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Under biochemical standard conditions, aerobic respiration produces approximately 38

Chapter 7, Problem 7.30

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QUESTION:

Under biochemical standard conditions, aerobic respiration produces approximately 38 molecules of ATP per molecule of glucose that is completely oxidized. (a) What is the percentage efficiency of aerobic respiration under biochemical standard conditions? (b) The following conditions are more likely to be observed in a living cell: \(p_{\mathrm{CO}_2}=5.3\times10^{-2}\mathrm{\ atm}\), \(p_{\mathrm{O}_2}=0.132\mathrm{\ atm}\), \([\text{ glucose }]=5.6\times10^{-2}\mathrm{\ mol\ }\mathrm{dm}^{-3}\), \([\mathrm{ATP}]=[\mathrm{ADP}]=\left[\mathrm{P}_{\mathrm{i}}\right]=1.0\times10^{-4}\mathrm{\ mol}\mathrm{\ dm}^{-3}\), \(\mathrm{pH}=7.4\), \(T=310\ \mathrm{K}\). Assuming that activities can be replaced by the numerical values of molar concentrations, calculate the efficiency of aerobic respiration under these physiological conditions. (c) A typical diesel engine operates between \(T_{\mathrm{c}}=873\mathrm{\ K}\) and \(T_{\mathrm{h}}=1923\mathrm{\ K}\) with an efficiency that is approximately 75 per cent of the theoretical limit of \(\left(1-T_{\mathrm{c}} / T_{\mathrm{h}}\right)\) (see Section 3.2). Compare the efficiency of a typical diesel engine with that of aerobic respiration under typical physiological conditions (see part b). Why is biological energy conversion more or less efficient than energy conversion in a diesel engine?

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QUESTION:

Under biochemical standard conditions, aerobic respiration produces approximately 38 molecules of ATP per molecule of glucose that is completely oxidized. (a) What is the percentage efficiency of aerobic respiration under biochemical standard conditions? (b) The following conditions are more likely to be observed in a living cell: \(p_{\mathrm{CO}_2}=5.3\times10^{-2}\mathrm{\ atm}\), \(p_{\mathrm{O}_2}=0.132\mathrm{\ atm}\), \([\text{ glucose }]=5.6\times10^{-2}\mathrm{\ mol\ }\mathrm{dm}^{-3}\), \([\mathrm{ATP}]=[\mathrm{ADP}]=\left[\mathrm{P}_{\mathrm{i}}\right]=1.0\times10^{-4}\mathrm{\ mol}\mathrm{\ dm}^{-3}\), \(\mathrm{pH}=7.4\), \(T=310\ \mathrm{K}\). Assuming that activities can be replaced by the numerical values of molar concentrations, calculate the efficiency of aerobic respiration under these physiological conditions. (c) A typical diesel engine operates between \(T_{\mathrm{c}}=873\mathrm{\ K}\) and \(T_{\mathrm{h}}=1923\mathrm{\ K}\) with an efficiency that is approximately 75 per cent of the theoretical limit of \(\left(1-T_{\mathrm{c}} / T_{\mathrm{h}}\right)\) (see Section 3.2). Compare the efficiency of a typical diesel engine with that of aerobic respiration under typical physiological conditions (see part b). Why is biological energy conversion more or less efficient than energy conversion in a diesel engine?

ANSWER:

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Respiration is an important chemical process occurring in all living organisms. It can occur in the presence or absence of oxygen. The two categories of respiration include aerobic respiration, and anaerobic respiration.

Aerobic respiration is a process that happens in the presence of oxygen to generate energy from food. This is found in several plants, animals, humans and other mammals. The end products produced in this process are water and carbon dioxide.

Anaerobic respiration occurs in the absence of oxygen. Glucose is broken down in the absence of oxygen to acquire energy. The fermentation of yeast is an example of anaerobic respiration.  

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