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For an enzyme catalyzed reaction, the reaction rate is half of its maximum value when
  • a)
    [S] = Km
  • b)
    [S] = 2Km
  • c)
    [S] = Km/2
  • d)
    [S] = (Km)2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
For an enzyme catalyzed reaction, the reaction rate is half of its max...
Explanation:
Enzymes are biological catalysts that speed up chemical reactions by lowering the activation energy required for the reaction to occur. The rate of an enzyme-catalyzed reaction is influenced by the concentration of the substrate, which is the molecule that the enzyme acts upon.

Michaelis-Menten Equation:
The relationship between the substrate concentration ([S]) and the reaction rate (v) for an enzyme-catalyzed reaction can be described by the Michaelis-Menten equation:

v = (Vmax * [S]) / (Km + [S])

Where:
- Vmax is the maximum reaction rate, which is reached when the enzyme is saturated with substrate.
- Km is the Michaelis constant, which is a measure of the affinity of the enzyme for the substrate. It represents the substrate concentration at which the reaction rate is half of its maximum value.

Understanding the Answer:
The question asks for the substrate concentration ([S]) at which the reaction rate is half of its maximum value. According to the Michaelis-Menten equation, this occurs when [S] is equal to Km.

When [S] = Km, the denominator of the equation becomes (Km + Km) = 2Km. Thus, the reaction rate is given by:

v = (Vmax * [S]) / (2Km)

Since Vmax is a constant value, the reaction rate is directly proportional to the substrate concentration [S]. Therefore, when [S] = Km, the reaction rate will be half of its maximum value. This corresponds to option 'A' as the correct answer.

Summary:
In an enzyme-catalyzed reaction, the reaction rate is half of its maximum value when the substrate concentration ([S]) is equal to the Michaelis constant (Km). This occurs because at Km, the enzyme is half-saturated with substrate, leading to a reaction rate that is half of its maximum value.
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For an enzyme catalyzed reaction, the reaction rate is half of its maximum value whena)[S] = Kmb)[S] = 2Kmc)[S] = Km/2d)[S] = (Km)2Correct answer is option 'A'. Can you explain this answer?
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