Import Question JSON

Current Question (ID: 11490)

Question:

Glycolysis is:

Options:

1. Oxidation of glucose to glutamate
2. Conversion of pyruvate to citrate
3. Oxidation of glucose to pyruvate
4. Conversion of glucose to haem

Solution:

Glycolysis is the metabolic pathway that converts glucose into pyruvate. It occurs in the cytoplasm of the cell and can proceed in the presence or absence of oxygen. In the presence of oxygen, pyruvate can enter the mitochondria and be further oxidized in the citric acid cycle and oxidative phosphorylation to generate more ATP. In the absence of oxygen, pyruvate is converted to lactate or ethanol in a process called fermentation. During glycolysis, glucose is broken down into two molecules of pyruvate, and a small amount of ATP is produced. The process involves a series of enzyme-catalyzed reactions that convert glucose into two molecules of glyceraldehyde 3-phosphate (G3P), which are then converted into two molecules of pyruvate. Along the way, energy is extracted from glucose and stored in the form of ATP and NADH. Overall, glycolysis is an important pathway for generating energy in cells, particularly during periods of high energy demand or when oxygen is limited.

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
}