An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol -min¹ when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme? Vmax = μmol-min-¹
An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol -min¹ when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme? Vmax = μmol-min-¹
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
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Q23:
![**Enzyme Kinetics Problem**
An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol·min⁻¹ when the substrate concentration is five times greater than the \( K_M \). What is the \( V_{max} \) of this enzyme?
\[ V_{max} = \, \_\_\_\_\_\_ \, \text{µmol·min⁻¹} \]
**Explanation:**
This problem involves calculating the maximum reaction velocity (\( V_{max} \)) of an enzyme using the Michaelis-Menten equation:
\[
v = \frac{V_{max} \cdot [S]}{K_M + [S]}
\]
Where:
- \( v \) is the initial reaction velocity (10 µmol·min⁻¹).
- \( [S] \) is the substrate concentration, which is five times the \( K_M \).
To solve for \( V_{max} \), plug in the given values:
\[
10 = \frac{V_{max} \cdot 5K_M}{K_M + 5K_M}
\]
Simplify the expression:
\[
10 = \frac{5V_{max} \cdot K_M}{6K_M}
\]
\[
10 = \frac{5V_{max}}{6}
\]
Multiply both sides by 6:
\[
60 = 5V_{max}
\]
Divide by 5:
\[
V_{max} = 12 \, \text{µmol·min⁻¹}
\]
So, \( V_{max} \) is 12 µmol·min⁻¹.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc94cb5d1-997a-49d0-b0ef-9ec02bf0a187%2Fbdc72ea7-eaa3-4599-a2de-cbcc79806521%2Fklrx2er_processed.png&w=3840&q=75)
Transcribed Image Text:**Enzyme Kinetics Problem**
An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol·min⁻¹ when the substrate concentration is five times greater than the \( K_M \). What is the \( V_{max} \) of this enzyme?
\[ V_{max} = \, \_\_\_\_\_\_ \, \text{µmol·min⁻¹} \]
**Explanation:**
This problem involves calculating the maximum reaction velocity (\( V_{max} \)) of an enzyme using the Michaelis-Menten equation:
\[
v = \frac{V_{max} \cdot [S]}{K_M + [S]}
\]
Where:
- \( v \) is the initial reaction velocity (10 µmol·min⁻¹).
- \( [S] \) is the substrate concentration, which is five times the \( K_M \).
To solve for \( V_{max} \), plug in the given values:
\[
10 = \frac{V_{max} \cdot 5K_M}{K_M + 5K_M}
\]
Simplify the expression:
\[
10 = \frac{5V_{max} \cdot K_M}{6K_M}
\]
\[
10 = \frac{5V_{max}}{6}
\]
Multiply both sides by 6:
\[
60 = 5V_{max}
\]
Divide by 5:
\[
V_{max} = 12 \, \text{µmol·min⁻¹}
\]
So, \( V_{max} \) is 12 µmol·min⁻¹.
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