kinetic values should be more than 10% different 2. The kinetics of an enzyme are measured as a function of substrate concentration in the presence andabsence of 100 uM inhibitor. Show calculations, include the table of values needed to make your graphsand provide full explanations for each of the following.Velocity (uM/minute)[S] (UM)No inhibitorInhibitor310.42.1514.52.91022.54.53033.86.89040.58.1i) Calculate the values of Km and Vmax in the absence of the inhibitor? Show all calculations. Circle yourfinal answers. Make sure units are included.{insert calculations here}ii) What are the values of Km and Vmax in the presence of the inhibitor? Show all calculations. Circleyour final answers. Make sure units are included.{insert calculations here}iii) What type of inhibition is this? Explain your reasoning{insert answer and explanation here}

Enzymes are catalysts that increase the rate of biochemical reactions. The inhibition of enzyme…

2. The kinetics of an enzyme are measured as a function of substrate concentration in the presence and absence of 100 uM inhibitor. Show calculations, include the table of values needed to make your graphs and provide full explanations for each of the following. Velocity (uM/minute) [S] (UM) No inhibitor Inhibitor 3 10.4 21 5 14.5 2.9 10 22.5 4.5 30 33.8 6.8 90 40.5 8.1

2. The kinetics of an enzyme are measured as a function of substrate concentration in the presence and absence of 100 uM inhibitor. Show calculations, include the table of values needed to make your graphs and provide full explanations for each of the following. Velocity (uM/minute) [S] (UM) No inhibitor Inhibitor 3 10.4 21 5 14.5 2.9 10 22.5 4.5 30 33.8 6.8 90 40.5 8.1

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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1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrate concentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in each experiment is 1.20 x 10 M. [1] = 5.00 x10 M. (Note: v initial rate or velocity.) [S], M 1.25E-3 2.50E-3 5.00E-3 10.0E-3 v, M/min, no inhibitor 0.480E-3 0.660E-3 0.810E-3 0.920E-3 v, M/min, 5.00x10 M 9.43E-5 1.41E-4 1.70E-4 2.10E-4 b). What kind of inhibition occurred. c). Calculate the value of Ki = a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and the values of Vmax and KM at 0.500 mM of I. it in 'How many molecules of
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062
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)
An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of Km tor this enzyme? Substrate added, umol/L Vo uM/sec 48 97 175 187 192 197 2 16 100 1000 198 OA 1 uM O8.2 pM OC 4 uM. OD. 8 uM OE 16 IM OF 1000 uM OG 97 uM/sec OH 175 UM/sec
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 are seen in Vmax and KM? Draw a line that approximates the result from addition of twice as much inhibitor to the reaction. 1/v (v in mM/min) 0.8 0.6 0.4 0.2 0 0.2 0.4 06 1/[s] ([S] in mM) 0.8
The kinetics of an enzyme were measured as a function of substrate concentration in the presence or absence of 10-4 M inhibitor. The following initial rate data were obtained: Substrate (So µM/L) 3 5 Velocity (Vo, µmol/L-min) Without inhibition 10 30 90 10.4 14.5 4.5 6.8 8.1 Is this a competitive or noncompetitive inhibition? Evaluate KM, KI, and Vmax. (HINT: use linear regression to first estimate kinetic parameters for uninhibited and then for the inhibited) With inhibition 2.1 2.9 22.5 33.8 40.5
Consider the given data for an enzyme-catalyzed reaction. Determine the Vm, Km and the type of inhibition based on the given data below Substrate concentration, uM 30 50 100 300 900 slope y-intercept Complete the table below (include correct units). Experiment A Vm Km Experiment A (Initial velocity without inhibitor, uM-min) Type of Inhibition: 10.4 14.5 22.5 33.8 40.8 Experiment B (Initial velocity with inhibitor, uM-min) 5.1 7.3 13.3 25.7 37.2 Experiment B
Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
Consider the Michaelis-Menten equation, below: Vmaz (S V. k + [S] %3D What is the relationship between changes in substrate concentration and velocity when the concentration of substrate, [S), is well below k7 The S terms cancel out in this equation, so there is no effect of changing substrate concentration. The S] term in the numerator is negligible, so there is no impact of changing substrate concentration. Because the enzyme has reached Vr there is no effect of changing substrate concentrations on enzyme velocity. OThe [S term in the denominator is negligible compared to k There is a linear relationship between substrate concentration and velocity.
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
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077