Question

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

Short answer

a) The pl of Ala is 6.11.

b) The pl of His is 7.68.

c) The pl of Glu is 3.08.

Step-by-step solution

Definition of p l

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

Calculate p l of Ala

1. pI of Ala is calculated as follows:

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

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.

Calculate pl of His

1. pI of His is calculated as follows:

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

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.

Calculate pl of Glu

1. pI of Glu is calculated as follows:

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

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.