Question

Consider the two \( pK_a \) values of valine as 2.32 and 9.62. The isoelectric point (pI) of this amino acid is ....... (rounded off to two decimal places)

Hint:

The isoelectric point (pI) is a key property of amino acids. It is calculated as the average of the \( pK_a \) values corresponding to the ionizable groups of the amino acid.

Answer 1

To calculate the isoelectric point (pI) of an amino acid, we use the formula: \[ pI = \frac{pK_a1 + pK_a2}{2} \] where \( pK_a1 \) and \( pK_a2 \) are the dissociation constants for the carboxyl and amino groups of the amino acid. For valine, the two \( pK_a \) values are given as: \[ pK_a1 = 2.32 \quad {and} \quad pK_a2 = 9.62. \] Now, applying the formula for the isoelectric point: \[ pI = \frac{2.32 + 9.62}{2} = \frac{11.94}{2} = 5.97. \] After rounding off to two decimal places, the isoelectric point of valine is: \[ pI = 5.97. \]