Enzymes and Enzyme Kinetics - CMBE

I. Immobilization of Enzymes1. MotivationConsider the following two designs for acontinuous enzyme process S PS, ESImmobilized ES, P, ES, PProcess IProcess II

Advantages of Process II• Enzyme is not lost after process and mighttherefore be used for greater duration.• Enzyme does not need to be separated fromproducts. (Conversely, products do not needto be separated from enzyme.)• Higher enzyme concentrations are attainablewithout sacrificing these advantages.• Some enzymes are more stable if they areimmobilized. (Many native enzymes are“immobilized” in the cell.)

Disadvantages of Process II• An additional process (and cost) isassociated with enzyme immobilization.• Enzymes may leak from immobilized state.• Diffusional limitations. Substrate(s) andproduct(s) must transfer across a boundaryto get to/from active site.• Can be difficult to control environmentimmediately affecting enzyme and its activity.• Some enzymes are less stable if they areimmobilized. (Structural change)

How to reduce structural changes and preventbinding at activity site:EESurfaceSurfacea. Mix enzyme with competitive inhibitorE+I

. Now immobilizeSurfacec. Wash the competitive inhibitor awaySurface

2. External Mass Transfer Effectsa. General Derivation• Enzymes are immobilized on surface ofuncharged, nonporous flat plate.• Entire surface is uniformly accessible tosubstrate in adjacent fluid.Consider S P as immobilized enzyme reactionCase I – Enzyme reaction is very slowSurface concentration (S SURF ) isidentical to bulk concentration (S BULK )Case II – Enzyme reaction is very fastSurface concentration (S SURF ) is zero

Case I – Enzyme reaction is very slowS BULKS0Distance from surfaceCase II – Enzyme reaction is very fastS BULKS0Distance from surface

Case I – Enzyme reaction is very slowRate of reaction is limited by enzyme and itsintrinsic reaction rateCase II – Enzyme reaction is very fastRate of reaction is limited by rate of mass transferAt steady-state, substrate and product will notaccumulate at the surface, and the rate of reactionis equal to the rate of mass transfer…rate of reaction = rate of mass transfer___________V MAX ′S SURFK M + S SURF= k L (S BULK –S SURF )

CommentsV___________MAX ′S SURF= k L (S BULK –S SURF )K M + S SURFThis equation applies to Michaelis-Mentenkinetics. If reaction has different kinetics, usedifferent expression for “rate of reaction”.Maximum surface rate of reaction = V MAX ′(occurs when S SURF is large)V MAX ′ E SURFMaximum rate of mass transfer = k L S BULK(occurs when S SURF = 0)

For convenience define a dimensionless groupDamkohler NumberDa =_____________________Max Rate of ReactionMax Rate of Mass TransferDa =V MAX ′______k L S BULKIf Da >> 1 system is limited by mass transfer“mass transfer limited regime”If Da ~ 1 mass transfer and reaction rate are ofsimilar magnitudeIf Da

___________V MAX ′S SURF= k L (S BULK –S SURF )K M + S SURFUnknown: S SURFParameters: S BULK , k L , V MAX ′, K MIntroduce dimensionless parameters:x =______S SURFS BULK =______K MS BULK(kappa) in many books;others define = S BULK /K M

___________V MAX ′S SURF= k L (S BULK –S SURF )K M + S SURFxS BULK = S SURFS BULK = K MThen Reaction Rate =_____________V MAX ′xS BULKS BULK + xS BULK_______V MAX ′x + xThen Mass Transfer Rate =k L (S BULK –xS BULK )k L S BULK (1 – x)

So…._______V MAX ′x + x= k L S BULK (1 – x)Remember….Da =V MAX ′______k L S BULKThus…._______x + x=(1 – x)_______DaWrite as a quadratic….x 2 + Dax + x – x – = 0x 2 + x – = 0Where…. = Da + -1

x 2 + x – = 0Solve Quadratic…._______√x = -1 ± 1 + 4/2)(Select sign so that x ≥ 0Note that when = 0x =__√

. Mass Transfer-Limited Regime_______√x = -1 ± 1 + 4/2)( = Da + -1x =______S SURFS BULK =______K MS BULKDa =V MAX ′______k L S BULKAs V MAX ′ ∞Da ∞ ∞ x 0S SURF 0

As V MAX ′ ∞ S SURF 0The faster the enzyme, the lower thesurface substrate concentration. If theenzyme is very fast, then…Reaction Rate = k L (S BULK –S SURF )Reaction Rate = k L S BULK0Vk LS BULK

Implications of Reaction Rate = k L S BULK(Mass Transfer Limited Regime)As long as reaction is fast enough…• Reaction rate is independent of V MAX ′• Reaction rate is independent of K M• Reaction rate is independent of presence ofinhibitors which affect K M and V MAX ′• Reaction rate is independent of (suboptimal)temperature and pH, which affect K M and V MAX ′

Mass transfer disguises the true kineticbehavior of enzymes.Studies aimed at determiningactivity/denaturation in immobilisedsystems should be conducted in reactionlimitedregime.Experimentally, to operate in reactionlimitedregime requires high fluid flow ratesto minimize mass transfer resistance. Highfluid flow rates can be problematic for shearsensitive enzymes.May be difficult to achieve reaction-limitedregime for very fast enzymes.

c. Reaction-Limited Regime_______x + x=(1 – x)_______DaAs V MAX ′ 0 Da 0 x 1 S SURF S BULKReaction Rate =Reaction Rate (x = 1) =_______V MAX ′x + x_______V MAX ′ + 1

d. External Effectiveness FactorAs noted previously, one way to determine theregime is by the Damkohler number.Another way is by the External EffectivenessFactor….External Effectiveness Factor = E =_______________________________________Observed Reaction RateReaction Rate without Mass Transfer Limitation

_______________________________________Observed Reaction RateReaction Rate without Mass Transfer Limitations=_______V MAX ′x + x_______V MAX ′ + 1______________ E =__________x ( + 1 )( + x )

E =__________x ( + 1 )( + x ) E is a measure of the influence ofexternal mass transfer resistance onthe observed reaction rate.if E

Example Problem 3.3 (see pp. 85-86)V MAX ′ = (6 × 10 -6 mol/s·mg)(1 × 10 -4 mg/cm 2 )= 6 × 10 -10 mol/cm 2·s= 6 × 10 -4 mol/cm 2·sK M = 2 × 10 -3 mol/L = 2 mol/cm 3k L = 4.3 × 10 -5 cm/sFind S SURF , V and E when …..

A) S BULK = 7 mol/cm 3 (3.5 × K M )Da = =V MAX′______k L S BULKK______ MS BULK=_____________(6 × 10 -4 )= 1.993(4.3 × 10 -5 )(7)=____(2)(7)= 0.2857 = Da + – 1 = 1.279_________ x = (-1 ± √1 + 4/2 ) = 0.1940S SURF = xS BULK = (0.1940)(7.0) = 1.36 mol/cm 3V TRUE =V_______ MAX ′x + x=________________(6 × 10 -4 )(0.1940)(0.2857 + 0.1940)= 2.43 × 10 -4 mol/cm 2 s________ x ( + 1) E = = + x________________(0.1940)(1.2857)(0.2857 + 0.1940)= 0.52

B) S BULK = 14 mol/cm 3Da = =V MAX′______k L S BULKK______ MS BULK=_____________(6 × 10 -4 )= 0.997(4.3 × 10 -5 ) (14)=____(2)(14)= 0.1429 = Da + – 1 = 0.1395_________ x = (-1 ± √1 + 4/2 ) = 0.3146S SURF = xS BULK = (0.3146)(14.0) = 4.40 mol/cm 3V TRUE =V_______ MAX ′x + x=________________(6 × 10 -4 )(0.3146)(0.1429 + 0.3146)= 4.13 × 10 -4 mol/cm 2 s________ x ( + 1) E = = + x________________(0.3146)(1.1429)(0.1429 + 0.3146)= 0.79

C) V MAX ′ = 12 × 10 -4 mol/cm 2 s (S BULK = 7 mmol/cm 3 )Da = =V MAX′______k L S BULKK______ MS BULK=_____________(12 × 10 -4 )= 3.986(4.3 × 10 -5 )(7)=____(2)(7)= 0.2857 = Da + – 1 = 3.272_________ x = (-1 ± √1 + 4/2 ) = 0.0851S SURF = xS BULK = (0.0851)(7.0) = 0.60 mol/cm 3V TRUE =V_______ MAX ′x + x=________________(12 × 10 -4 )(0.0851)(0.2857 + 0.0851)= 2.75 × 10 -4 mol/cm 2 s________ x ( + 1) E = = + x________________(0.0851)(1.2857)(0.2857 + 0.0851)= 0.30

D) S BULK = 7 mol/cm 3 but k L is 8.6 × 10 -5 cm/sDa =V MAX′______k L S BULK=_____________(6 × 10 -4 )= 0.997(8.6 × 10 -5 )(7) = 0.2857 = 0.2824x = 0.4116S SURF = xS BULK = (0.4116)(7.0) = 2.88 mol/cm 3V TRUE =V_______ MAX ′x + x=________________(6 × 10 -4 )(0.4116)(0.2857 + 0.4116)= 3.54 × 10 -4 mol/cm 2 s________ x ( + 1) E = = + x________________(0.4116)(1.2857)(0.2857 + 0.4115)= 0.76

E) V MAX ′ = 24 × 10 -4 mol/cm 2 s (S BULK = 7 mmol/cm 3 )Da =V MAX′______k L S BULK=_____________(24 × 10 -4 )= 7.97(4.3 × 10 -5 )(7) = 0.2857 = 7.2591x = 0.0391S SURF = 0.27 mol/cm 3V TRUE= 2.89 × 10 -4 mol/cm 2 s E = 0.15

Summary (units and exponents left out)S BULK V MAX ′ k L S SURF V TRUE EABCDE7 6 0.43 1.36 2.43 0.5214 6 0.43 4.40 4.13 0.797 12 0.43 0.60 2.75 0.307 6 0.86 2.88 3.54 0.767 24 0.43 0.27 2.89 0.15Compared to example A, increasing the enzyme level oractivity is not effective because system is already inmass transfer limited regime. (Da>1)