Enzyme kinetics

RIZWANABBAS3 23,623 views 18 slides Oct 22, 2017
Slide 1
Slide 1 of 18
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12
Slide 13
13
Slide 14
14
Slide 15
15
Slide 16
16
Slide 17
17
Slide 18
18

About This Presentation

Enzyme kinetics


Slide Content

Enzyme Kinetics
by Mahwish Kazmi

•Enzyme kinetics is the study of the mechanism of an
enzyme catalysed reation to determine the rate of the
reaction.
Study of enzyme kinetics is useful for measuring
•concentration of an enzyme in a mixture (by its catalytic
activity),
•its purity (specific activity),
•its catalytic efficiency and/or specificity for different
substrates
•comparison of different forms of the same enzyme in
different tissues or organisms,
•effects of inhibitors (which can give information about
catalytic mechanism, structure of active site, potential
therapeutic agents...)

Michaelis and Menton Equation
•A model for enzyme kinetics was propsed by Michaelis
and Menton in 1913.
•MM equation relates the initial rate of an enzyme
catalysed reaction to the substrate concentration and a
ratio of rate constants.
•It co-relates velocity with enzyme and substrate
concentration.
•It has been derived for a single substrate-enzyme-
catalysed reaction.

Michaelis and Menton Equation
•In MM expression
Total enzyme concentration= [E
T
] = [E] + [ES]
Free enzyme conc [E] = [E
T
] - [ES]
Substrate concentration = [S]
Initial velocity = V
o
, Velocity measured immediately after
mixing E + S, at beginning of reaction (initial velocity), is
called V
o
.
Maximum velocity = V
max

Half V
max
= K
m
(substrate concentration)
Km = substrate concentration that gives V
o
= 1/2 V
max
.

At very low [S]:
•V
o
is proportional to [S];
doubling [S] → double V
o
.
2. In mid-range of [S], V
o
is
increasing less as [S]
increases (where V
o
is
around 1/2 V
max
).
Km = [S] that gives V
o
= 1/2
V
max
.
3. At very high [S], V
o
is
independent of [S]:
V
o
= V
max
.
Enzyme-catalyzed reactions show a hyperbolic
dependence of V
o
on [S]

Derivation
•Initial velocity V
o
= k
2
[ES]
•Rate of formation of [ES] = k
1
[E][S]
= k
1([E
T]-[ES]).[S]
= k
1
[E
T
][S] - [ES][S]
•Rate of breakdown of [ES] =k
-1
[ES] + k
2
[ES]
= (k
-1
+k
2
)[ES]
•Steady state:
Rate of formation = Rate of breakdown
k
1
[E
T
][S] - [ES][S] = (k
-1
+k
2
)[ES]

•separation of rate constants
k
1
[E
T
][S] - [ES][S] = (k
-1
+k
2
)[ES]
k
1
[E
T
][S] = [ES][S] + (k
-1
+k
2
)[ES]
k
1
[E
T
][S] = {[S] + (k
-1
+k
2
)} [ES]
[ES] = k
1
[E
T
][S]
[S] + (k
-1
+k
2
)
[ES] = [E
T
][S]
[S] + k
-1
+k
2

k
1
k
-1
+k
2
= K
m
[ES] = [E
T
][S] ......... (i)
k
1
[S] + K
m

•Expressing Vo in term of [ES]:
multiply k2 on both side of eq (i)
k
2
. [ES] = k
2
[E
T
][S] ...... (ii)
[S] + K
m
As we know V
o
= k
2
[ES]
So, eq (ii) becomes
V
o
= k
2
[E
T
][S] .....(iii)
[S] + K
m
When [S] is greater, then V
o
becomes V
max
and
V
max
= k
2
[E
T
]
So, eq (iii) becomes
Vo = Vmax [S]
[S] + K
m

Michaelis-Menten equation
explains hyperbolic V
o
vs. [S]
curve:
1. At very low [S] ([S] << K
m
), V
o

approaches (V
max
/K
m
)[S]. V
max
and
K
m
are constants, so linear
relationship between V
o
and [S]
at low [S].
2. When [S] = K
m
, Vo = 1/2 V
max
3. At very high [S], ([S] >> K
m
), V
o

approaches V
max
(velocity
independent of [S])
Explaination of hyperbolic curve

Significance of MM equation
It describes
•kinetic behaviour of enzymes
•hyperbolic dependence of V
o
on [S]
•independance of number of steps involved
•different enzymes have different K
m
and V
max
.
•K
m and V
max may be influenced by pH, temperature.
•K
m
can be used as a relative measure of the affinity of the
enzyme for each substrate (smaller K
m
means higher
affinity)
•in a metabolic pathways, K
m
values may indicate the rate-
limiting step (highest K
m
means slowest step).
•V
max
is independent of [S] at saturation.

•Turnover number (k
cat
)
Number of substrate molecules converted into product
by one molecule of enzyme active site per unit time,
when enzyme is fully saturated with substrate.
•units of k
cat
is s
–1
Lysozyme: k
cat
= 0.5 s
–1
Catalase: k
cat
= 4 x 10
7
s
–1
•k
cat
/K
m
is the criterion of substrate specificity, catalytic
efficiency and "kinetic perfection”.
•units of kcat/Km = conc
–1
time
–1
.
•max. possible k
cat
/K
m
for an enzyme = ~ 10
8
–10
9
M
–1
sec
–1
.

Lineweaver-Burk plot
•A more convenient graphical representation of MM
equation
•It is a straight line plot, easier to evaluate than curves.
•Lineweaver-Burk plot is a double reciprocal plot obtained
by taking reciprocal of both sides of MM equation and
rearranging
1 = K
m
+ [S]
V [S] V
max
1 = K
m
1 + 1
V V
max
[S] V
max

•A plot of 1/V versus 1/[S] is a straight line having a slope
of K
max
/V
max
and an intercept of 1/V
max
on the y-axis

Inhibition
•Enzyme inhibition is one of the ways in which enzyme
activity is regulated experimentally and naturally.
•Most therapeutic drugs function by inhibition of a specific
enzyme.
•In body, some of the processes controlled by enzyme
inhibition are blood coagulation, blood clot dissolution,
complement activation, conective tissue turnover and
inflammatory reactions.
•It may be reversible or irreversible.

Reversible Inhibition
•It is further subdivided into competitive, noncompetitive
and uncompetitive types.
•In reversible inhibition, equilibrium exists between
inhibitor I and enzyme E as
E+I EI
•The eq. constant for the dissociation of EI complex,
called Ki is given by equation
Ki = [E][I]
[EI]
•Ki is the measure of affinity of the inhibitor for enzyme
simliar to Km.

Competitive inhibition
•The inhibitor is a structural analogue that competes with
the substrate for binding at active site.
•Because the inhibitor binds reversibly to the enzyme,so
when [S] far exceeds [I], the probability that an inhibitor
molecule will bind to the enzyme is minimized and the
reaction exhibits a normal V
max.

Noncompetitive inhibition
•Inhibitor does not usually bear any structural
resemblance to the subatrate and it binds to the enzyme
at a site distinct from the substrate binding site.
•No competition exists between inhibitor and substrate,
so inhibition cannot be overcome by increase of [S].
•Vmax is reduced by inhibitor but Km is unaffected
because the affinity of S for E is unchanged.

Uncompetitive Inhibition
•Inhibitor I combines with ES to form ESI complex
•It yields parallel line on double reciprocal plot and
intercepts on both x and y axes are altered by presence
of inhibitor.