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A biochemical experiment
Michaelis-Menten equation
160
V =
initial velocity
140
Vmax × C
K +C
120
100
V = initial velocity
80
C = concentration
60
Vmax = maximum velocity
0.0
0.2
0.4
0.6
0.8
1.0
K = rate constant
concentration
Linearize
V =
⇒
1
K +C
=
V
Vmax × C
=
⇒
Vmax × C
K +C
1
=
V
1
K
+
Vmax × C Vmax
1
Vmax
+
K
Vmax
1
×
C
Fit the line
Model:
0.020
1 / initial velocity
0.018
1
= β0 + β1
V
0.016
0.014
0.012
Intercept
Slope
0.010
1
+ error
C
0.00697
0.00022
0.008
0.006
0
10
20
30
40
50
V̂max = 1/Intercept = 1/0.00697
= 143
1 / concentration
K̂ = Slope × V̂max = 0.031
Residuals vs fitted values
Which is more reasonable?
0.002
1
= β0 + β1
V
Residuals
0.001
0.000
−0.001
V =
−0.002
0.008
0.010
0.012
0.014
Fitted values
0.016
0.018
1
+ error
C
Vmax × C
+ error
K+C
Nonlinear regression
We imagine that
Vi =
Vmax × Ci
+ ǫ,
K + Ci
ǫ ∼ iid normal(0, σ 2)
We estimate Vmax and K by the values for which
!2
X
V̂max × Ci
RSS =
Vi −
K̂ + Ci
i
is minimized.
−→ An iterative method; need “starting values”.
Nonlinear regression in R
> library(nls)
> nls.out <- nls(vel ˜ (Vm * conc) / (K + conc),
data=mydata,
start = c(Vm=143, K=0.031))
> summary(nls.out)$param
Vm
K
Est
160.28
0.048
SE
6.48
0.008
t-val
24.7
6.1
P-val
1.4e-09
1.7e-04
The fit
Residuals vs fitted values
Nonlinear regression
20
160
15
140
Residuals
initial velocity
10
120
Regr of 1/V on 1/C
100
80
5
0
−5
−10
60
−15
0.0
0.2
0.4
0.6
0.8
1.0
60
80
concentration
100
120
140
Fitted values
A second set of data
V =
initial velocity
200
(Vmax + ∆Vmax x) × C
+ error
(K + ∆K x) + C
150
x = 0/1 if cells were
untreated/treated.
100
treated
untreated
50
0.0
0.2
0.4
0.6
concentration
0.8
1.0
Estimation in R
> nls.outC <- nls(vel ˜ ((Vm +dV * x) * conc)/
(K + dK*x + conc),
data=puro,
start=c(Vm=160, K=0.048, dV=0, dK=0))
> summary(nls.outC)$param
Vm
K
dV
dK
Est
160.28
0.048
52.40
0.016
SE
6.90
0.008
9.55
0.011
t-val
23.2
5.8
5.5
1.4
P-val
2.0e-15
1.5e-05
2.7e-05
1.7e-01
K’s equal vs not
200
initial velocity
150
100
treated
untreated
K’s =
K’s not =
50
0.0
0.2
0.4
0.6
concentration
0.8
1.0
From last time...
800
y
600
400
y = β0 + β1x + β2x2 + β3x3 + ε
200
10
20
30
40
50
60
70
x
An alternative model
Linear “spline”
y=
R code:

 β0 + β1 x + ǫ
if x ≤ x0
 β + β x + ǫ if x ≥ x
0
1 0
0
> f <- function(x,b0,b1,x0)
ifelse(x<x0, b0+b1*x, b0+b1*x0)
> nls.out <- nls(y ˜ f(time,b0,b1,x0), data=mydata,
start=c(b0=200, b1=100, x0=30))
> summary(nls.out)$param
Est
SE t-val
b0
-146
71.3
-2.0
b1
39
4.3
9.0
x0
26
1.8
14.6
P-val
5.8e-02
2.1e-07
2.9e-10
80
Results
1200
1000
y
800
600
400
200
0
0
20
40
60
80
100
x
One last example...
0.48
y = a + be−ct + ε
chlorine
0.46
0.44
0.42
0.40
0.38
10
15
20
25
time
30
35
40
R code
> nls.out <- nls(chlorine ˜ a + b*exp(-c*time), data=cl,
start=c(a=0.05, b=0.49, c=0.1))
> summary(nls.out)$param
a
b
c
Est
0.390
0.219
0.099
SE
0.006
0.031
0.018
t-val
66.7
7.0
5.5
P-val
1.9e-43
1.7e-08
2.4e-06