Michaelis-Menten Kinetics: Calculating Enzyme Km
This problem requires calculating the Michaelis constant ($K_m$) for an enzyme using the Michaelis-Menten equation and the provided reaction parameters.
Given Parameters
- Maximum reaction velocity, $V_m$ = 75 nmol $l^{-1}$ min$^{-1}$
- Substrate concentration, $[S]$ = $1.0 \times 10^{-4}$ M
- Initial reaction velocity, $V_0$ = 60 nmol $l^{-1}$ min$^{-1}$
Michaelis-Menten Equation
The relationship between initial velocity ($V_0$), maximum velocity ($V_m$), substrate concentration ($[S]$), and the Michaelis constant ($K_m$) is given by the Michaelis-Menten equation:
$ V_0 = \frac{V_m [S]}{K_m + [S]} $
Calculating Km
To find $K_m$, we can rearrange the Michaelis-Menten equation:
- Multiply both sides by $(K_m + [S])$:
$ V_0 (K_m + [S]) = V_m [S] $
- Distribute $V_0$:
$ V_0 K_m + V_0 [S] = V_m [S] $
- Isolate the term containing $K_m$:
$ V_0 K_m = V_m [S] - V_0 [S] $
- Factor out $[S]$:
$ V_0 K_m = [S] (V_m - V_0) $
- Solve for $K_m$:
$ K_m = \frac{[S] (V_m - V_0)}{V_0} $
Substituting Values
Now, substitute the given values into the rearranged equation:
$ K_m = \frac{(1.0 \times 10^{-4} \text{ M}) (75 \text{ nmol } l^{-1} \text{ min}^{-1} - 60 \text{ nmol } l^{-1} \text{ min}^{-1})}{60 \text{ nmol } l^{-1} \text{ min}^{-1}} $
Final Calculation
Perform the calculation:
$ K_m = \frac{(1.0 \times 10^{-4} \text{ M}) (15 \text{ nmol } l^{-1} \text{ min}^{-1})}{60 \text{ nmol } l^{-1} \text{ min}^{-1}} $
$ K_m = (1.0 \times 10^{-4} \text{ M}) \times \frac{15}{60} $
$ K_m = (1.0 \times 10^{-4} \text{ M}) \times 0.25 $
$ K_m = 0.25 \times 10^{-4} \text{ M} $
$ K_m = 2.5 \times 10^{-5} \text{ M} $
The calculated $K_m$ value is $2.5 \times 10^{-5}$ M.
Important Questions from Enzyme Kinetics Michaelis Menten K_m V_{max}
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