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Biology: Cells, Organization, and Classification - Prokaryotic and Eukaryotic Cells

Grade 8IB

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

🔑Concepts

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Cells are the basic structural and functional units of all living organisms. They are broadly categorized into two types based on their internal structure: Prokaryotic and Eukaryotic.

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Prokaryotic Cells (e.g., Bacteria, Archaea): These cells lack a membrane-bound nucleus and other membrane-bound organelles. Their genetic material is located in a region called the nucleoid and often includes small circular rings of DNADNA called plasmids.

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Eukaryotic Cells (e.g., Animal, Plant, Fungi, Protists): These cells contain a defined nucleus enclosed by a nuclear membrane. They possess specialized organelles such as mitochondria for aerobic respiration and, in plants, chloroplasts for photosynthesis.

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Comparison of Size: Prokaryotic cells are generally much smaller, ranging from 0.1 to 5.0 μm0.1 \text{ to } 5.0 \ \mu\text{m} in diameter, whereas Eukaryotic cells range from 10 to 100 μm10 \text{ to } 100 \ \mu\text{m}.

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Biological Organization: Multicellular eukaryotic organisms show a hierarchy of organization: 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: To study cells, microscopes are used. The relationship between the actual size of an object (AA), the image size seen under the microscope (II), and the magnification (MM) is crucial for biological study.

📐Formulae

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

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

I=A×MI = A \times M

1 mm=1000 μm1 \text{ mm} = 1000 \ \mu\text{m}

1 μm=1000 nm1 \ \mu\text{m} = 1000 \ \text{nm}

💡Examples

Problem 1:

A student views a plant cell under a microscope. The image of the cell measures 40 mm40 \text{ mm} in length. If the magnification used is 800×800\times, calculate the actual length of the cell in micrometers (μm\mu\text{m}).

Solution:

  1. Convert the image size to micrometers: I=40 mm×1000=40000 μmI = 40 \text{ mm} \times 1000 = 40000 \ \mu\text{m}
  2. Use the formula for actual size: A=IMA = \frac{I}{M}
  3. Substitute the values: A=40000 μm800=50 μmA = \frac{40000 \ \mu\text{m}}{800} = 50 \ \mu\text{m}

Explanation:

First, the units must be consistent. Since the answer is required in μm\mu\text{m}, the image size in millimeters is multiplied by 10001000. Dividing the image size by the magnification power gives the actual physical size of the specimen.

Problem 2:

Identify whether a cell with the following features is prokaryotic or eukaryotic: Linear DNADNA, presence of mitochondria, and a cell wall made of cellulose.

Solution:

The cell is Eukaryotic (specifically a plant cell).

Explanation:

Linear DNADNA and membrane-bound organelles like mitochondria are defining characteristics of Eukaryotic cells. Prokaryotes have circular DNADNA and lack mitochondria. The cellulose cell wall further specifies it as a plant cell.