Two enzyme catalyzed reactions, RA (red) and Rg (blue), have kinetic profiles shown below. Report which reaction has the higher maximal rate and which reaction has the smaller Km: 1/[S] Higher Vm [Select ] Lower Km [ Select ]
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Q: kąk4 k2k1 E +S → ES → E + P (Vmax,f\ rax.f (Vmax,b [P] KM.f KM.b [S] [P] Vo 1+ KM.f Кмь KM,b ||
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Q: (Vmax.f [S] - (Vmax,b [P] KM,b [P] vo [S] 1+ KM.f KM.b ||
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- Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? a. As [S] increases, the initial velocity of reaction V0 also increases. b. At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. c. Km is the [S] at which V0 = 1/2 Vmax. d. The shape of the curve is a hyperbola. e. The y-axis is a rate term with units of μm/min.At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?Q is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound Q
- Enzyme A catalyzes the reaction S → P and has a KM of 50 μM and a Vmax of 100 nM s–1. EnzymeB catalyzes the reaction S → Q and has a KM of 5 mM and a Vmax of 120 nM s–1. When 100 μM ofS is added to a mixture containing equal amounts of enzymes A and B, which reaction product (Por Q) will be more abundant after 1 minute of reaction?Imagine a typical enzyme kinetic curve of [S] vs. V. The maximum [S] is about four times the Km. Lines are drawn at [S] = Km and V = Vmax. The asymptote is drawn at [S] = 0. 1. What effect does altering Km have on Vmax? Explain. 2. What effect does altering [E] have on Km? Vmax? Explain. 3. At what concentration of [S] is the initial velocity proportional to [E]? Explain.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.
- 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.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.Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
- Use the data below to determine the maximum velocity [in mM/s] of a certain enzyme-catalyzed reaction. v = 0.152 mM/s at [S] = 0.334 mM v = 0.190 mM/s at [S] = 0.450 mMa particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.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