If an E. coli cell has a surface area-to-volume (S/V) ratio of 4.5, and a Pelagibacter ubique has an S/V ratio of 22, which cell will be able to exchange nutrients and wastes with the environment more efficiently?

Question

Biology

Darion

26 September, 18:06

If an E. coli cell has a surface area-to-volume (S/V) ratio of 4.5, and a Pelagibacter ubique has an S/V ratio of 22, which cell will be able to exchange nutrients and wastes with the environment more efficiently?

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Answers (1)

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Backbone · Answer 1

The Pelagibacter ubique

Explanation:

If cell has a high surface to value ratio it means it will have enough surface area to intake nutrients for metabolism and also expelling the waste material out. One more important factor is that Pelagibacter ubique has small size, hence nutrient exchange will be fast. The smaller cell will have high metabolic activity.

Also, the cytoplasmic higher membrane to cytoplasm ratio corresponds to high metabolism. The smaller microorganism will also have higher membrane to cytoplasm ratio.