis of 2 phosphate molecules  = convert nucleoside triphosphate to nucleotide
      • Only added at 3′ end of nucleotide = DNA strands are organized in anti-paralllel way
      • Nucleoside/ Nucleoside monophosphate: sugar and a base
      • Okazaki fragments: lagging strands
          • enzyme DNA ligase: help join fragments together
  • Replication 
      • catalyzed by enzyme DNA polymerase III
      • needs a short sequence (primer) to start process   *primer: made of RNA and primase enzyme
      • The DNA polymerase I = enzyme that digests away RNA primer + replace with DNA


The Meselson-Stahl Experiment (p.63)

Data in Figure 14 suggests that replication is semi-conservative because a peak is not present at 1.710 in N-15. Because the DNA molecule built with only N0-14 did not have a peak, the replication can be stated to not be conservative. The figure is not dispersive because the second generation consists of 2 peaks instead of one. Also, if it is dispersive, the all the molecules would have the same ratio of N-14 to N-15. no matter the time/generation (last figure from figure 14). 

Meselson-Stahl Experiment (p.64)

(1) The sample that was pulsed for 10 seconds only consists of one high peak rather than the sample pulsed for 30 seconds, which had 2 peaks throughout. Also, the 3o second sample exhibited a higher radioactivity to the distance from the top,  compared to the sample pulsed for 10 seconds. 

(2) The sample pulsed for 30 seconds provides evidence for the presence of both a leading strand and many lagging strands because of the two apparent peaks in radioactivity. The high peak indicates the lagging strands for its steep increase and the lower peak shows the presence of leading strands due to the longer fragments.

(3) The sample pulsed for 60 seconds provides evidence for the activity of DNA ligase because of the number of molecules. The small molecules decreases and larger molecules would increase because the DNA ligase merge the lagging strands together.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s