yeast cells can catabolise glucose under both aerobic conditions (respiration) and anaerobic conditions (fermentation/glycolysis). In this experiment, we will study both processes using a culture of Saccharomyces cerevisiae which has been aerobically adapted. By relating the amount of ethanol produced to the amount of glucose consumed over the same period of time, we will attempt to determine which process is the most efficient provider of energy to the yeast cells. This assumes of course that the energy needs of the cells are similar under the two sets of conditions.
how can we determine which process is more efficient by calculating Moles ethanol produced per mole glucose catabolized ??
aerobic conditions (respiration)
C6H12O6 + 6O2 → 6CO2 + 6H2O
anaerobic conditions (fermentation/glycolysis)
C6H12O6 → 2C2H5OH + 2CO2
Ethanol is not the product of aerobic respiration. So, we can determine which process is more efficient by calculating moles of CO2 produced per mole of glucose catabolized.