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Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]
Correct answer is '0.001'. Can you explain this answer?
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Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10&ndash...
Vmax and 2.5 mM, respectively.
Vmax (maximum velocity) is the maximum rate of an enzyme-catalyzed reaction when the enzyme is saturated with substrate. KM (Michaelis constant) is the substrate concentration at which half of the enzyme's active sites are occupied with substrate.
In this case, the Vmax is 2.0, meaning that the enzyme can catalyze the reaction at a maximum rate of 2.0 units per minute (or whatever units are appropriate for the specific reaction). The KM is 2.5 mM, so when the substrate concentration is 2.5 mM, half of the enzyme's active sites are occupied with substrate.
Together, Vmax and KM can be used to calculate the enzyme's catalytic efficiency, or kcat/KM, which is a measure of how well the enzyme can convert substrate to product.
Vmax (maximum velocity) is the maximum rate of an enzyme-catalyzed reaction when the enzyme is saturated with substrate. KM (Michaelis constant) is the substrate concentration at which half of the enzyme's active sites are occupied with substrate.
In this case, the Vmax is 2.0, meaning that the enzyme can catalyze the reaction at a maximum rate of 2.0 units per minute (or whatever units are appropriate for the specific reaction). The KM is 2.5 mM, so when the substrate concentration is 2.5 mM, half of the enzyme's active sites are occupied with substrate.
Together, Vmax and KM can be used to calculate the enzyme's catalytic efficiency, or kcat/KM, which is a measure of how well the enzyme can convert substrate to product.
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Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer?
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Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer?.
Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vmax and KM for an enzyme catalyzed reactions are 2.0 × 10–3 Ms-1 and 1.0 × 10–6 M resp., The rate of reaction when substrate concentration is 1.0 × 10–6 M is? [up to three decimal places]Correct answer is '0.001'. Can you explain this answer?.
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