[S] << Km [S] = Km [S] >> Km Not true for any of these conditions
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![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. [S] = Km, and [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
Increasing [Ewall will increase Km.
Almost all active sites are empty.
[S] = Km
Answer Bank
[Enel is about equal to [Exotal].
[ES] is much higher than [Efree]
[S] >> Km
Reaction rate is independent of [S].
Not true for any of these
conditions
The rate is half of the maximum rate.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F51044012-1e7d-4f53-9311-c2b96c730020%2F053d3487-43c0-4fef-a00d-86eb4c8e18df%2F1hd9yt_processed.jpeg&w=3840&q=75)
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- For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.The following reaction sequence consists of two different substrates catalyzed by an enzyme:let's assume he described his reactions.;E + S1: ES1ES1 + S2: ES1S2ES1S2 → P + Ea.Derive the reaction velocity equation with Michaelis-Menten acceptance.b. Derives the rapid equality of S1 substrate concentration, rather than S2 substrate concentrationsimplify for reaction cards where it is higher.If a reaction has a ΔG°′ value of at least −30.5 kJ · mol−1, suffi -cient to drive the synthesis of ATP (ΔG°′ = 30.5 kJ · mol−1), can it still drive the synthesis of ATP in vivo when its ΔG is only −10 kJ · mol−1? Explain.
- Consider the following data for an enzyme-catalyzedhydrolysis reaction in the presence and absence ofinhibitor I: Using a Michaelis-Menten plot, determine Km for theuninhibited reaction and the inhibited reaction.Calculate the standard free-energy change of the reaction catalyzed by theenzyme phosphoglucomutase, given that, starting with 20 mM glucose 1-phosphate and no glucose 6- phosphate, the final equilibrium mixture at 25 °C and pH 7.0 contains 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate. Does the reaction in the direction of glucose 6-phosphate formation proceed with a loss or a gain of free energy?1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.