Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 8, Problem 42P
Interpretation Introduction
Interpretation:
The filled table which indicates precisely the effect of
Concept introduction:
The Michaelis-Menten equation relates the concentration of substrate with the velocity of the reaction. A constant which expresses the substrate’s concentration if the velocity of reaction equals to the half of the maximum velocity of the reaction is known as Michaelis-Menten constant. It is denoted by
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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?
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?
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?
For 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 M
Chapter 8 Solutions
Biochemistry
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- Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the Km of the reaction represented by Line (B).arrow_forwarda. 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). e. Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and Ky for the enzyme. c. (1) V. [(umol/(ml.s)] 7.6 (2) V- Τ (μmol/ (ml.s)] [S] (mM) (3) V. [(umol/(ml.s)] 6.6 2 4 14.6 26.6 45.8 4.4 8.6 16.4 29.8 11.4 17.8 24.6 28.2 16 24 60 40.8arrow_forwardBased on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the catalytic efficiency of the reaction represented by Line (A).arrow_forward
- Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Line (B) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.arrow_forwardBased on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the Vmax of the reaction represented by Line (C). Show all mathematical work, please.arrow_forwardBased on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. QUESTION: Line (A) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.arrow_forward
- Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Which of the following statements is true? Select any/all answers that apply. A. Both types of inhibitor mediate a slope effect on the Lineweaver-Burke plot. B. Both types of inhibitor decrease the apparent Vmax for this enzyme-catalyzed reaction. C. Both types of inhibitor alter the apparent Km of this enzyme-catalyzed reaction. D. Lines (A) and (C) share the same X-intercept, indicating that the noncompetitive inhibitor decreases the apparent Km of this enzyme-catalyzed reaction. E. Lines (A) and (C)…arrow_forwardConsider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?arrow_forwardmolecule A Plot of velocity versus substrate B Lineweaver-Burk plot 1/v Km 1 Vmax (S) Vmx 1 V max 1/2Vmax 1/Vmax -1/Km Km [S] 1/[S] fppt.com molecule Exercise The following data describe an enzyme-catalyzed reaction. Plot these results using the Lineweaver-Burk method, and determine values for KM and Vinax- The symbol mM represents millimoles per liter; 1 mM = 1 × 10 3 mol L. (The concentration of the enzyme is the same in all experiments.) Velocity (mM sec-) Substrate Concentration (тм) 2.5 0.024 5.0 0.036 10.0 0.053 15.0 0.060 20.0 0.061 fppt.comarrow_forward
- An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min−110 µmol⋅min−1 when the substrate concentration is five times greater than the ?M.KM. What is the ?maxVmax of this enzyme?arrow_forwardThe table below lists experimental conditions that can be applied to a reaction catalyzed by a hypothetical Michaelis–Menten enzyme. For each experimental condition described, complete the table to indicate as precisely as possible the effect (no change, half as large doubles, increases, decreases) on the maximal velocity, Vmax, and Michaelis constant, KM, of the hypothetical enzyme.arrow_forwardYou have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=arrow_forward
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