Question

Calculate the isoelectric point for each amino acid.

a. cysteine: ${\mathbf{pK}}_{\mathbf{a}}$ (COOH) = 2.05; ${\mathbf{pK}}_{\mathbf{a}}$$\left(\alpha -{\mathrm{NH}}_3^{+}\right)$=10.25

b. methionine: ${\mathbf{pK}}_{\mathbf{a}}$(COOH) = 2.28; ${\mathbf{pK}}_{\mathbf{a}}$$\mathbf{(}\mathbf{\alpha }\mathbf{-}{\mathbf{NH}}_{\mathbf{3}}^{\mathbf{+}}\mathbf{)}$=9.21

Short answer

a. 6.15

b. 5.75

Step-by-step solution

Step 1:Isoelectric point

The isoelectric point (PI) calculates the pH where charges are balanced. The balanced charges may acquire positive or negative charges in a particular medium.

Formula to calculate the PI

The formula to calculate the PI is shown as:

$\mathbf{PI}\mathbf{=}\frac{{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}{\mathbf{pK}}_{\mathbf{2}}\mathbf{+}{\mathbf{pK}}_{\mathbf{3}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}}\mathbf{+}{\mathbf{pK}}_{\mathbf{n}}}{\mathbf{n}}$

Step 3:Explanation

a. Isoelectric point of cysteine is calculated as:

$$\begin{gathered} \mathrm{PI}=\frac{{\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{COOH}\right)+{\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{\alpha }-{\mathrm{NH}}_3^{+}\right)}{2} \\ =\frac{2.05+10.25}{2} \\ =6.15 \\ \mathrm{b}.\mathrm{Isoelectric}\mathrm{point}\mathrm{of}\mathrm{methionine}\mathrm{is}\mathrm{calculated}\mathrm{as} \\ \mathrm{PI}=\frac{{\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{COOH}\right)+{\mathrm{pK}}_{\mathrm{a}}\left(\mathrm{\alpha }-{\mathrm{NH}}_3^{+}\right)}{2} \\ =\frac{2.28+9.21}{2} \\ =5.75 \end{gathered}$$