Final Exam – 104AMonday, May 108:00 – 11:00 am

100 NoyesAQD,AQE,AQF YuanAQA,AQL SedlacekAQI,AQK Smith

62 Krannert Art MuseumAQB,AQC ParkAQN GuptaAQG Phelan

Final Exam – 104CFriday, May 147:00 – 10:00 pm

228 Natural History Bldg.CQA,CQC RanderiaCQB,CQP PhelanCQI,CQJ Loman

100 NoyesCQD,CQE,CQK DokukinCQG,CQK Brea

Amino Acids

CH2N H

COOH

amine

carboxylic acid

R varies with amino acidR = H glycine

RH

CH2N H

CO

OH

R

Amino acids

glycine non-chiral

all other -amino acids in proteins

H

L-enantiomers

at neutral pH (pH = 7.0) zwitterion

H3N+O-

very high b.p. (> 200oC)

very soluble in water

R

acid-base chemistry

CH2N H

CO

OH

R

amino acids diprotic acids 2 pKa

low pH

amine and c.a.protonated

CH3N+ H

CO

OH

R

positive charge

neutral pHamine protonated

no net charge

CH2N H

CO

O-

Rhigh pH

amine and c.a.deprotonated negative charge

H3N+O-

isoelectric point pH = pHI

c.a. deprotonated

Titration of an amino acidalanine R = CH3

CH3N+ H

CO

OH

RCH3

pKa1 = 2.34

pKa2 = 9.69

Ka1 = [H+][A-][HA]

0.1 M [H+] [A-] [HA]

initial 0 0 0.1change +x +x -xequil. +x +x 0.1-x

Ka1 =

4.57 x 10-3 = x2

0.1 - x

X = 2.14 x 10-2 = pH = 1.67

10-2.34 = 4.57 x 10-3

[H+]

pH

equivalents of OH-0

1.02.03.04.05.06.07.08.09.0

×

net charge +1

C

CO

OHH3N+

CH3

H

C

CO

O-

H3N+

CH3

H

pH

equivalents of OH-0

1.02.03.04.05.06.07.08.09.0

×

net charge

+1

×

1/2

0.05 M

0.05 M

pH =[HA]

pKa + log [A-]0.05 M

0.05 MpH = pKa

pKa1 = 2.34

= 2.34

+1/2C

CO

OHH3N+

CH3

H

C

CO

O-

H3N+

CH3

H

pH

equivalents of OH-0

1.02.03.04.05.06.07.08.09.0

× ×

1/2

at equivalence point:

1

+1 +1/2

net charge0

isoelectric point

pH =2

pKa1 + pKa2

pH = (2.34 + 9.69)/2pH = pHI = 6.02

×

pKa1 = 2.34

C

CO

O-

H3N+

CH3

H

pKa2 = 9.69

pH

equivalents of OH-0

1.02.03.04.05.06.07.08.09.0

×

net charge

+1

×

×

1/2

+1/2at 2nd half-way point:

pKa2 = 9.69

pH = pKa2

0 -1/2

3/21

×

-1

= 9.69

C

CO

O-

H3N+

CH3

H

C

CO

O-

H2N

CH3

H

glutamic acid R = - CH2CH2COOH

CH3N+ H

CO

OH

R

pKa1 = 3.20

pKa2 = 4.25CH2CH2C =O

OH

(-COOH)

(R-COOH)pKa3 = 9.67

(-NH3)

It will take ___ equivalents to titrate glutamic acid3

1st group 2nd group 3rd group

equivalents OH-

pH

2

4

8

10

12CH3N+ H

CO

OH

CH2CH2C=O

OH

pKa1 = 3.20

×

pKa2 = 4.25

×

×

pKa3 = 9.67

3.2 + 4.252

= 3.7×

4.25 + 9.672

= 7.0

×

+1 0 -1 -2

pHI = 3.7

1 2 3

pHI

every amino acid has characteristic pHI

every protein also has characteristic pHI

pH = pHI

net charge

neutralpH < pHI pH > pHIpositive negative

protein molecules repel each other

protein molecules repel each other

protein molecules attract each other

precipitate

pH of milk = 6.3 pHI of casein = 4.7

Electrophoresis

- --

- -++

+ +

+ -

anode cathode

migration depends on charge and size

R-groups-Hglycine

alanine -CH3

valine -CH-CH3 CH3

leucine -CH2-CH-CH3

CH3

isoleucine-CH-CH2-CH3

CH3

phenylalanine

methionine -CH2-CH2-S-CH3

proline 2o amine

Nonpolar

CH2N H

CO2H

RN

H

COOH

CH2

Polar R-groupsserine -CH2-OH

threonine -CH-CH3OH

tyrosine

asparagine -CH2-C-NH2

Oglutamine -CH2-C-NH2

O-CH2

tryptophancysteine -CH2-SH

OH

CH2

NH

CH2

Acidic R-groups

glutamic acid -CH2-CH2-C =OOH

aspartic acid -CH2-C =OOH

Basic R-groups

lysine -CH2-CH2-CH2-CH2-NH2

arginine -CH2-CH2-CH2-NH-C-NH2=

NH

histadineN N

H

CH2