Import Question JSON

Current Question (ID: 12125)

Question:
In pteridophytes, antheridium are present in:
Options:
  • 1. Leafy gametophyte
  • 2. Prothallus
  • 3. Sporophyll
  • 4. Protonema
Solution:
Antheridia, which are male reproductive structures, are typically present in the prothallus of pteridophytes. The prothallus is a small, heart-shaped gametophyte that develops from the spore of a pteridophyte fern or lycopod. Antheridia produce and release sperm cells for fertilization of the egg cells produced by the archegonia, which are the female reproductive structures found on the same prothallus or nearby prothalli. Option 1 (Leafy gametophyte) is incorrect as pteridophytes do not have leafy gametophytes. The gametophyte generation of pteridophytes is typically small and independent, and is called a prothallus. Option 3 (Sporophyll) is incorrect as sporophylls are modified leaves that bear sporangia, which are structures that produce spores. Sporophylls are part of the sporophyte generation of pteridophytes, not the gametophyte generation. Option 4 (Protonema) is incorrect as protonema is a thread-like structure that is characteristic of mosses and not present in pteridophytes. Protonema is the stage in the life cycle of mosses that develops from a spore and gives rise to the leafy gametophyte.

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
}