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Biology - Cell Structure and Organization

Grade 8IGCSE

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

🔑Concepts

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The cell is the basic functional and structural unit of all living organisms.

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The Nucleus contains genetic material in the form of DNADNA and controls all cell activities.

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Cytoplasm is a jelly-like substance where metabolic chemical reactions take place.

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The Cell Membrane is a partially permeable layer that controls the entry and exit of substances.

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Mitochondria are the site of aerobic respiration, which releases energy in the form of ATPATP.

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Ribosomes are small organelles responsible for protein synthesis.

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Plant cells possess a Cell Wall made of cellulose for structural support and to prevent the cell from bursting.

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Chloroplasts in plant cells contain chlorophyll to absorb light energy for photosynthesis: 6CO2+6H2O→lightC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{light} C_6H_{12}O_6 + 6O_2.

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Large Permanent Vacuoles in plants store cell sap and help maintain turgor pressure.

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Specialized cells include Root Hair Cells (large surface area for osmosis), Xylem Vessels (lignified walls for water transport), and Red Blood Cells (contain HbHb or Haemoglobin for O2O_2 transport).

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Levels of Organization: Organelles →\rightarrow Cells →\rightarrow Tissues →\rightarrow Organs →\rightarrow Organ Systems →\rightarrow Organism.

📐Formulae

Magnification=Image SizeActual SizeMagnification = \frac{\text{Image Size}}{\text{Actual Size}}

Actual Size=Image Size (I)Magnification (M)Actual\ Size = \frac{\text{Image Size (I)}}{\text{Magnification (M)}}

1 mm=1000 μm1\text{ mm} = 1000\text{ }\mu\text{m}

💡Examples

Problem 1:

A student measures a diagram of a mitochondrion. The image length is 20 mm20\text{ mm} and the actual length of the mitochondrion is 2 μm2\text{ }\mu\text{m}. Calculate the magnification.

Solution:

M=20000 μm2 μm=10,000×M = \frac{20000\text{ }\mu\text{m}}{2\text{ }\mu\text{m}} = 10,000\times

Explanation:

First, convert 20 mm20\text{ mm} to micrometres: 20×1000=20,000 μm20 \times 1000 = 20,000\text{ }\mu\text{m}. Then, use the formula M=IAM = \frac{I}{A} to find the magnification.

Problem 2:

Calculate the actual width of a cell if the image width is 4.5 cm4.5\text{ cm} and the magnification is ×150\times 150. Provide the answer in μm\mu\text{m}.

Solution:

A=45000 μm150=300 μmA = \frac{45000\text{ }\mu\text{m}}{150} = 300\text{ }\mu\text{m}

Explanation:

Convert 4.5 cm4.5\text{ cm} to μm\mu\text{m} (4.5×10×1000=45,000 μm4.5 \times 10 \times 1000 = 45,000\text{ }\mu\text{m}). Divide the image size by the magnification.