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SURVEY OF BIOCHEMISTRY

Enzyme Kinetics and Inhibition

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Rates of Chemical Reactions

Enzyme kinetics is the study of rates of reactions catalyzedby enyzmes.

v = A Pk

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v =−d[A]

dt=d[P]

dt

The rxn rate (velocity, v) can be described in several ways:[1] disappearance of reactant, A[2] appearance of product, P

These eqn’s relate velocity to concentration of reactants and products.

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Rate Laws

Enzyme kinetics is the study of rates of reactions catalyzedby enyzmes.

v = A Pk

€

v =−d[A]

dt=d[P]

dt

A rate law is an equation describing the velocity of achemical reaction.

Differential Rate LawsIntegrated Rate Laws

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Differential Rate Laws

• Differential rate laws correspond to order of the reaction.

Order of Reaction Rate Law0 v = k

1 v = k [A]

2 v = k [A]2 or k[B]2 orv = k [A] x [B]

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Integrated Rate Laws:First Order

• Integrated rate laws express the rxn velocity in terms of time.

Rate of Disappearance

of A

- d[A]

dt= k [A] first order rxn

Rearranging…d[A]

[A]= - k dt

A products

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Integrated Rate Laws:First Order

Rearranging…d[A]

[A]= - k dt

Integrate onboth sides

of eqn:

€

∫[A]0

[A]t

€

1

[A]dA =

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∫- k dt

(ln [A]t + constant) - (ln [A]0 + constant) = - kt

ln [A]t - ln [A]0 = - kt

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Integrated Rate Laws:First Order

ln [A]t - ln [A]0 = - kt ln [A] = ln [A]0 - kt

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Integrated Rate Law:Other Versions of First Order

ln [A]t - ln [A]0 = - kt

Rearranging: ln [A]t

[A]0

= - kt

Take exponentof both sides:

[A]t

[A]0

= e -kt

[A]t = [A]0 e -kt first order rxn

first order rxn

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Integrated Rate Law: Second Order

• How does the integrated rate law change if the order of the reaction is second order?

2A products

Rate of Disappearance

of A

- d[A]

dt= k [A]2 second order rxn

Rearranging…d[A]

[A]2= - k dt

Show result on

board

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Michaelis-Menten Equation

Many enzymes obey Michaelis-Menten kinetics behavior:

E + S ES E + Pk1

k-1

k2

Rate limiting step

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v =d[P]

dt= k 2[ES]

Problem:[ES] is difficult to measure!

What can we do?

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Michaelis-Menten Equation

E + S ES E + Pk1

k-1

k2

I. Assume equilibrium is maintained in 1st stepII. Assume “steady state”

k1 [E] [S] - k-1 [ES] - k2 [ES] = 0

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d[ES]

dt=

Depletion of ESFormation of ES

See notes on board…

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v =d[P]

dt= k 2[ES]Recall

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Michaelis-Menten Kinetics

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v =V max[S]

Km +[S]

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Lineweaver-Burk Plot

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Enzyme Inhibition

• What is an inhibitor?

• Modes of Inhibition– Competitive

binds to same site in E as S– Uncompetitive– Noncompetitive– Mixed

bind to different site in E than S

Note: Text does not distinguish “non” and “mixed”

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Competitive Inhibition

• Competitive inhibitors bind to the same site on E as S

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Competitive Inhibition

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Competitive Inhibition

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Uncompetitive Inhibition

Uncompetitive inhibitors bind directly to the ES complex but not to the

free enzyme

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Uncompetitive Inhibition

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Mixed Inhibition

Mixed inhibitors can bind to E or ES complex

S cannot bind if I is already bound!

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Mixed Inhibition

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Noncompetitive Inhibition

Noncompetitive inhibitors can bind to E

or ES complex

S can bind even if I is already bound!

+ I

See board for plot

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Updates and Reminders

• Exam #2 in two weeks (June 26)– Chapter 7: Protein Function– Chapter 11: Enzyme Catalysis– Chapter 12: Kinetics & Inhibition– Chapter 8: Carbohydrates– Chapter 14: Introduction to Metabolism

• Suggested HW problems online this weekend

• Resources: What You Should Know more coming soon