Q14. Enzyme A聽catalyzes the reaction... [FREE SOLUTION]

Chapter 12: Q14. (page 399)

Enzyme $A$ catalyzes the reactions S to P and has a $K_{M}$ of $50 渭惭$ and a $V_{\max}$ of $100 {nMs}^{鈥� 1}$ . Enzyme $B$ catalyzes the reaction $S 鈫� Q$ and has a $K_{M}$ of $5 mM$ and a $V_{\max}$ of $120 {nMs}^{鈥� 1}$ . When $100 渭惭$ of $S$ is added to a mixture containing equivalent amounts of enzymes $A$ and $B$ , after $1$ minute which reaction product will be more abundant: P or Q?

Short Answer

Expert verified

The product P would be more abundant because the enzyme A has a lower $K_{M}$ .

Step by step solution

Introduction

The substrate concentration at which the reaction rate is 50% of V_max is known as Km (also known as the Michaelis constant). V_max shows the reaction of maximum velocity at which all enzymes are saturated with the substrate.

Product P will be abundantly produced

The enzyme A catalyzes the reaction $S 鈫� P$ ,

$K_{M} = 0.05 mM V_{\max} = 100 {nMs}^{鈥� 1}$

The enzyme B catalyzes the reaction $S 鈫� Q$ ,

$K_{M} = 5 mM V_{\max} = 120 {nMs}^{鈥� 1}$

It can be inferred from the data that V_maxis almost the same for both enzymes A and B. This depicts that the relative efficiency of the enzymes depends on their K_M values. Therefore, product P will be abundantly produced because enzyme A has a lower K_M value.

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Short Answer

Step by step solution

Introduction

Product P will be abundantly produced

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