c) 1.0 1.2 Estimate Vmax and KM for all cases. What type inhibitor is [I]? Estimate K₁ and/or Kı' depending on the type inhibitor. Estimate K.cat. Estimate Kcat/KM.
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![1. Consider an enzyme-catalyzed reaction giving the following results at a fixed
enzyme concentration of 1 x 10-6 M in the presence or absence of 5 mM or 10 mM
concentrations of an inhibitor I
[S] (mol/L)
0.0010
0.0025
0.0050
0.0100
0.0500
0.1000
a)
Control
0.00167
0.00333
0.00500
0.00667
0.00909
0.00952
v (m/s)
[I] = 5 mM
0.00090
0.00200 -
0.00333
0.00500
0.00833
0.00909
[I] = 10 mM
0.00062
0.00142
0.00250
0.00400
0.00769
0.00869
Estimate Vmax and KM for all cases.
What type inhibitor is [I]?
Estimate K₁ and/or Kı' depending on the type inhibitor.
Estimate Kcat.
Estimate Kcat/KM.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05dea9d6-4326-4c2b-a236-db1a7bc75ad8%2Ffd7b1fad-912b-4a4e-b3b4-26c6cf4dacbc%2Flo6gkq_processed.jpeg&w=3840&q=75)
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- 4. a. Use the data in the graph above to estimate a KM value for the enzyme in the presence of these metabolites, and enter them into the table below. b. Classify these metabolites as either activators or inhibitors, and explain your rationale below.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 µMFor an enzymatic reaction, the following data were obtained for two different initial enzyme concentrations: [S](g/L) v ([E0] = 0,015 g/L)(g/L.min) v ([E0] = 0,00875 g/L) (g/L.min) 40 2,28 1,34 20 1,74 1,02 13,4 1,4 0,82 10 1,18 0,68 8 1 0,58 (a)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.015 g/L. (b)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.00875 g/L. (c)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.015 g/L. (d)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.00875 g/L.
- 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.2. an enzyme-catalyzed reaction is studied in the presence and absence of an inhibitor. The following data was obtained. (S)[M] V (umol/min) V (+inhibitor)(umol/min) 6 x 10-6 20.8 12 1 x 10-5 29 15 2 x 10-5 45 20 6 x 10-5 67.6 24 1.8 x 10-4 87 28 plot 1/[S] as abscissa and 1/V as ordinate for both catalyzed reaction and reaction with inhibitor. Use the same graph for both plots. calculate the Km of enzyme in the reaction without inhibitor Km1 of the enzyme in the reaction with inhibitor Vmax of the uninhibited reaction Vmax of the inhibited reaction what kind of inhibitor was added to the enzyme-catalyzed reaction? explain your answer in terms of changes in Km and Vmax.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 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
- A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.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.
- Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMThe 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.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?