Question

Calculate the pI of each of the following amino acids:

a.asparagine b. arginine c. serine d. aspartate

Short answer

(a) pI = 5.43

(b) pI = 10.76

(c) pI = 5.68

(d) pI = 2.98

Step-by-step solution

Step-by-Step Solution  Step 1: Isoelectronic point (pI)

The isoelectric point is defined as the pH of the solution at which the net charge of protein becomes neutral or will carry no units of positive or negativecharge. Inother words, it is the pH at which the amount of positive charge on an amino acid or protein exactly balances the amount of negative charge:

pI = pH at which there is no net charge

Formula for Isoelectronic point (pI)

Isoelectric point (pI) can be calculated using the formula i.e.

$\mathrm{pI}=\frac{{\mathrm{pK}}_{\mathrm{a}}1+{\mathrm{pK}}_{\mathrm{a}}2}{2}$

The pI of most amino acids that have an ionizable side chainis the average

of the pKa values of the similarly ionizing groups (either positively charged groups ionizing to uncharged groups or uncharged groups ionizing to negatively charged groups).

The pI of an amino acid that does not have an ionizable side chain—such as alanine—is midway between its twopKavalues.

Calculate the pI of each of the following amino acids

a. Asparagine-

$\begin{array}{l}\mathrm{pI}=\frac{2.02+8.84}{2}\\ =\frac{10.86}{2}\\ =5.43\end{array}$

b. Arginine-

$\begin{array}{l}\mathrm{pI}=\frac{9.04+12.48}{2}\\ =\frac{21.52}{2}\\ =10.76\end{array}$

c. Serine-

$\begin{array}{l}pI=\frac{2.21+915}{2}\\ =\frac{11.36}{2}\\ =5.68\end{array}$

d. aspartate

$\begin{array}{l}\mathrm{pI}=\frac{2.09+3.86}{2}\\ =\frac{5.95}{2}\\ =2.98\end{array}$