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Current Question (ID: 12625)

Question:
Cell envelope in bacteria consists of
Options:
  • 1. Plasma membrane only
  • 2. Plasma membrane and mesosome
  • 3. Glycocalyx only
  • 4. Glycocalyx, cell wall and plasma membrane
Solution:
The cell envelope in bacteria consists of three layers - the glycocalyx, cell wall, and plasma membrane. The glycocalyx is a layer of polysaccharides that can be either a capsule or slime layer. It functions in protecting the cell from environmental stresses, and can also help the cell adhere to surfaces. The cell wall is a rigid structure made of peptidoglycan that provides support and protection to the cell. It also helps maintain the cell's shape. The plasma membrane is a phospholipid bilayer that regulates the flow of molecules in and out of the cell. It also contains various proteins that perform important functions such as transport, signaling, and energy production. Option 1 is incorrect as plasma membrane is only one of the layers of the cell envelope. Option 2 is incorrect as mesosomes are invaginations of the plasma membrane and are not a distinct layer of the cell envelope. Option 3 is incorrect as the glycocalyx is only one of the layers of the cell envelope and cannot constitute the entire envelope.

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
}