BOC334 (Proteomics)
Practical 1
Calculating the charge of proteins
Aliphatic amino acids (VAGLIP)
O
H2N
OH
O
H2N
H3C
CH3
Glycine, Gly, G
no charge
Hydrophobicity = 0.67
MW 57Da
pKa COOH = 2.35
pKa NH2 = 9.6
pI=5.97
Leucine, Leu, L
no charge
OH Hydrophobicity = 2.2
MW 113Da
pKa COOH = 2.36
pKa NH2 = 9.60
pI = 5.98
O
H2N
OH
CH3
Alanine, Ala, A
no charge
Hydrophobicity = 1.0
MW 71Da
pKa COOH = 2.34
pKa NH2 = 9.69
pI = 6.01
O
H2N
H3C
OH
CH3
O
H2N
CH3
Isoleucine, Ile, I
no charge
OH Hydrophobicity = 3.1
MW 113Da
pKa COOH = 2.36
CH3
pKa NH2 = 9.68
pI = 6.02
O
H
N
OH
Valine, Val, V
no charge
Hydrophobicity = 2.3
MW 99Da
pKa COOH = 2.32
pKa NH2 = 9.62
pI = 5.97
Proline, Pro, P
no charge
Hydrophobicity = -0.29
MW 97Da
pKa COOH = 1.99
pKa NH2 = 10.96
pI = 6.48
Aromatic amino acids (FYW)
O
H2N
Phenylalanine, Phe, F
no charge
OH Hydrophobicity = 2.5
Absorbs UV
MW 147Da
pKa COOH = 1.83
pKa NH2 = 9.13
pI=5.48
O
H2N
HO
O
H2N
NH
OH
Tryptophan, Trp, W
no charge
Hydrophobicity = 1.5
Absorbs UV
MW 186Da
pKa COOH = 2.38
pKa NH2 = 9.39
pI=5.89
OH
Tyrosine, Tyr, Y
weak charge
Hydrophobicity = 0.08
Absorbs UV
MW 163Da
pKa COOH = 2.20
pKa NH2 = 9.11
pI=5.66
Polar but uncharged (SNQT)
O
H2N
OH
HO
O
H2N
O
NH2
Serine, Ser, S
no charge
Hydrophobicity = -1.1
MW 87Da
pKa COOH = 2.21
pKa NH2 = 9.15
pI = 5.68
Asparagine, Asn, N
no charge
OH Hydrophobicity = -2.7
MW 114Da
pKa COOH = 2.02
pKa NH2 = 8.08
pI = 5.41
O
H2N
HO
OH
CH3
O
H2N
O
NH2
Threonine, Thr, T
no charge
Hydrophobicity = -0.75
MW 101Da
pKa COOH = 2.11
pKa NH2 = 9.62
pI = 5.87
Glutamine, Gln, Q
no charge
OH Hydrophobicity = -2.9
MW 128Da
pKa COOH = 2.17
pKa NH2 = 9.13
pI = 5.65
Sulphur containing (CM)
O
H2N
HS
OH
Cysteine, Cys, C
weak charge
Hydrophobicity = 0.17
MW 103Da
pKa COOH = 1.96
pKa NH2 = 8.18
pI = 5.07
O
H2N
S
H3C
OH
Methionine, Met, M
no charge
Hydrophobicity = 1.1
MW 131Da
pKa COOH = 2.28
pKa NH2 = 9.21
pI = 5.74
Charged (DEHKR)
Acidic
O
H2N
OH
O
OH
Aspartic acid, Asp, D
negative charge
Hydrophobicity = -3.0
MW 115Da
pKa COOH = 2.19
pKa NH2 = 9.60
pI = 2.77
O
H2N
HO
Glutamic acid, Glu, E
negative charge
OH Hydrophobicity = -2.6
MW 129Da
pKa COOH = 2.19
pKa NH2 = 9.67
pI = 3.22
O
Basic
H2N
HN
Histidine, His, H
O
Weak positive charge
Hydrophobicity = -1.7
OH MW 137Da
pKa COOH = 1.82
pKa NH2 = 9.17
pI = 7.59
O
H2N
O
Lysine, Lys, K
positive charge
OH Hydrophobicity = -4.6
MW 128Da
pKa COOH = 2.18
pKa NH2 = 8.95
pI = 9.47
H2N
HN
N
NH2
HN
NH2
Arginine, Arg, R
OH positive charge
Hydrophobicity = -7.5
MW 156Da
pKa COOH = 2.17
pKa NH2 = 9.04
pI = 10.76
Proteins have charges
Ribbon
Electrostatic surface
Histone H3 binds to DNA
Histone H3 (ribbon)
Histone H3 (electrostatic surface)
Nucleosome assembly protein (NAP1) binds to histones
NAP1 (ribbon)
NAP1 (electrostatic surface)
What charge is a protein?
•Depends on the pH
•The nature of the amino acids constituting the protein
•Each amino acid may contribute a specific fractional charge
at a given pH
•The charge of the amino acid is given by the degree of
dissociation of dissociable protons at the given pH
Titration of alanine
pH=2.6
H
|
+H N – C – COOH
3
|
CH3
+1
pH=10.0
H
|
+H N – C – COO3
|
CH3
0
H
|
H2N – C – COO|
CH3
-1
Titration of lysine
pH=2.2
H
|
+H N – C – COOH
3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
+2
pH=9.0
H
|
+H N – C – COO3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
+1
pH=10.5
H
|
H2N – C – COO|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
0
H
|
H2N – C – COO|
CH2
|
CH2
|
CH2
|
CH2
|
NH2
-1
The Henderson-Hasselbalch equation
+ + AHA 
H
K
(1)
Ka = [H+][A-]/[HA]
(2)
[H+] = Ka[HA]/[A-]
(3)
- log[H+] = - logKa - log[HA]/[A-]
(4)
pH = pKa + log[A-]/[HA]
(5)
a
pH = pKa + log(R)
(6)
pH - pKa = log(R)
(7)
10(pH - pKa) = R
(8)
Acid and base fractions in a titration
[A-]/[HA] = R
(1)
[A-] = [HA]R
(2)
We know that AT = [A-] + [HA]
(3)
Substitute eq. 2 in eq. 3
AT = [HA]R + [HA]
(4)
= [HA](1 + R)
(5)
1
[HA] = AT
(1 + R)
Substitute rearranged eq. 2 in eq. 6
[A-] = AT
R
(1 + R)
(6)
(7)
What is the charge of lysine at pH6.5?
H
|
+H N – C – COO3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
•Lysine has 3 dissociable protons
•Calculate the charge of each one at a time
Start with the carboxyl group
H
|
+H N – C – COO3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
pKa = 2.18
10(pH-pKa) = R
10(6.5-2.18) = 104.32
=20893
The dissociated COO- carries a charge
So use the equation for A[A-] = AT×R/(1+R)
[A-] = 1×20893 /(1+20893)
Charge = 0.999×-1 = -0.999
Next do the α-carbon amino
H
|
+H N – C – COO3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
pKa = 8.95
10(pH-pKa) = R
10(6.5-8.95) = 10-2.45
=0.004
The associated NH3+ carries a charge
So use the equation for HA
[HA] = AT/(1+R)
[HA] = 1 /(1+ 0.004)
Charge = +0.996
Finally, do the side-chain amino
H
|
+H N – C – COO3
|
CH2
|
CH2
|
CH2
|
CH2
|
NH3+
pKa = 10.53
10(pH-pKa) = R
10(6.5-10.53) = 10-4.03
=0.0009
The associated NH3+ carries a charge
So use the equation for HA
[HA] = AT/(1+R)
[HA] = 1 /(1+ 0.0009)
Charge = +0.999
Add the charges from the different groups
• At pH6.5:
– Carboxyl: -0.999
– α-amino: +0.996
– Side-chain amino: +0.999
– Total charge
– = -0.999+ 0.996+ 0.999
– = +0.996
Programs exist the calculate the charge
Calculate the charge of the di-peptide EK at pH 7.0
The di-peptide EK
pKa = 9.47
H
O
H3N+ C
C
CH2
CH2
COOpKa = 4.07
O
N C
C
pKa = 2.18
O-
H CH2
CH2
CH2
CH2
NH3+
pKa = 10.5
•A di-peptide has 4 dissociabale protons
•The charge at a given pH value can be calculated as demonstrated
Glutamic acid amino group
10(pH-pKa) = R
10(7.0-9.47) = 10-2.47
=0.003388
The associated NH3+ carries a charge
So use the equation for HA
[HA] = AT/(1+R)
[HA] = 1 /(1+ 0.003388)
Charge = 0.997
Glutamic acid carboxyl group
10(pH-pKa) = R
10(7.0-4.07) = 102.93
=851.14
The dissociated COO- carries a charge
So use the equation for A[A-] = AT×R/(1+R)
[A-] = 1. 851.14 /(1+ 851.14)
Charge = 0.999×-1 = -0.999
Lysine amino group
10(pH-pKa) = R
10(7.0-10.5) = 10-3.5
=0.000316
The associated NH3+ carries a charge
So use the equation for HA
[HA] = AT/(1+R)
[HA] = 1 /(1+ 0.000316)
Charge = 0.999
Lysine carboxyl group
10(pH-pKa) = R
10(7.0-2.18) = 104.82
=66,069.34
The dissociated COO- carries a charge
So use the equation for A[A-] = AT×R/(1+R)
[A-] = 1. 66,069.34 /(1+ 66,069.34)
Charge = 0.999×-1 = -0.999
Net charge
0.997-0.999+0.999-0.999 = -0.003
Calculate the net charge of the tri-peptide “DHR” at pH 5.2
Amino acid
D
H
R
COO2.1
1.8
2.2
NH3+
9.8
9.2
9.0
Side-chain
3.9
6.0
12.5
Deadline Monday 6 Feb by 5pm – BOC334 mailbox at
front door of Biotechnology Building
Late submissions will not be marked – No Excuses, No
Exceptions