Which cell is expected to carry out respiration at a more active rate based on size?

The smaller cell is expected to carry out respiration at a more active rate based on size due to its higher surface area-to-volume ratio, which is an important principle in cell biology. As cells increase in size, their volume grows at a faster rate than their surface area. This relationship affects how efficiently a cell can exchange materials, such as oxygen and nutrients, with its environment.

Smaller cells have a greater surface area relative to their volume, facilitating quicker absorption of essential substances and the disbursal of metabolic waste. As a result, these cells can perform cellular respiration more effectively, meeting energy demands more efficiently than larger cells, which might struggle with these exchanges due to their comparatively lower surface area-to-volume ratio.

In larger cells, the increased distance that oxygen and nutrients must travel to reach the mitochondria (where respiration occurs) can hinder the efficiency of respiration. Therefore, smaller cells, with their advantageous surface area-to-volume ratio, are better equipped to sustain a higher rate of respiration.