Chapter 7 In-Class Problems

Chapter 7

Opener and In-Class Problems

Enzyme:

HMG-CoA Reductase

• Substrate: HMG-CoA

• Product: Mevalonate

• Rate-limiting enzyme in the biosynthesis of cholesterol

• Inhibitor(s): Statins (lovastatin here)

• How is lovastatin inhibiting HMG-CoA reductase?

• You are determining the kinetics of an enzyme by keeping the enzyme concentration constant (5uM) and measuring the rate of product formation at different substrate concentrations. You obtain the data shown.

• Pair up with another student. Using graph paper, one student should make a Michaelis-Menten plot and the other should make a Lineweaver-Burk plot.

• From your graph(s), determine Km and Vmax. Compare your answers. Did you have the same answers? If not, how did they differ? Which type of graph was best for finding the answers? Why?

Michaelis-Menten

• Estimate Vmax

• Calculate 1/2 Vmax

• Interpolate point on X axis that corresponds to 1/2 Vmax

• This equals Km

Lineweaver-Burk (Vmax)

y=mx+b

Calculate y intercept by setting X=0

y=0.219X + 0.0226y=0+0.0226

y=0.0226

y=1/Vmax

Vmax = 1/y = 44.24778761 uM/min

Lineweaver-Burk (Km)

Calculate x intercept by setting y = 0

y = 0.219x + 0.0226

0=0.219x + 0.0226

0.219x = -0.0226

x = -0.0226/0.219 =

x = -1/Km

m = -1/x

Km =-1/-.103196347 = 9.690265487 uM

Continued

Calculate the turnover number and catalytic efficiency of this enzyme.

Kcat = Turnover Number

• Kcat = Vmax/Enzyme concentration

• Kcat = 44.2 uM/min ÷ 5 uM = 8.84 min-1

• A measure of how many substrate molecules are converted to product per min

Enzyme Efficiency

• Kcat/Km = Enzyme efficiency

• Kcat =8.84 min-1

• Kcat/Km = 8.84 min-1 ÷ 9.7 uM

• Efficiency = 0.91 uM-1 min-