Consider this intermediate in the derivation of the Michaelis-Menten equation.[E] [S][ES|k-1 + kzkmAssume that k is negligible compared to the other rate constants.If the k is very small, it suggests that the enzyme has a Select an optionaffinity for its substrate, while if the if the km is very large, itsuggests that the enzyme has a Select an option.affinity for its substrate.Select an optionSubmitYou have used 0 of highSavlowmoderate

Km : The concentration of substrate at which enzyme achieves half Vmax-maximum velocity

Answered: Consider this intermediate in the…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter23: Fatty Acid Catabolism

Section: Chapter Questions

Problem 21P: Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved...

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Similar questions

Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved residues in slightly different positions can change the catalytic rate of reaction.
Using the ActiveModel for aldose reductase, describe the structure of the TIM barrel motif and the structure and location of the active site.
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
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.
The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for an enzyme-catalyzed, single-substrate reaction E + S ES → E + P. The model can be more readily understood when comparing three conditions: [S] > Km- Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Not true for any of these conditions Almost all active sites will [ES] is much lower than [Efree]. be filled. The rate is directly proportional to Increasing [Etotal] will increase [S]. Km: Adding more S will not increase [Efree] is equal to [ES]. the rate.
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
For an enzyme obeying the Michaelis-Menten equation with Km = 5 µM, kcat = 10 s-¹ and a total enzyme concentration of 1 nM, Calculate Vmax. Calculate the substrate concentration at which v = 0.1 Vmax Calculate the substrate concentration at which v = 0.9 Vmax What fraction of the enzyme is bound to substrate when v = 0.9 Vmax? Sketch a graph showing the Michaelis-Menten plot. Make sure you label the axes on your plot. Label on the graph: i) the point on the graph at which S=Km ii) Vmax At low substrate concentration, v = Keat [Etot] [S] Km Circle on your graph where this equation applies. What name is used to refer to keat/Km? : Calculate keat/Km for this enzyme. How does this value compare with the fastest enzymes?
The 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.
The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V for an enzyme-catalyzed, single-substrate reaction E + SES →E + P. The model can be more readily understood when comparing three conditions: [S] > Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V, where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Reaction rate is independent of [S]. Not true for any of these conditions The rate is half of the maximum rate.
Suppose that you have isolated the enzyme sucrase (able to hydrolyze sucrose into glucose and fructose), and you wish to determine the nature of inhibitor A for this enzyme. You have prepared five different concentrations of substrate (sucrose), and five different concentrations of inhibitor A (plus the control, with zero mM of inhibitor A). The following Table lists the inhibitor A concentrations [I], substrate concentrations [S], and resulting enzyme velocities (Vo) for all six of these experiments: [I] [S] Vo 1/[S] 1/ Vo 0 mM 0.1 mM 0.333333333333 mM per minute 0 mM 0.2 mM 0.50 0 mM 0.3 mM 0.60 0 mM 0.4 mM 0.666666666667 0 mM 0.5 mM 0.714285714286 0.1 mM 0.1 mM 0.20 0.1 mM 0.2 mM 0.333333333333 0.1 mM 0.3 mM 0.428571428571 0.1 mM 0.4 mM 0.50 0.1 mM 0.5 mM 0.555555555556 0.20 mM 0.1 mM 0.142857142857…
when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
Consider the enzyme-catalyzed conversion of a substrate S into a product in the presence of an inhibitor I with the following properties (mixed inhibition): (i) The inhibitor competes with S to bind the active site of the enzyme with dissoci- ation constant K¡ for the complex (EI). (ii) The inhibitor is also capable of binding the enzyme at a secondary site of the enzyme with dissociation constant K{ for the complex (IES). The complex (IES) is unable to form P. iii) The dissociation constants of the (IES) and (EI) complexes satisfy the relation K{ = K1/2. What is the value of the ratio vo/(k2 [E]o) when the concentration of the substrate is [S]o = KM/3 and the concentration of the inhibitor is [I]o = K{/4? Recall that KM is the Michaelis constant.

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