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Unit 1-2 - Revi ew of U ni t 1-1 Exper i ments showi ng...

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Review of Unit 1-1 Experiments showing that DNA is the genetic material Griffith Avery, MacLeod & McCarty Hershey & Chase The nucleotide Pentose sugar Nitrogenous base Phosphate group
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Transformation of Streptococcus pneumoniae Smooth (S) cells > death Rough (R) cells > mouse lives Heated S cells > mouse lives Heated S + live R > death Fig 2.2
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Clicker Question (Example) The “factor” in the Griffith experiment that was able to “transform” avirulent rough cells into virulent smooth cells was A. The polysaccharide coat B. The genes that coded for enzymes necessary to generate the polysaccharide coat C. Protein and DNA D. RNA
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Clicker Answer (Example) The “factor” in the Griffith experiment that was able to “transform” avirulent rough cells into virulent smooth cells was A. The polysaccharide coat B.The genes that coded for enzymes necessary to generate the polysaccharide coat C. Protein and DNA D. RNA
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Fractionate “dead” S cells (Formerly Virulent) S Combine “treated” S cell fractions with avirulent R cells Inject mice No treatment +SI I I (degrades polysaccharides) +Protease (degrades protein) +RNase (degrades RNA) +DNase (degrades DNA) R No treatment +SIII +Protease +RNase +DNase
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Spin cells gently in centrifuge Where is the majority of radioactivity? Pellet (cells) Supernatant (phage coats) 35 S-labeled 32
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D. The nucleotide Nucleoside plus phosphate = nucleotide Note: to distinguish between sugar and base, positions in the sugar are designated with prime (’)
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Unit 1-2
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I I I. The Physical Structure of DNA
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Background Linear structure was known – phosphate bonds link sugars, base linked to sugar How could this apparently simple molecule hold enough information to be the genetic material? Does the 3-dimensional structure give us a clue?
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Chargaff's rule: Base ratio analysis of DNA indicated that DNA was not a random polymer An important piece of the puzzle…
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A: 29 30 26 T: 31 29 24 G: 18 18 21 C: 18 15 17 96 92 88 % recovered Human yeast bovine % of different bases in DNA samples from different sources
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The amount of Adenine = Thymine The amount of Guanine = Cytosine (A G or C, T C or G) G + C does not always equal A + T Base Pairing Rules
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Figure 2.13
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Maurice Wilkins and Rosalind Franklin at King’s College, London James Watson and Francis Crick at Cavendish Labs, Cambridge Who wants to know the structure?
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X-ray Diffraction Analysis of DNA Crystals: Determined Double Helix Structure of DNA Nobel Prize awarded for this discovery (1962) Structure determined In 1953 http:/ / www.chem.yale.edu/ ~chem125/ 125/ xray/ diffract.html#Franklin Figure 2.12
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Franklin’s x-ray diffraction pictures showed a simple, regularly repeating structure, likely to be a helix Suggested that DNA molecule was: 1. Long and thin with a 2 nm diameter 2. Coiled in a helix 3. Had repeating patterns every 0.34 nm and 3.4 nm (3.4 Å and 34 Å) 4. There might be 2 strands
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5. 2nm wide diameter: perfect for purine-pyrimidine bonds purine-purine would be too wide pyrimidine-pyrimidine would be too narrow both would lead to irregular widths
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Maurice Wilkins and Rosalind Franklin at King’s College, London James Watson and Francis Crick at Cavendish Labs, Cambridge Who wants to know the structure?
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