0 0

ln(2) 0.6930

1

2

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

0

1ln( )S A S

a

0

0

ln( ) ln( )

1 1ln( ) ln( )

i

i

A ai A

and

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

Ab

un

da

nce

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 i

AN N A A

A

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=8

x0=0.0625 slope = -y0 / x0 = -ln(8)/ln(0.0625)=0.75

The power or allometric function

abxy

or

bxay

)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

"All the branches of a tree at every stage of its height when put together are equal in thickness to the trunk„ (Leonardo da Vinci, Notebooks,

1510)

zz

x

x

x

x

a

a

y

y

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 bss

sbssaL

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 bss

sbssaA 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 r

A r

x

n n

K K

M V

M V

X = 0.75

a

n n

K K

V r

V 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(

10

1

35

3

z

cm

cm

cm

cmz

207.1

2

10

10010100 408

10100

35 cmcmcm

cmll

AAz

(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

Da

rk a

rea

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]

Me

an

de

nsi

ty /

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]

Mean

dens

ity pe

r bin

ary w

eight

class

A

0

20

40

60

80

100

120

0 0.01 0.1 1 10 100

Body weight class [mg]

Numb

er of

spec

ies

By = 1.11x0.73

0200400600800

10001200140016001800

0 5000 10000 15000 20000

Body weight [g]

Me

tab

olis

m r

ate

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 10000

Mean weight [mg]

To

tal b

iom

ass

[mg

] / m

2

00

ln ln( )a ii

NN N i a i

N

Species - area relationship

1/ 1/00

1ln( ) ln a a

ii

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

Ab

un

da

nce

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( 177.00

177.00 SSSE

474)543965(46 177.0 FRS

260)17970(46 177.0 KWS

184)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

Skk

SEkES

12

01

max 0

0

V E

V ES

SK

SVV

max

0

S

V0

Vmax

Vmax / 2

K

Michaelis-Menten equation

( )

( )

af xy

b f x

Monod function

[ ]

[ ][ ]

MbOK

Mb 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 y

const y p O const K p O

2

2

2

( ) ( )

( )

( )

n

n

n

yK

const y p O

K p Oy

const K p O

Hill equation of oxygen binding

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.2 0.4 0.6 0.8 1

p(O2)

y

n=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