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At what [s] is the velocity [V0] of an enzyme- catalized reaction is 25% of the Vmax ________ km.
Correct answer is '1/3'. Can you explain this answer?
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At what [s] is the velocity [V0] of an enzyme- catalized reaction is 2...
Explanation:
The Michaelis-Menten equation is used to describe the velocity of enzyme-catalyzed reactions. It is given as V = (Vmax * [S]) / (Km + [S]), where V is the reaction velocity, Vmax is the maximum velocity, [S] is the substrate concentration, and Km is the Michaelis constant.
When the substrate concentration is very high, the enzyme is said to be saturated, and the velocity is at its maximum value, Vmax.
At 25% of the Vmax, the velocity is 0.25 * Vmax.
Substituting this value into the Michaelis-Menten equation gives 0.25 * Vmax = (Vmax * [S]) / (Km + [S]).
Simplifying this equation gives [S] / (Km + [S]) = 0.75.
Multiplying both sides by (Km + [S]) gives [S] = 0.75 * Km + 0.75 * [S].
Solving for [S] gives [S] = 3 * Km.
Since the question asks for the value of [S] in terms of Km, we can divide both sides by Km to get [S]/Km = 3.
Therefore, [S] = Km/3, which means that the velocity is 25% of the Vmax at a substrate concentration of Km/3.
The value of [S] that gives a velocity of 25% of the Vmax is known as the quarter-saturation constant or the Michaelis constant.
Thus, the answer to the question is [S] = Km/3 or 1/3 Km.
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At what [s] is the velocity [V0] of an enzyme- catalized reaction is 25% of the Vmax ________ km.Correct answer is '1/3'. Can you explain this answer?
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