Lecture 2 and 3 Revise - Lecture 2 DNA structure pH and...

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1 Note: David Ballou Office hours - Mondays 2:00 - 3:30 PM Lecture 2: DNA structure, pH, and Amino Acids
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2 How is DNA held together in a Double Helix? Chemical bonds Electrostatic interactions Hydrogen bonds Van der Waal interactions Hydrophobic effects
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3 Forces involved in stabilizing the DNA helix Covalent Bonds Electrostatic interactions Hydrogen bonding Hydrophobic effect van der Waals interactions 3.4 Å > 10 Å
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Van der Waals Interactions – (0.5 – 1) kcal/mol
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5 Water surrounds nonpolar molecules in an ordered fashion When nonpolar molecules coalesce, water is released This results in more disorder of water => increased entropy Clathrate Hydrophobic Effect
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6 The base pairs are stacked on top of each other along the length of the DNA double helix Specific base pairing (hydrogen bonding) between the bases of the two strands stabilizes the double helix
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7 Water significantly weakens electrostatic and hydrogen bonds by competition
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8 Demonstration of Entropic Effect Voet and Voet, Wiley
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9 Energy Relationships G = H sys – T S system G < 0 => Process is favorable G > 0 => Process is unfavorable G = Gibbs free energy of a molecule or system H = Enthalpy (e.g., energy of the bonds) S = Entropy (~ amount of disorder) We measure the changes in these quantities from the beginning to the end of a reaction or process
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10 Thermodynamics of DNA structure First law : total energy of a system and its surroundings is constant. Energy released from formation of chemical bonds must be used to generate heat or to do work, or to form other new chemical bonds or both. Second law : the total entropy of a system and its surroundings always increases for a spontaneous process. G = H sys – T S system < 0
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11 DNA structure is dependent on pH Most biochemical structures are stable around neutral pH (7.0) Extremes in pH can alter macromolecular structures
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12 HA H + + A K a = [H + ][A - ]/[HA] (note) log( K a ) = log([H + ]) + log([A ]/[HA]) -log([H + ]) = -log( K a ) + log([A ]/[HA]) pH = -log[H + ] pKa = -log( K a ) pH = pK a + log([A ]/[HA]) Derive Henderson-Hasselbalch Equation
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13 Henderson-Hasselbalch Equation pH = pKa + log([A ]/[HA]) When pH = pK a 0 = log([A - ]/[HA] 1 = [A - ]/[HA] so [A - ] = [HA]
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14 Titration of Acetate CH 3 COOH CH 3 COO Ð + H + Titration of H 2 O
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15 Using the Henderson-Hasselbach equation pH = pK a + log([A-]/[HA]) For acetic acid, K a = 1.78 x 10 -5 pK = 4.75 At pH = 4.75, pH = pK a and [A ] = [HA] But at pH = 3.75, pH-pK a = -1 so [A-]/[HA] = 0.1 or ~90% of A is protonated Titration of Acetate
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16 Phosphoric Acid H 3 PO 4 H 2 PO 4 Ð HPO 4 Ð PO 4 Ð OH OH OH K 1 K 2 K 3 2 3 pKa = 2.12 pKa = 7.21 pKa =12.67
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17 Voet and Voet, Wiley pK 1 =2.12 pK 1 =7.21 pK 1 =12.67 Titration of Phosphoric Acid
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18 At a given pH: – A functional group with a pK
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Lecture 2 and 3 Revise - Lecture 2 DNA structure pH and...

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