Question
Enzyme A catalyzes the reaction $S \rightarrow P$ and has a $K_{M}$ of $50 \mu \mathrm{M}$ and $\mathrm{a} V_{\max }$ of $100 \mathrm{nM} \cdot \mathrm{s}^{-1} .$ Enzyme $\mathrm{B}$ catalyzes the reaction $\mathrm{S} \rightarrow \mathrm{Q}$ and hasa $K_{M}$ of $5 \mathrm{mM}$ and a $V_{\max }$ of $120 \mathrm{nM} \cdot \mathrm{s}^{-1} .$ When $100 \mu \mathrm{M}$ of $\mathrm{S}$ is added to a mixture containing equivalent amounts of enzymes $A$ and $B$, after 1 minute which reaction product will be more abundant: $\mathrm{P}$ or $\mathrm{Q}$ ?
Step 1
The equation is given by: \[v = \frac{{V_{max} \cdot [S]}}{{K_M + [S]}}\] where \(v\) is the reaction velocity, \(V_{max}\) is the maximum reaction velocity, \(K_M\) is the Michaelis constant, and \([S]\) is the substrate concentration. Show more…
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Enzyme A catalyzes the reaction S→P and has a Km of 50 µM and a Vmax of 100 nM s⁻¹. Enzyme B catalyzes the reaction S→Q and has a Km of 5 mM and a Vmax of 120 nM s⁻¹. When 100 µM 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?
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