Import Question JSON

Current Question (ID: 13745)

Question:

Cell junctions like tight, adhering, and gap junctions are mainly present between the neighbouring cells of:

Options:

1. Neural tissue
2. Connective tissue
3. Muscular tissue
4. Epithelial tissue

Solution:

Epithelial tissue is a type of tissue that covers the body's external and internal surfaces, and forms the lining of organs and cavities. Epithelial cells are closely packed and form continuous sheets, and they need to be held tightly together to maintain the integrity of the tissue. Various types of cell junctions, including tight junctions, adherens junctions, and desmosomes, are present between the neighbouring cells of epithelial tissue to provide structural support, maintain cell polarity, and regulate the movement of molecules across the epithelial layer. While cell junctions may also be present in other types of tissues such as neural, muscular, and connective tissues, they are not as prevalent as they are in epithelial tissue. Neural tissue, for example, contains specialized junctions called synapses, which allow for the transmission of electrical and chemical signals between neurons, while muscular tissue contains specialized junctions called intercalated discs, which allow for coordinated contraction of cardiac muscle cells. Connective tissue, meanwhile, contains various types of cells and fibers, but does not typically have the tightly packed sheet-like structure of epithelial tissue.

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
}