The isoelectric point of a molecule can be calculated from the pKa values of its ionisable groups. Calculate the isoelectric point for the molecule below. Give your answer to one decimal point.

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The isoelectric point (pI) is determined from the pKa values of ionizable groups that surround the species with zero net charge. In this molecule, the ionizable groups have pKa values of 1.8 (carboxyl), 6.1 (imidazole), and 9.3 (amino).

First, consider the charge state as pH increases:
- At pH < 1.8: the carboxyl is COOH (neutral), the amino is NH3+ (+1), and the imidazole is protonated (+1). Net charge ≈ +2.
- Between 1.8 and 6.1: COOH deprotonates to COO− (−1), amino remains +1, imidazole still +1. Net charge ≈ +1.
- Between 6.1 and 9.3: COO− (−1), amino +1, imidazole loses its proton becoming neutral (0). Net charge = 0.
- Above 9.3: COO− (−1), amino deprotonates to NH2 (0), imidazole 0. Net charge ≈ −1.

The pI is the average of the pKa values surrounding the zero-charge region, which are 6.1 and 9.3. Averaging them gives (6.1 + 9.3) / 2 = 7.7.

Therefore, the calculated isoelectric point is 7.7.

My LMS Subjects Quiz 3

The isoelectric point of a molecule can be calculated from the pKa values of its ionisable groups. Calculate the isoelectric point for the molecule below. Give your answer to one decimal point.

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Similar Questions

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