Question

Estimation of $V_{max}$ and ${\mathit{K}}_{\mathrm{m}}$ by Inspection Graphical methods are available for accurate determination of the $V_{max}$ and $K_{\mathrm{m}}$ of an enzyme-catalyzed reaction. However, these quantities can sometimes be estimated by inspecting values of $V_0$ at increasing [S]. Estimate the $V_{max}$ and $K_m$ of the enzyme-catalyzed reaction for which the data in the table were obtained. $$\text{Unknown environment 'tabular'}$$ $$\text{Unknown environment 'tabular'}$$

Short answer

$V_{max}\approx 140\mu M/min$; $K_m\approx 1\times {10}^{-5}M$.

Step-by-step solution

Understanding the context

We are tasked with estimating two parameters, $V_{max}$ and $K_m$, which are crucial in enzyme kinetics. $V_{max}$ is the maximum rate achieved by the system, while $K_m$ represents the substrate concentration at which the reaction rate is at half of $V_{max}$. We will use the data of substrate concentration $[S]$ and initial velocity $V_0$ provided in the table to make this estimation.

Identifying $V_{max}$

From the table, the initial velocities $V_0$ approach a maximum value as $[S]$ increases. We observe that $V_0$ increases significantly until $[S]$ reaches $4\times {10}^{-5}M$, beyond which the change in $V_0$ becomes very small. The highest values observed are 140 and 139 ($[S]=2\times {10}^{-3}M$ and $1\times {10}^{-2}M$, respectively), suggesting $V_{max}\approx 140\mu M/min$.

Estimating $K_m$

The $K_m$ value can be estimated by identifying the $[S]$ at which $V_0$ is approximately half of $V_{max}$. Since $V_{max}\approx 140$, half of this is 70. Checking the table, $[S]=1\times {10}^{-5}M$ produces a $V_0$ of 70, indicating $K_m\approx 1\times {10}^{-5}M$.

Conclusion

By inspecting the substrate concentrations and the corresponding initial velocities, we have estimated the enzyme's $V_{max}$ to be approximately $140\mu M/min$ and $K_m$ to be approximately $1\times {10}^{-5}M$. This approach gives an initial approximation without detailed graphical analysis such as Lineweaver-Burk plots.

Key concepts

Michaelis-Menten Equation

The Michaelis-Menten equation is a fundamental concept in enzyme kinetics, which explains how the rate of enzymatic reactions depends on substrate concentration. The equation is expressed as:$$V=\frac{V_{max}\cdot [S]}{K_m+[S]}$$where:

• $V$ is the rate of the reaction.
• $V_{max}$ is the maximum rate of the reaction.
• $[S]$ is the substrate concentration.
• $K_m$ is the Michaelis constant, which is the substrate concentration at half of $V_{max}$.

This equation helps us understand how changes in substrate concentration affect the reaction rate. When $[S]$ is very low, the reaction rate ($V$) increases linearly with $[S]$. However, as $[S]$ increases further, the reaction rate approaches $V_{max}$ and becomes less sensitive to changes in $[S]$. This behavior characterizes the saturation effect in enzyme kinetics.

Vmax

In the realm of enzyme kinetics, $V_{max}$ represents the maximum rate an enzyme-catalyzed reaction can achieve when the enzyme active sites are fully saturated with substrate. This value is crucial because it provides insight into the enzyme's catalytic efficiency and capacity.Estimating $V_{max}$ involves observing how the reaction rate ($V_0$) changes with increasing substrate concentration. As substrate concentration increases, $V_0$ continues to rise until a point is reached where further increases in substrate concentration result in little to no change in $V_0$. This plateau is identified as $V_{max}$.In the provided exercise, as substrate $[S]$ increases from $2.5\times {10}^{-6}M$ to $1\times {10}^{-2}M$, the $V_0$ values initially climb sharply. Beyond a certain concentration, the increases in $V_0$ become negligible, allowing us to approximate $V_{max}$ at about $140\mu M/min$. This estimation helps in understanding the upper limit of the enzyme's catalytic capability under saturation conditions.

Km

The Michaelis constant, $K_m$, is a crucial parameter in enzyme kinetics, providing insight into the enzyme's affinity for its substrate. Specifically, $K_m$ is the substrate concentration at which the reaction velocity is half of $V_{max}$. It helps us determine how efficiently the enzyme binds the substrate.A lower $K_m$ value indicates a high affinity between enzyme and substrate, meaning less substrate is needed to reach half-maximum velocity. Conversely, a high $K_m$ suggests a lower affinity, requiring more substrate to achieve the same rate.In the exercise, $K_m$ is estimated by finding the substrate concentration $[S]$ that results in a reaction velocity approximately half of the $V_{max}$. Given $V_{max}\approx 140\mu M/min$, $K_m$ occurs when $V_0$ is around $70\mu M/min$. From the data table, this happens at $[S]=1\times {10}^{-5}M$, thus $K_m\approx 1\times {10}^{-5}M$. Understanding $K_m$ helps clarify the enzyme's behavior in different substrate concentrations.

Substrate Concentration

Substrate concentration, denoted $[S]$, is a critical factor affecting the rate of enzyme-catalyzed reactions. The influence of $[S]$ on the reaction velocity is elegantly captured by the Michaelis-Menten equation.At the start, when the substrate concentrations are low, the reaction rate directly depends on the amount of substrate available. This phase exhibits a linear increase in rate with substrate concentration because there is enough active enzyme to bind to the increasing substrate molecules. As the substrate concentration rises, more enzyme active sites are occupied, accelerating the reaction velocity.However, as the substrate concentration becomes very high, the enzyme molecules become saturated—they are fully occupied, and the rate cannot increase further. This saturation point indicates the maximum velocity ($V_{max}$) of the reaction, beyond which any additional substrate does not influence the rate.By analyzing different $[S]$ values in kinetic studies, one can derive crucial parameters such as $V_{max}$ and $K_m$. This understanding helps in grasping how enzymes function in various biological settings and their potential efficiency under different conditions.