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pH and buffers
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Definition of pH pH= - log[H+] if [H+] = 1mM = 0.001M = 10-3 M, pH=3
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Dissociation of H2O, ion product of water (Kw)
H2O is also an acid, H2O ↔ H+ + OH- Ka= [H+] [OH-]/ [H2O] [H2O]=55.5 M (1000 g/l and Mw 18) Ka= [H+] [OH-] = Ka [H2O] = Kw (ion product of water)
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Calculating concentrations of H+ and OH-
[H2O] is high, can be said to be constant throughout ionization, so Kw is a constant: x 55.5 = 1 x [H+][OH-]= If pH=7 then [H+]=10-7 and [OH-] = 10-7
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HA H+ + A− pKa = -log(Ka) Dissociation of an acid Ka = [H+][A−]/[HA]
Acid dissociation constant pKa = -log(Ka) (-) log Ka = (-)log [H+] [A-] / [HA]
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(-) log Ka = (-)log [H+] [A-] / [HA] If pH = pKa, then pH = -log[H+] = -log [H+][A-] / [HA] [H+] = [H+] [A-] / [HA] 1 = [A-] / [HA] or [HA] = [A-] If pH=pKa, conc of acid is same as base
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pKa = - log (Ka) ACID-BASE EQUILIBRIA: Strength of an Acid
- Acids vary greatly in strength, resulting in a very large range of Ka values. It is often useful to express their acidity as pKa values: pKa = - log (Ka) Strong Acid HCl H+ + Cl- Ka > 103 pKa < -3 Weak Acid CH3COOH H+ + CH3COO- Ka = 1.8 x 10-5 pKa = 4.7 Water is itself a very weak acid H2O H+ + OH- Kw = 1.0 x 10-14 pKw = 14 Source: Stephen Kawai
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Addition of HCl to water or sodium acetate
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Henderson-Hasselbalch equation
Ka = [H+] [A-] / [HA] log Ka = log [H+] + log [A-] / [HA] pH = pKa + log [A-] / [HA] [A-] / [HA] = 10 pH-pKa
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1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add? Absence of buffer, pH7.4= = 5 x 10-8 pH7.3= = 4 x10-8 Difference 1 x 10-8
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1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add? [A-] / [HA] = 10 pH-pKa [HPO42-] / [H2PO4-] = = =1.55 [HPO42-] = 1.55 x [H2PO4-] Total concentration is 1mM, so [HPO42-] + [H2PO4-] = 1 [H2PO4-] = 1 – [HPO42-]
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1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add? [A-] / [HA] = 10 pH-pKa [HPO42-] / [H2PO4-] = = =1.55 [HPO42-] = 1.55 x [H2PO4-] Total concentration is 1mM, so [HPO42-] + [H2PO4-] = 1 [H2PO4-] = 1 – [HPO42-]
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1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add? [HPO42-] = 1.55 x (1-[HPO42-]) = 1.55 – 1.55 x [HPO42-] 2. 55 x [HPO42-] = 1.55 [HPO42-] = 1.55 / 2.55 = (mM) [H2PO4-] = 1 – = 0.392
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1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add? Same procedure for pH 7.3 gives [HPO42-] = and [H2PO4-] = 0.448 0.668 – = HPO42- is converted into H2PO4-, using up mM H+ (acid) With buffer mM without buffer 1 x 10-8 5600 times more acid consumed.
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Back to amino acids and peptides
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Amino acids are joined by peptide bonds
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N- and C-termini of polypeptides
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The polypeptide backbone
Main covalent bonding between amino acids
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Covalent bonding between cysteines
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Primary structure of bovine insulin
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Planar character of the peptide bond
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Explanation for planar character
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Typical bond lengths within a peptide unit
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Peptide bonds can be “cis” or “trans”
trans predominates
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Proline is an exception
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Rotation is possible around Ca bonds
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“Dihedral” or “torsion” angles define the degrees of rotation
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Only some conformations are allowed
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Amino acid sequence defines 2o and 3o structures
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For the tutorial quiz week 1:
-be able to do pH problems (simple ones) -ionization of groups on amino acids -amino acids-be able to a) recognize structures; b) know names, 1 letter code and 3 letter code; c) know properties of R groups For the laboratory week 1: -read the intro to the lab manual -come with required supplies and clothing -prepare a “pre-lab” (see page 5 of the manual)
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