The steady-state kinetics of an enzyme has been studied in the absence and presence of an inhibitor (Inhibitor A). The initial rate given as a function of substrate concentration in the following table: Substrate Concentration, [S] Initial rate, v, (mmol/L-min) Initial rate, v, (mmol/L-min) (mmol/L) No Inhibitor With Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38

The steady-state kinetics of an enzyme has been studied in the absence and presence of an inhibitor (Inhibitor A). The initial rate given as a function of substrate concentration in the following table: Substrate Concentration, [S] Initial rate, v, (mmol/L-min) Initial rate, v, (mmol/L-min) (mmol/L) No Inhibitor With Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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1. The enzyme you are studying uses the substrate pictured at right. An inhibitor box in the freezer contains two tubes labeled Inhibitor #1 and Inhibitor #2, along with the structures of the compounds contained within them and their respective Ki. Which one is more likely to be the non-competitive inhibitor? Inhibitor #1 K₁ = 10 μM m CH₂OH OH -0, $ OH OH Inhibitor #2 K₁ = 50 μM Briefly Justify: OH CH₂OH OH OH OH Substrate
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1. Start with the Michaelis-Menten equation and convert it to a double-reciprocal equation. Show how you would make a linear Lineweaver-Burk plot using this equation. 2. Based on th chart make a Lineweaver-Burk plot and later use the plot to complete the information in the table.
2. An enzyme-catalyzed reaction is carried out at several different substrate concentrations (listed below), either (1) with no inhibitor, or (2) in the presence of an inhibitor, I, at 0.63 g/L. Given the following data, determine: A. what type of inhibitor I is B. the value of Ki No Inhibitor: With Inhibitor: Substrate Concentration (g/L) 0.86 Reaction Velocity (g/L-s) Reaction Velocity (g/L-s) 0.179 0.161 2.35 3.80 4.90 6.10 18.4 0.398 0.567 0.666 0.726 0.950 0.322 0.407 0.443 0.495 0.639
3. b. If chymotrypsin is treated with disopropylfluorophosphate, catalysis cannot occur even in the presence of excess substrate concentration. Is this Inhibltion reversible or Irreversible?. Dilsopropylfluorophosphate Hyd
3. Enzyme specificity. To determine the specificity of substrate binding for a particular enzyme/protein, structurally related compounds may be used as potential substrates and Km values may be calculated. However, many compounds structurally related to the substrate may bind to the active site but cannot be converted to product. In these instances, the substrate analogs are used as potential competitive inhibitors of substrate binding. Low K, values indicate high affinity of the enzyme for the inhibitor, whereas high K,values indicate low binding affinity. Consider the enzyme xanthine oxidase, which catalyzes the formation of uric acid from the purine bases hypoxanthine or xanthine in humans. The Km for hypoxanthine is 15.0 μM and for xanthine it is 45.0 μM. A few compounds used as competitive inhibitors of the normal substrate hypoxanthine are listed in the table below with their K; values. Comparing the structures of hypoxanthine with the listed substrate analogs, what can you…
WHICH OPTIN IS TRUE QUICK I WILL RATE NO EXPLANATION NEEDED
1. ( . tion can be represented by the so-called Michaelis equation ) The velocity of an enzyme catalyzed reac- Vmax [S] Vo = Km Vo = max kcat [Eo] [So] (Км + [Sol) V max where [So] is the initial substrate concentration, [Eo] is the total enzyme concentration, kcat is the cataly- tic rate constant or turnover number, and KM repre- sents the Michaelis constant. The graph on the right illustrates the dependence of the velocity of the reaction as a function of substrate concentration. Ex- Km [S] (mm) plain why the maximal velocity can be represented by the expression Vmax = Kcat [Eo] at very high substrate concentrations. Vo (µm/min)
Bypass reaction in the anabolic pathway often occurs when A The ΔG of the reverse reaction in the catabolic pathway has a large negative value B The ΔG of the reverse reaction in the catabolic pathway has a large positive value C The ΔG of the reaction in the anabolic pathway is close to zero D The ΔG of the reaction in the anabolic pathway is positive E None of the above
The effect of an inhibitor I on the rate of a single-substrate, enzyme-catalyzed reaction was investigated and gave the following results (See table below). a. What is the mode of action (pure competitive, mixed competitive, uncompetitive or noncompetitive) of the inhibitor? Briefly explain your answer. b. Estimate the value of Km and Vmax for uninhibited reaction.