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Respiration in Plants - Respiratory quotient

Grade 11CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Respiratory Quotient (RQRQ) is defined as the ratio of the volume of CO2CO_2 evolved to the volume of O2O_2 consumed during respiration over a specific period of time.

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The value of RQRQ depends entirely on the type of respiratory substrate being oxidized in the cell.

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For Carbohydrates (e.g., Glucose), the RQRQ is 1.01.0 because the amount of CO2CO_2 evolved is equal to the amount of O2O_2 consumed during complete oxidation.

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For Fats (e.g., Tripalmitin), the RQRQ is less than 1.01.0 (typically around 0.70.7) because fats are oxygen-poor and require more external O2O_2 for oxidation.

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For Proteins, the RQRQ value is approximately 0.90.9.

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For Organic Acids (e.g., Malic acid), the RQRQ is greater than 1.01.0 because these compounds are relatively oxygen-rich.

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In Anaerobic Respiration, since O2O_2 is not consumed, the RQRQ is considered to be infinity (∞\infty).

📐Formulae

RQ=Volume of CO2 evolvedVolume of O2 consumedRQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}}

For Glucose: C6H12O6+6O2→6CO2+6H2O+Energy  ⟹  RQ=6CO26O2=1.0\text{For Glucose: } C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy} \implies RQ = \frac{6CO_2}{6O_2} = 1.0

For Tripalmitin: 2(C51H98O6)+145O2→102CO2+98H2O+Energy  ⟹  RQ=102CO2145O2=0.7\text{For Tripalmitin: } 2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O + \text{Energy} \implies RQ = \frac{102CO_2}{145O_2} = 0.7

💡Examples

Problem 1:

Calculate the RQRQ if a germinating seed consumes 145145 units of O2O_2 and releases 102102 units of CO2CO_2. Identify the likely substrate.

Solution:

RQ=102145≈0.7RQ = \frac{102}{145} \approx 0.7

Explanation:

Since the RQRQ value is 0.70.7, the respiratory substrate is likely a Fat (specifically Tripalmitin).

Problem 2:

What is the RQRQ of Malic acid (C4H6O5C_4H_6O_5) during complete oxidation?

Solution:

C4H6O5+3O2→4CO2+3H2OC_4H_6O_5 + 3O_2 \rightarrow 4CO_2 + 3H_2O RQ=4CO23O2=1.33RQ = \frac{4CO_2}{3O_2} = 1.33

Explanation:

Organic acids like malic acid have an RQ>1.0RQ > 1.0 because they contain more oxygen within their molecular structure compared to carbohydrates or fats.