Biological Chemistry Laboratory
Biology 3515/Chemistry 3515
Spring 2016
Lecture 6:
A Bit on Quiz 1 and Enzyme Kinetics II
c
David
P. Goldenberg, 2003
University of Utah
goldenberg@biology.utah.edu
From Quiz 1
Two possible methods for measuring hemoglobin in a human blood sample:
1
Direct measurement of absorbance at 420 nm, due to heme.
420 = 5×105 M−1 cm−1
2
Bradford dye-binding assay.
Requires calibration with samples of protein with known concentration.
Clicker Question # 1
Which method is better for this application?
1
Absorbance at 420 nm
2
Bradford assay
No wrong answers (for now).
Back to Enzyme Kinetics
E + S −−−→ E + P
E: Enzyme, S: Substrate (reactant), P: Product
Velocity (M/min)
Velocity (M/min)
The key observations:
[E] (M)
[S] (M)
Michaelis and Menten
Leonor Michaelis (1875–1949)
Maude Leonora Menten (1879–1960)
Michaelis, L. & Menten, M. L. (1913). Die kinetik der invertinwirkung. Biochem. Z., 49, 333–369.
Johnson, K. A. & Goody, R. S. (2011). The original Michaelis constant: Translation of the 1913 Michaelis-Menten
paper. Biochemistry , 50, 8264–8269. http://dx.doi.org/10.1021/bi201284u
A Simple Mechanism Accounts for Hyperbolic Kinetics
k
k
1
−−
−*
E+S−
ES −−cat
−→ E + P
)
−
−
k
2
V =
[S]Vmax
Km + [S]
The Individual Steps and Rate Constants
k
E + S −−−1→ ES
A second-order reaction
In the absence of other reactions:
d[ES]
dt
= k1 [E][S]
−1
The second-order rate constant, k1 has units of s M
−1
k
2
E + S ←−−
− ES
A first-order reaction
In the absence of other reactions:
d[ES]
dt
−1
or min M
= −k2 [ES]
The first-order rate constant, k2 has units of s−1 or min−1
−1
The Individual Steps and Rate Constants
k
cat
ES −−−
→E+P
A first-order reaction
In the absence of other reactions:
d[P]
dt
= kcat [ES]
The first-order rate constant, kcat has units of s−1 or min−1
The rate constants, k1 , k2 and kcat are konstant!
They do not change with concentration.
The fine print: This is an assumption that may break down at extreme concentrations.
The values of the rate constants generally depend on other factors, such
as temperature or other solution conditions.
Concentration (M)
Approaching Steady State
after Mixing Substrate and Enzyme
S
k
k
1
cat
−−
−*
E+S−
)
−− ES −−−→ E + P
k2
d[ES]
P
E.S
dt
= k1 [E][S] − k2 [ES] − kcat [ES]
Time (min)
At steady state:
d[ES]
dt
= k1 [E][S] − k2 [ES] − kcat [ES] = 0
Solving this equation gives the Michaelis-Menten equation.
The Michaelis-Menten Mechanism and Equation
k1
k
cat
−−
−*
E+S−
)
−− ES −−−→ E + P
k2
V =
[S]Vmax
Km + [S]
where:
Vmax = kcat ([E] + [E · S]) = kcat [E]T
Units : min−1 · M = M/min
and:
Km =
k2 + kcat
k1
min−1 + min−1
Units :
=M
min−1 M−1
[S] Km
V =
V ≈
[S]Vmax
Km + [S]
[S]Vmax
[S]
Velocity (M/min)
When [S] Km :
= Vmax
[S] (M)
[S] = Km
V =
V ≈
[S]Vmax
Km + [S]
[S]Vmax
Velocity (M/min)
When [S] Km :
Km
[S] (M)
[S] = Km
V =
V =
[S]Vmax
Km + [S]
[S]Vmax
2[S]
=
1
V
2 max
Velocity (M/min)
When [S] equals Km :
[S] (M)
Km is not equal to 1/2 Vmax !
The equation:
V =
[S]Vmax
Km + [S]
The units:
• Km has units of concentration, like [S].
•
Vmax has units of reaction velocity, e.g., M/min, like V .
Km equals the substrate concentration at which V = Vmax /2.
Velocity
Analysis of Data from the V versus [S] Experiment
Substrate Concentration
We want to fit the experimental data to the Michaelis-Menten Equation:
V =
[S]Vmax
[S] + Km
From the fit, we obtain estimates of Km and Vmax .
Clicker Question #2:
Estimate Vmax from the graph:
40
1
30 nM/s
2
40 nM/s
3
50 nM/s
4
80 nM/s
20
∗
0
0
2
4
6
∗
close enough for credit.
Clicker Question #3:
Estimate Km from the graph:
40
20
0
0
2
4
6
1
1 µM
2
2 µM
3
5 µM
4
10 µM
A Classic Method for Analyzing Enzyme Kinetics Data
Rearrangement of the Michaelis-Menten Equation:
V =
[S]Vmax
[S] + Km
[S] + Km
[S]
Km
1
=
=
+
V
[S]Vmax
[S]Vmax
[S]Vmax
Km
1
1
1
=
·
+
V
Vmax
[S] Vmax
A plot of 1/V versus 1/[S] should generate a straight line with a slope of Km /Vmax
and an intercept of 1/Vmax on the 1/V axis.
The Lineweaver-Burk Plot
0
0
If the data are perfect, this plot gives good estimates of Km and Vmax .
But, experimental error in V can lead to strange effects!
Experimental Error and Uncertainty
Error bars for rate measurements are of approximately constant size (e.g., ±0.005
A/min), rather than a constant percentage of the measurement.
Rate (A/min)
0.1
0.08
0.06
0.04
0.02
0
Enzyme Concentration
For 0.1 A/min, ±0.005 A/min = ±5%.
For 0.01 A/min, ±0.005 A/min = ±50%.
Least-squares fitting works well if the absolute uncertainties of all data points are
approximately equal.
What Happens When We Take Reciprocals?
V = 0.1 ± 0.005
1
= 9.52,
0.105
1
= 10.5,
0.095
1
= 10 ± 0.5
V
1
= 200,
0.005
1
= 100 ± 50
V
V = 0.01 ± 0.005
1
= 66.7,
0.015
The values of 1/V derived from small velocities can have huge absolute errors.
The Effects on a Lineweaver-Burk Plot
Errors in the least precise measurements (low V ) can cause large changes in the line
fit to the Lineweaver-Burk plot.
Clicker Question #4:
Which parameter is likely to be more sensitive to errors in a Lineweaver-Burk plot?
1
Km
2
Vmax