Lysine has three ionizable functional groups with the following pKa values:
α-amino group = 9.04
α-carboxylic group = 2.17
R group = 12.48
The R group for lysine is –CH2CH2CH2CH2NH2
(i) Write the equilibrium equations for its three ionizations and assign the proper pKa for each ionization. Show the net charge on the lysine molecule at each ionization stage.
(ii) Calculate the Isoelectric Point (pI):
The pI is the pH at which two amino acid has a net charge of zero (0).
1. Write out equations to show ionisation. Start with the amino acid in its fully protonated form. The pK values increase as you loose H-atoms so start the ionization using the group with the lowest pK value and then you increase as ionization continues.
2. Ensure that pk values are placed above reversible arrows.
3. Calculate the net charge of each molecule by adding the positive charges and subtracting the negative ones.
4. Calculate the pI using the equation below: ( pK1 value is the pk value to the left of the zero net charge and pK2 value is to the right of the net charge of zero).
pI = pk1 + pk2 ⁄ 2
pI = 9.04+12.48/2
pI = 10.76
All my Biochem Bloggers please note this titration steps for exam. It has been confirmed that it is going to come!!! Don’t say I didn’t warn you.lol.