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Biology: Cells, Organization, and Classification - Plant and Animal Cells

Grade 8IB

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

🔑Concepts

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The Cell Theory states that all living organisms are composed of one or more cells, the cell is the basic unit of structure and organization in organisms, and all cells come from pre-existing cells.

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Common Organelles: Both plant and animal cells contain a Nucleus (contains DNA), Cytoplasm (jelly-like substance where reactions occur), Cell Membrane (controls entry/exit), and Mitochondria (site of aerobic respiration where energy is released).

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Plant Cell Specifics: Plant cells have a rigid Cell Wall made of cellulose for support, Chloroplasts containing chlorophyll for photosynthesis, and a large Permanent Vacuole filled with cell sap.

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Animal Cell Specifics: Animal cells lack a cell wall and chloroplasts. They may have small, temporary vacuoles used for storage or transport.

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Levels of Organization: Life is organized in a hierarchical structure: 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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Magnification: In microscopy, the relationship between the image size, actual size, and magnification is expressed using the formula I=A×MI = A \times M. Unit conversion is essential, where 1 mm=1000 \mum1\text{ mm} = 1000\text{ \mu m} (micrometers).

📐Formulae

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

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

Total Magnification=Eyepiece Lens Magnification×Objective Lens Magnification\text{Total Magnification} = \text{Eyepiece Lens Magnification} \times \text{Objective Lens Magnification}

💡Examples

Problem 1:

A student views a plant cell under a microscope. The image of the cell measures 45 mm45\text{ mm}. If the actual size of the cell is 0.15 mm0.15\text{ mm}, calculate the magnification.

Solution:

M=45 mm0.15 mm=300M = \frac{45\text{ mm}}{0.15\text{ mm}} = 300

Explanation:

Using the formula M=IAM = \frac{I}{A}, we divide the image size by the actual size. Since both units are in millimeters, no conversion is necessary. The magnification is 300×300\times.

Problem 2:

An image of a mitochondrion is 20 mm20\text{ mm} long when viewed at a magnification of 10000×10000\times. What is the actual length of the mitochondrion in micrometers (μm\mu m)?

Solution:

A=IMA=20 mm10000A=0.002 mm0.002×1000=2 \mum\begin{array}{r} A = \frac{I}{M} \\ A = \frac{20\text{ mm}}{10000} \\ A = 0.002\text{ mm} \\ 0.002 \times 1000 = 2\text{ \mu m} \end{array}

Explanation:

First, use the formula A=IMA = \frac{I}{M} to find the actual size in millimeters. Then, convert millimeters to micrometers by multiplying by 10001000 because 1 mm=1000 \mum1\text{ mm} = 1000\text{ \mu m}.