krit.club logo

Cell Biology - Prokaryotic vs Eukaryotic Cells

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Cell Theory: All organisms are composed of one or more cells. The cell is the basic unit of life, and all cells come from pre-existing cells.

•

Prokaryotic Cells: These are simple, single-celled organisms (e.g., Bacteria and Archaea) that lack a membrane-bound nucleus and organelles. Their DNA is usually a single circular chromosome located in the nucleoid region.

•

Eukaryotic Cells: Complex cells (e.g., Animals, Plants, Fungi, and Protists) that contain a distinct nucleus and membrane-bound organelles such as mitochondria and the Golgi apparatus.

•

Ribosome Size: Prokaryotic cells contain smaller 70S70S ribosomes, whereas eukaryotic cells contain larger 80S80S ribosomes in their cytoplasm.

•

Cell Wall Composition: Most prokaryotes have a cell wall containing peptidoglycan. Among eukaryotes, plants have cellulose walls and fungi have chitin walls, while animal cells lack a cell wall.

•

Genetic Material: Prokaryotes may contain extra-chromosomal DNA called plasmids. Eukaryotic DNA is linear and associated with histone proteins to form chromatin.

•

Size Comparison: Prokaryotic cells typically range from 0.10.1 to 5.0μm5.0 \mu m in diameter, while eukaryotic cells are significantly larger, ranging from 1010 to 100μm100 \mu m.

•

Surface Area to Volume Ratio: As a cell increases in size, its volume (VV) increases faster than its surface area (SASA), leading to a decrease in the SAV\frac{SA}{V} ratio, which limits the efficiency of diffusion.

📐Formulae

Magnification=Size of ImageextActualSizeofSpecimen\text{Magnification} = \frac{\text{Size of Image}}{ ext{Actual Size of Specimen}}

Actual Size(A)=IM\text{Actual Size} (A) = \frac{I}{M}

Surface Area to Volume Ratio=SAV\text{Surface Area to Volume Ratio} = \frac{SA}{V}

Surface Area of a Cube=6s2\text{Surface Area of a Cube} = 6s^2

Volume of a Cube=s3\text{Volume of a Cube} = s^3

💡Examples

Problem 1:

A micrograph of a bacterial cell (prokaryote) shows the cell to be 4 cm4 \text{ cm} long. If the magnification is ×10,000\times 10,000, calculate the actual size of the cell in micrometers (μm\mu m).

Solution:

Using the formula A=IMA = \frac{I}{M}, first convert the image size to μm\mu m: 4 cm=40 mm=40,000μm4 \text{ cm} = 40 \text{ mm} = 40,000 \mu m. Then, A=40,000μm10,000=4μmA = \frac{40,000 \mu m}{10,000} = 4 \mu m.

Explanation:

The actual size of the prokaryote is 4μm4 \mu m, which fits within the typical range for bacterial cells (0.1−5.0μm0.1 - 5.0 \mu m).

Problem 2:

Compare the Surface Area to Volume ratio of a cubic model of a prokaryote (s=1μms = 1 \mu m) and a eukaryotic cell (s=10μms = 10 \mu m).

Solution:

For the prokaryote (1μm1 \mu m): SA=6(1)2=6μm2SA = 6(1)^2 = 6 \mu m^2, V=13=1μm3V = 1^3 = 1 \mu m^3, so SAV=6\frac{SA}{V} = 6. For the eukaryote (10μm10 \mu m): SA=6(10)2=600μm2SA = 6(10)^2 = 600 \mu m^2, V=103=1000μm3V = 10^3 = 1000 \mu m^3, so SAV=0.6\frac{SA}{V} = 0.6.

Explanation:

The smaller prokaryotic cell has a much higher SAV\frac{SA}{V} ratio (6:16:1) compared to the larger eukaryotic cell (0.6:10.6:1), allowing it to exchange materials with the environment more efficiently without specialized transport systems.