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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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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
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Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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3 years ago
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