The Invisible Living World: Beyond Our Naked Eye

Given: Labels to be placed on a cell diagram.

Concept: A typical plant cell has a cell wall (outermost), cell membrane (just inside the cell wall), cytoplasm (fluid filling the cell), nucleus (control centre, surrounded by nuclear membrane), and chloroplasts (green plastids for photosynthesis). A prokaryotic cell (like a bacterium) has a nucleoid instead of a true nucleus and lacks a membrane-bound nucleus.

Placement of labels:

| Label | Location in the diagram |
|---|---|
| Cell wall | Outermost rigid layer (plant cell) |
| Cell membrane | Layer just inside the cell wall |
| Cytoplasm | Fluid-filled space between cell membrane and nucleus |
| Nucleus | Round/oval structure in the centre of the cell (eukaryotic cell) |
| Chloroplast | Green oval structures in the cytoplasm (plant cell only) |
| Nucleoid | Region in the cytoplasm of a prokaryotic cell (e.g., bacterium) where DNA is located, without a membrane |

Key distinction: The nucleus (with a membrane) is found in plant and animal cells (eukaryotes), while the nucleoid (without a membrane) is found in bacteria (prokaryotes). Chloroplasts and cell wall are features of plant cells only.

Correct Option: (c) Yeast produced a gas inside the test tube B which inflated the balloon.

Justification: Yeast is a microorganism (fungus) that breaks down sugar in the presence of warmth through a process called fermentation. During fermentation, yeast converts sugar into alcohol and carbon dioxide (CO₂) gas. This CO₂ gas accumulates inside the test tube and inflates the balloon attached to test tube B. Test tube A has no yeast, so no gas is produced and its balloon remains uninflated. Options (a), (b), and (d) are incorrect because water evaporation, atmospheric warmth, and sugar alone cannot produce enough gas to inflate the balloon under these conditions.

Given: The balloon from test tube B (containing gas produced by yeast) is attached to a test tube of lime water and shaken.

Concept: Lime water (calcium hydroxide solution) turns milky/white when carbon dioxide (CO₂) is passed through it. This is a standard test for CO₂:
$$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O}$$
The white precipitate of calcium carbonate (CaCO₃) makes the lime water appear milky.

What Aanandi wants to find out: She wants to identify the gas produced by yeast during fermentation. Specifically, she wants to confirm whether the gas produced is carbon dioxide (CO₂). If the lime water turns milky after shaking, it confirms that the gas collected in the balloon is CO₂, which was released by yeast during the fermentation of sugar.

Expected observation: The lime water will turn milky, confirming that yeast produces carbon dioxide gas during fermentation.

Given: Wheat farmer adds nitrogen fertiliser; bean (legume) farmer does not.

Concept: Leguminous plants (beans, peas, lentils, etc.) have a special relationship with nitrogen-fixing bacteria called *Rhizobium*. These bacteria live in root nodules of leguminous plants.

Explanation:

- Wheat is a non-leguminous crop. Its roots do not have nitrogen-fixing bacteria. Therefore, the soil around wheat crops does not get naturally replenished with nitrogen. The farmer must add nitrogen-rich fertilisers externally to provide the nitrogen needed for healthy crop growth.

- Bean (legume) plants have *Rhizobium* bacteria living in nodules on their roots. These bacteria can trap (fix) atmospheric nitrogen (N₂) from the air and convert it into nitrogen compounds that the plant can use. This process is called biological nitrogen fixation. As a result, the soil around legume crops actually becomes richer in nitrogen over time.

Conclusion: The bean farmer does not need to add nitrogen fertiliser because the *Rhizobium* bacteria in the root nodules of her bean plants naturally fix atmospheric nitrogen and supply it to the plant, making the soil fertile. This is why leguminous crops are also grown in rotation with other crops to improve soil fertility.

Given: Pit A — fruit/vegetable peels mixed with dried leaves; Pit B — fruit/vegetable peels only (no dried leaves). Both covered with soil and observed after 3 weeks.

What Snehal is trying to test: Snehal is trying to test whether the presence of dried leaves (which contain microorganisms and provide additional organic matter and aeration) affects the rate of decomposition of fruit and vegetable peels.

Detailed explanation:
- Dried leaves contain microorganisms (bacteria and fungi) that help break down organic waste.
- Mixing dried leaves with the peels may also improve aeration and moisture balance in the pit, creating better conditions for microbial activity.
- By comparing pit A (with dried leaves) and pit B (without dried leaves), she is investigating whether adding dried leaves speeds up the decomposition process and leads to better/faster manure formation.

Expected observation: Pit A is likely to show faster and more complete decomposition (better manure formation) compared to pit B, because the dried leaves provide additional microorganisms and improve the conditions needed for decomposition.

Conclusion: Snehal is testing the role of dried leaves (and the microorganisms they carry) in enhancing the decomposition of organic waste into manure.

Concept: Different microorganisms have distinct characteristics and roles.

(i) I live in every kind of environment, and inside your gut.

Answer: Bacteria

Bacteria are found everywhere — in water, soil, air, extreme environments (hot springs, cold zones), and inside the human body, especially in the gut (intestine), where they help in digestion.

(ii) I make bread and cakes soft and fluffy.

Answer: Yeast (a type of fungus, e.g., *Saccharomyces cerevisiae*)

Yeast ferments sugar and releases carbon dioxide (CO₂) gas. The CO₂ bubbles get trapped in the dough, making it rise and become soft and fluffy. This is why yeast is used in making bread, cakes, and pastries.

(iii) I live in the roots of pulse crops and provide nutrients for their growth.

**Answer: *Rhizobium* (nitrogen-fixing bacteria)**

*Rhizobium* bacteria live in the root nodules of leguminous (pulse) crops such as peas, beans, and lentils. They fix atmospheric nitrogen into nitrogen compounds, which enrich the soil and provide nutrients to the plant, promoting healthy growth.

Aim: To test that microorganisms need optimal temperature, air, and moisture for their growth.

Materials required: 6 slices of bread (same type), 6 zip-lock/polythene bags, water, a refrigerator, a warm place (like near a window or in sunlight).

Experimental Design:

Set up 6 groups as follows:

| Group | Condition | Treatment |
|---|---|---|
| A | Optimal (control) | Moist bread, open to air, kept at room temperature (warm place) |
| B | No moisture | Dry bread, open to air, kept at room temperature |
| C | No air | Moist bread, sealed in airtight bag, kept at room temperature |
| D | Low temperature | Moist bread, open to air, kept in refrigerator |
| E | No moisture + low temperature | Dry bread, sealed bag, kept in refrigerator |
| F | All conditions absent | Dry bread, sealed bag, kept in refrigerator |

Procedure:
1. Take 6 slices of bread.
2. For groups A, C, D: Lightly sprinkle water on the bread to make it moist.
3. For groups B, E, F: Keep the bread dry.
4. Place each slice in a separate bag. Seal the bags for groups C, E, F. Leave bags open for A, B, D.
5. Keep groups A, B, C at room temperature (warm place).
6. Keep groups D, E, F in the refrigerator.
7. Observe all groups after 3–5 days.

Expected Observations:
- Group A (moist + air + warm): Maximum fungal/mould growth — confirms all three conditions support microbial growth.
- Group B (dry + air + warm): Little or no growth — lack of moisture inhibits growth.
- Group C (moist + no air + warm): Reduced growth — lack of air slows growth.
- Group D (moist + air + cold): Little or no growth — low temperature inhibits growth.
- Groups E and F: No growth — absence of multiple conditions prevents microbial growth.

Conclusion: Microorganisms grow best when optimal temperature, air (oxygen), and moisture are all present. Absence of any one of these conditions slows down or prevents their growth, proving that microorganisms need all three conditions for optimal growth.

Given: Slice 1 — near the sink (warm, moist, open to air); Slice 2 — in the refrigerator (cold, less moisture, closed).

Observations after three days:

| | Slice near the sink | Slice in the refrigerator |
|---|---|---|
| Appearance | Mould/fungal growth visible (green, black, or white patches) | Little or no mould growth; bread may be slightly dry |
| Smell | Musty/unpleasant smell | Normal or slightly stale smell |
| Texture | Soft, possibly soggy | Dry or slightly hard |

Reasons:

- Slice near the sink: The area near the sink is warm and humid (moist). Microorganisms (mainly fungi/mould spores) present in the air settle on the bread. The warm temperature, moisture, and availability of nutrients in the bread provide ideal conditions for the microorganisms to grow and multiply rapidly. Hence, mould grows on this slice within 3 days.

- Slice in the refrigerator: The refrigerator maintains a low temperature (around 4°C). At such low temperatures, the growth and reproduction of microorganisms is greatly slowed down or inhibited. Therefore, the bread in the refrigerator shows little or no mould growth even after 3 days.

Conclusion: Microorganisms need warmth and moisture to grow. Cold temperatures (as in a refrigerator) inhibit microbial growth, which is why refrigeration is used to preserve food.

Given: Curd left out at room temperature for a day becomes more sour.

Concept: Curd contains *Lactobacillus* bacteria, which are responsible for its formation and taste.

Two possible explanations:

Explanation 1 — Continued bacterial activity:
Curd contains live *Lactobacillus* bacteria. When left at room temperature, these bacteria continue to ferment the lactose (milk sugar) present in the curd. During fermentation, they produce more lactic acid. As the concentration of lactic acid increases over time, the curd becomes more and more sour.

$$\text{Lactose} \xrightarrow{\text{Lactobacillus}} \text{Lactic acid} + \text{other products}$$

Explanation 2 — Growth of additional microorganisms:
When curd is left open at room temperature, other microorganisms from the surrounding air (such as additional bacteria) may also settle on it and grow. These microorganisms may also produce acidic compounds during their metabolic activities, further increasing the sourness of the curd.

Conclusion: The increased sourness is primarily due to the continued production of lactic acid by *Lactobacillus* bacteria (and possibly other microbes) at warm room temperature, which is more favourable for their growth than cold temperatures.

Note: Based on the chapter context (Activity 2.8, Aanandi's experiment, and the description of yeast fermentation), Flask A contains sugar solution with yeast kept in a warm place, and it is connected via a delivery tube to test tube B containing lime water.

(i) What happens to the sugar solution in flask A?

Answer: Yeast (a microorganism/fungus) is present in the sugar solution in flask A. In the warm conditions, yeast ferments the sugar (glucose/sucrose). During fermentation, yeast breaks down the sugar and produces:
- Carbon dioxide (CO₂) gas — which bubbles out of the solution.
- Alcohol (ethanol) — which remains in the solution.
- A small amount of energy is also released.

As a result, the sugar solution in flask A undergoes fermentation, producing CO₂ gas and alcohol. The solution may also develop a slightly different (alcoholic) smell.

(ii) What do you observe in test tube B after four hours? Why do you think this happened?

Answer — Observation: The lime water in test tube B turns milky (white/cloudy).

Reason: The CO₂ gas produced during fermentation in flask A travels through the delivery tube into test tube B containing lime water. Carbon dioxide reacts with calcium hydroxide (lime water) to form calcium carbonate (CaCO₃), which is insoluble and appears as a white precipitate, turning the lime water milky:

$$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O}$$

This confirms that the gas produced by yeast during fermentation is carbon dioxide.

(iii) What would happen if yeast was not added in flask A?

Answer: If yeast was not added to flask A, no fermentation would take place. Without yeast, the sugar solution would remain unchanged — no CO₂ gas would be produced. As a result:
- No gas would pass through the delivery tube into test tube B.
- The lime water in test tube B would remain clear (would not turn milky).
- There would be no change in the smell or composition of the sugar solution in flask A.

This confirms that yeast is essential for fermentation and for the production of CO₂ gas.