Chapter 3.2: Respiration in Plants - Numerical Problems

Respiration in plants involves the biochemical breakdown of organic compounds (primarily glucose) to release energy. For Class 9, numerical problems focus on comparing ATP yields between aerobic and anaerobic respiration and analyzing experimental data (like respirometers).

📊 Visualizing Respiration Pathways

Glucose breakdown pathway:

Glucose (C₆H₁₂O₆) ➔ (Glycolysis) ➔ Pyruvate

If O₂ Present (Aerobic): ➔ Mitochondria ➔ (Krebs Cycle & ETC) ➔ 6 CO₂ + 6 H₂O + 38 ATP

If No O₂ Present (Anaerobic): ➔ Cytoplasm ➔ (Fermentation) ➔ 2 C₂H₅OH (Ethanol) + 2 CO₂ + 2 ATP

Problem 1: ATP Yield Comparison (Low Difficulty)

Question: A plant cell requires exactly 114 molecules of ATP to carry out a specific repair process.

Calculate how many glucose molecules must be broken down aerobically to produce this energy.
If the roots are waterlogged and deprived of oxygen, how many glucose molecules would be required to produce the same 114 ATP anaerobically?

Step-by-Step Solution:

Aerobic Glucose Requirement:
- In aerobic respiration, 1 molecule of glucose yields 38 molecules of ATP.
- $\text{Glucose required (aerobic)} = \frac{\text{Total ATP needed}}{\text{ATP per glucose}}$
- $\text{Glucose required (aerobic)} = \frac{114}{38} = \mathbf{3 \text{ molecules}}$ .
Anaerobic Glucose Requirement:
- In anaerobic respiration (fermentation in plants), 1 molecule of glucose yields only 2 molecules of ATP.
- $\text{Glucose required (anaerobic)} = \frac{114}{2} = \mathbf{57 \text{ molecules}}$ .

Insight: Plants must consume their food reserves 19 times faster (57 vs 3) under anaerobic conditions just to survive!

Problem 2: Analyzing Respirometer Data (Medium Difficulty)

Question: A student sets up a respirometer to measure the respiration rate of germinating pea seeds.

Setup A: Contains seeds and a small cotton plug soaked in Potassium Hydroxide (KOH).
Setup B: Contains seeds and a small cotton plug soaked in water (no KOH).

Over 2 hours, the colored fluid in the capillary tube connected to Setup A moves inward towards the chamber, indicating a volume decrease of $12 \text{ mL}$ . Over the same 2 hours, the fluid in the tube connected to Setup B does not move at all.

What is the exact physiological purpose of KOH in Setup A?
Calculate the rate of oxygen consumption in $mL/\text{hour}$ .

Step-by-Step Solution:

Role of KOH:
- KOH is a chemical that strongly absorbs Carbon Dioxide ( $CO_2$ ) gas.
- As seeds respire, they consume $O_2$ from the air and release $CO_2$ . Because KOH immediately absorbs the released $CO_2$ , the only gas remaining in the chamber continuously decreases in volume as $O_2$ is consumed. This drop in internal pressure pulls the fluid inward.
Rate of $O_2$ Consumption:
- The volume decrease in Setup A exactly represents the volume of $O_2$ consumed.
- Total $O_2$ consumed = $12 \text{ mL}$ .
- $\text{Rate} = \frac{12 \text{ mL}}{2 \text{ hours}} = \mathbf{6 \text{ mL/hour}}$ .

Problem 3: The "Day vs. Night" Net Exchange (High Difficulty / Concept Trap)

Question: A potted plant is placed in a sealed glass bell jar.

During the day (in sunlight), gas sensors detect a net increase of oxygen by $15 \text{ mL/hour}$ .
At night (in complete darkness), the sensors detect a net decrease of oxygen by $3 \text{ mL/hour}$ .

Assuming the temperature remains constant, what is the plant's actual rate of aerobic respiration during the day?

Common Trap: Students often assume plants do not respire during the day because photosynthesis takes over, and thus incorrectly answer $0 \text{ mL/hour}$ or $15 \text{ mL/hour}$ .

Step-by-Step Solution:

Understand Constant Respiration: Plants respire continuously, 24/7. Since temperature is constant, the respiration rate at night ( $3 \text{ mL/hour}$ of $O_2$ consumed) is exactly the same as during the day.
Analyze Daytime Net Change: During the day, the net $O_2$ increase ( $15 \text{ mL}$ ) is the result of Gross Photosynthetic $O_2$ produced minus the Respiration $O_2$ consumed.
Conclusion: The actual respiration rate during the day remains $3 \text{ mL/hour}$ of $O_2$ consumption. (This means the gross photosynthesis was actually producing $18 \text{ mL/hour}$ of $O_2$ , out of which $3 \text{ mL}$ was used by the plant for respiration, leaving a net gain of $15 \text{ mL}$ ).

Problem 4: Misinterpreting Anaerobic Products (Trap Question)

Question: A student is asked to write the balanced chemical equation for anaerobic respiration occurring in the deeply waterlogged roots of a mango tree. They write: $C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 \text{ (Lactic Acid)} + \text{Energy (2 ATP)}$

Identify the critical error in this equation and provide the correct one.

Step-by-Step Solution:

Identify the Error: The student wrote the equation for animal muscle fatigue (lactic acid fermentation).
The Correct Biology: Plants undergo alcoholic fermentation under anaerobic conditions. They do not produce lactic acid.
The Correct Equation: The correct equation must produce ethanol and carbon dioxide: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH \text{ (Ethanol)} + 2CO_2 + \text{Energy (2 ATP)}$ Note: Forgetting that $CO_2$ is a byproduct of plant anaerobic respiration is a critical and very common error!

Chapter 3.2: Respiration in Plants - Numerical Problems

📊 Visualizing Respiration Pathways

Glucose breakdown pathway:

Glucose (C₆H₁₂O₆) ➔ (Glycolysis) ➔ Pyruvate

If O₂ Present (Aerobic): ➔ Mitochondria ➔ (Krebs Cycle & ETC) ➔ 6 CO₂ + 6 H₂O + 38 ATP

If No O₂ Present (Anaerobic): ➔ Cytoplasm ➔ (Fermentation) ➔ 2 C₂H₅OH (Ethanol) + 2 CO₂ + 2 ATP

Problem 1: ATP Yield Comparison (Low Difficulty)

Question: A plant cell requires exactly 114 molecules of ATP to carry out a specific repair process.

Calculate how many glucose molecules must be broken down aerobically to produce this energy.
If the roots are waterlogged and deprived of oxygen, how many glucose molecules would be required to produce the same 114 ATP anaerobically?

Step-by-Step Solution:

Aerobic Glucose Requirement:
- In aerobic respiration, 1 molecule of glucose yields 38 molecules of ATP.
- $\text{Glucose required (aerobic)} = \frac{\text{Total ATP needed}}{\text{ATP per glucose}}$
- $\text{Glucose required (aerobic)} = \frac{114}{38} = \mathbf{3 \text{ molecules}}$ .
Anaerobic Glucose Requirement:
- In anaerobic respiration (fermentation in plants), 1 molecule of glucose yields only 2 molecules of ATP.
- $\text{Glucose required (anaerobic)} = \frac{114}{2} = \mathbf{57 \text{ molecules}}$ .

Insight: Plants must consume their food reserves 19 times faster (57 vs 3) under anaerobic conditions just to survive!

Problem 2: Analyzing Respirometer Data (Medium Difficulty)

Question: A student sets up a respirometer to measure the respiration rate of germinating pea seeds.

Setup A: Contains seeds and a small cotton plug soaked in Potassium Hydroxide (KOH).
Setup B: Contains seeds and a small cotton plug soaked in water (no KOH).

What is the exact physiological purpose of KOH in Setup A?
Calculate the rate of oxygen consumption in $mL/\text{hour}$ .

Step-by-Step Solution:

Role of KOH:
- KOH is a chemical that strongly absorbs Carbon Dioxide ( $CO_2$ ) gas.
- As seeds respire, they consume $O_2$ from the air and release $CO_2$ . Because KOH immediately absorbs the released $CO_2$ , the only gas remaining in the chamber continuously decreases in volume as $O_2$ is consumed. This drop in internal pressure pulls the fluid inward.
Rate of $O_2$ Consumption:
- The volume decrease in Setup A exactly represents the volume of $O_2$ consumed.
- Total $O_2$ consumed = $12 \text{ mL}$ .
- $\text{Rate} = \frac{12 \text{ mL}}{2 \text{ hours}} = \mathbf{6 \text{ mL/hour}}$ .

Problem 3: The "Day vs. Night" Net Exchange (High Difficulty / Concept Trap)

Question: A potted plant is placed in a sealed glass bell jar.

During the day (in sunlight), gas sensors detect a net increase of oxygen by $15 \text{ mL/hour}$ .
At night (in complete darkness), the sensors detect a net decrease of oxygen by $3 \text{ mL/hour}$ .

Assuming the temperature remains constant, what is the plant's actual rate of aerobic respiration during the day?

Common Trap: Students often assume plants do not respire during the day because photosynthesis takes over, and thus incorrectly answer $0 \text{ mL/hour}$ or $15 \text{ mL/hour}$ .

Step-by-Step Solution:

Understand Constant Respiration: Plants respire continuously, 24/7. Since temperature is constant, the respiration rate at night ( $3 \text{ mL/hour}$ of $O_2$ consumed) is exactly the same as during the day.
Analyze Daytime Net Change: During the day, the net $O_2$ increase ( $15 \text{ mL}$ ) is the result of Gross Photosynthetic $O_2$ produced minus the Respiration $O_2$ consumed.
Conclusion: The actual respiration rate during the day remains $3 \text{ mL/hour}$ of $O_2$ consumption. (This means the gross photosynthesis was actually producing $18 \text{ mL/hour}$ of $O_2$ , out of which $3 \text{ mL}$ was used by the plant for respiration, leaving a net gain of $15 \text{ mL}$ ).

Problem 4: Misinterpreting Anaerobic Products (Trap Question)

Identify the critical error in this equation and provide the correct one.

Step-by-Step Solution:

Identify the Error: The student wrote the equation for animal muscle fatigue (lactic acid fermentation).
The Correct Biology: Plants undergo alcoholic fermentation under anaerobic conditions. They do not produce lactic acid.
The Correct Equation: The correct equation must produce ethanol and carbon dioxide: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH \text{ (Ethanol)} + 2CO_2 + \text{Energy (2 ATP)}$ Note: Forgetting that $CO_2$ is a byproduct of plant anaerobic respiration is a critical and very common error!

Numerical Problems - Respiration in Plants

Chapter 3.2: Respiration in Plants - Numerical Problems

📊 Visualizing Respiration Pathways

Problem 1: ATP Yield Comparison (Low Difficulty)

Problem 2: Analyzing Respirometer Data (Medium Difficulty)

Problem 3: The "Day vs. Night" Net Exchange (High Difficulty / Concept Trap)

Problem 4: Misinterpreting Anaerobic Products (Trap Question)

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Numerical Problems - Respiration in Plants

Chapter 3.2: Respiration in Plants - Numerical Problems

📊 Visualizing Respiration Pathways

Problem 1: ATP Yield Comparison (Low Difficulty)

Problem 2: Analyzing Respirometer Data (Medium Difficulty)

Problem 3: The "Day vs. Night" Net Exchange (High Difficulty / Concept Trap)

Problem 4: Misinterpreting Anaerobic Products (Trap Question)

On this page