Triose phosphate isomerase catalyzes the conversion of glyceraldehyde-3-phosphate to dihydroxyacetone phosphate. Glyceraldehyde-3-P \(\rightleftarrows\) dihydroxyacetone-P The \(K_{m}\) of this enzyme for its substrate glyceraldehyde- -phosphate is \(1.8 \times 10^{-5} M .\) When [glyceraldehydes-3-phosphate ] \(=30 \mu M\), the rate of the reaction, \(v,\) was \(82.5 \mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{sec}^{-1}\) a. What is \(V_{\max }\) for this enzyme? b. Assuming 3 nanomoles per mL of enzyme was used in this experiment \((\left[E_{\text {total }}\right]=3 \text { nanomol/mL), what is } k_{\text {cat }}\) for this enzyme? c. What is the catalytic efficiency \(\left(k_{\mathrm{ca}} / K_{m}\right)\) for triose phosphate isomerase? d. Does the value of \(k_{\mathrm{ca}} / K_{m}\) reveal whether triose phosphate isomerase approaches "catalytic perfection"? e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that imposes the physical limit on kinetic perfection?

Short Answer

Expert verified
\(V_{max}\), \(k_{\text{cat}}\) and catalytic efficiency (i.e., \(k_{\text{cat}/K_m}\)) can be calculated from the given values using the Michaelis-Menten equation and the provided relations. If \(k_{\text{cat}/K_m}\) is close to \(10^9 M^{-1}s^{-1}\), we can say the enzyme is working with 'catalytic perfection'. The ultimate speed limit of an enzyme-catalyzed reaction is determined by the diffusion rate of enzyme and substrate molecules.

Step by step solution

01

Calculation of \(V_{\max }\)

Use the Michaelis-Menten equation \(v = \frac{{V_{max}[S]}}{{K_m + [S]}}\) where \(v\) is the rate of reaction, \([S]\) is the substrate concentration, \(V_{max}\) is the maximum enzymatic reaction rate, and \(K_m\) is the Michaelis constant. Given that when \([S] = 30\mu M\), \(v = 82.5\mu mol\cdot mL^{-1}\cdot sec^{-1}\) and \(K_m = 1.8 \times 10^{-5} M\). Solve for \(V_{max}\).
02

Calculation of \(k_{\text{cat}}\)

Use the relation \(k_{\text {cat}} = \frac{{V_{max}}}{{[E_{total}]}}\), where \([E_{total}]\) is the total enzyme concentration. Given that \([E_{total}] = 3\text { nanomol/mL}\). Solve for \(k_{\text{cat}}\).
03

Calculation of catalytic efficiency

Use the relation \(k_{\text{cat}/K_m}\), which is a measure of the catalytic efficiency of the enzyme, Here, \(k_{\text{cat}}\) is the catalytic constant obtained from step 2 and \(K_m\) is given as \(1.8 \times 10^{-5} M\).
04

Check for catalytic perfection

The limit for 'catalytic perfection' is often considered to be when \(k_{\text{cat}/K_m}\) approaches \(10^9 M^{-1}s^{-1}\), the maximum rate at which enzymes can encounter substrate in an aqueous solution. Compare the efficiency obtained in step 3 with this value.
05

Identify the factor determining the speed limit of an enzyme-catalyzed reaction

The ultimate speed limit of an enzyme-catalyzed reaction is determined by the rate at which the enzyme and substrate molecules diffuse together, i.e., the diffusion limit. This is the physical limit on the kinetic perfection of enzymes.

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Key Concepts

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

Understanding the Michaelis-Menten Equation
The Michaelis-Menten equation is a cornerstone of enzymology, describing the relationship between the rate of an enzymatic reaction and the concentration of substrate available. At its core, this mathematical model helps us understand how enzymes work to catalyze reactions.

When an enzyme binds to its substrate, it forms an enzyme-substrate complex, which can then be converted into the product of the reaction. The Michaelis-Menten equation, which is expressed as \(v = \frac{V_{max}[S]}{K_m + [S]}\), reflects how the rate of the reaction (\(v\)) changes with increasing substrate concentration (\([S]\)). Here, \(V_{max}\) represents the maximum rate achieved by the system at maximum (saturating) substrate concentration, and \(K_m\) is the substrate concentration at which the reaction rate is half of \(V_{max}\).

This model assumes the formation of the enzyme-substrate complex is at a steady state and that the concentration of this complex does not change over time. The Michaelis-Menten equation is particularly useful in determining key characteristics of an enzyme, such as its maximum velocity and its affinity for the substrate, which are critical for understanding how efficiently an enzyme functions in catalysis. The equation is pivotal in designing experiments and interpreting kinetic data for various enzymes.
Exploring Catalytic Efficiency
Catalytic efficiency is an expression of how adept an enzyme is at converting a substrate into product. It's defined by the ratio of the catalytic constant, \(k_{cat}\), to the Michaelis constant, \(K_m\), and denoted as \(k_{cat}/K_m\).

The \(k_{cat}\) value represents the number of substrate molecules converted into product per enzyme molecule per unit of time, assuming the enzyme is fully saturated with the substrate. Essentially, this is a measure of an enzyme's productivity. The higher the \(k_{cat}\), the more product an enzyme can generate in a given time frame. On the other hand, \(K_m\) reflects the affinity of the enzyme for its substrate - a low \(K_m\) indicates high affinity, meaning the enzyme effectively binds substrate at low concentrations.

Combine these two concepts, and you get a measure of catalytic efficiency. A high \(k_{cat}/K_m\) value suggests a highly efficient enzyme, as it can not only process a large amount of substrate quickly (high \(k_{cat}\)) but also does so effectively at low substrate concentrations (low \(K_m\)). This ratio is key in understanding enzyme kinetics and is often used to compare the performance of different enzymes.
The Role of Triose Phosphate Isomerase in Enzyme Kinetics
Triose phosphate isomerase is a highly efficient enzyme that plays a crucial role in the glycolytic pathway, allowing the interconversion of glyceraldehyde-3-phosphate to dihydroxyacetone phosphate. This reaction is pivotal because it ensures a continuous flux of substrates through the pathway, contributing to efficient energy production in cells.

Based on kinetic measurements, triose phosphate isomerase has a remarkably high catalytic efficiency, approaching what is often termed 'catalytic perfection'. This enzyme operates at speeds close to the diffusion limit, meaning it catalyzes the reaction as fast as it encounters its substrate. The comparison of its observed \(k_{cat}/K_m\) ratio with the theoretical maximum is an example of how enzymes can be finely tuned to their biological roles.

In education and research, triose phosphate isomerase serves as a model enzyme due to its high efficiency and critical role in a fundamental biological process. By studying its kinetics, students and scientists gain insights into enzyme mechanism and evolution.
The Diffusion Limit of Enzyme Kinetics
An enzyme's efficiency is not limitless. There's a ceiling to how fast an enzyme can catalyze a reaction, and this upper boundary is known as the diffusion limit. This limit represents the rate at which enzyme and substrate molecules collide in the solution, influenced by their concentration and how quickly they can move towards each other in the solvent.

The diffusion limit often defines the theoretical maximum rate of an enzyme-catalyzed reaction. Because enzymes work by binding substrates that diffuse to them, the collision frequency between enzyme and substrate molecules constrains the ultimate rate at which the reaction can occur, regardless of how the enzyme might otherwise be optimized.

The concept of the diffusion limit helps explain why some enzymes, like triose phosphate isomerase, are so efficient that they operate near this limit - effectively catalyzing every collision with a substrate into a reaction. Such enzymes are usually referred to as being kinetically 'perfect' or near 'catalytic perfection' because their performance is as good as it can be, given the physical constraints of their environment.

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Most popular questions from this chapter

Liver alcohol dehydrogenase (ADH) is relatively nonspecific and will oxidize ethanol or other alcohols, including methanol. Methanol oxidation yields formaldehyde, which is quite toxic, causing, among other things, blindness. Mistaking it for the cheap wine he usually prefers, my dog Clancy ingested about \(50 \mathrm{mL}\) of windshield washer fluid (a solution \(50 \%\) in methanol). Knowing that methanol would be excreted eventually by Clancy's kidneys if its oxidation could be blocked, and realizing that, in terms of methanol oxidation by ADH, ethanol would act as a competitive inhibitor, I decided to offer Clancy some wine. How much of Clancy's favorite vintage \((12 \% \text { ethanol })\) must he consume in order to lower the activity of his ADH on methanol to \(5 \%\) of its normal value if the \(K_{m}\) values of canine ADH for ethanol and methanol are 1 millimolar and 10 millimolar, respectively? (The \(K_{1}\) for ethanol in its role as competitive inhibitor of methanol oxidation by ADH is the same as its \(K_{m \cdot}\) ) Both the methanol and ethanol will quickly distribute throughout Clancy's body fluids, which amount to about 15 L. Assume the densities of \(50 \%\) methanol and the wine are both \(0.9 \mathrm{g} / \mathrm{mL}\).

Acetylcholinesterase catalyzes the hydrolysis of the neurotransmitter acetylcholine: Acetylcholine \(+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) acetate \(+\) choline The \(K_{m}\) of acetylcholinesterase for its substrate acetylcholine is \(9 \times 10^{-5} M .\) In a reaction mixture containing 5 nanomoles/mL of acetylcholinesterase and \(150 \mu M\) acetylcholine, a velocity \(v_{\mathrm{o}}=\) \(40 \mu \mathrm{mol} / \mathrm{mL} \cdot\) sec was observed for the acetylcholinesterase reaction. a. Calculate \(V_{\max }\) for this amount of enzyme. b. Calculate \(k_{\text {cat }}\) for acetylcholinesterase. c. Calculate the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for acetylcholinesterase. d. Does acetylcholinesterase approach "catalytic perfection"?

Equation 13.9 presents the simple Michaelis-Menten situation where the reaction is considered to be irreversible ([P] is negligible). Many enzymatic reactions are reversible, and \(P\) does accumulate. a. Derive an equation for \(v\), the rate of the enzyme-catalyzed reaction \(\mathrm{S} \rightarrow \mathrm{P}\) in terms of a modified Michaelis-Menten model that incorporates the reverse reaction that will occur in the presence of product, \(\mathbf{P}\) b. Solve this modified Michaelis-Menten equation for the special situation when \(v=0\) (that is, \(S \rightleftharpoons P\) is at equilibrium, or in other words, \(\left.K_{\mathrm{eq}}=[\mathrm{P}] /[\mathrm{S}]\right)\) (J. B. S. Haldane first described this reversible Michaelis-Menten modification, and his expression for \(K_{\mathrm{eq}}\) in terms of the modified M-M equation is known as the Haldane relationship.)

The citric acid cycle enzyme fumarase catalyzes the conversion of fumarate to form malate. $$\text { Fumarate }+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { malate }$$ The turnover number, \(k_{\mathrm{cat}},\) for fumarase is \(800 / \mathrm{sec} .\) The \(K_{m}\) of fumarase for its substrate fumarate is \(5 \mu M\) a. In an experiment using 2 nanomole/L of fumarase, what is \(V_{\max } ?\) b. The cellular concentration of fumarate is \(47.5 \mu M .\) What is \(v\) when [fumarate] \(=47.5 \mu M ?\) c. What is the catalytic efficiency of fumarase? d. Does fumarase approach "catalytic perfection"?

Enzyme A follows simple Michaelis-Menten kinetics. a. The \(K_{m}\) of enzyme A for its substrate \(\mathrm{S}\) is \(K_{m}^{\mathrm{S}}=1 \mathrm{m} M .\) Enzyme \(\mathrm{A}\) also acts on substrate \(\mathrm{T}\) and its \(K_{m}^{\mathrm{T}}=10 \mathrm{m} M .\) Is \(\mathrm{S}\) or \(\mathrm{T}\) the preferred substrate for enzyme A? b. The rate constant \(k_{2}\) with substrate \(\mathrm{S}\) is \(2 \times 10^{4} \mathrm{sec}^{-1}\); with substrate \(\mathrm{T}, k_{2}=4 \times 10^{5} \mathrm{sec}^{-1} .\) Does enzyme A use substrate S or substrate T with greater catalytic efficiency?

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