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1 = 2, for sufficiently large values of 1. The energetic equivalent of two molecules of ATP is used to activate an amino acid, yet only one molecule of ATP is used. Explain
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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 AG for ATP hydrolysis at 37 °C under these condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]The ΔG°′ for hydrolytically removing a phosphoryl group from ATP is about twice as large as the ΔG°′ for hydrolytically removing a phosphoryl group from AMP (−14 kJ · mol−1). Explain the discrepancy.2+ The activity of the Ca 2+ -ATPase is regulated reversibly under normal conditions to maintain homeostatic concentrations of Ca²- inside the sarcomere. However, in a rare genetic disorder, irreversible activation of the Ca 2+ -ATPase can occur. Assuming 37 °C, pH = 7.4, and the steadystate concentrations for ATP, ADP Pi, and Ca2+ (out) given in part (a), calculate the minimum [Ca2+] inside a sarcomere that has irreversibly activated Ca 2+ -ATPase (i.e., the Ca2+-ATPase activity is always “on”). Express your answer to two significant figures and include the appropriate units.
- A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.Calculate the actual, physiological AG for the reaction Phosphocreatine + ADP = creatine + ATP at 37 °C, as it occurs in the cytosol of neurons, where [phosphocreatine] = 4.7 mM, [creatine] = 1.0 mM, [ADP] = 0.73 mM, and [ATP] = 2.6 mM. Standard Free Energies of Hydrolysis of Some Phosphorylated Compounds Phosphorylated compound AGʻ° (kJ/mol) phosphoenolpyruvate -61.9 phosphocreatine -43.0 ADP (→ AMP + P;) -32.8 ATP (→ ADP + P;) -30.5Consider the complete oxidation of one mole of simple TAG containing behenic acid residues (22:0). I. For one mole of the fatty acid residue, determine the following: d. What is the number of net ATP yield of ATP in the complete oxidation of the fatty acid residue
- A particular enzyme has a ΔΔG‡ of -22.1 kJ mol-1 at 37.0 °C. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.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.
- 28. If 20 mM solution of Acetyl-phosphate is transformed to acetate by incubating with a catalytic amount of acetate kinase until equilibrium is reached. Acetyl-phosphate -> Acetate + Pi At equilibrium, the concentration of Acetyl-phosphate is 8 mM. Determine Keq and AG" for the hydrolysis of Acetyl-phosphate reaction. (R = 0.008314 kJ/mol K) .The isomerization of dihydroxyacetone phosphate (DHAP) to glyceraldehyde 3-phosphate (GAP) is catalyzed by triose phosphate isomerase. In the cell, the concentration ratio of DHAP/GAP = 5.5. Calculate [DHAP] (in M) when [GAP] = 0.00002An enzyme has a rate enhancement of 1.3x106. Calculate the value of ΔΔG‡ at 25.0 °C in kJ mol-1. (Hint: be sure to pay attention to units and signs.) (R = 8.3145 J mol-1 K-1)