lecture 15 - university of michiganessen/html/powerpoints/lecture... · lecture 15 . today’s...

31
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 15

Upload: others

Post on 30-May-2020

5 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of

chemical reactions and the design of the reactors in which they take place.

Lecture 15

Page 2: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Today’s lecture   Enzymes

 Michealis-Menten Kinetics  Lineweaver-Burk Plot  Enzyme Inhibition  Competitive   Uncompetitive

2

Page 3: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Last lecture

3

Page 4: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

4

Last lecture

Page 5: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Enzymes Michaelis-Menten Kinetics. Enzymes are protein like substances with catalytic properties.

Enzyme unease. [From Biochemistry, 3/E by Stryer, copywrited 1988 by Lubert Stryer. Used with permission of W.H. Freeman and Company.]

5

Page 6: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Enzymes It provides a pathway for the substrate to proceed at a faster rate. The substrate, S, reacts to form a product P.

A given enzyme can only catalyze only one reaction. Example, Urea is decomposed by the enzyme urease.

Slow S P

Fast

6

Page 7: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

7

Page 8: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

8

Page 9: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

A given enzyme can only catalyze only one reaction. Urea is decomposed by the enzyme urease, as shown below.

9

Page 10: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

The corresponding mechanism is:

10

Page 11: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Michaelis-Menten Kinetics

11

Page 12: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Michaelis-Menten Kinetics

12

Page 13: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Vmax=kcat* Et

Turnover Number: kcat Number of substrate molecules (moles) converted to product in a given time (s) on a single enzyme molecule

(molecules/molecule/time) For the reaction

40,000,000 molecules of H2O2 converted to product per second on a single enzyme molecule.

13

H2O2 + E →H2O + O + E kcat

Page 14: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Summary

14

Page 15: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

(Michaelis-Menten plot) Vmax

-rs

S1/2

Solving:

KM=S1/2

therefore KM is the concentration at which the rate is half the maximum rate

CS

Michaelis-Menten Equation

15

Page 16: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Inverting yields

Lineweaver-Burk Plot

slope = KM/Vmax

1/Vmax

1/S

1/-rS

16

Page 17: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Types of Enzyme Inhibition

Competitive

Uncompetitive

Non-competitive

17

Page 18: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

18

Competitive Inhibition

Page 19: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

1) Mechanisms:

Competitive Inhibition

19

Page 20: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

2) Rates:

Competitive Inhibition

20

Page 21: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Competitive Inhibition

21

Page 22: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

From before (no competition):

Intercept does not change, slope increases as inhibitor concentration increases

Competitive Inhibition

No Inhibition

Competitive Increasing CI

22

Competitive

Page 23: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

23

Page 24: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Uncompetitive Inhibition Inhibition only has affinity for enzyme-substrate complex

Developing the rate law

(1)

(2) 24

Page 25: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Adding (1) and (2)

From (2)

25

Page 26: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Total enzyme

26

Page 27: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Slope remains the same but intercept changes as inhibitor concentration is increased

Lineweaver-Burk Plot for uncompetitive inhibition 27

1−rS

=1

Vmax S( )KM + S( ) 1+

I( )KI

⎝ ⎜

⎠ ⎟

⎝ ⎜

⎠ ⎟

1−rS

=KM

Vmax1S( )

⎝ ⎜

⎠ ⎟ +

1Vmax

1+I( )KI

⎝ ⎜

⎠ ⎟

Page 28: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

28

Page 29: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

Both slope and intercept changes

Increasing I

No Inhibition

E + S E·S P + E

(inactive)I.E + S I.E.S (inactive) +I +I -I -I

Noncompetitive Inhibition (Mixed)

29

Page 30: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

30

Page 31: Lecture 15 - University of Michiganessen/html/powerpoints/lecture... · Lecture 15 . Today’s lecture ... 14 (Michaelis-Menten plot) V max-r s S 1/2 Solving: K M =S 1/2 therefore

End of Lecture 15

31