Using linear regression analysis, determine the values of Vmax and KM of the enzyme inthe PRESENCE of inhibitor: (Express your answer in 3 decimal places, do NOT include the units)VmaxKM =!3!What is the MODE of inhibition exerted by DEDS? You were asked to determine the mode of inhibition exerted by Inhibitor DEDS to a newlydiscovered enzyme known as BILISTASE. The kinetics data are shown below:[Substrate, uM)Vo with DEDS (uM/min)Vo without DEDS (uM/min)1.6671.6003.5442.5001.9234.2593.3332.1394.73714.0002.2685.0195.0002.4105.336Using linear regression analysis, determine the values of Vmax and KM of the enzyme inthe ABSENCE of inhibitor: (Express your answer in 3 decimal places, do NOT include the units)VmaxKM

Michaelis menten constant, Km is the substrate concentration required to produce half maximum…

Answered: Using linear regression analysis,…

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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Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minute
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM
Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?
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
A biochemist wants to determine the effect of an inhibitor on a certain enzyme. The data are shown below: Vo (mM/min) (without inhibitor) Vo (mM/min) (with inhibitor) [Substrate] (mM) 15 1.11 0.67 30 1.81 1.17 45 2.28 1.56 60 2.62 1.88 90 3.09 2.36 Using linear regression analysis, determine the values of Vmax and Km of the enzyme in the ab- sence of an inhibitor. - y-int = - slope = - vmax = - Km =
The 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 =
Which of the following statements are true about the relationships of [S], KM, and Vmax? (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.
Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitor competitive or noncompetitive? [S] (M) 10,000 1 x 104 2.000-5 x 104 666673 1.5 x 10-3 400- 2.5 x 10-³ 2005x103 V, No Inhibitor (Arbitrary units) 0.026 38,461 0.092 10.87 37100 0.136 7.35 Wak 0.150 -67 0.165 6.06 L slope: 0033 yint=5057 полесте um.x=0.198 km 6.5x 10-4 0.010 0.040 25 10.086 11.63 0.120 8.33 0.142 7.04 2500 V, Inhibitor Present (Arbitrary Units) 100 ↓ $op.= .009 yint: 5.21 900x12 km: 1.73 y/o conf. V10-3 321000
You design an enzyme assay and choose a substrate concentration equal to the Km of the enzymatic reaction. You measure the rate of the reaction at this substrate concentration to be 75.0 μmol/min. Calculate the Vmax of this reaction in μmol/min. Express your answer with one decimal place. Your previous attempt suffered from low signal-to-noise ratio when you used a substrate concentration equal to the Km. So you decide to increase the substrate concentration to 0.45 mM. You remeasure the rate of the reaction at this new substrate concentration and find it to be 135.0 μmol/min. Using the same Vmax value found above, calculate the Km value of this enzymatic reaction. Express your answer in μM with one decimal place. Considering the previous two questions, what would be the rate of the reaction at a substrate concentration of 0.01 mM? Express your answer in μmol/min with one decimal place.
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
For an enzyme that has a Km of 25 mM, Vmax of 50 mM/s and kcat of 250 s-1, how long does a single reaction take? Select one: 50 ms 250 ms 4 ms 25 s 4 s
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Vmax of the inhibited enzyme reaction?

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Using linear regression analysis, determine the values of Vmax and KM of the enzyme in the PRESENCE of inhibitor: (Express your answer in 3 decimal places, do NOT include the units)