Question

The tetrapeptide, NRMD, exhibits four pK_a values: pK₁ = 1.88, pK₂ = 3.65,
pK₃ = 8.80, and pK₄ = 12.48.

What is the isoelectric point for this peptide?

A. pI=2.8

B. pI=5.3

C. pI=6.2

D. pI= 8.1

E. pI=10.6

Answer 1

Isoelectric point of peptide is that pH where the peptide has no net charge or the number of positive charge is equal to the number of the negative charge.

It is to be remembered that acid groups in peptide deprotonates at pH near 3 to 4 and basic groups in peptide deprotonates at the pH of 8 to 11.

pI = (pK_a1 + pK_a2)/2 where pK_a1 is that pH from where the net zero charged peptides starts to exist and pK_a2 is that pH upto where the net zero charged peptides do exist.

In the given question net zero charged pepdite exist between the pH of 3.65(pK_a2) and 8.80(pK_a3)

pI = (3.65+8.80)/2=6.225