Many metabolites are maintained at steady‑state concentrations that are far from equilibrium. A comparison of ?′eq and ? , the mass‑action ratio, can determine whether a metabolic reaction is far from equilibrium... Many metabolites are maintained at steady-state concentrations that are far from equilibrium. A comparison of Kg and Q,the mass-action ratio, can determine whether a metabolic reaction is far from equilibrium.The equation for this equilibrium is,fructose 6-phosphate + ATP = fructose 1,6-bisphosphate + ADPCalculate Kea for this reaction at T = 25.0 °C.AG'-14.2 kJ/molKeg =Calculate the mass-action ratio, Q , from the approximate physiological concentrations for rat heart tissue shown in the table.MetaboliteConcentration (µM)fructose 6-phosphate84.0fructose 1,6-bisphosphate25.0ATP12,100ADP1,280Select the true statements about the PFK-1 reaction.The PFK-1 reaction in heart tissue does not reach equilibrium.In heart tissue, the PFK-1 reaction will be driven toward product formation.In heart tissue, the PFK-1 products are more abundant than reactants.Under standard conditions, the PFK-1 reaction reaches equilibrium when the concentrations of all products and reactantsare equal.II

In this question, a metabolic reaction is given. The given equation is : This question has 3 parts : (a) Calculation of K'eq at temperature 25 °C (b) Calculation of mass action ratio Q for which the given concentrations are : (c) Determination of true statement from the given statements.Equilibrium constant (Keq) : Keq is the equilibrium constant which can be calculated by using the following formula : ∆G'° = - RT ln K'eq ...................... (1) Where, ∆G'° = Standard gibbs free energy R = Universal gas constant (J/mol . K) T = Temperature K'eq = Equilibrium constant Now, the given values are : ∆G'° = - 14.2 kJ/mol = -14200 J/mol Temperature = 25 °C Put these values in equation (1) to get the value of K'eq : ∆G'° = - RT ln K'eq-14200 J/mol = - 8.314 Jmol K × 298 K × ln K'eq-14200 J/mol = -2477.57 Jmol ln K'eqln K'eq = -14,200 -2477.57ln K'eq = 5.73 Thus, the value of K'eq is 5.73 .Calculation of mass action ratio (Q) : Mass action ratio is also called reaction quotient which is the ratio of the concentration of products to the concentration of reactants at any time of the reaction. For a general reaction, aA + bB → cC + dD Q = Cc DdAa Bb .............. (1) Where, a, b, c and d are the coefficients of the species A, B, C and D. [A], [B], [C] and [D] are the concentrations of the species A, B, C and D. The given values of concentration are : The given reaction is : Now, calculating the value of Q using formula (1) Q = fructose-1,6-biphosphate ADPfructose-6-phosphate ATPQ = 25.0 μM 1,280 μM84.0 μM 12,100 μMQ = 0.0315 Thus, the value of Q is 0.0315.The direction of progress of reaction can be determined by observing the values of K'eq and Q : If K'eq = Q, the reaction is at equilibrium If K'eq > Q, the reaction favors the forward direction as there is less product formed relative to reactant. If K'eq < Q, the reaction favors the backward direction as there is more product formed relative to reactant. In the given reaction, the calculated value of K'eq is 5.73 and the calculated value of Q is 0.0315. Since, the value of Q is less than the value of K'eq , then there is less product formed as compared to reactant and the reaction does not reach at the equilibrium and the proceeds in the forward direction. Thus, the correct options are : The PFK-1 reaction in heart tissue does not reach equilibrium. In heart tissue, the PFK-1 reaction will be driven toward product formation.(a) The value of K'eq is 5.73 . (b) The value of Q is 0.0315. (c) The correct options are : The PFK-1 reaction in heart tissue does not reach equilibrium. In heart tissue, the PFK-1 reaction will be driven toward product formation.