When 100% carbon is oxidized to `CO_(2)`, the efficiency of such a respiration is approximately

Options

A. 0.4
B. 0.6
C. 0.9
D. 1

Answer

To determine the efficiency of respiration when 100% carbon is oxidized to carbon dioxide (CO₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy Yield from ATP**:
- Each molecule of ATP releases approximately 7.3 kcal of energy.

2. **Calculate Total ATP Produced**:
- During complete oxidation of glucose in aerobic respiration, 38 ATP molecules are produced.

3. **Calculate Total Energy from ATP**:
- Multiply the number of ATP molecules by the energy per ATP:
\[ \text{Total Energy} = 38 \, \text{ATP} \times 7.3 \, \text{kcal/ATP} = 277.4 \, \text{kcal} \]

4. **Determine Energy Content of Glucose**:
- The energy content of one molecule of glucose is approximately 686 kcal.

5. **Calculate Efficiency of Respiration**:
- Efficiency can be calculated using the formula:
\[ \text{Efficiency} = \frac{\text{Energy produced from ATP}}{\text{Energy content of glucose}} = \frac{277.4 \, \text{kcal}}{686 \, \text{kcal}} \approx 0.4 \]

6. **Final Result**:
- The efficiency of respiration when 100% carbon is oxidized to CO₂ is approximately 0.4 or 40%.