BOC334 (Proteomics) Practical 1. Calculating the charge of proteins

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1 BC334 (Proteomics) Practical 1 Calculating the charge of proteins

2 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 = MW 97Da pk a CH = 1.99 pk a NH 2 = pi = 6.48

3 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

4 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 = 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

5 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

6 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 = NH 2 HN NH 2

7 Proteins have charges Ribbon Electrostatic surface

8 Histone H3 binds to DNA Histone H3 (ribbon) Histone H3 (electrostatic surface)

9 Nucleosome assembly protein (NAP1) binds to histones NAP1 (ribbon) NAP1 (electrostatic surface)

10 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

11 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

12 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

13 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)

14 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)

15 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

16 Start with the carboxyl group H H 3 N C C - NH 3 pka = (ph-pka) = R 10( ) = =20893 The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [A-] = /(120893) Charge = =

17 Next do the α-carbon amino pka = 8.95 H H 3 N C C - NH 3 10 (ph-pka) = R 10( ) = =0.004 The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /( ) Charge = 0.996

18 Finally, do the side-chain amino H H 3 N C C - NH 3 pka = (ph-pka) = R 10( ) = = The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /( ) Charge = 0.999

19 Add the charges from the different groups At ph6.5: Carboxyl: α-amino: Side-chain amino: Total charge = = 0.996

20 Programs exist the calculate the charge

21 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

22 Glutamic acid amino group 10 (ph-pka) = R 10( ) = = The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /( ) Charge = 0.997

23 Glutamic acid carboxyl group 10 (ph-pka) = R 10( ) = = The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [A-] = /( ) Charge = =

24 Lysine amino group 10 (ph-pka) = R 10( ) = = The associated NH 3 carries a charge So use the equation for HA [HA] = A T /(1R) [HA] = 1 /( ) Charge = 0.999

25 Lysine carboxyl group 10 (ph-pka) = R 10( ) = =66, The dissociated C - carries a charge So use the equation for A - [A - ] = A T R/(1R) [A-] = 1. 66, /(1 66,069.34) Charge = =

26 Net charge =

27 Calculate the net charge of the tri-peptide DHR at ph 5.2 Amino acid C- NH3 Side-chain D H R Deadline Monday 6 Feb by 5pm BC334 mailbox at front door of Biotechnology Building Late submissions will not be marked No Excuses, No Exceptions

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