MathAnswers.pdf

Werner Ulrich

Contact: ulrichw@uni.torun.pl

Mathematics for biologists

Questions and Answers

Lecture 1: Some basic mathematics

Basic questions

1. Interpret the following mathematical symbols and construct terms with them

∑ ∏ ∼

; ;;;;;;;;;;;;;;;;;; | |

∝≈≤≥∧∨¬∀∃∈∪∩ ∆

e x x

2. Write down the complete Greek alphabet (small and large letters).

For advanced readers

1. Look for the axiomatic system of Euclid. Try to find out why the axiom number 5, the axiom about

parallels stirred such a controversy.

Solution: Look at http://math.youngzones.org/Non-Egeometry/

Lecture 2: Why nature loves logarithms

Basic questions

1. You have two data points: a (7;3) and b (0;1). Derive slope and intercept of the linear function through

a and b.

Solution: y = 0.2857x + 1

2. Convert ln(e), ln(10), and ln(2) into decimal and binary logarithms.

log 10 (e) = ln(e)/ln(10); log 10 (10) = ln(10)/ln(10); log 10 (2) = ln(e)/ln(2);

log 2 (e) = ln(e)/ln(2); log 2 (10) = ln(10)/ln(2); log 2 (2) = ln(2)/ln(2);

3. Two data points a (1:1) and b (2;8) are assumed to be part of a power function. Compute this function

from a and b. What is if a would have the coordinates a (-2; 8) or a (-2;-8)?

Solution: y = x 3

In the second case power functions cannot be constructed.

4. Take the values of the third question and compute slopes and intercepts for an exponential function.

Solution: y = 2exp(-xln(2)) =

In the second case an exponential function cannot be constructed.

Solution: log ( ) log ( ) * log ( )

5. Plot N! against the respective estimations of Stirlings’s

approximation using a spreadsheet program.

Plot the ε of the Stirling equation against N. Discuss the results.

0.995

0.99

0.985

0.98

Solution: You see that above n = 10 ε becomes nearly 1. This even

improves the approximation because you can include √(ε/12n) with ε

= 1.

0.975

0.97

6. Compute the roots of the following equations

yx x

yyx x

=+→ =±−

+= −→ =±

yx x x

=−− → = +

5 10

16.25 2.5

yxex e

=+→−

=+−→=−

=−→−

ln( 1) 1

x e

ye x

= +

ln( )

1 does not exist

ln(

−

)

→=

For advanced readers

1. Assume a spherical bacterial cell with radius r. At what amount does its surface grow if its volume

becomes 50% larger?

Solution:

V r A r

π π

;

1.5*3

→=

 

4.5

new

 

2. The energy an insect needs for flying scales proportionally to its volume. The ability of this insect to

produce energy scales proportional to its surface. Is there an upper limit for the body size of that insect?

Solution: For flying energy production E P should be larger or equal to the energy use E U . Therefore E P > E U .

E U = aV; E P = bV 2/3 . a and b are physiological constants and the ratio of b / a will be V 1/3 , hence b must be larger

than a. As the body size growths by a factor of k > 1 this ratio will change b /a = k 1/3 V 1/3 . Hence, b the cell

specific rate of energy production must increase because a is assumed to be a constant b can only increase if

either energy is more and more efficiently produced or the ratio of volume to surface changes with body size.

This seems to be impossible above certain body sizes.

3. Solve the following equations for y

If you think that there are no explicit solutions possible, try to approximate y for certain values of x.

Solution:

 

yx y y x

=→=

−

2/ .5

xy yx x

xy y

+= →= −

+=→

1.5

ln( 1)

no simple solution

1 1

1 no simple solution, for x and y

large or small an approximation is possible by

dropping y or y .

+= + → = −+

+= + →

y x y x x

y y

ab b

ab a b a

ab a b a b a

aba

−=−

−=+ −

(

) (

)(

)

Where is the error?

Solution: From the definition follows that a = b, In the fourth line you divided through a - b hence through 0.

This is not allowed.

Lecture 3: Proportionalities

Basic questions

1. Assume you have 300 ml of a solution of 50% methanol in water. For your experiment you need 200 ml

of a solution of 40%. What to do if you have only solutions of 30% and 90%? Develop a general solution.

Solution: a) You have to use two equations, one for the whole solution and one for Methanol. Hence,

x + y = 200 and x*0.5 + y*0 = 200*0.4 x is the amount of solution and y the amount of water you need.

Hence x = 160 ml and y = 40 ml.

b) x + y = 200 and x*0.3 + y * 0.9 = 200*0.4. Hence y = 33.3 ml and x = 166.7 ml

2. A balloon has at ground a diameter of 2 m. What diameter has the balloon at 3000 m if the atmospheric

pressure there is 68.5 kPa? Temperature at ground was 20 ° C, at 3000m 5 ° C.

Solution: V 1 = 4/3πr 3 = 4.189 m 3 . Ground pressure is assumed to be 101.3 kPa. You need the law of Avogadro

p 1 *V1/T 1 =p 2 V 2 /T 2 . Hence the diameter is 2.24 m.

3. Azotobacter chroococcum takes his nitrogen demand from air. How much air do you need (at 32 ° C) for

2.5 dm 3 bacterial culture if the culture has a density of 0.84 mg cm -3 and the dry mass contains 7% N?

Solution: 2.5 dm 3 is 2500 cm 3 . Hence this volume should contain 0.84*2500 mg giving 2100*7% = 147 mg N =

73.5 mg N 2 . The mol weight of N 2 is 14 g mol -1 . Hence you need 0.00525 mol. You need a volume of

0.00525*8.314*305.15/101325 [mol * JK -1 mol -1 *K/(Nm -2 )]= 0.0001315 m -3 N2 = 0.00101 m 3 /0.78 = 0.0001685

m 3 air = 168,5 cm 3 .

Solution:

4. Look at the following ‘proof’. Assume we have two numbers a = b

4. How much heavier are CO 2 and CO in relation to air?

Solution: 1 mol air contains approximately 78% N 2 = 0.78*14 g, 21% O 2 = 0.21*16 g, and 1% Ar = 0.01*39.9g

= 14.68g. 1 mol CO 2 = 44g, 1 mol CO = 28g.

For advanced readers

1. The oxidation of 20g of an unknown chemical yielded 12.94 g CO 2 and 1.48 g water. What had been

oxidized?

Solution: You have to apply the Gauß scheme

n 1 C x O y H z + n 2 O 2 → n 3 CO 2 + n 4 H 2 O

n 1 y + n 2 = 2n 3 +n 4

n 1 x = n 3

n 1 z = n 4

Additionally

20 / 12.94 = (n 1 x12+n 1 y16+n 1 z) / (n 3 44)

20 / 1.48 = (n 1 x12+n 1 y16+n 1 z) / (n 4 18)

This system has to be solved for natural numbers to yield

C 2 H 5 OH + 3O 2 → 2CO 2 + 3H 2 0

Notice that there is no single solution.

2. An anaerobic bacterium that was isolated from waste water produces an unknown gas. A probe of this

gas needs 491 s to pass through a manometer. At identical temperature and pressure the same amount of

nitrogen needs 650 s to pass. What gas might this bacterium produce?

Solution: We are dealing in both cases with identical conditions. Hence, in both cases we are dealing with n mol.

1 mol nitrogen weights under standard condition 14 g. From physics you know that the speed of a gas is inverse

proportional to the square of its weight (s = 1/2mv 2 ). Therefore 491/650 = (14 / x) 2 → x = 16.1. Probably, the gas

is Methan (CH 4 ).

Lecture 4: Some basic functions and their application in biology

Basic questions

1. What is an algebraic function?

= ∑

Solution: An algebraic function is a function of the general form

y a x

2. Discuss the difference between a function and a relation.

Solution: A function has for every value of x or combination of values x i one and only one value of y. A relation

might have several solutions.

3. Radioactive decay is described by an exponential function y = y 0 e -kt . Compute the half-time of the decay

process in dependence of k.

Solution: N 1/2 = ln(2) / k

4. A biological process is assumed to follow an exponential function. Compute this function from the data

points a (1;1) and b (5;5).

Solution: y = 0.67exp(0.4)

5. C 14 has a half time of 5568 years. How long would it take until of one mol C 14 only one atom remains? Is

this a realistic question?

−

ln( 2 )

5568

ln(2) ln(6.0210 )

We have to solve

1 6.0210

→ =

7.5*10

years

5568

Our universe exists approximately 15 billion = 1.5x10 10 years.

6. Assume that relative abundances of two communities of plants follow log-series with the same slope of -

0.5. One community has in the mean 10 species per m 2 , the other 20. What is the quotient of species

numbers S 1 / S 2 at 10 m 2 .

Solution: The species - area relations of both communities should follow logarithmic functions. Hence

10 0.5 ln(10)

0.527

20 0.5 ln(10)

7. What is a heavy tail?

Solution: A heavy tail is the end of a power function where it can be approximated by a linear regression.

8. The species – area relation of butterflies in Europe is described by S = 45A 0.1 [km 2 ]. For parasitic

Hymenoptera this relation is S = 1500A 0.15 [km 2 ]. How many butterfly and hymenopteran species do you

expect for Poland [312685 km 2 ]?

Solution: Butterflies 159, the observed number is 162. Hymenoptera: 10000, the real number is unknown but the

estimate seems to be quite realistic.

9. The mean number of bird species in Poland is about 320, the total European [10500000 km 2 ] species

number is about 500. How many species do you expect for the Czech Republic [78866 km 2 ] and for France

[543965 km 2 ]? Would it make sense to estimate the species number of Luxemburg [2586 km 2 ]? What

about Kujaw-Pommern [17970 km 2 ]?

Solution: Czech: 269 species, France 343, Luxemburg 174. The last value for Luxemburg is not reliable because

we extrapolated far beyond the areas used for computing the power function. Kujaw Pommern should have 223

species; the real number is about 200 nesting species. We see that extrapolations far beyond the measured data

points might be misleading.

10. The mean number of bee species per km 2 in Poland is 110, the total number of Polish bees is 463.

Estimate the number of bees in the district of Kujaw-Pommern [17970 km 2 ].

Solution: 334 species. This number seems to be quite realistic.

For advanced readers

1. Every year man destroys about 0.1 to 1% of the tropical forests. If this rate would be constant, how long

would it take that only half of the earth’s tropical forests remain?

Solution: You have to solve

t AAe −

ln(1

rate t

)

. For 0.1% yearly destruction it would take 690 years, for 1%

yearly destruction 69 years.

= . For this equation the number of individuals is

divided into binary logarithmic individuals classes N (so-called octaves), so that N 1 has 1 individual, N 2 2,

N 3 4, N 4 8…Plot the resulting species – octave curve. How to interpret the constants µ and S 0 ? Compute

the upper individual class N max . What is the total number of species if S 0 is 20 species and a has a value of

0.2?

SN Se

− −

(

)

Solution: You have to take octaves from 1 to 8 with a mean µ of 4. At this mean the number of species is

highest. 8 is the maximum number of octave because half species do not exist. The total number of species is

174.

2. In ecological communities the number of species can often be approximated by the following equation,

the so-called lognormal distribution

()

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