1. Interpretation of Michaelis Menten Equation

2. Michaelis-Menten  Graphically representation:

3. Interpetation of Mechales Menten equation  There are three main points  K m=[s]  K m>>>[s]  [s]>>> K m

4. Case 1: When  Km=[s] Thus  Km describe that substrate concentration that is needed to achieve the reaction rate is exactly half of the Vmax V o= V max[s] [s]+[s] V o= V max[s] 2[s] V o= V max 2

5. Case 2 In the bbegninng When concentration of substrate is very low  Km>>>[s] Thus at this point  The rate of reaction is directly proportional to the substrate concentration inccreasing [s] increases the rate of reaction V o= V max[s] K m

6. Case 3 At the end  [s]>>>km So  so  increasing the concentration of substrate does not effect rate of reaction V o= V max[s] [s] V o= V max

7. Lineweaver–Burk Plot It is also known as double reciprocal plot. The plot provides a useful graphical method for analysis of the Michaelis– Menten equation :

8. Taking the reciprocal gives

9. Graph:

10. Interpretation of k m with respect to k 2 and k -1 Case 1 k2 ˂ ˂ k-1  km and kd Case 2 k2 ˃ ˃ k-1  km

11. Interpretation of V max  V max depends on rate limiting step:  If k 2 is rate limiting then V max = k 2 [Et]  If k3 is rate limiting then V max = k 3 [E t ]

12.  When several steps are partially rate limiting  kcat  First order rate constant  Turnover number

13. Comparing Catalytic Mechanism And Efficienies

14.  The kinetic parameter Km and Kcat are useful for study and Comparasion of different enzymes.  Km is Michaels constant  Kcat is 1st order rate constant or turnover number  Values of kcat and km → cellular environment.  Substrate concentration encountered in vivo by enzyme and chemistry of reaction.  Evaluate kinetic efficiency of enzymes.  Two enzymes catalyzing different reactions→same K cat

15. Rate of catalyzed and uncatalyzed reactions is different  K m depends on substrate concentration Specificity constant: to compare the catalytic efficiency of different enzymes or K cat of different substrate by same enzyme is to compare ratio for two reactions → rate constant is conversion of E + P. when [S] << K V 0 = k cat /k m [E t ] [S]