Enzymes crash course

Biocatalysts that increase rate of biochemical reactions. Chemical natureChemical nature :

1. Proteins 2. RNA(RIBOZYMES) 3. Abzymes (Ab + enzymes)

CofactorsCofactors- Additional chemical component {inorganic ions & organic molecules (coenzymes)}

HoloenzymeHoloenzyme- Complete catalytically active enzyme + coenzyme (Prosthetic gp) Apoprotein or apoenzyme- protein part of holoenzyme

Prosthetic group – Coenzyme/metal ion that is very Prosthetic group – Coenzyme/metal ion that is very tightly/covalently tightly/covalently

ATP + D-glucose → ADP + D-glucose-6-phosphate

ATP:glucose phosphotransferase

E.C. no: 2.7.1.1.2 – Transferases7 – Phosphotransferases1 – Phosphotransferase with a hydroxyl group as an acceptor1 – D-glucose as the phosphoryl group acceptor

Active site- catalytic site of an enzyme

Substrate- molecule bound in the active site & acted upon by the enzyme

Function of enzyme: To increase the rate of reaction by reducing activation energy Does not disturb the reaction equilibrium

Ground state- starting point for forward or reverse reaction

Reaction intermediate: any species on the reaction pathway that has a transient existence [ES and EP complexes]

Rate-limiting step- step (or steps) with the highest activation energy

The equilibrium between S and Pequilibrium between S and P reflects the difference in thedifference in the free energies of their ground free energies of their ground statesstates.

When the free energy of the ground state of P is lower than that of S, then ∆∆G’G’oo for the for the reaction is negativereaction is negative and the equilibrium favors P. Note:- The position and direction

of equilibrium are not affected by any catalyst.

Transition stateTransition state: point at which decay to the S or P

state is equally probable Activation energy, Activation energy, GG‡‡.. : :

The difference between the energy levels of the ground state and the transition state.

energy barriers to chemical reactions

Comparison between Comparison between enzyme catalyzed enzyme catalyzed

&& uncatalyzed reactionuncatalyzed reaction

Relationship between rate constant & activation energy

ENZYME KINETICS

Substrate Concentration Affects the Rate of Enzyme-

Catalyzed ReactionsInitially The conc of Substrate = [S]Initial Conc of Enzyme = [E] in

nanomolesInitial Rate or velocity = VoIn the initial phase of enzyme

catalysed reaction, there is no significant change in [S]

Hence it can be considered as constant

When the enzyme conc is held constant then, The effect of Vo on varying [S] is represented graphically as follows

Direct plot

Michaelis-Menten equation- the rate equation for a one-substrate enzyme-catalyzed reaction.

Relationship Between[S] & Vmax

Km

• Km is equivalent to the substrate concentration at which V0 is one-half Vmax.

Interpreting Vmax and Km:

Double reciprocal plot or Lineweaver-Burk equation

Kinetic parameters to compare enzyme activities

-Interpretation of KKmm

- Interpretation of V- Interpretation of Vmaxmax in terms of K in terms of Kcatcat

-Comparison of Catalytic mechanisms & efficiencies of enzymes using KKcatcat & KKmm

Interpreting Km

• vary greatly from enzyme to enzyme• used (often inappropriately) as an indicator of

the affinity of an enzyme for its substrate

Interpreting VInterpreting Vmaxmax in in terms of Kterms of Kcatcat

Kcat = rate constant of rate limiting step

In the first reaction K2 = KcatIn IInd reaction K3 = KcatHence,

Vmax = kcat[Et]Hence the Michelis Menten eq

becomes

• The constant kcat is a first-order rate constant and hence has units of reciprocal time. It is also called the turnover number. – equivalent to the number of substrate

molecules converted to product in a given unit of time on a single enzyme molecule when the enzyme is saturated with substrate.

Comparing Catalytic mechanisms Comparing Catalytic mechanisms & efficiencies using Kcat & km& efficiencies using Kcat & km• Each enzyme has different values of kcat and Km

which depends upon– the cellular environment, – the concentration of substrate normally

encountered in vivo by the enzyme– the chemistry of the reaction being catalyzed

• Values of Kcat & Km are needed to evaluate the kinetic efficiency of enzymes (Both the parameter alone is insufficient for estimation)

– Lower Km : Enz acts even on low substrate Lower Km : Enz acts even on low substrate conc.conc.

– Higher Km : Enz acts on high substrate conc.Higher Km : Enz acts on high substrate conc.

• The ratio kcat/Km can be used to compare catalytic efficiency of different enzymes or turnover no. of diff. substrates by the same Enz.– This parameter, is called the specificity

constant• rate constant for the conversion of E + S

to E + P.

When [S] << Km

Becomes

• second-order rate equation and the constant kcat/Km is a second-order rate constant with units of M-1s-1.

Reactions with Two or More Substrates

• Enzymatic reactions with two substrates usually involve transfer of an atom or a functional group from one substrate to the other

• MM equation can be used for Bi substrate reaction– Rates of reaction steps are measured

independently

Enzyme Inhibition

ENZYME INHIBITIONENZYME INHIBITION:

Any ligand that reduces the velocity of an enzyme catalyzed reaction

Commonly used in every day life as drugs, antibiotics, toxins, poisons etc.

Major regulatory mechanism of living cells

Tells us about the specificity of the enzyme and architecture of the active site

2 Types: Reversible Inhibition

Irreversible Inhibition

INHIBITOR :INHIBITOR :

Reversible Inhibition- Competitive, Uncompetitive and mixed

Competitive Inhibition:• Competitive inhibitor competes with the

substrate for the active site of an enzyme

– inhibitors resemble the substrate– reduce the efficiency of the enzyme– inhibitor binds reversibly to the enzyme

• Can be overcome by increasing substrate conc.

Uncompetitive Inhibition

• Uncompetitive inhibitor binds at a site distinct from the substrate active site– Inhibitor binds only to the ES

complex

Mixed inhibit ion• mixed inhibitor binds

at a site distinct from the substrate active site, but it binds to either E or ES.

• usually affects both Km and Vmax

Inhibit ion Formula Effect on Vmax

Effect on Km

Competit ive same increases

Uncompetit ive decreases decreases

Mixed decreases Increases/decreases

Noncompetit ive When α = α’ decreases same

Irreversible Inhibition• Irreversible inhibitors

– bind covalently with enzyme– destroy a functional group on an enzyme

that is essential for the enzyme’s activity, or those that form a particularly stable non-covalent association.

• Suicide inactivators– special class of irreversible inhibitors– relatively unreactive until they bind to the

active site of a specific enzyme– converted to a very reactive compound

when combined irreversibly with the enzyme.

– also called mechanism-based inactivators,

• hijack the normal enzyme reaction mechanism to inactivate the enzyme

• play a significant role in rational drug design

REGULATION OF ENZYME ACTIVITY

• Feedback regulationFeedback regulation

• Allosteric regulationAllosteric regulation

• Regulation by Reversible covalent Regulation by Reversible covalent modificationmodification

• Regulation by Proteolytic activationRegulation by Proteolytic activation

• Regulation by enzyme synthesis and Regulation by enzyme synthesis and breakdownbreakdown

• Regulatory enzymesRegulatory enzymes

– exhibit increased or decreased catalytic activity in response to certain signals

• Allosteric enzymes Allosteric enzymes

– Functions through reversible, noncovalent binding of regulatory compounds called allosteric modulators or allosteric effectors, (small metabolites or Cofactors)

Feedback regulation

• enzyme is inhibited by an end-product of the metabolic pathway in which it is involved

• often takes place at the committed stepcommitted step in the pathway

– first step to produce an intermediate which is unique to the pathway in question,

Allosteric regulation

• Regulation of allosteric enzymes

• Binding of substrates is cooperative

• Allosteric enzymes – multi-subunit proteins, with one

or more active sites on each subunit

– binding of substrate at one active site induces a conformational change in the protein that is conveyed to the other active sites, altering their affinity for substrate molecules

– controlled by effector or modulator molecules

• Activators and Inhibitors– bind to the enzyme at a site

other than the active site (either on the same subunit or on a different subunit),

– causes a change in the conformation of the active site

– alters the rate of enzyme activity

• An allosteric activator increases the rate of enzyme activity, while

• An allosteric inhibitor decreases the activity of the enzyme.

• Methylation – Methyl accepting chemotaxis protein of bacteria Transmembrane sensor protein in bacteria

• ADP-ribosylation – Diptheria toxin – ADP-ribosylation of eEF2

(inhibition of protein synthesis) Cholera toxin - ADP-ribosylation of G

protein (inhibition of signaling pathway)

• Phosphorylation -

Covalent modifications

Regulation by Proteolytic activation

• Needed to activate large inactive enzymes or zymogens

• involves irreversible hydrolysis of one or more peptide bonds

• E.g. trypsin, chymotrypsin and elastase

• A plot of V versus V/S is generated for an enzyme A plot of V versus V/S is generated for an enzyme catalyzed reaction, and a straight line is obtained. catalyzed reaction, and a straight line is obtained. Indicate the information that can be obtained from the Indicate the information that can be obtained from the plotplot

a. Vmax and turnover no. Km can be obtained only from a plot of 1/V v/s 1/S

b. Km/Vmax from the slopec. Vmax, Km and turnover numberd. Only Km and turnover number

Eadie Hofstee Plot

Multiply with Vmax on both sides

Vmax / Vo = Km/S +1

-[Km/S] = 1 – Vmax/Vo

-[Km/S] Vo = Vo – Vmax

Vo = -[Km/S] Vo + Vmax

Slope: -Km ; x intercept: Vmax/Km; y intercept : VmaxAns: c

• Phosphoglucomutase is added to 0.1 M Glucose -6-Phosphoglucomutase is added to 0.1 M Glucose -6-Phosphate (G-6-P). The standard change of the Phosphate (G-6-P). The standard change of the reaction, G-6-P G-1-P is 1.8 kcal/mole at 25reaction, G-6-P G-1-P is 1.8 kcal/mole at 25ººC. The C. The equilibrium concentrations of G-6-P and G-1-P, equilibrium concentrations of G-6-P and G-1-P, respectively, are: respectively, are:

a. 96 mM, 45mMb. 100 mM, 0 mMc. 45 mM, 96 mMd. 0 mM, 100 mM

Ans: a

• Michaelis and Menten derived their equation using Michaelis and Menten derived their equation using which of the following assumption?which of the following assumption?

a. Rate limiting step in the reaction is the breakdown of ES complex to product and free enzyme

b. Rate limiting step in the reaction is the formation of ES complex

c. Concentration of the substrate can be ignoredd. Non-enzymatic degradation of the substrate is the

major step

Ans: a

Ans: 2

Ans: 3

74. The hydrolysis of pyrophosphate to orthophosphate is important for several biosynthetic reactions. In E. coli, the molecular mass of the enzyme phosphatase is 120 kD, and it consists of 6 identical subunits. The enzyme activity is defined as the amount of enzyme that hydrolyses 10 μmol of pyrophosphate in 15 mins at 37ºC under standard assay condition. The purified enzyme has a Vmax of 2800 units per mg of the enzyme. How many moles of the substrate are hydrolyzed per second per milligram of the enzyme when the substrate concentration is much greater than Km?

a. 0.05 μmolb. 62 μmolc. 31.1 μmold. 1 μmol

Ans: c

1 unit of Enz. activity = Amt. of Enz. required to convert 10 μmoles of substrate/15 mins

2800 units of Enz. Activity = Amt. of Enz. required to convert 2800*10 μmoles of substrate converted/15 minsOr 28000 μmoles /900 seconds

At Vmax, Enz. Activity = 2800 units / mg of Enz

i.e. 1 mg of Enz. can hydrolyze 28000 μmoles Sub./900 seconds

i.e. 28000/900 = 31.1 μmoles

Enzymes have an optimum temperature at which they work best. Temperatures above and below this optimum will decrease enzyme activity. Which graph best illustrates the effect of temperature on enzyme activity?

a. Ab. Bc. Cd. D

Ans: B

You are trying to reproduce experimental data from the previous student in the lab. He/she had reported that the enzyme under investigation has a kcat of 1500 s-1 and a catalytic efficiency of 7.5 107 M-1 s-1. What is the KM of this enzyme?

a.2 * 10 -5 Mb.2 * 10 -6 Mc.20 * 10 -5 M/sd.20 * 10 -6 M/s

Ans: a

An enzyme has a Km for substrate (S) of 10 mM and Vmax of 5 μmol.L-1.sec-1 at a total enzyme concentration of 1 nM. At [S] = 10 mM, kcat is:

A. 2500 per M per sec.B. 5000 per M per sec.C. 2500 per sec.D. 5000 per sec.E. 1250 per M per sec.

Ans: D

In an enzyme mediated reaction, the inhibitor I reduces the rate of product formation from substrate. If velocity/substrate data are plotted in Lineweaver-Burk form, the inhibitor is competitive when:

A. Both the y- and x-intercepts are decreased.B. The slope of the plot is decreased.C. The x-intercept is decreased, the y-intercept is unchanged and the slope of the plot increases.D. The slope of the plot is increased and the y-intercept decreasesE. The x-intercept increased and the y-intercept is unchanged.

Ans: E

Competitive inhibition

Uncompetitive inhibition

Noncompetitive inhibition

Irreversible inhibition

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