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Cell: Structure and Function - Cell Theory and Cell as the Basic Unit of Life

Grade 11ICSEBiology

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

🔑Concepts

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The cell is defined as the fundamental structural and functional unit of all living organisms. It is the smallest entity capable of independent existence and performing the essential functions of life.

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Cell Theory was primarily proposed by Matthias Schleiden (1838), a German botanist, and Theodore Schwann (1839), a British zoologist. They concluded that all plants and animals are composed of cells.

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Rudolf Virchow (1855) later refined the theory with the statement 'Omnis cellula e cellulaOmnis\ cellula\ e\ cellula', meaning that all new cells arise from the division of pre-existing cells.

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Modern Cell Theory includes: 1. All living organisms are composed of cells and cell products. 2. The cell is the structural and functional unit of life. 3. New cells arise from pre-existing cells. 4. Genetic information (DNADNA) is passed from parent to daughter cells.

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Cells exhibit a wide range of sizes: MycoplasmaMycoplasma is the smallest (0.3 μm0.3\ \mu m), while the Ostrich egg is the largest isolated single cell (15−18 cm15-18\ cm). In humans, RBCsRBCs are approximately 7 μm7\ \mu m in diameter.

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The Surface Area to Volume ratio (SA:VSA:V) limits cell size. As a cell grows, its volume (VV) increases as the cube of the radius (r3r^3), while its surface area (SASA) increases only as the square (r2r^2), making transport of nutrients like O2O_2 and glucoseglucose less efficient in very large cells.

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Exceptions to Cell Theory include Viruses (non-cellular entities), Coenocytic organisms like RhizopusRhizopus (multinucleate mass), and mature mammalian RBCsRBCs and Sieve tube cells (lack a nucleus).

📐Formulae

Magnification=Size of ImageActual Size of ObjectMagnification = \frac{\text{Size of Image}}{\text{Actual Size of Object}}

Surface Area of a spherical cell=4πr2Surface\ Area\ of\ a\ spherical\ cell = 4\pi r^2

Volume of a spherical cell=43πr3Volume\ of\ a\ spherical\ cell = \frac{4}{3}\pi r^3

Surface Area to Volume Ratio (SA:V)=4πr243πr3=3r\text{Surface Area to Volume Ratio } (SA:V) = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}

💡Examples

Problem 1:

Calculate the magnification if a cell with an actual size of 10 μm10\ \mu m is drawn with a size of 50 mm50\ mm.

Solution:

First, convert all units to the same scale. 50 mm=50,000 μm50\ mm = 50,000\ \mu m. Using the formula: Magnification=50,000 μm10 μm=5,000×Magnification = \frac{50,000\ \mu m}{10\ \mu m} = 5,000\times

Explanation:

Magnification is a dimensionless ratio that indicates how many times larger the image appears compared to the real object.

Problem 2:

Determine the Surface Area to Volume ratio for a spherical cell where r=1 μmr = 1\ \mu m and compare it to a cell where r=3 μmr = 3\ \mu m.

Solution:

For r=1 μmr = 1\ \mu m: SA:V=31=3 μm−1SA:V = \frac{3}{1} = 3\ \mu m^{-1}. For r=3 μmr = 3\ \mu m: SA:V=33=1 μm−1SA:V = \frac{3}{3} = 1\ \mu m^{-1}.

Explanation:

The smaller cell (r=1 μmr = 1\ \mu m) has a higher SA:VSA:V ratio (3 vs 13\ vs\ 1), allowing for more efficient exchange of materials like CO2CO_2 and wastewaste products across the plasma membrane.

Cell Theory and Cell as the Basic Unit of Life Revision - Class 11 Biology ICSE