(a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph. (b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
(a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph. (b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
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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Michaelis-Menten Plot Problem Help
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![**Michaelis-Menten Plot Analysis of WT and Mutant (V105A) Enzymes**
Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium *Staphylococcus aureus*. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars.
### Graph Description:
- **Axes:** The x-axis represents substrate concentration \([S]\) in mM, ranging from 0 to 500 mM. The y-axis represents the reaction velocity \(v\) in \(\mu \text{mol} \, \text{s}^{-1} \, \text{mg}^{-1}\) of enzyme, ranging from 0.00 to 1.50 \(\mu \text{mol} \, \text{s}^{-1} \, \text{mg}^{-1}\).
- **Curves:**
- The black curve represents the WT enzyme.
- The red curve represents the mutant V105A enzyme.
- **Data Points:** Each curve has data points with error bars indicating variability in the measurements.
### Questions:
(a) **Estimate the \(K_m\) and \(V_{max}\) for the wild-type and mutant enzyme from the graph.**
(b) **Calculate the \(K_{cat}\) and \(K_{cat}/K_m\) for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 \(\mu \text{mol/mg}\).**
(c) **Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.**
### Analysis Approach:
- **Estimate \(V_{max}\):** Look for the plateau of each curve to identify \(V_{max}\).
- **Estimate \(K_m\):** Find the substrate concentration where the reaction velocity is half of \(V_{max}\).
- **Calculate \(K_{cat}\):** Use the formula \(K_{cat} = \frac{V_{max}}{[E_t]}\), where \([E_t]\) is the total enzyme concentration.
- **Determine Efficiency:** Assess the catalytic efficiency using \(K_{cat}/K_m\) and compare between the WT and mutant enzymes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff800a840-0309-4834-92a3-4244363b7afa%2F29cbd316-83b0-41e4-8907-9fc0b5a41f14%2Frrh3oyom_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Michaelis-Menten Plot Analysis of WT and Mutant (V105A) Enzymes**
Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium *Staphylococcus aureus*. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars.
### Graph Description:
- **Axes:** The x-axis represents substrate concentration \([S]\) in mM, ranging from 0 to 500 mM. The y-axis represents the reaction velocity \(v\) in \(\mu \text{mol} \, \text{s}^{-1} \, \text{mg}^{-1}\) of enzyme, ranging from 0.00 to 1.50 \(\mu \text{mol} \, \text{s}^{-1} \, \text{mg}^{-1}\).
- **Curves:**
- The black curve represents the WT enzyme.
- The red curve represents the mutant V105A enzyme.
- **Data Points:** Each curve has data points with error bars indicating variability in the measurements.
### Questions:
(a) **Estimate the \(K_m\) and \(V_{max}\) for the wild-type and mutant enzyme from the graph.**
(b) **Calculate the \(K_{cat}\) and \(K_{cat}/K_m\) for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 \(\mu \text{mol/mg}\).**
(c) **Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.**
### Analysis Approach:
- **Estimate \(V_{max}\):** Look for the plateau of each curve to identify \(V_{max}\).
- **Estimate \(K_m\):** Find the substrate concentration where the reaction velocity is half of \(V_{max}\).
- **Calculate \(K_{cat}\):** Use the formula \(K_{cat} = \frac{V_{max}}{[E_t]}\), where \([E_t]\) is the total enzyme concentration.
- **Determine Efficiency:** Assess the catalytic efficiency using \(K_{cat}/K_m\) and compare between the WT and mutant enzymes.
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