Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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- a 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.arrow_forward2. Enzyme-catalyzed reactions. Answer the following with true or false. If false, explain why. (a) The initial rate of an enzyme-catalyzed reaction is independent of substrate concentration. (b) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. (c) The Michaelis constant Km equals the substrate concentration at which velocity (v) = Vmax/2. (d) The Km for a regulatory enzyme varies with enzyme concentration. (e) If enough substrate is added, the normal Vmax of an enzyme-catalyzed reaction can be attained even in the presence of a noncompetitive inhibitor. (f) The Km of some enzymes may be altered by the presence of metabolites structurally unrelated to the substrate. (g) The rate of an enzyme-catalyzed reaction in the presence of a rate-limiting concentration of substrate decreases with time. (h) The sigmoidal shape of the v versus [S] curve for some regulatory enzymes indicates that affinity of the enzyme for the…arrow_forward5. 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 aarrow_forward
- a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your workarrow_forwardBy what factor is sample A2 more concentrated than sample A1?arrow_forwardi) 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-¹arrow_forward
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