Import Question JSON

Current Question (ID: 14642)

Question:

The majority of water and salt filtered into Bowman's capsule is reabsorbed by:

Options:

1. the brush border of the transport epithelia of the proximal tubule
2. diffusion from the descending limb of the loop of Henle into the hypertonic interstitial fluid of the medulla
3. active transport across the transport epithelium of the thick upper segment of the ascending limb of the loop of Henle
4. selective secretion and diffusion across the distal tubule

Solution:

The majority of water and salt filtered into Bowman's capsule is reabsorbed by option 1, which is the brush border of the transport epithelia of the proximal tubule. After filtration, the fluid enters the proximal tubule where it passes through the brush border of the transport epithelia, which increases the surface area for reabsorption. The brush border contains numerous microvilli that provide a large surface area for the reabsorption of water and solutes. Most of the reabsorption of water, electrolytes, and nutrients occurs in the proximal tubule, where nearly 60-70% of the filtered Na+ and Cl- ions are reabsorbed along with 100% of the glucose and amino acids. Option 2 describes the mechanism of water reabsorption in the descending limb of the loop of Henle, where water moves out of the tubule by osmosis due to the high concentration of solutes in the surrounding interstitial fluid. Option 3 describes the mechanism of active transport across the transport epithelium of the thick ascending limb of the loop of Henle, which is responsible for reabsorption of Na+, K+, and Cl- ions. Option 4 describes the process of selective secretion and diffusion across the distal tubule, which is responsible for fine-tuning of the composition of urine by secreting or reabsorbing specific ions and molecules.

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
}