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Current Question (ID: 12435)

Question:

Identify the correct statement regarding collenchyma in plants:

Options:

1. They occur in layers below the epidermis of monocotyledonous plants.
2. Cells are much thickened at the corners due to the deposition of lignin.
3. Collenchymatous cells do not contain chloroplasts.
4. They provide mechanical strength to the growing parts of the plant.

Solution:

Collenchyma is a type of ground tissue in plants that provides mechanical support to the growing parts of the plant, such as stems, petioles, and leaf veins. Collenchyma cells have thickened cell walls, especially at the corners, due to the deposition of cellulose, hemicellulose, and pectin, which provide structural rigidity to the cells. Statement 1 is incorrect, as collenchyma cells can occur in both monocotyledonous and dicotyledonous plants and are not limited to layers below the epidermis. Statement 2 is incorrect, as the thickening in collenchyma cells is primarily due to cellulose, hemicellulose, and pectin, not lignin. Statement 3 is incorrect, as some collenchyma cells may contain chloroplasts, although they are not fully specialized for photosynthesis like mesophyll cells. Chloroplast-containing collenchyma cells are known as chlorenchyma and can participate in photosynthesis to some extent.

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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
}