Biocatalysts such as enzymes increase the rate of biochemical reactions. Enzymes are typically proteins but can also be RNA or artificial molecules called abzymes. Enzymes may require cofactors such as metal ions or organic coenzymes to be catalytically active. The document discusses enzyme kinetics including Michaelis-Menten kinetics, factors that affect reaction rate, kinetic parameters such as Km and Vmax, enzyme inhibition types including competitive, noncompetitive and irreversible inhibition, and regulation of enzyme activity through feedback, allosteric regulation and covalent modification.

2.  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 veryProsthetic group – Coenzyme/metal ion that is very
tightly/covalentlytightly/covalently

3. ATP + D-glucose → ADP + D-glucose-6-phosphate
ATP:glucose phosphotransferase
E.C. no: 2.7.1.1.
2 – Transferases
7 – Phosphotransferases
1 – Phosphotransferase with a hydroxyl group as an acceptor
1 – D-glucose as the phosphoryl group acceptor

4.  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

5.  The equilibrium between S and Pequilibrium between S and P
reflects the difference in thedifference in the
free energies of their groundfree 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 thefor 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

6. Comparison betweenComparison between
enzyme catalyzedenzyme catalyzed
&&
uncatalyzed reactionuncatalyzed reaction

7. Relationship between
rate constant &
activation energy

8. ENZYME KINETICS
Substrate Concentration
Affects the Rate of Enzyme-
Catalyzed Reactions
Initially The conc of Substrate = [S]
Initial Conc of Enzyme = [E] in
nanomoles
Initial Rate or velocity = Vo
In 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

9. Michaelis-Menten equation- the rate equation for a one-substrate
enzyme-catalyzed reaction.
Relationship
Between
[S] & Vmax

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

12. Interpreting Vmax and Km:
Double reciprocal plot or Lineweaver-Burk equation

13. Kinetic parameters to compare
enzyme activities
-Interpretation of KKmm
- Interpretation of V- Interpretation of Vmaxmax in terms of Kin terms of Kcatcat
-Comparison of Catalytic mechanisms & efficiencies of enzymes using KKcatcat & KKmm

14. Interpreting Km
• vary greatly from enzyme to enzyme
• used (often inappropriately) as an indicator of
the affinity of an enzyme for its substrate

15. Interpreting VInterpreting Vmaxmax inin
terms of Kterms of Kcatcat
Kcat = rate constant of rate limiting
step
In the first reaction K2 = Kcat
In IInd reaction K3 = Kcat
Hence,
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.

16. Comparing Catalytic mechanismsComparing 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 substrateLower 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-1
s-1.

17. Reactions with Two or More
Substrates

18. • 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

19. Enzyme Inhibition

20. 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 :

21. 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.

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

23. Mixed inhibition
• 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

24. Inhibition Formula Effect on
Vmax
Effect on
Km
Competitive same increases
Uncompetitive decreases decreases
Mixed decreases Increases/
decreases
Noncompetitive When α = α’ decreases same

25. 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

26. REGULATION OF ENZYME ACTIVITY
• Feedback regulationFeedback regulation
• Allosteric regulationAllosteric regulation
• Regulation by Reversible covalentRegulation by Reversible covalent
modificationmodification
• Regulation by Proteolytic activationRegulation by Proteolytic activation
• Regulation by enzyme synthesis andRegulation by enzyme synthesis and
breakdownbreakdown
• Regulatory enzymesRegulatory enzymes
– exhibit increased or decreased
catalytic activity in response to
certain signals
• Allosteric enzymesAllosteric enzymes
– Functions through reversible,
noncovalent binding of
regulatory compounds called
allosteric modulators or
allosteric effectors, (small
metabolites or Cofactors)

27. 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,

28. 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.

32. • 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

34. 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

35. • A plot of V versus V/S is generated for an enzymeA 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 theIndicate 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 slope
c. Vmax, Km and turnover number
d. Only Km and turnover number

36. 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 : Vmax
Ans: c

37. • 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 thePhosphate (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. TheC. 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, 45mM
b. 100 mM, 0 mM
c. 45 mM, 96 mM
d. 0 mM, 100 mM
Ans: a

38. • Michaelis and Menten derived their equation usingMichaelis 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 ignored
d. Non-enzymatic degradation of the substrate is the
major step
Ans: a

39. Ans: 2

40. Ans: 3

41. 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 μmol
b. 62 μmol
c. 31.1 μmol
d. 1 μmol
Ans: c

42. 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 mins
Or 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

43. 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. A
b. B
c. C
d. D
Ans: B

44. 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
M
b.2 * 10 -6
M
c.20 * 10 -5
M/s
d.20 * 10 -6
M/s
Ans: a

45. 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

46. 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
decreases
E. The x-intercept increased and the y-intercept is
unchanged.
Ans: E

47. Competitive inhibition

48. Uncompetitive inhibition

49. Noncompetitive inhibition

50. Irreversible inhibition

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