1. Structure, Replication and Recombination of DNA

2. Information Flow From DNA DNA RNA transcription Protein translation replication

3. DNA Structure Primary Structure Chain of Nucleotides Secondary Structure Double Helix

4. DNA Structure Nucleotide = building block of DNA Three components of a nucleotide Three components of a nucleotide 1. Nitrogen-containing base 1. Nitrogen-containing base purine or pyrimidine purine or pyrimidine 2. 5-carbon sugar 2. 5-carbon sugar 3. Phosphate group 3. Phosphate group

5. DNA Structure Purine bases Adenine (A) Guanine (G) Pyrimidine bases Cytosine (C) Thymine (T) 5-carbon sugar Deoxyribose Phosphate PO 4 PO 4

6. P---OH + OH---C  P---O---C + H 2 O

8. Chemical Bonding Covalent Bond Strong Atoms Share Electrons Formation of a Nucleotide HydrogenBondWeak Atoms “Share” a Hydrogen Pairing of Nucleotide Bases

9. Hydrogen bonds hold pairs of bases together.

10. 3’5’ 3’ 5’

11. DNA Secondary Structure: The Double Helix Two polynucleotide chains are wound togetherTwo polynucleotide chains are wound together Bases are located inside the helixBases are located inside the helix Sugar-phosphate groups are on the outside as a “backbone”Sugar-phosphate groups are on the outside as a “backbone” Bases are arranged like rungs on a ladder, perpendicular to the “backbone”Bases are arranged like rungs on a ladder, perpendicular to the “backbone”

12. DNA Secondary Structure: The Double Helix Hydrogen bonding between bases holds the chains together:Hydrogen bonding between bases holds the chains together: A pairs with T G pairs with C Polynucleotide chains have opposite polarityPolynucleotide chains have opposite polarity One is 5’  3’ One is 5’  3’ other is 3’  5’ other is 3’  5’ 10 base pairs per turn of the helix10 base pairs per turn of the helix

13. Applying Your Knowledge In the DNA double helix, which base is paired with adenine? 1.Adenine 2.Cytosine 3.Guanine 4.Thymine 5.Uracil

14. DNA Replication: An Overview

15. DNA Replication DNA replication is semiconservative. Each strand is used as a template to produce a new strand. AGCTAGCTAGCT TCGATCGATCGA TCGATCGATCGA new AGCTAGCTAGCT new AGCTAGCTAGCT AGCTAGCTAGCT  AGCTAGCTAGCT old TCGATCGATCGA TCGATCGATCGAold  TCGATCGATCGA old

16. DNA Replication 5’—A G C T — 3’ 5’—A G C T — 3’ 3’—T C G A—5’ A —5’ G C G 3’— T C T— 3’ 5’— A DNA replication requires 1. DNA polymerase, an enzyme that adds 1. DNA polymerase, an enzyme that adds nucleotides in a 5’  3’ direction. nucleotides in a 5’  3’ direction. 2. Nucleoside triphosphates 2. Nucleoside triphosphates 3. Energy: release of diphosphate 3. Energy: release of diphosphate

17. Origin of Replication origin of replication replication fork DNA replication begins at a replication origin and proceeds bidirectionally, creating two replication forks for each origin. Eukaryotic chromosomes have multiple origins of replication.

18. Continuous and Discontinuous Synthesis DNA Polymerase builds a new strand in a 5’  3’ direction. This leads to continuous synthesis on the strand oriented 3’  5’ 3’  5’ and discontinuous on the strand oriented 5’  3’. movement of fork 5’3’ 3’ 5’ Lagging strand Discontinuous Leading strand Continuous Okasaki Fragments

19. Applying Your Knowledge Discontinuous synthesis occurs on the DNA template oriented 1.3’  5’ 2.5’  3’ 3.Either 3’  5’ or 5’  3’ 4.Both 3’  5’ and 5’  3’ 5.Neither 3’  5’ or 5’  3’

20. Steps in DNA Replication (Bacterial) 1.Initiation Initiator Proteins bind to replication origin and cause a small section to unwind.

21. Steps in DNA Replication (Bacterial) 2.Unwinding Helicase molecules further unwind helix. Single-stranded binding proteins keep helix from reforming. DNA gyrase reduces supercoils ahead of replication fork.

22. Steps in DNA Replication (Bacterial) 3.Elongation Primase synthesizes a short RNA strand = primer. DNA polymerase III adds nucleotides to the primer in a 5’  3’ 5’  3’ direction.

23. Steps in DNA Replication (Bacterial) 3.Elongation DNA polymerase I replaces primer RNA with DNA nucleotides. DNA ligase seals gaps in sugar- phosphate backbone.

24. Steps in DNA Replication (Bacterial) 4.Termination Termination occurs when two replication forks meet. E. coli coli cells have a protein called Tus that binds to termination sequences and blocks helicase movement.

25. Accuracy of DNA Replication 1.Nucleotide 1.Nucleotide Selection 2.DNA 2.DNA proofreading: 3’  5’ 3’  5’ exonuclease activity of DNA polymerase 3.Mismatch 3.Mismatch Repair: repair enzymes

26. Modes of Replication

27. Differences for Eukaryotic DNA Replication LicensingLicensing Factor attaches to each origin, initiator protein only recognizes “licensed” origins ThirteenThirteen polymerases function in replication, recombination, repair –Alpha: –Alpha: synthesizes primer and a short stretch of DNA –Delta: –Delta: continues replication on both leading and lagging strands

28. Applying Your Knowledge A new DNA strand is synthesized 1.From 3’  5’ 2.From 5’  3’ 3.Either from 3’  5’ or 5’  3’ 4.Both from 3’  5’ and 5’  3’ 5.Neither from 3’  5’ or 5’  3’

29. Applying Your Knowledge Which enzyme produces a short stretch of RNA nucleotides used as a starting point for DNA synthesis? Which enzyme produces a short stretch of RNA nucleotides used as a starting point for DNA synthesis? 1.Helicase 2.DNA polymerase I 3.Single strand binding protein 4.DNA polymerase III 5.Primase

30. Applying Your Knowledge Which enzyme can remove an incorrectly-inserted nucleotide? Which enzyme can remove an incorrectly-inserted nucleotide? 1.Helicase 2.DNA Gyrase 3.Single strand binding protein 4.DNA polymerase 5.Primase

31. Recombination

32. Holliday Model of Recombination  Single  Single strand breaks occur at the same position on homologous DNA helices.  Single-stranded  Single-stranded ends migrate into the alternate helix.

33. Holliday Model of Recombination  Each  Each migrating strand joins to the existing strand, creating a Holliday junction.  Branch  Branch point can migrate, increasing the amount of heteroduplex DNA.

34. Holliday Model of Recombination Resolving the Holliday Intermediate  Separation  Separation of the duplexes requires cleavage in either the horizontal or vertical plane.

35. Holliday Model of Recombination Resolving the Holliday Intermediate  Cleavage  Cleavage in the vertical plane, followed by rejoining of nucleotide strands, produces crossover recombinant products.