Which of the following would increase the rate of an enzyme-catalyzed reaction?

The following section shows the optimized conditions for the lipase-catalyzed transesterification of rac-4 with two diastereomers using Meito QLM, including the quantity of water, solvent, and concentration. The reaction improved markedly while maintaining high selectivity. When this method was adopted for other zerumbone derivatives, the stereoselective reaction also occurred efficiently.

We have previously studied the recognition behavior of the threo and erythro forms of 4 with a lipase.5 This showed that the lipase completely recognizes the stereochemistry not at the hydroxyl position but at the 2-methyl position. The desired (2R)-form of 4 was obtained in high yield by a lipase-catalyzed transesterification using a diastereomeric mixture of rac-413 as the starting material without a complicated procedure to separate the diastereomers.

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B9780444637437000147

Supplement 2

Thierry Lalot, ... Ernest Maréchal, in Comprehensive Polymer Science and Supplements, 1989

2.4 Thermodynamic Aspects

Most enzyme-catalyzed reactions are equilibria which must be shifted towards the formation of the desired products for good yields. This result can be achieved in several ways. Yamada et al.41 studied the lipase-catalyzed macrolactonization of hydroxy acid esters in anhydrous isooctane (Scheme 2). To shift the equilibrium the authors trapped the methanol produced with molecular sieves.

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Scheme 2.

Another example reports the use of an unstable low molar mass product (enol) which is eliminated in the form of the corresponding tautomer. This technique was used by Nicolosi et al.49 for acetylating phenols in the presence of lipase; the acyl donor is vinyl acetate and the evolving enol is eliminated as ethanal (Scheme 3).

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Scheme 3.

The selective dissolution (or precipitation) of substrates, reactants and products is an efficient technique.

The effect of organic solvents on the equilibrium position is particularly shown in the enzymatic estenfication of glycerol by decanoic acid, which leads to a mixture of three isomeric acylglycerols.74 The reaction is carried out in a biphasic mixture of water with a solvent insoluble (or sparingly soluble) in water. The yields of the three isomers can be connected to the hydrophobicity or to the solubility of the water in the organic phase; both scales are equivalent, with a few exceptions. The position of the equilibrium can be determined by controlling the hydrophilicity of the medium.

More recent work accounts for the shift of the equilibrium by the elimination of a product of the reaction by nitrogen bubbling63 or by using a high-boiling solvent under vacuum.60

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B978008096701100241X

A critical review of the 2005 literature preceded by two chapters on current heterocyclic topics

Benito Alcaide, Pedro Almendros, in Progress in Heterocyclic Chemistry, 2007

4.4 FUSED POLYCYCLIC β-LACTAMS

The enzyme-catalyzed reactions involved in formation of the bicyclic clavam and carbapenem nuclei, including β-amino acid and β-lactam formation, have been reviewed and compared with those involved in penicillin and cephalosporin biosynthesis <05CC4251>. The synthesis of deuterium labelled l- and d-glutamate semialdehydes, their evaluation as substrates for carboxymethylproline synthase, and their implications for carbapenem biosynthesis have been investigated <05CC1155>. The totally synthetic bicyclic β-lactam 38 related to salinosporamide A and omuralide has showed potent proteasome inhibition activity <05JA15386>. It has been reported that ceftriaxone 39, in addition to its classical antibacterial activity, offers neuroprotection by increasing glutamate transporter expression <05NAT73>. The crystal structure of the Stenotrophomonas maltophilia L1 enzyme in complex with the hydrolysis product of the 7α-methoxyoxacephem, moxalactam has been reported <05JA14439>. Ab initio quantum mechanical/molecular mechanical (QM/MM) calculations, augmented by extensive molecular dynamics simulations, describing the serine acylation mechanism for the class A TEM-1 β-lactamase with penicillanic acid as substrate have been reported <05JA15397>. The complete mechanism of acylation with benzylpenicillin, using a combined quantum mechanical and molecular mechanical (QM/MM) method (B3LYP/6-31G + (d)//AM1-CHARMM22) has been modelled <05JA4454>. Structure-activity correlations have been deduced in an investigation of the hydrolytic cleavage of penicillin G mediated by different dinuclear zinc complexes <05CEJ5343>. The mechanism of action of C6-(N1-methyl-1,2,3-triazolylmethylene)penem 40 toward class C β-lactamases has been investigated and the crystal structure of Enterobacter cloacae 908R β-lactamase complexed with 40 has been reported <05JA3262>. Hybrid QM/MM molecular dynamics simulation and density functional theoretical (DFT) calculations of the dizinc metallo-β-lactamase CcrA from Bacteroides fragilis in a complex with the cephalosporin nitrocefin showed that the substrate β-lactam is directed towards the active site dizinc center through the interactions of aminocarbonyl and carboxylate with the two active site zinc ions and two conserved residues, Lys167 and Asn176 <05JA4232> .

Which of the following would increase the rate of an enzyme-catalyzed reaction?

The 3-chloromethyl cephalosporin 41 has been used for the preparation of nitrocefin 42 via Finkelstein and Wittig reactions <05JOC367>. The bicyclic β-lactam 43, a simple C3 homologue of sulbactam, has been prepared and evaluated as an improved inhibitor of class C β-lactamases <05JOC4510> .

Which of the following would increase the rate of an enzyme-catalyzed reaction?

The acid-base properties of cephalosporins, such as cefalotin, cefazoline, and cefalexin, have been studied <05RJGC1513>. Vancomycin-cephalosporin hybrids have been achieved using chemoenzymatic strategies <05OL1513>. A practical synthesis for the large-scale production of the new carbapenem antibiotic ertapenem sodium has been developed <05JOC7479>. The new iminosugars 1-oxabicyclic β-lactam disaccharides have been synthesized as inhibitors of elongating α-d-mannosyl phosphate transferase <05OBC1043>. A new approach to the synthesis of tazobactam 44, which belongs to a class of penicillanic acid sulfones, has been achieved using an organosilver compound <05S442>. 6-Aminopenicillanates have been N-acylated via a domino addition–Wittig alkenylation sequence to give the corresponding 6-(E-2′-alkenoyl)amides 45 <05TL1127>. The intramolecular nitrone-alkene cycloaddition reaction using 2-azetidinone-tethered alkenylaldehydes as starting materials has been introduced as an efficient route to prepare bridged tricyclic β-lactams 46 <05EJO1680>. Bicyclic aza-β-lactam intermediates have been postulated on the ring expansion of pyroglutamates <05OL1117> .

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Different-sized fused bicyclic β-lactams of non-conventional structure such as 47, have been obtained from 4-oxoazetidine-2-carbaldehydes through a novel carbonyl–bromoallylation/Heck reaction sequence <05JOC2713>. Six-, seven-, or eight-membered bicyclic 2-azetidinones have been prepared through triphenyltin hydride-promoted intramolecular free radical cyclization of β-lactam-tethered bromodienes <05S2335>. The reaction of 4-acetoxy-2-azetidinones with organoindium reagents generates 4-(1-substituted allenyl)-2-azetidinone derivatives, which have been cyclized to the corresponding bicyclic β-lactams by Au(III) catalysis <05AG(E)1840>. o-Halogenophenyl- and o-halogenobenzyl-4-alkenyl-β-lactams have been used for the regio- and stereoselective preparation of benzofused tricyclic β-lactams 48, including benzocarbapenems and benzocarbacephems, via intramolecular aryl radical cyclization <05T2767>. 2-Azetidinone-tethered haloarenes have proved to be appropriate substrates for the synthesis of fused or not fused β-lactam-biaryl hybrids by aryl-aryl radical cyclization and/or rearrangement <05T7894>. The microwave-induced Staudinger reaction between ketenes generated from α-diazoketones and cyclic imines has been shown as a suitable method for the synthesis of polycyclic β-lactams <05JOC334>. Synthetic and computational studies on intramolecular [2 + 2] sulfonyl isocyanate-olefin cycloadditions to yield bicyclic β-lactam-sulfonamide hybrids has been documented <05T5615> .

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/S0959638007800080

Fundamentals of Biochemical Reaction Engineering

Shaofen LiProfessor, in Reaction Engineering, 2017

11.4.2.2 Continuous Stirred Tank Reactor

1.

Enzyme-catalyzed reaction. In a CSTR, if the reaction is controlled by enzyme catalysis and the kinetics equation follows the Michaelis–Menten equation, substituting the kinetics equation directly into Eq. (3.42) gives the residence time

(11.46)τ=VrQ0=(cs0−cs)(Km+cs)rmaxcs=(cs0−cs)(Km+cs)k2cE0cs

Substituting cs=cs0(1−Xs) into the above equation and simplifying gives

(11.47)τ=1k2cE0(cs0Xs+KmXs1−Xs)

When inhibitor is present, we need to substitute the corresponding kinetics equation into Eq. (3.42).

2.

Microorganism reaction. In a CSTR, assuming there is no biomass in the feed material, then at steady-state, biomass growth rate in the reactor equals biomass outflow rate from the reactor, i.e.,

(11.48)Q0cx=rxVr=μcxVτ

The ratio of feed volumetric flow rate to culture solution volume is defined as dilution rate, i.e., D=Q0/Vt. Substituting it into Eq. (11.48) yields

(11.49)µ=D

D represents the extent that feed material was diluted in the reactor, and the dimension is (time)−1.

It can be inferred from Eq. (11.49) that if a cell is cultured in a CSTR, at steady-state, cell-specific growth rate equals the dilution rate of the reactor. This is an important feature for cell cultures in CSTRs. We can use this feature to change cell-specific growth rate at steady-state by controlling the feeding rate of the substrate. Therefore, CSTR is also called chemostat when it is used for cell culture. Cell growth characteristics can be conveniently investigated with a chemostat.

In a CSTR, the limiting substrate concentration and biomass concentration are related to the dilution rate. If biomass growth follows the Michaelis–Menten equation, then

(11.50)D=μ=μmaxcsKs+cs

Therefore, in the reactor substrate concentration and dilution rate has the following relationship

(11.51)cs=KsDμmax−D

Assuming the limiting substrate is only used for cell growth, then at steady-state

(11.52)Q0(cs0−cs)=rs⋅Vr

While

(11.53)rs=rxYx/s=μ⋅csYx/s

Substituting Eq. (11.52) into the above equation and combining with Eq. (11.49), we get the cell concentrations in the reactor

(11.54)cx=Yx/s(cs0−cs)

Substituting Eq. (11.51) gives the relationship between cell concentration and dilution rate, i.e.,

(11.55)cx=Yx/s(cs0−Ks⋅Dμmax−D)

From Eq. (11.51) we can see that, as D increases, cs in reactor also increases; when D is large enough to make cs=cs0, the dilution rate becomes the critical dilution rate, i.e.,

(11.56)Dc=μc=μmaxcs0Ks+cs0

The dilution rate of the reactor must be less than the critical dilution rate. When D>Dc, cell concentration in the reactor will become smaller and smaller, and the cell will be finally “washed” out from reactor, which is definitely not allowed.

Cell yield rate Px is also cell’s growth rate, i.e.,

(11.57)Px=rx=μcx=DCx=DYx/s(cs0−KsDμmax−D)

Fig. 11.16 shows cell concentration, limiting substrate concentration and cell formation rate as a function of dilution rate at cs0=10 g/L, µmax=1 h−1, Yx/s=0.5, and Ks=0.2 g/L. There is a maximum value in the cell formation curve. Setting dPx/dD=0 gives the optimal dilution rate Dopt (Fig. 11.17).

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Figure 11.16.

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Figure 11.17.

(11.58)Dopt=μmax(1−Ks(Ks+cs0))

Then, cell concentration in the reactor is

(11.59)cx=Yx/s(cs0+Ks−Ks(Ks+cs0))

The maximum cell formation rate, Px,max is

(11.60)Px,max=Yx/sμmaxcs0(1+Kscs0−Kscs0)2

When cs0>>Ks

(11.61)Dopt≈μmax

(11.62)Px,max≈Yx/sμmaxcs0

In a CSTR, the relationship between the product formation rate and dilution rate can be derived by performing material balance over the reactor based on the type of product formation and kinetics equation.

If product formation belongs to type I, i.e., growth-associated product formation, performing material balance over product,

(11.63)Q0cp−Q0cp0=rp⋅VL

Because

(11.64)rp=qp⋅cx=Yp/xμcx

Substituting into Eq. (11.63) and assuming there is no product in the feeding material, rearranging Eq. (11.63) gives product concentration

(11.65)cp=qp⋅cxD

Example 11.5

Escherichia coli were cultured in a 10 L CSTR at 30°C. Its kinetics equation follows the Monod equation, where µmax=1.0 h−1 and Ks=0.2 g/L. Glucose’s feed concentration is 10 g/L, the feed volumetric flow rate is 4 L/h, and Yx/s=0.5

1.

Determine cell concentration and growth rate in the reactor.

2.

Calculate the optimal feed rate at maximum cell yield and the maximum cell yield.

Solution

1.

Dilution rate of CSTR is

D=Q0/Vr=4/10=0.4 h−1

So, cell-specific growth rate is

µ=D=0.4 h−1

From Eq. (11.51), we obtain substrate concentration in the reactor

cs=KsDμmax−D=0.2×0.41.0−0.4=0.133g/L

And cell concentration

cx=Yx/s(cs0−cs)=0.5⋅(10−0.133)=4.93g/L

Px=rx=μcx=Dcx=0.4×4.93=1.97g/L

2.

The optimal feed rate

Dopt=μmax(1−KsKs+cs0)=1.0⋅(1−0.20.2+10)=0.86h−1

Q0=Dopt⋅Vr=0.86×10=8.6L/h

Cell concentration in the reactor

cx=Yx/s(cs0+Ks−Ks(Ks+cs0))=0.5(10+0.2−0.2(0.2+10))=4.39g/L

Maximum cell formation rate

Px,max=Dopt⋅cx=0.8/6×4.39=3.78g/L⋅h

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B9780124104167000112

Biochemical Reactions and Enzyme Kinetics

John D. Enderle PhD, in Introduction to Biomedical Engineering (Third Edition), 2012

8.6.2 Allosteric Activators and Inhibitors

Next, consider the noncompetitive allosteric modifier that binds with an enzyme on a regulatory site and a substrate that binds with the enzyme on the active site. The effect of a modifier on the reaction is to either increase or decrease the activity of the enzyme.

Consider an enzyme catalyzed reaction between a substrate, S, modifier, M, and enzyme, E, synthesizing product, P, as

(8.124)

Which of the following would increase the rate of an enzyme-catalyzed reaction?

where C1=SE,C2=ME,andC3=SMEare the intermediate complexes.6 Note that we have eliminated the reverse reaction from the product to the intermediate complex. Another variation of allosteric reaction eliminates the product from forming through C3.

The equations that describe the system in Eq. (8.124) are

(8.125)q˙S=K−1qC1+K−4qC3−K1qSqE−K4qSqC2q˙M=K−3qC2+K−6qC3−K3qMqE−K6qMqC1q˙C1=K1qSqE+K−6qC3−(K−1+K2)qC1−K6qMqC1q˙C2=K3qMqE+(K−4+K5)qC3−K−3qC2−K4qSqC2q˙C3=K4qSqC2+K6qMqC1−(K−4+K5+K−6)qC3q˙P=K2qC1+K5qC3

We eliminate qEfrom Eq. (8.125) by using qE=E0−qC1−qC2−qC3, giving

(8.126)q˙S=(K−1+K1qS)qC1+(K−4+K1qS)qC3−K1qSE0−(K4−K1)qSqC2q˙M=(K−3+K3qM)qC2+(K−6+K3qM)qC3−K3qME0−(K6−K3)qMqC1q˙C1=K1qSE0−K1qSqC2+(K−6−K1qS)qC3−(K−1+K2+K6qM+K1qS)qC1q˙C2=K3qME0−K3qMqC1+(K−4+K5−K3qM)qC3−(K3qM+K−3+K4qS)qC2q˙C3=K4qSqC2+K6qMqC1−(K−4+K5+K−6)qC3q˙P=K2qC1+K5qC3

with nonzero initial conditions of qS(0)=S0,qM(0)=M0,andqE(0)=E0.

Quasi-Steady-State Approximation to the Reaction Rate

The quasi-steady-state approximation is found from Eq. (8.126) with q˙C1=q˙C2=q˙C3=0, yielding

(8.127)0=K1qSE0−K1qSqC2+(K−6−K1qS)qC3−(K−1+K2+K6qM+K1qS)qC10=K3qME0−K3qMqC1+(K−4+K5−K3qM)qC3−(K3qM+K−3+K4qS)qC20=K4qSqC2+K6qMqC1−(K−4+K5+K−6)qC3

Equation (8.127) can be solved for qC1andqC3using the D-Operator method or the graph theory method from Rubinow, giving

(8.128)qC1=qSE0{K1K3K6iqS+K3i(K1K4K6i+K1K6KMi)+K6iqM}DqC3=E0qMqS{K1K3qS+K1K4K3i+K1K6KMs+qM}D

where

D=qS2K1K3(K6i+qM)+qS{K1K3K6iKMs+K3i(K1K4K6i+K1K6KMi)+qM(K6i+K1K3KMi+K1K4K3i+K1K6KMs)+qM2}+KMsK3i(K1K4K6i+K1K6KMi)+qM{KMs(K1K4K6i+K1K6K3i)+K3iKMi}+KMiqM2

KMs=K−1+K2K1KMi=K−4+K5K4K3i=K−3K3

K6i=K−6K6

The reaction rate is

(8.129)V=q˙P=K2qC1+K5qC3=qS[K2E0{K1K3K6iqS+K3i(K1K4K6i+K1K6KMi)+K6iqM}+K5E0qM{K1K3qS+K1K4K3i+K1K6KMs+qM}]D

This reaction rate is quite complex compared with the others derived previously. The reaction velocity from Eq. (8.129) is plotted in Figure 8.28. As observed, as the quantity of the allosteric modifier increases, it reduces Vmaxand the reaction rate.

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Figure 8.28. Velocity of product appearance for a substrate, enzyme, and allosteric modifier using Eq. (8.129). The parameters used are: K1=4,K−1=2,K2=5,K3=1,K−3=10,K4=0.3,K−4=0.03,K5=2,K6=2,K−6=4,and E0=15.

Reaction Rate from the True Steady-State

We will consider a simpler allosteric modifier model7 to further investigate the velocity of the reaction using the true steady-state rather than the quasi-steady-state approximation in the analysis. Here, the product is synthesized only from the intermediate complex C1, and a single reaction rate is used for SandMas follows:

(8.130)

Which of the following would increase the rate of an enzyme-catalyzed reaction?

The equations describing the system are given as

(8.131)q˙S=K−1(qC1+qC3)−K1qS(qE+qC2)q˙M=K−3(qC2+qC3)−K3qM(qE+qC1)q˙C1=K−3qC3+K1qSqE−(K−1+K2+K3qM)qC1q˙C2=K3qMqE+K−1qC3−(K−3+K1qS)qC2q˙C3=K1qSqC2+K3qMqC1−(K−1+K−3)qC3q˙P=K2qC1

where the variables and initial conditions are given as before. As before, qEis eliminated from Eq. (8.131) by using qE=E0−qC1−qC2−qC3, and a steady-state analysis is used to determine the reaction rate—that is, we let q˙S=q˙M=q˙C1=q˙C2=q˙C3=0.Keener and Sneyd give the reaction rate as

(8.132)V=Vmax(1+K3qMK−3)(1+K−1K1qS)

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B9780123749796000083

Enzymes

Shijie Liu, in Bioprocess Engineering (Second Edition), 2017

7.5 Temperature Effects

The rate of enzyme-catalyzed reactions increases with temperature up to a certain limit. Above a certain temperature, enzyme activity decreases with temperature because of enzyme denaturation. Fig. 7.16 depicts the variation of reaction rate with temperature and the presence of an optimal temperature. The ascending part of Fig. 7.16 is known as temperature activation. The rate varies according to the Arrhenius equation in this region

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Fig. 7.16. Effect of temperature on the activity of an enzyme. Here we have assumed a value of Ea = 46 kJ mol− 1, Ed = 301 kJ mol− 1 and kd0=1.0×1049h−1. The increase in maximum rate is due to the increase in the activity via Arrhenius law, while the descending of the curve is due to the dominance of the thermal denaturation. The enzyme activity or relative maximum rate is averaged for a total exposure of 1-h, 10-h, and 60-h to the temperature, which is shown in Eq. (7.91).

(7.85)rmax=kcE

(7.86)kc=kc0exp−EaRT

where Ea is the activation energy (kJ mol− 1) and E is the active enzyme concentration.

The descending part of Fig. 7.16 is known as temperature inactivation or thermal denaturation. The kinetics of thermal denaturation can be expressed as

(7.87)−dEdt=kdE

(7.88)E=E0e−kdt

where [E]0 is the initial enzyme concentration and kd is the denaturation constant. kd also varies with temperature according to the Arrhenius equation.

(7.89)kd=kd0exp−EdRT

where Ed is the deactivation energy (kJ mol− 1). Consequently,

(7.90)rmax=kc0E0exp−EaRTe−kdt

The activity or average maximum rate for a total exposure time of t is thus given by

(7.91)r¯max=∫0trmaxdtt=kc0E0exp−EaRT1−e−kdtkdt

The activation energies of enzyme-catalyzed reactions are within the 15–85 kJ mol− 1 range (mostly about 46 kJ mol− 1). Deactivation energies Ed vary between 170 and 550 kJ mol− 1 (mostly about 300 kJ mol− 1). That is, enzyme denaturation by temperature is much faster than enzyme activation. A rise in temperature from 30° to 40°C results in a 1.7-fold increase in enzyme activity, but a 45-fold increase in enzyme denaturation. Variations in temperature may affect both rmax and Km values of enzymes. Fig. 7.16 is plotted using Eq. (7.91).

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B9780444637833000071

Enzyme Mechanisms

Stephen J. Benkovic, Ann M. Valentine, in Encyclopedia of Physical Science and Technology (Third Edition), 2003

IV Origins of the Catalytic Efficiency of Enzymes

The source of the stereospecificity of enzyme-catalyzed reactions is clearly revealed by the fit of the substrate to the enzyme's active site that spatially then directs the stereochemical course of the chemical events. The speed of these reactions has been attributed to the lowering of the activation energy for the process by the greater affinity of the enzyme for the transition state than that for the substrate. Although this proposal is an adequate rationale, it is often a necessary thermodynamic statement that does not offer insights into how the activation barrier is actually lowered.

The preorganization of substrate and active site residues within a protein cavity converts an intermolecular process to intramolecular and may have both an enthalpic and an entropic advantage. The active site provides an environment in which the enzyme·substrate complex is populated with cofactors that are poised for reaction. These structures, or NACs (near attack conformers), are similar in structure to the transition state so that only slight changes in bond distances and angles within the structures through the normal dynamic motions of the protein are sufficient to trigger the crossing of the reaction barrier. The enzyme's active site is also preorganized in the sense that the locus of general acids/bases, nucleophiles, solvents, dipoles, hydrogen bonds, and so forth are fixed by the NAC to interact with the transition state. Molecular dynamic calculations sampling several enzyme classes suggest that the affinity of these enzymes for their transition states is little changed from that for the substrate. The enzyme-catalyzed reaction also benefits in many cases due to the nonaqueous interaction of the active site cavity, which can often accelerate, by large factors, the reaction over that in aqueous media.

For DHFR in particular, molecular dynamics calculations, NMR measurements of solution structure, and kinetics measurements of mutant forms of the enzyme appear to support the importance of dynamic motions of the protein fold to trigger the reaction of an enzyme-substrate NAC. The mutations in question (for example Gly120 in Fig. 7) are well removed from the active site and underscore the role of the entire protein fold. The contribution of dynamic motions to the overall catalytic rate remains to be elucidated for the majority of enzymes. Their existence may explain why more rigid molecules such as imprinted polymers and catalytic antibodies do not generally exhibit the large rate accelerations noted with enzymes despite the fact that they too have converted an intermolecular process to an intramolecular process.

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B0122274105002313

Glycolysis Overview☆

Robert A. Harris, Justin S. Johnson, in Reference Module in Biomedical Sciences, 2019

Abstract

Glycolysis is a linear metabolic pathway of enzyme-catalyzed reactions that converts glucose into two molecules of pyruvate in the presence of oxygen or two molecules of lactate in the absence of oxygen. The latter pathway, anaerobic glycolysis, is believed to be the first process to have evolved in nature to produce adenosine triphosphate (ATP). In some cells—notably in mature red blood cells—glycolysis is the only means of ATP production because of the lack of mitochondria. In most cells glycolysis converts glucose to pyruvate which is subsequently oxidized to carbon dioxide and water by mitochondrial enzymes. Obligate ATP production via glycolysis also occurs in the absence of oxygen whether mitochondria are present or not. Overproduction of lactic acid by anaerobic glycolysis can lead to lactic acidosis, a life-threatening medical condition. Finally, even when both mitochondria and oxygen are present, cancer cells preferentially produce ATP by the conversion of glucose to lactate by aerobic glycolysis. The robust flux of glycolysis in cancer cells maintains high levels of intermediates required for the synthesis of macromolecules required for rapid growth and protection against reactive oxygen species.

Read moreNavigate Down

View chapterPurchase book

Read full chapter

URL: https://www.sciencedirect.com/science/article/pii/B978012801238311342X

Engineering Fundamentals of Biotechnology

K. Miyanaga, H. Unno, in Comprehensive Biotechnology (Second Edition), 2011

2.05.2.1 Michaelis–Menten Equation

In the usual single-substrate-enzyme-catalyzed reaction, the relationship between initial rate of reaction and the substrate concentration assumes the form of saturation curve. A mathematical model to describe the kinetics was developed by Henri in 1902 [8] and Michaelis and Menten in 1913 [9]. Furthermore, Briggs and Haldane [5] generalized this model using the quasi-steady-state assumption in 1925. A simple schematic diagram is described in Figure 1.

Which of the following would increase the rate of an enzyme-catalyzed reaction?

Figure 1. Schematic diagram of enzyme reaction.

Henri and Michaelis and Menten assumed a production reaction of enzyme–substrate (ES) complex derived from free enzyme and substrate rapidly reaches equilibrium state. In equilibrium state, the following equation is derived:

[1]k1[E][S]=k−1[ES]

The dissociation constant (Ks) of ES complex is

[2]KS=k−1k1=[E][S][ES]

On the other hand, in the quasi-steady-state, it is assumed that the initial substrate excessively exists in the batch system compared with the initial enzyme, and the concentration of ES complex rapidly reaches steady state under the condition where substrate concentration [S] is fixed. It means that time course of ES complex concentration is assumed as zero. In addition, we must be conscious of the limitation that this perspective can be applicable only under the condition where ES complex concentration is sufficiently low.

What would increase the rate of an enzyme

The rate of an enzyme-catalyzed reaction increases with an increase in the concentration of an enzyme. At low temperatures, an increase in temperature increases the rate of an enzyme-catalyzed reaction. At higher temperatures, the protein is denatured, and the rate of the reaction dramatically decreases.

Which of the following increases the rate of chemical reaction in enzymes?

Catalysts lower the activation energy for reactions. The lower the activation energy for a reaction, the faster the rate. Thus enzymes speed up reactions by lowering activation energy.

What are 4 things that could affect the rate of a reaction catalyzed by an enzyme?

Several factors affect the rate at which enzymatic reactions proceed - temperature, pH, enzyme concentration, substrate concentration, and the presence of any inhibitors or activators.

Which of the ff will increase the rate of an enzyme

Temperature can have an important effect on enzyme activity and reaction rates. At low temperature, warming usually increases the rate of an enzyme catalyzed reaction because the reactants have more energy, and can more readily achieve the activation energy level.