1. Chapter 16 Section 1 – Pt. 2 Pgs. 305 – 310 Objective: I can thoroughly describe the structure of DNA and all of its parts.

2.  Rosalind Franklin – X-ray diffraction  James Watson and Francis Crick  Put it all together

3.  DNA is 1 of 4 types of macromolecules  All macromolecules are types of polymers: chains of repeating units  Repeating unit is called monomer  For DNA:  Type of macromolecule = nucleic acid  Monomer = nucleotide  Polymer = polynucleotide chain

4.  Nucleotide is made of 3 parts  5-carbon sugar called Deoxyribose  Has 1 less oxygen than ribose  Phosphate Group  Nitrogenous Base  4 Types: A, T, C, G  Nucleotides “polymerize” through Sugar-Phosphate (covalent)bonds  2 chains (strands) join via bases  Twist into double helix shape

5.  The Sugar-Phosphate Backbone  “The sides of the ladder”  Formed from bonds between deoxyribose sugar and phosphate  Strongest part of DNA:  Phosphodiester Bond  These bonds were made via (hint: polymerization)  Dehydration Synthesis

6.  4 Nitrogenous Bases  2 categories  Purines (larger: 2 rings)  Adenine and Guanine  Pyrimidines (smaller: 1 ring)  Thymine and Cytosine  Base Pairing: A – T C – G  Purine + Pyrimidine (2 rings + 1 ring)  Otherwise, gap or bulge

7.  So why can’t A - C or T - G?  Nitrogenous Bases from opposite strands held by hydrogen (H) bonds  # of H-bonds SPECIFIC A and T form 2 H-bonds C and G form 3 H-bonds

8.  Because A-T and C-G  # of A = # of T  # of C = # of G  Measured as % of Nitrogen Bases  THUS, if know the amount of ONE base, will know the amount of the others…  If 1 DNA molecule is 22% A…  Then know 22% T  Know C & G = 56%  So, C = 28% and G = 28% % of A & T = 44% Actually, Chargaff discovered in reverse!!!

9.  Dexoyribose has an orientation  5 carbons arranged in a pentagon  5 th carbon hanging off pentagon – will bond with phosphate (same nucleotide)  1 st carbon will bond with nitro. base  3 rd carbon will bond with next nucleotide 5’ 4’ 3’2’ 1’

10.  5’ end is the phosphate side of nucleotide  3’ end is the sugar side of nucleotide 5’ 3’ 5’ 3’ 5’ 3’ 5’ If the nucleotide sequence of one strand of DNA is 5’-C-A-T-3’, then the complementary strand would be A.5’-G-T-A-3’ B.5’-A-T-G-3’ C.5’-C-A-T-3’ D.5’-T-A-C-3’ E.3’-A-T-G-5’

11.  Two strands of DNA run anti-parallel to each other: opposite directions  5’ to 3’ & 3’ to 5’  Nucleotide cannot “flip” due to 3D structure - can only “turn” upside down