Question

Calculate a best approximation for the pI of Peptide (KFYV). Show all mathematical work and justify the pK _a values you choose to use.

Answer 1

K = Lysine

F = Phenylalanine

Y = Tyrosine

V = Valine

Here, 'K' is the N-terminal residue, for which pK = 9.0 (NH₃⁺ NH₂)

The side chain of 'K' has pK = 10.5 (NH₃⁺ NH₂)

The side chain of 'Y' has pK = 10.5 (OH O^-)

'V' is the C-terminal residue, for which pK = 3.5 (COOH COO^-)

The isoelectric point (pI) is the pH at which the net charge on the peptide molecule is zero.

Therefore, pI = (9.0 + 10.5)/2 = 9.75

Explanation: At pH = 9.75, the first equilibrium shifts to right side, then the corresponding charge = 0

At pH = 9.75, the second equilibrium shifts to left side, then the corresponding charge = +1

At pH = 9.75, the third equilibrium shifts to left side, then the corresponding charge = 0

At pH = 9.75, the fourth equilibrium shifts to right side, then the corresponding charge = -1

Therefore, the overall or net charge = 0 +1 +0 -1 = 0

Hence, pI = 9.75