1 / 76

Enzymes:

This article discusses how enzymes increase reaction rates, their specificity, regulation, and localization. It also explores the thermodynamics and kinetics of reactions, including activation energy and the effect of substrate concentration on enzyme activity.

mfullilove
Télécharger la présentation

Enzymes:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Enzymes: • increase the rates ofreactions • are highly specific for their preferred substrate • Can be regulated • can be localized in certain organelles • Can be organized into pathways.

  2. Rate Enhancement

  3. The Thermodynamics of a Reaction Reaction performed Under standard conditions: 25oC and 1 atmosphere air pressure Free Energy (G) Substrate (1 M) Progress of the reaction

  4. The Thermodynamics of a Reaction Reaction performed Under standard conditions: 25oC and 1 atmosphere air pressure Free Energy (G) Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  5. The Thermodynamics of a Reaction Reaction performed Under standard conditions: 25oC and 1 atmosphere air pressure Free Energy (G) A negative DG0 Makes the reaction Thermodynamically favourable Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  6. The Thermodynamics of a Reaction Reaction performed Under standard conditions: 25oC and 1 atmosphere air pressure Free Energy (G) The DGo is related to the Keq. The exact relationship is as follows: DGo = -RTlnKeq Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  7. The Kinetics of a Reaction Free Energy (G) Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  8. The Kinetics of a Reaction Activation Energy DG# Free Energy (G) Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  9. The Kinetics of a Reaction The enzyme lowers the activation energy DG# Free Energy (G) Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  10. The Kinetics of a Reaction The enzyme lowers the activation energy DG# DG# Free Energy (G) Substrate (1 M) DG0 Product (1 M) Progress of the reaction

  11. Recapping… • Enzymes CANNOT change the thermodynamics of a reaction: DGo or Keq • They CANNOT change the direction of a reaction or the position of the equilibrium. • They DO increase the rate of the reaction by lowering the activation energy.

  12. Reaction Rate Measurements • The rate of a reaction is measured as the #moles of product produced per unit time. • The most user friendly units are mmol/min. • The term ASSAY is used in Biochemistry to describe a reaction that measures something; enzyme activity or the concentration of a metabolite.

  13. Measuring the rate of a reaction [product] Time (min)

  14. Measuring the rate of a reaction The initial linear rate Is used for all enzyme kinetics measurements [product] dP/dt Time (min)

  15. Calculating Velocity • Using the Alkaline Phosphatase Experiment, suppose you had a change in Absorbance per min (DA/min) of 0.3….. • Step 1Convert to Dconcentration/min by dividing the DA/min by the extinction coefficient (e), say 15 mM-1cm-1 giving Dconc/min = 0.02 mM/min • Step 2Convert to #nmoles/min in the assay. The assay is 1 mL so 0.02 mM/min = 0.02 mmol/mL/min = 20 nmol/min/assay.

  16. The Effect of [substrate] on a simple first order reaction without an enzyme Slope = k, the rate constant Reaction rate First order means the reaction rate is dependent on the concentration of only one reactant. [substrate]

  17. The Effect of [substrate] on a simple first order reaction with an enzyme All the available enzyme is saturated with substrate Reaction rate [substrate]

  18. The Effect of [substrate] on a simple first order reaction with an enzyme Vmax Reaction rate The KM is the [S] at ½Vmax [substrate] KM

  19. The Effect of [substrate] on a simple first order reaction with an enzyme Vmax The KM describes the shape of the curve Reaction rate The KM is the [S] at ½Vmax [substrate] KM

  20. Vmax: What is the maximum speed the car can go at?

  21. KM: how much petrol do you need to travel at 60 kph? Maybe the little car is more efficient?

  22. The Lineweaver-Burk Plot: A double reciprocal plot used to find Vmax and KM 1/Vmax 1/v 1/[S] -1/KM

  23. The KM Vmax Two different isoenzymes with different KMs for the same substrate. Which has the higher affinity for the substrate? Reaction rate KM KM [substrate]

  24. The KM Vmax Higher affinity because it takes less substrate to attain Vmax. Reaction rate KM KM [substrate]

  25. The KM Vmax Higher affinity means a lower KM Reaction rate KM KM [substrate]

  26. The Progress of the Reaction in more detail. E + S ES EX# E + P EX# Free Energy (G) DG# S DG0 P + E ES Progress of the reaction

  27. The KM and the Vmax E + S ES E + P KM Kcat = Vmax/[E] Measures the affinity of the enzyme and substrate Measures how fast the reaction can go

  28. Rate: ES E + S KM = Rate: E + S ES The KM In most cases: OR Rate of dissociation KM = Rate of association

  29. The Significance of KM • The [S] which gives ½ Vmax • A measure of the affinity the enzyme has for the substrate • A low KM means high affinity and vice versa a high KM means low affinity • The KM is independent of the [E]

  30. What is Vmax? • Vmax is measured in Units (U). • 1 Unit (U) is the amount of enzyme required to release 1 mmole of product (P) in 1 minute under Vmax conditions. • You measure the rate of the reaction over a short time (min).

  31. Vmax and [E] The Vmax can be used practically to measure the amount of active enzyme in a sample e.g. serum. You will use this in the gene expression prac. Vmax [enzyme]

  32. kcat • The number of molecules of substrate converted to product by 1 molecule of enzyme in 1 second. • This is equivalent to the # mmoles of product produced per sec per mmole of enzyme OR the #nmoles P/sec/nmol E • Units are seconds-1

  33. Calculating kcat • Begin with the Vmax • In the Alkaline Phosphatase experiment the Vmax is usually around 30 nmol/min. • This works out to 0.5 nmoles P/sec (/60) • Now all we need to know is how many nmoles of Enzyme produced this rate. • To calculate this we need to know how much enzyme we added to the assay and the molecular weight of the enzyme.

  34. Calculating kcat • In the Alkaline Phosphatase experiment you added 20 L of enzyme solution containing 50 g/mL enzyme. This means we have 20*50 ng = 1 g. • So now our Vmax rate is: 0.5 nmoles P/sec/1 000 ng E

  35. Calculating kcat • The Vmax rate is: 0.5 nmoles P/sec/1 000 ng E • If the molecular weight of Alkaline Phosphatase is 100,000 then there is 1/100 (1 000/100 000) of a nmole of Enzyme in the assay.

  36. Calculating kcat • If 1/100 th of a nmole of Enzyme catalyses the formation of 0.5 nmoles of product in 1 second then…. • 1 nmole of enzyme will catalyse 0.5*100 = 50 nmoles of product in 1 second • The Kcat = 50 nmol sec-1nmol-1 • The Kcat = 50 sec-1

  37. The Steady State Assumption • Used to explain the shape of the hyperbola • A level of ES, [ES], is established very early in the reaction • This [ES] level is dependent on the [S] • This [ES] remains constant throughout the reaction.

  38. [conc] Time (min) The Steady State Assumption Product Substrate

  39. Total Enzyme added = [ES] + [Efree] [conc] [ES] [Enzymefree] Time (min) The Steady State Assumption

  40. The Michaelis Menten Plot Vmax Reaction rate The KM is the [S] at ½Vmax [substrate] KM

  41. Prac Results:Alkaline Phosphatase

  42. Prac Results:Alkaline Phosphatase

  43. The Michaelis Menten Relationship • The equation below describes the hyperbola. • Velocity = Vmax * [S] ([S] + Km)

  44. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship The steady state assumption means that the rate of formation of ES = the rate of breakdown of ES

  45. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship • Mathematically the steady state assumption means that the rate of formation of ES = the rate of breakdown of ES…. • Rate of formation = k1*[E]*[S] • Rate of breakdown = k-1*[ES] + k2*[ES] • Therefore: k1*[E]*[S] = k-1*[ES] + k2*[ES]

  46. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship k1*[E]*[S] = k-1*[ES] + k2*[ES] Simplifying k1*[E]*[S] = [ES] (k-1+ k2) Now KM = (k-1+ k2)/ k1 So [E]*[S] = [ES]* KM

  47. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship [E]*[S] = [ES]* KM Now the Total amount of Enzyme ET = Efree + ES E = Efree = ET – ES • [ET] – [ES] = [ES]* KM/[S] • [ET] = [ES]* KM/[S] + [ES]

  48. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship • [ET] = [ES]* KM/[S] + [ES] Now Vmax = [ET]*k2 and velocity (v) = [ES]*k2 So multiplying both sides by k2 [ET]* k2 = [ES]* k2(KM/[S] + 1) Vmax = v *(KM/[S] + 1)

  49. k2 k1 E + S ES E + P K-1 Deriving the Michaelis Menten Relationship Vmax = v *(KM+[S])/[S] ) Cross multiplying 1/v = (KM + [S]) Inverting V = Vmax * [S] (KM+[S]) Vmax*[S]

More Related