Include your Lineweaver-Burk plots from question 75 in Section 2.2, and provide

Question

Include your Lineweaver-Burk plots from question 75 in Section 2.2, and provide the following:
a. Labels for each plot.

b. Label the positions on the plot, and corresponding equations, that provide the Vmax values.
c. Label the positions on the plot, and corresponding equations, that provide the Km values.

75. The following table provides the rates of reaction for an enzyme measured inthe absence and presence of an inhibitor. According to this information, what is the mechanism of inhibition? (Hint, make Lineweaver-Burk plots and see Chapter 7 of the textbook.) A. competitive B. uncompetitive C. noncompetitive D. allosteric E. irreversible inhibition [S] (mM) Rate w/o Inhibitor (HM/s) Rate w/ Inhibitor (HM/s) 1.17 1.3 2.5 2.6 2.1 6.3 6.5 7.6 5.9 13 26

Explanation / Answer

we know that

V= Vmax*S/(KM+S)

for drawing Lineweaver-burk plot, the equation can be written as

1/V= (KM/Vmax)*1/S + 1/Vmax, so a plot of 1/V 1vs 1/S gives a straight line. the intercept is 1/Vmax and slope is KM/Vmax

the data point of 1/S ( S in M) and 1/V ( in M/s) is tabulated and the plot is shown below

from the plots 1/V ( no inhibitor)= 98665, V= 1/98665 M/s =1.013*10-5 M/s

with inhibitor, 1/V= 1/98665= 1.013*10-5 M/s

KM/Vmax ( no inhibitor)= 392.1, KM= 392.1*1.013*10-5 = 0.003974 M with inhibitor KM/Vmax= 982.4, KM= 982.4*1.013*10-5 = 0.009957 M

since V remain the same, the inhibition is competitive inhibition.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---