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Cell Biology - Introduction to Cells

Grade 11IBBiology

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 smallest unit of life, and cells arise from pre-existing cells.

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Exceptions to Cell Theory: Striated muscle fibers (multinucleated and very long), Giant Algae (AcetabulariaAcetabularia - large but unicellular), and Aseptate fungi (lack septa, continuous cytoplasm).

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Functions of Life: All organisms carry out Metabolism, Reproduction, Homeostasis, Growth, Response, Excretion, and Nutrition (MR H GREN).

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Surface Area to Volume Ratio: As a cell increases in size, its volume (V∝r3V \propto r^3) increases faster than its surface area (SA∝r2SA \propto r^2). This limits cell size as the metabolic rate exceeds the rate of exchange of materials and heat.

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Emergent Properties: Multicellular organisms show properties that arise from the interaction of their cellular components; 'the whole is greater than the sum of its parts'.

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Stem Cells: Undifferentiated cells that can divide and differentiate along different pathways. They are used to treat Stargardt's disease (replacing retinal cells) and Leukemia (replacing bone marrow cells).

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Differentiation: Involves the expression of some genes and not others in a cell’s genome, leading to specialized cell functions.

📐Formulae

Magnification=Size of ImageActual Size of SpecimenMagnification = \frac{\text{Size of Image}}{\text{Actual Size of Specimen}}

ActualSize=Size of ImageMagnificationActual Size = \frac{\text{Size of Image}}{\text{Magnification}}

SA:V Ratio=Surface AreaVolumeSA:V \text{ Ratio} = \frac{\text{Surface Area}}{\text{Volume}}

1 mm=103 μm=106 nm1\text{ mm} = 10^3\text{ }\mu\text{m} = 10^6\text{ nm}

💡Examples

Problem 1:

A student measures the length of a cell on a micrograph as 45 mm45\text{ mm}. If the magnification of the image is ×1500\times 1500, calculate the actual length of the cell in micrometers (μm\mu\text{m}).

Solution:

30 μm30\text{ }\mu\text{m}

Explanation:

First, convert the image size to micrometers: 45 mm×1000=45,000 μm45\text{ mm} \times 1000 = 45,000\text{ }\mu\text{m}. Use the formula ActualSize=ImageSizeMagnificationActual Size = \frac{Image Size}{Magnification}. Therefore, ActualSize=45,0001500=30 μmActual Size = \frac{45,000}{1500} = 30\text{ }\mu\text{m}.

Problem 2:

Calculate the surface area to volume ratio for a spherical cell with a radius (rr) of 2 μm2\text{ }\mu\text{m}.

Solution:

1.5 μm−11.5\text{ }\mu\text{m}^{-1}

Explanation:

Using the formulae for a sphere: SA=4πr2=4π(2)2=16πSA = 4\pi r^2 = 4\pi (2)^2 = 16\pi and V=43πr3=43π(2)3=323πV = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2)^3 = \frac{32}{3}\pi. The ratio is 16π32π/3=16×332=32=1.5\frac{16\pi}{32\pi/3} = 16 \times \frac{3}{32} = \frac{3}{2} = 1.5.

Introduction to Cells - Revision Notes & Key Diagrams | IB Grade 11 Biology