1. Calculate the energy yield of glucose oxidation:
* Standard energy yield: The complete oxidation of 1 mole of glucose yields about 2,870 kJ of energy.
* Energy yield of ATP: The hydrolysis of 1 mole of ATP releases about 30.5 kJ of energy.
2. Calculate the energy efficiency for the mutant organism:
* Mutant ATP yield: 29 moles of ATP produced per mole of glucose.
* Total energy captured by mutant: 29 moles ATP * 30.5 kJ/mole = 884.5 kJ
* Efficiency: (Energy captured / Total energy released) * 100%
= (884.5 kJ / 2870 kJ) * 100%
= approximately 30.8%
3. Compare to normal efficiency:
* Normal ATP yield: 36-38 moles of ATP per mole of glucose.
* Normal efficiency: (36 * 30.5 kJ / 2870 kJ) * 100% to (38 * 30.5 kJ / 2870 kJ) * 100%
* Normal efficiency range: Approximately 38% to 40%
Conclusion:
The mutant organism has an approximate efficiency of 30.8% for cellular respiration, compared to the normal range of 38% to 40%. This mutant is less efficient at capturing energy from glucose oxidation.
