Concept explainers
To review:
The concentration of phosphate at equilibrium is to be calculated.
Introduction:
Glucose-6-phosphatase is the enzyme thatcarries out the hydrolysis of glucose-6-phosphate. The end product of the hydrolysis is a phosphate group and a free glucose molecule. This hydrolysis reaction is an instrumental step in the production of glucose through gluconeogenesis.∆Go’ for the hydrolysis of glucose-6-phosphate is –13.8 kJ/mol (kilojoule per mole).
Explanation of Solution
The hydrolysis reaction of glucose-6-phosphate is shown below:
In case the pH (potential of hydrogen) is 7, the change in Gibbs free energy is denoted by ∆Go’. ∆Go’ for the hydrolysis of glucose-6-phosphate is –13.8 kJ/mol (kilojoule per mole) and the concentration of glucose-6-phosphate is4mM (millimolar).
Change in Gibb’s free energy
At equilibrium,
Calculate the value of equilibrium constant by substituting the values of
The hydrolysis reaction of glucose-6-phosphate is as follows:
The equilibrium constant is related to equilibrium concentrations of the reactants and products as
Here, the concentration of glucose is ignored because glucose is exported out of the cell as soon as it is produced through the hydrolysis reaction. Thus, the above expression becomes:
Substitute
Therefore, it can be concluded that the final concentration of inorganic phosphate is
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Chapter 4 Solutions
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