Past Paper Question:
Consider an enzyme $E$ with four binding sites at which a substrate $S$ can (reversibly) bind to form a complex $C_1$, $C_2$, $C_3$, $C_4$ (with $1$, $2$, $3$ or $4$ binding sites occupied respectively) and hence (irreversibly) produce a product $P$. One of the chemical reactions is therefore given by the pathway: $$S+E \rightleftharpoons^{4k_1}_{k_-1} C_1 \rightarrow^{k_2}_{}P+E$$ Assuming there is no cooperativity with the binding substrate, but that the presence of two bound substrates renders the third and fourth sites inaccessible, give the remaining chemical reactions.
My Attempt:
Since the question says "no cooperativity" it wouldn't be something like this: $$C_1+S \rightleftharpoons^{2k'_1}_{2k'_-1} C_2 \rightarrow^{2k'_2}_{}P+C_1$$ $$C_2+S \rightleftharpoons^{k''_1}_{3k''_-1} C_3 \rightarrow^{3k''_2}_{}P+C_2$$ Since the question says "renders the third and fourth sites inaccessible" I'm thinking its more like a reaction with some modifier M: $$E+M \rightleftharpoons^{k_3}_{k_-3} C_M$$ $$C_M+S \rightleftharpoons^{k'_1}_{k'_-1} C_{SM} \rightarrow^{k'_2}_{}C_M+P$$ $$C_1+M \rightleftharpoons^{k'_3}_{k'_-3} C_{SM}$$
My Question:
How do you use this to find the correct chemical reaction paths ways?