Consider the reaction catalyzed by hexokinase:ATP $+$ glucose $\rightleftharpoons$ ADP $+$ glucose- 6 -phosphateA mixture containing $40\mathrm{mM}$ ATP and $20\mathrm{mM}$ glucose was incubated with hexokinase at $\mathrm{pH}7$ and $25^{\circ }\mathrm{C}$. Calculate the equilibrium concentrations of the reactants and products (see Table $16-3$ ).

Step by Step Solution: Step 1. Determine the equilibrium constant K for the given reaction at 25°C and pH 7 using the table 16-3. Step 2. Write the equilibrium expression for the reaction K = [ADP][Gluc-6-P]/[ATP][Gluc]. Step 3. Let's assume the concentration change at equilibrium be x M for both glucose and ADP. Step 4. Therefore, at equilibrium, the concentration of ATP will be (40-x) M, and the concentration of Glucose-6-phosphate will be (20+x) M. Step 5. Substitute these values in the equilibrium expression and simplify to get the expression in terms of x: K = (20+x)^2 * x / (40-x). Step 6. Solve this quadratic equation for x using the quadratic formula to get the value of x = 4.39 M. Step 7. Substitute this value of x in the concentrations expressions to calculate the values at equilibrium. Step 8. At equilibrium, [ADP] = x = 4.39 M, [ATP] = 35.61 M, [glucose] = 15.61 M, [glucose-6-phosphate] = 24.39 M.