Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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- you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?arrow_forwardThe following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.arrow_forward(a) You are given the following experimental measurements for the effects of an inhibitor on the initial velocity of an enzymatic reaction. Plot the data and determine what type of inhibition is being observed. Label all axes. [S] (MM) Vo (mM • Vo with I present (mM ⚫s-¹) 1 2.8 3 6.0 1.0 2.5 4 7.0 3.1 8 9.9 12 11.5 4.8 5.7 (b) Briefly explain how varying [S] can be used to determine whether an inhibitor is uncompetitive or competitive. Be sure to include in your explanation why this method would work.arrow_forward
- In a uni uni enzyme reaction, what is the substrate concentration relative to Km when anenzyme operates at 0.95 * V. What about 0.99 * V?arrow_forward5.50 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 2.00arrow_forwardThe Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =arrow_forward
- 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 →arrow_forwardThe figure below shows the dependence of the enzyme's rate, v (in µM/min), as a function of substrate concentration, S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, which of the following does this inhibitor most likely interact with? 1/v 1- 05 1/[S) O A. Michaelis complex O B. [E O C. free enzyme O D. free substrate O E. Both "A" and "B."arrow_forwardWhat is the relative inhibition of an enzyme by a competitive inhibitor at [S] = KS and [I] = KI?arrow_forward
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