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In a calorimeter, the combustion of 1 mole of glucose produces 690 kcal. a.What

ID: 1035982 • Letter: I

Question

In a calorimeter, the combustion of 1 mole of glucose produces 690 kcal.

a.What percentage of ATP energy is produced when 7 moles of glucose react in glycolysis? (7.3 kcal/mole ATP)

b.What percentage of ATP energy is produced when 7 moles of glucose are oxidized into acetyl CoA? (7.3 kcal/mole ATP)

c.What percentage of ATP energy is produced when 7 moles of glucose react in the citric acid cycle? (7.3 kcal/mole ATP)

d.What percentage of ATP energy is produced when 7 moles of glucose are completely oxidized to CO2 and H2O? (7.3 kcal/mole ATP)

Explanation / Answer

Ans. Given, energy released from combustion of glucose = 690 kcal/mol

So,

            Total energy released from combustion of 7 mol glucose =

                                                Molar enthalpy of combustion x Moles of glucose

                                                = (-690 kcal/mol) x 7 mol

                                                = -4830.0 kcal

Note: The –ve sign indicates release of energy.

#a. Total ATP produced in glycolysis = ATP per glucose in glycolysis x Mol of Glu

                                                = (2 ATP/ glucose) x 7 ATP

                                                = 14 ATP

# Total energy conserved in ATP = No. of ATP x dG0’ of ATP hydrolysis

                                                = 14 ATP x (-7.3 kcal mol-1 ATP-1)

                                                = -102.2 kcal

Now,

            % energy conserved in ATP =

                        (Energy conserved in ATP / Total energy released from glucose) x 100

                        = [-102.2 kcal / (-4830 kcal)] x 100

                        = 2.12 %

#b. Glycolysis yields 2 pyruvates + 2 ATP from 1 glucose. Each pyruvate is oxidized to acetyl-CoA by pyruvate dehydrogenase complex without any ATP production.

SO, oxidation of glucose to acetyl-CoA produces 2 ATP per glucose.

So, Total ATP produced till acetyl-CoA formation =

                                                ATP per glucose till acetyl-CoA formation x Mol of Glu

                                                = (2 ATP/ glucose) x 7 ATP

                                                = 14 ATP

# Total energy conserved in ATP = No. of ATP x dG0’ of ATP hydrolysis

                                                = 14 ATP x (-7.3 kcal mol-1 ATP-1)

                                                = -102.2 kcal

Now,

            % energy conserved in ATP =

                        (Energy conserved in ATP / Total energy released from glucose) x 100

                        = [-102.2 kcal / (-4830 kcal)] x 100

                        = 2.12 %

#c. Note: ATP produced only during TCA (1 substrate level phosphorylation per pyruvate. Thus, 2 ATP per glucose) is accounted.

Total ATP produced in TCA = ATP per glucose in TCA x Mol of Glu

                                                = (2 ATP/ glucose) x 7 ATP

                                                = 14 ATP

# Total energy conserved in ATP = No. of ATP x dG0’ of ATP hydrolysis

                                                = 14 ATP x (-7.3 kcal mol-1 ATP-1)

                                                = -102.2 kcal

Now,

            % energy conserved in ATP =

                        (Energy conserved in ATP / Total energy released from glucose) x 100

                        = [-102.2 kcal / (-4830 kcal)] x 100

                        = 2.12 %

#d. Complete oxidation of glucose includes the following processes-

            Glycolysis ----Glycolysis------> 2 Pyruvate + 2 ATP + 2 NADH

            2 Pyruvate ----PDH-------------> Acetyl-CoA + 2 NADH

            2 Acetyl-CoA ---------TCA-----> 2 CO2 + 2 ATP + 6 NADH + 2FADH2

            10 NADH + 2FADH2 -----ETC--> 28 ATP + H2O

That is, complete oxidation of 1 glucose produces 32 ATP/ glucose.

# Total ATP produced upon complete oxidation = ATP per glucose x Mol of Glu

                                                = (32 ATP/ glucose) x 7 ATP

                                                = 224 ATP

# Total energy conserved in ATP = No. of ATP x dG0’ of ATP hydrolysis

                                                = 224 ATP x (-7.3 kcal mol-1 ATP-1)

                                                = -1635.2 kcal

Now,

            % energy conserved in ATP =

                        (Energy conserved in ATP / Total energy released from glucose) x 100

                        = [-1635.2 kcal / (-4830 kcal)] x 100

                        = 33.86 %