Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
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Chapter 8, Problem 17P
Initial rate data for an enzyme that obeys Michaelis-Menten kinetics are shown in the following table. When the enzyme concentration is 3 nmol ml-1, a Lineweaver-Burk plot of this data gives a line with a y-intercept of 0.00426 (µmol- ml s).
- Calculate kcat for the reaction.
- Calculate KM for the enzyme.
- When the reactions in part (b) are repeated in the presence of 12 µM of an uncompetitive inhibitor, the y-intercept of the Lineweaver-Burk plot is 0.352 (µmol-1 ml s). Calculate K’I for the inhibitor.
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Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
From a kinetics experiment, kcat was determined to be 295sec-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (µM∙min-1) for the enzyme and catalytic efficiency ( in M-1sec-1) for the enzyme. The Km for the enzyme was determined to be 2.55 x 10-2M.
An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reaction
velocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ?
Express your answer in micromolar to three significant figures.
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Chapter 8 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Ch. 8 - Prob. 1PCh. 8 - The enzyme urease catalyzes the hydrolysis of urea...Ch. 8 - An enzyme contains an active site aspartic acid...Ch. 8 - The folding and unfolding rate constants for a...Ch. 8 - In some reactions, in which a protein molecule is...Ch. 8 - Would you expect an “enzyme” designed to bind to...Ch. 8 - The initial rate for an enzyme-catalyzed reaction...Ch. 8 - a. If the total enzyme concentration in Problem 7...Ch. 8 - Prob. 9PCh. 8 - Prob. 10P
Ch. 8 - The following data describe the catalysis of...Ch. 8 - At 37 oC, the serine protease subtilisin has kcat...Ch. 8 - The accompanying figure shows three...Ch. 8 - The steady-state kinetics of an enzyme are studied...Ch. 8 - The same enzyme as in Problem 14 is studied in the...Ch. 8 - Enalapril is an anti-hypertension “pro-drug"...Ch. 8 - Initial rate data for an enzyme that obeys...Ch. 8 - Prob. 18PCh. 8 - Suggest the effects of each of the following...Ch. 8 - The inhibitory effect of an uncompetitive...Ch. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - In kinetics experiments, the hydrolysis of the...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.arrow_forwardThe turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?arrow_forwardLineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?arrow_forward
- An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar to three significant figures..arrow_forwardFor an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K Marrow_forwardAlthough graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:arrow_forward
- An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?arrow_forwardThe Lineweaver-Burk plot, which illustrates the reciprocal of the reaction rate (1/v) versus the reciprocal of the substrate concentration (1/[S]), is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, Vmax, and the Michaelis constant, Km, which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants Vmax and Km. What term is represented by C? Linewegver-Burk Pt 1/Vmax O A. В. -1/Km Km/Vmax C.arrow_forwardMake a v vs. [S] plot for an enzyme that obeys Michaelis-Menten kinetics with a Vmax of 150 mM/min and Km of 1.5 mM.arrow_forward
- columns) and absence (second column = control) of enzyme inhibitor. Both inhibitors were added in each reaction at a concentration of 2 mM. The enzyme concentration was similar in all and was approximately 0.001 Им. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? Find, KI, for the inhibitor binding to the enzyme, for experiments (2) and (3). d. Calculate the reaction Kcat for the Control in experiment (1). Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and Ky for the enzyme a. C. e. [S] (mM) (1) V. {(umol/(ml.s)] (2) V. [[umol/(ml.s)]| 4.4 (3) V. umol/{ml.s] 6.6 11.4 76 14.6 8.6 26.6 16.4 29 8 17.8 24.6 28.2 16 45.8 24 60 40.8arrow_forwardAn enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :arrow_forwardA bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 umol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM? (Please take note of the units)arrow_forward
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