Calculate the equilibrium constant for the reaction glucose-1-phosphate + H2O → glucose + H2PO4− at pH 7.0 and 25°C (ΔG°′ = −20.9 kJ · mol−1).

The ΔG°′ value for glucose-1-phosphate is -20.9 kJ/mol. If glucose and phosphate are both at 4.8 mM, what is the equilibrium concentration of glucose-1-phosphate?

Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places

From a kinetics experiment, the Vmax was determined to be 450µM∙min-1.  For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole.  Assume a reaction volume of 700µL.  Calculate the kcat  (in sec-1) for the enzyme.

During glycolysis, glucose is converted into fructose-6- phosphate in two successive reactions: glucose + ATP glucose 6-phosphate + ADP glucose 6-phosphate fructose 6-phosphate The AG' for the overall reaction is: +15.0 kJ/mol. -18.4 kJ/mol. -16.7 kJ/mol. → -15.0 kJ/mol. +18.4 kJ/mol. AG¹⁰ = -16.7 kJ/mol AG¹⁰ = +1.7 kJ/mol

A particular reaction has a ΔG‡ of 30.0 kJ mol-1 at 25.0 °C. In the presence of an enzyme, the same reaction has a ΔG‡ of 1.50 kJ mol-1 at the same temperature. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)

For a Michaelis-Menten reaction, k₁=5 x 107/M-s, k-1-2 x 104/s, and k2=4 x 10²/s. Calculate the Ks and KM for this reaction. Does substrate binding achieve equilibrium or steady state?

Please provide an explanation for how to work through this practice problem:   Using table 12.1, calculate the free energy change for the synthesis of ATP from cAMP and inorganic phosphate. (Note: cAMP is hydolyzed to AMP, and the free engery of hydrolysis for ATP and ADP is approximately equal.)   Table 12.1: (Compound: Change in Free Energy in kJ/mol)   cAMP: -50.4 Creatine phosphate: -43.3 ATP: -30.5 Glucose 6-phosphate: -13.9 AMP: -9.2

A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = H+ (aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.220 M, [H+] = 3.00 × 10−4 M, and [A¯] = 3.00 × 10−4 M. Calculate the value of pKa for the acid HA. pKa =

Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 23.7, and the temperature is 37.0 °C? AG" for the reaction is -16.7 kJ/mol. The constant R = 8.3145 J/(mol-K) AG= Number kJ/mol