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Ilia_Sergeevich [38]
3 years ago
14

Chymotrypsin is a serine protease enzyme that we talked about in class. In order for chymotrypsin to function, His-57 in the act

ive 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.
Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0

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 <em>(1)</em>

If total active sites are 100%

100 = His-57H⁺ + His-57 <em>(2)</em>

Replacing (2) in (1)

His-57H⁺ = 9,09%

<em>His-57 = 90,91%</em>

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

I hope it helps!

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