Answer to Question #160112 in Biochemistry for Sena

Question

A particular enzyme at a research facility is being studied by a group of graduate students. This enzyme has a Km value of 5.0 X 10-6 M. The students study this enzyme with an initial substrate concentration of 0.055 M. At one minute, 7 μM of product was made. What is the amount of product produced after 5 minutes?What is the Vmax?(Hint: unit of velocity is μM/min)

Answer 1

V_o=V_max[Substrate]/{K_m+[Substrate]}

7.0*10^-6M=V_max(0.055M)/{5.0*10^-6M +0.055M}

V_max=7.1*10^-6M/min

At 5 minutes, the amount of product formed is:

[P]=V_max*t

7.1*10^-6M/min*5min

[P]=3.55*10^-5M