Class 09 Biology - Introduction to Life (The Building Block of Life)

Activities

Activity 2.1: Estimating the Size of a Cell

Aim/Objective: To estimate the actual size of a microscopic object (onion peel cell) using a light microscope.

Materials Required:

• Compound light microscope
• Transparent ruler with mm markings
• Prepared slide of onion peel cells
• Calculator

Procedure:

1. Place the transparent ruler on the microscope stage.
2. Measure the diameter of the circular field of view in millimetres (mm) and convert it to micrometres (µm) (1 mm = 1000 µm).
3. Replace the ruler with the onion peel slide.
4. Count the number of cells arranged in a straight line along the diameter of the field of view.
5. Calculate the size of one cell using the formula: Size = (Diameter of field of view in µm) / (Number of cells).

Observation:

• If the field of view is 5 mm (5000 µm) and 25 cells are seen across it, each cell is approximately 200 µm in size.

Explanation:

• Microscopes use lenses to magnify images. The Total Magnification is the product of the objective lens and the eyepiece (e.g., 10x * 10x = 100x).
• Understanding the scale is crucial in cell biology because most cells are far below the resolution limit of the human eye (0.1 mm). Micrometres (µm) and nanometres (nm) are the standard units used to measure these microscopic structures.

Conclusion:

• Cells have a measurable size that can be determined by relating the magnified image to a known scale.

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Activity 2.2: Demonstration of Osmosis in Potato Pieces

Aim/Objective: To observe the movement of water across a selectively permeable membrane (osmosis) using potato tissue.

Materials Required:

• One large potato
• Knife
• Plain water
• 20% salt or sugar solution
• Two beakers (A and B)
• Weighing balance

Procedure:

1. Cut a potato into two pieces of equal size.
2. Record the initial weight of both pieces.
3. Place one piece in Beaker A (plain water) and the other in Beaker B (20% salt/sugar solution).
4. Leave them undisturbed for one hour.
5. Record the final weight and observe any change in size or texture.

Observation:

• The potato piece in Beaker A swells and its weight increases.
• The potato piece in Beaker B shrinks, becomes limp, and its weight decreases.

Explanation:

• This is a demonstration of osmosis: the net movement of water from a region of higher water concentration (dilute) to a region of lower water concentration (concentrated) through a selectively permeable membrane (the cell membrane).
• Beaker A contains a hypotonic solution relative to the potato cells, so water enters the cells. Beaker B contains a hypertonic solution, causing water to leave the cells.

Conclusion:

• Water moves across cell membranes based on the concentration gradient of the surrounding medium.

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Activity 2.3: Comparison of Plant and Animal Cells (Onion vs. Cheek Cells)

Aim/Objective: To identify the structural differences between plant and animal cells, specifically the presence of a cell wall.

Materials Required:

• Onion peel, Toothpick (for cheek cells)
• Slides, Coverslips, Microscope
• Safranin (for onion), Methylene blue (for cheek cells)
• 20% Sugar solution

Procedure:

1. Prepare a temporary mount of onion peel and observe it.
2. Prepare a temporary mount of human cheek cells and observe it.
3. Add a few drops of 20% sugar solution to both slides and observe the changes after 30 minutes.

Observation:

• Onion cells are box-shaped with a clear boundary (cell wall). Cheek cells are irregularly shaped.
• In sugar solution, the content of the onion cell shrinks away from the wall (plasmolysis), but the outer shape remains the same. The cheek cell shrinks entirely.

Explanation:

• The cell wall (made of cellulose) provides structural rigidity to plant cells, allowing them to withstand environmental stress and osmotic pressure without bursting or losing shape.
• Animal cells lack a cell wall and are surrounded only by a flexible plasma membrane, allowing for the flexibility required in animal tissues but making them more susceptible to changes in osmotic pressure.

Conclusion:

• The cell wall is a distinguishing feature that provides mechanical support and shape to plant cells.

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Activity 2.4: Classification of Prokaryotic and Eukaryotic Cells

Aim/Objective: To distinguish between prokaryotic and eukaryotic cell structures.

Materials Required:

• Diagrams of a bacterial cell, plant cell, and animal cell.

Procedure:

1. Compare the three diagrams based on the presence of a nucleus and organelles.
2. Tabulate the findings (presence of cell wall, cytoplasm, membrane-bound organelles, etc.).

Observation:

• Bacterial cells lack a nuclear membrane (have a nucleoid) and membrane-bound organelles.
• Plant and animal cells have a well-defined nucleus and various organelles like mitochondria and ER.

Explanation:

• Prokaryotes (like bacteria) are simpler, smaller, and represent an earlier evolutionary stage. Their cellular processes occur directly in the cytoplasm.
• Eukaryotes are more complex. The presence of membrane-bound organelles (compartmentalization) allows different chemical reactions to occur simultaneously and efficiently in separate areas of the cell.

Conclusion:

• Cells are classified into two broad categories based on their internal complexity and organization of genetic material.

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Activity 2.5: Observation of Cell Division (Mitosis) in Onion Root Tips

Aim/Objective: To observe the various stages of mitosis in actively dividing plant tissue.

Materials Required:

• Onion with fresh roots
• Aceto-alcohol (fixative), 70% Ethanol
• Dilute HCl, Aceto-carmine stain
• Spirit lamp, Slides, Coverslips, Microscope

Procedure:

1. Grow onion roots in water for 5-6 days.
2. Fix the root tips in aceto-alcohol for 24 hours.
3. Soften the tissue with dilute HCl.
4. Stain with aceto-carmine and gently heat the slide.
5. Squash the root tip under a coverslip to spread the cells.
6. Observe under the microscope and identify cells in different stages.

Observation:

• Different cells show different arrangements of chromosomes: some are scattered, some aligned in the middle, and some moving to opposite poles.

Explanation:

• The root tip contains meristematic tissue, where cells divide rapidly by mitosis to increase root length.
• Mitosis is a process of equational division where the DNA is replicated and then distributed equally into two daughter cells. This ensures that every new cell has the same genetic information as the parent cell, which is essential for growth and repair.

Conclusion:

• Growth in multicellular organisms is achieved through the regulated and orderly process of mitotic cell division.