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"?

Short Answer

Expert verified
The actual values will depend upon the output of previous calculations. Following the calculation process, you will be able to determine Vmax, kcat, and catalytic efficiency values for the given enzyme and substrate concentration. To assess catalytic perfection, compare the enzyme's catalytic efficiency with the diffusion limit, which is approximately \(10^9 M^{-1}*sec^{-1}\). If the enzyme's catalytic efficiency approaches or exceeds this value, it can be considered 'catalytically perfect'.

Step by step solution

01

Calculate the maximum reaction velocity (Vmax)

According to the Michaelis-Menten equation, \(Vmax = Vo((Km + [S]) / [S])\), where Vo is the initial velocity, Km is the Michaelis constant, and [S] is the substrate concentration. Inserting the given values, we get: \(Vmax = 40 ((9 *10^{-5} + 150 * 10^{-6}) / 150 * 10^{-6}) \mu mol/ mL * sec\)
02

Calculate the turnover number (kcat)

kcat is the ratio of maximum reaction velocity to enzyme concentration. So, \(kcat = Vmax / [E]\). We calculated Vmax in the previous step, and [E] is given as 5 nmoles/mL or 5 * 10^-6 moles/mL. Thus, \(kcat = (Vmax from Step 1) / (5 * 10^{-6}) sec^{-1}\)
03

Calculate catalytic efficiency

Catalytic efficiency is the ratio of kcat to Km. So, \(k_{\mathrm{cat}} / K_{m} = (kcat from Step 2) / (9 * 10^{-5}) M^{-1} sec^{-1}\)
04

Evaluate Catalytic Perfection

An enzyme approaches 'catalytic perfection' when its catalytic efficiency is close to or above the diffusion limit, which is approximately \(10^9 M^{-1}*sec^{-1}\). Compare the catalytic efficiency obtained in Step 3 to this value.

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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 cornerstone of enzyme kinetics that bridges the behavior of enzymes and their substrates. Enzymes catalyze biochemical reactions by converting reactants, known as substrates, into products. The Michaelis-Menten equation describes how the reaction rate (\(v_0\)), also referred to as velocity, depends on the concentration of the substrate ([S]) and the enzyme's affinity for the substrate.

It is mathematically represented as:\[\begin{equation}\ v_0 = \frac{V_{max} [S]}{K_m + [S]}\end{equation}\]
, Where:
  • [V_max] is the max velocity which occurs when all enzyme active sites are saturated with substrate.
  • [K_m] is the Michaelis constant, indicative of the substrate concentration at which the velocity is half of [V_max].
In the problem involving acetylcholinesterase, this equation helps calculate [V_{max}] when the enzyme's initial velocity [v_0] and the substrate concentration are known. Remember, a low [K_m] value highlights a high affinity between enzyme and substrate, while a high [V_max] implies an efficient catalytic process at saturating substrate levels.

Understanding this relationship not only allows for the calculation of the maximum speed of the enzymatic reactions but also offers insights into how enzymes operate under different concentrations of substrate.
Catalytic Efficiency
Catalytic efficiency is a measure of how effectively an enzyme converts a substrate into a product. It combines the concepts of enzyme affinity and turnover by relating [k_{cat}], or turnover number, to the [K_m] value. The higher the catalytic efficiency, the better the enzyme performs.

\[\begin{equation}\text{Catalytic efficiency} = \frac{k_{cat}}{K_m}\end{equation}\]
It is important to understand that this ratio emphasizes both the enzyme's ability to work quickly (as indicated by [k_{cat}]) and its effectiveness at lower substrate concentrations (as indicated by [K_m]). In our acetylcholinesterase example, we calculate this efficiency to gauge how well the enzyme functions with the neurotransmitter acetylcholine. The importance of catalytic efficiency lies in its ability to predict reaction rates in living organisms where enzymes often work under substrate concentrations that are well below their [K_m] values.
Turnover Number ([k_{cat}])
The turnover number, or [k_{cat}], is a vital kinetic parameter often used to describe the number of substrate molecules converted to product per enzyme molecule per unit of time when the enzyme is fully saturated with the substrate.

\[\begin{equation}\ k_{cat} = \frac{V_{max}}{[E]}\end{equation}\]
Here, [V_{max}] is the reaction's maximum velocity, and [E] is the enzyme concentration. This value helps in determining the efficiency of an enzyme by giving a clear picture of its catalytic power without the influence of enzyme concentration or substrate availability.

For acetylcholinesterase, we calculate [k_{cat}] to understand its proficiency in breaking down acetylcholine. The larger the turnover number, the more molecules of substrate an enzyme can process in a second, which is characteristic of highly efficient enzymes.
Enzyme-Substrate Complex
The enzyme-substrate complex is an intermediate formed during the catalytic action of an enzyme. It represents the physical binding of the enzyme to its substrate before the substrate transforms into the product. The stability and formation of this complex are critical for the enzyme's catalytic action and are central to understanding enzyme kinetics.

The affinity between an enzyme and its substrate can be inferred from the [K_m] value, as it represents the substrate concentration at which the rate of formation of the enzyme-substrate complex is equal to the rate of its breakdown.

Through the study of acetylcholinesterase and its interaction with acetylcholine, we gain insights into how enzymes bind to their specific substrates to facilitate a chemical reaction. The tighter the binding (lower [K_m]), the more likely the enzyme will catalyze the reaction at lower substrate concentrations, indicating high affinity and efficiency.

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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 \(0.9 \mathrm{g} / \mathrm{mL}\).

Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results: \(k_{1}=2 \times 10^{8} M^{-1} \sec ^{-1}\) \(k_{-1}=1 \times 10^{3} \sec ^{-1}\) \(k_{2}=5 \times 10^{3} \mathrm{sec}^{-1}\) a. What is \(K_{\mathrm{S}},\) the dissociation constant for the enzyme- substrate complex? b. What is \(K_{m},\) the Michaelis constant for this enzyme? c. What is \(k_{\text {cat }}\) (the turnover number) for this enzyme? d. What is the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for this enzyme? e. Does this enzyme approach "kinetic perfection"? (That is, does \(k_{\mathrm{cat}} / K_{m}\) approach the diffusion-controlled rate of enzyme association with substrate? f. If a kinetic measurement was made using 2 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give \(v=0.75 V_{\max } ?\) h. If a kinetic measurement was made using 4 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? What would \(K_{m}\) equal under these conditions?

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"?

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?

The following kinetic data were obtained for an enzyme in the absence of any inhibitor \((1),\) and in the presence of two different inhibitors (2) and (3) at \(5 \mathrm{m} M\) concentration. Assume \(\left[\mathrm{E}_{T}\right]\) is the same in each experiment. $$\begin{array}{cccc} & (1) & (2) & (3) \\ {[\mathrm{S}]} & v(\mu \mathrm{mol} / & v(\mu \mathrm{mol} /) & v(\mu \mathrm{mol} /) \\ (\mathrm{m} M) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec} \\ 1 & 12 & 4.3 & 5.5 \\ 2 & 20 & 8 & 9 \\ 4 & 29 & 14 & 13 \\ 8 & 35 & 21 & 16 \\ 12 & 40 & 26 & 18 \end{array}$$ Graph these data as Lineweaver-Burk plots and use your graph to find answers to a. and b. a. Determine \(V_{\max }\) and \(K_{m}\) for the enzyme. b. Determine the type of inhibition and the \(K_{1}\) for each inhibitor.

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