Compare the following two organisms:

Organism A is a single-celled organism with a surface area-to-volume ratio of 4.5.
Organism B is a multicellular organism with a surface area-to-volume ratio of 3.2.
Which organism’s surface area to volume ratio will better support its cellular interior(s), and why?

Organism B because its lower surface area to volume ratio means it can better maintain a constant internal environment.
Organism A because it has the higher surface area to volume ratio.
Organism B because it is multicellular.
Organism A because it is a single-celled organism.

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
7 days ago

Organism A because it has the higher surface area to volume ratio.

The surface area-to-volume ratio is crucial for the exchange of materials (such as nutrients and waste) between the organism and its environment. A higher surface area to volume ratio (like that of Organism A at 4.5) can facilitate more efficient exchange, as there is more surface area available relative to the volume that needs to be serviced. This can be particularly beneficial for single-celled organisms, where all necessary functions must occur in a limited space.

In contrast, Organism B has a lower surface area to volume ratio (3.2), which typically indicates that the multicellular organism is less efficient in exchanging materials per unit of volume compared to Organism A. Therefore, Organism A's higher surface area-to-volume ratio better supports its cellular interior by promoting efficient exchange processes essential for survival.