Question

What is the fraction of total tyrosine that will be in the complete deprotonated form at pH 9.6?

Given pKa -COO = 2.2, pKa -NH3 = 9.11, pKa -OH = 10.07

Answer 1

We want to completely deprotonate tyrosine; hence, we must remove the proton from the –OH group to form an anion.

The pH is 9.6 which is close to pK (-OH) = 10.07; hence, we shall use the value of pK as 10.07.

Use the Henderson-Hasslebach equation.

pH = pK + log [A^-]/[HA] where HA is the protonated form and A^- is the fully deprotonated form. Plug in values.

9.6 = 10.07 + log [A^-]/[HA]

===> -0.47 = log [A^-]/[HA]

===> [A^-]/[HA] = antilog (-0.47) = 0.3388

Thus, the ratio of the fully deprotonated and protonated forms of tyrosine at pH 9.6 is 0.3388:1. The fraction of the fully deprotonated form is given by

Fraction A^- = [A^-]/([A^-] + [HA]) = 0.3388/(0.3388 + 1) = 0.3388/1.3388 = 0.2531 ≈ 0.25 (ans).