Question

1. Calculate the charge on the side chain of the amino acid histidine using the Henderson-Hasselbach equation pH = pKa + log (A/HA+). Use pH =7 and a pKa = 6 and solve to 2 significant figures.

2. Cysteine proteases have unusually reactive cysteine side chains. For one of these enzymes in solution at pH 7 you are able to measure the amount of the deprotonated sulfhydryl (ie, the negatively charged species) as 40% of the total. Calculate the pKa of this side chain to 2 significant figures using the Henderson-Hasselbach equation pH = pKa + log (A-/HA)

Answer 1

Ans. #1. Putting the values in HH equation –

            7 = 6 + log ([A] / [AH⁺])

            Or, ([A] / [AH⁺]) = antilog (7-6) = 10¹ =

            Or, [A] = 10 [AH⁺]                 - equation 1

Now, let the total amount of chemical species, i.e. [A] + [AH⁺] = 1 (on a fraction scale)

            So, [A] + [AH⁺] = 1               - equation 2

Comparing equation 1 and 2 –

            10 [AH⁺] + [AH⁺] = 1

            Or, [AH⁺] = 1 / 11 = 0.091

Hence, relative charge on the side chain of histidine = + 0.091

#2. Reaction-             C-SH ------> C-S^- + H⁺

            Where, C-SH = cysteine side chain

Given- [C-S^-] = 40 % = 0.4 when the total amount of amino acid = 100% or 1.

Note that Cyst side chain acts as an acidic group (whereas that of His acted as a basic group).

So, HH equation is: pH = pKa + log ([A^-] / [AH])

# [AH] = 1.0 - [A-] = 1.0 - 0.4 = 0.6

Putting the values in above equation-

            7 – log (0.4 / 0.6) = pKa

            Or, 7 – (-0.18) = pKa

            Hence, pKa = 7.18 = 7.20