An enzyme has a $K_m$ of $4.7 \times 10^{-5} \text{ M}$ and $V_m$ is $22 \text{ micro moles per litre per min}$. The enzyme reaction is carried out at a substrate concentration of $2 \times 10^{-4} \text{ M}$. The initial reaction velocity for this enzyme catalyzed reaction will be
Enzyme Kinetics: Calculating Initial Velocity
This problem involves calculating the initial reaction velocity ($v$) of an enzyme-catalyzed reaction using the Michaelis-Menten equation. We are given the Michaelis constant ($K_m$), the maximum velocity ($V_{max}$), and the substrate concentration ($[S]$).
Michaelis-Menten Equation
The Michaelis-Menten equation relates the initial reaction velocity ($v$) to the maximum velocity ($V_{max}$), the substrate concentration ($[S]$), and the Michaelis constant ($K_m$):
$v = \frac{V_{max}[S]}{K_m + [S]}$
Given Values
- Michaelis constant, $K_m = 4.7 \times 10^{-5} \text{ M}$
- Maximum velocity, $V_{max} = 22 \text{ micromoles per litre per min}$
- Substrate concentration, $[S] = 2 \times 10^{-4} \text{ M}$
Calculation Steps
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Substitute values into the Michaelis-Menten equation:
$v = \frac{(22 \text{ micromoles per litre per min}) \times (2 \times 10^{-4} \text{ M})}{(4.7 \times 10^{-5} \text{ M}) + (2 \times 10^{-4} \text{ M})}$
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Calculate the numerator:
$V_{max}[S] = 22 \times (2 \times 10^{-4}) = 44 \times 10^{-4} = 4.4 \times 10^{-3}$
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Calculate the denominator ($K_m + [S]$):
Convert $K_m$ to the same power of 10 as $[S]$ for easier addition:
$K_m = 4.7 \times 10^{-5} \text{ M} = 0.47 \times 10^{-4} \text{ M}$
$K_m + [S] = (0.47 \times 10^{-4}) + (2 \times 10^{-4}) = 2.47 \times 10^{-4} \text{ M}$
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Calculate the initial velocity ($v$):
$v = \frac{4.4 \times 10^{-3}}{2.47 \times 10^{-4}} \text{ micromoles per litre per min}$
$v \approx 17.81 \text{ micromoles per litre per min}$
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Round to the nearest option:
The calculated velocity is approximately $17.8 \text{ micromoles per litre per min}$.
Important Questions from Enzyme Kinetics Michaelis Menten K_m V_{max}
- An enzyme following Michaelis-Menten kinetics, catalyses a reaction with an initial velocity ($V_0$) of $2\ \mu\text{M s}^{-1}$ at the substrate concentration of $10\ \mu\text{M}$. If the turnover number ($k_{\text{cat}}$) of the enzyme for the given substrate is $500\ \text{s}^{-1}$ and the enzyme concentration in the reaction is $0.01\ \mu\text{M}$, then the value of the Michaelis-Menten constant ($K_m$) would be__________ $\times\ 10^{-6}\ \text{M}$ (in integer).
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