Advanced Numerical Problems: Respiration in Plants

This section covers high-level numerical problems dealing with cellular respiration in plants. The problems explore nuances such as differing P/O ratios, mixed substrate quotients, proton motive force calculations, and fermentation mass balances.

1. Complex ATP Yield & Shuttle Systems

Question: A eukaryotic plant cell completely oxidizes 3 molecules of sucrose. Assuming the cytosolic NADH utilizes the glycerol-3-phosphate (G3P) shuttle and the modern P/O ratios are applied (NADH = 2.5 ATP, FADH₂ = 1.5 ATP), calculate the net total ATP generated. Contrast this with the theoretical maximum yield (older P/O ratios: NADH=3, FADH₂=2) utilizing the malate-aspartate shuttle.

Solution

Step 1: Understand the Substrate Sucrose is a disaccharide ( $C_{12}H_{22}O_{11}$ ). Complete hydrolysis of 3 sucrose molecules yields: $3 \text{ Sucrose} \rightarrow 6 \text{ Glucose/Fructose (Hexoses)}$

Step 2: Pathway Breakdown per Hexose Molecule For 1 molecule of Hexose (Glucose/Fructose):

Glycolysis: 2 net ATP (substrate-level), 2 cytosolic NADH
Link Reaction (2 Pyruvate $\rightarrow$ 2 Acetyl CoA): 2 mitochondrial NADH, 2 CO₂
Krebs Cycle (2 Acetyl CoA): 6 mitochondrial NADH, 2 FADH₂, 2 ATP (substrate-level), 4 CO₂

Step 3: Modern Yield with G3P Shuttle

Substrate-level ATP: 2 (Glycolysis) + 2 (Krebs) = 4 ATP
Cytosolic NADH (via G3P shuttle): Enters ETC as FADH₂. $2 \times 1.5 = \mathbf{3 \text{ ATP}}$
Mitochondrial NADH (Link + Krebs): $8 \times 2.5 = \mathbf{20 \text{ ATP}}$
Mitochondrial FADH₂ (Krebs): $2 \times 1.5 = \mathbf{3 \text{ ATP}}$
Total per Hexose = 4 + 3 + 20 + 3 = 30 ATP
Total for 6 Hexoses (3 Sucrose) = $30 \times 6 = \mathbf{180 \text{ ATP}}$

Step 4: Older Theoretical Yield with Malate-Aspartate Shuttle

Substrate-level ATP: 4 ATP
Cytosolic NADH (via Malate-Aspartate): Enters ETC as NADH. $2 \times 3 = \mathbf{6 \text{ ATP}}$
Mitochondrial NADH: $8 \times 3 = \mathbf{24 \text{ ATP}}$
Mitochondrial FADH₂: $2 \times 2 = \mathbf{4 \text{ ATP}}$
Total per Hexose = 4 + 6 + 24 + 4 = 38 ATP
Total for 6 Hexoses (3 Sucrose) = $38 \times 6 = \mathbf{228 \text{ ATP}}$

[!WARNING] Common Pitfalls

Substrate Trap: Forgetting that sucrose is a disaccharide. Calculating for 3 hexoses instead of 6 halves your answer.

Shuttle Confusion: The G3P shuttle transfers electrons to FAD, effectively turning cytosolic NADH into FADH₂. The malate-aspartate shuttle transfers them to NAD⁺.

2. Respiratory Quotient (RQ) of Mixed Substrates

Question: A germinating seed respires a mixture of carbohydrates (starch) and fats (arachidic acid, $C_{20}H_{40}O_{2}$ ) such that the ratio of oxygen consumed by carbohydrates to oxygen consumed by fats is 3:1. Calculate the apparent Respiratory Quotient (RQ) of the seed.

Solution

Step 1: Determine RQ for Individual Substrates

Carbohydrates: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$ $RQ_{carb} = \frac{6 \text{ CO}_2}{6 \text{ O}_2} = 1.0$
Arachidic Acid: $C_{20}H_{40}O_2 + 29O_2 \rightarrow 20CO_2 + 20H_2O$ $RQ_{fat} = \frac{20 \text{ CO}_2}{29 \text{ O}_2} \approx 0.6896$

Step 2: Apportion the Oxygen Consumption Let the total volume (or moles) of $O_2$ consumed be $4x$ .

$O_2$ consumed by carbohydrates = $3x$
$O_2$ consumed by fats = $1x$

Step 3: Calculate CO₂ Evolved

$CO_2$ from carbohydrates: Since $RQ = 1.0$ , $CO_2$ evolved = $O_2$ consumed = $3x$
$CO_2$ from fats: Since $RQ = \frac{20}{29}$ , $CO_2$ evolved = $\frac{20}{29} \times 1x \approx 0.6896x$
Total $CO_2$ evolved = $3x + 0.6896x = 3.6896x$

Step 4: Calculate Apparent RQ $Apparent RQ = \frac{\text{Total } CO_2}{\text{Total } O_2} = \frac{3.6896x}{4x} \approx \mathbf{0.922}$

[!CAUTION] Trap Question Alert! Many students simply average the two RQs: $(1.0 + 0.6896) / 2 = 0.845$ . This is incorrect! You must weight the $CO_2$ production based on the given ratio of oxygen consumption, not a simple arithmetic mean.

3. Chemiosmosis and Proton Motive Force

Question: In the inner mitochondrial membrane, the ETC pumps protons ( $H^+$ ) to create a gradient. Assume:

10 $H^+$ are pumped per matrix NADH oxidized.
6 $H^+$ are pumped per matrix FADH₂ oxidized.
4 $H^+$ must flow through ATP synthase to synthesize 1 ATP.

A mutant plant cell has "leaky" mitochondrial membranes, causing 20% of the pumped protons to diffuse back into the matrix without passing through ATP synthase. Under these conditions, calculate the net ATP yield from the complete oxidation of 1 molecule of Pyruvate (assume no other losses).

Visual Aid: Oxidative Phosphorylation Model

Electron Transport Chain & Chemiosmosis Flow:

Electron Donors:
- NADH (Matrix) donates $e^-$ to Complex I $\rightarrow$ Pumps 4 $H^+$ to IMS.
- FADH₂ (Matrix) donates $e^-$ to Complex II.
Electron Transport:
- Electrons from Complex I & II go to Coenzyme Q.
- Coenzyme Q transfers $e^-$ to Complex III $\rightarrow$ Pumps 4 $H^+$ to IMS.
- Complex III transfers $e^-$ to Cytochrome c.
- Cytochrome c transfers $e^-$ to Complex IV $\rightarrow$ Pumps 2 $H^+$ to IMS.
Proton Motive Force & ATP Synthesis (from IMS):
- 80% Flow: Through ATP Synthase ( $4 H^+$ /ATP) $\rightarrow$ ATP Produced.
- 20% Leak: Diffuses back into the Matrix.

Solution

Step 1: Yield from 1 Pyruvate The complete oxidation of 1 Pyruvate through the Link Reaction and Krebs Cycle yields:

Link Reaction: 1 NADH
Krebs Cycle: 3 NADH, 1 FADH₂, 1 ATP (substrate-level)
Total: 4 NADH, 1 FADH₂, 1 ATP

Step 2: Calculate Protons Pumped

From 4 NADH: $4 \times 10 H^+ = 40 H^+$
From 1 FADH₂: $1 \times 6 H^+ = 6 H^+$
Total Protons Pumped: $46 H^+$

Step 3: Account for Membrane Leakage Due to the 20% leak, only 80% of the protons contribute to ATP synthesis:

Effective Protons = $46 H^+ \times 0.80 = \mathbf{36.8 H^+}$

Step 4: Calculate Oxidative ATP

ATP from synthase = $36.8 H^+ / 4 H^+ \text{ per ATP} = \mathbf{9.2 \text{ ATP}}$

Step 5: Final ATP Yield Don't forget the substrate-level phosphorylation from the Krebs cycle!

Net ATP Yield = $9.2 \text{ (oxidative)} + 1 \text{ (substrate-level)} = \mathbf{10.2 \text{ ATP}}$

[!TIP] While ATP is discrete (you can't have 0.2 of an ATP molecule), chemiosmotic calculations often yield fractional values because the proton gradient is a continuous energy store. The "10.2" represents a statistical average per pyruvate oxidized.

4. Alcoholic Fermentation Mass Balance

Question: A batch of yeast is grown anaerobically in a closed bioreactor containing 180g of glucose. Assuming 100% conversion via alcoholic fermentation, calculate the volume of $CO_2$ gas produced at Standard Temperature and Pressure (STP, 22.4 L/mol) and the final mass of ethanol formed.

Solution

Step 1: The Balanced Chemical Equation $C_6H_{12}O_6 \rightarrow 2C_2H_5OH (\text{Ethanol}) + 2CO_2$

Step 2: Calculate Moles of Substrate

Molar mass of Glucose ( $C_6H_{12}O_6$ ): $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \text{ g/mol}$
Moles of Glucose = $180 \text{ g} / 180 \text{ g/mol} = \mathbf{1 \text{ mol}}$

Step 3: Calculate Mass of Ethanol

From the equation, 1 mole of glucose yields 2 moles of ethanol.
Molar mass of Ethanol ( $C_2H_5OH$ ): $(2 \times 12) + (6 \times 1) + (1 \times 16) = 46 \text{ g/mol}$
Mass of Ethanol = $2 \text{ mol} \times 46 \text{ g/mol} = \mathbf{92 \text{ g}}$

Step 4: Calculate Volume of CO₂

From the equation, 1 mole of glucose yields 2 moles of $CO_2$ .
Volume at STP = $2 \text{ mol} \times 22.4 \text{ L/mol} = \mathbf{44.8 \text{ L}}$

[!WARNING] Common Pitfalls Confusing the products of different anaerobic pathways! If this were lactic acid fermentation (e.g., in overworked muscle cells or Lactobacillus), the reaction would be $C_6H_{12}O_6 \rightarrow 2C_3H_6O_3$ . Zero $CO_2$ is produced in lactic acid fermentation.

Advanced Numerical Problems: Respiration in Plants

1. Complex ATP Yield & Shuttle Systems

Solution

Step 2: Pathway Breakdown per Hexose Molecule For 1 molecule of Hexose (Glucose/Fructose):

Glycolysis: 2 net ATP (substrate-level), 2 cytosolic NADH
Link Reaction (2 Pyruvate $\rightarrow$ 2 Acetyl CoA): 2 mitochondrial NADH, 2 CO₂
Krebs Cycle (2 Acetyl CoA): 6 mitochondrial NADH, 2 FADH₂, 2 ATP (substrate-level), 4 CO₂

Step 3: Modern Yield with G3P Shuttle

Substrate-level ATP: 2 (Glycolysis) + 2 (Krebs) = 4 ATP
Cytosolic NADH (via G3P shuttle): Enters ETC as FADH₂. $2 \times 1.5 = \mathbf{3 \text{ ATP}}$
Mitochondrial NADH (Link + Krebs): $8 \times 2.5 = \mathbf{20 \text{ ATP}}$
Mitochondrial FADH₂ (Krebs): $2 \times 1.5 = \mathbf{3 \text{ ATP}}$
Total per Hexose = 4 + 3 + 20 + 3 = 30 ATP
Total for 6 Hexoses (3 Sucrose) = $30 \times 6 = \mathbf{180 \text{ ATP}}$

Step 4: Older Theoretical Yield with Malate-Aspartate Shuttle

Substrate-level ATP: 4 ATP
Cytosolic NADH (via Malate-Aspartate): Enters ETC as NADH. $2 \times 3 = \mathbf{6 \text{ ATP}}$
Mitochondrial NADH: $8 \times 3 = \mathbf{24 \text{ ATP}}$
Mitochondrial FADH₂: $2 \times 2 = \mathbf{4 \text{ ATP}}$
Total per Hexose = 4 + 6 + 24 + 4 = 38 ATP
Total for 6 Hexoses (3 Sucrose) = $38 \times 6 = \mathbf{228 \text{ ATP}}$

[!WARNING] Common Pitfalls

Substrate Trap: Forgetting that sucrose is a disaccharide. Calculating for 3 hexoses instead of 6 halves your answer.

Shuttle Confusion: The G3P shuttle transfers electrons to FAD, effectively turning cytosolic NADH into FADH₂. The malate-aspartate shuttle transfers them to NAD⁺.

2. Respiratory Quotient (RQ) of Mixed Substrates

Solution

Step 1: Determine RQ for Individual Substrates

Carbohydrates: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$ $RQ_{carb} = \frac{6 \text{ CO}_2}{6 \text{ O}_2} = 1.0$
Arachidic Acid: $C_{20}H_{40}O_2 + 29O_2 \rightarrow 20CO_2 + 20H_2O$ $RQ_{fat} = \frac{20 \text{ CO}_2}{29 \text{ O}_2} \approx 0.6896$

Step 2: Apportion the Oxygen Consumption Let the total volume (or moles) of $O_2$ consumed be $4x$ .

$O_2$ consumed by carbohydrates = $3x$
$O_2$ consumed by fats = $1x$

Step 3: Calculate CO₂ Evolved

$CO_2$ from carbohydrates: Since $RQ = 1.0$ , $CO_2$ evolved = $O_2$ consumed = $3x$
$CO_2$ from fats: Since $RQ = \frac{20}{29}$ , $CO_2$ evolved = $\frac{20}{29} \times 1x \approx 0.6896x$
Total $CO_2$ evolved = $3x + 0.6896x = 3.6896x$

Step 4: Calculate Apparent RQ $Apparent RQ = \frac{\text{Total } CO_2}{\text{Total } O_2} = \frac{3.6896x}{4x} \approx \mathbf{0.922}$

[!CAUTION] Trap Question Alert! Many students simply average the two RQs: $(1.0 + 0.6896) / 2 = 0.845$ . This is incorrect! You must weight the $CO_2$ production based on the given ratio of oxygen consumption, not a simple arithmetic mean.

3. Chemiosmosis and Proton Motive Force

Question: In the inner mitochondrial membrane, the ETC pumps protons ( $H^+$ ) to create a gradient. Assume:

10 $H^+$ are pumped per matrix NADH oxidized.
6 $H^+$ are pumped per matrix FADH₂ oxidized.
4 $H^+$ must flow through ATP synthase to synthesize 1 ATP.

Visual Aid: Oxidative Phosphorylation Model

Electron Transport Chain & Chemiosmosis Flow:

Electron Donors:
- NADH (Matrix) donates $e^-$ to Complex I $\rightarrow$ Pumps 4 $H^+$ to IMS.
- FADH₂ (Matrix) donates $e^-$ to Complex II.
Electron Transport:
- Electrons from Complex I & II go to Coenzyme Q.
- Coenzyme Q transfers $e^-$ to Complex III $\rightarrow$ Pumps 4 $H^+$ to IMS.
- Complex III transfers $e^-$ to Cytochrome c.
- Cytochrome c transfers $e^-$ to Complex IV $\rightarrow$ Pumps 2 $H^+$ to IMS.
Proton Motive Force & ATP Synthesis (from IMS):
- 80% Flow: Through ATP Synthase ( $4 H^+$ /ATP) $\rightarrow$ ATP Produced.
- 20% Leak: Diffuses back into the Matrix.

Solution

Step 1: Yield from 1 Pyruvate The complete oxidation of 1 Pyruvate through the Link Reaction and Krebs Cycle yields:

Link Reaction: 1 NADH
Krebs Cycle: 3 NADH, 1 FADH₂, 1 ATP (substrate-level)
Total: 4 NADH, 1 FADH₂, 1 ATP

Step 2: Calculate Protons Pumped

From 4 NADH: $4 \times 10 H^+ = 40 H^+$
From 1 FADH₂: $1 \times 6 H^+ = 6 H^+$
Total Protons Pumped: $46 H^+$

Step 3: Account for Membrane Leakage Due to the 20% leak, only 80% of the protons contribute to ATP synthesis:

Effective Protons = $46 H^+ \times 0.80 = \mathbf{36.8 H^+}$

Step 4: Calculate Oxidative ATP

ATP from synthase = $36.8 H^+ / 4 H^+ \text{ per ATP} = \mathbf{9.2 \text{ ATP}}$

Step 5: Final ATP Yield Don't forget the substrate-level phosphorylation from the Krebs cycle!

Net ATP Yield = $9.2 \text{ (oxidative)} + 1 \text{ (substrate-level)} = \mathbf{10.2 \text{ ATP}}$

[!TIP] While ATP is discrete (you can't have 0.2 of an ATP molecule), chemiosmotic calculations often yield fractional values because the proton gradient is a continuous energy store. The "10.2" represents a statistical average per pyruvate oxidized.

4. Alcoholic Fermentation Mass Balance

Solution

Step 1: The Balanced Chemical Equation $C_6H_{12}O_6 \rightarrow 2C_2H_5OH (\text{Ethanol}) + 2CO_2$

Step 2: Calculate Moles of Substrate

Molar mass of Glucose ( $C_6H_{12}O_6$ ): $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \text{ g/mol}$
Moles of Glucose = $180 \text{ g} / 180 \text{ g/mol} = \mathbf{1 \text{ mol}}$

Step 3: Calculate Mass of Ethanol

From the equation, 1 mole of glucose yields 2 moles of ethanol.
Molar mass of Ethanol ( $C_2H_5OH$ ): $(2 \times 12) + (6 \times 1) + (1 \times 16) = 46 \text{ g/mol}$
Mass of Ethanol = $2 \text{ mol} \times 46 \text{ g/mol} = \mathbf{92 \text{ g}}$

Step 4: Calculate Volume of CO₂

From the equation, 1 mole of glucose yields 2 moles of $CO_2$ .
Volume at STP = $2 \text{ mol} \times 22.4 \text{ L/mol} = \mathbf{44.8 \text{ L}}$

[!WARNING] Common Pitfalls Confusing the products of different anaerobic pathways! If this were lactic acid fermentation (e.g., in overworked muscle cells or Lactobacillus), the reaction would be $C_6H_{12}O_6 \rightarrow 2C_3H_6O_3$ . Zero $CO_2$ is produced in lactic acid fermentation.

Numerical Problems - Respiration in Plants

Advanced Numerical Problems: Respiration in Plants

1. Complex ATP Yield & Shuttle Systems

Solution

2. Respiratory Quotient (RQ) of Mixed Substrates

Solution

3. Chemiosmosis and Proton Motive Force

Visual Aid: Oxidative Phosphorylation Model

Solution

4. Alcoholic Fermentation Mass Balance

Solution

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Numerical Problems - Respiration in Plants

Advanced Numerical Problems: Respiration in Plants

1. Complex ATP Yield & Shuttle Systems

Solution

2. Respiratory Quotient (RQ) of Mixed Substrates

Solution

3. Chemiosmosis and Proton Motive Force

Visual Aid: Oxidative Phosphorylation Model

Solution

4. Alcoholic Fermentation Mass Balance

Solution

On this page