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Final Exam – 104A Monday, May 10 8:00 – 11:00 am 100 Noyes AQD,AQE,AQFYuan AQA,AQLSedlacek AQI,AQKSmith 62 Krannert Art Museum AQB,AQCPark AQNGupta AQGPhelan.

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Presentation on theme: "Final Exam – 104A Monday, May 10 8:00 – 11:00 am 100 Noyes AQD,AQE,AQFYuan AQA,AQLSedlacek AQI,AQKSmith 62 Krannert Art Museum AQB,AQCPark AQNGupta AQGPhelan."— Presentation transcript:

1 Final Exam – 104A Monday, May 10 8:00 – 11:00 am 100 Noyes AQD,AQE,AQFYuan AQA,AQLSedlacek AQI,AQKSmith 62 Krannert Art Museum AQB,AQCPark AQNGupta AQGPhelan

2 Final Exam – 104C Friday, May 14 7:00 – 10:00 pm 228 Natural History Bldg. CQA,CQCRanderia CQB,CQPPhelan CQI,CQJLoman 100 Noyes CQD,CQE,CQKDokukin CQG,CQKBrea

3 Amino Acids C H2NH2NH C O OH amine carboxylic acid  Rvaries with amino acid R = H glycine R H

4 C H2NH2N H C O OH R Amino acids glycine non-chiral all other  -amino acids in proteins H L-enantiomers at neutral pH (pH = 7.0) zwitterion H3N+H3N+ O-O- very high b.p. (> 200 o C) very soluble in water R

5 acid-base chemistry C H2NH2N H C O OH R amino acidsdiprotic acids 2 pK a low pH amine and c.a. protonated C H3N+H3N+ H C O OH R positive charge neutral pH amine protonated no net charge C H2NH2N H C O O-O- R high pH amine and c.a. deprotonated negative charge H3N+H3N+ O-O- isoelectric point pH = pH I c.a. deprotonated

6 Titration of an amino acid alanine R = CH 3 CH3N+H3N+ H C O OH R CH 3 pK a1 = 2.34 pK a2 = 9.69 K a1 =[H + ][A - ] [HA] 0.1 M [H + ][A - ][HA] initial 0 00.1 change+x -x equil. +x 0.1-x K a1 = 4.57 x 10 -3 =x2x2 0.1 - x X = 2.14 x 10 -2 = pH = 1.67 10 -2.34 = 4.57 x 10 -3 [H + ]

7 pH equivalents of OH - 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 × net charge +1

8 pH equivalents of OH - 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 × net charge +1 × 1/2 0.05 M pH = [HA] pK a +log [A - ] 0.05 M pH = pK a pK a1 = 2.34 = 2.34 +1/2

9 pH equivalents of OH - 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 × × 1/2 at equivalence point : 1 +1 +1/2 net charge 0 isoelectric point pH = 2 pK a1 + pK a2 pH = (2.34 + 9.69)/2 pH = pH I = 6.02 × pK a1 = 2.34 pK a2 = 9.69

10 pH equivalents of OH - 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 × net charge +1 × × 1/2 +1/2 at 2nd half-way point : pK a2 = 9.69 pH = pK a2 0-1/2 3/2 1 × = 9.69

11 glutamic acid R = - CH 2 CH 2 COOH CH3N+H3N+ H C O OH R pK a1 = 3.20 pK a2 = 4.25 CH 2 CH 2 C = O OH (  -COOH) (R-COOH) pK a3 = 9.67 (  -NH 3 ) It will take ___ equivalents to titrate glutamic acid 3 1st group2nd group3rd group

12 equivalents OH - pH 2 4 8 10 12 C H3N+H3N+ H C O OH CH 2 CH 2 C = O OH pK a1 = 3.20 × pK a2 = 4.25 × × pK a3 = 9.67 3.2 + 4.25 2 = 3.7 × 4.25 + 9.67 2 = 7.0 × +1 0 -2 pH I = 3.7 123

13 pH I every amino acid has characteristic pH I every protein also has characteristic pH I pH = pH I net charge neutral pH < pH I pH > pH I positivenegative protein molecules repel each other protein molecules repel each other protein molecules attract each other precipitate pH of milk = 6.3pH I of casein = 4.7

14 Electrophoresis - - - - - + + + + + - anode cathode migration depends on charge and size

15 R-groups -H glycine alanine-CH 3 valine -CH-CH 3 CH 3 leucine -CH 2 -CH-CH 3 CH 3 isoleucine -CH-CH 2 -CH 3 CH 3 phenylalanine methionine-CH 2 -CH 2 -S-CH 3 proline 2 o amine Nonpolar

16 Polar R-groups serine -CH 2 -OH threonine -CH-CH 3 OH tyrosine asparagine-CH 2 -C-NH 2 O glutamine -CH 2 -C-NH 2 O -CH 2 tryptophan cysteine-CH 2 -SH

17 Acidic R-groups glutamic acid -CH 2 -CH 2 -C = O OH aspartic acid -CH 2 -C = O OH

18 Basic R-groups lysine -CH 2 -CH 2 -CH 2 -CH 2 -NH 2 arginine-CH 2 -CH 2 -CH 2 -NH-C-NH 2 = NH histadine


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