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Biology: Life and Systems - Cell Theory and Organelles

Grade 8IB

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

🔑Concepts

The Cell Theory: 1) All living organisms are composed of one or more cells. 2) The cell is the basic unit of structure and organization in organisms. 3) Cells arise from pre-existing cells.

Prokaryotic vs. Eukaryotic Cells: Prokaryotes lack a membrane-bound nucleus and organelles (e.g., bacteria), while Eukaryotes have a distinct nucleus and specialized organelles (e.g., plants and animals).

Organelles: Specialized structures within a cell. Key examples include the Nucleus (contains DNA), Mitochondria (site of aerobic respiration), and Ribosomes (site of protein synthesis).

Plant vs. Animal Cells: Plant cells possess a rigid cell wall made of cellulose and chloroplasts for photosynthesis, whereas animal cells lack these but may have small, temporary vacuoles and centrioles.

Cellular Respiration: The process by which mitochondria convert glucose into energy, represented by the chemical equation: C6H12O6+6O26CO2+6H2O+ATPC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{ATP}.

Photosynthesis: The process in chloroplasts that converts light energy into chemical energy: 6CO2+6H2O+lightC6H12O6+6O26CO_2 + 6H_2O + \text{light} \rightarrow C_6H_{12}O_6 + 6O_2.

Surface Area to Volume Ratio: As a cell increases in size, its volume (VV) increases faster than its surface area (SASA), which limits the efficiency of diffusion across the cell membrane.

📐Formulae

M=IAM = \frac{I}{A}

A=IMA = \frac{I}{M}

C6H12O6+6O26CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}

SA:V=6s2s3SA:V = \frac{6s^2}{s^3}

💡Examples

Problem 1:

A student observes a biological specimen under a microscope. The image size of a cell is measured to be 40 mm40\text{ mm}, and the actual size of the cell is 0.08 mm0.08\text{ mm}. Calculate the magnification used.

Solution:

M=IA=40 mm0.08 mm=500xM = \frac{I}{A} = \frac{40\text{ mm}}{0.08\text{ mm}} = 500x

Explanation:

Magnification (MM) is calculated by dividing the measured image size (II) by the actual size of the specimen (AA). Ensure both measurements are in the same units before dividing.

Problem 2:

A cube-shaped model of a cell has a side length (ss) of 2 \mum2\text{ \mu m}. Calculate its Surface Area to Volume ratio (SA:VSA:V).

Solution:

SA=6×s2=6×(2)2=24 \mum2SA = 6 \times s^2 = 6 \times (2)^2 = 24\text{ \mu m}^2, V=s3=23=8 \mum3V = s^3 = 2^3 = 8\text{ \mu m}^3, Ratio=248=3:1\text{Ratio} = \frac{24}{8} = 3:1

Explanation:

The SA:VSA:V ratio is critical for cell survival. A higher ratio (found in smaller cells) allows for faster exchange of nutrients and waste through the plasma membrane.

Cell Theory and Organelles - Revision Notes & Key Formulas | IB Grade 8 Science