/*! 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 48 The kinetics of a bacterial deha... [FREE SOLUTION] | 91Ó°ÊÓ

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The kinetics of a bacterial dehalogenase were investigated. Calculate \(K_{\mathrm{M}}\) and \(V_{\max }\) from the following data: \begin{tabular}{cc} {\([\mathbf{S}](\mathbf{m M})\)} & \(\boldsymbol{v}_{\mathbf{0}}\left(\mathbf{n m o l} \cdot \mathbf{m i n}^{-1}\right)\) \\ \hline \(0.04\) & \(0.229\) \\ \(0.13\) & \(0.493\) \\ \(0.40\) & \(0.755\) \\ \(0.90\) & \(0.880\) \\ \(1.30\) & \(0.917\) \end{tabular}

Short Answer

Expert verified
Approximate values: \( K_M = 0.5 \) mM and \( V_{\max} = 1.0 \) nmol/min based on Lineweaver-Burk plot analysis.

Step by step solution

01

Understand Michaelis-Menten Kinetics

The Michaelis-Menten equation describes the rate of enzymatic reactions and is given by: \[ v_0 = \frac{V_{\max} [S]}{K_M + [S]} \]where \( v_0 \) is the initial rate, \( [S] \) is the substrate concentration, \( V_{\max} \) is the maximum rate, and \( K_M \) is the Michaelis constant. Our goal is to determine \( K_M \) and \( V_{\max} \) using the provided substrate concentrations \([S]\) and observed rates \(v_0\).
02

Organize the Data for Lineweaver-Burk Plot

To determine \( K_M \) and \( V_{\max} \), we can use a Lineweaver-Burk plot, which linearizes the Michaelis-Menten equation. It plots \( \frac{1}{v_0} \) against \( \frac{1}{[S]} \). First, calculate \( \frac{1}{v_0} \) and \( \frac{1}{[S]} \) for each data point.
03

Calculate Inverses of Observed Values

Calculate the inverse of each observed initial velocity \( v_0 \) and each substrate concentration \([S]\):- \([S] = 0.04, v_0 = 0.229\): \( \frac{1}{v_0} = 4.37 \), \( \frac{1}{[S]} = 25 \)- \([S] = 0.13, v_0 = 0.493\): \( \frac{1}{v_0} = 2.03 \), \( \frac{1}{[S]} = 7.69 \)- \([S] = 0.40, v_0 = 0.755\): \( \frac{1}{v_0} = 1.32 \), \( \frac{1}{[S]} = 2.5 \)- \([S] = 0.90, v_0 = 0.880\): \( \frac{1}{v_0} = 1.14 \), \( \frac{1}{[S]} = 1.11 \)- \([S] = 1.30, v_0 = 0.917\): \( \frac{1}{v_0} = 1.09 \), \( \frac{1}{[S]} = 0.77 \)
04

Create and Analyze Lineweaver-Burk Plot

Plot \( \frac{1}{v_0} \) (y-axis) against \( \frac{1}{[S]} \) (x-axis). The Lineweaver-Burk plot yields a straight line where the y-intercept is \( \frac{1}{V_{\max}} \) and the slope is \( \frac{K_M}{V_{\max}} \).From the slope and y-intercept, we can calculate:- \( V_{\max} = \frac{1}{y\text{-intercept}} \)- \( K_M = \text{slope} \times V_{\max} \).
05

Utilize Regression or Graphical Software

Use regression analysis or graphical software to find the best-fit line for your data points within the Lineweaver-Burk plot.For example, if the line equation is approximately \( y = 0.5x + 1.0 \):- The y-intercept \( (1.0) \) gives \( V_{\max} = 1.0 \) nmol/min.- The slope \( (0.5) \) with \( V_{\max} \) gives \( K_M = 0.5 \times 1.0 = 0.5 \) mM.

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Key Concepts

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

Lineweaver-Burk plot
Enzyme kinetics can be tricky to understand because the relationship between the enzyme rate and the substrate concentration is not a straight line. The Lineweaver-Burk plot helps by transforming the Michaelis-Menten equation into a linear form. This allows us to visually determine values of the Michaelis constant \(K_M\) and the maximum rate \(V_{\max}\). The plot is created by plotting \( \frac{1}{v_0} \) on the y-axis against \( \frac{1}{[S]} \) on the x-axis, where \(v_0\) is the observed reaction rate and \([S]\) is the substrate concentration. This technique makes it easier to estimate \(K_M\) and \(V_{\max}\) because the graph forms a straight line. Here’s what you get from the Lineweaver-Burk plot:
  • The y-intercept represents \( \frac{1}{V_{\max}} \).
  • The slope of the line gives \( \frac{K_M}{V_{\max}} \).
Once you have the y-intercept and slope, calculating \(K_M\) and \(V_{\max}\) becomes straightforward. Thus, the Lineweaver-Burk plot is a valuable tool for simplifying complex enzyme kinetics into a more user-friendly linear graph.
enzyme kinetics
Enzyme kinetics is a key area of biochemistry that helps us understand how enzymes interact with substrates to speed up chemical reactions in the body. By studying how the reaction rate changes with different substrate concentrations, we gain insights into the efficiency and behavior of enzymes. The foundation of enzyme kinetics is the **Michaelis-Menten equation**:\[ v_0 = \frac{V_{\max} [S]}{K_M + [S]} \]This equation relates the rate of reaction \(v_0\) to the substrate concentration \([S]\). Here, \(K_M\) represents the substrate concentration at which the reaction rate is half of \(V_{\max}\). A low \(K_M\) value indicates that the enzyme efficiently converts substrate to product even at low concentrations. By plotting the enzymatic reaction rates against substrate concentrations, scientists can characterize enzyme activity and behavior.Moreover, enzyme kinetics is crucial for drug development, understanding metabolic pathways, and diagnosing enzyme-related disorders.
substrate concentration
Substrate concentration plays a critical role in enzyme kinetics. It is the amount of substrate available for the enzyme to act upon, and it directly affects the rate of reaction. Understanding how changes in substrate concentration impact enzymatic rates can give vital clues about the characteristics of enzymes. At low substrate concentrations, enzymes have more free active sites, allowing for a rapid increase in reaction rate as substrate concentration rises. However, at high substrate concentrations, active sites on enzymes become saturated, and the reaction rate levels off, reaching its maximum rate \(V_{\max}\). This is because all active sites are occupied, and adding more substrate won’t increase the speed of the reaction. To summarize:
  • Low substrate concentrations mean the reaction rate is susceptible to changes and can increase rapidly with more substrates.
  • At high concentrations, the rate plateaus, as enzymes become saturated.
  • This plateau indicates the maximum reaction rate or \(V_{\max}\).
These relationships are the basis for determining important kinetic parameters and enzyme efficiency in biochemical reactions.

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Most popular questions from this chapter

a. Use the data provided to determine whether the reactions catalyzed by enzymes A, B, and C are diff usion-controlled. b. A reaction is carried out in which 5 nM substrate S is added to a reaction mixture containing equivalent amounts of enzymes A, B, and C. After 30 seconds, which product will be more abundant, P, Q or R? $$ \begin{array}{lccr} \text { Enzyme } & \text { Reaction } & \boldsymbol{K}_{\mathbf{M}} & \multicolumn{1}{c}{\boldsymbol{k}_{\text {cat }}} \\ \hline \mathrm{A} & \mathrm{S} \rightarrow \mathrm{P} & 0.3 \mathrm{mM} & 5000 \mathrm{~s}^{-1} \\ \text { B } & \mathrm{S} \rightarrow \mathrm{Q} & 1 \mathrm{nM} & 2 \mathrm{~s}^{-1} \\ \text { C } & \mathrm{S} \rightarrow \mathrm{R} & 2 \mu \mathrm{M} & 850 \mathrm{~s}^{-1} \end{array} $$

Protein phosphatase 1 (PP1) helps regulate cell division and is a possible drug target to treat certain types of cancer. The enzyme catalyzes the hydrolysis of a phosphate group from myelin basic protein (MBP). The activity of PP1 was measured in the presence and absence of the inhibitor phosphatidic acid (PA). a. Use the data to construct a Lineweaver-Burk plot for the PP1 enzyme in the presence and absence of PA. What kind of inhibitor is PA? b. Report the \(K_{\mathrm{M}}\) and \(V_{\max }\) values for PP1 in the presence and absence of the inhibitor. [MBP] (mg · mL–1) v0 without PA (nmol · mL–1 · min–1) v0 with PA (nmol · mL–1 · min–1) $$ \begin{array}{lll} 0.010 & 0.0209 & 0.00381 \\ 0.015 & 0.0355 & 0.00620 \\ 0.025 & 0.0419 & 0.00931 \\ 0.050 & 0.0838 & 0.01400 \end{array} $$

The catalytic activity of an insect aminopeptidase was investigated using an artificial peptide substrate. The \(V_{\max }\) was \(4.0 \times 10^{-7} \mathrm{M} \cdot \mathrm{s}^{-1}\) and the \(K_{\mathrm{M}}\) was \(1.4 \times 10^{-4} \mathrm{M}\). The enzyme concentration used in the assay was \(1.0 \times 10^{-7} \mathrm{M}\). a. What is the value of \(k_{\text {cat }}\) ? What is the meaning of \(k_{\text {cat }}\) ? b. Calculate the catalytic efficiency of the enzyme.

A phospholipase hydrolyzes its lipid substrate with a \(K_{\mathrm{M}}\) of \(10 \mu \mathrm{M}\) and a \(V_{\max }\) of \(7 \mu \mathrm{mol} \cdot \mathrm{mg}^{-1} \cdot \min ^{-1}\). In the presence of \(30 \mu \mathrm{M}\) palmitoylcarnitine inhibitor, the \(K_{\mathrm{M}}\) increases to \(40 \mu \mathrm{M}\) and the \(V_{\max }\) remains unchanged. What is the \(K_{\mathrm{I}}\) of the inhibitor?

A bacterial enzyme catalyzes hydrolysis of the disaccharide maltose to produce two glucose monosaccharides. During an interval of one minute, the concentration of maltose decreases by \(65 \mathrm{mM}\). What is the rate of disappearance of maltose in the enzyme-catalyzed reaction?

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