For a Michaelis-Menten reaction, k₁=5 x 107/M-s, k-1-2 x 104/s, and k2=4 x 102/ Calculate the Ks and KM for this reaction. Does substrate binding achieve equilibrium or steady state?
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- The rate constants of an enzyme-catalyzed reaction, obeying the Michaelis-Menten kinetics, have been determined : E + S K₁ = 2 x 108 M-¹ S-¹ -1 -1 -1 K-₁= 1 x 10³ S K₂ = 5 x 10³ S-1 K₁ 1 K-1 ES K₂ E +P 1- Determine the Michaelis constant Km of the enzyme. 2- Determine the catalytic constant (kcat) of the enzyme. 3- Determine the catalytic efficiency of the enzyme.Write out the full redox reaction that happens when both of these reactions are added to the same reaction where there is 1M concentrations of each reactant and products, pH7, and temeprature is 290K. Identify which compounds will be reduced and which compounds will be oxidized. redox reaction 1: Pyruvate¯ + 2H+ + 2e¯ → lactate¯ E'o(V) = -0.185 redox reaction 2: NAD+ + H+ + 2e¯ → NADH E'o(V) = -0.320a 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.
- Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?Certain bacteria can respire in anoxic environments using arsenic (V) as electron acceptor. The relevant unbalance half reactions are: H₂ AsO+H → H¸AsO +H₂O, logK = 10.84, AG = -14.5kca I m I ol-e CH₂O+H₂OH + CO2 (g), logK = 1.2, AG° = -1.63kca — ol-e 1) Balance the two half reactions 2) What is m the overall respiration reaction, standard free energy change \Delta GO 3) Is this process energetically more or less favorable than sulfate reduction? (\Delta GO for the reduction of sulfate to HS- is -5.78 kcal/mol - e-) 4) If [H2AsO4-] = [ H3ASO3] = 0.5 mM and pH = 7, estimate pe of the systemThe typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
- Use the Michaelis-Menten equation to complete the enzyme kinetic data set, whenKm is known to have a value of 1 mmol L−1 [S] (mmol L−1) Vo (μmol L−1 min−1) 0.5 50 1.0 2.0 3.0 10.0Given the following equations of the line, a) determine the KM of the substrate that binds the strongest with the enzyme. b) determine which of the following substrates binds the least with the enzyme5. 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 equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a
- Compound A is the substrate for two enzymes, El and E2, their reaction rates, r1 and r2,Determine the Km and rmax for both enzymes, with respect to the concentration of A. Which set of data is more likely to be for El and which for E2, and why? Concentration of 0.2 0.6 1.2 3 4 5 6 8 12 15 A (mM) Reaction rate (r.) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 4.88 4.9 4.92 4.94 4.95 4.96 4.97 Reaction rate (r,) (mmol/L*min) 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.801) 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)i) Re-arrange the Michaelis Menten equation so it involves the ratio [S]. Show all steps beginning Km noting any assumptions or required conditions. Km ii) Calculate the ratio [lo for the case when the rate of product formation is 68% of Vmax and the substrate is in great excess. d[P] dt : k₂ with = [E],[S] Km+[S]' [S]o Km iii) Explain, in a few sentences, why the ratio determines the ratio V Vmax V Vmax Begin by explaining the meaning of stating simply "it's the ratio...." is not sufficient. Include in your explanation the factors that effect v and Vmax. Consider what factors make v different from or equal to Vmax. Consider what Km represents concerning processes involving ES. " iv) Calculate KM at 310K at given the following rate constant information: k₁ = 17 s-¹M-1 at 300K with A = 7300 s-¹M-1 K-1₁ 6 s¹ at 300K with A = 14500 s -1 k₂ = 31 s¹ at 300K with A = 600 s-¹