2. The kinetic effect of purely competitive inhibitor of an enzyme:(A) Increases K without affecting Vmax (B) Decreases Km without affecting Vmax(C) Increases Vmax Without affecting K (D) Decreases Vmax without affecting Km3. In reversible non-competitive enzyme activity inhibition:(A) Vma is increased(C) Vmax is decreased4. In enzyme kinetics Vmax reflects:(A) The amount of an active enzyme(C) Half the substrate concentration5. In non-competitive enzyme activity inhibition, inhibitor:(A) Increases Km(C) Does not affect Km(B) K is increased(D) Concentration of active enzyme is reduced(B) Substrate concentration(D) Enzyme substrate complex(B) Decreases Km(D) Increases Km

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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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For an enzyme-catalyzed reaction, the presence of 5 nM of a reversible inhibitor yields a Vmax value that is 80% of the value in the absence of the inhibitor. The KM value is unchanged. (a) What type of inhibition is likely occurring? (b) What proportion of the enzyme molecules have bound inhibitor? (c) Calculate the inhibition constant.
1. A sigmoidal plot of substrate concentration ([S]) verses reaction velocity (V) may indicate: (A) Michaelis-Menten kinetics (C) Competitive inhibition 2. The kinetic effect of purely competitive inhibitor of an enzyme: (A) Increases K without affecting Vma (B) Decreases K without affecting Vmas (C) Increases Vma without affecting Km (D) Decreases Vmas without affecting Km 3. In reversible non-competitive enzyme activity inhibition: (A) Vma is increased (C) Vma is decreased 4. In enzyme kinetics Vma reflects: (A) The amount of an active enzyme (C) Half the substrate concentration 5. In non-competitive enzyme activity inhibition, inhibitor: (A) Increases Km (C) Does not affect Km (B) Co-operative binding (D) Non-competitive inhibition (B) K is increased (D) Concentration of active enzyme is reduced (B) Substrate concentration (D) Enzyme substrate complex (B) Decreases K (D) Increases Km
(a) If the total enzyme concentration 1 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute?(b) Calculate kcat/KM for the enzyme reaction. Is this a fairly efficient enzyme? (as shown.)
2. 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…
The same enzyme as studied in the presence of a different inhibitor (inhibitor B). In this case, two different concentrations of inhibitor are used. Data are as follows: (a) What kind of inhibitor is inhibitor B?(b) Determine the apparent Vmax and KM at each inhibitor concentration.(c) Estimate K1 from these data.
B) Uncompetitive inhibition: Where do these inhibitors bind? To what mechanistic form of the enzyme do these inhibitors bind? What are changes to Km and Vmax?
Shown below is Lineweaver-Burk plot for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions: 1/v (sec/mM) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.2 0.4 y = 0.997 +3.01x (+1) 0.6 1/[S] (mM-¹) y = 0.09999 + 3.01x 0.8 (-1) 1 (c) What is the type of inhibition mechanism? A. competitive inhibition B. uncompetitive inhibition (substrate-dependent) C. mixed inhibition (a) Does this enzyme obey Michaelis-Menten kinetics? (yes or no) Explain: D. noncompetitive inhibition (substrate-independent) E. allosteric inhibition 1.2 (b) What are the apparent values of Vmax and KM for each experiment (with and without inhibitor)? d) If the concentration of the inhibitor is 0.1 mM, what is the value of KI and/or K'I (whichever is relevant)?
24. Researcher X did a series of experiments and determined the following about Enzyme A and its substrate (S). KM - 4 mM Vmax 100 mM/s axis. (. a) Draw the Michealis-Menton plot with all the above information shown. Be sure to label Vmax Rate of neation (mm) rate of react in mm. supstratiate concentration MM/S b) Convert the Michealis-Menton plot into a Lineweaver-Burk plot below, Label the axis. Vmax Rate of reaction (MM) 8 100 90 Go 100 80 - tmax 30 201 c) Using a dotted line, add in how á uncompetitive inhibitor how the rate would look with (S) (MM/S) x 1|S an uncompetitive inhibitor added. 100 KOMI Ke Vmax Vmar Sun competiti 6 • substante (MMIS)
A) Competitive inhibition: Where do these inhibitors bind? To what mechanistic form of the enzyme do these inhibitors bind? What are changes to Km and Vmax?
QUESTION 5 Consider the graph below. Reaction rate (v) Vmax 0 Effector 3 mM Effector 1 mM. Control 0 mM Ko.5 (3) Ko.5 (1) Ko.5 (0) [S] Which of the following conclusion(s) can be drawn from the data illustrated in the figure?
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)
4. Graphing the results from kinetics experiments with enzyme inhibitors The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [ET] is the same in each experiment. [S] (mM) 1 2 4 8 12 1 2 4 8 12 [S] 1/[S] (mm) a. Determine Vmax and Km for the enzyme. b. Determine the type of inhibition and the Kifor each inhibitor. v (1) M-1 μmol/mL.se Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a. and b. 1000 500 250 125 83.3 Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding v, we will calculate 1/[S] and 1/v. Please explain the formula (process) used to get the Red undelined numbers, example each Row, please. Thanks!! C 12 20 29 35 40 1/v (1) v (μmol/mL sec 12 20 29 35 40 -400 Condition No inhibitor → 20 x 104 1/v (1) mL sec/μmol 10 x 104 8.33x104 5.00x104 3.45x104 2.86x104 2.50x104 5…

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