8)plot of enzyme activity with and without an inhibitor present gave the following plot.What type of inhibitor is present? How does this inhibitor function? What changes areseen in Vmax and KM? Draw a line that approximates the result from addition of twice asmuch inhibitor to the reaction.1/v (v in mM/min)0.80.60.40.200.20.4061/[s] ([S] in mM)0.8

Reversible enzyme inhibition is a type of enzyme regulation in which the inhibitor can bind and…

Answered: 3) A plot of enzyme activity with and…

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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An inhibitor "I" is added to the enzymatic reaction at a concentration of 1.0 g/L. The data obtained are shown in the table below. No Inhibitor With Inhibitor [S](g/L) v ([E0] = 0,015 g/L) (g/L.min) v ([E0] = 0,015 g/L)(g/L.min) 40 2,28 1,818 20 1,74 1,316 13,4 1,4 0,986 10 1,18 0,76 8 1 0,62 (a)What kind of inhibition occurred?(b)Determine Ki (in g/L) for the inhibition of the previous exercise.
The Lineweaver - Burk plot (Figure 1) shows an enzyme-catalyzed reaction in the absence and presence of 0.1µM inhibitor (ketoconazole). O Estimate Vmax and Km in the absence and presence of the inhibito.. (ii) Determine the type of inhibition shown by the inhibitor. Explain. 0.1- 0.08 0.06 - With inhibitor 0.04 0.02 Without inhibitor 0.4 -0.2 -0.02 0.2 0.4 0.6 0.8 1.0 1/{S\(&M-1) 1/vo (pmol-11 min)
Identify the enzymes in Table, whose catalytic effi ciencies are near the diff usion-controlled limit.
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In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM The inhibition occurred at 37.0°C. Assuming that the temperature increased to 37.5°C, what will happen to the Vmax and Km in the chemical system involving Inhibitor XYZ? Why do you say so? Provide illustrations using appropriate dimensions in the x and y axes. C.
5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00
The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
D 1. 0.25 Short answer questions, write answers on blank paper. A plot of enzyme activity with and without an inhibitor present gave the following plot. 0.2 0.15 1/V v in mM/s 0.1 0.05 0 0 0.05 0.1 y=0.67x+0.02 0.15 y=0.2x+0.02 0.2 0.25 0.3 1/[S]. [S] in mM Without the inhibitor, what are the values of Vmax and Km? In the presence of the inhibitor, which of these items has changed and what is (are) the new value(s)? What type of inhibitor is present (explain why you have arrived at this conclusion)? Be sure to include units as necessary--answers that do not contain required units will be marked incorrect
At body temperature (37°C), k of an enzyme-catalyzed reaction is 2.3X10¹⁴ times greater than k of the uncatalyzed reaction. Assuming that the frequency factor A is the same for both reactions, by how much does the enzyme lower the Eₐ?
The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →
a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your work
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM

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