Chymotrypsin is a serine protease enzyme that we talked about in class. in order for chymotrypsin to function, his-57 in the active site must be deprotonated. at ph = 7, approximately what percentage of chymotrypsin proteins are active? use the henderson-hasselbalch equation provided and assume the pka of his is 6.

Percentage of chymotrypsin proteins that are active is 90,91%

Explanation:

The active site, His-57, is in equilibrium, thus:

His-57H⁺ ⇄ His-57 + H⁺

Where His-57H⁺ is the active site in its protonated form.

For this equilibrium, Henderson-Hasselbalch formula is:

pH = pka + log₁₀ (His-57/His-57H⁺)

If pH is 7 and pka is 6:

7 = 6 + log₁₀ (His-57/His-57H⁺)

1 = log₁₀ (His-57/His-57H⁺)

10 = His-57/His-57H⁺

10 (His-57H⁺) = His-57 (1)

If total active sites are 100%

100 = His-57H⁺ + His-57 (2)

Replacing (2) in (1)

His-57H⁺ = 9,09%

His-57 = 90,91%

Thus, percentage of chymotrypsin proteins that are active is 90,91%

I hope it helps!