TEKNIK REAKSIOLEH :

AHMADI FACHRYM.ADLAN ADAM

HAFIZH MAULANA

to the binding energy of the substrate to the enzyme through multiple bonds with the specific

functional groups on the enzyme (amino side chains, metal ions). The interactions that stabilize

the enzyme-substrate complex are hydrogen bonding and hydrophobic, ionic, and London van

der Waals forces. If the enzyme is exposed to extreme temperatures of pH environment (i.e..

both high and low pH values), it may unfold losing its active sites. When this occurs, the enzyme

is said to be denatured. See Problems P7-15B.

There are two models for substrate-enzyme interactions: the lock anda key model and the

induced fit model, both of which are shown in Figure 7-6. For many years the lock and the lock

and key model was preferred because of the stereospecific effects of one enzyme acting on one

substrate. Howefer, the induced fit model is the more useful model. In the induced fit model both

the anzyme molecule a d substrate molecules are distorted. These changes in conformation

distort one or more of substrate bonds, thereby stressing and weakening the bond to make the

molecule more susceptible to rearrangement or attachment.

GAMBAR (a-b)

GAMBAR (substrate (o-glucose))

There are six classes of enzymes and only six:

1. Oxidoreductases AH2 + B + E A + BH2 +E

2. Transferases AB + C + E AC + B + E

3. Hydrolases AB + H2O + E AH + BOH + E

4. Isomerases A + E isoA + E

5. Lyases AB + E A + B + E

6. Ligases A + B +E AB + E

More information about enzyme can be found on the following two web sites:

http://us.expasy.org/enzyme/and www.chem.qmw:ac:uk/iubmb/enzyme

These sites also give information about enzymatic reactions in general.

7.2.2 Mechanisms

In developing some of the elementary principles of the kinetics of enzyme reactions, we shall

discuss an enzymatic reactions that has been suggested by Levine and LaCourse as part of a

system that would reduce the size of an artificial kidney. The desired result is the production of

an artificial kidney that could be worn by the patient and would incorporate a replaceable unit for

the elimination of the nitrogenous waste produts such as uric acid and creatinine. In the

microencapsulation scheme proposed by Levine and LaCourse, the enzyme urease would be

used in the removal of urea from the bloodstream. Here, the catalytic action of urease would

cause urea to decompose into ammonia and carbon dioxide. The mechanism of the reaction is

believed to procced by the following sequence of elementary reactions :

1. The enzyme urease (E) reacts with the substrate urea (S) to form an enzyme-

substrate complex (E•S).

NH2CONH2 + Urease [NH2CONH2 •Urease] (7-13)

2. This complex (E•S) can decompese back to urea (S) and urease (E):

[NH2CONH2 • Urease] Urease + NH2CONH2 (7-14)

3. Or it can react with water (W) to give the products (P) ammonia and carbon

dioxide, and recover the enzyme urease (E)

[NH2CONH2 • Urease] + H2O 2NH3 + CO2 + Urease (7-15)

We see that some of the enzyme added to the solution binds to the urea, and some remains

unbound. Although we can easily measure that total concentration of enzyme, (E), it is difficult

to measure the conversation of free enzyme, (E).

Letting E, S, W, E•S, and P represent the enzyme, substrate, water, the enzyme-substrate

complex, and the reaction products, respectively, we can write reactions (7-13). (7-14), and (7-

15) symbolically in the forms

S+E E•S (7-16)

E•S E+S (7-17)

E•S + W P+E (7-18)

Here P = 2NH3 + CO2

The corresponding rate laws for Reaction (7-16), (7-17), and (7-18) are

r1s= -k1 (E)(S) (7-16A)

r2s= -k2 (E•S) (7-17A)

r3P= -k3 (E•S)(W) (7-18A)

The net rate of disappearance of substrate, -rs. is

r1s= -k1 (E)(S)- k2 (E•S) (7-19)

This rate law is of not much use to us in making reaction engineering calculations

because we cannot measure the concentration of enzyme substrate complex (E•S) in terms of

measured variables. The net rate of formation of the enzyme-substrate complex is

rE•S = -k1 (E)(S)- k2 (E•S) - k3 (W) (E•S) (7-20)

Using the PPSH. rE•S=0, we solve equation (7-20) for (E•S)

(E•S) = K1 ( E )(S )

K 2+ K3(W )(7-21)

and substitute for (E•S) into [Equation (7-19)]

-rs= k1 (E)(S) - k2 K1 ( E )(S )

K2+ K3(W )

Simplifying

-rs= K1+K 3 ( E )(S)(W )

K2+K 3(W ) (7-22)

We still cannot use this rate law because we cannot measure the unbound enzyme concentration (E): however, we can measure the total enzymw concentration. Et

In the absence of enzyme denaturization , the total concentration of the enzyme in the system, (Et), is constant and equal to the sum of the concentrations of the free or unbounded enzyme, (E), and the enzyme-substrate complex (E•S):

(Et) = (E) + (E•S) (7-23)

Substituting for (E•S)

(Et)= (E) + K1 ( E )(S )

K2+ K3(W )

Solving for (E)

(E) = ( Et )(k¿¿1+k3(W ))

k2+k3 (W )+k1(S)¿

substituting for (E) in equation (7-22), the rate law for substrate consumption is

(7-24)-rs = K 1+K 3 (W )(Et)(S)k1(S )+k2+k3(W )

Note : Throughout, Et = (Et) = total concentration of enzyme with typical units (kmol/m3 or g/dm3).

7.2.3 Michaelis-Menten Equation

Because the reaction of urea is carried out in aquaeous solution. water is, of coarse, in excess, and the concentration of water is therefore considered constant. Let

kcat = k3 (W) and KM = K cat+k2

k1

Dividing the numerator and denominator of Equation (7-24) by k1, we obtain a form of the Michaelis-Menten equation :

-rs = k cat ( E t )(S)

( S )+K m

(7-25)

The parameter kcat is also referred to as theturnover number. It is the number of substrate molecules converted to product in a given time on a single-enzyme molecule when the enzyme is saturated with substrate . For example, turnover number for decomposition H2O2 by the enzyme catalase is 40 x 106 s-i. That is. 40 million molecules of H2O2 are decomposed every second on a single-enzyme molecule saturated with H2O2. The constant KM (mol/dm3) is called the Michaelis constant and for simple systems is measure of the attraction of enzyme for its substrate, so it's also called the affinity constant. The Michaelis constant, KM. For the decomposition of H2O discussed earlier is 1.1 M while that for chymotrypsin is 0.1 M.

If, in addition, we let Vmax represent the maximum rate of reaction for a given total enzyme concentration.

Vmax= Kcat (Et)

The michaelis-Menten equation takes the familiar form

(7-26)-rs = V max (S )Km+(S)

For a given enzyme concentration, a sketch of the rate of disappearance of the substrate is shown as a function of the substrate concentration in Figure 7-7.

Grafik Figure 7-6 Michaelis-Menten plot identifying the parameters Vmax and Km

A plot of this type is sometimes called a Michaelis-Menten plot. At low substrate concentration. KM > (S)

-rs = V max(S)

Km

and the reaction is apparent first order in the substrate concentration, at high substrate concentrations,

(S) > Vmax

Consider the case when the substrate concentration is such that the reaction rate is equal to one-half the maximum rate.

rs=V max

2

then

V max

2 = V max (S 1

2)

Km+(S12

) (7-27)

Solving equation (7-27) for the Michaelis constant yields

(7-28)

The Michaelis constant is equal to the substrate concentration at which the rate of reaction is equal to one half the maximum rate.

The parameters Vmax and Km characterize the enzymatic reactions tht are described by Michaelis-Menten kinetics. Vmax is dependent on total enzyme concentration, whereas KM is not.

Two enzymes may have the same values for kcat but have different reaction rates because of different values of KM. One way to compare the ratio kcat/KM. When this ratio approaches 108

to 109 (dm3/mol/s) the reaction rate approaches becoming diffusion-limited. That is, it takes a long time for the enzyme and substrate to find each other. But once they do they reach immediately. We will discuss diffusion-limited reactions in Chapters 11 and 12.

Example 7-3

l

−rurea =

lV max

+ KM

V max ( lCurea ) (E7-3.2)

A plot of the reciprocal reaction rate versus the reciprocal urea concentration should be a straight line with an intercept 1/Vmax and slope KM/Vmax. This type of plot is called a Lineweaver-Bark plot. The data in Table E7-3.1 are presented in Figure E7-3.1 in the form of a Lineweaver-Bark plot. The Intercept is 0.75, s0

lV max

= 0.75

m3.s/kmol

TABLE E7-3.1 RAW AND PROCESSED DATA

KM = (S1/2)

Curea

(kmol/m3)-rurea

(kmol/m3..s)1/Curea

(m3/kmol)1/-rurea

(m3..s/ kmol)0.20 1.08 5.0 0.930.02 0.55 50.0 1.82

0.01 0.38 100.0 2.63

0.005 0.20 200.0 5.00

0.002 0.09 500.0 11.11

Figure E7-3.1 (a)Michaelis-Menten plot; (b)lineweaver-Burk plot

Therefore, the maximum rate of reaction is

Vmax= 1.33 kmol/m3.s = 1.33 mol/dm3.s

From the slope, which is 0.02 s we can calculate the Michaelis constant. KM:

KM

V max = slope = 0.02 s

KM = 0.0266 kmol/m3

Subtituting Km and Vmax into Equation (7-26) give us

(E7-3.3)-r = 1.33 Curea

0.0266+Curea

where Curea has units of kmol/m3 and -rs has units of kmol/m3.s. Levine and LaCourse suggest that the total concentration of urease, (Et), corresponding to the value of Vmax above is approximately 5 g/dm3.

In addition to the Lineweaver-Burk plot, one can also use a Hanes-Woolf plot or and Eadie-Hofstee plot. Here S ≡ Curea and -rs ≡ -rurea equation (7-26)

-rs = V max(S)KM+(S ) (7-26)

Can be rearranged in the following form. For the Eadie-Hofstee form,

(E7-3.4)

For the Hanes-Wolf form. we have

(E7-3.5)

For the Eadie-Hofstee model we plot -rs as a function of (-rs/S) and for the Hanes-Woolf model, we plot [(S)/-rs] as a function of (S). The Eadie-Hofstee plot does not bias the points at low substrate concentrations, while the Hanes-Woolf plot gives a more accurate evaluation of Vmax in Table E7-3.2, we add two columns to Table E7-3.1 to generate these plots (Curea≡ S¿

Table E7-3.2 RAW AND PROCESSED DATA

S(kmol/m3)

-rs

(kmol/m3•s)1/S

(m3/kmol)1/-rs

(m3•s/kmol)Sl-rs

(s)-rs/S(1/s)

0.20 1.08 5.0 0.93 0.185 5.40.02 0.55 50.0 1.82 0.0364 27.50.01 0.38 100.0 2.63 0.0263 380.005 0.20 200.0 5.00 0.0250 400.002 0.09 500.0 11.11 0.0222 45

Plotting the data in Table E7-3.2, we arrive at figures E7-3.2 and E7-3.3.

Figure E7-3.2 Hanes-Woolf plot Figure E7-3.3 Eadie-Holfstee plot

-rs = Vmax - KM (−r s

(S) )

(S )−rs

=KM

V max+ 1

V max(S )

Regression

Equation (7-26) was used in the regression program of Polymath with the following results for Vmax and Km

Nonlinear regression (L-M)

Model : rate = Vmax*Curea/(Km+Curea)

Variable Ini guess Value 95% confidence

Vmax 1 1.2057502 0.0598303 Km 0.02 0.0233322 0.003295

Nonlinear regression settings Vmax=1.2 mol/dm3•s Max # iterations = 64 KM =0.003295 mol/dm3

PrecisionR^2 = 0.9990611R^2adj = 0.9987481Rmsd = 0.0047604Variance = 1.888E-04

The product-Enzyme Complex

In many reaction the enzyme and product complex (E•P) is formed directly from the enzyme substrate complex (E•P) according to the sequence

E+S E•S P+E

Applying the PPSH, we obtain

−r s=V max(C s−C p KC)CS+K max+K PCP

(7-29)

which is often referred to as the Briggs-Haldane Equation (see Problem P7-10) and the application of the PPSH to enzyme kinetics often called the Briggs-Haldane approximations

7.2.4 Batch Reactor Calculations for Enzyme Reactions

A mole balance on urea in the bach reactor gives

−dNurea

dt=−rurea V

Because this reactions is liquid phase, the mole balance can be put in the following form:

−dN urea

dt=−rurea (7-30)

The rate low for urea decomposition is

−rurea=V max Curea

KM+Curea (7-31)

Substituting Equation (7-31) into Equation (7-30) and then rearranging and integrating. We get

t= ∫Curea

Curea 0 dC urea

−rurea= ∫

Curea

Curea 0 K M+Curea

V max CureadC urea

t=KM

V maxln

Curea0

C urea+

Curea0−Curea

V max (7-32)

We can write Equation (7-32) in terms of conversion as

Curea=Curea0(1−x)

t=KM

V maxln 1

1−X+

Curea0 XV max

(7-32)

The parameters KM and Vmax can readily be determined from batch reactor data by using the integral method of analysis. Dividing both sides of Equation (7-32) by tKm/Vmax and rearranging yields

1t

ln 11−x

=¿V max

K M−

Curea0 XK M T

¿

We see than KM and Vmax be determined from the slope and intercept of a plot of 1/t ln [1/t(1-x)] versus X/t. We could also express the Michaelis – Menten equation in terms of the substrate concentration S :

1t

lnS0

S=¿

V max

K M−

S0−SK M T

¿ (7-33)

Where S0 is the initial concentration of substrate. In cases similar to Equation (7-33) where there is no possibility of confusion. We shall not bother to enclose the substrate or other species in parentheses to represent concentration [i.e..Cs ≡ (S) ≡ S ]. The corresponding plot in terms of substrate concentration is shown in Figure 7-8.

Figure 7-7 Evaluating Vmax and KM.

Example 7-4 Batch Encymatic Reactors

Calculate the time needed to convert 99% of the urea to ammonia and carbon dioxide in a 0.5dm3

bacth reactor. The initial concentration of urea is 0.1 mol/dm3, and the urease concentration is

0.001 g/dm3. The reaction is to be carried out isothermally at the same temperature at which the data in Table E7-3.2 were obtained.

Solution

We can use Equation (7-32).

t=KM

V maxln 1

1−X+

Curea0 XV max

(7-32)

Where KM = 0.0266 mol/dm3. X = 0.99 and Curea 0 = 0.1 mol/dm3. Vmax was 1.33 mol/dm3-s. However, for the conditions in the batch reactor, the enzyme concentration is only 0.001 g/dm3

compared with 5 g in Example 7-3. Because Vmax = E1.k3. , Vmax for the second enzyme concentration is

V max 2=E t 2

E t 1V max 1=

0.0015

x 1.33=2.66 x 10−4 mol /s−¿ dm3¿

Km = 0.0266 mol/dm3 and X = 0.99

Subtituting into Equation (7-32)

t= 2.66 x 10−2 mol/dm3

2.66 x10−4 mol dm3/sln( 1

0.01 ¿)+(0.1 mol

dm3 )(0.99)

2.66 x10−4 mol dm3/ss ¿

= 460s + 380s

= 840s (14 minutes)

Effect of Temperature

The effect of temeprature on enzymatic reactions is very complex. If the enzyme structure would remain unchanged as the temperature is increased, the rate would probably follow the Arrhenius temperature dependence. However, as the temperature increases, the enzyme can unfold and/or become denatured and lose it catalytic activity. Consequently, as the temperature increases, the reaction rate, -rs, increases up to a maximum with increasing temperature and then decreases as the temperature is increased further. The descending part of this curve is called temeprature inactivation or thermal denaturizing.10 Figure 7-8 shows an example of this optimum in enzyme activity.11

Figure 7-8 Catalytic breakdown rate of H2O2 depending on temperature. Courtesy of S. Aiba, A.E.. Humphrey and N.F Mills. Biochemical Engineering.Academic Press (1973)

10M.L..Shuller and F.Kargi. Bioprocess Engineering Basic Concepts. 2nded. (Upper Saddle River. N.J.: Prentice Hall. 2002). P. 77

11S. Aiba. A.E. Humphrey. and N.F. Mills. Biochemical Engineering. (New York: Academic Press. 1973). P. 47

Side note : Lab-on-a-chip. Enzyme-catalyzed polymerization of nucleotides is a key step in DNA identification. The microfluidic device shown in Fig-ure SN7.1 is used to identify DNA strands. It was developed by Professor Mark Burns’s group at the University of Michigan.

Figure SN7.1 Microfluidic device to identify DNA. Courtesy of Science, 282, 484 (1998)

In order to identify DNA, its concentration must be raised to a level that can be easily quantified. This increase is typically accomplished by replicating the DNA in the following manner. After a biological sanple (e.g., purified saliva, blood) is injected ito the micro device. It is heated and the hydrogen bonds connecting the DNA strands are broken. After breaking, a primer attaches to the DNA to form a DNA primer complex. DNA*. An enzyme E then attaches to this pair forming the DNA* enzyme complex, DNA*∙ E. Once this complex is formed a polymerization reaction occurs as nucleotides (dNTPs-dATP, dGTP, dCTP, and dTTP-N) attach to the primer one molecule at a time as shown in Figure SN7.2. the enzyme interacts with the DNA strand to add the proper nucleotide in the proper order. The addition continues as the enzyme moves down the strand attaching the nucleotides until the order end of DNA strand is reached. At this point the enzyme drops off the strand and duplicate. Double-stranded DNA molecule is formed. The reaction sequence is

Figure SN7.2 Replication Sequence

The schematic in Figure SN7.2 can be written in terms of singles-step reaction where N is one of the four nucleotides.

Complex Formation :

DNA + Primer → DNA*

DNA* + E DNA * ∙ E

Nucleotide addition/polymerization

DNA * ∙ E + N → DNA * ∙ N1∙ E

DNA * ∙ N1∙ E+ N → DNA * ∙ N2∙ E

The process then continues much like a zipper as the enzyme moves along the strand to add more nucleotides to extend the primer. The addition of the last nucleotide is

DNA * ∙ Ni-1∙ E+ N → DNA * ∙ Ni∙ E

Where I is the number of nucleotide molecules on the originalDNA minus the nucleotides in primer. Once a complete double-stranded DA is formed, the polymerization stops, the enzme drops off, and separation occurs,

DNA * ∙ N1∙ E → 2DNA + E

Here 2DNA strands really represents one double-stranded DNA helix. Once replicated in the device, the length of DNA molecules can be analyzed by electrophoresis to indicate relevant genetic information.

7-3 Inhibition of Enzyme Reactions

In addition to temperature and solution pH, another factor that greatly influences the rates of enzyme-catalyzed reactions is presenceof an inhibitor. Inhibitors are species that interact with enzymes and render the enzymes ineffective to catalyze its specific reaction. The most dramatic consequence of enzyme inhibition are found in living organisms where the inhibition of any particular enzyme involved in a primary metabolic pathway will render the entire pathway inoperative, resulting in either serious damage or deathof the organisms. For example, the inhibition of single enzyme, cytochrome oxidase by cyanide will cause the aerobic oxidation process to stop : death occurs in a very few minutes. There are also beneficial inhibitor such as the one used in the treatment of leukimia and other neoplastic diseases.aspirin inhibits the enzyme that catalyzes the synthesis of prostaglandin involved in the pain-producing process.

The three most common types of reversible inhibition occuring in enzymatic reaction are competitive, uncompetitive, and noncompetitive. The enzyme molecule is analogous to a heterogeneous catalytic surface in that in contains active sites. When competitive inhibition occurs, the substrate and inhibitor are usually similar molecules that compete for the samesite on the enzyme. Uncompetitive inhibition occurs when the inhibitor deactivates the enzyme-substrate complex, sometimes by attaching it self to both the substrate and enzyme molecules of the complex. Noncompetitive inhibition occurs with the enzymes containing at least two different types f sites. The substrate attaches only to one type of site, and the inhibitor attaches only to the other to render the enzyme inactive.

7.3.1 Competitive Inhibition

Competitive inhibition is of particular importance in pharmacokinetics (drug therapy). If a patient were administered two or more drugs that react simultaneosly within the body with a

common enzyme, cofactor, or active species, this interaction could lead to competitive inhibition in the formation of the respective metabolites and produce serious consequence.

In competitive inhibition another substance, I. competes with the substrate for the enzyme molecules to form an inhibitor-enzyme complex, as shown here.

Reaction steps Competitive Inhibition Pathway

(1) E + S k1 E ∙ S

(2) E ∙ S k2 E + S

(3) E ∙ S k3 P + E

(4) 1 + E k4 E ∙ 1 (inactive)

(5) E ∙ 1 k5 E + 1

In addition to three Michaelis-Menten reaction steps. There are two additional steps as the inhibitor reversely ties up the enzyme as shown in reaction steps 4 and 5.

The rate law for the formation of product is the same [ef. Equation (7-18A)] as it was before in the absence of inhibitor.

r1 = k3 ( E ∙ S) (7-34)

Reaction Steps Uncompetitive Pathway

(1) E + S k1 E ∙ S

(2) E ∙ S k2 E + S

(3) E ∙ S k3 P + E

(4) 1 + E ∙ S k4 1 ∙ E ∙ S (inactive)

(5) 1 ∙ E + S k5 1 ∙ E ∙ S

Starting with equation for rate of formation of product. Equation (7-34), and then applying the pseudo-steady-state hypothesis to the intermediate (1 ∙ E ∙ S) . we arrive at the rate law for uncompetitive inhibition

−r s=r p=

V max(S)

K M+(S )(1+( I )K1

)where K1=

k3

k4 (7-40)

Rearranging

1

−rs=

1+ KM

(S ) V max+ 1

V max(1+

( I )K1

) (7-41)

The Lineweaver-Burk plot is shown in Figure 7-11 for different inhibitor concentrations. The slope (KM/Vmax) remains the same as the inhibition (I) concentration is increased, whilethe intercept (1+ (I)/ K1) increases.

Figure 7-11 Lineweaver – Burk plot for uncompetitive inhibition

In noncompetitive inhibition, also called mixed inhibition, the substrate and inhibitor molecules react with different types of sites on the enzyme molecule. Whenever the inhibitor is attached to the enzyme it is inactive and cannot form products. Consequently, the deactivating complex (1 ∙ E ∙ S) can be formed by two reversible reaction paths.

1. After a substrate molecule attaches to the enzyme molecule at the substrate site, the inhibitor molecule attaches to the enzyme at the inhibitor site.

2. After the inhibitor molecule attaches to the enzyme molecule at the inhibitor site, the substrate molecule attaches to the enzyme at the substrate site.

These paths, along with the formation of the product. P, are shown here. In noncompetitive inhibition, the enzyme can be be tied up in its inactive form either, before or after forming the enzyme substrate complex as shown in steps 2,3 and 4.

Reaction Steps Noncompetitive Pathway

(1) E + S k1 E ∙ S

(2) E + 1 k2 1 ∙ E (inactive)

(3) 1 + E ∙ S k3 1 ∙ E ∙ S (inactive)

(4) S + 1 ∙ E k4 1 ∙ E ∙ S (inactive)

(5) E ∙ S k5 P

Again starting with the rate law for the rate of formation of product and then applying the PSSH to the complexes (1 ∙ E) and (1 + E ∙ S) we arrive at the rate of the law for the noncompetitive inhibition.

−r s=

V max (S )

KM +( S )(1+( I )K1

) (7-42)

The derivation of the rate law is given in the Summary Notes on the web and CD-ROM. Equation (7-42) is in the form of the rate law that is given for an enzymatic reaction exhibiting noncompetitive inhibition. Heavy metal ions such as Pb2+, Ag+, and Hg2+ as well as inhibitors that react with the enzyme to form chemical derivatives, are typical examples of noncompetitive inhibitors.

Rearranging

1−rs

=1

V max (1+( I )K1 )+

1+K M

(S )V max(1+

( I )K1 ) (7-43)

Figure 7-12 Lineweaver-Burk plot for noncompetitive enzyme inhibition

For noncompetitive inhibition, we see in Figure 7-12 that both the slope [ K M

V max (1+( I )K1 )]

and intercept [ 1V max (1+

( I )K1 )] increase with increasing inhibitor concentration. In practice,

uncompetitive inhibition and mixed inhibition are observed anly for enzymes with two ormore substrate. S1 and S2.

The three types of inhibition are compared with a reaction in which no inhibitors are present on the Lineweaver-Burk plot shown in Figure 7-13

Figure 7-13 Lineweaver-Burk plots are three types of enzyme inhibition