Question

Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available.$$\begin{array}{ccc}\hline \begin{array}{l}\text { [S] } \\\\\text { (mM) }\end{array} & \begin{array}{c}V_{\text {, No lnhibitor }} \\\\\text { (mmol min }^{-1} \text {) }\end{array} & \begin{array}{c}V_{\text {, Inhibitor }} \text { Present } \\\\\text { (mmol min }^{-1} \text {) }\end{array} \\\\\hline & & \\\3.0 & 4.58 & 3.66 \\\5.0 & 6.40 & 5.12 \\\7.0 & 7.72 & 6.18 \\\9.0 & 8.72 & 6.98 \\\11.0 & 9.50 & 7.60 \\\\\hline\end{array}$$.What are the \(K_{\mathrm{y}}\) and \(V_{\mathrm{max}}\) values for the inhibited and uninhibited reactions? Is the inhibitor competitive or noncompetitive?

Short answer

Calculate reciprocals, plot \(1/V\) vs \(1/[S]\), determine slopes and intercepts, calculate \(K_m\) and \(V_{max}\), and identify inhibition type.

Step-by-step solution

- Calculate the reciprocals

For each value of substrate concentration \([S]\) and velocity \(V\), calculate the reciprocals \(1/[S]\) and \(1/V\). This includes both the data with no inhibitor and with the inhibitor present.

- Organize data into a table

Create a table organizing the \([S]\), \(V_{\text{No Inhibitor}}\), and \(V_{\text{Inhibitor Present}}\) values alongside their corresponding reciprocal values \(1/[S]\), \(1/V_{\text{No Inhibitor}}\), and \(1/V_{\text{Inhibitor Present}}\).

- Plot the Lineweaver-Burk plots

Plot \(1/V\) against \(1/[S]\). Use the values for \(1/V_{\text{No Inhibitor}}\) to create one plot and \(1/V_{\text{Inhibitor Present}}\) to create another plot. The linearized Lineweaver-Burk equation is \[ \frac{1}{V} = \frac{K_m}{V_{max}} \times \frac{1}{S} + \frac{1}{V_{max}} \], which resembles a linear equation \(y = mx + b\).

- Determine the slopes and intercepts

Find the slopes and y-intercepts of each plot. The slope gives \(K_m/V_{max}\) and the y-intercept gives \(1/V_{max}\).

- Calculate K_m and V_{max}

Using the slope and y-intercept values from each plot, calculate \(K_m\) and \(V_{max}\) for both the inhibited and uninhibited reactions.

- Determine the type of inhibition

Compare the plots to determine if the inhibitor affects \(V_{max}\) or \(K_m\). For competitive inhibition, \(V_{max}\) remains unchanged but \(K_m\) increases. For non-competitive inhibition, \(V_{max}\) decreases but \(K_m\) remains unchanged.

Key concepts

enzyme kinetics

Enzyme kinetics is the study of the rates of chemical reactions that are catalyzed by enzymes. In these reactions, the enzyme binds to a substrate to form an enzyme-substrate complex, which then converts into the product. Understanding enzyme kinetics helps in determining how different factors influence the rate of enzymatic reactions.
To analyze enzyme kinetics, we often measure the velocity (V) of the reaction at various substrate concentrations \( [S] \). This information can be graphically represented and analyzed to determine kinetic parameters like \( K_m \) (Michaelis constant) and \( V_{max} \) (maximum velocity). These parameters give insight into the enzyme's efficiency and how it interacts with substrates.

enzyme inhibition

Enzyme inhibition refers to the process in which a molecule (inhibitor) binds to an enzyme and decreases its activity. This can occur in different ways:

• Competitive Inhibition: The inhibitor competes with the substrate for the active site of the enzyme.
• Noncompetitive Inhibition: The inhibitor binds to a different site on the enzyme, which alters the enzyme's shape and reduces its activity.

Inhibitors play crucial roles in regulating enzyme activity within cells and can also be important in drug design. Understanding how inhibitors work can help in manipulating enzyme functions and creating treatments for various diseases.

Michaelis-Menten equation

The Michaelis-Menten equation is a mathematical model that describes the kinetics of enzyme-catalyzed reactions. It is given by:

\[ V = \frac{V_{max} [S]}{K_m + [S]} \]

Where:

• \(V \) = Reaction velocity
• \(V_{max} \) = Maximum reaction velocity
• \(K_m \) = Michaelis constant (substrate concentration at which the reaction velocity is half of \( V_{max} \) )
• \( [S] \) = Substrate concentration

This equation helps in understanding how reaction velocity changes with varying substrate concentration. \( K_m \) and \( V_{max} \) are key parameters that provide insight into the affinity of the enzyme for the substrate and the enzyme's maximum catalytic efficiency.

competitive inhibition

Competitive inhibition occurs when an inhibitor molecule competes with the substrate for binding to the enzyme's active site. This type of inhibition can be identified using Lineweaver-Burk plots.

Key characteristics include:

• \( V_{max} \) remains unchanged. This is because the high substrate concentration can outcompete the inhibitor, restoring the maximum velocity.
• \( K_m \) increases. This is because a higher substrate concentration is needed to reach half of \( V_{max} \).

In a Lineweaver-Burk plot, competitive inhibition is demonstrated by a steeper slope and a shift in the x-intercept, but the y-intercept (representing \( 1/V_{max} \) ) stays the same.

noncompetitive inhibition

Noncompetitive inhibition occurs when the inhibitor binds to a site on the enzyme other than the active site, causing a conformational change that reduces enzymatic activity.

Key characteristics include:

• \( V_{max} \) decreases. This is because the inhibitor binding alters the enzyme's structure, hindering its catalytic ability regardless of the substrate concentration.
• \( K_m \) remains unchanged. The affinity of the enzyme for the substrate is not affected by the inhibitor binding.

In a Lineweaver-Burk plot, noncompetitive inhibition is shown by an increased y-intercept (representing \( 1/V_{max} \)) while the x-intercept (representing \( -1/K_m \)) remains the same, indicating a parallel increase in the slope.