The document discusses various concepts of enzyme kinetics, including reaction orders, Michaelis constant (Km), turnover number (kcat), and steady-state approximation. It also outlines the conditions under which Km equals the rate constant Ks, as well as the effects of different factors on enzyme activity such as substrate concentration and inhibitors. Moreover, the document features problem-solving examples related to enzyme activity and kinetics, ultimately aiding in the understanding of enzyme catalysis.

1. ENZYME TECHNOLOGY –
SOLVED PROBLEMS
Dr.Anant Achary and Dr.S.Karthikumar
Kamaraj College of Engineering and Technology
S.P.G.C.Nagar, K.Vellakulam, Madurai Dist. TN, INDIA
skarthikumar@gmail.com

2. What is meant by the order of a reaction?
Q.1

3. • Reactions can be independent of the concentration of substrate (0-
order), directly dependent on substrate (1st-order) or dependent on
substrate concentration raised to some higher power (2, etc.).
A.1

4. Define Km
Q.2

5. Km
• Km = Michaelis constant. It is the substrate concentration that gives
1/2 Vmax. That is, it is the concentration of substrate at which half
the active sites are filled. Km is also related to rate constants of the
individual steps in the reaction.
A.2

6. Define Ks
Q.3

7. A.3

8. Define kcat
Q.4

9. • kcat = the turnover number and is equal to k2 in the Michaelis
Menten mechanism.
• The turnover number of an enzyme is equal to the number of moles
of substrate converted to product per minute per mole of enzyme
present when the enzyme is fully complexed with substrate.
A.4

10. • Km is all too frequently equated with Ks. In fact, in most reactions
there is an appreciable disparity between the values for Km and Ks.
For the reaction A  B define conditions under which Km = Ks.
Describe conditions under which this is not true
Q.5

11. A.5

12. • What is the steady-state approximation, and under what conditions is
it valid?
Q.6

13. The steady state approximation assumes that the concentrations of the
intermediates in a reaction do not change while the rate of product
formation is being measured. This holds for the early stages of a
reaction, after the ES complex has formed and before appreciable
changes have occurred in either the substrate or product
concentrations.
A.6

14. True or False
a. At saturating levels of substrate, the rate of an enzyme catalysed
reaction is proportional to the enzyme concentration.
b. The Michaelis constant Km equals the substrate concentration at which
v = 1/2 Vmax.
c. The Km for a regulatory enzyme varies with enzyme concentration.
d. If enough substrate is added, the normal Vmax of an enzyme catalysed
reaction can be attained even in the presence of a noncompetitive
inhibitor.
Q.7

15. e. The Km of some enzymes may be altered by the presence of
metabolites structurally unrelated to the substrate.
f. The rate of an enzyme-catalyzed reaction in the presence of a rate-
limiting concentration of substrate decreases with time.
g. The sigmoidal shape of the v versus (S) curve for some regulatory
enzymes indicates that the affinity of the enzyme for substrate
decreases as the substrate concentration is increased.
Q.7 Cont.

16. • a. True. Vmax = k2[E]t
• b. True
• c. False. The value of Km is independent of enzyme concentration for almost all enzymes
• d. False. A non-competitive inhibitor cannot be overcome by substrate concentration.
• e. True. This occurs in regulatory enzymes.
• f. True.
• g. False. The initial increasing slope of the curve shows that binding of the first substrate
molecule increases the affinity of the enzyme for subsequent substrate molecules.
A.7

17. 1. The _____________ of a reaction is the numerical relationship between
substrates and products
2. The rate constant ____________ of an enzyme-catalysed reaction is a measure
of the catalytic efficiency at saturating
levels of substrate.
3. _______________ inhibitors do not alter the Vmax of an enzyme-catalyzed
reaction.
4. The sigmoidal shape of the v versus [S] curve for some regulatory enzymes
results from a _______________ effect of substrate on the substrate binding sites.
5. For an enzyme whose Km can be regulated, the presence of a _____________
effector increases the level of substrate required to attain a given reaction rate.
Q.8

18. • 1. Stoichiometry
• 2. kcat
• 3. Competitive
• 4. homotropic
• 5. negative
A.8

19. Assume that an enzyme catalyzed reaction follows Michaelis Menten
kinetics with a Km of 1 x 10-6 M. If the initial reaction rate is 0.1
μmol/min at 0.1 M, what would it be at 0.01 M, 10-3 M, and 10-6 M?
Q.9

20. A.9

21. A more general form of an equation for an enzyme catalyzed
reaction is:
Consider the essentially irreversible reaction represented by the free energy
diagram below.
A. Using the letters indicated in the diagram, relate each of the rate
constants in the Equation above to the energy-level difference that
determines it.
B. Which rate constant limits the rate of formation of product?
C. Does Km approximately equal Ks for this enzyme?
Q.10

22. A. The rate constant for each step is inversely related to the difference between the energy level of
the reactants and the highest energy barrier between the reactants and the products of that step. In
terms of the letters in the Figure,
k1 is determined by b-a
k-1 is determined by b-c
k2 is determined by d-c
k-2 is determined by d-e
k3 is determined by f-e, and
k-3 is determined by f-g.
B. k2 corresponds to a much higher energy barrier than the other forward rate constants and
therefore must limit the rate of product formation.
c. Since k2 is small relative to k-1 , Km approximates Ks for this enzyme. Therefore, Km is a measure
of affinity for substrate
A.10

23. To study the dependence of the rate of an enzyme-
catalyzed reaction on the substrate concentration, a
constant amount of enzyme is added to a series of
reaction mixtures containing different
concentrations of substrate (usually expressed in
mol/L).
The initial reaction rates are determined by
measuring the number of moles (or μmoles) of
substrate consumed (or product produced) per
minute. Consider such an experiment in which the
initial rates in Table were obtained at the indicated
substrate concentrations.
Initial rates at various substrate concentrations for a
hypothetical enzyme-catalyzed reaction
a. What is Vmax for this reaction?
b. Why is v constant above substrate concentrations of 2.0 x
10-3 M?
c. What is the concentration of free enzyme at 2.0 x 10-2 M
substrate concentration?
Q.11

24. a. Vmax = 60μmol/min
b. v is constant because it has reached Vmax; the enzyme is saturated
with substrate.
c. The concentration of free enzyme is negligible because all of the
enzyme is in the ES form.
A.11

25. To study the dependence of the rate of an enzyme-catalyzed
reaction on the substrate concentration, a constant amount of
enzyme is added to a series of reaction mixtures containing
different concentrations of substrate (usually expressed in mol/L).
Total reaction mixture volume is 10mL
The initial reaction rates are determined by measuring the
number of moles (or μmoles) of substrate consumed (or product
produced) per minute. Consider such an experiment in which the
initial rates in Table were obtained at the indicated substrate
concentrations.
A. What is Vmax for this concentration of enzyme?
B What is the Km of this enzyme?
C. Show that this reaction does or does not follow simple
Michaelis-Menten kinetics.
D. What are the initial rates at [S] = 1.0 x 10-6 M and at
[S] = 1.0 x 10-1 M?
E. Calculate the total amount of product made during the first
five minutes when [S] = 2.0 x 10-3 M. Could you make the
same calculation at [S] = 2.0 X 10-6 M?
F. Suppose that the enzyme concentration in each reaction
mixture were increased by a factor of 4. What would be the
value of Km? of Vmax? What would be the value of v at [S]
= 5.0 x 10-6 M?
Q.12

26. A. Vmax = 0.25 μmol/min
B. For a reaction obeying Michaelis-Menten kinetics, Vmax and Km are
simply constants relating v to [S]. Km can be calculated by substituting
Vmax and any pair of v and [S] values at v < Vmax. For example, at [S] =
5.0 x 10-5 M and v = 0.20 μmol/min the equation becomes
A.12

27. C. If the reaction follows simple Michaelis-Menten kinetics, then the
Michaelis-Menton equation should relate v to [S] over a wide range of [S].
This can be tested by determining whether the equation yields the same
value of Km at several different values of [S] and v < Vmax. Under the
conditions of this problem, the same value, Km = 1.3 x 10-5 M, is obtained at
[S] = 5.0 x 10-6 M, v = 0.071 μmol/min and at [S] = 5.0 x 10-7 M, v= 0.0096
μmol/min. Therefore, Michaelis-Menten kinetics are obeyed
A.12. Cont.

28. • D.
A.12 Cont.

29. E. At [S] = 2.0 x 10-3 M, v = Vmax = 0.25 μmol/min. Since 0.25 μmole is much less than the amount
of substrate present (2.0 x 10-3 mole/liter x 10-2 L x 106 μmol/mol = 20 μmol) the reaction can
proceed for five minutes without significantly changing the substrate concentration.
Thus, 0.25 μmol/min x 5 min = 1.25 μmol
At [S] = 2.0 x 10-6 M,
During 5 minutes at this rate, 0.033 μmol/min x 5 min = 0.17 μmol of product would be produced.
However, this value exceeds the total amount of substrate present (2.0 x 10-6 mol/L x 10-2 L x 106
μmol/mol = 0.020 μmol). Clearly, during the-5 min reaction, [S] and therefore v would decrease
significantly. Calculation of the exact amount of product made would require integration of a
differential equation; this amount obviously cannot exceed 0.020 μmole
A.12 Cont.

30. F. Km is independent of enzyme concentration, since a change in [E]
does not affect the three rate constants, k1, k2, and k3.
Hence Km would remain equal to 1.25 x 10-5 M.
Since Vmax = k3[E]o, increasing the enzyme concentration by a factor
of 4 increases Vmax by a factor of 4. Therefore, Vmax = 1.0 μmole/min.
At [S] = 5.0 x 10-6 M,
A.12 Cont.

31. The Km of a certain enzyme is 1.0 X 10-5 M in a reaction that is described by Michaelis-Menten kinetics. At a
substrate concentration of 0.10 M, the initial rate of the reaction is 37 μmol/min for a certain concentration of
enzyme. However, you observe that at a lower substrate concentration of 0.010 M the initial reaction rate remains 37
μmoles/min.
• a. Using numerical calculations, show why this tenfold reduction in substrate concentration does not alter the
initial reaction rate.
• b. Calculate v as a fraction of Vmax for [S] = 0.20 Km, 0.50 Km, 1.0 Km, 2.0 Km, 4.0 Km, and 10 Km.
• c. From the results in (b), sketch the curve relating v/Vmax to [S]/Km. What is the best range of [S] to use in
determining Km or investigating the dependence of v on [S]?
Q.13

32. a. Since both substrate concentrations are well above Km, you
can assume that Vmax = 37 μmol/min. Then
Therefore, at [S] 1.0 x 10-2 M, v still is equal to Vmax.
A.13

33. b. From the Michaelis-Menten
equation, the following
relationships can be calculated:
A.13 Cont.

34. c. When you plot these values you will be able to see that the best
range of [S] for studying the dependence of v on [S] is in the
neighborhood of Km or below it, since changes in [S] below Km cause
greater changes in v than do changes in [S] above Km. Therefore, when
using graphic methods to determine Km and Vmax, several
measurements should be made at [S] well below Km.
A.13 Cont.

35. The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic
reactions such as the synthesis of DNA. This hydrolytic reaction is catalyzed in E. coli by a
pyrophosphatase that has a mass of 120 kDa and consists of six identical subunits.
Purified enzyme has a Vmax of 2800 units per milligram of enzyme. For this enzyme, a unit of
activity is
defined as the amount of enzyme that hydrolyzes 10 μmol of pyrophosphate in 15 minutes at 37°C
under standard assay conditions.
A. How many moles of substrate are hydrolyzed per second per milligram of enzyme when the
substrate concentration is much greater than Km?
B. How many moles of active site are there in 1 mg of enzyme? Assume that each subunit has one
active site.
C. What is the turnover number of the enzyme?
Q.14

36. a.1 'unit' here = 10 μmole/15 min at 37ºC
= 10/15 μmole min-1
= 10/15.60 μmole s-1
When [S] >> Km,
Vo = Vmax
Here Vmax = 2800 units mg-1
= 2800 x 10/15.60
= 31.3 μmol s-1 mg-1.
NOTE. THIS IS BY DEFINITION THE SP. ACT.
A.14

37. b. 1mg of enzyme = 10-3/MW moles = 103/MW μmoles
= 103/120000
= 1/120 μmoles.
But each mole of enzyme has 6 active sites. Therefore 1 mg of enzyme
=1/120 x 6
= 1/20
= 0.05 μmoles active site.
A.14 Conti.

38. c. The units of sp. act. are, μmoles (unit time)-1 (mg enzyme)-1. Here μmoles s-1
mg-1 is used, and the value is 31.3 μmole s- 1 mg-1.
The turnover number, kcat, is effectively the activity in terms of (μmole active site)-
1, instead of (mg enzyme)-1 as in sp. act., and its units are,
μmoles s-1 (μmole active site)-1 = s-1.
We know from '(b)' that there are 0.05 μmoles active site per
mg of enzyme, i.e.,
μmole of active site = 1/0.05
= 20 mg of enzyme, and that the sp. act. (activity per mg of enzyme)
= 31.3 μmoles s-1 mg-1.
Therefore the kcat (activity per mmole of active site, i.e. its activity per 20 mg of
enzyme) is = 31.3 x 20
= 626 s-1.
A.14 Conti.

39. In the cases of severe liver damage, an enzyme EL
is released into the blood. After severe exercise,
an isozyme from muscle, EM, is found in the
blood. EL and EM can be differentiated since they
have different kinetic constants. The Km of the
liver enzyme is 3 x 10-4 M; the Km of the muscle
enzyme is 7 x 10-5 M.
Data from assays on an unconscious patient's
blood are given below. Ten microliters of blood
was used in each assay.
a. Is the patient likely to be suffering from a liver disease or had
she been exercising too strenuously?
b. Explain your reasoning for your answer to part a.
Q.15

40. • a. Liver
• b. The [S]Vmax/2 is close to the Km of the liver enzyme
A.15

41. Define the following
A. a unit of enzyme activity
B. steady-state conditions
c. oligomeric enzyme
Q.16

42. A. The amount of enzyme which converts 1 μmole of substrate to
product per min (at 25/37oC)
B. Under steady state conditions, S is converted into P at a constant
rate, the [S] and [P] vs time plots are linear, and [ES] is Constant
C. An enzyme with a quaternary structure, i.e., made up of more than
one subunit (protomer, monomer). The subunits may be the same
(e.g., a homodimer), or different (e.g., a heterodimer).
A.16

43. Under what experimental conditions does an enzyme-catalysed
reaction follow zero-order kinetics?
Q.17

44. • Saturating substrate ([S] >> Km)
A.17

45. • Indicate the effects of substrate concentration, enzyme
concentration, temperature, inhibitors or activators on enzyme
activity by labeling correctly both axes of the graphs given shown
below
Your choice for axes are: Energy, [E], Temperature, [S], 1/[S], 1/v, v.
(You may use the same label on more than one graph.)
Q.18

46. A.18

47. An enzyme-catalyzed reaction was assayed at several substrate
concentrations. Two data points which fell on the Lineweaver- Burk plot
are v = 41.7 μmol S/min when [S] = 5 x 10-4 M and v = 16.7 μmol S/min
when [S] = 5 x 10-6 M. Place the two points on a line on the
accompanying graph.
A. Determine the value of Km and Vmax in the correct units. When an
inhibitor was added, the velocities fell to 1/2 their uninhibited values.
B. Plot the inhibited line on the graph.
C. Is the inhibitor competitive or non-competitive? On what evidence
did you base your decision?
Q.19

48. • a. Km = 8.3 x 10-6 M, Vmax = 45.5 μmoles min-1
• c. Non-competitive (Km same, Vmax smaller)
A.19

49. Fumarase (L-malate hydrolase) catalyzes the reversible hydration of fumarate to L-malate. The fumarase from pig heart has been
crystallized (M.W. 197,000 Da) and consists of four subunits that can be dissociated into inactive monomers under relatively mild
conditions. Substrate can induce reformation of tetramers with complete recovery of the activity. The subunits have a molecular
weight of 48,500 Da and each contains three free –SH groups. Fumarase requires no cofactors. Kinetic studies implicate the
participation of a pair of groups on the enzyme (one acidic, one basic) with pKa's of 6.2 and 6.8. These groups have been postulated
to be two imidazole groups of histidine residues, one in the imidazole form and one in the imidazolium form. The reverse reaction is
stereospecific for L-malate and only Lmalate is produced from fumarate.
A. For fumarase, an intact ___________ structure is necessary.
B. Predict graphically the effect of pH on the activity of fumarase. At what pH
would you expect maximal enzyme activity?
C. Malonate, -OOC-CH2-COO- is an inhibitor of fumarase. What type of
inhibitor would you expect malonate to be?
D. Using a Lineweaver-Burk plot, graphically illustrate the effect of malonate
on the kinetics of fumarase. Be sure to label all parts of the graph as well as
the inhibitor data-
Q.20

50. A. Quaternary (oligomeric)
B. pH 6.5
C. Competes with fumarate because of similar structure, therefore
a competitive inhibitor
A.20

51. A. What is meant by an allosteric enzyme?
B. What is the difference between the active site and the regulatory
site of an allosteric enzyme?
C. How does an allosteric inhibitor produce its effect on an enzyme?
Q.21

52. A. They are enzymes whose kinetic properties cannot be accounted for
by the Michaelis-Menton model.
B. The active site is where the substrate binds to the regulatory
enzyme; the regulatory site is where the effector molecules bind. These
two sites are different.
C. An allosteric inhibitor typically binds and stabilizes the enzyme in an
inactive or less active (conformation) state.
A.21

53. A. The activation energy for a non enzyme-catalyzed reaction is
_____________ than the activation energy for the same reaction
catalyzed by an enzyme.
B. Why would increasing the temperature of an enzyme-catalysed
reaction from 25°C to 37°C increase the rate of the reaction?
C. Why would increasing the temperature to 60°C probably cause a
decrease in the rate of the enzyme-catalyzed reaction?
Q.22

54. A. Greater
B. Yes
C. Yes
A.22

55. A. If the concentration of substrate is 10-3 M and the concentration of
enzyme is 10-8 M, what would be the effect of the observed rate of
production of product if the enzyme concentration were doubled?
(Assume Km is 10-6 M.)
B. What would be the effect on the observed rate if the substrate
concentration were doubled to 2 x 10-3 M but the enzyme
concentration remained at 10-8 M?
Q.23

56. A. [S] >> Km
Therefore, vo = Vmax Therefore, the velocity is doubled when [E]tot is
doubled
B. None, because the enzyme is already saturated with substrate
A.23

57. Answer the following with true or false; justify your answer in each
case.
A. The initial rate of an enzyme-catalyzed reaction is independent of
substrate concentration.
B. If enough substrate is added, the normal Vmax of an enzyme
catalyzed reaction can be attained even in the presence of a
noncompetitive inhibitor.
C. The rate of an enzyme-catalyzed reaction in the presence of a rate-
limiting concentration of substrate decreases with time.
D. The sigmoid shape of the v-versus-[S] curve for some regulatory
enzymes indicates that the affinity of the enzyme for substrate
decreases as [S] is increased
Q.24

58. A. F
B. F
C. T
D. F (positive Km-type)
A.24

59. Two forms of isocitric dehydrogenase exist in mammals. One which is NAD+ specific and found
only in mitochondria and a NADP+ specific enzyme found in both cytosol and mitochondria
NADP+ specific isocitric dehydrogenase catalyzes the decarboxylation by formation of an
unstable enzyme bound chelate of Mn2+ and a a-keto acid intermediate. The free energy change
for formation of aketo glutaric acid under physiological conditions is: DG° = -5 kcal/mol. AMP
regulates the enzymatic activity by reducing Km for isocitrate by 10 fold. Only {isocitrate2-} was
found to be the substrate form that binds to the enzyme.
Q.25

60. A. Mn2+ is an example of a(an)
B. NAD+ is an example of a(an); The reaction velocity was found to be 4th order with respect to isocitric acid indicating
high cooperativity.
C. The fact that "high cooperativity "is found indicates that this enzyme is a(an) ______________ enzyme. The NAD+
specific enzyme has a molecular weight of 330,000 Daltons and is made up of 8 identical subunits.
D. The NAD+ specific enzyme is an example of a(an) ______________ protein and the level of structural organization for
the 8 identical subunits is termed the structure of the protein.
E. AMP is said to act as a(an) Reaction velocity is decreased in presence of ATP which acts by binding at the same site
that NAD+ binds.
F. ATP is said to act as a(an) ____________
G. Production of NADH during the course of the reaction would be expected to ____________ the reaction velocity and
NADH would be an example of a(an) ____________.
H. What is the significance of DGo = -5 kcal/mole in terms of:
1. The desire of the reaction to go as written? (One sentence answer).
2. The activation energy of the reaction?
I. If isocitrate ionization was the only important controlling factor for the pH-activity profile of the reaction. (See
characteristics at the beginning of this exam). Using the information given, construct an accurate pH-activity profile for
the enzyme catalyzed reaction. Use graph paper.
Q.25 cont.

61. A. Cofactor
B. Coenzyme
C. Allosteric (positive Km-type)
D. Oligomeric
E. Quaternary
F. +ve heterotropic effector
G. -ve heterotropic effector
H. Equal, coenzyme
I.
1. If NAD+/NADP+ and NADH /NADPH and isocitrate and a-ketoglutarate all at 1 M concentration are
mixed at 1 atm pressure, since the free energy change under standard conditions for the forward (L R)
reaction is negative, NAD+/NADP+ and isocitrate will be converted into NADH/NADPH and a-glutaric acid.
(Also, at equilibrium, the ratio of products to reactants (Keq), will be greater than
1, i.e., the equilibrium lies to the right.)
2. The standard free energy change in the L R direction is the difference between the standard free
energies of activation in the forward and reverse directions (DG0’ = DG*for - DG*rev)
A.25

62. Zero order kinetics in an enzyme catalyzed reaction only occurs when
we have:
a. a high specific activity
b. an isozyme present
c. high substrate concentration
d. a high Km
e. high enzyme concentration
f. a high turnover number
Q.26

63. • Answer: C
A.26

64. The redox pairs NAD+ NADH and NADP+ NADPH are well suited for use
in coupled clinical enzyme assay systems:
a. because of their acid-base properties.
b. because of their distinctly different absorbance properties of their
oxidized and reduced species.
c. because of their occurrence in cells.
d. because of their ability to take the place of enzyme reactions.
e. because they are coenzymes.
Q.27

65. • Answer : b
A.27

66. The success of a measurement of an enzyme activity using a coupled
enzyme assay depends on having
a. The NAD+ NADH dehydrogenase reaction.
b. the second reaction as non-limiting.
c. A non-buffered reaction medium.
d. a colorimeter adequate in the visible region of the spectrum.
e. a trained M.D. to supervise.
Q.28

67. • Answer : b
A.28

68. In measuring the rate of a coupled reaction one must know:
a. how to control the humidity in the sample chamber.
b. where the lag phase ends.
c. the maximum absorbance of the unknown enzyme.
d. the rate of absorbance change during the preincubation period.
e. the total absorbance change throughout the assay.
Q.29

69. • Answer : b
A.29

70. The levels of LDH isozymes in the blood are indicative of certain disease states: Some of the
isozymes present will react with a particular substrate while others will not. Thus, one can
experimentally measure the activity of these particular isozymes while other isozymes are also
present in the sample. In particular, this assay makes particular use of:
a. the colligative properties of the solution.
b. the substrate specificity of the isozyme of interest.
c. the preincubation phase of the isozyme of interest.
d. the tertiary structure of the isozymes of interest.
e. the total activity of all enzymes species.
Q.30

71. • Answer : b
A.30

72. Let's say the following assay was established to measure E1 levels in
serum.
a. List the solution conditions you would have to control to make this a valid assay.
b. List the species to equation 1 and 2 whose concentrations you would have to manipulate in order to
make a valid assay.
c. One of your technicians ran the assay under conditions which, with normal serum levels of E1, the
Vmax for equation
(1) turned out to be about equal to Vmax for equation
(2). State what is wrong with the assay.
d. In one sentence state what you would do to rectify the problem.
Q.31

73. a. The standard conditions of assay for E1 must be met (pH, ionic strength, cofactors,
temperature, etc), so the results may be compared to the range of normal values.
b. [A] must be at a sufficiently high concentration to saturate E1. [NAD+] must be at a
sufficiently high concentration to saturate E2. E2 must be present at sufficiently high levels
of activity for it not to be a limiting factor in the assay
c. The rate of the second reaction must be much greater than that of the first reaction (E2
at high activity), so that only the amount of E1 limits the rate of the reaction, i.e., Vmax2
must be >> Vmax1.
d. Increase the activity of E2 present in the assay medium, so that it is much greater than
that expected for E1.
A.31

74. Clinical data for enzymes are often reported in terms of international
units. What is an enzyme unit of activity and what relationship does it
bear to specific (enzyme) activity?
Q.32

75. The IU is the amount of enzyme converting 1 μmole of S into P per min
(at a given temperature, pH etc). The specific activity = IU per mg total
protein (it increases as the enzyme is purified)
A.32

76. What is meant by the term "zero order" reaction as it pertains to
enzymes
Q.33

77. When the enzyme is saturated with substrate,
v0 = Vmax = kcat Et = a constant for that assay,
i.e., the rate = a constant = k, therefore it is not dependent on [S], and
zero order kinetics in [S} are observed.
A.33

78. Why is the preincubation phase of a coupled enzyme assay necessary?
Q.34

79. To take into account:
Breakdown of NADH/NADPH due to
a. Its intrinsic chemical instability
b. The effect of other factors in our serum sample on NADH levels
besides the enzyme we want to measure.
A.34

80. Which one of the following
relationships best describes how
reaction velocity can be used to
indicate the level of enzyme in
blood serum as developed for
clinical labs?
Q.35

81. d. Vmax = k [Eo] (i.e., Vmax = kcat [E]tot)
A.35

82. THANK YOU