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- 2. a. Label the following structures as ATP, ADP, AMP, P₁, PPi, or Adenosine. Place your answers in the boxes provided. HO H OH H OH OH O N. H H OH + ATP NH₂ . + ADP 01210 ++++++ O-AMP O O Adenosine *+-+ O 0=210 HO b. Put an "X" in the box for the reaction that is thermodynamically most favorable. OH OPP OPP Adenine + AMP + AdenosineWhich of the following substances does not participate in the CalvinBenson cycle? a. ATP b. NADPH c. RuBP d. PGAL e. O2 f. CO27. A study of hypothetical reaction 2A + B -----> C + D gave these results: Experiment Initial Concentration (mol/L) Initial Rate (mol L-1 s-1) [A] [B] 1 0.10 0.050 6.0 × 10-3 2 0.20 0.050 1.2 × 10-2 3 0.30 0.050 1.8 × 10-2 4 0.20 0.150 1.1 × 10-1 a) What is the rate law for this reaction? b) Calculate the rate constant k and express it in appropriate units. 8. The reaction 2 NO(g) + 2 H2(g) -------> N2(g) + 2 H2O(g) is found to be first-order in H2(g). Which rate equation cannot be correct? a) Rate = k[NO]2[H2] b) Rate = k[H2] c) Rate = k[NO]2[H2]2 9. When phenacyl bromide and pyridine are both dissolved in methanol, they react to form phenacylpyridinium bromide. When equal concentrations of reactants were mixed with methanol at 35°C, these data where obtained: Time (min) [Reactant] (mol/L) Time (min) [Reactant] (mol/L) 0 0.0385 500 0.0208 100 0.0330 600 0.0191 200 0.0288 700 0.0176 300 0.0255 800 0.0163 400 0.0220 1000 0.0143 a) What is…
- 2. (a) H2O (g) obeys the truncated viral equation of state given below at 400 K and up to 100 bar pressure. nRT n'RTB(T) P= + V Calculate the amount of work done as 2.00 mol of H2O (g) expands reversibly and isothermally from 2.00 to 50.0 dm³ at 400 K with B(T) = B(400 K) = -356 cm³ mol-'. (b) An ideal gas is expanded reversibly and isothermally. Decide which of q, w, AU, and AH are 0, positive, or negative.4. Make a conclusion about the practical reversibility of the reaction at 298K by calculating the constant for the reaction of ATP + AMP 2ADP, AGº=-2.10 kJ / mol.Paper electrophoresis of Asn, Ala, Asp, Lys and Ser mixtures at pH = 7 shows the fastest movement towards the anode The fully saponified 10g oil sample consumed 1.87g KOH, and the iodine value of the oil was 28.3. So how many unsaturated double bonds are there in each oil molecule on average? (lodine is known to have an atomic weight of 127) Given that 100g cellulose sample is completely hydrolyzed to obtain 78g glucose, then the percentage of cellulose in the sample is _% Given that the Km of catalase is 20mmol/L and the substrate concentration is 80mmol/L, then the percentage of catalase bound to the substrate is %
- 1. Bomb calorimeters are used to determine the heat of combustion of various samples like fuels. Benzoic acid (C6H5COOH) is typically used as a standard in calibrating bomb calorimeters . In a typical reaction 1.20 g of benzoic acid was burned in excess oxygen gas and releases 31.723 kJ of energy . The temperature of the water bath of the calorimeter is 24.6 degrees Celsius . A. Calculate heat B. Calculate change in internal energy C. Calculate work D. What is its change in enthalpy1. Penicillin is hydrolyzed and thereby rendered inactive penicillinase, an enzyme present in some penicillin-resistant bacteria. The molecular weight of this enzyme in Staphylococcus aureus is 29.6 kilo Daltons (29.6 kg/mole or 29,600 g/mole or 29,600 ng/nmole) The amount of penicillin hydrolyzed in 2 minute in a 10-mL solution containing 109 g (1 ng) of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. [Hint: Convert everything to the same concentration terms] Show all calculations and include spreadsheets and graphs to determine Km, Vmax and kcat for this enzyme. Make sure your final answers have correct units. [Another hint: Note that [S] and amount hydrolyzed are already in concentration terms. So you don't need to worry about the volume for calculating [S] and V. ] Penicillin concentration (microM) Amount hydrolyzed (nanoM) 1 110 3 250 5 340 10 450 30 580…12. Explain why those biological reactions that have their equilibria shifted towards theproducts have negative values for ΔGo of reactions. Explain how equilibria relates toGibbs free energy.
- 13. Calculate the equilibrium constant K'eg, for each of the following reactions at pH 7.0 and 25°C. glucose + Pi a. Glucose 6-phosphate + H20 enz. Glucose 6-phosphatase; AG'O=-13.8kJ/mol1. Calculate the initial velocity (Vo) of a Michaelis-Menten reaction as a fraction of Vmax when [S] = 10Km, when [S] = 0.1Km.2.For question number 1 if a mixture was prepared containing 1 M Glucose 6-Phosphate and 0.001 M Glucose 1-Phosphate the ∆G for this reaction is: included question 1 however, need help with 2 and provided the option for the answer 1.What is the Keq for the conversion of Glucose 1-Phosphate to Glucose 6-Phosphate if the phosphate transfer potential for Glucose 1-Phosphate and Glucose 6-Phosphate are 20.9 kJ/mol and 13.8 kJ/mol respectively?