In the scheme for enzymatic catalysed reaction proposed by Michaelis and Menten, the steps involve reversible formation of enzyme-substrate (ES) complex followed by conversion of the complex to the product (P) a) Derive the rate equation for enzymatic process. State an assumption made in this derivation. b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibition
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In the scheme for enzymatic catalysed reaction proposed by Michaelis and Menten, the steps involve reversible formation of enzyme-substrate (ES) complex followed by conversion of the complex to the product (P)
a) Derive the rate equation for enzymatic process. State an assumption made in this derivation.
b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibition
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- In the scheme for enzymatic catalysed reaction proposed by Michaelis and Menten, the steps involve reversible formation of enzyme-substrate (ES) complex followed by conversion of the complex to the product (P). a) Derive the rate equation for enzymatic process. State an assumption made in this derivation. Terbitkan persamaan kadar tindak balas bagi proses enzim. b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibitionFor the following two scenarios, sketch the complete reaction free energy diagram for an enzyme-catalyzed conversion of a single substrate (S) into product (P), where the reaction is spontaneous in the forward direction. For each of these, overlay the free energy diagram for the uncatalyzed reaction and indicate AAG" on your sketch: a). Substrate binding is rate limiting b). The chemical step is rate limitingConsider this intermediate in the derivation of the Michaelis-Menten equation. [E] [S] [ES| k-1 + kz km Assume that k is negligible compared to the other rate constants. If the k is very small, it suggests that the enzyme has a Select an option affinity for its substrate, while if the if the km is very large, it suggests that the enzyme has a Select an option. affinity for its substrate. Select an option Submit You have used 0 of high Sav low moderate
- Most of the enzyme reactions followed the mathematical kinetic plots suggested by Michaelis-Menten plots: (a) Is high or low KM preferable for an enzyme reaction? Give reason. (b) Draw the Lineweaver-Burk plot and on the same plot, draw a graph to indicate competitive inhibition. (c) Draw another Lineweaver-Burk plot and on the same plot, draw a graph to indicate non-competitive inhibition.a) Based on the data shown in the image, what are the Km and Vmax for the enzyme with L-DOPA and D-DOPA? Show any relevant analyses or calculatins you did to determine these values. ( HINT a graph might be helpful here! ) b) Based on your answer to part a, briefly describe how the kinetics of the enzyme differs for the two substrates. Which Substrate has better binding affinity to the enzymeThe 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
- Consider the enzyme-catalyzed conversion of a substrate S into a product in the presence of an inhibitor I with the following properties (mixed inhibition): (i) The inhibitor competes with S to bind the active site of the enzyme with dissoci- ation constant K¡ for the complex (EI). (ii) The inhibitor is also capable of binding the enzyme at a secondary site of the enzyme with dissociation constant K{ for the complex (IES). The complex (IES) is unable to form P. iii) The dissociation constants of the (IES) and (EI) complexes satisfy the relation K{ = K1/2. What is the value of the ratio vo/(k2 [E]o) when the concentration of the substrate is [S]o = KM/3 and the concentration of the inhibitor is [I]o = K{/4? Recall that KM is the Michaelis constant.After purifying alkaline phosphatase, you perform enzyme kinetic experiments with and without an inhibitor to obtain the following plot: With inhibitor Without inhibitor 1/V (1/mM min?) 0.3 0.2 -150 -100 -50 50 100 150 200 250 300 1/s (1/mM) a) What is the name given to this type of enzymatic plot? b) Using the graph, calculate K_ and V_ inhibitor. Show all calculations. for the enzyme, with and without the maxGiven 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.
- A substrate is converted to a product via the reaction sequence (1) E +S ES k2 (2) ES + S ES2 k4 (3) ES2 ES +P (4) ES , E +P (a). Using pseudo-steady state hypothesis for various forms of enzyme active sites, obtain relations between rates of individual steps in the mechanism. (b). Relating concentrations of vacant active sites to concentrations of occupied active sites, obtain expressions for the rate of consumption of S, (-Rs), and the rate of formation of P, Rp. (c). What are the maximum values of (-Rs) and Rp and under which conditions are these attained?Compare and contrast Bound Fraction equation in ligand binding and Michaelis-Menten equation in enzyme kinetics, including their double-reciprocal forms. Discuss what Km is important for and what Vmax (or kcat) is important for? Under what (substrate) conditions is Km more important than Vmax, and under what (substrate) conditions is Vmax more important than Km? Based on the discussions in question 2, explain what type of inhibitors works best under (a) high substrate concentration and (b) low substrate concentration.The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for an enzyme-catalyzed, single-substrate reaction E + S ES → E + P. The model can be more readily understood when comparing three conditions: [S] > Km- Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Not true for any of these conditions Almost all active sites will [ES] is much lower than [Efree]. be filled. The rate is directly proportional to Increasing [Etotal] will increase [S]. Km: Adding more S will not increase [Efree] is equal to [ES]. the rate.