▲ [S] = 0.2 mM "[S] = 0.4 mM I/Vav 8 9 5 y=0.01357x+3.6976 3 y=0.0795x+2.7693 2 1 -30 -20 -10 0 10 20 30 [I](M) Mixed inhibition, Ki= -21.2 UM ○ Mixed inhibition, Ki= 1/16.5 uM Competitive inhibition, Ki= 16.5 UM ○ Competitive inhibition, Ki= -1/16.5 UM None of the above
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Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex?
- Mixed inhibition, Ki= -21.2 uM
- Mixed inhibition, Ki= 1/16.5 uM
- Competitive inhibition, Ki= 16.5 uM
- Competitive inhibition , Ki= -1/16.5 uM
- None of the above
![▲ [S] = 0.2 mM
"[S] = 0.4 mM
I/Vav
8
9
5
y=0.01357x+3.6976
3
y=0.0795x+2.7693
2
1
-30
-20
-10
0
10
20
30
[I](M)
Mixed inhibition, Ki= -21.2 UM
○ Mixed inhibition, Ki= 1/16.5 uM
Competitive inhibition, Ki= 16.5 UM
○ Competitive inhibition, Ki= -1/16.5 UM
None of the above](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca7515a8-b44b-40fa-83f2-449e6ef6ad98%2F9b715a20-687a-4818-aa45-8fb5dc59f079%2Ftei9otj_processed.jpeg&w=3840&q=75)
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- An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.A schematic representation of the enzyme IspD complexed to inhibitor 3, and a series of inhibitors 3-5 are shown below. Ala202 lle240 mwww NH NH Val263 ОН www HN N- lle177 HN 'N' CI 3 X = N 4 X = C-CN 5 X = C-COO IC50 274 µM IC50 140 nM IC50 35 nM NH2 HN Val266 N -N O-H---- N HN %3D Arg157 HN wwww lle265 Explain why structure 4 is a more potent inhibitor (lower IC50 value) than inhibitor 3 and why structure 5 is a much weaker inhibitor (higher IC50 value) than 3 and 4.The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.
- The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.Chymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. UreaPlease note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
- Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.The rate constant for the uncatalyzed reaction of two molecules of glycine ethyl ester to form glycylglycine ethyl ester is 0.6 M- 1s - 1. In the presence ofCo2 +, the rate constant is 1.5 x 106 M- 1s - 1. What rate enhancement does the catalyst provide?
- 5) In an experiment to investigate the inhibition of the enzyme-glucosidase the following data for the rates of reaction with glucopyranoside for various substrate concentrations was obtained. By constructing a Leaver-Burk plot, determine the value of the Michaelis constant. [S]/ (10-6 mol dm-3) v/ (10-3 mol dm-3 s-1) 1.00 2.00 3.00 4.00 16.7 33.3 41.1 49.8For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.142A 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.