An enzyme with concentration of 3.0 uM is added to a flask containing 125 uM of substrate. You know the Km is equal to 0.054 mM, and has a Vmax of 1.4 uM/s. What is the observed velocity? O 0.98 uM/s 0.074 uM/s O 5.2 um/s 1.4 uM/s
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- If Km = 2 mM, and Vo = 100 umol.s when [S] = 2 mM, what is the velocity Vo for the reaction when [S] = 18 mM? 360 umol.s 180 umol.s 90 umol.s 200 umol.sAn enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.Plot a line weaver-burk graph for Km=6.30mM and Vmax=360uM/min when an experiment has 5 tubes with concentrations of substrates of 1.0mM, 10mM, 50mM, and 100mM.
- An airplane flying directly eastward at a constant speed travels 293 km in 2.0 h. (a) what is the average velocity of the plane? (b) what is the instantaneous velocity?Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMThe equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)
- From the equation of a lineweaver-burke plot, calculate Km and Vmax. Show your work and include units. The equation is y=393.4x +8.4337What is the initial rate of reaction show by the data in the graph below? A. 0.6 mmol per litre / second B. 6 mmol per litre / second C. 10 mmol per litre / second D. 60 mmol per litre / secondConsider the following reaction at 25°C with the ΔG°’ = +1800 J/mol for the forward reaction.The molar concentrations at the beginning of the reaction were [A] = 19 mM and [B] = 10 mM.After 1 hour, the concentrations were [A] = 16 mM and [B] = 13 mM. Calculate the ΔG of the reaction at the 1 hour timepoint. Please round to 1 decimal point.Gas constant = 8.315 J/mol K
- If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?You are given as following : 20 µl pure LDH on ice, 2.0ml of 6mM NAD+, 2.0ml of 150mM lactate, and 0.14M CAPS buffer. LDH reaction cocktail has final concentration of 1mM NAD+ and 25mM lactate in 0.14M CAPS buffer. LDH activityis measured by mixing 10 µl of LDH sample and 990 µl LDH reaction cocktail before getting ∆A340/min reading on spectrometer. (a). Describe in detail how you would prepare for your LDH reaction cocktail including how to make dilutions.Two solutions, 250.0 mL of 1.00 M CaCl2(aq) and 250.0 mL of 1.00 M K2SO4(aq), are combined, and the temperature decreased by 2.40 degrees C. Determine qrxn per mole of CaSO4(s) formed in the reaction. A) +12.0 kJ/mol B) -12.0 kJ/mol C) +6.00 kJ/mol D) -6.00 kJ/mol