Import Question JSON

Current Question (ID: 14768)

Question:

During muscular contraction, which of the following events occur? (i) H-zone disappears (ii) A band widens (iii) I band shortens (iv) Width of A band is unaffected (v) M line and Z line come closer

Options:

1. (i),(iii),(iv) and (v)
2. (i),(ii) and (v)
3. (ii),(iv) and (v)
4. (i),(ii) and (iii)

Solution:

Each myofibril has alternate dark and light bands on it. The light bands contain actin and are called I-band or Isotropic band, whereas the dark band called A or Anisotropic band contains myosin. In the centre of each I band is an elastic fibre called Z line which bisects it. The thin filaments are firmly attached to the Z line. The thick filaments in the A band are also held together in the middle of this band by a thin fibrous membrane called M line. Utilizing the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge. This pulls the attached actin filaments towards the centre of A band. The Z line attached to these actions are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. Hence the events occurring in muscular contraction are H-zone disappears, I-band reduces in width, The width of A band is unaffected and M-line and Z-line come closer.

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
}