Question #a0b90

1 Answer
Dr. Hayek
Oct 5, 2015

S_2S2 will be preferably enzymatically converted in product

Explanation:

K_mKm 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.

Related questions
See all questions in Factors that Affect Enzyme Activity
Impact of this question
2811 views around the world
You can reuse this answer
Creative Commons License