1 / 62

Structure (chapter 10, pages 266 – 278) and Replication of DNA (chapter 12, pages 318 – 334)

Structure (chapter 10, pages 266 – 278) and Replication of DNA (chapter 12, pages 318 – 334). Structure of DNA. Designate the Nucleotides Purines Guanine = G Adenine = A Pyrimidines Thymine = T Cytosine = C. Structure of DNA.

mbilodeau
Download Presentation

Structure (chapter 10, pages 266 – 278) and Replication of DNA (chapter 12, pages 318 – 334)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Structure (chapter 10, pages 266 – 278)and Replication of DNA (chapter 12, pages 318 – 334)

  2. Structure of DNA • Designate the Nucleotides • Purines • Guanine = G • Adenine = A • Pyrimidines • Thymine = T • Cytosine = C

  3. Structure of DNA • Nucleotides join together, forming a polynucleotide chain, by phosphodiester bonds • The phosphate attached to the 5’ carbon on one sugar • Attaches to the 3’ hydroxyl (OH) group on the previous nucleotide

  4. 5’-phosphate of last nucleotide chemically bonded to the 3’-hydroxyl of the next-to-last nucleotide A phosphodiester bond

  5. Structure of DNA • DNA is a double helix (two strands) held together by hydrogen bonds • Adenine (A) and thymine (T) are paired • Guanine (G) and cytosine (C) are paired • Always a purine pairs with a pyrimidine

  6. 5’-end 3’-end (free 3’-OH) The two polynucleotide strands (the backbones) in the double helix run in opposite directions, and are said to be anti-parallel 5’-end (free 5’- phosphate) 3’-end

  7. 5’-end 3’-end (free 3’-OH) Because of the pairing (A-T; G-C), one polynucleotide chain is always complementaryto the base sequence of the other strand 5’-end (free 5’- phosphate) 3’-end

  8. It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material. J. D. Watson and F. H. C. Crick, 1953

  9. Matthew Meselson and Franklin Stahl, 1958 entirely new AND entirely old DNA molecules present ALL DNA molecules are made up of both old and new DNA entirely new DNA molecules present BUT not entirely old DNA molecules

  10. Meselson and Stahl • Experiment • Grew E. coli in a growth medium containing only 15N(heavy nitrogen) (note: Normal isotope is 14N lighter nitrogen) • Did this for many generations so that all of the bacterial DNA would be “heavy”

  11. Meselson and Stahl • Experiment • Then grew the bacteria with 15N incorporated in their DNA in medium containing only 14N (would be incorporated into the new DNA) This way they could differentiate the original DNA from newly incorporated DNA

  12. Meselson and Stahl • Experiment • At each generation • Isolated the DNA • Looked at the density of the DNA in a CsCl gradient

  13. Matthew Meselson and Franklin Stahl, 1958

  14. What Meselson and Stahl expected if semiconservative replication

  15. entirely new DNA molecules present BUT not entirely old DNA molecules

  16. First generation results helped them rule out one of the three possible mode of replication

  17. Matthew Meselson and Franklin Stahl, 1958 entirely new AND entirely old DNA molecules present ALL DNA molecules are made up of both old and new DNA entirely new DNA molecules present BUT not entirely old DNA molecules

  18. Clincher evidence! Why? In dispersive model lighter DNA band should not have formed

  19. Great test question: Predict what the cesium chloride gradients would look like for conservative and dispersive replication!

  20. Should be able to draw something like this for conservative and dispersive

  21. So the DNA double helix unwinds and each strand acts as a template for replication of the new half Meselson and Stahl showed that the semiconservative pattern of replication is what was found

  22. Replication General features: 1. There is a specific site where replication begins (origin) which must be recognized 2. The two strands of DNA must be separated 3. The original strand becomes the template for the new DNA strand 4. A primer molecule must be added on which the new DNA chain can be built 5. New nucleotides must be added complementary to the template strand 6. The newly synthesized DNA must be edited and joined into one continuous molecule

  23. Replication Origin of replication • Where synthesis of new DNA begins • A specific location with a specific sequence of nucleotides

  24. In some organisms a specific location that can be mapped

  25. Multiple and random origins in eukaryotes

  26. Origins of replication In bacteria and virusesone origin of replication In eukaryotes there can be thousands of replication origins

  27. Recognizes the Origin of Replication Initiator Proteins Start to denature the DNA so each strand can act as a template

  28. Replication • DNA is unwound by a helicase • Separates the double helix by breaking the hydrogen bonds

  29. Helicase

  30. The separated (single strand DNA) is combined with single-strand binding proteins • Protects DNA from degradation • Keeps the complementary strands from rejoining

  31. Replication • As DNA is unwound it will tangle and knot, called supercoiling (from the unwinding of the helix)

  32. The supercoiling must be relaxed (the DNA unknotted) This is done by a class of enzymes called topisomerases (gyrase)

  33. Review: Proteins and their function in the early stages of replication 4 3 2 1 1 = initiator proteins 2 = single strand binding proteins 3 = helicase 4 = topoisomerase (gyrase)

  34. Replication DNA polymerases • Enzymes that synthesize new DNA

  35. 5’ triphosphates of the four nucleotides must be present (dATP, dGTP, dTTP, dCTP) • Two of the phosphates are cleaved-off, providing energy to run the reaction

  36. The preexisting single strand of DNA is the template strand Phosphates cleaved to provide energy for the reaction Nucleotide monophosphates are then joined to the 3’OH group Complementary base to the template strand

  37. Replication • DNA Polymerase • Must have a free 3’-OH group onto which to add the new nucleotides • No known DNA polymerase is able to initiate chains; thus, requires a primer to start synthesis

  38. DNA is always polymerized in a 5’ to 3’ direction and antiparallel to the template strand

  39. KNOW ALL TERMS!

  40. Replication • DNA Polymerase • Must have a free 3’-OH group onto which to add the new nucleotides • No known DNA polymerase is able to initiate chains; thus, requires a primer to start synthesis

  41. Must have a primer (which is an RNA molecule) The primer is synthesized by the enzyme primase (RNA polymerase)

More Related