Enzyme Kinetics: Answers to Problems

This page provides answers to Problems for Classroom Discussion

Gale Rhodes
Department of Chemistry
University of Southern Maine
Portland, Maine 04104-9300

Revised 2006/07/14

Enyzme Kinetics and Inhibition: Example for Analysis in Class

(Data taken from Lindquist, R.N., Problems and Solutions Guide to Accompany Rawn's Biochemistry, Neil Patterson Publishers, 1990.)

The table gives enzyme-catalyzed reaction rates (initial rate, V₀) measured at various substrate concentrations in solutions with [E] = 1.2 x 10-4 mmol/L.

1. Use the data from Experiment 1 to calculate V_max, K_m, and k_cat for this enzyme-catalyzed reaction.
   
2. Use the data from Experiment 2 to determine the apparent V_max and K_m in the presence of the inhibitor. From this information, determine the type of inhibition (competitive, noncompetitive or uncompetitive) and calculate the dissociation constant K_I for the inhibitor. For help, see Summary of Kinetic Effects of Reversible Inhibitors.

Answers

Results obtained by Lineweaver-Burke plot and linear least squares line fitting with Excel ("trendline, linear"); if you graph by hand or use other programs, your answers may differ slightly. Units of all quantities in calculations are shown in brackets.

Sorry about the chopped legends. Excel excelling as usual.

Experiment 1:

• V_max is the inverse of the (1/V₀) -intercept on the Lineweaver-Burke plot:
  
  V_max = 1/(4.6155 [L-min/mmol S]) = 0.217 mmol S/L-min
  
  
• The slope of the Lineweaver-Burke plot is equal to K_m/V_max, so K_m = slope x V_max:
  
  K_m = (0.378 [L-min/mmol S]/[L/mmol S]) x (0.217 [mmol S/L-min]) = 0.0820 mmol S/L
  
  
• V_max = k_cat x [E]_total so k_cat = V_max/[E]_total
  
  k_cat = (0.217 [mmol S/L-min]/(1.2 x 10-4 [mmol E/L] = 1.81 x 10³ [(mmol S/mmol E)/min] x (1 min/60 sec)
  
  k_cat = 30.1 sec^-1
  

Experiment 2:

Using the same methods as in experiment 1,

• V_max(apparent) = 0.183 mmol S/L.min
  
  
• K_m(apparent) = 0.152 mmol S/L
  

Inhibition is competitive, but note that with realistic data like this, there is some uncertainty in this conclusion. I conclude that inhibition is competitive because the inhibitor raises K_m by almost 85% and lowers V_max by only 16%. Given the poor data (look at the R² values), the change is Vmax is probably not significant. You may judge otherwise.

• From the Summary of Kinetic Effects of Reversible Inhibitors, for competitive inhibition,
  
  K_I = K_m[I]/(K_m^app - K_m)
  
  K_I = (0.0820 [mmol S/L] x 0.033 [mmol I/L])/(0.152 [mmol S/L] - 0.0820 [mmol S/L])
  
  K_I = 0.039 mM
  
  Compare K_I with K_m (not K_m^app). The inhibitor binds about twice as tightly as the substrate.

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