Question

Question #a0b90

Answer 1

`S_2` will be preferably enzymatically converted in product

Explanation:

`K_m` by definition is the concentration of substrate at half maximum velocity (`1/2V_max`).

A smaller `K_m` implies that the enzyme gets saturated at a low maximum velocity (`V_max`).

Therefore, since `K_(m1) < K_(m2)` and both substrates have the same initial concentrations, I would say substrate `S_2` will be preferably enzymatically converted in product (will be consumed faster).

I hope this helps.