28. What is the intracellular glucose concentration if the AG for the following reaction is -20.1 kl/mol at 37°C and concentrations for glucose-6-phosphate and phosphate are both 1 mM? glucose-6-phosphate - glucose + P AG" = -13.8 kl/mol a. 2.0 uM 12 uM c. 2.0 mM d. 27 mM e. 87 mM
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- 13. 0.9% (m/v) NaCI solution and 5% (m/v) glucose solution are both isotonic to red blood cells. SHOW your work and watch sig figs & units. c. convert the concentration from M to % (m/v) for a 0.342 M NaC solution. (HINT: convert to g/ml and then multiple by 100%)Determine the values of KM and Vmax for the decar-boxylation of a β-keto acid given the following data. You have to plot the graph by using excel. Please include slope Substrate Concentration (mol L1) Velocity (mM min1)2.500 0.5881.000 0.5000.714 0.4170.526 0.3700.250 0.256The 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.
- Calculate the free energy change for glucose entry into cells when the extracellular concentration is 4.5 mM and the intracellular concentration is 2.1 mM at 37oC. Express your answer in kJ/mol.Calculate glucose concentration. Na (sodium) and glucose secondary active transport. Na transport (which drive glucose import) G=R*T*In(Na in/ Na out)+Z*F*Y(psi symbol) Na in=14mM Na out=145MM Z=+1 F=96.5 KJ/V*mol Y(psi)=-0.05V What is the glucose in and out concentration? Please be very through when explaining this calculation. (I am stuck at why the energy sign changes from negative to positive when using the calculated energy from sodium to glucose)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 49.8
- Calculcate Kcat for PNP substrate for both enzyme concentrations. enzyme volume: 20 ul Bovine Intensince Alkaline phosphatase molecular weight: 140,000 Bovine intenstine Alkaline phosphatase activity: 300 units/ml and 14 units/mg extinction coefficient PNP: 18.5 abs (mM-1 cm-1) Vmax: 0.332 moles/sec a) enzyme 1 concentration: undiluted b) enzyme 2 concentration: 1:1 dilutionCalculate all the glucose data into cmol. What is the number of cmol glucose at timepont 10.42 (19.65 g/L)?Determine the values of KM and Vmax for the decar-boxylation of a β-keto acid given the following data. Substrate Concentration (mol L1) Velocity (mM min1)2.500 0.5881.000 0.5000.714 0.4170.526 0.3700.250 0.256
- From your Lineweaver-Burk plot,the vlaues are: Km Vmax Uninhibited 0.09 mmol/L 3.02 min/mmol Inhibited 6.22 mmol/L 9.98 min/mmol By describing the potential changes in the kinetic parameters, identify and justify the type of inhibitor that was inhibiting the acid phosphatase activity.1. Hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase is 99% complete at equilibrium (i.e., 1% of the substrate remains). Which of the following statements is most nearly correct? (R= 8.315 J/mol-K; T = 298 K) A AG" is-11 kl/mol. B) AG" is +5 kJ/mol. C) AG" is 0 kJ/mol. DAG" is +11 kJ/mol. E) AG" cannot be determined from the information given. Explain your choice in one line: -2.16x298 (2) - 11386.15 -> 11.4 КТ 5 The high state conculto alle My negative and st" positive (either for D) then onditions, oneCalculate what the intracellular glucose concentration should be if to transport glucose in active form from the external medium (c1 = 5 mM) at 37 ºC, a energy input of + 2970 J / mol.