6. (15) The enzyme invertase catalyzes the hydrolysis of sucrose. The initial re

Question

6. (15) The enzyme invertase catalyzes the hydrolysis of sucrose. The initial reaction rates vo as a function of initial sucrose substrate (S) concentrations are given below: initial [S]o (M) initial rate vo (M s1) 0.029 0.059 0.088 0.182 0.266 0.310 0.330 0.372 0.175 Prepare a Lineweaver-Burk plot (1/vo vs. 1/SJo), an Eadie-Hoftse plot (vo vs. vo[Slo), and a Dixon plot ([SJo/vo vs. [S]o) of the above data. For each plot, calculate the values of the maximum velocity Vm and the Michaelis constant Kw from the slope and intercept of each graph. Submit the three graphs with this assignment

Explanation / Answer

Lineweaver-burk plot is 1/V= (KM/Vmax)*1/S + 1/Vmax

so a plot of 1/V vs 1/S gives slope of KM/Vmax and intercept of 1/Vmax. The plot along with data points is shown below

From the plot, 1/Vmax= 2.1491, the intercept, Vmax= 1/2.1491= 0.465 M/s

slope is KM/Vmax= 0.0967, KM= 0.0967*0.465 =0.044996 M

2. Eddie=Hoftsee plot is given by V= -Km*(V/S)+ Vmax

So a plot of V vs V/S gives straight line whose intercept is Vmax and slope is -KM

the plot is shown below

from the plot, intercept is Vmax=0.4639 M/s and KM=0.0446 M.

Dixon plot is S/V= S/Vmax+ KM/Vmax so a plot of S/V vs S gives straight line whose slope is 1/Vmax and intercept is KM/Vmax

from the plot, slope =1/Vmax=2.1496, Vmax=1/2.1496=0.465 M/s, Km/Vmax=0.0968, KM=0.465*0.0968=0.045 M

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