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You are treating a cell with Rotenone or MG132 and measuring the molar ratio between NADH/NAD+ (see figure). What can you conclude about Rotenone? * p<0.01
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- Combining 0.304 mol Fe2O30.304 mol Fe2O3 with excess carbon produced 11.6 g Fe.11.6 g Fe. Fe2O3+3C⟶2Fe+3COFe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles? What is the theoretical yield of iron in moles? What is the percent yield?Determine the Ki for the inhibitor at 30 °Cand decide what type of inhibitor is being used. Eo T I S V (g/I) (°C) (mmol/ml) (mmol/ml) (mmol/ml-min) 1.6 30 0.1 2.63 1.6 30 0.033 1.92 1.6 30 0.02 1.47 1.6 30 0.01 0.96 1.6 30 0.005 0.56 1.6 49.6 0.1 5.13 1.6 49.6 0.033 3.70 1.6 49.6 0.01 1.89 1.6 49.6 0.0067 1.43 1.6 49.6 0.005 1.11 0.92 30 0.1 1.64 0.92 30 0.02 0.90 0.92 30 0.01 0.58 0.92 30 0.6 0.1 1.33 0.92 30 0.6 0.033 0.80 0.92 30 0.6 0.02 0.57What would be the standard free energy change if FADH 2 is used to reduce O 2 to water? Standard reduction potential values to be used for calculations.
- This process requires 151 kJ for every 1.00 mol of Fe3O4 reduced. How much energy (in kilojoules) is required to produce 55 g of iron?Consider the following equilibrium at 298 K. R=8.314 J/K-mol Acetyl-CoA + Oxaloacetate + H20 Citrate + COASH + H* Which one of the following is the correct value for K'eg ifAG" = -32.2 kJ/mol? O 1.11 0.987 1.01 x 10-13 13.0 9.92 x 1012 2480 4.42 x 105 2.27 x 10-6 1.01 -2480 2.56 -13.0 1.08 -0.0130 0.0130 0.926Convert the experimental glucose concentration from mM (millimolar) to g/100 mL. 2.340 mM Include all steps please for the above question.
- CONVERT THE FOLLOWING 8₁₁.. 1 Gm = 82. 1.25 Gm = 83. 500 mg = 24. 1 mg = 85. 4000mcg = _mg _mg Gm _mcg GmWhat would be the standard free energy change if FADH 2 is used to reduce O 2 to water? Standard reduction potential values to be used for calculations. BIU (.a >5 €. Σ FMCalculate the equilibrium constant (K'eg) for each of the three reactions at pH 7.0 and 25 °C, using the AGʻ° values given. (a) glucose 6-phosphatase glucose 6-phosphate + H,O glucose + P; AG'° = –13.8 kJ/mol K'eq= (b) B-galactosidase lactose + H,O glucose + galactose AG'O = -15.9 kJ/mol K'eq=
- Use Beer Lamberts law to calculate the concentration of oxidised DCPIP A= 600 L = 1cm K = 11800 M^-1You make reaction progress curve by plotting absorbance vs time (seconds) and find the equation of the line to be y = -0.00235x + 0.7129. Calculate the U/µL and U/mL of lactate dehydrogenase activity in this fraction. The LDH activity is done identical to what is indicated in the lab manual. Show each step of the calculation from AU/time to M/min, to mol/min, to µmol/min to µmol/min/µL (=U/µL).Refer to the figure shown here, and determine the value of E for the overall oxidation/reduction reaction (refer to the book/lecture slides if you need help with the overall reaction). 121/202 + 2H+ + 2e → H₂0 NAD + H* + 2e → NADH O-1.136 volts O 0.496 volts O+1.136 volts voltsm -0.496 volts EU (volts) +0.816 - 0.320