High School

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An enzyme catalyzes the reaction \( A \rightarrow B \). The initial rate of the reaction was measured as a function of the concentration of \( A \). The following data were obtained:

| [A] (mM) | \( V_0 \) (nM/min) |
|------------|---------------------|
| 0.05 | 0.08 |
| 0.1 | 0.16 |
| 0.5 | 0.79 |
| 10 | 7.3 |
| 50 | 13 |
| 100 | 40 |
| 500 | 53 |
| 1000 | 73 |
| 5000 | 76 |
| 10000 | 79 |
| 20000 | 80 |

a) What is the \( K_m \) of the enzyme for the substrate \( A \)?

b) What is the value of \( V_0 \) when [A] = 43?

c) What is the value of the y-intercept of the line?

d) What is the value of the x-intercept of the line?

Show all calculations and use correct units and labels.

Answer :

The Km of the enzyme for substrate A is approximately 0.5 nM, V₀ at [A] = 43 is about 80 nM/min, the y-intercept is 0.0125 min/nM, and the x-intercept is -2 nM⁻¹.

Let's solve the given enzyme kinetics problem step-by-step.

  1. From the data, we know that the reaction velocity (V₀) approaches a maximum value (Vmax) of 80 nM/min. The substrate concentration at which the reaction rate is half of Vmax (i.e., Vo = 40 nM/min) is the Michaelis constant (Km). From the data, this concentration is approximately at 0.5 nM.
  2. Given that [A] = 43 is relatively small compared to the other entries in the data table, we look at the trend in V₀. The approximate value can be interpolated using Michaelis-Menten kinetics.
  3. Using V₀ = (Vmax * [A]) / (Km + [A])
  4. Vo = (80 * 43) / (0.5 + 43) ≈ 80 nM/min.
  5. The y-intercept in a Michaelis-Menten plot (Lineweaver-Burk plot) is 1/Vmax. Here, Vmax = 80 nM/min.
  6. Thus, y-intercept = 1 / 80 = 0.0125 (min/nM).
  7. The x-intercept in a Lineweaver-Burk plot is -1/Km. From our earlier calculation, Km ≈ 0.5 nM.
  8. Thus, x-intercept = -1 / 0.5 = -2 (nM⁻¹).

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