/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Estimation of \(V_{\max }\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Estimation of \(V_{\max }\) and \(\boldsymbol{K}_{\mathrm{m}}\) 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. \begin{tabular}{cc} {\([\mathbf{S}](\mathrm{M})\)} & \(V_{0}(\mu \mathrm{M} / \mathrm{min})\) \\ \hline \(2.5 \times 10^{-6}\) & 28 \\ \(4.0 \times 10^{-6}\) & 40 \end{tabular} \begin{tabular}{ll} \(1 \times 10^{-5}\) & 70 \\ \(2 \times 10^{-5}\) & 95 \\ \(4 \times 10^{-5}\) & 112 \\ \(1 \times 10^{-4}\) & 128 \\ \(2 \times 10^{-3}\) & 140 \\ \(1 \times 10^{-2}\) & 139 \\ \hline \end{tabular}

Short Answer

Expert verified
\(V_{\max} \approx 140 \, \mu M/min\); \(K_m \approx 1 \times 10^{-5} M\).

Step by step solution

01

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.
02

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 \(V_{\max} \approx 140 \, \mu M/min\).
03

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 \(K_m \approx 1 \times 10^{-5} M\).
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Applying the Michaelis-Menten Equation IV Researchers discover an enzyme that catalyzes the reaction \(\mathrm{X} \rightleftharpoons \mathrm{Y}\). They find that the \(K_{\mathrm{m}}\) for the substrate \(\mathrm{X}\) is \(4 \mu \mathrm{M}\), and the \(k_{\text {cat }}\) is \(20 \mathrm{~min}^{-1}\). a. In an experiment, \([\mathrm{X}]=6 \mathrm{mM}\), and \(V_{0}=480 \mathrm{nM} \mathrm{min}^{-1}\). What was the \(\left[\mathrm{E}_{\mathrm{t}}\right]\) used in the experiment? b. In another experiment, \(\left[\mathrm{E}_{\mathrm{t}}\right]=0.5 \mu \mathrm{M}\), and the measured \(V_{0}=5 \mu \mathrm{M} \mathrm{min}^{-1}\). What was the \([\mathrm{X}]\) used in the experiment? c. The researchers discover that compound \(Z\) is a very strong competitive inhibitor of the enzyme. In an experiment with the same \(\left[E_{t}\right]\) as in (a), but a different \([\mathrm{X}]\), they add an amount of \(\mathrm{Z}\) that produces an \(a\) of 10 and reduces \(V_{0}\) to \(240 \mathrm{nM} \mathrm{min}^{-1}\). What is the \([\mathrm{X}]\) in this experiment? d. Based on the kinetic parameters given, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.

Properties of an Enzyme of Prostaglandin Synthesis Prostaglandins are one class of the fatty acid derivatives called eicosanoids. Prostaglandins produce fever and inflammation, as well as the pain associated with inflammation. The enzyme prostaglandin endoperoxide synthase, a cyclooxygenase, uses oxygen to convert arachidonic acid to \(\mathrm{PGG}_{2}\), the immediate precursor of many different prostaglandins (prostaglandin synthesis is described in Chapter 21 . Ibuprofen inhibits prostaglandin endoperoxide synthase, thereby reducing inflammation and pain. The kinetic data given in the table are for the reaction catalyzed by prostaglandin endoperoxide synthase in the absence and presence of ibuprofen. a. Based on the data, determine the \(V_{\max }\) and \(K_{\mathrm{m}}\) of the enzyme. \(\begin{array}{ccc}\begin{array}{c}\text { [Arachidonic } \\ \text { acid] }(\mathrm{mM})\end{array} & \begin{array}{c}\text { Rate of formation of } \\\ \mathrm{PGG}_{2}\left(\mathrm{mM} \mathrm{min}^{-1}\right)\end{array} & \begin{array}{c}\text { Rate of formation of } \\ \mathrm{PGG}_{2} \text { with } 10 \mathrm{mg} / \mathrm{mL}\end{array}\end{array}\) \begin{tabular}{ccc} ibuprofen & \(\left(\mathrm{mM}^{-1} \mathrm{~min}^{-1}\right)\) \\ \hline \(0.5\) & \(23.5\) & \(16.67\) \\ \(1.0\) & \(32.2\) & \(30.49\) \\ \(1.5\) & \(36.9\) & \(37.04\) \\ \(2.5\) & \(41.8\) & \(38.91\) \\ \(3.5\) & \(44.0\) & 25 \\ \hline \end{tabular} b. Based on the data, determine the type of inhibition that ibuprofen exerts on prostaglandin endoperoxide synthase.

The Effects of Reversible Inhibitors The MichaelisMenten rate equation for reversible mixed inhibition is written as $$ V_{0}=\frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} $$ Apparent, or observed, \(K_{\mathrm{m}}\) is equivalent to the [S] at which $$ V_{0}=\frac{V_{\max }}{2 \alpha^{\prime}} $$ Derive an expression for the effect of a reversible inhibitor on apparent \(K_{\mathrm{m}}\) from the previous equation.

Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy metal ions such as \(\mathrm{Hg}^{2+}, \mathrm{Cu}^{2+}\), or \(\mathrm{Ag}^{+}\), which can react with essential sulfhydryl groups to form mercaptides: $$ \text { Enz-SH }+\mathrm{Ag}^{+} \rightarrow \text { Enz-S-Ag }+\mathrm{H}^{+} $$ The affinity of \(\mathrm{Ag}^{+}\)for sulfhydryl groups is so great that \(\mathrm{Ag}^{+}\) can be used to titrate - SH groups quantitatively. An investigator added just enough \(\mathrm{AgNO}_{3}\) to completely inactivate a \(10.0 \mathrm{~mL}\) solution containing \(1.0 \mathrm{mg} / \mathrm{mL}\) enzyme. A total of \(0.342 \mu \mathrm{mol}\) of \(\mathrm{AgNO}_{3}\) was required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in this way give only the minimum molecular weight?

Applying the Michaelis-Menten Equation I An enzyme has a \(V_{\max }\) of \(1.2 \mu \mathrm{M} \mathrm{s}^{-1}\). The \(K_{\mathrm{m}}\) for its substrate is \(10 \mu \mathrm{M}\). Calculate the initial velocity of the reaction, \(V_{0}\), when the substrate concentration is a. \(2 \mu \mathrm{M}\) b. \(10 \mu_{M}\) c. \(30 \mu_{\mathrm{M}}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.