not true about the Michaelis-Menten equation?The equation that gives the rate, v, of anthe substrate concentration [S] is the Michaelis-Menten equation= Vmax[S]/(Km + [S]), where V,enzyme-catalyzed reaction for all values ofmax and Km are constants. Which of the following isa)for [S] << Km, V = Vmaxapplies to most enzymes, but allosteric enzymes have different kineticswhen [S] = Km, then v =Vmax/2gives the rate when the enzyme concentration, temperature, pH, and ionicstrength are constantfor very high values of [S], v approaches Vmaxe)Which is correct about the constant Km in the Michaelis-Menten equation?also called the catalytic constant or turnover numberequal to the number of product molecules produced per unit time when theenzyme is saturated with substrateit 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 is½ the maximum rate (2 Vmax)a)b)c)d)k[A][B]

Enzymes are biocatalysts which accelerate the rate of a biochemical reaction. Enzymes catalyze a…

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Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter26: Synthesis And Degradation Of Nucleotides

Section: Chapter Questions

Problem 13P

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The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.
Initial 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.
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?
Many enzymes obey simple Michaelis–Mentenkinetics, which are summarized by the equationrate = Vmax [S]/([S] + Km)where Vmax = maximum velocity, [S] = concentration ofsubstrate, and Km = the Michaelis constant.It is instructive to plug a few values of [S] into theequation to see how rate is affected. What are the rates for[S] equal to zero, equal to Km, and equal to infinite concen-tration?
multiple choice, choose the correct answer Under which of the following conditions would an enzyme fail to be accurately described by the M-M equation:1. kcat >> k1 or k-12. In the presence of a negative allosteric effector3. At low concentrations, the reaction is first order with respect to substrate concentration4. At high concentrations the reaction is zero order with respect to substrate
Compare and contrast Bound Fraction equation in ligand binding and Michaelis-Menten equation in enzyme kinetics, including their double-reciprocal forms. Discuss what Km is important for and what Vmax (or kcat) is important for? Under what (substrate) conditions is Km more important than Vmax, and under what (substrate) conditions is Vmax more important than Km? Based on the discussions in question 2, explain what type of inhibitors works best under (a) high substrate concentration and (b) low substrate concentration.
Effects of Changing Metabolite Concentrations on Glycolysis In an erythrocyte undergoing glycolysis what would be the effect of a sudden increase in the concentration of a. AÎ¤P? b. AMP? c. fructose-1.6-bisphosphate? d. fructose-2, 6-bisphosphate? e. citrate? f. glucose-6-phospthate?
Regulation of Glutamine Synthetase by Covalent Modification Suppose at certain specific metabolite concentrations in vivo the cyclic cascade regulating E. coli glutamine synthetase has reached a dynamic equilibrium where the average state of GS adenylylation is poised atn=6. Predict what change in nwill occur if: [ ATP ] increases, PIIA/PIID increases, [ -KG ]/[ Gln ] increases, [ Pi ] decreases.
Modeling the Regulation of AcetyI-CoA Carboxylase Based on the information presented in the text and in Figures 24.4 and 24.5, suggest a model for the regulation of acetyl-CoA carboxylase. Consider the possible roles of subunit interactions, phosphorylation, and conformation changes in your model.
If an enzyme catalyzed reaction has a KM of 5mM and a Vmax of 60 nm/sec, the substrate concentration at 30 nM/sec is? Thank you.
Recall that phosphonacetyl L - aspartate (PALA) is a potent inhibitor of ATCase because it mimics the two physiological substrates. However, in the presence of substrates, low concentrations of this unreactive bisubstrate analog increase the reaction velocity. On the addition of PALA, the reaction rate increases until an average of three molecules of PALA are bound per molecule of enzyme. This maximal velocity is 17-fold greater than it is in the absence of PALA. The reaction rate then decreases to nearly zero on the addition of three more molecules of PALA per molecule of enzyme. Why do low concentrations of PALA activate ATCase?
The KMof the enzyme for the substrate adenosine is 3 × 10ꟷ5M. The product inosine acts as an inhibitor of the reaction, with an inhibition constant (KI, the dissociation constant for enzyme-inhibitor binding) of 3 × 10ꟷ4M. However, a transition state analog,Inhibits the reaction with KIof 1.5 × 10ꟷ13M. Explain why 1,6-dihydroinosine serves as a better inhibitor of adenosine deaminase than inosine. Elaborate on your answe

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