Final answer:
The surface area-to-volume ratio helps determine a cell's efficiency in exchanging materials with its environment. With the provided formulas for a sphere's surface area and volume, one can calculate this ratio for spherical cells and predict the relative rates of nutrient procurement and waste elimination by diffusion.
Step-by-step explanation:
To calculate the ratio of cellular surface area to cell volume, also known as the surface area-to-volume ratio, you can use the formulas for a sphere's surface area and volume, given that the cells are assumed to be spherical. The surface area (A) of a sphere is calculated by 4πr², and the volume (V) is calculated by (4/3)πr³, where r is the radius of the sphere. Therefore, the ratio of surface area to volume (A/V) would be 3/r, meaning this ratio is inversely proportional to the radius of the sphere.
The above principle explains why as a cell grows in size, its surface area-to-volume ratio decreases, which can impact the cell's ability to procure nutrients or eliminate wastes by diffusion. A higher surface area-to-volume ratio means a cell can more efficiently exchange substances with its environment. This concept is critical when considering the rate of diffusion, as cells with larger surface area-to-volume ratios are generally able to diffuse substances more effectively.
Calculating the Ratio for Microbes in the Human Gut
To calculate the cellular surface area of the microbes in the human gut, use the following steps:
- Compute the surface area for one microbe using the formula A = 4πr² with r = 0.001 mm.
- Multiply the surface area by the number of microbes, which is 10ⁱ⁴, to get the total surface area.
- With the total surface area known, you can infer the relative rates of procurement of nutrients by comparing it to the host cells' available intestinal lining surface area.
Predictions Based On Surface Area-to-Volume Ratios
To predict which cell might eliminate wastes or procure nutrients faster by diffusion, compare the surface area-to-volume ratios. A cell with a higher ratio will likely procure nutrients and eliminate waste more effectively than a cell with a lower ratio.