Chapter 3.2: Respiration in Plants - Numerical Problems

Respiration in plants involves the biochemical breakdown of organic compounds (primarily glucose) to release energy. Mastering numerical problems in this chapter requires a strong grasp of chemical stoichiometry, energy equivalents (ATP), and gas exchange dynamics in both aerobic and anaerobic conditions.

Below are advanced, step-by-step problems designed to test deep conceptual understanding, moving beyond simple formula memorization.

📊 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

🟢 Level 1: Stoichiometry and ATP Yields

Question 1: The Pasteur Effect and Glucose Demand A rapidly growing apical meristem of a plant requires exactly $1.14 \times 10^4$ ATP molecules per second to sustain its cellular activities.

Calculate the minimum number of glucose molecules that must be completely oxidized (aerobically) per second to meet this demand.
If the soil becomes waterlogged, cutting off oxygen supply to the roots, the root cells switch entirely to anaerobic respiration. How many glucose molecules per second would be required to maintain the exact same ATP production rate?
Calculate the ratio of glucose consumption in anaerobic conditions compared to aerobic conditions.

Step-by-Step Solution:

Aerobic Glucose Requirement:
- In aerobic respiration, 1 molecule of glucose yields 38 molecules of ATP.
- $\text{Glucose required (aerobic)} = \frac{\text{Total ATP needed}}{\text{ATP per glucose}}$
- $\text{Glucose required (aerobic)} = \frac{11,400}{38} = \mathbf{300 \text{ molecules/second}}$ .
Anaerobic Glucose Requirement:
- In anaerobic respiration (alcoholic fermentation in plants), 1 molecule of glucose yields only 2 molecules of ATP.
- $\text{Glucose required (anaerobic)} = \frac{11,400}{2} = \mathbf{5,700 \text{ molecules/second}}$ .
Consumption Ratio:
- $\text{Ratio} = \frac{\text{Anaerobic requirement}}{\text{Aerobic requirement}} = \frac{5700}{300} = \mathbf{19:1}$ .
- Insight: Plants must consume sugar 19 times faster under anaerobic conditions just to survive, leading to rapid depletion of their stored food reserves.

🟡 Level 2: Energy Efficiency and Biomass

Question 2: Calculating Thermodynamic Efficiency The complete combustion of one mole of glucose releases $2880 \text{ kJ}$ of energy. The synthesis of one mole of ATP from ADP and inorganic phosphate ( $P_i$ ) captures approximately $30.5 \text{ kJ}$ of energy.

Calculate the energy efficiency of aerobic respiration in plant cells.
Calculate the energy efficiency of anaerobic respiration in plant cells.
Determine the amount of energy (in kJ) lost as heat when one mole of glucose is aerobically respired.

Step-by-Step Solution:

Aerobic Efficiency:
- Total ATP produced aerobically = 38 moles.
- Energy captured in ATP = $38 \times 30.5 \text{ kJ} = 1159 \text{ kJ}$ .
- $\text{Efficiency} = \left( \frac{\text{Energy Captured}}{\text{Total Energy Available}} \right) \times 100$
- $\text{Efficiency} = \left( \frac{1159}{2880} \right) \times 100 = \mathbf{40.24\%}$ .
Anaerobic Efficiency:
- Total ATP produced anaerobically = 2 moles.
- Energy captured in ATP = $2 \times 30.5 \text{ kJ} = 61 \text{ kJ}$ .
- $\text{Efficiency} = \left( \frac{61}{2880} \right) \times 100 = \mathbf{2.12\%}$ .
Heat Lost (Aerobic):
- $\text{Heat Lost} = \text{Total Energy} - \text{Energy Captured}$
- $\text{Heat Lost} = 2880 \text{ kJ} - 1159 \text{ kJ} = \mathbf{1721 \text{ kJ}}$ . (This heat helps maintain plant temperature or is passively dissipated into the environment).

Question 3: Gas Volumes at STP A batch of germinating gram seeds respires aerobically in a sealed container. Analysis shows they have consumed $4.5 \times 10^{-3} \text{ moles}$ of oxygen gas ( $O_2$ ). (Assume standard conditions (STP) where 1 mole of gas occupies 22.4 Liters. Molar mass of glucose = 180 g/mol)

Calculate the volume of $CO_2$ gas evolved in milliliters (mL).
Calculate the exact mass (in milligrams) of glucose consumed by the seeds.

Step-by-Step Solution:

Volume of $CO_2$ Evolved:
- Balanced equation: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}$ .
- The stoichiometric ratio of $O_2$ consumed to $CO_2$ evolved is 6:6, which simplifies to 1:1.
- Moles of $CO_2$ evolved = Moles of $O_2$ consumed = $4.5 \times 10^{-3} \text{ moles}$ .
- $\text{Volume} = \text{Moles} \times \text{Molar Volume (at STP)}$
- $\text{Volume} = 4.5 \times 10^{-3} \times 22.4 \text{ L} = 0.1008 \text{ L} = \mathbf{100.8 \text{ mL}}$ .
Mass of Glucose Consumed:
- According to the equation, 6 moles of $O_2$ are required to break down 1 mole of glucose.
- Moles of glucose = $\frac{4.5 \times 10^{-3}}{6} = 0.75 \times 10^{-3} \text{ moles}$ .
- $\text{Mass} = \text{Moles} \times \text{Molar Mass}$
- $\text{Mass} = (0.75 \times 10^{-3} \text{ moles}) \times 180 \text{ g/mol} = 0.135 \text{ g} = \mathbf{135 \text{ mg}}$ .

Level 3: Respirometer and Respiratory Quotient (RQ)

Question 4: Analyzing Respirometer Data A student sets up a respirometer to measure the respiration rate of germinating pea seeds.

Setup A: Contains seeds and a small vial of Potassium Hydroxide (KOH).
Setup B: Contains seeds and a small vial of water (no KOH).

Over 2 hours, the fluid in the manometer 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 manometer connected to Setup B does not move at all.

What is the physiological purpose of KOH in Setup A, and why did the fluid move?
Calculate the rate of oxygen consumption in $cm^3/\text{hour}$ .
Using the data from Setup B, calculate the Respiratory Quotient (RQ). Identify the predominant biochemical substrate the seeds are utilizing.

Step-by-Step Solution:

Role of KOH:
- KOH strongly absorbs carbon dioxide gas ( $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$ ).
- 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 manometer 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 = $12 \text{ mL}$ ( $1 \text{ mL} = 1 \text{ cm}^3$ ).
- $\text{Rate} = \frac{12 \text{ cm}^3}{2 \text{ hours}} = \mathbf{6 \text{ cm}^3/\text{hour}}$ .
Respiratory Quotient (RQ) and Substrate:
- In Setup B (no KOH), the fluid did not move. This means the pressure inside remained constant.
- Therefore, the volume of $CO_2$ evolved exactly replaced the volume of $O_2$ consumed ( $12 \text{ cm}^3$ ).
- $RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}} = \frac{12}{12} = \mathbf{1.0}$ .
- An RQ of 1.0 indicates that the seeds are predominantly respiring Carbohydrates (like starch or glucose).

Trap Questions & Common Pitfalls

Trap 1: The "Day vs. Night" Fallacy

The Scenario: A potted plant is placed in a sealed glass bell jar. During the day, gas sensors detect a net increase of oxygen by $15 \text{ mL}/\text{hour}$ . At night (in complete darkness), the sensors detect a net decrease of oxygen by $3 \text{ mL}/\text{hour}$ .
The Question: What is the plant's actual rate of aerobic respiration during the day?
The Trap: Students often assume plants do not respire during the day, only photosynthesize, and thus incorrectly answer $0 \text{ mL}/\text{hour}$ or $15 \text{ mL}/\text{hour}$ .
The Solution: Plants respire continuously, 24/7. Assuming the temperature is constant, the respiration rate at night ( $3 \text{ mL}/\text{hour}$ $O_2$ consumed) is approximately the same as during the day. 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. Therefore, the actual respiration rate during the day remains $3 \text{ mL/hour}$ of $O_2$ consumption. (The gross photosynthesis was actually producing $18 \text{ mL/hour}$ ).

Trap 2: Misinterpreting Anaerobic Products

The Scenario: Write the balanced chemical equation for anaerobic respiration occurring in the waterlogged roots of a mango tree.
The Trap: Writing the equation for animal muscle fatigue: $C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 \text{ (Lactic Acid)} + \text{Energy}$ .
The Solution: Plants and yeast undergo alcoholic fermentation, not lactic acid fermentation. The correct equation must produce ethanol and carbon dioxide: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH \text{ (Ethanol)} + 2CO_2 + \text{Energy (2 ATP)}$ . Forgetting that $CO_2$ is a byproduct of plant anaerobic respiration is a critical error!

Chapter 3.2: Respiration in Plants - Numerical Problems

Below are advanced, step-by-step problems designed to test deep conceptual understanding, moving beyond simple formula memorization.

📊 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

🟢 Level 1: Stoichiometry and ATP Yields

Question 1: The Pasteur Effect and Glucose Demand A rapidly growing apical meristem of a plant requires exactly $1.14 \times 10^4$ ATP molecules per second to sustain its cellular activities.

Calculate the minimum number of glucose molecules that must be completely oxidized (aerobically) per second to meet this demand.
If the soil becomes waterlogged, cutting off oxygen supply to the roots, the root cells switch entirely to anaerobic respiration. How many glucose molecules per second would be required to maintain the exact same ATP production rate?
Calculate the ratio of glucose consumption in anaerobic conditions compared to aerobic conditions.

Step-by-Step Solution:

Aerobic Glucose Requirement:
- In aerobic respiration, 1 molecule of glucose yields 38 molecules of ATP.
- $\text{Glucose required (aerobic)} = \frac{\text{Total ATP needed}}{\text{ATP per glucose}}$
- $\text{Glucose required (aerobic)} = \frac{11,400}{38} = \mathbf{300 \text{ molecules/second}}$ .
Anaerobic Glucose Requirement:
- In anaerobic respiration (alcoholic fermentation in plants), 1 molecule of glucose yields only 2 molecules of ATP.
- $\text{Glucose required (anaerobic)} = \frac{11,400}{2} = \mathbf{5,700 \text{ molecules/second}}$ .
Consumption Ratio:
- $\text{Ratio} = \frac{\text{Anaerobic requirement}}{\text{Aerobic requirement}} = \frac{5700}{300} = \mathbf{19:1}$ .
- Insight: Plants must consume sugar 19 times faster under anaerobic conditions just to survive, leading to rapid depletion of their stored food reserves.

🟡 Level 2: Energy Efficiency and Biomass

Calculate the energy efficiency of aerobic respiration in plant cells.
Calculate the energy efficiency of anaerobic respiration in plant cells.
Determine the amount of energy (in kJ) lost as heat when one mole of glucose is aerobically respired.

Step-by-Step Solution:

Aerobic Efficiency:
- Total ATP produced aerobically = 38 moles.
- Energy captured in ATP = $38 \times 30.5 \text{ kJ} = 1159 \text{ kJ}$ .
- $\text{Efficiency} = \left( \frac{\text{Energy Captured}}{\text{Total Energy Available}} \right) \times 100$
- $\text{Efficiency} = \left( \frac{1159}{2880} \right) \times 100 = \mathbf{40.24\%}$ .
Anaerobic Efficiency:
- Total ATP produced anaerobically = 2 moles.
- Energy captured in ATP = $2 \times 30.5 \text{ kJ} = 61 \text{ kJ}$ .
- $\text{Efficiency} = \left( \frac{61}{2880} \right) \times 100 = \mathbf{2.12\%}$ .
Heat Lost (Aerobic):
- $\text{Heat Lost} = \text{Total Energy} - \text{Energy Captured}$
- $\text{Heat Lost} = 2880 \text{ kJ} - 1159 \text{ kJ} = \mathbf{1721 \text{ kJ}}$ . (This heat helps maintain plant temperature or is passively dissipated into the environment).

Calculate the volume of $CO_2$ gas evolved in milliliters (mL).
Calculate the exact mass (in milligrams) of glucose consumed by the seeds.

Step-by-Step Solution:

Volume of $CO_2$ Evolved:
- Balanced equation: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}$ .
- The stoichiometric ratio of $O_2$ consumed to $CO_2$ evolved is 6:6, which simplifies to 1:1.
- Moles of $CO_2$ evolved = Moles of $O_2$ consumed = $4.5 \times 10^{-3} \text{ moles}$ .
- $\text{Volume} = \text{Moles} \times \text{Molar Volume (at STP)}$
- $\text{Volume} = 4.5 \times 10^{-3} \times 22.4 \text{ L} = 0.1008 \text{ L} = \mathbf{100.8 \text{ mL}}$ .
Mass of Glucose Consumed:
- According to the equation, 6 moles of $O_2$ are required to break down 1 mole of glucose.
- Moles of glucose = $\frac{4.5 \times 10^{-3}}{6} = 0.75 \times 10^{-3} \text{ moles}$ .
- $\text{Mass} = \text{Moles} \times \text{Molar Mass}$
- $\text{Mass} = (0.75 \times 10^{-3} \text{ moles}) \times 180 \text{ g/mol} = 0.135 \text{ g} = \mathbf{135 \text{ mg}}$ .

Level 3: Respirometer and Respiratory Quotient (RQ)

Question 4: Analyzing Respirometer Data A student sets up a respirometer to measure the respiration rate of germinating pea seeds.

Setup A: Contains seeds and a small vial of Potassium Hydroxide (KOH).
Setup B: Contains seeds and a small vial of water (no KOH).

What is the physiological purpose of KOH in Setup A, and why did the fluid move?
Calculate the rate of oxygen consumption in $cm^3/\text{hour}$ .
Using the data from Setup B, calculate the Respiratory Quotient (RQ). Identify the predominant biochemical substrate the seeds are utilizing.

Step-by-Step Solution:

Role of KOH:
- KOH strongly absorbs carbon dioxide gas ( $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$ ).
- 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 manometer 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 = $12 \text{ mL}$ ( $1 \text{ mL} = 1 \text{ cm}^3$ ).
- $\text{Rate} = \frac{12 \text{ cm}^3}{2 \text{ hours}} = \mathbf{6 \text{ cm}^3/\text{hour}}$ .
Respiratory Quotient (RQ) and Substrate:
- In Setup B (no KOH), the fluid did not move. This means the pressure inside remained constant.
- Therefore, the volume of $CO_2$ evolved exactly replaced the volume of $O_2$ consumed ( $12 \text{ cm}^3$ ).
- $RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}} = \frac{12}{12} = \mathbf{1.0}$ .
- An RQ of 1.0 indicates that the seeds are predominantly respiring Carbohydrates (like starch or glucose).

Trap Questions & Common Pitfalls

Trap 1: The "Day vs. Night" Fallacy

The Scenario: A potted plant is placed in a sealed glass bell jar. During the day, gas sensors detect a net increase of oxygen by $15 \text{ mL}/\text{hour}$ . At night (in complete darkness), the sensors detect a net decrease of oxygen by $3 \text{ mL}/\text{hour}$ .
The Question: What is the plant's actual rate of aerobic respiration during the day?
The Trap: Students often assume plants do not respire during the day, only photosynthesize, and thus incorrectly answer $0 \text{ mL}/\text{hour}$ or $15 \text{ mL}/\text{hour}$ .
The Solution: Plants respire continuously, 24/7. Assuming the temperature is constant, the respiration rate at night ( $3 \text{ mL}/\text{hour}$ $O_2$ consumed) is approximately the same as during the day. 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. Therefore, the actual respiration rate during the day remains $3 \text{ mL/hour}$ of $O_2$ consumption. (The gross photosynthesis was actually producing $18 \text{ mL/hour}$ ).

Trap 2: Misinterpreting Anaerobic Products

The Scenario: Write the balanced chemical equation for anaerobic respiration occurring in the waterlogged roots of a mango tree.
The Trap: Writing the equation for animal muscle fatigue: $C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 \text{ (Lactic Acid)} + \text{Energy}$ .
The Solution: Plants and yeast undergo alcoholic fermentation, not lactic acid fermentation. The correct equation must produce ethanol and carbon dioxide: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH \text{ (Ethanol)} + 2CO_2 + \text{Energy (2 ATP)}$ . Forgetting that $CO_2$ is a byproduct of plant anaerobic respiration is a critical error!

Numerical Problems - Respiration in Plants

Chapter 3.2: Respiration in Plants - Numerical Problems

📊 Visualizing Respiration Pathways

🟢 Level 1: Stoichiometry and ATP Yields

🟡 Level 2: Energy Efficiency and Biomass

Level 3: Respirometer and Respiratory Quotient (RQ)

Trap Questions & Common Pitfalls

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Numerical Problems - Respiration in Plants

Chapter 3.2: Respiration in Plants - Numerical Problems

📊 Visualizing Respiration Pathways

🟢 Level 1: Stoichiometry and ATP Yields

🟡 Level 2: Energy Efficiency and Biomass

Level 3: Respirometer and Respiratory Quotient (RQ)

Trap Questions & Common Pitfalls

On this page