The maximum velocity of an enzyme at a substrate concentration is 20 mM s-1 at an enzyme concentration of 5 mM. The value of turnover number of enzyme will be_____________ s-1 [Answer in integer]Correct answer is '4.0'. Can you explain this answer?

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The maximum velocity of an enzyme at a substrate concentration is 20 mM s^-1 at an enzyme concentration of 5 mM. The value of turnover number of enzyme will be_____________ s^-1 [Answer in integer]

Correct answer is '4.0'. Can you explain this answer?

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The maximum velocity of an enzyme at a substrate concentration is 20 m...

From the law of mass action and assumption in Michaelis Menten equation
V_max = K_cat x [E_t]
Hence K_cat or turnover number = V_max /E_t
= 20/5 = 4.0 s^-1

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The maximum velocity of an enzyme at a substrate concentration is 20 mM s-1 at an enzyme concentration of 5 mM. The value of turnover number of enzyme will be_____________ s-1 [Answer in integer]Correct answer is '4.0'. Can you explain this answer?

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