PLANT PHYSIOLOGY
RESPIRATION IN PLANTS
Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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2870 kJ
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1870 kJ
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3870 kJ
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3370 kJ
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Detailed explanation-1: -The theoretical maximum yield of ATP for the oxidation of one molecule of glucose during aerobic respiration is 38. In terms of substrate-level phosphorylation, oxidative phosphorylation, and the component pathways involved, briefly explain how this number is obtained.
Detailed explanation-2: -Aerobic respiration About 3000 kJ mol-1 of energy is released. Burning glucose in air would release this amount of energy in one go.
Detailed explanation-3: -Glycolysis is the process by which glucose is broken down within the cytoplasm of a cell to form pyruvate. Under aerobic conditions, pyruvate can diffuse into mitochondria, where it enters the citric acid cycle and generates reducing equivalents in the form of NADH and FADH2.
Detailed explanation-4: -Irrespective of the path (aerobic or anaerobic) taken, glycolysis results in a net gain of two molecules of ATP per molecule of glucose.