5. Let's revisit the question from last week's problem set with the cellulase enzyme discovered by Colby students. A small molecule was also identified that inhibits the newly discovered cellulase. The data are shown below. [S] 2 x 10 M 4 x 10-6 1 x 10-5 2 x 10-5 4 x 10-5 1 x 104 2 x 10-3 1 x 10-² 4 x 10-² KM = Vo (μm/sec) 28 40 Vmax= A. Use the data in the table above to estimate KM and Vmax for cellulase in the presence and absence of inhibitor. KM (+ Inhibitor) = Vmax (+Inhibitor) = 75 95 112 128 139 140 141 Vo (μm/sec) + Inhibitor (2 µM) 16 27 45 58 72 85 89 89 90 C. What is the Ki of the inhibitor? B. What is the mechanism of inhibition? Briefly explain.
![5. Let’s revisit the question from last week’s problem set with the cellulase enzyme discovered by Colby students. A small molecule was also identified that inhibits the newly discovered cellulase. The data are shown below.
| [S] (M) | V₀ (µM/sec) | V₀ (µM/sec) + Inhibitor (2 µM) |
|-------------|-------------|-------------------------------|
| 2 x 10⁻⁶ | 28 | 16 |
| 4 x 10⁻⁶ | 40 | 27 |
| 1 x 10⁻⁵ | 75 | 45 |
| 2 x 10⁻⁵ | 95 | 58 |
| 4 x 10⁻⁵ | 112 | 72 |
| 1 x 10⁻⁴ | 128 | 85 |
| 2 x 10⁻³ | 139 | 89 |
| 1 x 10⁻² | 140 | 89 |
| 4 x 10⁻² | 141 | 90 |
**A. Use the data in the table above to estimate Kₘ and Vₘₐₓ for cellulase in the presence and absence of inhibitor.**
≈Kₘ =
≈Kₘ (+ Inhibitor) =
≈Vₘₐₓ =
≈Vₘₐₓ (+ Inhibitor) =
**B. What is the mechanism of inhibition? Briefly explain.**
**C. What is the Kᵢ of the inhibitor?**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F424482ac-ef50-4214-a1e3-aaaeb325792a%2F637bc330-32d2-4bce-88cd-d7261f43c6ba%2Fhqzpnag_processed.png&w=3840&q=75)
Parameters such as Km and Vmax are used for comparing enzyme activities. If we know the initial rate of reaction and substrate concentration, Km and vax can be calculated from the Michaelis Menton equation ;
vo = (vmax · [S]) / (Km +[S])
Where, vo is the rate of formation of product, [S] is the substrate concentration, Km is called the Michaelis-Menton constant which represents the enzyme's affinity for the substrate and vmax is the maximum rate at which the enzyme can catalyze the reaction.
vo and [S] can be estimated experimentally but Km and vmax are estimated from the Michaelis Menton plot.
Km = [S] when vo= vmax/2
The result is a curve when we plot vo vs [S] data on a graph. Accurate values of Km and vmax cannot be determined from a curve so we transform the Michaelis Menton Equation into Lineweaver Burk Equation by taking the reciprocal of vo and [S] and plot a 1/vo vs 1/[S] which gives us a straight line.
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