ENZYME IN CSIR POINT OF VIEW

1. ENZYME

2. • DEFINITION : Enzymes are protenecious biocatalyst (except
ribozyme)which work by lowering the activation energy
and remain unchanged after reaction.
• PROPERTIES OF ENZYME :
• Enzymes are protein in nature except ribozymes
• Enzymes are highly specific .They are specialized protein
and have high degree of specificity for their substrate.
• Enzyme exhibit enormous catalytic power. It increases the
rate of a reaction by lowering the activation energy

3. ENZYME STRUCTURE

4. TERMS
 APOENZYME
 COFACTOR,
 HOLOENZYME
 PROSTHETIC GROUP

5. What is activation energy?
• In chemistry, activation energy is the energy
which must be available to a chemical system
with potential reactants to result in a
chemical reaction. Activation energy may also
be defined as the minimum energy required
to start a chemical reaction.

6. Q. ENZYME LOWER THE ACTIVATION ENERGY , BUT WHERE
DOES THE ENERGY TO LOWER THE ACTIVATION ENERGY?
The binding energy released due to the
interaction between enzyme and substrate
lowers the activation energy.
Only the correct substrate can form maximum
interactions with the enzyme and thus maximized
binding energy .
Furthermore, the full complement of such
interactions is formed only when the enzyme
facilitates the formation of transition state.
Transition state is the point of highest free
energy.

7. Different bonds at the active site of
an enzyme with substrate

9. CO-enzyme
• Organic cofactor is called an co-enzyme.
• Co-enzymes which are tightly associated with
the protein covalently or non –covalently
called prosthetic group.

11. Naming and classification of enzyme
IUB Classification
Oxido reductase
2. Transferase
3hydrolase
4.lyases
Isomerases
Ligases.

12. HOW enzyme work?
An enzyme accelerates the rate of a chemical reaction several times as
compared to uncatalyzed reaction in water.
It increases the rate of a chemical reaction by lowering the activation energy.

13. • However, an enzyme does not change the free
energies of the initial and final states.
• The free energy of reaction delta G , remain
unchanged in the presence of an enzyme, so the
relative amounts of reactant and products at
equilibrium are unchanged .
• In other word enzyme does not change the
position of equilibrium in a chemical reaction.
• It only accelerate the attainment of equilibria but
do not shifts their position.

14. • The substrate bind to the active site by
multiple weak interactions
• Free energy released in the formation of a
large number of weak interactions between
the enzyme and the substrate is termed
binding energy.

15. model of enzyme action

16. INDUCED FIT MODEL

17. • Induced fit model enzymes are flexible and
shape of the active sites can be markedly
modified by the binding of substrate.

18. ENZYME KINETICS
Kinetics of enzyme catalyzed reaction
TON calculation
Line weaver- Burk plot
Enzyme inhibition : a. competetive
 b. Non competetive
 C. uncompetetive.

23. Problem SET

24. • Which of the following statements about Michaelis-
Menten kinetics is correct?
• a) Km, the Michaelis constant, is defined as the
concentration of substrate required for the reaction
to reach maximum velocity.
• b) Km, the Michaelis constant, is defined as the
dissociation constant of the enzyme-substrate
complex.
• c) Km, the Michaelis constant, is expressed in
terms of the reaction velocity.
• d) Km, the Michaelis constant, is a measure of the
affinity the enzyme has for its substrate.

25. • Which of the following statements about the
mechanism of allosteric control of enzyme activity is
correct?
• a) Allosteric enzymes are typically single-subunit
enzymes.
• b) Allosteric enzymes show greater sensitivity to
changes in substrate concentration compared to
classical type enzymes with hyperbolic kinetics.
• c) Allosteric enzymes show Michaelis-Menten kinetics.
• d) Allosteric enzymes show reduced sensitivity to
changes in substrate concentration compared to
classical type enzymes with hyperbolic kinetic

26. • Which of the following statements about Michaelis-
Menten kinetics are correct? Please select all that
apply.
• a) A high Michaelis constant (Km) indicates a high
affinity of an enzyme for its substrate.
• b) A low Michaelis constant (Km) indicates a high
affinity of an enzyme for its substrate.
• c) The Michaelis constant (Km) of an enzyme
increases when the enzyme concentration is
increased.
• d) The Michaelis constant (Km) of an enzyme is
unchanged when the enzyme concentration is
increased.

28. • The turnover number and specific activity of
an enzyme ( molecular weight 40,000 D) in a
reaction (Vmax= 4 mol of substrate reacted/
min, enzyme amount = 2 g) are
• 1. 80,000/min, mol substrate/min
• 2. 80,000/min, mol substrate/second
• 3. 40,000/min, mol substrate/min
• 4. 40,000/min, mol substrate/min

29. • 23. Enzymes accelerate a reaction by which
one of the following strategies?
• 1. Decreasing energy required to form the
transition state.
• 2. Increasing kinetic energy of the substrate.
• 3. Increasing the free energy difference
between substrate and the product.
• 4. Increasing the turn over number of enzymes.

30. SPECIFIC ACITIVITY
• TON: turn over number: is the number of
moles of substrate transformed per minute
per mole of enzyme under optimum
condition.

31. • Specific activity[edit]
• The specific activity of an enzyme is another common unit. This is the activity of an enzyme
per milligram of total protein (expressed in μmol min−1mg−1). Specific activity gives a
measurement of enzyme purity in the mixture. It is the moles of product formed by
an enzyme in a given amount of time (minutes) under given conditions per milligram of total
proteins. Specific activity is equal to the rate of reaction multiplied by the volume of reaction
divided by the mass of total protein. The SI unit is katal kg−1, but a more practical unit is
μmol mg−1 min−1. Specific activity is a measure of enzyme processivity, at a specific
(usually saturating) substrate concentration, and is usually constant for a pure enzyme. For
elimination of errors arising from differences in cultivation batches and/or misfolded
enzyme etc. an active site titration needs to be done. This is a measure of the amount of
active enzyme, calculated by e.g. titrating the amount of active sites present by employing
an irreversible inhibitor. The specific activity should then be expressed as μmol
min−1 mg−1 active enzyme. If the molecular weight of the enzyme is known, the turnover
number, or μmol product per second per μmol of active enzyme, can be calculated from the
specific activity. The turnover number can be visualized as the number of times each
enzyme molecule carries out its catalytic cycle per second.

32. • Q. One microgram of a pure enzyme
(MW=92,000) catalyzed a rection at a rate of
0.50micromole /min under optimum condition .
Calculate a . Specific activity of the enzyme in
terms of units/mg protein and units /moles and
TON of the enzyme ?
• ANS: SA = Vmax/mg= 0.5micromole/mim
• =500units/mg protein
• 10^-3

33. • Specific enzyme activity
(usually stated simply as 'specific activity')
is the number of enzyme units per ml
divided by the concentration of protein in
mg/ml. Specific activity values are
therefore quoted as units/mg or
nmol/min/mg (if unit definition B
isapplied).

35. Determination of vmx ,km

36. Double reciprocal plot :

37. ENZYME INHIBITION
• Inhibition are 3 type
1. Competetive :.
2. Non competetive :
3. Uncompetetive :

38. Diagram of the inhibition

39. COMPETITIVE INHIBITION
• Definition: The structure of a competitive
inhibitor is closely resembles that of the
enzyme normal substrate.
• Because of its structure , a competitive
inhibitor binds reversibly to the enzyme active
site.
• In competitive inhibition : V0= decrease,
Vmax= remain same , Km = Increase.

40. Competitive inhibitor graph

41. EXAMPLE OF COMPETITIVE
INHIBITION
• An example of competitiveinhibition cou
ld be malonic acid which competes with
succinate for active sites of succinic
dehydrogenase, an important enzyme in
the Krebs cycle.

43. Example of competitive inhibition

46. NON COMPETITIVE INHIBITON
• Definition: In non competetive inhibition , the
inhibitor binds to the enzyme at a site other
than the active site.
• Inhibitor binding alter the enzyme 3D
configuration.
• In non competitive inhibition
• V0= decrease Vmax= decrease Km = remain
same.

49. Example of Non competitive inhibition

50. NON –COMPETITIVE INHIBITION

52. UNCOMPETITIVE INHIBITION
• Definition: Inhibitor binds at a distance site
from the substrate . However,an
uncompetitive inhibitor will bind only to the
ES complex.
• In uncompetitive inhibition
• V0 = decrease Vmax =decrease Km =Decrease

55. Graph for Uncompetitive inhibition

56. EXAMPLE OF UNCOMPETITIVE
INHIBITION