Import Question JSON

Current Question (ID: 12150)

Question:
Pinus seeds cannot germinate and establish without fungal association. This is because:
Options:
  • 1. its seeds contain inhibitors that prevent germination.
  • 2. its embryo is immature.
  • 3. it has an obligate association with mycorrhizae.
  • 4. it has a very hard seed coat.
Solution:
Pinus, which is a type of gymnosperm, has a mutualistic association with certain species of fungi known as mycorrhizae. These fungi colonize the roots of Pinus seedlings and form a symbiotic relationship, where both the plant and the fungus benefit. The fungus helps the Pinus seedling by enhancing its nutrient uptake, particularly phosphorus, from the soil, and the plant provides the fungus with carbohydrates that it produces through photosynthesis. Pinus seeds are typically coated with a waxy layer that prevents them from germinating and establishing without the assistance of mycorrhizal fungi. The mycorrhizal association is necessary for the successful establishment of Pinus seedlings, as it helps the seedlings overcome the inhibitory effects of the waxy seed coat and enhances their nutrient uptake capabilities, especially phosphorus, which is often limiting in many soils. Option (1) is incorrect because inhibitors in the seed are not the main reason why Pinus seeds require fungal association for germination. Option (2) is incorrect because the immaturity of the embryo is not the main reason why Pinus seeds require fungal association for germination. Option (4) is incorrect because while the hard seed coat of Pinus seeds may be a physical barrier to germination, it is the obligate association with mycorrhizae that is the primary reason why Pinus seeds cannot germinate and establish without fungal association.

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
}