Question #78557

A reaction is catalyzed by an enzyme with parameters k=103 s-1, KM=61·10-6 M and concentration [E]0= 3.5·10-6 M. The initial concentration of the substrate is [S]0= 3.6·10-5 M. Estimate initial reaction rate. Find the substrate’s concentration providing 2-fold decreasing of the reaction rate.

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Answer on Question #78557 - Chemistry - Physical Chemistry

Question: A reaction is catalyzed by an enzyme with parameters k=103k=103 s-1, KM=61·10-6 M and concentration [E]0= 3.5·10-6 M. The initial concentration of the substrate is [S]0= 3.6·10-5 M. Estimate initial reaction rate. Find the substrate's concentration providing 2-fold decreasing of the reaction rate.

Solution:


v0=k[xˉ]s[zˉ]s[zˉ]s+Km,v_0 = \frac{k \cdot [\bar{x}]_s \cdot [\bar{z}]_s}{[\bar{z}]_s + K_m},v0=1033.51063.6105/(3.6105+61106)=1.3108/9.7105=1.3104 M/s.v_0 = 103 \cdot 3.5 \cdot 10^{-6} \cdot 3.6 \cdot 10^{-5} / (3.6 \cdot 10^{-5} + 61 \cdot 10^{-6}) = 1.3 \cdot 10^{-8} / 9.7 \cdot 10^{-5} = 1.3 \cdot 10^{-4} \text{ M/s}.


Answer: 1.3104 M/s1.3 \cdot 10^{-4} \text{ M/s}.

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