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The rate of a simple enzyme reaction is given by the standard Michaelis-Menten equation: \\[\text { rate }=V_{\max }[\mathrm{S}] /\left(K_{\mathrm{M}}+[\mathrm{S}]\right)\\] If the \(V_{\max }\) of an enzyme is 100 \mumole/sec and the \(K_{M}\) is \(1 \mathrm{mM},\) at what substrate concentration is the rate \(50 \mu \mathrm{mole} / \mathrm{sec} ?\) Plot a graph of rate versus substrate \((\mathrm{S})\) concentration for \([\mathrm{S}]=0\) to \(10 \mathrm{mM}\). Convert this to a plot of 1/rate versus \(1 /[\mathrm{S}] .\) Why is the latter plot a straight line?

Short Answer

Expert verified
The rate is 50 \(\mu\text{mole/sec}\) at a substrate concentration of 1 mM.

Step by step solution

01

Understanding the Michaelis-Menten Equation

The Michaelis-Menten equation is \[ \text{rate} = \frac{V_{\max}[S]}{K_M + [S]} \]. Here, \(V_{\max}\) is the maximum rate of the reaction, \(K_M\) is the Michaelis constant, and \([S]\) is the substrate concentration. You need to substitute the given values and solve for \([S]\) when the rate is \(50\ \mu \text{mole/sec}\).
02

Substitute Given Values and Solve

Given \(V_{\max} = 100\ \mu\text{mole/sec}\) and \(K_M = 1\ \text{mM}\), and the rate is \(50\ \mu\text{mole/sec}\). Substitute into the equation: \[ 50 = \frac{100 [S]}{1 + [S]} \]. Solve for \([S]\) by cross-multiplying to get \(50(1 + [S]) = 100[S]\) which simplifies to \(50 + 50[S] = 100[S]\). Further simplify to find \([S] = 1\ \text{mM}\).
03

Plotting Rate Versus Substrate Concentration

Create a plot of rate versus substrate concentration from \([S] = 0\ \text{mM}\) to \([S] = 10\ \text{mM}\). Use the equation \(\text{rate} = \frac{100[S]}{1 + [S]}\) to calculate the rate for several values of \([S]\) and plot these data points on a graph to visualize the relationship.
04

Convert to Linear Plot

To create a linear plot of \(1/\text{rate}\) versus \(1/[S]\), use the Lineweaver-Burk transformation: \[ \frac{1}{\text{rate}} = \frac{K_M}{V_{\max}[S]} + \frac{1}{V_{\max}} \]. This transformation of the data points will give you a straight line because it linearizes the Michaelis-Menten equation, allowing parameters to be easily determined.
05

Explain Why the Plot is a Straight Line

The plot of \(1/\text{rate}\) versus \(1/[S]\) is a straight line because this is the rearrangement of the Michaelis-Menten equation in a linear form, known as a Lineweaver-Burk plot. The linearization is akin to the equation of a line, \(y = mx + b\), making the slope and intercept simpler to analyze.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enzyme Kinetics
Enzyme kinetics is a fascinating field that studies how enzymes bind to substrates and convert them into products. Understanding these kinetics is essential in biochemistry and pharmaceutical industries, as it aids in the design of drugs that can either inhibit or enhance enzyme activity.
Enzymes are biological catalysts that speed up chemical reactions without being consumed in the process. The Michaelis-Menten equation, which is central to enzyme kinetics, helps describe the rate of enzymatic reactions.
  • The equation: \[ \text{rate} = \frac{V_{\max}[S]}{K_M + [S]} \] provides a means to calculate how fast a reaction proceeds based on substrate concentration \([S]\).
  • \(V_{\max}\) is the maximum reaction rate achieved by the system, at maximum (saturating) substrate concentrations.
  • The Michaelis constant \(K_M\) is the substrate concentration at which the reaction rate is half of \(V_{\max}\). It provides insight into the enzyme's affinity for the substrate; a low \(K_M\) value means high affinity, as less substrate is needed to reach half of \(V_{\max}\).
By analyzing these parameters, scientists can determine how effectively an enzyme interacts with a substrate and adjust conditions to optimize reactions.
Lineweaver-Burk Plot
The Lineweaver-Burk plot is a graphical representation of enzyme kinetics, designed to linearize the hyperbolic equation of Michaelis-Menten into a straight line. This transformation is helpful for precisely calculating kinetic parameters for enzyme reactions.
The equation used in Lineweaver-Burk transformation is: \[ \frac{1}{\text{rate}} = \frac{K_M}{V_{\max}[S]} + \frac{1}{V_{\max}} \]This formula is reminiscent of the line equation, \(y = mx + b\), where:
  • The slope \(m\) of the line is \(\frac{K_M}{V_{\max}}\).
  • The y-intercept \(b\) is \(\frac{1}{V_{\max}}\).
  • The x-intercept, which can be calculated as \(-\frac{1}{K_M}\), provides insights into the binding affinity.
Creating this plot involves measuring the reaction rates at various substrate concentrations, inverting these values, and plotting \(1/\text{rate}\) against \(1/[S]\). This plot becomes a straight line, allowing for more straightforward analysis and determination of kinetic constants.
Substrate Concentration
Substrate concentration is a crucial factor in influencing the rate of an enzyme-catalyzed reaction. In the Michaelis-Menten model, substrate concentration \([S]\) directly affects how quickly an enzyme can convert substrates into products.
At low substrate concentrations, the rate of reaction increases sharply as more substrate molecules become available to interact with enzyme active sites. However, as substrate concentration continues to rise, the rate of reaction begins to plateau, approaching \(V_{\max}\).
  • This plateau occurs because the enzyme becomes saturated with substrate, and adding more substrate cannot increase the rate further - the enzyme's activity is operating at full capacity.
  • This relationship is described by the Michaelis-Menten equation: \( \text{rate} = \frac{V_{\max}[S]}{K_M + [S]} \).
In enzymology studies, determining substrate concentration at various rates helps in understanding enzyme efficiency and optimal conditions for reactions, illustrating key biochemical principles in metabolic pathways.

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