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.

Short Answer

Expert verified
Vmax and Km will be determined from the Lineweaver-Burk plot constructed for experiment (1) by analyzing the y and x intercepts respectively. The type of inhibition for enzyme inhibitors will be identified by plotting data from experiments (2) and (3) onto the same plot and observing the intersection points. Lastly, calculations will be made for Ki, depending on the type of inhibition.

Step by step solution

01

Construct Lineweaver-Burk Plot

The first step will involve plotting the Lineweaver-Burk plot for the given data in the absence of any inhibitor. It's a double reciprocal plot with 1/v (where v is the reaction velocity) on the y-axis and 1/[S] (where [S] is the substrate concentration) on the x-axis. Plot the points for 1/v against 1/[S] for each of the substrate concentration given in the table for experiment (1). Draw a straight line through these points.
02

Find Vmax and Km

Intercepts on the y and x-axis on the Lineweaver-Burk plot represent 1/Vmax and -1/Km respectively. Estimate the values from the plot for experiment (1).
03

Construct Lineweaver-Burk Plot with inhibitors

Repeat step 1 for the data given for experiments (2) and (3) with inhibitors. Draw the lines onto the same graph.
04

Determine the type of inhibition

The type of inhibition can be determined from the comparison between the Lineweaver-Burk plots of uninhibited reaction and reactions with inhibitors. If the lines intersect at the y-axis, it suggests competitive inhibition. If the lines intersect at the x-axis or above, it suggests noncompetitive inhibition. If the lines have the same slope, it suggests uncompetitive inhibition. Make a note of the type of inhibition for each inhibitor.
05

Determine Ki

The Ki (Inhibition constant) for each inhibitor can be calculated from the equations depending on the type of inhibition. For competitive inhibition, Ki=Km(1-[I]/Km'), for uncompetitive inhibition, Ki= Km'/([I]-Km'), and for noncompetitive inhibition, Ki= Km'([I]/Km'-1). Where Km' is the Km in the presence of the inhibitor and [I] is the concentration of the inhibitor.

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is the study of how enzymes bind to substrates and turn them into products. The kinetics of an enzyme reaction involves several important parameters that characterize the enzyme's efficiency and behavior.

One key plot used in enzyme kinetics is the Lineweaver-Burk plot, also known as the double-reciprocal plot. This graphical representation plots the reciprocal of the reaction velocity (1/v) against the reciprocal of the substrate concentration (1/[S]). It is particularly useful for analyzing enzyme inhibition and for determining two fundamental constants of enzyme kinetics: the maximum velocity (( V_{max} )) and the Michaelis constant (( K_{m} )).

Through such analysis, researchers can obtain valuable insights into how enzymes work, which can inform drug development, among other applications.
Vmax (Maximum Velocity)
The ( V_{max} ) of an enzyme-catalyzed reaction represents the maximum rate at which the enzyme can convert substrate into product. This occurs when all enzyme active sites are saturated with substrate.

On the Lineweaver-Burk plot, ( V_{max} ) can be determined from the y-intercept, as the plot is of 1/v against 1/[S]. The y-intercept is equal to 1/( V_{max} ), and thus, by taking the reciprocal, one can find the actual ( V_{max} ) value. This parameter is crucial as it provides a ceiling for how fast a reaction can proceed under specific conditions.
Km (Michaelis Constant)
The Michaelis constant, denoted as ( K_{m} ), is another fundamental parameter in enzyme kinetics. It represents the substrate concentration at which the reaction velocity is half of its maximum value (( V_{max} )).

( K_{m} ) is a measure of the affinity of an enzyme for its substrate—low values of ( K_{m} ) indicate high affinity. In the Lineweaver-Burk plot, ( K_{m} ) can be determined from the x-intercept which corresponds to -1/( K_{m} ), and just like with ( V_{max} ), the actual ( K_{m} ) value can be found by taking the negative reciprocal of this intercept.
Enzyme Inhibition
Enzyme inhibitors are molecules that reduce the activity of the enzyme, slowing down the reaction rate. Inhibitors can bind to enzymes in different ways, leading to various types of inhibition: competitive, noncompetitive, and uncompetitive.

Competitive inhibitors resemble the substrate and bind to the active site, preventing the substrate from binding. Noncompetitive inhibitors bind to an enzyme at a site other than the active site and can bind regardless of whether the substrate has already bound. Uncompetitive inhibitors bind only when the enzyme-substrate complex has formed.

The Lineweaver-Burk plot is invaluable for distinguishing these inhibition types, as each produces characteristic changes in the plot's lines concerning the uninhibited reaction.
Ki (Inhibition Constant)
The inhibition constant, or ( K_{i} ), measures the potency of an inhibitor, indicating how tightly an inhibitor binds to an enzyme. A low ( K_{i} ) means a strong inhibitor that binds effectively even at low concentrations.

Depending on the type of inhibition, the ( K_{i} ) can be calculated from changes in ( K_{m} ) or ( V_{max} ). For example, for competitive inhibition, ( K_{i} ) can be derived using the modified Michaelis constant in the presence of an inhibitor (( K_{m}' )). The Lineweaver-Burk plot helps to visualize and calculate ( K_{i} ) by examining how the presence of an inhibitor alters the intercepts and slopes of the graph.

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Most popular questions from this chapter

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"?

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.)

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"?

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?

According to the Michaelis-Menten equation, what is the \(v / V_{\max }\) ratio when \([\mathrm{S}]=4 K_{\text {w? }} ?\)

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