Consider the following mechanistic scheme for an enzyme catalyzed reaction E + S
ID: 530105 • Letter: C
Question
Consider the following mechanistic scheme for an enzyme catalyzed reaction E + S K__rightarrow_k_1 ES k_2_rightarrow EP k_3_rightarrow E + P Show step by step how the following equation can be derived: V = k_2 k_3/k_2 + k_3 [E]_0 [S]/[k_3/k_2 + k_3]. K_1/k_1 + [s] Make the following assumptions: The last step shall be rate limiting, so that v = k_3 [EP] The steady state approximation is valid for EP There is a rapid pre-equilibrium between ES and the reactants E and s. s shall be present in large excess, so that [S]_o= [s], but [E]_o= [E] + [ES] +[EP]. What happens if [S] is really large? Write out an expression for a rate law that an experimentalist should observe upon working with a large excess of substrate. what happens if [s] is tiny? What type of rate law would an experimentalist observe in this case?Explanation / Answer
when [S] is relativey large sum of two terms in denominator would be nearly equal to [S] since ( large term + small term gives value to nearly large term)
Hence we get rate = [(k2k3) /( k2+k3) ] [Eo] [S] / [S]
= [(k2k3)/(k2+k3) ] [Eo]
When [S] is very small we get sum of two terms iin denominator nearly equal to (k3/k2+k3)
Hence rate becomes = [(k2k3)/( k2+k3) ] [Eo] [S] / ( k3/k2+k3)
= k2 [Eo] [S]