Enzymes are often described as following a two-step mechanism: Where E = enzyme, S = substrate,...
- Enzymes are often described as following a two-step mechanism:
Where E = enzyme, S = substrate, ES = enzyme-substrate complex, and P = product.
- Write the balanced equation for the overall reaction.
- Identify and intermediates in the reaction mechanism.
- Derive an expression for the rate law.
Homework Answers
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Enzymes are often described as following a two-step
mechanism:
Where E = enzyme, S = substrate,...
Enzymes are often described as following the two-step mechanism: E+S⇌ES(fast) ES→E+P(slow) Where E = enzyme, S...
Enzymes are often described as following the two-step mechanism: E+S⇌ES(fast) ES→E+P(slow) Where E = enzyme, S = substrate, ES = enzyme-substrate complex and P = product. Question: Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors.Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor. Express your answer as a chemical equation.
2. Consider this two step mechanism for a reaction… Step 1 NO2 + O3 --> NO3 ...
2. Consider this two step mechanism for a reaction… Step 1 NO2 + O3 --> NO3 + O2 slow; rate determining step Step 2 NO3 + NO2 --> N2O5 fast a. What is the overall reaction? b. Identify the intermediates in the mechanism. c. Write the rate law expression for each step of the mechanism including any reversible reactions. c. What is the predicted rate law expression? Be sure to only list reactants from the overall equation and not intermediates...
#13 Interpreting Mechanisms 1. Consider the following mechanism: Step 1 Br2  
#13 Interpreting Mechanisms 1. Consider the following mechanism: Step 1 Br2 2Br fast Step 2 Br + H2 H2Br fast Step 3 H2Br + Br 2HBr slow a. What is the overall equation? b. Identify the intermediate(s) if any. c. What is the molecularity and the rate law for each step including any reversible steps? d. What is the predicted rate law expression for this reaction. Be sure to only list reactants from the overall equation and...
Enzyme E bind to substrate S to form ES complex leading to product formation. However, the ES com...
Enzyme E bind to substrate S to form ES complex leading to product formation. However, the ES complex also undergoes suicidal inactivation that would result in the loss of the total enzyme activity. In addition, after the binding of S to E there exposes another binding site in E for uncompetitive inhibitor I. The ESI complex does not have the capacity to form product and would not undergo inactivation. The equilibriums, reactions and the kinetic parameters involved can be represented...
Question B1. The binding of a substrate (S) to an enzyme (E) can be described by...
Question B1. The binding of a substrate (S) to an enzyme (E) can be described by the equilibrium E + S 4 ES At 273 K the equilibrium constant for this reaction has been found to be 0.134 for a particular substrate and enzyme. (a) Calculate the standard Gibbs energy change for the binding reaction at 273 K. [1 mark] (b) Calculate the free energy change for the reaction when the concentrations of the species at 273 K are: [E]...
1. A proposed mechanism for a reaction is: 1221 (rapid equilibrium) I + H2 = H2...
1. A proposed mechanism for a reaction is: 1221 (rapid equilibrium) I + H2 = H2 (rapid equilibrium) H21 + 1 = 2 HI (slow) a. Write the overall balanced equation for this reaction. (4 pts.) b. Indicate which step is the rate determining step. (1 pt.) c. Based on the mechanism proposed, write the rate law for the overall reaction. (2 pts.) d. Using ki, ki, ka, etc., write the value for the overall rate constant k. (2 pts.)...
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S...
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E and ends with the release of product P from the enzyme. E+S →ES → EP E+P Under many circumstances, KM = [E] [S] / [ES] What proportion of enzyme molecules are bound to substrate when [S] = KM? Why? Recall that when [S] = KM, the reaction rate is Vmax/2. Does your answer to Part A make sense in light of...
step Suppose the reaction between nitric oxide and oxygen proceeds by the following mechanism: elementary reaction...
step Suppose the reaction between nitric oxide and oxygen proceeds by the following mechanism: elementary reaction rate constant NO(g) +0,() ► NO,()+(2) ky 2 NO(g) +0 (2) NOG) kz Suppose also , ekz. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates x 5 ?
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the...
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the mechanism: E+S →ES, rate constant = ki, ES → E+S, rate constant = k-1, ES → E+P, rate constant = k2, E + EI, rate constant = k3. EI → E + I, rate constant = k-3. Assuming that the concentrations of S and I are much larger than the total enzyme concentration, derive an expression for the initial rate of appearance of product,...
Problem 7 (15 mark) For many enzymes the mechanism of action involves the formation of two...
Problem 7 (15 mark) For many enzymes the mechanism of action involves the formation of two intermediates: ka E+S →ES ES → E+S ES →ES ko ES' → E + P Show that the rate of formation of product has the form: V = ko[E]o / (1 + (Km/[S].)). with Km given by Km = kc(ka’+kb)/(ka(kb+kc))
Enzyme E bind to substrate S to form ES complex leading to product formation. However, the ES complex also undergoes suicidal inactivation that would result in the loss of the total enzyme activity. In addition, after the binding of S to E there exposes another binding site in E for uncompetitive inhibitor I. The ESI complex does not have the capacity to form product and would not undergo inactivation. The equilibriums, reactions and the kinetic parameters involved can be represented...
Question B1. The binding of a substrate (S) to an enzyme (E) can be described by the equilibrium E + S 4 ES At 273 K the equilibrium constant for this reaction has been found to be 0.134 for a particular substrate and enzyme. (a) Calculate the standard Gibbs energy change for the binding reaction at 273 K. [1 mark] (b) Calculate the free energy change for the reaction when the concentrations of the species at 273 K are: [E]...
1. A proposed mechanism for a reaction is: 1221 (rapid equilibrium) I + H2 = H2 (rapid equilibrium) H21 + 1 = 2 HI (slow) a. Write the overall balanced equation for this reaction. (4 pts.) b. Indicate which step is the rate determining step. (1 pt.) c. Based on the mechanism proposed, write the rate law for the overall reaction. (2 pts.) d. Using ki, ki, ka, etc., write the value for the overall rate constant k. (2 pts.)...
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E and ends with the release of product P from the enzyme. E+S →ES → EP E+P Under many circumstances, KM = [E] [S] / [ES] What proportion of enzyme molecules are bound to substrate when [S] = KM? Why? Recall that when [S] = KM, the reaction rate is Vmax/2. Does your answer to Part A make sense in light of...
step Suppose the reaction between nitric oxide and oxygen proceeds by the following mechanism: elementary reaction rate constant NO(g) +0,() ► NO,()+(2) ky 2 NO(g) +0 (2) NOG) kz Suppose also , ekz. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates x 5 ?
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the mechanism: E+S →ES, rate constant = ki, ES → E+S, rate constant = k-1, ES → E+P, rate constant = k2, E + EI, rate constant = k3. EI → E + I, rate constant = k-3. Assuming that the concentrations of S and I are much larger than the total enzyme concentration, derive an expression for the initial rate of appearance of product,...
Problem 7 (15 mark) For many enzymes the mechanism of action involves the formation of two intermediates: ka E+S →ES ES → E+S ES →ES ko ES' → E + P Show that the rate of formation of product has the form: V = ko[E]o / (1 + (Km/[S].)). with Km given by Km = kc(ka’+kb)/(ka(kb+kc))
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