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Enzyme Kinetics

Biochemical catalysts are helpful to speed up the biochemical reactions in body and made up of proteins. In the absence of enzymes, biochemical reactions proceed slowly as enzymes catalyzed reactions which are essential to sustain life.

As enzymes are act as catalyst, they do not consume during reaction and regenerate again at the end of the reactions. The minimum amount of energy required to initiate a reaction is called as activation energy. Enzymes help to reduce the activation energy of reaction and increase the rate of reaction.

Kinetics Enzyme

Fundamentals of Enzyme Kinetics

  1. Enzymes are proteins in which one or more polypeptide chains are folded together and form active site for substrate of biochemical reactions. On active sites of enzymes, substrates bind to form products.
  2. Biochemical reactions catalyzed by enzymes are effected by pH, temperature, medium of reaction system as well as enzyme and substrate concentrations. The favorable pH for any biochemical reaction is known as optimum pH.
  3. Enzymes are active only at optimum pH and get breakdown at high or low pH which is called as denaturation of enzyme. After denaturation of enzyme, there will be np more active site on enzyme for bonding with substrate.
  4. Similarly other factors like temperature and presence of other substrates. The optimum temperature for most of the enzymes is 298 K. Substrate fits into the active site forms a complex with enzyme to initiate any biochemical reaction.
  5. The rate of reaction effects with the concentration of enzyme as well as substrates. As the concentration of enzyme and substrate increases, the rate of reaction increases and reached at maximum velocity (Vmax).
  6. After Vmax, there will be no increment in rate of reaction as all the active sites of enzyme get bonded with substrates to form Enzyme-substrates complex. Further enzyme-substrates complex decompose to product and enzyme which can bonded with another substrates.
  7. For example, the hydrolysis of sucrose is catalyzed by Sucrase enzyme to form glucose and fructose.

Enzyme Kinetics

Related Calculators
Calculate Kinetic Energy Calculate Kinetic Friction

Enzyme Inhibition Kinetics

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Any chemical substance which can bind with active sites of enzymes and inhibit their reactivity as no more active site is available for substrates. Such substances are called as enzyme inhibitors and process is known as enzyme inhibition. Enzyme inhibitors bonded at active site of enzymes and hinder the enzyme from catalyzing its reaction.
Many drug molecules act as enzyme inhibitors and can be identified by their specificity and their potency. There are many natural enzyme inhibitors which can be involved in the regulation of metabolism. The process of enzyme inhibition can be reversible or irreversible in nature. Irreversible inhibitors form covalent bond with enzymes and change it chemically which hinder enzymatic activity.

However, reversible inhibitors bind non-covalently with enzyme and inhibit the formation of enzyme-substrate complex.

1. Reversible inhibitors

When an inhibitor bonds with enzyme by non-covalent bonds like hydrophobic interactions, hydrogen bonds or ionic bonds, the interaction will be reversible inhibition. Since the bonding between inhibitors and enzyme is weak in nature, therefore there will be no chemical reaction between inhibitor and enzyme and can be easily removed by dilution or dialysis.

On the basis of the effects of varying the concentration of the enzyme's substrate on the inhibitor reversible enzyme inhibitors can be classified in four types.

(a) Competitive inhibition: In the presence of inhibitor which has similar structure to the real substrate, there will be a competition between the substrate and inhibitor to bind with enzyme, as both cannot bond at the same time. If inhibitor has an affinity to bind with enzyme, it will compete with substrate for access to the enzyme's active site and called as competitive inhibition. The high concentration of substrate overcomes the inhibitions.
Enzyme Velocity

As in competitive inhibition, inhibitor can bind to enzyme (E) not with enzyme-substrate complex( ES), hence the value of Km(Michaelis-Menten constant) increases as the inhibitor interferes with substrate binding, but the maximum velocity (Vmax) remained same.

Hence the observed Km will be

Where Ki = dissociation constant for inhibitor binding

The curve between the concentration of inhibitor, and observed Km will be linear with Km intercept at Y-axis and negative Ki at X-axis.
Competitive Inhibition

(b) Non-competitive inhibition: It is a type of mixed inhibition only which involves the binding of the inhibitor with enzyme to reduce its activity but does not affect the binding of substrate. Hence the extent of inhibition depends on the concentration of the inhibitor.

Non Competitive Inhibition

Since non-competitive inhibitors have identical affinities for enzyme (E) and enzyme-substrate-complex (ES), hence Ki =Kii where; Ki is the constant for binding of inhibitor with enzyme and Kii is with enzyme-substrate complex. There is no change in Km hence it does not affect substrate binding but decreases the maximum velocity (Vmax).
Non-Competitive Inhibition
(c) Uncompetitive inhibition: In such type of inhibition, the inhibitor binds with substrate-enzyme complex which causes maximum velocity to decreases due to removal of activated complex. The value of Km (Michaelis-Menten constant) decreases in such type of inhibition because of better binding efficiency. Since inhibitor bonds with enzyme-substrate complex, the kinetic of uncompetitive can be shown as below.
Uncompetitive Inhibition
Hence Km becomes to Km/(1 + [I]/Kii) and Vmax changed to Vmax/(1 + [I]/Kii)


(d) Mixed inhibition: It is an intermediate condition of competitive and uncompetitive inhibition i.e. the inhibitor can bind either with enzyme or with enzyme's substrate. Mixed inhibition can be reduced by increasing concentrations of substrate but not overcome completely.

In such type of inhibition, inhibitors generally bind at a different site on an enzyme known as allosteric site which changes the tertiary structure or three-dimensional shape of the enzyme and inhibit its activity.
Since in mixed- inhibition, inhibitors can be bond with both enzyme (E) and enzyme-substrate complex (ES) but with different affinities, therefore Ki ≠Kii .

Mixed Inhibition

2. Irreversible inhibitors

  1. Because of covalent bond between inhibitor and enzyme, some inhibitions can be irreversible. Many reactive functional groups involve in covalent bond formation like nitrogen mustards, haloalkanes, fluorophosphonates, aldehydes and alkenes. These reactive functional groups bonded with amino acid side chains of enzymes and form covalent adducts.
  2. Compare to reversible inhibitors; irreversible inhibitors are generally specific for enzyme, therefore they do not inactivate all proteins. They are specific for active sites of enzymes; hence do not destroy the protein structure but alter the active sites only.
  3. Generally enzymes contain sulfhydral (-SH), alcohol, or acid groups on their active sites which can be easily react with many chemical like compounds contain Ag+, Hg2+, Pb2+ ions and these chemical act an irreversible inhibitor.
  4. For example, Nerve gases like diisopropylfluorophosphate (DFP) can bond on the active site of acetylcholine esterase due to the presence of hydroxyl group of serine.
  5. Similarly oxalic and citric acid form complexes with calcium ions to inhibit the blood clotting as Ca2+ ions are necessary for the enzyme metal ion activator.
  6. Irreversible inhibitors(I) can form a reversible non-covalent complex with the enzyme (E) or enzyme-substrate complex to form EI or ESI which further reacts to form a covalently modified "dead-end complex" (EI*) and the rate of the formation of this complex is known as inactivation rate or kinact.
  7. The formation of enzyme-substrate complex competes with the formation of enzyme-inhibitor complex; therefore the irreversible inhibition can be prevented by competition with substrate or with a second reversible inhibitor.
Irreversible Inhibitors

Enzyme Kinetics Problems

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Below you could see problems

Solved Examples

Question 1: The standard free energy change for an UN-catalyzed reaction (S « P) is +120 kJ/mol, and a specific enzyme enhances the rate of the reaction by 109 fold. What will be the direction of reaction under standard conditions in the presence of the enzyme?
The presence of catalyst does not change Keq, as Enzymes only accelerate the rate of biochemical reaction reaches equilibrium but the position of the equilibrium will remain same.

Question 2: The value of Km and Vmax for two alternative substrates A and B for the same enzyme are as follow.

Substrate Em
4.0mM 25mMol/sec
0.5mM 15mMol/sec

Which substrate will react most rapidly at low substrate concentration (<< Km)?
Since at low substrate concentration [S], Vmax = kcat [Et]. As we have Vmax, which is directly proportional to kcat at a given enzyme concentration, so comparison of Vmax/Km gives idea about enzyme's efficiency with substrates. The value of substrate B = Vmax/Km =30 (µmol/sec/mM) And for A = 6.25 (µmol/sec/mM). Hence at low concentration substrate B will be used more efficiently react compare to substrate A.

Question 3: The following table shows the rate of reaction of substrate to product in the presence of enzyme; v (mol/sec) under different conditions;
  1. Reaction 1: Reaction of substrate with enzyme to form product in the absence of inhibitor.
  2. Reactions 2 and 3: Reaction takes place in the presence of two different inhibitors, each with 10mM concentration.

[S] mM
v, reaction 1
(μ mol/sec)
v, reaction 2
(inhibitor A)
(μ mol/sec)
v, reaction 3
(inhibitor B)
(μ mol/sec)
2.5 1.17 0.77

Let’s take same amount of enzyme in each case and determine Km and Vmax for the enzyme (in absence of inhibitors) and determine Ki and the type of inhibition for each inhibitor.
Km = 3.3 mM, Vmax = 10 mmol/sec
  1. Inhibitor A (competitive inhibition), there will be no change in Vmax , but Kmapp = Km{1 + ([I]/KI)}-1/Kmapp = -0.13mM-1 K mapp = 7.7mM K mapp /Km = 7.7mM/3.3mM= 2.33 = 1 +[I]/ KI = 1+ [10mM]/ KI KI (inhibitor-A) =10mM/ 2.33 – 1 = 10mM/ 1.33 = 7.52 mM
  2. Inhibitor B (pure non-competitive) there will be no change in Km; but Vmaxapp = Vmax / {1 + ([I]/ KI )} 1/ Vmaxapp = 0.3(μ m o l / s e c )-1 Vmaxapp = 3.33 μmo l / s e c Vmax/ Vmaxapp = 10 μmol/sec/ 3.33 μm o l / s e c = 3 = 1 +[I]/KI = 1 +[10mM]/ KI KI (inhibitorB) =10mM / 3 - 1 = 10mM/2 = 5.0 mM

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