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Current Question (ID: 14643)

Question:

The proximal convoluted tubule is lined by the:

Options:

1. Simple cuboidal epithelium
2. Simple columnar epithelium
3. Simple cuboidal brush bordered epithelium
4. Simple columnar brush bordered epithelium

Solution:

The proximal convoluted tubule (PCT) is the first segment of the renal tubule in the kidney nephron. It is responsible for reabsorbing various substances from the filtrate back into the bloodstream. The PCT is lined by a single layer of cells with a brush border on their apical surface. The cells lining the PCT have microvilli on their apical surface, which give them a brush-like appearance. These microvilli greatly increase the surface area of the cells and enhance their reabsorptive capabilities. The epithelial cells lining the PCT are simple cuboidal in shape. They are specialized for reabsorption and have numerous mitochondria to support active transport processes. Option (1) - Simple cuboidal epithelium - is partially correct. The PCT is lined by simple cuboidal cells, but these cells also have a brush border on their apical surface. Option (2) - Simple columnar epithelium - is incorrect. The PCT is not lined by simple columnar cells. Option (4) - Simple columnar brush bordered epithelium - is incorrect. The PCT is not lined by simple columnar cells.

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
}