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Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate ΔG for ATP hydrolysis at 37oC under these conditions. The energy charge is the concentration of ATP plus half the concentration of ADP divided by the total adenine
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Biochemistry: Concepts and Connections (2nd Edition)
- The formation of glutamine from glutamate and ammonium ions requires 14.2 kJ mol-1 of energy input. It is driven by the hydrolysis of ATP to ADP mediated by the enzyme glutamine synthetase. {a) Given that the change in Gibbs energy for the hydrolysis of ATP corresponds to ΔG = -31 kJ mol-1, under the conditions prevailing in a typical cell , can the hydrolysis drive the formation of glutamine? (b) What amount {in moles) of ATP must be hydrolysed to form 1 mol glutamine? (c) Suppose that the radius of a typical cell is 10 μm and that inside it 106 ATP molecules are hydrolysed each second. What is the power density of the cell in watts per cubic metre (1W = 1 Js- 1)? (d) A computer battery delivers about 15 Wand has a volume of 100 cm3 Which has the greater power density, the biological cell or the battery?arrow_forwardThe drug troglitazone was used to treat diabetes but was withdrawn from the market when patientswho took the drug suffered from severe side effects. The data below show the activity of an enzymein the steroid biosynthetic pathway in the presence and absence of 10 µM troglitazone.[S] (µM) v0 (pmol min-1)Create a plot of 1/v0 vs. 1/[S]. Calculate KM and vmax in presence and absence of inhibitor. Whattype of inhibitor is troglitazone? Clearly support your answer.arrow_forwardThe enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose 6-phosphate. After the reactants and products were mixed and allowed to reach equilibrium at 25°C, the concentration of glucose 1-phosphate was 4.5 mM and that of glucose 6-phosphate was 86 mM. Calculate Keq' and AG for this reaction. The reaction coordinate diagram for an enzyme-catalyzed reaction is shown below. How many transition states and intermediates are in the reaction? Is the reaction thermodynamically favorable? Which step is the rate-determining step of the reaction? G Reaction coordinatearrow_forward
- 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?arrow_forwardCoupled reactions occur where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. This approach is common in biological settings. Determine if ATP could be generated by this biochemical reaction. You have calculated that cell potential is +0.637V. An example of a coupled reaction is the first step of glycolysis, the phosphorylation of glucose to form glucose-6-phosphate shown below. kJ/mol The net AG° for this reaction is 1 2 3 н он H. H- H- H H H H 4 6. H. glucose phosphate anion glucose-6-phosphate AG = +14.0 kJ/mol 7 8 9. АТР ADP phosphate anion AG = -30.5 kJ/mol +/- LOarrow_forwardThe protein catalase is an enzyme that catalyzes the decomposition of hydrogen peroxide:2 H2O2 (aq) → 2 H2O (l) + O2 (g)and has a Michaelis-Menten constant of 25 × 10-3 mol·dm-3 and a turnover number of 4.0×107s-1.The total enzyme concentration is 0.016×10-6 mol·dm-3 and the initial substrate concentration is4.32×10-6 mol·dm-3 Calculate the maximum reaction rate (????) for this enzyme, and the initial rateof this reaction. Note that catalase has a single active site.arrow_forward
- 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.arrow_forwardIn a major metabolic pathway involving the monosaccharide glucose, one of the reactions involve the conversion of glucose to glucose-6-phosphate summarized below together with accompanying free energy change: glucose + phosphate = glucose-6-phosphate + H,0 AG = 13.8 kJ · mol-1 (Reaction 1) In cells, the production of G6P (Reaction 1 above) is coupled to a reaction that involves the hydrolysis of ATP to ADP (shown below, Reaction 2) which makes the overall reaction much more favorable to the production of glucose-6-phosphate. ATP + H,O = ADP + phosphate (Reaction 2) Why do you think coupling the production of glucose-6-phosphate to the hydrolysis of ATP makes the overall reaction spontaneous? What can you say about the free energy change accompanying the hydrolysis of ATP (Reaction 2)?arrow_forwardThe reaction in which glucose 6-phosphate (G6P) is converted to fructose 6-phosphate (F6P) is an example of an isomerization reaction. Notice that both molecules have the molecular formula C6H13O9P. The isomerization is catalyzed by glucose 6-phosphate isomerase. This is a reversible reaction in cells, with a moderately negative ΔG, but a very high activation energy. If you have a flask that contains a solution containing cellular concentrations of G6P and F6P, but no enzyme, how will the concentration of each molecule change over time?arrow_forward
- a) (1) Calculate the physiological AG (not AG.) for the reaction: Phosphocreatine + ADP - creatine + ATP Given; Phosphocreatine + H;0 - creatine + Pi ADP + Pi → ATP + H;0 AG.' -43 kJ/mol AG.- +30.5 kJ/mol at 25°C as it occurs in the cytosol of neurons, in which phosphocreatine is present at 4.7 mM, creatine at 1.0 mM, ADP at 0.20 mM, and ATP at 2.6 mM. (R = 8.315 JK-' mol-) (ii) Caleulate the free energy change at standard conditions for the following reaction: Acetaldehyde + NADH + H* + Ethanol + NAD* The half- reactions are: Acetaldehyde + 2H + 2e + Ethanol E°- - 0.20V NAD-+ 2H- + 2e ++ NADH + H- E=-0.32V (F= 96.485 kJ/V/mol)arrow_forwardGiven 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.arrow_forwardCalculate the actual, physiological ΔG for the reaction at 37 °C, as it occurs in the cytosol of neurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6 mM.arrow_forward
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