Question #a0b90

1 Answer
Dr. Hayek
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 (#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.

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