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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Question
Chapter 8, Problem 36P
Interpretation Introduction
Interpretation:
The value 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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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
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?
Vmax [S]
Km + [S]
Vo =
Eadie-Hofstee plot
Lineweaver-Burk (L-B) plot
v=Vm-Km
[S]
Km 1
Vm Vm [S]
1
The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation
becomes
Vo-Vmax(S)/(Km+a'[S))
Put the Minchaelis-Menten equation for uncompetitive inhibition in the Lineweaver-Burk format. The slope of this plot will be
Chapter 8 Solutions
Biochemistry
Ch. 8 - Prob. 1PCh. 8 - Prob. 2PCh. 8 - Prob. 3PCh. 8 - Prob. 4PCh. 8 - Prob. 5PCh. 8 - Prob. 6PCh. 8 - Prob. 7PCh. 8 - Prob. 8PCh. 8 - Prob. 9PCh. 8 - Prob. 10P
Ch. 8 - Prob. 11PCh. 8 - Prob. 12PCh. 8 - Prob. 13PCh. 8 - Prob. 14PCh. 8 - Prob. 15PCh. 8 - Prob. 16PCh. 8 - Prob. 17PCh. 8 - Prob. 18PCh. 8 - Prob. 19PCh. 8 - Prob. 20PCh. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - Prob. 24PCh. 8 - Prob. 25PCh. 8 - Prob. 26PCh. 8 - Prob. 27PCh. 8 - Prob. 28PCh. 8 - Prob. 29PCh. 8 - Prob. 30PCh. 8 - Prob. 31PCh. 8 - Prob. 32PCh. 8 - Prob. 33PCh. 8 - Prob. 34PCh. 8 - Prob. 35PCh. 8 - Prob. 36PCh. 8 - Prob. 37PCh. 8 - Prob. 38PCh. 8 - Prob. 39PCh. 8 - Prob. 40PCh. 8 - Prob. 41PCh. 8 - Prob. 42PCh. 8 - Prob. 43PCh. 8 - Prob. 44PCh. 8 - Prob. 45PCh. 8 - Prob. 46PCh. 8 - Prob. 47PCh. 8 - Prob. 48PCh. 8 - Prob. 49P
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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?arrow_forwardReaction rate 0.35 0.30 0.25 0.20 0.15- 0.10 0.05 0.00 0 1000 2000 Enzyme total is 5mM 3000 4000 1. Substrate concentration 1. The above graph is an enzyme reaction, what is the vMax, KM, and Kcat of this enzymatic reaction? What does vMax, kM and Kcat mean, explain? 2. What are the three forms of regulation that follow the Michaelis Menten kinetics? What are the mechanisms by which these inhibitors can regulate the enzyme?arrow_forwardVmax [S] Vo = Km+ [S] Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For competitive inhibition The Michaelis-Menten equation becomes Vo=Vmax[S]/(aKm+[S]) Put the Michaelis-Menten equation for competitive inhibition in the Eadie-Hofstee format. The Y-intercept of this plot will bearrow_forward
- not true about the Michaelis-Menten equation? The equation that gives the rate, v, of an the substrate concentration [S] is the Michaelis-Menten equation = Vmax[S]/(Km + [S]), where V, enzyme-catalyzed reaction for all values of max and Km are constants. Which of the following is a) for [S] << Km, V = Vmax applies to most enzymes, but allosteric enzymes have different kinetics when [S] = Km, then v = Vmax/2 gives the rate when the enzyme concentration, temperature, pH, and ionic strength are constant for very high values of [S], v approaches Vmax e) Which is correct about the constant Km in the Michaelis-Menten equation? also called the catalytic constant or turnover number equal to the number of product molecules produced per unit time when the enzyme is saturated with substrate it is the constant in the first order rate equation v = k[A] it is the constant in the second order rate equation v = equal to the substrate concentration at which the velocity or rate of a reaction is ½ the…arrow_forwardSubstrate KM (M) N-Acetylvaline ethyl ether 8.8 X 10 -2 N-Acetyltyrosine ethyl ether 6.6 X 10-4 Which substrate has the higher apparent affinity for the enzyme? Explain. Which substrate is likely to give a higher value for Vmax?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-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions? 2.75 min^-1? 3,300,000 min^-1? 330,000 s^-1? 19,800,000 min^-1? 5,500 s^-1?arrow_forwardWhat is the catalytic efficiency of Catalase ? Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Carbonic anhydrase Substrate CO2 HCO3 KM (M) 1.2 x 10-2 2.6 x 10-2 Kcat (s-1) 1.0 x 106 4.0 x 105 Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 O A. 4 x 108 M-s-1 O B. 4 x 108 M-1.s-1 OC25x 10-9 M-s1 D. 2.5 x 102 M-1.s-1 OE 1.0 x 107 s1arrow_forwardInitial rate data for an enzyme that obeys Michaelis–Menten kinetics areshown in the following table. When the enzyme concentration is 3 nmolml-1, a Lineweaver–Burk plot of this data gives a line with a y-intercept of0.00426 (μmol-1 ml s). (a) Calculate kcat for the reaction.(b) Calculate KM for the enzyme.(c) When the reactions in part (b) are repeated in the presence of 12 μM ofan uncompetitive inhibitor, the y-intercept of the Lineweaver–Burk plotis 0.352 (μmol-1 ml s). Calculate K′I for this inhibitor.arrow_forward
- Velocity (mmol/minute) [S], (mM) No inhibitor Inhibitor 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8 The kinetics of an enzyme are measured as a function of substrate in the presence and the in absence of 2mM inhibitor (I). What are the values of Vmax and KM in the absence of inhibitor? In its presence? In its presence? What is the type of inhibition?arrow_forward1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…arrow_forward6-25 substrate-band enzyme concentrations. The the turnover number is equal to umax- b) V=Umax •57(Km+S) anstont For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, V (as a percentage of Vmax), observed at the following values? a) [S] = KM C) d) e) [S] = 0.5KM [S] = = 0.1KM [S] = 2KM [S] = 10KM w reactores -maximumrate of reaction boteles conc. Would you expect the structure of a competitive inhibitor of a given enzyme to be similar to that of its substrate?arrow_forward
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