Question #a0b90

Oct 5, 2015

${S}_{2}$ will be preferably enzymatically converted in product

Explanation:

${K}_{m}$ by definition is the concentration of substrate at half maximum velocity ($\frac{1}{2} {V}_{\max}$).

A smaller ${K}_{m}$ implies that the enzyme gets saturated at a low maximum velocity (${V}_{\max}$).

Therefore, since ${K}_{m 1} < {K}_{m 2}$ 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.