BC334 (Proteomics) Practical 1 Calculating the charge of proteins
Aliphatic amino acids (VAGLIP) N H 2 H Glycine, Gly, G no charge Hydrophobicity = 0.67 MW 57Da pk a CH = 2.35 pk a NH 2 = 9.6 pi=5.97 CH 3 H Alanine, Ala, A no charge Hydrophobicity = 1.0 MW 71Da pk a CH = 2.34 pk a NH 2 = 9.69 pi = 6.01 H 3 C CH 3 H Valine, Val, V no charge Hydrophobicity = 2.3 MW 99Da pk a CH = 2.32 pk a NH 2 = 9.62 pi = 5.97 C H 3 CH 3 H Leucine, Leu, L no charge Hydrophobicity = 2.2 MW 113Da pk a CH = 2.36 pk a NH 2 = 9.60 pi = 5.98 CH 3 CH 3 H Isoleucine, Ile, I no charge Hydrophobicity = 3.1 MW 113Da pk a CH = 2.36 pk a NH 2 = 9.68 pi = 6.02 H N H Proline, Pro, P no charge Hydrophobicity = -0.29 MW 97Da pk a CH = 1.99 pk a NH 2 = 10.96 pi = 6.48
Aromatic amino acids (FYW) N H 2 H Phenylalanine, Phe, F no charge Hydrophobicity = 2.5 Absorbs UV MW 147Da pk a CH = 1.83 pk a NH 2 = 9.13 pi=5.48 H H Tyrosine, Tyr, Y weak charge Hydrophobicity = 0.08 Absorbs UV MW 163Da pk a CH = 2.20 pk a NH 2 = 9.11 pi=5.66 NH H Tryptophan, Trp, W no charge Hydrophobicity = 1.5 Absorbs UV MW 186Da pk a CH = 2.38 pk a NH 2 = 9.39 pi=5.89
Polar but uncharged (SNQT) H H Serine, Ser, S no charge Hydrophobicity = -1.1 MW 87Da pk a CH = 2.21 pk a NH 2 = 9.15 pi = 5.68 H CH 3 H Threonine, Thr, T no charge Hydrophobicity = -0.75 MW 101Da pk a CH = 2.11 pk a NH 2 = 9.62 pi = 5.87 N H 2 NH 2 H Asparagine, Asn, N no charge Hydrophobicity = -2.7 MW 114Da pk a CH = 2.02 pk a NH 2 = 8.08 pi = 5.41 N H 2 NH 2 H Glutamine, Gln, Q no charge Hydrophobicity = -2.9 MW 128Da pk a CH = 2.17 pk a NH 2 = 9.13 pi = 5.65
Sulphur containing (CM) HS H Cysteine, Cys, C weak charge Hydrophobicity = 0.17 MW 103Da pk a CH = 1.96 pk a NH 2 = 8.18 pi = 5.07 H 3 C S H Methionine, Met, M no charge Hydrophobicity = 1.1 MW 131Da pk a CH = 2.28 pk a NH 2 = 9.21 pi = 5.74
Charged (DEHKR) Acidic H H Aspartic acid, Asp, D negative charge Hydrophobicity = -3.0 MW 115Da pk a CH = 2.19 pk a NH 2 = 9.60 pi = 2.77 H H Glutamic acid, Glu, E negative charge Hydrophobicity = -2.6 MW 129Da pk a CH = 2.19 pk a NH 2 = 9.67 pi = 3.22 Basic HN N H Histidine, His, H Weak positive charge Hydrophobicity = -1.7 MW 137Da pk a CH = 1.82 pk a NH 2 = 9.17 pi = 7.59 H Lysine, Lys, K positive charge Hydrophobicity = -4.6 MW 128Da pk a CH = 2.18 pk a NH 2 = 8.95 pi = 9.47 HN N H 2 H Arginine, Arg, R positive charge Hydrophobicity = -7.5 MW 156Da pk a CH = 2.17 pk a NH 2 = 9.04 pi = 10.76 NH 2 HN NH 2
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 ph=10.0 H H 3 N C CH CH 3 H H 3 N C C - CH 3 H C C - CH 3 1 0-1
Titration of lysine H H 3 N C CH NH 3 ph=2.2 ph=9.0 ph=10.5 H H 3 N C C - NH 3 H C C - NH 3 H C C - NH 2 2 1 0-1
The Henderson-Hasselbalch equation HA H A - (1) K a K a = [H ][A - ]/[HA] (2) [H] = K a [HA]/[A - ] (3) - log[h ] = - logk a - log[ha]/[a - ] (4) ph = pk a log[a - ]/[HA] (5) ph = pk a log(r) (6) ph - pk a = 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 A T = [A - ] [HA] (3) Substitute eq. 2 in eq. 3 A T = [HA]R [HA] (4) = [HA](1 R) (5) [HA] = A T 1 (1 R) (6) Substitute rearranged eq. 2 in eq. 6 [A - ] = A T R (1 R) (7)
What is the charge of lysine at ph6.5? H H 3 N C C - NH 3 Lysine has 3 dissociable protons Calculate the charge of each one at a time
Start with the carboxyl group H H 3 N C C - NH 3 pka = 2.18 10 (ph-pka) = R 10(6.5-2.18) = 10 4.32 =20893 The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [A-] = 1 20893 /(120893) Charge = 0.999-1 = -0.999
Next do the α-carbon amino pka = 8.95 H H 3 N C C - NH 3 10 (ph-pka) = R 10(6.5-8.95) = 10-2.45 =0.004 The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /(1 0.004) Charge = 0.996
Finally, do the side-chain amino H H 3 N C C - NH 3 pka = 10.53 10 (ph-pka) = R 10(6.5-10.53) = 10-4.03 =0.0009 The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [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 3 N H pka = 2.18 C C N C C - H pka = 4.07 C - NH 3 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 NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /(1 0.003388) Charge = 0.997
Glutamic acid carboxyl group 10 (ph-pka) = R 10(7.0-4.07) = 10 2.93 =851.14 The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [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 NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /(1 0.000316) Charge = 0.999
Lysine carboxyl group 10 (ph-pka) = R 10(7.0-2.18) = 10 4.82 =66,069.34 The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [A-] = 1. 66,069.34 /(1 66,069.34) Charge = 0.999-1 = -0.999
Net charge 0.997-0.9990.999-0.999 = -0.003
Calculate the net charge of the tri-peptide DHR at ph 5.2 Amino acid C- NH3 Side-chain D 2.1 9.8 3.9 H 1.8 9.2 6.0 R 2.2 9.0 12.5 Deadline Monday 6 Feb by 5pm BC334 mailbox at front door of Biotechnology Building Late submissions will not be marked No Excuses, No Exceptions