Problem 15 - Campbell Biology Concepts & Connections

Question

Imagine a spherical cell with a radius of \(10 \mu \mathrm{m}\). What is the cell's surface area in \(\mu \mathrm{m}^{2}\) ? Its volume, in \(\mu \mathrm{m}^{3}\) ? (Note: For a sphere of radius \(r,\) surface area \(=4 \pi r^{2}\) and volume \(=4 / 3 \pi r^{3}\). Remember that the value of \(\pi\) is \(3.14 .\) ) What is the ratio of surface area to volume for this cell? Now do the same calculations for a second cell, this one with a radius of \(20 \mu \mathrm{m}\). Compare the surface-to-volume ratios of the two cells. How is this comparison significant to the functioning of cells?

Step-by-Step Solution

Answer

The first cell's surface-to-volume ratio is 0.3, and the second cell's is 0.15. A higher surface-to-volume ratio is advantageous for cellular efficiency.

Step 1: Calculate Surface Area of First Cell

Use the formula for the surface area of a sphere: \[ A = 4 \pi r^{2} \]. Plug in the radius of the first cell: \[ A_1 = 4 \pi (10)^{2} \]. Simplify: \[ A_1 = 4 \pi \times 100 = 400 \pi \]. Substitute the value of \pi as 3.14: \[ A_1 = 400 \times 3.14 = 1256 \mu \mathrm{m}^{2} \].

Step 2: Calculate Volume of First Cell

Use the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^{3} \]. Plug in the radius of the first cell: \[ V_1 = \frac{4}{3} \pi (10)^{3} \]. Simplify: \[ V_1 = \frac{4}{3} \pi \times 1000 = \frac{4000}{3} \pi \]. Substitute the value of \pi as 3.14: \[ V_1 = \frac{4000}{3} \times 3.14 = 4186.67 \mu \mathrm{m}^{3} \].

Step 3: Calculate Surface-to-Volume Ratio of First Cell

Use the results from Steps 1 and 2 to find the ratio: \[ \text{Ratio}_1 = \frac{A_1}{V_1} = \frac{1256}{4186.67} \]. Simplify: \[ \text{Ratio}_1 \approx 0.3 \].

Step 4: Calculate Surface Area of Second Cell

Use the same surface area formula: \[ A = 4 \pi r^{2} \]. Plug in the radius of the second cell: \[ A_2 = 4 \pi (20)^{2} \]. Simplify: \[ A_2 = 4 \pi \times 400 = 1600 \pi \]. Substitute the value of \pi as 3.14: \[ A_2 = 1600 \times 3.14 = 5024 \mu \mathrm{m}^{2} \].

Step 5: Calculate Volume of Second Cell

Use the same volume formula: \[ V = \frac{4}{3} \pi r^{3} \]. Plug in the radius of the second cell: \[ V_2 = \frac{4}{3} \pi (20)^{3} \]. Simplify: \[ V_2 = \frac{4}{3} \pi \times 8000 = \frac{32000}{3} \pi \]. Substitute the value of \pi as 3.14: \[ V_2 = \frac{32000}{3} \times 3.14 = 33493.33 \mu \mathrm{m}^{3} \].

Step 6: Calculate Surface-to-Volume Ratio of Second Cell

Use the results from Steps 4 and 5 to find the ratio: \[ \text{Ratio}_2 = \frac{A_2}{V_2} = \frac{5024}{33493.33} \]. Simplify: \[ \text{Ratio}_2 \approx 0.15 \].

Step 7: Compare Surface-to-Volume Ratios

The surface-to-volume ratios of the first and second cells are approximately 0.3 and 0.15, respectively. The first cell has a higher surface-to-volume ratio, which is significant for cell functioning because a higher ratio allows more efficient nutrient uptake and waste removal, crucial for smaller cells to maintain their metabolism.

Key Concepts

surface area calculation

The surface area of a cell is crucial for understanding how a cell interacts with its environment. Surface area determines the cell's ability to absorb nutrients and expel waste products. For a spherical cell, the surface area is calculated using the formula: ewline \(A = 4 \pi r^2\). To perform this calculation, we need to know the cell's radius. In our exercise, we have a radius of \(10 \mu m\). Here’s a step-by-step breakdown: ewline \begin{enumerate.first}\item Plug the radius into the formula: \(A_1 = 4 \pi (10)^2\)
\item Calculate the square of the radius: \(100\)
\item Multiply by \(4 \pi\): \(400 \pi\)
\item Substitute \(\pi\) with \(3.14\): \(400 \times 3.14\)
\item Result: \(1256 \mu m^2\)ewline Let’s do a similar comparison for a cell with a radius of \(20 \mu m\). Following the same steps, we find its surface area: \(5024 \mu m^2\). Understanding this aids in visualizing how larger cells have proportionally less surface area for their volume, meaning their surface area to volume ratio is smaller.

volume calculation

Volume is equally important as surface area but serves a different purpose. Volume gives us an idea of the cell's internal environment, which affects metabolic activities. The formula for the volume of a sphere is: ewline\(V = \frac{4}{3} \pi r^3\). Here's how we can calculate it for our cell with a radius of \(10 \mu m\): ewline \begin{enumerate.first}\item Substitute the radius into the formula: \(V_1 = \frac{4}{3} \pi (10)^3\)
\item Compute the cube of the radius: \(1000\)
\item Multiply by \(\frac{4}{3} \pi\): \(\frac{4000}{3} \pi\)
\item Plug in \(\pi\) as \(3.14\): \(\frac{4000}{3} \times 3.14\)
\item Result: \(4186.67 \mu m^3\)ewline Similarly, for the second cell with a radius of \(20 \mu m\), the volume is calculated to be \(33493.33 \mu m^3\). A larger cell has a greater volume, meaning there’s more space internally which requires more resources for maintenance, metabolism, and waste management.

cell metabolism

Cell metabolism refers to the chemical processes that occur within a cell to maintain life. It includes nutrient uptake, energy production, and waste elimination. The surface area to volume ratio (SA:V ratio) is a critical factor in cell metabolism. A high SA:V ratio, found in smaller cells, allows:

• More efficient nutrient absorption
• Faster waste removal
• Quick response to environmental changes

Conversely, larger cells with a lower SA:V ratio face challenges like:

• Slower nutrient uptake
• Inefficient waste expulsion
• Difficulty maintaining homeostasis

In our exercise, comparing a cell with a radius of \(10 \mu m\) to one with \(20 \mu m\), the first cell's SA:V ratio of \(0.3\) illustrates more metabolic efficiency compared to the second cell’s ratio of \(0.15\). This difference underscores why cells stay small or adopt shapes that increase surface area without a large increase in volume, optimizing their metabolic functions.