Import Question JSON

Current Question (ID: 14716)

Question:
What three physical barriers must filtrate cross within the renal corpuscle?
Options:
  • 1. podocytes, pedicels, and filtration slits
  • 2. macula densa, vascular pole, and parietal layer of glomerular capsule
  • 3. glomerular endothelium, basement membrane, and podocytes
  • 4. proximal convoluted tubules, connecting tubules, and papillary ducts
Solution:
The glomerular endothelium is the first barrier that the filtrate encounters as it enters the renal corpuscle. It is a highly fenestrated capillary endothelium that allows the passage of water and solutes but prevents the passage of blood cells and large plasma proteins. The basement membrane is the second barrier and is made up of extracellular matrix molecules such as collagen and proteoglycans. It provides an electrostatic barrier that prevents the passage of negatively charged molecules such as large proteins. The podocytes and their pedicels (foot processes) are the final barrier that the filtrate must cross. They form the visceral layer of the glomerular capsule and wrap around the capillaries. The pedicels interdigitate to form filtration slits, which are covered by a thin diaphragm. The size of the filtration slits limits the passage of macromolecules and helps maintain the selectivity of the filtration barrier.

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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
}