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An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate of 20 µM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is _____ µM/min.
    Correct answer is between '29.0,31.0'. Can you explain this answer?
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    An enzyme catalyzes the conversion of 4 × 10−4 M substrate ...
    Calculating Vmax

    Step 1: Use the Michaelis-Menten equation to find Vmax.

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

    Step 2: Plug in the given values.

    V = 20 M/min
    [S] = 4 x 10^4 M
    Km = 2 x 10^4 M

    20 = Vmax(4 x 10^4)/(2 x 10^4 + 4 x 10^4)

    Step 3: Simplify the equation.

    20 = Vmax(4 x 10^4)/6 x 10^4

    20 x 6 x 10^4 = Vmax(4 x 10^4)

    Vmax = (20 x 6 x 10^4)/4 x 10^4

    Vmax = 30 M/min

    Step 4: Round to the nearest tenth.

    Vmax = 30.0 M/min

    Therefore, the value of Vmax is between 29.0 and 31.0 M/min.
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    An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate of 20 µM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is _____ µM/min.Correct answer is between '29.0,31.0'. Can you explain this answer?
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    An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate of 20 µM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is _____ µM/min.Correct answer is between '29.0,31.0'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate of 20 µM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is _____ µM/min.Correct answer is between '29.0,31.0'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An enzyme catalyzes the conversion of 4 × 10−4 M substrate into product at a rate of 20 µM/min. If the Km value for the enzyme is 2 × 10−4 M, the value of Vmax is _____ µM/min.Correct answer is between '29.0,31.0'. Can you explain this answer?.
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