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MBB 407/511 Lecture 21: Eukaryotic DNA Replication

Nov. 29, 2005. MBB 407/511 Lecture 21: Eukaryotic DNA Replication. Outline:. V. Telomerase. VI. Packaging of Eukaryotic DNA. VII. DNA Topology. Circular dsDNAs can be replicated completely: E. Coli Eukaryotes • Plasmids • mitochrondrial DNA chromosome

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MBB 407/511 Lecture 21: Eukaryotic DNA Replication

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  1. Nov. 29, 2005 MBB 407/511 Lecture 21: Eukaryotic DNA Replication

  2. Outline: V. Telomerase VI. Packaging of Eukaryotic DNA VII. DNA Topology

  3. Circular dsDNAs can be replicated completely: E. Coli Eukaryotes • Plasmids • mitochrondrial DNA chromosome • E. coli Chromosome • circular dsDNA viruses (e.g., SV40) Replication of linear dsDNAs (e.g., eukaryotic chromosomes) poses an “end replication problem.”

  4. Primer gap Primer gap Primer gap The End Replication Problem

  5. Adding Telomeres to the Leading Strand Allows for Addition of a New Okazaki Fragment to the Lagging Strand

  6. Packaging of Eukaryotic DNA Chromatin is ~50% DNA/50% protein by weight

  7. Chromatin Proteins

  8. DNA Wraps Around Histone Octomers to Form Nucleosomes 2 H2A 2 H2B 2 H3 2 H4 “Beads on a String” Core DNA is protected from DNases Nucleosomes • Two each of H2A, H2B, H3, H4 = histone octomer (core) • 146 bp core DNA wraps around a histone core • Between each nucleosome is 20-60 bp linker DNA bound by a molecule of H1 histone • Spaced ~200 bp apart (146 bp core + 20-60 linker DNA) = Regular spacing between nucleosomes Why are histones positively charged?

  9. Histone H1 Induces Tighter DNA Wrapping Around the Nucleosome Histone H1 binds two DNA helices H1 is the Linker Histone: Binds the Linker DNA 30-nm Fiber

  10. H1 A Model for Chromosome Structure DNA exists in chromatin form during interphase Solenoid = cylindrical coil DNA is most compact in chromosome form during metaphase of mitosis

  11. What is Supercoiling? 10.5 bp Supercoiling occurs in nearly all chromosomes (circular or linear) Relaxed vs Supercoiled DNA Relaxed DNA has no supercoils Negatively supercoiled DNA is underwound (favors unwinding of the helix) (circular DNA isolated from cells is always negatively supercoiled) Positively supercoiled DNA is overwound

  12. L = T + W Linking Number (L or Lk) = number of times the two strands are intertwined Twists (T or Tw) = number of helical turns For a 2,000 bp DNA duplex, T = 200 (2,000 bp  1 turn/10 bp = 200 turns) Writhes (W or Wr) = number of times the duplex crosses itself (only topologically constrained DNA molecules can have writhe) A relaxed DNA molecule has zero writhes. ( For a relaxed DNA molecule, L = T) Writhes = Supercoils Over- or underwinding results in writhes instead of twists Why? Because the strain of writhes is less than the strain of twists

  13. Additional Terms Used To Describe Topology The Linking Number Difference (DL) is the difference between the linking number of a DNA molecule (L) and the linking number of its relaxed form (L0). The equation is DL = L – Lo. DL is a measure of the number of writhes For a relaxed molecule: DL = 0 The Superhelical Density (s) is a measure of supercoiling that is independent of length. The equation is s = DL / Lo. s is a measure of the ratio of writhes to twists For a relaxed molecule: s = 0 DNA in cells has a s of –0.06 (for circular molecules purified from bacteria and eukaryotes)

  14. Sample Linking Number Questions 1) A. L = T + W; for relaxed molecule W = 0  L0 = T; 5500 bp X 10 bp/turn = 550 turns B. L = T + W = 550 + (– 50) = 500 C. DL = L – L0 = 500 – 550 = – 50 D. s = DL / L0 = – 50 / 550 = – 0.09

  15. 2) Instead of treating the relaxed 5,500 bp plasmid DNA molecule above with DNA gyrase, you transfer it from aqueous solution to 50% ethanol. Under these conditions, the structure changes from B-DNA to A-DNA due to the relatively lower water concentration. (A-DNA has 11 bp/turn). A. What is the linking number after transfer to 50% ethanol? B. How many helical turns will there be after transfer to 50% ethanol? C. How many writhes will there be after transfer to 50% ethanol? A. L = 550 (linking # stays the same because no bonds are broken) B. 5500 X 11 bp/turn = 500 helical turns C. L = T + W; 550 = 500 + W; W = +50

  16. Type I and II Topoisomerases (usually relax supercoiled DNA) Rule #1: They change the linking number by changing the # of writhes. Rule #2: The change the linking number by breaking one or both strands of the DNA molecule, winding them tighter or looser, then rejoining the ends. Rule #3: They work only on topologically constrained DNA molecules because only topologically constrained DNA molecules can have writhe.

  17. Type I Topoisomerases Topo I of E. coli 1) acts to relax only negative supercoils 2) increases linking number by +1 increments Topo I of eukaryotes 1) acts to relax positive or negative supercoils 2) changes linking number by –1 or +1 increments

  18. Relaxation of SV40 DNA by Eukaryotic Topo I Maximum supercoiled 3 min. Topo I 25 min. Topo I

  19. Type II Topoisomerases They relax or underwind DNA by cutting both strands then sealing them. They change the linking number by increments of +2 or –2.

  20. E. Coli vs. Eukaryotic Type II Topoisomerases: The strain of underwinding DNA is relieved by: Negative supercoils or Local disruption of base pairs DNA Gyrase DNA Helicase Euk. Topo II Topo II of E. coli (DNA Gyrase) 1) Acts on both neg. and pos. supercoiled DNA 2) Increases the # of neg. supercoils by increments of –2 3) Requires ATP Topo II of Eukaryotes 1) Relaxes only negatively supercoiled DNA 2) Increases the linking number by increments of +2 3) Requires ATP

  21. All Topoisomerases Cleave DNA Using a Covalent Tyrosine-DNA Intermediate

  22. The Role of Topoisomerases in DNA Replication 1) Topoisomerases remove positive supercoils that normally form ahead of the growing replication fork DNA gyrase E. Coli  DNA gyrase (adds neg. supercoils) Eukaryotes  Topo I (relaxes pos. supercoils)

  23. 2) Replicated circular DNA molecules are separated by type II topoisomerases Catenated (linked) Type II topoisomerases decatenate and catenate

  24. A Review of the Different Topoisomerases Topo Type E. coli Eukaryotic I Topo I Topo I Cleaves Relaxes only – supercoils Relaxes – and + supercoils 1 strand (nicks) Changes linking # by +1 Changes linking # by +1 or –1 & reseals Requires no cofactors Requires no cofactors II Topo II (DNA Gyrase) Topo II Cleaves Acts on – and + supercoils Relaxes only – supercoils 2 strands (ds cut) Changes linking # by Changes linking # by & reseals increments of -2 increments of -2 Catenates and decatenates DNA Catenates and decatenates DNA Requires ATP Requires ATP Eukaryotic topoisomerases cannot introduce net supercoils, Therefore, how can eukaryotic DNA become negatively supercoiled? Introduces net neg. supercoils

  25. How Does Eukaryotic DNA Become Negatively Supecoiled? Plectonemic supercoils Solenoidal (Toroidal) supercoils Q: What will happen if you remove the histone core?  DNA wrapping around histone cores leads to net negative supercoils! A: The solenoidal supercoil will adopt a plectonemic conformation

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