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To measure the rate of organismal respiration, a 10 mL microrespirometer was constructed. These measure relative volume as oxygen is consumed by respiring organisms.

As oxygen gas is consumed during aerobic respiration, it is replaced by $CO_2$ gas at a ratio of one molecule of $CO_2$ for each molecule of $O_2$. In the following protocol, the $CO_2$ produced is removed by a small piece of cotton treated with potassium hydroxide ($KOH$).

$KOH$ reacts with $CO_2$ to form the solid potassium carbonate ($K_2CO_3$). A control microrespirometer was filled with 5 mL of glass beads in one setup while 5 mL of germinating pea seeds were placed in another.

Over the course of 20 minutes, changes in the volume were measured by following the distance of a red soap bubble placed on the end of a 40 uL capillary tube at time = 0.

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In one alternate experiment, instead of using germinating peas, 5 mL of Drosophila melanogaster were placed in the microrespirometer.

Which of the following would this change have on the results?


A faster consumption of oxygen would be recorded because animals carry out aerobic respiration faster than plants.


A slower consumption of oxygen would be recorded because animals only carry out respiration, not photosynthesis; therefore, the carbon dioxide would be absorbed by the $KOH$.


No change in oxygen consumption would occur because 5 grams of living matter in peas is comparable to 5 grams of living matter in animals.


No answer can be given because not enough information is available.

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