Below is the set of kinetics data of a velocity of an enzymatic reaction without inhibitor (Vo) and with an inhibitor (Voi):Which one is correct about the inhibition?SVoVoi10.2860.1330.50.220.080.330.180.0570.250.150.0440.0360.13O Inhibitor decreases Vmax of the enzyme0.2O Inhibitor is most likely structurally similar to a substrateO Inhibition of the enzyme is irreversibleO Inhibitor can only bind to ES complex

Inhibitors are primarily of 2 types; reversible and irreversible inhibitors. Reversible inhibitors…

Answered: Below is the set of kinetics data of a…

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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From the following data of an enzymatic reaction, determine the type of inhibition, the KM for the substrate and Ki for the enzyme-inhibitor complex. [S] mM 2.0 3.0 4.0 10.0 15.0 [P] mM/h (without inhibitor) 139 179 213 313 370 [P] mM/h ([I]=6mM) 88 121 149 257 313
Which of the following is true under the following conditions: an enzyme displaying Michaelis-Menten kinetics where the enzyme concentration is 10 nM, the substrate concentration is 45 mM, and the Km is 50 µM? a) The enzyme has low catalytic efficiency for the substrate. b)The rate of catalysis is near half-maximal velocity. c)The enzymatic reaction is near maximal velocity. d)Halving the substrate concentration has little effect on the catalytic rate. e) There is not enough information provided.
In an enzyme experiment, when the enzyme is added only to the substrate, the Vmax value is 0.828 and the Km value is 17.76. When the substrate and activator are added to this enzyme, the Vmax is 0.964 and the Km value is 10.6066. When substrate and inhibitor are added to the enzyme, the Vmax value is 0.550 and the Km value is 11.34. Estimate the effect of the substance added. Are these substances suitable activators/inhibitors for the laccase enzyme?
For an enzyme that obeys Michaelis-Menten kinetics, in order for the reaction velocity (v) to be 80% of the maximal velocity (Vmax), the substrate concentration must be…. Choice 1 of 5:½ Km Choice 2 of 5:2 Km Choice 3 of 5:4 Km Choice 4 of 5:8 Km Choice 5 of 5:not enough information is given
The Km value of an enzyme-catalyzed reaction and its Vmax is 70 mmol/min. What is the rate of the reaction of the reaction when the substrate concentration is 7 × 10−2 mmol/min? a. 35 mmol/min b. 50 mmol/min c. 60 mmol/min d. 70 mmol/min
A mixed inhibitor of an enzyme (sometimes called a mixed non-competitive inhibitor) can decrease the rate of a reaction by any of the following EXCEPT by: a) binding to a site other than the active site of the enzyme. b) binding to the active site of the enzyme, preventing substrate binding. c) decreasing kcat. d) Increasing KM.
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.
Consider an enzyme-catalyzed reaction that follows Michaelis-Menten kinetics with KM =3.0 mmol·dm-3. What concentration of a competitive inhibitor with KI = 20 umol·dm-3 willreduce the rate of formation of product by 50% when the substrate concentration is held at 0.10mmol·dm-3.
Under the following conditions, fill in the blanks. Then, describe why this inhibitor is the type of inhibitor you identified it as. If you were to add 5nM of a reversible inhibitor, the Km for the measured enzyme catalyzed reaction would ______ (Increase, Decrease, Stay the same) to ______µM (choose appropriate value) and Vmax would _______ (Increase, Decrease, Stay the same) to ______µMs-1. So, this inhibitor is a ______ (Competitive, Uncompetitive, Mixed) inhibitor. Conditions: kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM
In a plot of 1/v vs 1/[S] for an enzyme-catalyzed reaction, presence of a non-competitive inhibitor will alter the: * Intercept on the 1/v axis. Intercept on the 1/[S] axis. Curvature of the plot. Km
A researcher has measured the initial rate of an enzyme-catalyzed reaction as a function of substrate concentration in the presence and absence of 0.001μM inhibitor. She obtains the following data. What is the Vmax and Km for the "no inhibitor" and "+ inhibitor" experiments. Ans, What would be value of V0 for the "+inhibitor" at [S] = 0.4μM and the inhibitor is most likely a competitive, uncompetitive or mixed inhibitor.
Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? a. As [S] increases, the initial velocity of reaction V0 also increases. b. At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. c. Km is the [S] at which V0 = 1/2 Vmax. d. The shape of the curve is a hyperbola. e. The y-axis is a rate term with units of μm/min.

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