In a reaction system, the concentrations of Enzyme-Substrate complex [ES], free enzyme [E] and free substrate [S] are 5mM, 2mM and 45 mM respectively. If the enzyme has 5 identical binding sites for this substrate, then calculate the value of Equilibrium Association constant (Ka). BIU@G == HEE = E ¶ ¶ Σ 5 (10
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- The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?The total concentration of enzyme in a reaction, [E], is made up of the concentration of enzyme bound to substrate, [ES], and the concentration of enzyme still free in solution, [Ef]. Similarly, the total amount of substrate is made up of [Sf] and [ES]. We can assume that the concentration of enzyme is much less than that of the substrate, [E] << [S]. Assuming the steady state condition and the relationships between [E], [Ef] and [ES], and similar ones for S, given in lectures, derive an expression for the saturation factor, , in terms of [S] and . (Note that [E] and [S] denote the total amounts of enzyme and substrate added to the reaction, respectively. You may assume that [S]>>[E].)An enzyme is present at a concentration of 1 nM and has a Vmax of 2 µM s-'. The Km for its primary substrate is 4 µM. Calculate kcat- kcat Calculate the apparent Vmax and apparent Km of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2. Assume that the enzyme concentration remains at 1 nM. apparent Vmax µM s-1 apparent Km = µM
- From the following data of an enzymatic reaction, determine the type of inhibition, the KM for the substrate and Ki for the enzyme-inhibitor complex. [S] mM 2.0 3.0 4.0 10.0 15.0 [P] mM/h (without inhibitor) 139 179 213 313 370 [P] mM/h ([I]=6mM) 88 121 149 257 313A generalized enzyme active site is shaped like a hemisphere with a radius of 45Å. The active site holds the following amino acids in a homeostatic solution (pH = 7.38): -HAVARILKHAVARILKHAVARILK- Assuming the charge is distributed uniformly along the hemisphere, determine the force at which this active site acts upon a single ATP molecule at the center of the hemisphere.By using Excel or GoogleSheets, graph the Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and V values for the inhibited and uninhibited reactions? Is the inhibitor competitive or max noncompetitive? [S] (mM) V, No Inhibitor (mmol min-') V, Inhibitor Present (mmol min-') 1× 10-4 5 × 10-4 1.5 × 10-3 2.5 × 10-3 5 × 10-3 0.026 0.092 0.136 0.150 0.010 0.040 0.086 0.120 0.165 0.142 Activate
- 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-1For 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 Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM
- You find in the literature that an enzyme purified very close to homogeneity has a specific activity of 12,300 U/mg. What is likely to be the largest source of error in the quoted value? U (unit) typically is defined as the amount of enzyme that is required to consume 1.00 umoles of substrate per minute under some given set of conditions. Units are typically determined spectrophotometrically by measuring the rate of formation of product.When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?The initial velocities of two different enzyme-catalyzed reactions were measured over a series of substrate concentrations. The following results were obtained: Enyme A: KM = 1.5 mM, Vmax = 10 μM s-1 Enyme B: KM = 5.0 mM, Vmax = 85 µM s-1 (a) Which enzyme binds to its substrate more tightly (assume k.1 >> k₂ in the Michaelis-Menten model)? (b) Calculate the initial velocities of each reaction when the substrate concentration is 2.5 mM. (c) Calculate the Kcat of each enzyme if the total enzyme concentration is 100 nM. (d) Which enzyme is the more efficient catalyst? Explain your answer. The enzyme carbonic anhydrase is strongly inhibited by the drug acetazolamide. A plot of the initial reaction velocity (as a percentage of Vmax) in the absence and presence of the inhibitor is shown below. What type of inhibition is taking place? Explain your reasoning. V (% of Vmax) 100 50 0.2 0.4 No inhibitor Acetazolamide [S] (MM) 0.6 0.8 1