Calculate the plof (a) Ala, (b) His, and (c) Glu.

The pH value at which a molecule has no electrical charge is known as the isoelectric point.

$pl=\frac{1}{2}\left(p{K}_i+p{K}_j\right)$

Here, K_i and K_j are the protonated and deprotonated values of the amino acid Ala.

The K_i and K_j values of alanine obtained from the referred table are 2.35 and 9.87.

Thus,

$$\begin{gathered} pl=\frac{1}{2}\left(p{K}_i+p{K}_j\right) \\ pl=\frac{1}{2}\left(2.35+9.87\right) \end{gathered}$$

Therefore, pl of Ala is 6.11.

$pl=\frac{1}{2}\left(p{K}_1+p{K}_2\right)$

Here, K₁and K₂ are the protonated and deprotonated values of the amino acid His.

The K₁and K₂ values of His obtained from the referred table are 6.04 and 9.33.

Thus,

$$\begin{gathered} pl=\frac{1}{2}\left(p{K}_1+p{K}_2\right) \\ pl=\frac{1}{2}\left(6.04+9.33\right) \end{gathered}$$

Therefore, pl of His is 7.68.

$pl=\frac{1}{2}\left(p{K}_a+p{K}_b\right)$

Here, K_a and K_b are the protonated and deprotonated values of the amino acid Glu.

The K_a and K_bvalues of Glu obtained from the referred table are 2.10 and 4.07.

Thus,

$$\begin{gathered} pl=\frac{1}{2}\left(p{K}_a+p{K}_b\right) \\ pl=\frac{1}{2}\left(2.10+4.07\right) \end{gathered}$$

Therefore, pl of Glu is 3.08.

Hence, the pI values of Ala, His and Glu are 6.11, 7.68 and 3.08 respectively.