6-25substrate-band enzyme concentrations. The theturnover number is equal to umax-b)V=Umax •57(Km+S)anstontFor an enzyme that displays Michaelis-Menten kinetics, what isthe reaction velocity, V (as a percentage of Vmax), observed atthe following values?a)[S] = KMC)d)e)[S] = 0.5KM[S] = = 0.1KM[S] = 2KM[S] = 10KMwreactores-maximumrate of reactionboteles conc.Would you expect the structure of a competitive inhibitor of agiven enzyme to be similar to that of its substrate?

For a one-substrate enzyme-catalyzed reaction, the Michaelis-Menton equation shows the quantitative…

Answered: 6-25 For Umax 57(Kr+S) constant an…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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(a) At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?(b) Determine the fraction of Vmaxthat would be obtained at the following substratecon centrations [S]: ½Km, 2Km, and 10Km.(c) An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species.The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
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
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.
Sketch on one reaction rate vs. substrate concentration graph & sketch on one Lineweaver-Burk type plot the following:a) A Michaelis-Menten enzyme with a Vmax = 60 1/s and a KM = 125 M.b) An uncompetitive inhibitor of the enzyme described in a).c) An allosteric enzyme with the same Vmax as the enzyme described in a) and follows the sequential model
Assume that an enzyme-catalyzed reaction follows the scheme shown: E + S ES E + P k₁ = 1 x 10%/M-s k-1 = 2.5 x 10%/s k2= 3.4 x 107s What is the dissociation constant for the enzyme-substrate, Ks? What is the Michaelis constant, Km, for this enzyme? What is the turnover number, Kcat, for this enzyme? What is the catalytic efficiency for the enzyme? If the initial Et concentration is 0.25mM, what is Vmax?
An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?
At what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
1) Is the ratio of the forward rate constant and reverse rate constant changed by the presence of an enzyme catalyst? 2) 4) What is the simplest mathematical relationship between substrate concentrations [S], initial velocity (Vo) of an enzyme catalyzed reaction, the maximal velocity of the reaction (Vmax), and the ½ maximal Vo (i.e. Km) ? (Hint: what is the Michaelis-Menten Equation? 3) Graphically illustrate the most useful derivation of the Michaelis-Menton equation (that darn linearized, double-reciprocal)
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.
Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.
An enzyme catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?

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