Chapter 13, Problem 103FP

Interpretation Introduction

(a)

Interpretation:

The maximum number of moles of ATP that could form from ADP and phosphate if all the energy of combustion of 1 mol of glucose could be utilized should be calculated.

Concept introduction:

The major source of energy in animals is glucose. The energy released by the combustion of glucose is consumed to produce ATP.

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6({\text{aq) + 6 O}}_2\text{(g) }\to {\text{ 6 CO}}_2{\text{(g) + 6 H}}_2\text{O(l) }{\Delta }_r{G}^0=-2870\text{ kJ/mol}$

In cells this reaction is coupled to the synthesis of ATP from ADP and phosphate.

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6({\text{aq) + 6 O}}_2\text{(g) + 36 phosphate + 36 ADP }\to {\text{ 6 CO}}_2{\text{(g) + 6 H}}_2\text{O(l) + 36 ATP}$

ATP synthesis reaction,

${\text{ADP + P}}_i\to {\text{ATP + H}}_2\text{O }{\Delta }_r{G}^0=+34.5\text{ kJ/mol}$

Interpretation Introduction

(b)

Interpretation:

Efficiency of energy conversion of the cell should be calculated.

Concept introduction:

Energy released by the combustion of glucose is not completely goes into the production of ATP. Only a fraction of energy is participating. Some energy is lost as heat.

Interpretation Introduction

(c)

Interpretation:

Δ_rG⁰ for the conversion of 1 mol ADP to ATP and Δ_rG⁰ for oxidation of 1 mol of glucose under the given conditions should be calculated.

Concept introduction:

${\Delta }_{r}G={\Delta }_r{G}^0+RT\ln Q$

Interpretation Introduction

(d)

Interpretation:

The efficiency of energy conversion for the cell under the conditions given in part (c) should be calculated.

Concept introduction:

$\text{Efficiency = }\frac{\text{No}\text{. of moles of ATP }\times \text{ Energy required to produce 1 mole of ATP}}{\text{Energy released by oxidation of 1 mole of glucose}}\times 100\%$