Import Question JSON

Current Question (ID: 14192)

Question:

What will happen if the secretion of parietal cells of gastric glands is blocked with an inhibitor?

Options:

1. Gastric juice will be deficient in chymosin
2. Gastric juice will be deficient in pepsinogen
3. In the absence of HCl secretion, inactive pepsinogen is not converted into the active enzyme pepsin
4. Enterokinase will not be released from the duodenal mucosa and so trypsinogen is not converted to trypsin

Solution:

Gastric glands are numerous microscopic tubular glands formed by the epithelium of stomach. The parietal cells (oxyntic cells) are large and most numerous on the side walls of gastric glands. They secrete hydrochloric acid and castle intrinsic factor. The peptic cells (zymogen) cells of gastric glands secrete gastric digestive enzymes as proenzymes-pepsinogen and prorennin and small amount of gastric amylase and gastric lipase. The hydrochloric acid maintains a strongly acidic pH of about 1.5 to 2.5 in the stomach. HCl converts pepsinogen and prorennin to pepsin and rennin respectively. The secretion of intestinal glands is called succus entericus. It contains many enzymes, maltase, lactase, sucrase, enterokinase, nucleotidase etc. The enterokinase converts proenzyme trypsinogen into active trypsin.

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
}