Import Question JSON

Current Question (ID: 12449)

Question:

Function of sieve tubes are controlled by

Options:

1. Nucleus of sieve tube
2. Nucleus of companion cell
3. Nucleus of parenchyma
4. No function as they are dead cells

Solution:

The function of sieve tubes, which are specialized cells in the phloem responsible for the long-distance transport of organic nutrients in plants, is controlled by the nucleus of the companion cell. Sieve tubes are composed of sieve tube elements, which lack nuclei and other organelles, and they rely on companion cells for various physiological functions. Companion cells are adjacent nucleated cells closely associated with sieve tube elements and are responsible for providing energy, nutrients, and regulatory functions to support the activities of sieve tube elements. The nucleus of the companion cell plays a crucial role in controlling the functions of sieve tubes, including the regulation of nutrient transport, signal transduction, and other metabolic activities necessary for phloem transport. In contrast, sieve tube elements themselves are considered to be metabolically inactive and do not have functional nuclei.

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
}