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0 0ln(2) 0.693
012
xx x
xC C C e C e
0kx
xC C e
1 0 (1 )kxxC C e
The exponential function is a general description of a random process, where the probability of a certain event is independent of time
C0/2 C0/2
C1/2 C1/2
C2/2 C2/2
0
0.1
0.2
0.3
0.4
0.5
0.6
0 5 10 15
x
Cx
Lecture 2Some basic functions and their application in biology
2/1
2/1
)2ln(
)2ln(
xk
kx
C14 has a half time of 5568 years. How long would it take until of one mol C14 only one atom remains?
One mol contains 6.0210*1023 atoms
ktt eNN 0
kte 23100210.615568
)2ln()2ln(
2/1
t
k
5568/)2ln(23100210.61 te
439841)2ln(/)100210.6ln(5568
5568/)2ln()100210.6ln(23
23
t
t
The age of fossilized organic matter can be determined by the C14-method of radioactive decay. The half-live of C14 is 5568 years. The equilibrium content of C14
in living plants is about 10-6 ppm (parts per million). How old is a fossilized plant with a C14 content of 1.5*10-7 ppm?
tt
t eNN 2/1
)2ln(
0
te 5568
)2ln(67 10105.1
15239)2ln(
5568)]10ln()105.1[ln( 67
t
0ai
iN N e
The logarithmic function
0
ln
iN aiN
01 ln( )S A Sa
0
0
ln( ) ln( )
1 1ln( ) ln( )
i
i
A ai Aand
i A Aa a
Log-series relative abundance distribution
0.0001
0.001
0.01
0.1
1
0 2 4 6 8 10
Species
Abu
ndan
ce
2
2
2 2
2 2
N N individuals / m
10 individuals / m
0.1 individuals / m 1 individual / 10m
0.01 individuals / m 1 individual / 100m
1/N A
0 00
1// /
1/i
i iA
N N A AA
Species – area relation
y = 8x0.75
0.1
1
10
100
1000
10000
100000
0.001 0.1 10 1000 100000
Body weight [kg]
Bra
in w
eigh
t [g] Man
y0=8x0=0.0625 slope = -y0 / x0 = -ln(8)/ln(0.0625)=0.75
The power or allometric function
abxy
orbxay
)ln()ln(ln
The Sierpinski triangle
1. Start with a triangle, 2. shrink to 1/2 size, 3. make three copies 4. arrange the three copies in quadrants 2,3, and
4 5. goto (2).
Self similar objects
zz
xx
xx
aa
yy
2
1
2
1
2
1zy ax
Self similarity
How life makes a complex pattern
1000 km500 km
y = 157000x0.30
1000
10000
100000
1000000
0.001 0.01 0.1 1
Scaling factor
Leng
th o
f Eur
ope
[km
]Scaling factor = 1 / unit of measurement
1000 km500 km
Scaling factor = 1 / unit of measurement = magnification
dddd
assll
L
1
a is called the normalization constant
dddd
ssasll
A
111
1
)(1
Ruler length l
sbblLEuclid
1
What is the value of a?
If our object is classical Euclidean d = 0
11)( ddd bsss
bssaL
Both equations match if b = cs1)1(11 dd csscsLNow we consider an area. The area scales to the square of the ruler length
22 1
sbblAEuclid
112
1 1)( ddd bsss
bssaA Both equations match if b = cs2
1)2(12 dd csscsA
1)3(13 dd csscsV
dddd
ssasll
L )(1
33 1
sbblVEuclid
Euclidean dimension E=1
Euclidean dimension E=2
Euclidean dimension E=3
. Euclidean dimension E=0
33 dscsV
22 dscsA
11 dscsL 1)( dEcsY
E+d takes always values between the actual Euclidean dimension and the next higher one. It is commonly termed the fractal dimension of an object.
0 < d < 1
An important class of fractal objects are self similar objects. We describe them by power functions.
rK2LK
rK
LKrK2LK
rK
LK
b
n n
K K
A rA r
x
n n
K K
M VM V
X = 0.75
an n
K K
V rV r
What are the relation between radius, volumen and surface in such a branching pattern?
M A
B V0.75M B
A branching pattern
Calculate the total leaf area of this fern
We have to measure two leafs at different scale to get the scaling exponent of the area - length relationship
zlAA 0z
ll
AA
2
1
2
1
Let the average length of the smallest leaflets be 1 cm and its area 3 cm2. At the next higher scale leaflet length might be 10 cm and the respective area 35 cm2. The whole fern is 1 m long.
07.1)10ln()1ln()35ln()3ln(
101
353
z
cmcm
cmcm z
207.1
2
10
10010100 408
1010035 cm
cmcmcm
llAA
z
(2 ) 1( ) 2.28dA s bs D
1)()( dDbssL
How should population density scale to body weight?0 5 10 15 20 25
020406080
100120140160
f(x) = 3 x^1.28
Magnification
Dar
k ar
ea
y = 4.36x-0.89
R2 = 0.81
0.001
0.1
10
1000
100000
10000000
1E-06 0.0001 0.01 1 100 10000
Mean weight [g]
Mea
n de
nsity
/ m2
How should population density scale to body weight?
sL /1 VN 2/3AV 28.1sA 4/3WM
64.092.13/192.12/328.1 )()( WWssV
33 slW MN /1
39.175.064.0 WWWN
Population density is proportional to available space and to available energy.
75.0WM 0.75 0.75z zMD W W W
zD W
0.01
0.1
1
10
0 0.01 0.1 1 10 100
Body weight class [mg]
Mea
n de
nsity
per
bina
ry w
eigh
t class
A
020406080
100120
0 0.01 0.1 1 10 100
Body weight class [mg]
Num
ber o
f spe
cies
By = 1.11x0.73
0200400600800
10001200140016001800
0 5000 10000 15000 20000
Body weight [g]
Met
abol
ism
rate
MD is proportional to total population biomass
What is if z is about 0.75?
Energy equivalence (Damuth’s) rule(for poikilotherms: equal biomass
hypothesis)
0.75 0.75 0MD W W const
1
10
100
1000
10000
100000
1E-06 0.0001 0.01 1 100 10000Mean weight [mg]
Tota
l bio
mas
s [m
g] /
m2
00
ln ln( )a ii
NN N i a iN
Species - area relationship
1/ 1/00
1ln( ) ln a ai
i
Ai i A A
a A
0zS S A
0.0012 0.012 0.12 1.2 12 1201
10
100
f(x) = 20.5898381843181 x^0.128687366991925R² = 0.618393525360921
Area [km2]
S
1 10 1001
10
100
f(x) = 14.6440097473 x^-0.8577832199R² = 0.935292824287407
Species rank order
Abu
ndan
ce N
The mean number of bee species per km2 in Poland [312685 km2] is 110, the total number of Polish bees is 463. Estimate the number of bees in the district of Kujaw-Pommern [17970 km2].
The mean number of bird species in Poland is about 430, the total European [10500000 km2] species number is about 800. How many species do you expect for France [543965 km2]?
Would it make sense to estimate the species number of Luxembourg [2586 km2]? What about Kujaw-Pommern [17970 km2]?
0zS S A 11.0
)312685ln()110ln()463ln()312685(110463 zz
334)17970(110 11.0 KPS
0.0001 0.01 1 100 100001
10
100
f(x) = 20.5898381843181 x^0.128687366991925R² = 0.618393525360921
Area [km2]
S
177.0)1050000ln()312685ln(
)800ln()430ln(1050000312685
800430
z
z
46)10500000(
800)10500000( 177.00177.0
0 SSSE
474)543965(46 177.0 FRS
260)17970(46 177.0 KWS184)2568(2.64 127.0 LS
Observed: 530
Observed: 250
Observed: 262
Observed: unknown
The inverse hyperbola
1 0 2( )E S ES
k S E ES k ES
SkkSEk
ES12
01
max 0
0
V EV ES
SKSV
V
max0
S
V0
Vmax
Vmax / 2
K
Michaelis-Menten equation
( )( )
af xyb f x
Monod function
[ ][ ][ ]MbOKMb O
Haemoglobin or myoglobin bind oxygen according to the partial pressure of O2
Denoting y for [MbO], p(O2) for the partial pressure of oxygen and using [MbO] + [Mb] = const we get
2
2 2
( )( ) ( ) ( )
K p OyK yconst y p O const K p O
2
2
2
( ) ( )
( ) ( )
n
n
n
yKconst y p O
K p Oyconst K p O
Hill equation of oxygen binding
00.10.20.30.40.50.60.70.8
0 0.2 0.4 0.6 0.8 1
p(O2)
yn=1
n=2n=3
n=4
Home work and literatureRefresh:
• Fractal geometry• Self similarity• Branching processes• Logarithmic transformations • Species – area relationships• Radioactive decay
Prepare to the next lecture:
• Vectors• Vector operations (sum, S-product,
scalar product)• Scalar product of orthogonal vectors• Distance metrics (Euclidean,
Manhattan, Minkowski)• Cartesian system, orthogonal vectors• Matrix• Types of matrices• Basic matrix operations (sum, S-
product, dot product)
Literature:
Mathe-onlineFractal geometry: http://classes.yale.edu/fractals/Fractals: http://en.wikipedia.org/wiki/Fractal
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