Class 10 Biology - Life Processes

Activities

Activity 5.1: Variegated Leaf Starch Test

Aim/Objective: To demonstrate that chlorophyll is essential for photosynthesis by testing for the presence of starch in variegated leaves.

Materials Required:

A potted plant with variegated leaves (e.g., money plant or croton)
Dark room
Sunlight
Beaker
Alcohol
Water bath
Dilute iodine solution
Boiling water
Paper for tracing

Procedure:

Take a potted plant with variegated leaves and keep it in a dark room for three days to de-starch it.
Keep the plant in sunlight for about six hours.
Pluck a leaf from the plant and mark the green areas (where chlorophyll is present) by tracing them on a sheet of paper.
Dip the leaf in boiling water for a few minutes to kill the cells.
Immerse the leaf in a beaker containing alcohol and place the beaker in a water bath.
Heat until the alcohol begins to boil and the leaf loses its green colour (chlorophyll is dissolved in alcohol).
Dip the leaf in a dilute solution of iodine for a few minutes.
Rinse off the iodine solution and observe the colour change. Compare this with the original tracing.

Observation:

The areas of the leaf that were originally green turn blue-black when treated with iodine.
The non-green areas do not show this colour change and remain yellowish-brown.

Explanation:

Photosynthesis is the process by which green plants produce glucose (which is stored as starch) using sunlight, water, and CO2. This process requires chlorophyll, the green pigment found in chloroplasts.
In a variegated leaf, only the green parts contain chlorophyll. Therefore, photosynthesis and the subsequent production of starch only occur in these green regions. Iodine is an indicator that turns blue-black in the presence of starch.
The fact that only the previously green areas turned blue-black proves that starch was produced only where chlorophyll was present.

Conclusion:

Chlorophyll is essential for the process of photosynthesis.

Activity 5.2: Role of Carbon Dioxide in Photosynthesis

Aim/Objective: To demonstrate that carbon dioxide is essential for photosynthesis.

Materials Required:

Two healthy potted plants of nearly the same size
Dark room
Two glass plates
Potassium hydroxide (KOH)
Watch-glass
Two bell-jars
Vaseline
Sunlight
Materials for starch test (iodine, alcohol, water bath)

Procedure:

Keep both plants in a dark room for three days to de-starch them.
Place each plant on a separate glass plate.
Place a watch-glass containing potassium hydroxide next to one of the plants (Plant A).
Cover both plants with separate bell-jars.
Use vaseline to seal the bottom of the jars to the glass plates to make the setup air-tight.
Keep the plants in sunlight for about two hours.
Pluck a leaf from each plant and perform the starch test using iodine solution.

Observation:

The leaf from Plant B (without KOH) turns blue-black, indicating the presence of starch.
The leaf from Plant A (with KOH) does not turn blue-black, indicating the absence of starch.

Explanation:

Potassium hydroxide (KOH) has the property of absorbing carbon dioxide from the surrounding air. In the bell-jar covering Plant A, the KOH absorbed all available CO2, making it unavailable for the plant.
Since Plant A could not perform photosynthesis without CO2, no starch was produced. Plant B, having access to CO2 in its bell-jar, successfully performed photosynthesis and produced starch.

Conclusion:

Carbon dioxide is essential for photosynthesis to occur.

Activity 5.3: Action of Saliva on Starch

Aim/Objective: To demonstrate the digestion of starch by the enzyme salivary amylase present in human saliva.

Materials Required:

1% Starch solution
Two test tubes (A and B)
Human saliva
Dilute iodine solution
Dropper

Procedure:

Take 1 mL of 1% starch solution in both test tubes A and B.
Add 1 mL of saliva to test tube A.
Leave both test tubes undisturbed for 20-30 minutes.
Add a few drops of dilute iodine solution to both test tubes and observe the colour change.

Observation:

Test tube B (starch only) turns blue-black upon adding iodine.
Test tube A (starch + saliva) does not turn blue-black or shows a much lighter colour.

Explanation:

Saliva contains an enzyme called salivary amylase (also known as ptyalin). This enzyme breaks down complex starch molecules into simpler sugars like maltose.
In test tube A, the salivary amylase digested the starch into sugar during the 20-30 minute incubation period. Since starch was no longer present, the iodine test (which specifically detects starch) gave a negative result. In test tube B, the starch remained unchanged and reacted with iodine.

Conclusion:

Saliva contains enzymes that begin the chemical digestion of carbohydrates (starch) in the mouth.

Activity 5.4: Detection of Carbon Dioxide in Exhaled Air

Aim/Objective: To show that exhaled air contains a higher concentration of carbon dioxide than atmospheric air.

Materials Required:

Freshly prepared lime water [Ca(OH)2]
Two test tubes
A syringe or pichkari
A glass tube or straw

Procedure:

Take freshly prepared lime water in two test tubes.
In the first test tube, use a syringe or pichkari to bubble atmospheric air through the lime water.
In the second test tube, blow air from your mouth (exhaled air) into the lime water using a straw.
Note the time taken for the lime water in each tube to turn milky.

Observation:

The lime water in the second tube (exhaled air) turns milky almost immediately.
The lime water in the first tube (atmospheric air) takes a much longer time to turn milky.

Explanation:

Lime water reacts with carbon dioxide to form calcium carbonate, which is insoluble in water and appears as a milky white precipitate.
Exhaled air is a byproduct of cellular respiration, where glucose is oxidized to produce energy, water, and a significant amount of CO2. Atmospheric air contains only about 0.04% CO2, whereas exhaled air contains about 4%. The higher concentration of CO2 in exhaled air causes the rapid reaction with lime water.

Conclusion:

Human beings release carbon dioxide as a waste product during the process of respiration.

Activity 5.5: Carbon Dioxide Production during Fermentation

Aim/Objective: To demonstrate the production of carbon dioxide during the anaerobic respiration (fermentation) of yeast.

Materials Required:

Fruit juice or sugar solution
Yeast
Test tube
One-holed cork
Bent glass tube
Freshly prepared lime water

Procedure:

Take fruit juice or sugar solution in a test tube and add some yeast to it.
Fit the test tube with a one-holed cork and a bent glass tube.
Dip the other end of the glass tube into another test tube containing fresh lime water.
Observe the changes in the lime water over time.

Observation:

Bubbles of gas are seen escaping from the yeast mixture.
The lime water in the second test tube turns milky.

Explanation:

Yeast is a unicellular fungus that performs anaerobic respiration (fermentation) in the absence of oxygen. It breaks down sugar into ethanol and carbon dioxide while releasing a small amount of energy.
The CO2 gas produced during this process travels through the glass tube and reacts with the lime water, turning it milky (forming calcium carbonate).

Conclusion:

Carbon dioxide is a product of fermentation by yeast.

Activity 5.6: Comparative Breathing Rates

Aim/Objective: To compare the breathing rates of aquatic and terrestrial organisms.

Materials Required:

An aquarium with fish
A stopwatch or watch

Procedure:

Observe the movement of the mouth and gill-slits (operculum) of a fish in an aquarium.
Count the number of times the fish opens and closes its mouth in one minute.
Count your own breathing rate (number of breaths per minute) while resting.
Compare the two rates.

Observation:

The breathing rate of the fish is much faster than the breathing rate of a human being.

Explanation:

Aquatic organisms like fish obtain oxygen dissolved in water. The concentration of dissolved oxygen in water is much lower than the concentration of oxygen in the air (which is about 21%).
To compensate for the low availability of oxygen, aquatic organisms must process a larger volume of water through their respiratory organs (gills) in a shorter time, leading to a higher breathing rate compared to terrestrial animals who breathe air.

Conclusion:

The breathing rate of an organism is influenced by the availability of oxygen in its environment.

Activity 5.7: Variation in Haemoglobin Levels

Aim/Objective: To investigate the variations in haemoglobin content among humans and animals based on age and gender.

Materials Required:

Data from a health centre and a veterinary clinic

Procedure:

Visit a local health centre and collect data on the normal range of haemoglobin for children, adult men, and adult women.
Visit a veterinary clinic and collect similar data for animals like cows or buffaloes (calves, males, and females).
Compare the findings.

Observation:

Haemoglobin levels vary: men generally have higher levels than women; adults have different levels than children; and animals have their own specific ranges.

Explanation:

Haemoglobin is a respiratory pigment found in red blood cells that carries oxygen to tissues.
Differences in levels can be due to physiological needs (e.g., higher muscle mass in males requiring more oxygen), hormonal differences (menstruation in females), or species-specific metabolic rates. Calves or children have different levels due to rapid growth phases.

Conclusion:

Haemoglobin levels are not uniform and are adapted to the specific physiological and metabolic requirements of the individual and species.

Activity 5.8: Demonstration of Transpiration Pull

Aim/Objective: To show that plants lose water through their leaves (transpiration).

Materials Required:

Two small pots of same size with same amount of soil
A small plant in one pot
A stick of the same height as the plant in the other pot
Plastic sheets
Sunlight

Procedure:

Take the two pots (one with a plant, one with a stick).
Cover the soil in both pots with plastic to prevent evaporation from the soil surface.
Cover both the plant and the stick with separate plastic sheets.
Place both setups in bright sunlight for about half an hour.
Observe the inner surface of the plastic sheets.

Observation:

Droplets of water are seen on the inner surface of the plastic sheet covering the plant.
No such droplets (or significantly fewer) are seen on the sheet covering the stick.

Explanation:

Transpiration is the process by which plants lose water in the form of vapour through the stomata on their leaves.
When the plant is covered, the transpired water vapour cannot escape into the atmosphere and condenses into liquid droplets on the cool inner surface of the plastic. The stick does not perform any biological process to release water, confirming that the water in the first case came from the plant's life processes.

Conclusion:

Plants actively lose water through their aerial parts via transpiration.

Class 10 Biology - Life Processes

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