A peptide has the sequence Glu-His-Trp-Ser-Gly-Leu-Arg-Pro-Gly

(a) What is the net charge of the molecule at pH 3, 8, and 11? (Use pKa values for side chains and terminal amino and carboxyl groups as given in Table 3-1.)

(b) Estimate the pl for this peptide.

The correct answer and explanation is :

Let’s address each part of the question step by step.

Part (a): Net charge of the peptide at pH 3, 8, and 11

We are given the peptide sequence:
Glu-His-Trp-Ser-Gly-Leu-Arg-Pro-Gly

To calculate the net charge at different pH values, we need to consider the ionizable groups (the amino groups, carboxyl groups, and side chains of the amino acids) and their pKa values. Each group will either be protonated or deprotonated depending on the pH.

Amino acids involved:

Glu (Glutamic acid): Side chain pKa ~ 4.25 (negative charge at pH > 4.25)
His (Histidine): Side chain pKa ~ 6.0 (positive charge at pH < 6.0)
Arg (Arginine): Side chain pKa ~ 12.5 (positive charge at pH < 12.5)
N-terminal amino group (Amino terminus): pKa ~ 9.0 (positive charge at pH < 9.0)
C-terminal carboxyl group (Carboxyl terminus): pKa ~ 2.0 (negative charge at pH > 2.0)

Now, let’s evaluate the net charge at different pH values.

At pH 3:

N-terminal (amino): pKa ~ 9.0 → protonated, thus +1 charge.
Glu (side chain): pKa ~ 4.25 → protonated, thus 0 charge.
His (side chain): pKa ~ 6.0 → protonated, thus +1 charge.
Arg (side chain): pKa ~ 12.5 → protonated, thus +1 charge.
C-terminal (carboxyl): pKa ~ 2.0 → protonated, thus 0 charge.

Total net charge at pH 3 = +1 (N-terminal) + 0 (Glu) + +1 (His) + +1 (Arg) + 0 (C-terminal) = +3.

At pH 8:

N-terminal (amino): pKa ~ 9.0 → deprotonated, thus 0 charge.
Glu (side chain): pKa ~ 4.25 → deprotonated, thus -1 charge.
His (side chain): pKa ~ 6.0 → deprotonated, thus 0 charge.
Arg (side chain): pKa ~ 12.5 → protonated, thus +1 charge.
C-terminal (carboxyl): pKa ~ 2.0 → deprotonated, thus -1 charge.

Total net charge at pH 8 = 0 (N-terminal) + (-1) (Glu) + 0 (His) + +1 (Arg) + (-1) (C-terminal) = -1.

At pH 11:

N-terminal (amino): pKa ~ 9.0 → deprotonated, thus 0 charge.
Glu (side chain): pKa ~ 4.25 → deprotonated, thus -1 charge.
His (side chain): pKa ~ 6.0 → deprotonated, thus -1 charge.
Arg (side chain): pKa ~ 12.5 → deprotonated, thus 0 charge.
C-terminal (carboxyl): pKa ~ 2.0 → deprotonated, thus -1 charge.

Total net charge at pH 11 = 0 (N-terminal) + (-1) (Glu) + (-1) (His) + 0 (Arg) + (-1) (C-terminal) = -3.

Part (b): Estimating the pI (Isoelectric Point)

The pI is the pH at which the peptide has a net charge of zero. To estimate the pI, we need to consider the pKa values of the ionizable groups, especially those that contribute to the charge reversal near neutral pH.

From our calculations:

The peptide has a net charge of +3 at pH 3, -1 at pH 8, and -3 at pH 11.
The pI generally lies between the pH values where the peptide’s charge changes sign (i.e., from positive to negative).

Looking at the data, the net charge changes from positive at pH 3 to negative at pH 8. Thus, the pI is estimated to be between 3 and 8.

To be more precise, it’s common to take the average of the pH values where the net charge changes sign:

[
\text{pI} = \frac{3 + 8}{2} = 5.5
]

Thus, the estimated pI of the peptide is around 5.5.

Summary

Net charge at pH 3 = +3
Net charge at pH 8 = -1
Net charge at pH 11 = -3
Estimated pI = 5.5