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Cell Biology - Cell Organelles

Grade 9IB

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

🔑Concepts

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Cell Theory: All living organisms are composed of cells, the cell is the basic unit of life, and cells arise from pre-existing cells.

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Prokaryotic vs. Eukaryotic Cells: Prokaryotes (e.g., bacteria) lack a membrane-bound nucleus and organelles, while Eukaryotes contain a defined nucleus and specialized structures like mitochondria.

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The Nucleus: The 'control center' containing genetic material in the form of DNADNA (deoxyribonucleic acid) and the nucleolus where ribosomes are produced.

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Mitochondria: The site of aerobic cellular respiration, generating energy in the form of ATPATP (adenosine triphosphate). They contain a double membrane with internal folds called cristaecristae.

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Ribosomes: Small structures composed of RNARNA and protein responsible for protein synthesis. They can be 'free' in the cytoplasm or attached to the Rough Endoplasmic Reticulum (RERRER).

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Endoplasmic Reticulum (ER): The Rough ER is studded with ribosomes for protein transport; the Smooth ER is involved in lipid synthesis and detoxification.

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Golgi Apparatus: A series of flattened sacs that modify, sort, and package proteins and lipids for secretion or delivery to other organelles.

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Chloroplasts: Found in plant cells and algae, these contain chlorophyllchlorophyll and are the site of photosynthesis, converting light energy into chemical energy stored as C6H12O6C_6H_{12}O_6.

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Cell Membrane: A semi-permeable phospholipid bilayer that regulates the movement of substances in and out of the cell via passive and active transport.

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Surface Area to Volume Ratio (SA:VSA:V): As a cell increases in size, its volume (VV) grows faster than its surface area (SASA), which limits the efficiency of diffusion and waste removal.

📐Formulae

Magnification=Image size (I)Actual size (A)Magnification = \frac{\text{Image size (I)}}{\text{Actual size (A)}}

C6H12O6+6O2→6CO2+6H2O+Energy(ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + Energy (ATP)

6CO2+6H2O→lightC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{light} C_6H_{12}O_6 + 6O_2

SA:V=Total Surface AreaTotal VolumeSA:V = \frac{\text{Total Surface Area}}{\text{Total Volume}}

1μm=10−3mm=10−6m1 \mu m = 10^{-3} mm = 10^{-6} m

💡Examples

Problem 1:

A student views a plant cell under a light microscope. The actual width of the cell is 25μm25 \mu m. If the image of the cell in the student's drawing measures 50mm50 mm, calculate the magnification used.

Solution:

First, convert units so they match. 50mm=50,000μm50 mm = 50,000 \mu m. Then, Magnification=IA=50,000μm25μm=2000Magnification = \frac{I}{A} = \frac{50,000 \mu m}{25 \mu m} = 2000.

Explanation:

The magnification is 2000×2000\times. It is crucial to ensure both 'Image' and 'Actual' sizes are in the same units (typically μm\mu m) before dividing.

Problem 2:

Compare the Surface Area to Volume ratio of two cubical cells: Cell A with a side length of 2μm2 \mu m and Cell B with a side length of 4μm4 \mu m.

Solution:

For Cell A: SA=6×(22)=24μm2SA = 6 \times (2^2) = 24 \mu m^2, V=23=8μm3V = 2^3 = 8 \mu m^3. SA:V=3:1SA:V = 3:1. For Cell B: SA=6×(42)=96μm2SA = 6 \times (4^2) = 96 \mu m^2, V=43=64μm3V = 4^3 = 64 \mu m^3. SA:V=1.5:1SA:V = 1.5:1.

Explanation:

Cell A has a higher SA:VSA:V ratio (3:13:1) compared to Cell B (1.5:11.5:1). This explains why smaller cells are more efficient at exchanging materials with their environment.