Import Question JSON

Current Question (ID: 11875)

Question:
$\text{If any organism has non cellulosic cell wall where can it be placed according to whittaker's system of classification?}$
Options:
  • 1. $\text{Monera}$
  • 2. $\text{Protista}$
  • 3. $\text{Fungi}$
  • 4. $\text{Both (1) and (3)}$ (Correct)
Solution:
$\text{Whittaker's system of classification categorized organisms into five kingdoms: Monera, Protista, Fungi, Plantae, and Animalia.}$ $\text{Monera includes prokaryotic organisms, such as bacteria, which can have non-cellulosic cell walls.}$ $\text{Fungi, on the other hand, are eukaryotic organisms that typically have chitin as their cell wall component, which is non-cellulosic.}$ $\text{Therefore, an organism with a non-cellulosic cell wall could be placed in both Monera and Fungi, as per Whittaker's system of classification.}$ $\text{Protista is a kingdom that includes eukaryotic organisms with diverse characteristics and may have different types of cell walls.}$ $\text{Plantae includes plants, which have cellulose as their main cell wall component.}$ $\text{Animalia includes animals, which do not have cell walls.}$ $\text{Analysis of each kingdom:}$ $\text{- Monera: Prokaryotic organisms with peptidoglycan (non-cellulosic) cell walls}$ $\text{- Fungi: Eukaryotic organisms with chitin (non-cellulosic) cell walls}$ $\text{- Protista: Variable cell wall composition, some may have non-cellulosic walls}$ $\text{- Plantae: Cellulosic cell walls}$ $\text{- Animalia: No cell walls}$ $\text{Since both Monera and Fungi have non-cellulosic cell walls as their characteristic feature, the correct answer is option (4) Both (1) and (3).}$

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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
}