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Cell: The Unit of Life - Cell theory

Grade 11CBSEBiology

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

🔑Concepts

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The cell is the fundamental structural and functional unit of all living organisms. Antonie van Leeuwenhoek first saw and described a live cell.

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Matthias Schleiden (18381838) observed that all plants are composed of different kinds of cells which form the tissues of the plant.

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Theodore Schwann (18391839) studied different types of animal cells and reported that cells had a thin outer layer which is today known as the 'plasma membrane'. He also concluded that the presence of a cell wall is a unique character of plant cells.

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Rudolf Virchow (18551855) first explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e cellulaOmnis\ cellula\text{-}e\ cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape.

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Modern Cell Theory states: (i) All living organisms are composed of cells and products of cells. (ii) All cells arise from pre-existing cells.

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Cells vary greatly in size: Mycoplasmas, the smallest cells, are only 0.3 μm0.3\ \mu m in length; bacteria could be 33 to 5 μm5\ \mu m; human red blood cells are about 7.0 μm7.0\ \mu m in diameter.

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Viruses are considered an exception to the cell theory as they are acellular and do not possess a cellular machinery of their own, remaining inert outside a host cell.

📐Formulae

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

1 μm=10−6 m=10−3 mm1\ \mu m = 10^{-6}\ m = 10^{-3}\ mm

1 nm=10−9 m=10−7 cm1\ nm = 10^{-9}\ m = 10^{-7}\ cm

1 A˚=10−10 m=0.1 nm1\ \text{\AA} = 10^{-10}\ m = 0.1\ nm

💡Examples

Problem 1:

A student observes a human red blood cell (RBC) under a microscope. If the RBC has a diameter of 7.0 μm7.0\ \mu m and the microscope provides a magnification of 1000×1000\times, what is the apparent diameter of the cell in millimeters (mmmm)?

Solution:

The apparent diameter is 7.0 mm7.0\ mm.

Explanation:

Using the formula Magnification=Image SizeActual SizeMagnification = \frac{\text{Image Size}}{\text{Actual Size}}, we can rearrange it to find Image Size=Actual Size×Magnification\text{Image Size} = \text{Actual Size} \times Magnification. Given Actual Size=7.0 μm\text{Actual Size} = 7.0\ \mu m and Magnification=1000Magnification = 1000, the Image Size=7.0 μm×1000=7000 μm\text{Image Size} = 7.0\ \mu m \times 1000 = 7000\ \mu m. Since 1000 μm=1 mm1000\ \mu m = 1\ mm, the apparent size is 7.0 mm7.0\ mm.

Problem 2:

Given that a typical bacterial cell is 3 to 5 μm3\ to\ 5\ \mu m and a Mycoplasma is 0.3 μm0.3\ \mu m, calculate the ratio of the length of the smallest bacterium (3 μm3\ \mu m) to the length of a Mycoplasma.

Solution:

The ratio is 10:110:1.

Explanation:

The ratio is calculated as Length of BacteriumLength of Mycoplasma=3 μm0.3 μm=30.3=10\frac{\text{Length of Bacterium}}{\text{Length of Mycoplasma}} = \frac{3\ \mu m}{0.3\ \mu m} = \frac{3}{0.3} = 10. Therefore, the bacterium is 1010 times larger than the Mycoplasma.

Cell theory - Revision Notes & Key Diagrams | CBSE Class 11 Biology