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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...
Enzyme Catalyzed Reaction

Enzymes are biological catalysts that increase the rate of chemical reactions without being consumed or permanently altered. The rate of an enzyme-catalyzed reaction is influenced by several factors, including substrate concentration ([S]), enzyme concentration, temperature, and pH.

Michaelis-Menten Kinetics

The Michaelis-Menten kinetics is a mathematical model that describes the relationship between the reaction rate (v) and the substrate concentration ([S]) in an enzyme-catalyzed reaction. According to this model, the reaction rate initially increases with increasing substrate concentration until it reaches a maximum value called Vmax.

The Michaelis-Menten equation is given by:

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

where v is the reaction rate, Vmax is the maximum reaction rate, [S] is the substrate concentration, and Km is the Michaelis constant.

The Michaelis constant (Km) represents the substrate concentration at which the reaction rate is half of its maximum value. It is a measure of the affinity between the enzyme and the substrate.

Half of the Maximum Reaction Rate

To find the substrate concentration at which the reaction rate is half of its maximum value, we can set v equal to half of Vmax in the Michaelis-Menten equation:

(Vmax [S]) / (Km + [S]) = Vmax / 2

By simplifying the equation, we get:

2[Km] = [S]

Therefore, the substrate concentration at which the reaction rate is half of its maximum value is equal to twice the value of the Michaelis constant ([Km]). Thus, the correct answer is option 'A' ([S] = Km).
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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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