Import Question JSON

Current Question (ID: 12457)

Question:

Which of the following is not the component of secondary phloem?

Options:

1. Protophloem
2. Bast fibers
3. Companion cells
4. Phloem parenchyma

Solution:

Protophloem is a component of primary phloem, not secondary phloem. Primary phloem is the first phloem tissue that develops during plant growth and is typically composed of sieve elements, companion cells, phloem parenchyma, and sometimes fibers. Protophloem is the first-formed phloem and is characterized by having larger sieve elements and thinner cell walls compared to the later-formed, more mature phloem known as metaphloem. Secondary phloem, on the other hand, is the phloem tissue that is produced by the vascular cambium during secondary growth in plants. It is formed towards the outside of the stem or root, opposite to the secondary xylem, and contributes to the increase in girth or diameter of the plant. Components of secondary phloem can include sieve elements, companion cells, phloem fibers (bast fibers), and phloem parenchyma. Therefore, options 2, 3, and 4 (Bast fibers, Companion cells, and Phloem parenchyma) are all components of secondary phloem, while option 1 (Protophloem) is not.

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
}