Import Question JSON

Current Question (ID: 11570)

Question:

When proteins are used as respiratory substrates, the respiratory quotient would be about:

Options:

1. 1.2
2. 1.0
3. 0.9
4. 0.7

Solution:

The respiratory quotient (RQ) is the ratio of the amount of carbon dioxide produced to the amount of oxygen consumed during respiration. The value of RQ depends on the type of respiratory substrate being used. When proteins are used as respiratory substrates, the respiratory quotient would be about 0.9. This is because proteins are broken down into amino acids, which can then be converted to glucose, fatty acids, or other compounds that can enter the Krebs cycle. The Krebs cycle generates carbon dioxide as a byproduct, which is released during respiration. Oxygen is used as a reactant in the Krebs cycle, which generates water as a byproduct. Therefore, the ratio of carbon dioxide produced to oxygen consumed is approximately 0.9. Option 3, i.e., 0.9, is the correct answer. Option 1 is incorrect because an RQ greater than 1.0 indicates that the organism is producing more carbon dioxide than oxygen consumed, which is not possible. Option 2 is incorrect because an RQ of 1.0 indicates that the respiratory substrate being used is pure carbohydrate. Option 4 is incorrect because an RQ of 0.7 indicates that the respiratory substrate being used is primarily fat.

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}