If a michaelis-type enzyme has an apparent Km of 2*10^-8 and apparent Vmax of 1*10^-4 moles/min in the presence of 1*10^-3 M uncompetitive inhibitor (Ki=1*10^-7 M), what is the true Km of the enzyme
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A: Given Values: Vmax = 12 u s-1ml-1 Total enzyme concentration = 0.016 U ml-1
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A: The Michaelis- Menten Equation for Enzyme kinetics is - V= Vmax[S]/(km+[S]) By putting the above…
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Q: [Substrate, uM] Vo with DEDS (µM/min) Vo without DEDS (µM/min) 3.333 0.774 1.196 4.000 0.877 1.316…
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Q: Vo with Vo without [Substrate, µM] DEDS DEDS (µM/min) (µM/min) 3.333 0.774 1.196
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- 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 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.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.
- Given the following data, calculate Keq for the denaturation reaction of the protein β-lactoglobin at 25oC: ΔH° = –88 kJ/mol ΔS° = 0.3 kJ/mol. The free energy of hydrolysis of ATP in systems free of Mg2+ is −35.7 kJ/mol. When the concentration of this ion is 5 mM, ΔG°observed is approximately −31 kJ/mol at pH 7 and 38°C. Suggest a possible reason for this effect.The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1A particular enzyme has a ΔΔG‡ of -22.1 kJ mol-1 at 37.0 °C. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)
- 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.The effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.a) Determine kcat (in units of sec-1) for a particular enzyme, given the following information: Vo = 144 mmol/min; [S] = 2 mM; Km = 0.5 mM; Enzyme Molecular weight = 40,000 mg/mmole; 8 mg of enzyme used in assay generating this data. b) In general, explain how the total enzyme concentration affects turnover number and Vmax?
- 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 minuteShown below is a proposed mechanism for the cleavage of sialic acid by the viral enzyme neuraminidase. The kcat for the wild-type enzyme at pH =6.15, 37 °C is 26.8 s-1.(a) Describe the roles of the following amino acids in the catalytic mechanism: Glu117, Tyr409, and Asp149. List all of the following that apply:general acid/base catalysis (GABC), covalent catalysis, electrostaticstabilization of transition state.(b) Based on the information shown in the scheme, would you expect mutation of Glu 117 to Ala to have a greater effect on KM or kcat?(c) For the R374N mutant at pH = 6.15, 37 °C, kcat is 0.020 s-1, and KMis relatively unaffected. Based on this result, it seems that R374 is morecritical for catalysis than for substrate binding. Explain how R374 stabilizesthe reaction transition state more than the substrate (i.e., what feature of this reaction would explain tighter binding to the transition state vs. substrate?).Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained: