Import Question JSON

Current Question (ID: 14669)

Question:

Juxtaglomerular apparatus is a special sensitive region formed by cellular modifications in the:

Options:

1. DCT and Efferent arteriole at the point of their contact
2. DCT and the Afferent arteriole at the point of their contact
3. PCT and Afferent arteriole at the point of their contact
4. PCT and Efferent arteriole at the point of their contact

Solution:

The juxtaglomerular apparatus (JGA) is a specialized structure located in the kidney, where the afferent arteriole and the distal convoluted tubule (DCT) come into contact. The JGA includes three cell types: the macula densa cells, the juxtaglomerular cells (JG cells), and the extraglomerular mesangial cells. The macula densa cells are specialized cells in the wall of the DCT that monitor the composition of the fluid in the tubule. The JG cells are modified smooth muscle cells in the wall of the afferent arteriole that produce and release the enzyme renin. The extraglomerular mesangial cells are located between the macula densa cells and the JG cells and are involved in signaling between these cells. The JGA plays an important role in the regulation of renal blood flow and glomerular filtration rate (GFR) by releasing renin in response to changes in blood pressure or sodium levels.

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
}