Problem 17 - Biochemistry: Concepts and Connections

Question

Initial rate data for an enzyme that obeys Michaelis-Menten kinetics are shown in the following table. When the enzyme concentration is 3 nmol \(\mathrm{ml}^{-1}\), a Lineweaver-Burk plot of this data gives a line with a \(y\)-intercept of \(0.00426\left(\mu \mathrm{mol}^{-1} \mathrm{ml} \mathrm{s}\right)\). \begin{tabular}{cc} {\([\mathbf{S}] \boldsymbol{\mu} \mathbf{M}\)} & \(\boldsymbol{v}_{\mathbf{0}}\left(\boldsymbol{\mu} \mathbf{m o l} \mathbf{m l}^{-1} \mathbf{s}^{-1}\right)\) \\ \hline 320 & 169 \\ 160 & 132 \\ \(80.0\) & \(92.0\) \\ \(40.0\) & \(57.2\) \\ \(20.0\) & \(32.6\) \\ \(10.0\) & \(17.5\) \\ \hline \end{tabular} (a) Calculate \(k_{\text {cat }}\) for the reaction. (b) Calculate \(K_{\mathrm{M}}\) for the enzyme. (c) When the reactions in part (b) are repeated in the presence of \(12 \mu \mathrm{M}\) of an uncompetitive inhibitor, the \(y\)-intercept of the Lineweaver-Burk plot is \(0.352\left(\mu \mathrm{mol}^{-1} \mathrm{ml} \mathrm{s}\right)\). Calculate \(K^{\prime}{\underline{\phantom{xx}}}_{\mathrm{I}}\) for this inhibitor.

Step-by-Step Solution

Answer

(a) \(k_{\text{cat}} = 78,247 \, \text{s}^{-1}\); (b) \(K_M \approx 80 \, \mu\text{M}\); (c) \(K'_I = 0.151 \, \mu\text{M}\).

Step 1: Understanding the Lineweaver-Burk Plot

The Lineweaver-Burk plot is a double reciprocal plot of the Michaelis-Menten equation, where we plot \(\frac{1}{v_0}\) against \(\frac{1}{[S]}\). The \(y\)-intercept of the line provides the reciprocal of \(V_{\text{max}}\), i.e., \(\frac{1}{V_{\text{max}}} = 0.00426 \, \mu \text{mol}^{-1} \text{ml} \text{s}\). From the \(y\)-intercept, we can calculate \(V_{\text{max}}\).

Step 2: Calculate \(V_{\text{max}}\)

Using the reciprocal relation from the \(y\)-intercept of the Lineweaver-Burk plot:\[V_{\text{max}} = \frac{1}{0.00426} = 234.74 \, \mu \text{mol} \, \text{ml}^{-1} \, \text{s}^{-1}\].

Step 3: Calculate the Turnover Number \(k_{\text{cat}}\)

The turnover number \(k_{\text{cat}}\) is given by \(V_{\text{max}}/[E]_0\) where \([E]_0\) is the enzyme concentration:\[k_{\text{cat}} = \frac{234.74 \, \mu \text{mol} \, \text{ml}^{-1} \, \text{s}^{-1}}{3 \, \text{nmol} \, \text{ml}^{-1}} = 78,247 \, \text{s}^{-1}\].

Step 4: Calculate the Michaelis Constant \(K_M\)

The Lineweaver-Burk plot also provides the slope, which is \(\frac{K_M}{V_{\text{max}}}\). To find \(K_M\), we must use the initial rate data and find \(K_M\) using the fact that the slope is the ratio of \(K_M\) to \(V_{\text{max}}\). However, without slope, we use paired experiment values to estimate it, assuming substrate saturation at high concentrations and \(K_M \approx [S]\) at half \(V_{\text{max}}\), using following method: \[K_M = [S] \text{ when } v_0 = \frac{V_{\text{max}}}{2}\]. Using data given, we estimate \(K_M\) is near \([S] \) when it's between 80 and 160 where \( v_0 \approx \frac{V_{\text{max}}}{2} \), so \(K_M \approx 80 \, \mu\text{M}\).

Step 5: Calculate the \(K'_I\) for Uncompetitive Inhibition

The intercept in presence of the inhibitor is a measure of effective \(V_{\text{max}}\) due to inhibitor and changes by the factor \(1 + \frac{[I]}{K'_I}\). Using the new \(y\)-intercept \(0.352\) and original \(0.00426\), solve \[ \frac{1}{(1 + \frac{[I]}{K'_I})} = \frac{0.00426}{0.352} \]. This simplifies to \(1 + \frac{12 \, \mu\text{M}}{K'_I} = \frac{0.352}{0.00426} = 82.63\). Solving, \( K'_I = \frac{12}{82.63 - 1} = 0.151 \, \mu\text{M}\).

Key Concepts

Lineweaver-Burk plot

When it comes to understanding enzyme reactions, the Lineweaver-Burk plot is a valuable tool. You might wonder why. It's because this graph helps convert the nonlinear Michaelis-Menten equation into a straight line. This makes it much easier to analyze enzyme kinetics.

In this plot, we graph the reciprocal of the initial reaction velocity, \(\frac{1}{v_0}\), against the reciprocal of substrate concentration, \(\frac{1}{[S]}\). The main advantage here is the clearer determination of key constants like \(V_{\text{max}}\) and \(K_M\) from a straight line.

• The y-intercept gives \(\frac{1}{V_{\text{max}}}\) because it reflects the velocity when substrate concentration is infinitely high.
• The x-intercept indicates \(-\frac{1}{K_M}\), providing insights into the affinity of the enzyme for its substrate.
• The slope of the line is \(\frac{K_M}{V_{\text{max}}}\).

This conversion might sound complex initially, but with practice, interpreting the graph will become intuitive.

enzyme kinetics

Enzyme kinetics is all about understanding how enzymes work and how different factors affect their activity. Enzymes are biological catalysts that speed up chemical reactions without being consumed in the process. To analyze enzyme kinetics, we often refer to the Michaelis-Menten model. It's a mathematical description of the rate of enzyme-catalyzed reactions.

• The model assumes the formation of an enzyme-substrate complex as an intermediate step.
• The maximum reaction rate, \(V_{\text{max}}\), occurs when the enzyme is saturated with substrate.
• The Michaelis constant, \(K_M\), is the substrate concentration at which the reaction rate is half of \(V_{\text{max}}\). It reflects the affinity of the enzyme for the substrate.
• Another critical parameter is the turnover number, \(k_{\text{cat}}\), which represents the number of substrate molecules converted to product per enzyme molecule per unit of time when the enzyme is fully saturated.

The interplay between these parameters defines how efficiently an enzyme operates. Recognizing this helps in studying enzyme behavior in different environments and designing experiments to optimize enzyme use.

uncompetitive inhibition

Uncompetitive inhibition is a particular type of enzyme inhibition where the inhibitor only binds to the enzyme-substrate complex, not the free enzyme. This distinguishes it from other forms such as competitive or noncompetitive inhibition.

• In uncompetitive inhibition, since the inhibitor binds to the enzyme-substrate complex, both \(K_M\) and \(V_{\text{max}}\) are affected.
• The inhibitor decreases the apparent \(V_{\text{max}}\) and \(K_M\) by the same factor, \(1+\frac{[I]}{K'_I}\), where \(K'_I\) is the equilibrium constant for the inhibitor binding.
• The Lineweaver-Burk plot in the presence of an uncompetitive inhibitor shows parallel lines to the control (without inhibitor), reflecting the change in \(V_{\text{max}}\) and \(K_M\) without altering the slope.

Understanding uncompetitive inhibition is crucial for developing drugs that target specific enzyme functions, as it can regulate enzyme activity under specific conditions. By manipulating \(K'_I\), scientists and researchers can minimize harmful enzyme activity during therapeutic interventions.