Import Question JSON

Current Question (ID: 14006)

Question:

Given below is the graph showing the effect of substrate concentration on enzyme activity. In the presence of a competitive inhibitor, when the concentration of the substrate is progressively increased:

Options:

1. The Km value increases but the reaction will not achieve Vmax
2. The Km value increases but the reaction can ultimately achieve Vmax
3. The Km value decreases but the reaction will not achieve Vmax
4. The Km value decreases but the reaction can ultimately achieve Vmax

Solution:

Competitive inhibition is a form of enzyme inhibition where binding of an inhibitor prevents binding of the target molecule of the enzyme, also known as the substrate. This is accomplished by blocking the binding site of the substrate - the active site - by some means. The Vmax indicates the maximum velocity of the reaction, while the Km is the amount of substrate needed to reach half of the Vmax. Km also plays a part in indicating the tendency of the substrate to bind the enzyme. Competitive inhibition can be overcome by adding more substrate to the reaction; therefore, increasing the chances of the enzyme and substrate binding. As a result, this alters only the Km, leaving the Vmax the same.

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
}