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Current Question (ID: 11355)

Question:

Production of NADPH in a chloroplast takes place during

Options:

1. cyclic photophosphorylation
2. non-cyclic photophosphorylation
3. series photophosphorylation
4. substrate level photophosphorylation

Solution:

(2) Formation of ATP in chloroplasts in the presence of light is known as photophosphorylation. It takes place in two ways – (i) Cyclic photophosphorylation : In this process, electrons released from P₇₀₀ return back to P₇₀₀ in cyclic manner and neither oxygen nor NADPH is produced. (ii) Non-cyclic photophosphorylation : In this process high energy electrons released from P₆₈₀ do not return to P₆₈₀ and both oxygen and NADPH are synthesized. NCERT Reference: Page 213 XI NCERT

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
}