Enzyme Kinetics: Calculating Km
This question involves enzyme kinetics, specifically using the Michaelis-Menten model. We are given the relationship between the initial reaction velocity ($v_0$) and the maximum velocity ($v_{max}$), along with the substrate concentration ([S]). We need to find the Michaelis constant ($K_m$).
Michaelis-Menten Equation
The Michaelis-Menten equation relates the initial reaction velocity ($v_0$) to the maximum velocity ($v_{max}$) and the substrate concentration ([S]):
$ v_0 = \frac{v_{max}[S]}{K_m + [S]} $
Applying Given Conditions
We are told that the initial reaction velocity is one fourth of the maximum velocity:
$ v_0 = \frac{1}{4} v_{max} $
Substitute this into the Michaelis-Menten equation:
$ \frac{1}{4} v_{max} = \frac{v_{max}[S]}{K_m + [S]} $
Solving for Km
Divide both sides by $v_{max}$:
$ \frac{1}{4} = \frac{[S]}{K_m + [S]} $
Cross-multiply:
$ K_m + [S] = 4[S] $
Rearrange to solve for $K_m$:
$ K_m = 4[S] - [S] $
$ K_m = 3[S] $
Calculating Km with Substrate Concentration
The substrate concentration is given as [S] = $3.0 \times 10^{-3}$ mM.
First, convert the substrate concentration from millimolar (mM) to micromolar ($\mu$M), knowing that 1 mM = 1000 $\mu$M:
$ [S] = (3.0 \times 10^{-3} \text{ mM}) \times (1000 \, \mu\text{M} / \text{mM}) $
$ [S] = 3.0 \, \mu\text{M} $
Now, substitute this value into the equation $K_m = 3[S]$:
$ K_m = 3 \times (3.0 \, \mu\text{M}) $
$ K_m = 9.0 \, \mu\text{M} $
Conclusion
The value of $K_m$ is $9.0 \, \mu$M.
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).
The graph below shows the activity of enzyme pepsin in the presence of inhibitors aliphatic alcohols (P) or N-acetyl-1-phenylalanine (Q). Which ONE of the following represents the nature of inhibition by P and Q, respectively?
The following plot represents the Lineweaver-Burk equation of an enzymatic reaction both in the presence and the absence of inhibitor. Here, V is the velocity of reaction and S is the substrate concentration.
The nature of inhibition shown in the plot is
- For an enzyme catalyzed reaction, the plot that correctly represents the relationship between the rate and temperature is
- The Lineweaver-Burk plot of an enzymatic reaction shows $V_{max}$ of $160 \, \mu mol/l.min$ and $k_m$ of $60 \, \mu mol/l$. For a substrate concentration of $40 \, \mu mol/l$, the velocity of the reaction is estimated to be __________ $\mu mol/l.min$.
