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Cell Biology - Plant vs Animal Cells

Grade 9IB

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, cells are the basic unit of life, and all cells come from pre-existing cells.

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Eukaryotic Cells: Both plant and animal cells are eukaryotic, meaning they contain a membrane-bound nucleus and specialized organelles.

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Shared Organelles: Both cell types contain a Cell Membrane, Nucleus (containing DNA), Cytoplasm, Mitochondria (site of aerobic respiration to produce ATPATP), and Ribosomes (80S80S type for protein synthesis).

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Plant Cell Specifics: Characterized by a rigid Cell Wall made of cellulose for structural support, Chloroplasts containing chlorophyll for photosynthesis, and a large permanent Central Vacuole for turgor pressure.

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Animal Cell Specifics: Generally lack a cell wall and chloroplasts. They possess Centrioles (involved in cell division) and often have small, temporary vacuoles.

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Surface Area to Volume Ratio (SA:VSA:V): As a cell grows, its volume (VV) increases faster than its surface area (SASA). This limits the cell size as the rate of diffusion becomes insufficient to support the cell's metabolic needs.

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Magnification: The relationship between the size of an image and the actual size of the specimen, often expressed using the I=A×MI = A \times M triangle.

📐Formulae

Magnification(M)=Image Size (I)Actual Size (A)Magnification (M) = \frac{\text{Image Size (I)}}{\text{Actual Size (A)}}

ActualSize(A)=Image Size (I)Magnification (M)Actual Size (A) = \frac{\text{Image Size (I)}}{\text{Magnification (M)}}

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

Surface Area of a Cube=6×s2\text{Surface Area of a Cube} = 6 \times s^2

Volume of a Cube=s3\text{Volume of a Cube} = s^3

💡Examples

Problem 1:

A micrograph of a leaf cell shows a chloroplast that measures 15 mm15 \text{ mm} in length. If the magnification of the micrograph is 2000×2000 \times, calculate the actual length of the chloroplast in micrometers (μm\mu\text{m}).

Solution:

A=IMA = \frac{I}{M} A=15 mm2000A = \frac{15 \text{ mm}}{2000} A=0.0075 mmA = 0.0075 \text{ mm} A=0.0075×1000μm=7.5μmA = 0.0075 \times 1000 \mu\text{m} = 7.5 \mu\text{m}

Explanation:

First, use the formula for actual size (A=I/MA = I/M). The image size is 15 mm15 \text{ mm} and magnification is 20002000. Dividing gives the actual size in millimeters (0.0075 mm0.0075 \text{ mm}). To convert millimeters to micrometers, multiply by 10001000.

Problem 2:

Compare the SA:VSA:V ratio of two cubic cells: Cell A with a side length of 2μm2 \mu\text{m} and Cell B with a side length of 4μm4 \mu\text{m}.

Solution:

Cell A: SA=6×22=24μm2, V=23=8μm3  ⟹  SA:V=248=3\text{Cell A: } SA = 6 \times 2^2 = 24 \mu\text{m}^2, \text{ V} = 2^3 = 8 \mu\text{m}^3 \implies SA:V = \frac{24}{8} = 3 Cell B: SA=6×42=96μm2, V=43=64μm3  ⟹  SA:V=9664=1.5\text{Cell B: } SA = 6 \times 4^2 = 96 \mu\text{m}^2, \text{ V} = 4^3 = 64 \mu\text{m}^3 \implies SA:V = \frac{96}{64} = 1.5

Explanation:

As the side length of the cell doubles from 2μm2 \mu\text{m} to 4μm4 \mu\text{m}, the surface area to volume ratio decreases from 33 to 1.51.5. This illustrates why cells must remain small to maintain efficient transport of materials.