2. Reaction Mechanism and Molecularity Step 1 (slow): O3+ NO2 NO3 + O2 Step 2 (fast): NO, + NO: NO Your answer
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![2.
Reaction Mechanism and Molecularity
Step 1 (slow): O, + NO:
NO3 + O2
Step 2 (fast): NO, + NO:
NO,Os
Your answer
а.
What is the reaction intermediate
Your answer
b. What is the rate determining step?
Your answer
C.
What is the molecularity of rate determining step?
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- 2. An enzyme-catalyzed reaction has a Km = 120 µM. At what substrate concentration would this reaction proceed at: i) 0.25 Vmax, ii) 0.5 Vmax, and iii) 0.75 Vmax? (show your calculations; hint - think of vi/Vmax).3. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained: [S] (M) 2.5 x 10-6 4.0 x 106 1х 105 2 x 105 4х 10-5 1 x 104 2 x 10-3 1 x 102 Vo (µM / min) 28 40 70 95 112 128 139 140 A. Vmax = 140 and Km=1 x 10-5 B. Vmax = 70 and Km = 1 x 103 C. Vmax = 140 and Km=1 x 10-2 D. Vmax = 70 and Km = 1 x 10-² %3D1. Consider the following parameters related to an enzyme that follows Michaelis-Menten kinetics for the reaction: k(1) k(2) S ----> ES ----> P k(-1)
- 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?Why is understanding reaction rates significant? Indicate at least 3 key importance of understanding reaction rates.5. 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 Kwax values for the inhibited and uninhibited reactions? Is the inhibitor competitive or noncompetitive? [S] (mM) V, No Inhibitor (mmol min-) V, Inhibitor Present (mmol min-') 1 × 10-4 5 × 10-4 1.5 x 10-3 2.5 x 10-3 5 x 10-3 0.026 0.010 0.092 0.136 0.040 0.086 0.150 0.120 0.165 0.142
- How does the average reaction rate differ from an instantaneous reaction rate? © A. The average reaction rate is how quickly the reaction proceeds over time. An instantaneous reaction rate is how quickly the reaction proceeds at a specific time. B. The average reaction rate is how quickly the reaction proceeds at a specific time. An instantaneous reaction rate is how quickly the reaction proceeds over time. C. The average reaction rate is how quickly the reaction proceeds over time considering the reactants. An instantaneous reaction rate is how quickly the reaction proceeds at a specific time considering the products. D. The average reaction rate is how quickly the reaction proceeds over time. An instantaneous reaction rate is how quickly the reaction proceeds compared to another reaction.7. Which of the following statements is true about enzyme-catalyzed reactions? The reaction is faster than the same reaction in the absence of the enzyme b. The free energy change of the reaction is opposite from the reaction without the enzyme The reaction always goes in the direction toward chemical equilibrium d. Enzyme-catalyzed reactions require energy to activate the enzyme e. Enzyme-catalyzed reactions release more free energy than noncatalyzed reactions The following questions are based on the reaction A+B C+D Free Energy- A+B Progress of the Reaction C.Don When testing the effect of substrate concentration on the reaction rate, there comes a point at which increasing the substrate concentration has no effect on the reaction rate. That is, the reaction rate remains the same and no longer increases. Which of the statements below would explain this? Select one: O A. The enzyme is denatured OB. The substrate is denatured OC. All available enzymes are saturated or full with substrates. OD. The enzyme concentration is too high
- 1. For enzymatic reaction, a mechanism was proposed by Michaelis and Menten as follows: ES k, and k,' ES à E and P k,. E + S a. Use steady state assumption, derive expression for the reaction rate. Where E is concentration of enzyme, S substrate, ES complex of E and S, E = E, – ES. (If you have difficulty in doing it, please consult lecture note) b. Assume K = 0.038 mol.L' at 25 °C, when the substrate concentration is 0.156 Mol.L', the rate of the reaction is 1.21 m mol/L.s. The maximum rate of conversion reaction is reached at high substrate concentrations. Calculate the maximum rate of this enzyme catalyzed reaction.an enzyme acts on a substrate X. The enzyme exists in four different forms, with different catalytic efficiencies. The table shows the kcatand KM values for each form of the enzyme. If the concentration of substrate X in a solution is 5 µM, which of the four forms of the enzyme is the most efficient? Form of Enzyme kcat (s-1) KM (µM) A 50 10 B 50 1 C 100 4 D 1000 100 a. Form A b. Form B c. Form D d. Form CIn enzyme and chemical kinetics A -->P, what is the instantaneous appearance or disappearance of A called?: 1) Velocity v = d[E]/dt 2) Velocity v = -d[A]/dt 3) Velocity v = d[P]/dt 4) 2&3 Can be correct.
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