Amount of energy available per mole of oxygen used in biological oxidation is
A. 114 kcal
B. 686 kcal
C. 256 kcal
D. 60 kcal
Answer
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Hint: Our cells – and those of other living life forms – are great at reaping energy from glucose and other natural particles, for example, fats and amino acids.
Complete answer:
The oxidative breakdown of respiratory substrates with the assistance of oxygen is known as vigorous breath. It includes the total breakdown of substrates into carbon dioxide and water delivering energy. The energy delivered per mole of glucose or in 6 moles of oxygen is 686 Kcal or 2870 kJ.
\[{C_6}{H_{12}}{O_{6}} + 6{O_{2}} + 6{H_2}O + 38ADP + 38Pe6C{O_2} + 12{H_2}O + 38ATP + 420Kcal\]
Absolute free energy change during reaction: 686kcal
ATP synthesised (net addition): 36 ATP mol
The above condition tells around 6 moles of oxygen utilized which is equivalent to 686 kcal. Consequently for one mole, it will be $\dfrac{686}{6}$ = 114 kcal.
Hence, the correct answer is option (A).
Note: The reactions that separate energy from particles like glucose are called catabolic reactions. That implies they include breaking a bigger molecule into little pieces. For instance, when glucose is separated within the sight of oxygen, it's changed over into six carbon dioxide atoms and six water particles. In a cell, this general reaction is separated into numerous littler advances. Energy contained in the bonds of glucose is delivered in little blasts, and some of it is caught as adenosine triphosphate (ATP), a little particle that powers reaction in the cell. A significant part of the energy from glucose is disseminated as warmth, yet enough is caught to keep the metabolism of the cell running.
Complete answer:
The oxidative breakdown of respiratory substrates with the assistance of oxygen is known as vigorous breath. It includes the total breakdown of substrates into carbon dioxide and water delivering energy. The energy delivered per mole of glucose or in 6 moles of oxygen is 686 Kcal or 2870 kJ.
\[{C_6}{H_{12}}{O_{6}} + 6{O_{2}} + 6{H_2}O + 38ADP + 38Pe6C{O_2} + 12{H_2}O + 38ATP + 420Kcal\]
Absolute free energy change during reaction: 686kcal
ATP synthesised (net addition): 36 ATP mol
The above condition tells around 6 moles of oxygen utilized which is equivalent to 686 kcal. Consequently for one mole, it will be $\dfrac{686}{6}$ = 114 kcal.
Hence, the correct answer is option (A).
Note: The reactions that separate energy from particles like glucose are called catabolic reactions. That implies they include breaking a bigger molecule into little pieces. For instance, when glucose is separated within the sight of oxygen, it's changed over into six carbon dioxide atoms and six water particles. In a cell, this general reaction is separated into numerous littler advances. Energy contained in the bonds of glucose is delivered in little blasts, and some of it is caught as adenosine triphosphate (ATP), a little particle that powers reaction in the cell. A significant part of the energy from glucose is disseminated as warmth, yet enough is caught to keep the metabolism of the cell running.
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