Question

1.) Calculate the isoelectric point for glu (glutamic acid) and arg (argenine).

Draw structures of the following amino acids and show the form that predominates at pH 4 :

his, arg, trp, pro, tyr.

What is the ratio of acetate/acetic acid in an acetate buffer at pH 5.00?

Answer 1

There are three separate questions, right? They are not related to each other, right?

1) Note down the pK values of the α-COOH group, α-NH₂ group and any polar charged/uncharged side chain that the amino acids contain. Put them in a tabular manner.

Amino Acid	pK (α-COOH)	pK (α-NH2)	pK (side chain)
Glu	2.10	9.47	4.07
Arg	1.82	8.99	12.48

The isoelectric point, pI of an amino acid is defined as the pH where the amino acid exists as a Zwitterion, i.e, a molecule with opposite charges so as to be electrically neutral. The amino acid doesn’t migrate toward the positive or negative electrode at its pI.

For a simple amino acid like Glu, the isoelectric point is the average of the two pK values. However, for polar charged amino acids like Glu or Arg, the pI is the average of the two pK values that are in close proximity. Thus, the pI for Glu is the average of pK (α-COOH) and pK (side chain). Therefore, pI = ½*(2.10 + 4.07) = 3.085 ≈ 3.08 (ans).

The pI of Arg is the average of pK (α-NH₂) and pK (side chain), since these pK values are close to each other. Therefore, pI = ½*(8.99 + 12.48) = 10.735 ≈ 10.73 (ans).