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Biology: Cells, Organization, and Classification - Diversity and Five-Kingdom Classification

Grade 8IB

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

🔑Concepts

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Cell Theory: All living organisms are composed of one or more cells. The cell is the basic structural and functional unit of life. All cells arise from pre-existing cells.

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Prokaryotes vs. Eukaryotes: Prokaryotic cells (e.g., Monera) lack a membrane-bound nucleus and organelles. Eukaryotic cells (e.g., Protista, Fungi, Plantae, Animalia) contain a defined nucleus and specialized organelles like mitochondria and chloroplasts.

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Levels of Biological Organization: The hierarchy of life follows the order: Cell→Tissue→Organ→Organ System→Organism\text{Cell} \rightarrow \text{Tissue} \rightarrow \text{Organ} \rightarrow \text{Organ System} \rightarrow \text{Organism}.

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Five-Kingdom Classification: Proposed by Robert Whittaker, it classifies life into Monera (unicellular prokaryotes), Protista (mostly unicellular eukaryotes), Fungi (multicellular decomposers with chitin walls), Plantae (autotrophic multicellular eukaryotes), and Animalia (heterotrophic multicellular eukaryotes without cell walls).

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Taxonomic Hierarchy: The classification levels move from broad to specific: Kingdom→Phylum→Class→Order→Family→Genus→Species\text{Kingdom} \rightarrow \text{Phylum} \rightarrow \text{Class} \rightarrow \text{Order} \rightarrow \text{Family} \rightarrow \text{Genus} \rightarrow \text{Species}.

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Binomial Nomenclature: A two-part naming system using the GenusGenus and speciesspecies names. For example, the scientific name for humans is HomoHomo sapienssapiens.

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Surface Area to Volume Ratio (SA:VSA:V): As a cell grows larger, its volume increases faster than its surface area. This ratio limits the size of a cell because efficient diffusion requires a high SA:VSA:V.

📐Formulae

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

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

Surface Area of a Cube=6l2\text{Surface Area of a Cube} = 6l^2

Volume of a Cube=l3\text{Volume of a Cube} = l^3

Surface Area to Volume Ratio=6l2l3=6l\text{Surface Area to Volume Ratio} = \frac{6l^2}{l^3} = \frac{6}{l}

💡Examples

Problem 1:

A biological specimen has an actual length of 0.05 mm0.05\text{ mm}. If it is viewed under a microscope with a magnification of 400×400\times, what will be the size of the image seen through the lens in mm?

Solution:

Image Size=Actual Size×Magnification\text{Image Size} = \text{Actual Size} \times \text{Magnification} Image Size=0.05 mm×400=20 mm\text{Image Size} = 0.05\text{ mm} \times 400 = 20\text{ mm}

Explanation:

To find the image size, we multiply the actual dimensions of the cell by the magnification power of the microscope.

Problem 2:

Compare the SA:VSA:V ratio of two cubical cells: Cell A with a side length of 1 \mum1\text{ \mu m} and Cell B with a side length of 3 \mum3\text{ \mu m}. Which cell is more efficient at nutrient exchange?

Solution:

RatioA=61=6 \mum−1\text{Ratio}_A = \frac{6}{1} = 6\text{ \mu m}^{-1} RatioB=63=2 \mum−1\text{Ratio}_B = \frac{6}{3} = 2\text{ \mu m}^{-1}

Explanation:

Cell A has a higher SA:VSA:V ratio (6>26 > 2), meaning it has more surface area relative to its volume, making it more efficient for the diffusion of nutrients and waste.

Problem 3:

Identify the Kingdom of an organism that is multicellular, eukaryotic, has a cell wall made of chitin, and absorbs nutrients from decaying organic matter.

Solution:

Kingdom Fungi

Explanation:

The presence of a chitinous cell wall and saprotrophic (decomposer) nutrition are the primary identifying features of the Kingdom Fungi.