Answer:
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!
