The efficiency of respiration is approximately

Options

A. 45%
B. 50%
C. 90%
D. 30%

Answer

To determine the efficiency of respiration in plants, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Cellular Respiration**:
- Cellular respiration is the process by which glucose is oxidized to produce energy in the form of ATP (adenosine triphosphate).

2. **ATP Production**:
- During cellular respiration, one glucose molecule can yield approximately 38 ATP molecules.

3. **Energy Yield from ATP**:
- Each ATP molecule produces about 34 kilojoules (kJ) of energy.
- Therefore, the total energy produced by 38 ATP molecules can be calculated as:
\[ \text{Total Energy from ATP} = 38 \text{ ATP} \times 34 \text{ kJ/ATP} = 1292 \text{ kJ} \]

4. **Energy Yield from Glucose**:
- The complete oxidation of one glucose molecule releases approximately 686 kilocalories (kcal).
- To convert this energy into kilojoules, we use the conversion factor (1 kcal = 4.184 kJ):
\[ \text{Energy from Glucose} = 686 \text{ kcal} \times 4.184 \text{ kJ/kcal} = 2870 \text{ kJ} \]

5. **Calculate Efficiency**:
- Efficiency of respiration can be calculated by dividing the energy produced by ATP by the energy released from glucose and then multiplying by 100 to express it as a percentage:
\[ \text{Efficiency} = \left( \frac{\text{Energy from ATP}}{\text{Energy from Glucose}} \right) \times 100 \]
\[ \text{Efficiency} = \left( \frac{1292 \text{ kJ}}{2870 \text{ kJ}} \right) \times 100 \approx 44.6\% \]

6. **Final Result**:
- The efficiency of respiration is approximately 45%.