Table of contents

1. Introduction to Anatomy & Physiology5h 43m
2. Cell Chemistry & Cell Components12h 36m
3. Energy & Cell Processes10h 6m
4. Tissues & Histology10h 3m
5. Integumentary System2h 20m
6. Bones & Skeletal Tissue2h 16m
7. The Skeletal System2h 35m
8. Joints2h 17m
9. Muscle Tissue2h 33m
10. Muscles1h 11m
11. Nervous Tissue and Nervous System1h 35m
12. The Central Nervous System1h 6m
- Introduction to the Central Nervous System
  18m
- The Cerebrum
  48m
13. The Peripheral Nervous System1h 26m
14. The Autonomic Nervous System1h 38m
15. The Special Senses2h 41m
16. The Endocrine System2h 48m
17. The Blood3h 22m
18. The Heart3h 42m
19. The Blood Vessels3h 35m
20. The Lymphatic System3h 16m
21. The Immune System14h 37m
22. The Respiratory System3h 20m
23. The Digestive System2h 5m
24. Metabolism and Nutrition4h 0m
25. The Urinary System2h 39m
26. Fluid and Electrolyte Balance, Acid Base Balance37m
27. The Reproductive System2h 5m
28. Human Development1h 21m
29. Heredity3h 32m

2. Cell Chemistry & Cell Components

Prokaryotic & Eukaryotic Cells

Anatomy & Physiology

2. Cell Chemistry & Cell Components

Prokaryotic & Eukaryotic Cells

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2:12 minutes

Problem 15

Textbook Question

Imagine a spherical cell with a radius of 10 μm. What is the cell's surface area in μm2? Its volume, in μm3? (Note: For a sphere of radius r, surface area = 4πr2 and volume = 4/3πr3. Remember that the value of π 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 μm. Compare the surface-to-volume ratios of the two cells. How is this comparison significant to the functioning of cells?

Verified step by step guidance

Step 1: Calculate the surface area of the first cell using the formula for the surface area of a sphere: \( \text{Surface Area} = 4\pi r^2 \). Substitute \( r = 10 \mu m \) into the formula.

Step 2: Calculate the volume of the first cell using the formula for the volume of a sphere: \( \text{Volume} = \frac{4}{3}\pi r^3 \). Substitute \( r = 10 \mu m \) into the formula.

Step 3: Determine the surface area to volume ratio for the first cell by dividing the surface area by the volume.

Step 4: Repeat Steps 1 to 3 for the second cell with \( r = 20 \mu m \). Calculate its surface area, volume, and surface area to volume ratio.

Step 5: Compare the surface area to volume ratios of the two cells and discuss the significance. Consider how a larger surface area to volume ratio might affect cellular processes such as nutrient uptake and waste removal.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Surface Area and Volume Formulas

The surface area and volume of a sphere are calculated using specific mathematical formulas: surface area = 4πr² and volume = 4/3πr³. These formulas are essential for determining the physical properties of spherical objects, such as cells, which are often modeled as spheres in biological contexts. Understanding these formulas allows for the calculation of how much surface area is available for processes like nutrient absorption and waste elimination.

Surface-to-Volume Ratio

The surface-to-volume ratio (SA:V) is a critical concept in biology that describes the relationship between the surface area of a cell and its volume. A higher SA:V ratio indicates that a cell has more surface area relative to its volume, which is advantageous for efficient exchange of materials with the environment. This ratio influences cellular functions, growth, and the ability to maintain homeostasis, making it a key factor in cell size and shape.

Biological Significance of Cell Size

Cell size is significant in biology because it affects metabolic rates, nutrient uptake, and waste removal. Smaller cells typically have a higher surface-to-volume ratio, facilitating more efficient transport of substances across their membranes. As cells grow larger, their SA:V ratio decreases, which can limit their ability to function effectively, leading to the necessity for specialized structures or division to maintain optimal cellular function.