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Cell Biology - Unicellular and Multicellular Organisms

Grade 6IB

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

🔑Concepts

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Unicellular organisms consist of a single cell that carries out all life processes, such as BacteriaBacteria, AmoebaAmoeba, and ParameciumParamecium.

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Multicellular organisms are composed of many cells that are specialized to perform specific functions, such as humans, oak trees, and FungiFungi.

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Biological organization in multicellular organisms follows a hierarchy: 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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Cell specialization (differentiation) allows multicellular organisms to be more complex and efficient than unicellular ones.

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The Surface Area to Volume ratio (SAV\frac{SA}{V}) limits cell size; as a cell grows, its volume (VV) increases faster than its surface area (SASA), making nutrient exchange like O2O_2 and CO2CO_2 less efficient.

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Magnification (MM) is used to calculate the size of microscopic organisms where II is the image size and AA is the actual size.

📐Formulae

M=IAM = \frac{I}{A}

A=IMA = \frac{I}{M}

SA:V ratio=Total Surface AreaTotal VolumeSA:V \text{ ratio} = \frac{\text{Total Surface Area}}{\text{Total Volume}}

1 mm=1,000μm1 \text{ mm} = 1,000 \mu\text{m}

💡Examples

Problem 1:

An image of a plant cell measures 40 mm40 \text{ mm} in a textbook. If the actual size of the cell is 0.08 mm0.08 \text{ mm}, calculate the magnification.

Solution:

M=IA=40 mm0.08 mm=500×M = \frac{I}{A} = \frac{40 \text{ mm}}{0.08 \text{ mm}} = 500\times

Explanation:

To find the magnification, we divide the image size (I=40 mmI = 40 \text{ mm}) by the actual size (A=0.08 mmA = 0.08 \text{ mm}).

Problem 2:

Calculate the SAV\frac{SA}{V} ratio for a cube-shaped model cell with a side length (ss) of 2 cm2 \text{ cm}.

Solution:

SA=6s2=6(22)=24 cm2SA = 6s^2 = 6(2^2) = 24 \text{ cm}^2, V=s3=23=8 cm3V = s^3 = 2^3 = 8 \text{ cm}^3, SAV=248=3 cm−1\frac{SA}{V} = \frac{24}{8} = 3 \text{ cm}^{-1}

Explanation:

The surface area is calculated by 6×side26 \times \text{side}^2 and volume by side3\text{side}^3. The ratio shows how much membrane area is available per unit of volume.

Problem 3:

Convert a cell measurement of 0.05 mm0.05 \text{ mm} into micrometers (μm\mu\text{m}).

Solution:

0.05 mm×1,000=50μm0.05 \text{ mm} \times 1,000 = 50 \mu\text{m}

Explanation:

Since 1 mm=103μm1 \text{ mm} = 10^3 \mu\text{m}, we multiply the value in millimeters by 1,0001,000 to get the size in micrometers.