DATA HANDLING SKILLS, O Wolkenhauer

Tags: number line, expression, experimental data, logarithm, Cartesian plane, mathematics, cognitive skills, Scientific notation, log10, Albert Einstein, variables, Negative numbers, algebraic expression, natural logarithm, decimal places, exponential function, plot, product concentration, rate of reaction, Michaelis-Menten equation, substrate concentration, Michaelis-Menten, population model, experimental conditions, logarithmic function, differential equation, significant figures, maximum velocity, mathematical model, mathematical modelling
Content: DATA HANDLING SKILLS BSc Biochemistry BSc Biological Science BSc Bioinformatics September 24, 2002 Olaf Wolkenhauer Department of Biomolecular Sciences and Department of Electrical Engineering and Electronics Address: Control Systems Centre UMIST Manchester M60 1QD, UK Tel/Fax: +44 (0)161 200 4672 e-mail: [email protected] http://www.umist.ac.uk/csc/people/wolkenhauer.htm
Contents
1 Introduction
3
2 Simple Powers
4
3 Fractions
8
4 Percentages
12
5 Real Numbers
14
6 Logarithms
16
7 Variables and Constants
18
8 Algebraic Expressions and Equations
19
9 Measurement and Units
25
10 Concentrations
28
11 Accuracy
29
12 Relations and Functions
30
13 More on Functions
39
14 Proportionality
43
15 Differential Equations
44
16 Rate Equations
48
17 Michaelis-Menten Modelling
51
18 Supplementary Material
55
19 More Exercises
57
20 More Exercises: Solutions
67
2
1 Introduction We make the world around us comprehensible through our cognitive skills, making use of our senses and the mind to reason about phenomena and to answer questions. As animals we are limited in our perception and conception and for anything that goes beyond common experience, in the areas of science and engineering, we complement common sense with technologies and methodologies. For example, molecular systems are usually not directly observable and properties not directly measurable. Experimentation, helped by instrumentation, generates data we can analyse to explain the phenomenon under consideration. Raw data, i.e., a list of numbers usually does not reveal a relationship or pattern by itself. Mathematics helps us to reveal, explain and represent (model) principles for which the data may provide evidence. Like for anything else it takes time to learn the tricks of the trade. You should enjoy learning math as you like to acquire skills with the latest technology. The advantages over technology are obvious: it is cheap ­ paper and pencil to start with; it is safe ­ equations don't bite but most of all it is generic ­ it works in different contexts, it lasts a lifetime. Data handling, the analysis of data for the purpose of model building, hypothesis testing and decision making is central to all sciences. To be a good scientist, being able to answer questions, you need to be able to ask questions. Questioning is as important in learning math as it is in the natural sciences. For this course, the most important advice I can give to you is that if you get stuck, treat the math like a wet-lab experiment: · Ask why? and how? questions! · Help yourself with a pen and paper ­ visualise the question, ­ keep a record of the answering process. · Remember: Repeated trials are essential to gain confidence... These lecture notes have extra large margin to encourage you to make comments. Finally, for the given exercises, if you compare your results with those of others, please make sure that you don't just compare the final result but also the steps that lead to the solution. For most problems in mathematics there are several valid ways of reaching a solution! This should not confuse you but by comparing and "playing" with alternatives you will improve your problem-solving skills. I also recommend that you quickly read or "scan" the notes before the lecture. This will help you memorise new ideas and concepts. Some time after the lecture you should again read the notes but this time try to understand all the material and do the exercises. ENJOY YOUR EXPERIMENTS IN MATHEMATICS!
Use this space!
3
2 Simple Powers Most people who have no particular interest in mathematics find it difficult that in mathematics many things can be written, or arrived at, in different ways. Mathematicians enjoy this aspect, because, like using a natural language, this allows for creativity and by introducing new concepts, notation or representations one can enrich the `vocabulary'. You should not get confused by notation and alternative representations. Although not always obvious a lot of equivalent representations are for convenience. Honestly! Positive powers are a convenient shorthand for repeated multiplications. For example, for 4 · 4 we write 42 ("square") or in general an = a · a · a · . . . · a n times where a is called the base and n is referred to as the exponent, power or index. Let us see what happens when we multiply powers of the same base together: 62 · 63 = (6 · 6) · (6 · 6 · 6) = 65 or 62 · 63 = 62+3 = 65 . It does not matter whether the base is a number or letter x3 · x5 = (x · x · ·x) · (x · x · x · x · x) or x3 · x5 = x3+5 = x8 . An example for powers occurring in biology are processes that double in time, for instance, cell division. Imagine you start with two cells, at the next stage you have 2 · 2 cells and thereafter 2 · 2 · 2 etc. PCR amplification is an important biological tool that serves as an example. PCR produces an amount of DNA that doubles in each cycle of DNA synthesis. For example, three PCR cycles of reactions produce 8(= 2·2·2) DNA chains. We are going to consider mathematical representations of such "exponentially increasing processes" in greater detail in Sections 12 to 16. PRACTICE. The rule is: When multiplying powers of the same base, add the powers. Try simplifying the following: 1. 72 · 75 · 79 =
Multiplication of powers: an · am = an+m
2. 42 · 162 =
3. a4 · a5 · a7 =
4. a2 · b2 =
If we replace a and/or n by zero and one, we obtain a few special cases 1n = 1 , a1 = a , a0 = 1
4
and some familiar examples: 102 = 100 103 = 1000 106 = 1 000 000
hundred thousand million.
Let us see what happens when we divide powers of the same base:
35 32
=
3
·
3·3·3 3·3
·
3
=
3
·
3
·
3
=
33
The same result could be obtained by subtracting the indices
35 32
=
35-2
=
33
Likewise
712 75
=
712-7
=
77
a4 a3
=
a4-3
=
a1
=
a
.
PRACTICE. The rule is: When dividing powers of the same base, subtract the powers. Try the following without a calculator.
57 1. 53 =
Division of powers: an = an-m am
73 · 74 · 78 2. 75 · 76 =
a8 3. a5 =
y2 · y7 4. y4 =
Next we try to answer what the value of 33 2 is. One way is to Powers of powers:
proceed as follows: 33 2 = 33·33 = 33+3 = 36 similarly
22 3 = 22·22·22 = 22+2+2 = 26 .
ab c = a(b·c)
The same result could be obtained if the indices were multiplied together, i.e., 33 2 = 33·2=36 22 3 = 22·3=26 .
PRACTICE. The rule is: When raising the power of a number to a power, multiply the indices together.
5
1. 57 3 =
2. x2 4 =
3. a-2 3 =
4. x-1 -2 =
5. 32 · 74 3 =
57 4 6. 35 3 =
Negative indices arise when we simplify expression such as
43
4·4·4
1
46 = 4 · 4 · 4 · 4 · 4 · 4 = 4 · 4 · 4
Applying the division rule
43 46
=
4(3-6)
=
4-3
Hence
4-3
=
1 43
A negative index therefore indicates a reciprocal. We return to this in the next section. Other examples are
2-3
=
1 23
,
a-5 =
1 a5
,
x-1 = 1 . x
PRACTICE. Rewrite the following expressions.
1. 3-1 =
2. 5 · 2-2 =
3. x-2 =
6
4. 3 · a-2 =
Our number system uses base ten and each digit in decimal numbers such as 9023.4056 can therefore be represented by its place value. The place value of each digit in a base ten number is determined by its position with respect to the decimal point. Each position represents multiplication by a power of ten. For example, in 324, the 3 means 300 because it is 3 times 102 (102 = 100). The 2 means 20 because it is 2 times 101 (101 = 10), and the 4 means 4 times one because it is 4 times 100 (100 = 1). There is an invisible decimal point to the right of the 4. In 5.82 the 8 means 8 times one tenth because it is 8 times 10-1 (10-1 = 0.1). Summarised for the number 9023.4056 we have:
9023 . 4056
Thousands, 103
Ten-thousands, 10-4
Hundreds, 102
Thousands, 10-3
Tens or Units, 101
Hundredth, 10-2
Ones or Units, 100
Tenth, 10-1
Powers of ten are a convenient way to manipulate very large or very small numbers. Molecular biologists frequently deal with `extreme numbers', they handle milliliters (10-3L) and count in there a million (106) bacteria. (More on this when we introduce measurement units in Section 9). For example,
30 000 · 0.02 0.006 · 10
=
3Ч104 · 2Ч10-2 6Ч10-3 · 1Ч10
=
6Ч102 6Ч10-2
=
102 · 102
=
104
Scientific notation (or standard form) is the convention used for plac- ing a decimal after the first non-zero digit of a number, and then multi- plying it by the appropriate power of 10. In this way, large numbers such as 3, 690, 000, 000 are more conveniently represented as 3.69 Ч 109. Likewise, very small numbers such as 0.00000573 are reported as 5.73 Ч 10-6.
PRACTICE. Try these questions without a calculator.
0.0006
1.
=
2000
Powers of fractions:
a c ac
=
b
bc
Scientific notation: a Ч 10n
0.05 · 200
2.
=
0.002
3. 0.0009 · 1 · 4.9Ч105 = 7000 30 0.1
4. (0.0005)2 = 7
5. Although biologists deal with big numbers, astro-physicists probably break the records. For example, the speed of light in empty space is approximately 301, 000, 000 m/s. Stars are so far away that their distance from Earth is measured in terms of how long the light has taken to reach us. The light from our nearest star, Alpha Centauri takes 4.3 years to reach us. How far is this in meters? Negative exponents are dealt with in the next section and before then we only briefly note that for very large numbers, much greater than one, we obtain the scientific notation as follows. To express 4 500 000 in standard form a Ч 10n, we must first identify the value of a which is a number between 1 and 10. In this example a = 4.5 and 4 500 000 = 4.5 Ч 1 000 000 = 4.5 Ч 106. An alternative method is first to consider the position of the decimal point. For example, write 2 756 000 000 in standard form. Place the decimal point and to find n count the number of places to the right. So n = 9 and hence 2 756 000 000 = 2.756 Ч 109. To practise the standard notation, invent three examples! Remark: Powers are useful for a number of reasons. In biology, cell division or the growth of microbiological cultures are examples. On the other hand, our number system is "base ten", that is, a number like 25.3 is actually a short form of 2 · 101 + 5 · 100 + 3 · 10-1. For subsequent sections, I try to encourage you to treat the exercises like a laboratory experiment. In a laboratory environment it is natural to try ("experiment") with different tools and to ask questions when something doesn't work. For maths, many of us feel that a solution should appear instantly or never. This is a mistake and the ability to question and experiment in maths is in fact a sign of skill rather than weakness. The exercises required some knowledge of how to deal with fractions and bracketed expressions. We will address these issues in the following sections.
3 Fractions
A fraction is a mathematical concept used to describe proportions, ratios, and rates. Fractions consist of two parts, the numerator and the denominator:
num
3
, for example
den
4
(read this as `three over four') .
An equivalent way to write the example is
3 · 1 (`three fourth') 4
which highlights the fact that a whole is divided into 4 parts and what we have is 3 parts or "three quarters" in this particular case. Probably the most frequently used case are `hundreds' or as it is more commonly known as: percent. It even has its own symbol: %. More about percentages below. Sometimes fractions are equivalent, even this is not obvious at
Equivalent fractions
8
first glance:
3 6 9 12 15 == = = 4 8 12 16 20
To find a fraction in its lowest terms (with numbers on top and bottom lines that are as small as possible) we cancel down: 27 ч9 3 = 63 ч9 7 Dividing numerator and denominator by the same number, simplifies the expression but doesn't change its value. To check this, we write the operation out in detail:
27 63
=
27 9 63 9
=
27 · /9 63 · /9.
24 Try this: = 54
Cancelling fractions involves the division of fractions. In general, to divide two fractions you multiply the first fraction by the reciprocal of the second fraction:
3 8 1 4
=
3 8
·
4 1
=
3·4 8·1
=
12 8
=
(cancel)
The reciprocal of a fraction is therefore found by turning the fraction upside down. The reciprocal of a/b is b/a. To multiply two fractions we multiply the numerators and denominators. The following example shows how cancelling down, before multiplication, can simplify the operation:
7 41
·
12 5
3
=
7 1
·
3 5
=
21 5
To subtract or add two fractions, we first have to ensure that the denom-
inators of both fractions are the same. A simple way to find a common
multiple is to multiply the denominators. To find the common denomi-
nator you can use the lowest common multiple of the two denominators.
To
work
out
7 8
-
1 4
,
we
multiply
the
numerator
and
denominator
with
the factor that makes the denominator equivalent to the lowest common
multiple 8 · 4 = 32:
7 Ч4 28 = 8 Ч4 32 such that
1 Ч8 8 = 4 Ч8 32
28 -
8
28 - 8 20
=
==
32 32 32 32
(cancel)
Note: Although multiplying or dividing the numerator and denominator with the same number does not change the value of the fraction, this is not true for adding or subtracting numbers:
a+3
a
is not the same as .
b+3
b
Similar
a2 a b2 = b !
Cancelling fractions
Division of fractions:
a b c d
=
a b
·
d c
Multiplication of fractions:
a b
·
c d
=
a·c b·d
Substraction of fractions:
a b
-
c d
=
a·d - c·b b·d
Addition of fractions:
a c a·d + c·b
+= bd
b·d
9
Comparing fractions, it is often difficult to see the difference, say, whether 4/5 is greater or less than 1/2 by just looking at the numbers. However, basic fractions can be "visualised" by `sharing a cake' or by colouring a rectangle. For example, use the following rectangles to fill in the fraction next to it. (Do it NOW):
3
2
8
4
1
3
2
6
If we consider the basic fraction a/b, we can think of some special cases, setting b = a, a or b equal to one or zero. This leads to the following rules:
a a= , 1
0 =0, a
a =1, a
1 = a-1 (a 1) a
For 1/a and a 1 the fraction gives a number which is smaller than one. In Section 2, on positive exponents such as an and n 1, we observed that a number such as 250 can be expressed as 2.5Ч102 and since 102 = 100, the exponent 2 means effectively that we shift the decimal point by two positions to the right to get 250. Negative exponents are very much the same only that we move in the opposite direction. For example, 0.025 is the same as 2.5 Ч 10-2. We can therefore generalise the case 1/a = a-1 to any negative exponent:
1 = a-n an For a = 10 we obtain something familiar:
100 = 1/1
=1
10-1 = 1/10
= 0.1
10-2 = 1/100
= 0.01 (one hundredth or a percent)
10-3 = 1/1000 = 0.001
10-4 = 1/10 000 = 0.0001
10-5 = 1/100 000 = 0.00001
10-6 = 1/1 000 000 = 0.000 001
Describing one portion as 4/5 and another as 1/2 is not very convenient if we want to compare the two. Percentages are fractions with a denominator of 100. Dividing a cake, hundred percent, written 100%, is the whole, while 25% is a quarter. We are usually familiar with percentages, i.e., have an intuition about them and therefore it is often useful to convert a fraction to a percentage we multiply a fraction by 100:
4 · 100 = 80
therefore
4 = 80%
5
5
More on calculating with percentages follows further on. In the list above, we found that the fraction 1/100 in decimals is 0.01. To convert any decimal, say 0.473, to a fraction we notice that from above
47 3
400 70
3
473
0.473 = + +
=
++
=
10 100 1000 1000 1000 1000 1000
Negative indices:
1 an
=
a-n
Converting decimals to fractions
10
Convert the following decimals NOW ­ can you see the pattern? 0.12 =
0.3044 =
A ratio is a comparison between two like quantities (e.g. `you receive two, I keep three') while a rate is a comparison of two unlike quantities (`the limit is 25 miles per hour). A ratio is often written 2 : 3, read `two to three'. Receiving `two' out of five means you get 2/5. Notice the relation between a ratio and the fraction: the denominator of the fraction must be the total number of parts involved. Rates are similar to ratios. The units show the quantities being compared. For example, distance speed = time For instance, take a car travelling at an average speed of 30 mph (miles per hour). It will take 20 mins to go 10 miles. If the average speed increases to 40 mph, how long will it take to cover 10 miles? The car travels
40 miles in 1 hour = 60 mins 60 1 mile in mins 40 10 miles in 60 · 10 mins 40 = 15 mins
As the speed increases, the time will decrease (inverse proportion).
PRACTICE. Try these questions without a calculator.
1. Which two fractions are equivalent?
3 4 24 (a) 10 9 54
6 12 12 (b) 7 13 14
2. Write down 20p as a fraction of Ј1.
3.
Work
out
2 7
of
105
millimeters.
4.
Convert
4 5
to
a
percentage.
5. Convert 34% to a fraction.
6.
Out
of
24,
how
many
percent
are
1 3
and
5 8
.
7.
12 5
+
3 4
=
8.
2 5
·
1 3
+
2 5
=
a
9.
b c
·
e f
=
10. Divide 0.00034 by 0.7.
Ratios vs rates
11
11. If a car can travel 35 miles per gallon of petrol, how much petrol (to the nearest gallon) would it need to go 100 miles? 12. What is the length of 3/4 of a piece which is 2/3 metres long? Remark: If you wonder how you can best remember all the concepts and ideas mentioned so far, I recommend you go through all framed boxes, most of which are in the margin, and devise an example for each `rule'. Probably the best way of learning math is `learning by doing'.
4 Percentages
A percentage represents a fraction of 100. Percent therefore means "out of 100" or "per 100". We describe quantities in % because they are more intuitive than decimal numbers. Percents can be written as fractions by placing a number over 100 and simplifying or reducing. For example,
30% = 30 ·
1
30 ==
3
100 100 10
If
we
say
that
3 4
("three
quarters")
of
a
given
material
is
used,
we
mean
that
3 Ч 100% = 3 · 100 = 300 = 75%
4
41
4
of the available material is used. Therefore, fractions can be changed to percents by writing them with denominators of 100. The numerator is then the percent number:
3 3 · 20 60 5 = 5 · 20 = 100 = 60%
If we ask: What is 7% of 30?
7%
=
7 100
so
7% Ч 30
=
7 100
·
30 1
=
7 10
·
3 1
100 and 30 have the common factor 10.
=
7·3 10
=
21 10
=
2.1
To change a percent to a decimal number move the decimal point 2 places to the left because percent means "out of 100" and decimal numbers with two digits behind the decimal point also mean "out of 100". For example,
45% = 0 . 4 5 , 125% = 1 . 2 5
6% = 0 . 0 6 , 3.5% = 0 . 0 3 5
Note that because the 5 was already behind the decimal point and therefore does not count as one of the digits in the "move two places". To change a decimal number to a percent move the decimal point two places to the right:
0.47 = 4 7 % , 3.2 = 3 2 0 % , 0.205 = 2 0 . 5 %
To convert a decimal or fraction to a percentage, multiply it by 100. Conversion of % to decimal.
12
Although they are related to fractions and decimals, percentages are used and manipulated in different ways. Typical problems involving percentages can be illustrated using our `shopping instincts': Imagine you are asked to purchase the material needed for your experiments. Prices in catalogues are often given excluding VAT (Value Added Tax), so that must be added to find the actual cost. For example, your cultures are priced at Ј2000 + VAT (at 17.5%). The actual cost can be found in more than one way:
VAT = 17.5% of Ј2000
Ј2000 ·
17.5 100
=
Ј350
Final price = Ј2000 + 350 = Ј2350
Price = 117.5% of Ј2000
Ј2000 ·
117.5 100
=
Ј2350
A colleague tells you that he paid Ј1950 with a 15% reduction. What was the original price? You should not find it necessary to use the calculator's % function. (I can never remember how to use it). However, in case you want a kind of algorithm: Any problems that are or can be stated with percent and the words "is" and "of" can be solved using this formula:
% "is" number = 100 "of" number or "of" means multiply and "is" means equals.
Example 1: What percent of 125 is 50?
n 50 =. 100 125 Or n · 125 = 50, in either case the percent = 40%
Example 2: What number is 125% of 80?
125 n =. 100 80 Or 1.25 · 80 = n. In either case the number = 100.
If you are not confident with the rearrangement of algebraic equations, return after you read Section 8.
PRACTICE. Try these questions without a calculator.
1. What is 15% of 70?
2. A DNA fragment of 35 kilobases is digested by an exonuclease. The enzyme degrades seven kilobases. What percentage of the DNA is degraded?
3. You counted 10, 000 cells three weeks ago and 12, 000 today, what is the % increase?
"IS" and "OF" formula : % "is" number = 100 "of" number
13
Problem Solving Strategies Read the following section in your own time and reflect upon questions questions related to percentages. It is important to acknowledge that, in general, there is more than one way to obtain a solution. The following example is taken from [9] and illustrates strategies for mathematical problem solving. One of the main conclusions is: getting STUCK is a natural state of affairs: A company offers you a 30% discount but you must pay 15% tax. Which would you prefer to have calculated first, discount or tax? The natural response is to start by trying some specific cases. Say, we try it with an item priced at Ј100. DO SO NOW IF YOU HAVE NOT ALREADY Surprised by the result? Now, will the same happen for a price of say Ј120? TRY IT AND SEE! Write down your calculations and your insights. It is the only way to develop your thinking skills. There should be a pattern in the special cases you have tried. TRY EXAMPLES UNTIL YOU ARE SURE! Specialisation, which means turning to examples to learn, is as important as trying to vary your way of thinking. With any luck you will have found that 1. subtracting 20% from a price is the same as paying 80%, that is, you pay 0.8 times the price. 2. adding 15% to a price is the same as paying 115% of it, that is you pay 1.15 times the price. Then for any initial price of say Јx, calculating discount first: you pay 1.15 · (0.80 · Јx) tax first: you pay 0.80 · (1.15 · Јx) By writing the calculation in this form you can see that the order of calculation does not matter and, indeed, you pay the same in both cases.
Reflective thinking
5 Real Numbers
Like biologists classify species, mathematicians distinguish different kinds
of numbers. (Well, it is not quite the same but there are certainly some
unusual species on the real line...) A rational number is any number
which
can
be
expressed
exactly
as
a
fraction
a b
where
a
and
b
are
integers
(`whole numbers', i.e., . . . - 3, -2, , 0, 1, 2, 3 . . .). An irrational number
is any number which cannot be expressed exactly as a fraction in this
14


form. For example,
4
=
2 1
is
rational
while
2 = 1.4142135 . . . is
not. positive integers {1, 2, 3, . . .} are also called natural numbers.
Measurements are often assumed to be real numbers, i.e., any numerical
value we can represent on the number line:
e = 2.718 . . . = 3.141 . . .
-4 -3 -2 -1 0 1 2 3 4
Negative numbers are written to the left of zero. The further a number is to the right, the bigger it is. Addition indicates that you move to the right, while substraction takes you to the left:
-1 - 3 = -4
-2 + 5 = 3
-4 -3 -2 -1 0 1 2 3 4 Use the number line below to demonstrate the following properties of real numbers: 1. Addition (a) is associative, i.e., (a + b) + c = a + (b + c) (b) is commutative, i.e., a + b = b + a 2. Subtraction (a) is NOT commutative, i.e, a - b = b - a (b) is NOT associative, e.g., (5 - 2) - 1 = 3 - 1 = 2 but 5 - (2 - 1) = 4 - 1 = 3
Use drawings and examples to illustrate and memorise abstract concepts or rules.
-4 -3 -2 -1 0 1 2 3 4
What we cannot illustrate so easily with the number line is that
1. Division
(a)
is NOT
commutative:
a b
=
b a
,
i.e.,
we
cannot
divide
in
any
order and get the same answer.
(b)
is NOT
associative:
e.g.,
12 6 2
=
2 2
= 1 but
12 6 2
=
12 3
= 4.
2. Multiplication
(a) is commutative: a · b = b · a
15
(b) is associative: (a · b) · c = a · (b · c)
We should also remember that multiplying or dividing numbers of the same sign gives a positive number, while if any one of the two numbers is negative, the result is negative. Take the following examples:
3 · 4 = (+3) · (+4) = 12 (-4) · (-5) = 20
3 · (-4) = -12
-9 3
=
-9 3
=
9 -3
=
-3
6 Logarithms
A logarithm is the exponent or power to which a base must be raised to yield a given number. In the previous section we considered powers like 102 = 100 or more generally given the base, 10, and index n, we asked 10n =? On the other hand, given a number, say 100, and a base, say 10, we can ask for index n. In other words, what is the power that 10 is raised to, to give 100? This question has its own notation: log10 100 = 2 such that 102 = 100 . Try the following for yourself:
log10 10 000 =
log3 9 =
log2 64 =
log10 0.0001 =
There are three bases that are most relevant to science and engineering, base 10 (common logarithm), base 2 and base e. The latter is called natural logarithm1 (or Napierian log), denoted loge or ln: loge a = n such that en = a .
In general, i.e., for logarithms with any base, we find that
log 1 = 0 since a0 = 1 .
PRACTICE. Use a calculator to solve the following: 1. log10 3 =
2. log10 30 =
The logarithm of a number gives an exponent: logb a = n where bn = a Logarithm of a product: log(a · b) = log a + log b Logarithm of a fraction: log a = log a - log b b Logarithm of a power: log ab = b · log a
3. log10 300 =
4. log10 3000 =
5. log10 0.3 =
1Many calculators will use log to denote the logarithm to base 10 while ln is used to denote the natural logarithm.
16
6. log10 0.03 =
7. log10 0.003 =
There are many uses for logarithms. In the biosciences or specifically for the analysis of experimental data, it is often more convenient to use a log-scale of the number line:
log10 -4 -3 -2 -1 0 1 2 3 4
We notice that the distances between tick marks are equal although the difference in value increases rapidly. Obvious? Write the difference out on the number line above! Another way to see what the logarithm does to a given number is to produce a table:
Number 1/10, 000 1/1, 000 1/100 1/10 1/1 10 100 1, 000 10, 000 100, 000
log10 -4 -3 -2 -1 0 1 2 3 4 5
The table above is the first step towards the concept of a function. The example also suggests a method to use logarithms for "scaling" a graph (cf. Fig. 12.8). The use and usefulness of logarithms becomes clearer when we introduce this concept fully in Section 12.
PRACTICE. A number of rules for operating with logarithms are shown in the margins. Below are exercises to practice these rules. More exercises can be found in Section 19. Try the following questions (without a calculator or computer). 1. Show that 10(a-b) · 10(b-a) = 1
Notation: ln x loge x loge(x) Create tables to experiment with a new concept. If log10(a) = n, then 10n = a
2. Show by example that ab c = a(b·c)
3. Simplify 10a · 10a · 10a · 10a · 10a · 10a = 4. Simplify (10a)3 · (10a)-4
17
5. log (a · b2) - 2 · log b =
6. log
a2 b2
+2·b =
7. log (32) - log 3 - log 18 = Remark: Logarithm were invented by the Scotsman John Napier in the late sixteenth century. Because multiplication of large numbers is cognitively more complicated than to add them, Napier developed this concept. How products turn into sums is clear from the rule log(a · b) = log a + log b. Our parents are more appreciative of his efforts: the slide rule they used before pocket calculators came to our rescue, work on this principle.
7 Variables and Constants
When
we
introduced
fractions,
we
generalised
an
example,
say
3 4
·
2 5
=
3·2 4·5
by replacing the integers with letters:
a b
·
c d
=
a·c b·d
In dealing with more complex scientific equations, a quantity that varies is said to be a variable. If the symbol represents a fixed value and is not multiplied by any variables, then this is termed a constant. Coefficients are numbers that are multiplied by one or more variables. In -4xy, -4 is a coefficient and in -4xy + 3, `3' is a constant. Which letter is used for variables and which for coefficients or constants is arbitrary. For example, in -4xy + b, b may but must not be a constant. What is what, must be stated or in scientific equations is often obvious. To introduce new mathematical ideas, we usually make use of commonly used letters such as x, y to denote variables, f to denote a function. However, for applications of mathematical formula in the biosciences, we may prefer other conventions. For example in the following (biochemical) expression
v = Vmax · [S] , Km + [S]
the letter v denotes a variable (`velocity'), Vmax and Km denote constants and [S] is another variable (`substrate concentration', where the square brackets are used to denote concentrations). This may initially be confusing but you will hopefully find that notation is as useful as it is a nuisance. The equation above will be discussed in greater detail in Section 17 Generalisation through variables is an extremely important concept. In science, a measured set of experimental data is the evidence we use to establish or validate a principle (or `natural law'). The data are only one particular example from which we try to generalise. This usually leads to an equation (or set of equations), consisting of variables and
Vocabulary
18
constants as the one shown above. This representation of an observed phenomena is called a (mathematical) model. If we are fortunate and clever, the model will be a reasonable approximation to the natural system under consideration. If this is the case, we can make simulations and predictions. biological systems are much more complex than what engineers, physicists and applied mathematicians can successfully deal with in their field and therefore interdisciplinary collaborations have become very important in the life sciences.
Remark: The most famous equation of all is Albert Einstein's E = m·c2
by which he showed that energy (E) is equivalent to mass (m). In the
equation, E and m are variables and c is a constant: the speed of light
which in a vacuum is the enormous (and yet finite..) 299 792 456.2 ±
1.1
m s
.
Einstein's
life,
the
revolutions
in
physics
that
took
place
during
his time, how the equation E = m · c2 came about and how on earth
they measured the speed of light, is a fascinating story. If you want to know more, a readable book on the story of E = m · c2 is [2].
8 Algebraic Expressions and Equations
Algebra is a branch of elementary mathematics that generalises arithmetic by using variables to range over numbers. The following vocabulary is in use. Terms are constants or variable expression. Arbitrary examples are: 3a; -5c4d4; 25mp3; 7 are all terms. Algebraic expressions are terms that are connected by either addition or substraction. For example, 2s + 4a2 - 6 is an algebraic expression with three terms. Algebraic equations are statements of equality between at least two terms. Examples: 4z = 28 and 3(a - 4) + 6a = 10 - a. As can be seen from the examples, in algebraic expressions we use letters or symbols to represent a quantity. Take for instance the demonstration of the commutative law of addition:
a+b=b+a ,
or the law of association:
5b + a - b = 5b - b + a = 4b + a .
In the equation above, we simplified 5·b to 5b. If not stated otherwise, one uses only single letters and xy denotes the product x·y. This should only be done if no confusion is possible. To make associations clear we often will use brackets in forming expressions. Although this is done to make things clearer, brackets are also a source for errors when manipulating equations. Make sure that whatever quantity or symbol is found adjacent to the left-hand side of the brackets must multiply the contents of the brackets, including the addition and substraction sign. Take for example the expression
a - (b + c) = a - b - c .
(8.1)
Short forms and brackets
19
Note that the sign in front of c has changed with the removal of the brackets. On the other hand (a - b) + c = a - b + c
Browse back through the notes and remind yourself of the rules in Section 5! PRACTICE. Try these examples, NOW. Simplify the following expression where possible: 1. a - (2a + c)
2. a + p - c
3. xy + 2x - y + 4yx
4. -2(3 - y) For evaluation, rearrangement and simplification of expressions it is important to take care of the order in which operations are performed. Thus, for example, 5 · 3 + 2 · 4 - 3 is 20, not 25 or 65 because multiplication must be done before addition or substraction. Have a look at the following priority rules:
Evaluate expressions within brackets first Evaluate `inner' brackets before `outer' Take powers before multiplying or dividing
2 · (5 + 3) = 2 · 8 = 16 5 · (6 + 3 · (3 + 4)) = 5 · (6 + 3 · 7) = 5 · (6 + 21) = 5 · 21 = 135 5 · 32 = 5 · 9 = 45
not 10 + 3 = 13 not 5 · 9 · 7 = 315 not 82 = 64
With powers within powers work down from the top
x-2y2 means x(-2y2)
not (x-2y)2
Multiply and divide before adding and subtracting
3 · 4 + 8/2 = 12 + 4 = 16
not 3 · 12/2 = 18
Multiplication and division are done in the order in which they are found going left to right; that is, if division comes first going from left to right then it is done first. Similarly, addition and subtraction are done in the order in which they are found going left to right; that is, if subtraction comes first, going left to right, then it is done first. Note that the order or priority rules apply algebraic expression with variables and constant as well as for expression with numbers only. If in doubt with the result of a manipulation, use substitution, i.e., replacing symbols with numerical values. For example, considering equation (8.1) from above, let a = 6, b = 3 and c = 2:
a - (b + c) = a - b - c
6 - (3 + 2) =? 6 - 3 - 2
6 - 5 =! 1
Priority rules Substitution
20
which shows that both sides of the equation are balanced. Now try the same equation but with a = 6, b = 3 and c = 5:
a - (b + c) = a - b - c =? =!
Another example of expanding brackets is the following:
3(x - 2y) = 3x + 3 · (-2y) = 3x - 6y
or (x + 1)(x - 2y) = x2 - 2xy + x - 2y
Try this one:
(2x - y)(x + y)y =
In factorisation we express a number or expression in terms of a product. For example, the number 16 can be expressed in terms of its factor four (16 = 4 · 4). Many of us use factorisation in mental multiplication. For example, to solve 34 · 7, we could do the following:
34 · 7 = (30 + 4) · 7 = 30 · 7 + 4 · 7 = 210 + 28 = 238 .
We can apply the same idea to algebraic expressions. Consider the following expression and factorise it in x:
2x - 2xy + 3zx + x = x(2 - 2y + 3z + 1) = x(3 - 2y - 3z)
Factorisation often allows the cancellation of common factors and thereby simplifies an expression:
x xy - 3x x = x(y - 3) 1 = y-3 .
factorise xy - 3x, cancel common factors,
PRACTICE. Try simplifying the following expressions.
1. a · b + a · c =
Expansion: (a + b)(c + d) = ac + ad + bc + bd Factorisation: ab + ac = a(b + c) Use an extra sheet of paper for more space: Keep track of all steps.
2. a2 · b + a · c =
21
ac a3d 3. b + a2 =
2ab
4.
=
ab + 3ab
5. 2 - 6 = a+b b
Factorisation and expansion are two concepts frequently used to rearrange equations. In rearranging an equation we need to remember that we have to maintain the `balance' on both sides, i.e., we must preserve the equality of both sides. If you subtract, add, multiply and divide one side, the same has to happen to the other. Look at the following example, in which we try to isolate a to the left-hand side of the equation:
a·b+c·d = 1
a · b + cd - cd = 1 - c · d
a·b=1-c·d
a·b 1-c·d
=
b
b
1-c·d
a=
.
b
subtract (c · d) from both sides, divide both sides by b,
Normally you would not write the terms that cancel out and the second step in this example would therefore usually not be written out but is here included for completeness. In the example, I also wrote a · b which, here, is equivalent to ab. We could also have written (a·b)+(c·d) but this is not necessary because of the order in which operations are performed: multiplication first, addition afterwards. I said above that an equation is a statement that two quantities are equal, for instance 1 metre = 1000 mm. More often an equation contains an unknown quantity which we desire to find. In the equation 5x-7 = 23, x is the unknown quantity. There is only one value of x such that the left hand side (LHS) of the equation is equal to the right hand side (RHS). This value is x = 6. When we have calculated this value of x we have solved the equation and the value of x obtained is called the solution to the equation. In this example, the solution is x = 6. In the process of solving an equation the appearance of the equation may be considerably altered but the values on both sides must remain the same. A common problem is to isolate the variable. For example, to express x in terms of y and z:
xy - 2x = 3zx - 1
xy - 2x - 3zx = -1
x(y - 2 - 3z) = -1
x
=
- y
-
1 2-
3z
isolate all terms in x on one side, factorise, in terms of x, divide both sides by (y - 2 - 3z)
Rearranging equations
22
Try solving the following equation: 9x + 3 = 7x + 21
Solution: x = 9 It is important to check your result by inserting the solution into the original equation and checking the balance: when x = 9, LHS = 9·9+3 = 84. RHS = 7· 9 + 21 = 84. Correct. When an equation contains brackets, remember to remove them first and then solve: 5(2x + 6) = 10
Solution: x = -2 Check: when x = -2, LHS = 5 · (2 · x - 2 + 6) = 5 · 2 = 10 = RHS. Hence the solution is correct. When an equation contains fractions, multiply each term by the Lowest Common Multiple (LCM) of the denominators. The LCM of a set of numbers is the smallest number into which each of the given numbers will divide exactly (See Section 3). For example, solve
x+3 =3· x-2
45
2
52 Solution: x = = 2.08 25 Check:
Solve
x - 4 - 2x - 1 = 4
3
2
23
Solution: x = -29 = -7.25 4 Check:
PRACTICE. Rearrange the following equations. 1. Express b in terms of a, c and d (isolate b to the left-hand side): bc + ad = d2 - ad
2. Express a in terms of c, b and d: ad2 - ba = ca2
b-1
3. Express c in terms of a and b:
=a+b
c+1
4. Express y in terms of x: y2 + 2y + 1 = x
An important task in solving practical problems is to be able to translate information into symbols and thereby making up an algebraic expression. For example, find an expression which will give the total mass of a box containing x articles if the box has a mass of 8kg and each article has a mass of 400g. Produce a table in which you caculate for 0, 1, 2, . . . articles the total mass of the box. Then translate the table into a graph:
Remark: This process of building (hypothesizing) an equation, producing a table and a graph is central to mathematical modelling in the biosciences. We return to this issue in Section 12. The equatioin you derived here will re-appear in a general form in Section 13. If the price of an article is reduced from x pence to y pence, make an expression giving the number of extra articles that can be bought for Ј1: 24
If x sweets can be bought for 60 pence, what is the cost of y sweets?
To summarise this section, 1. To solve an equation the same operation must be performed on both sides. Thus the same amount can be added or subtracted from each side, or both sides can be multiplied or divided by the same amount. 2. After an equation has been solved the solution should be checked by substituting the result into the original equation. If each side of the equation has the same value, the solution is correct. 3. To construct a simple equation the quantity to be found is represented by a symbol. Then using the given information the equation is formed. Note that both sides of the equation must be in the same units. Remark: The word `algebra' comes from a ninth century book by Al-Khwarizmi (an Arab mathematician) called `Hisab al-jabr w'almuqabala' meaning `Calculation by Restoration and Reduction'. Algebra is a highly compact and efficient set of tools for solving practical and theoretical problems.
9 Measurement and Units
The basic idea of measurement is to compare and order. In Section 5 we used the number line to identify the order of real numbers and found that for any two numbers, the one on the right is greater than the other. An important concept for ordering is transitivity. This says that if I know that b is greater than a, and c is greater than b, then c must also be greater than a. To allow the comparison of attributes we require measurement or ratio scales such as length, area, volume, mass, weight, time interval, angle and many others. Official standards are in metric or SI (Systґeme Internationale d'Unitґes) units:
Attribute length mass time temperature amount of substance
SI unit metre kilogram second kelvin mole
Abbreviation m kg s K mol
Derived or compound units are
SI units
25
Quantity Energy Force Pressure Area Volume Capacity
Unit joule newton pascal square metres cubic metres litre
Symbol J N Pa m2 m3 L
Definition m2kg s-2 m kg s-2 m-1kg s-2 = N m-2
Note that the definitions become clearer when written as fractions. For
example,
the
pressure
is
defined
as
the force
per
square
meter:
1Pa
=
1N 1m2
.
A great advantage of the SI system is that it uses a base 10 system, with
standard prefixes to indicate the size of units:
giga mega kilo
G 109 M 106 k 103
deci centi milli
d 10-1 c 10-2 m 10-3
micro nano pico
µ 10-6 n 10-9 p 10-12
The system makes it easy to change from one size to another. So to express 2.11m in centimeters, it is only necessary to multiply by 100, so 2.11m = 211cm because 1m = 102cm = 100cm. Obviously there are again some rules to follow with units:
1. If you are dealing with measurements, never forget to write down the units and `carry' them through your calculations.
2. Only quantities which have the same units can be added or subtracted.
3. In ratios units can cancel. For example:
6m = 0.5 12 m
but 6 m kg s-2 = 0.5 m s-2 12 kg
A typical lab experiment (and exam question) involves conversions. For example, convert 55 millilitres to litres:
Since 1000 millilitres = 1 Litre
then
1 millilitre
=
1 1000
Litre
and
55 millilitres
=
55 ·
1 1000
Litres
=
55 1000
=
11 200
Litre
Converting 2.11m into cm, we can use the knowledge that 1m =
100cm and therefore only need to multiply by 100: 2.11 · 100 = 211.0 or
2.11m = 211cm. Notice that to multiply by one hundred we move the
decimal point two places to the right:
2.11 · 100=211.0
Similar, to convert 2.11m into mm, we know that 1cm = 10mm and 1m = 100cm = 1000mm. Multiplying by 1000 we move the decimal point three places to the right:
2.11m =
mm
Changing prefixes = moving the decimal point.
26
PRACTICE. Try the following and observe the movement of the decimal point:
2.11m =
km
23mL =
L
23mL =
µL
0.054g =
kg
0.054g =
µg
0.054g =
mg
The conversion of units and rewriting of numbers with different prefixes is a common source of errors. I would recommend you "explain" the following interconversions of non-SI and SI units of volume, by using the concept of powers with base ten, introduced in Section 2. Make a drawing to illustrate how the decimal point moves as you make the conversion!
1 litre (L) 1 millilitre (mL) 1 microlitre (µL) 1 nanolitre (nL)
103mL 1mL 10-3mL 10-6mL
= 1 dm3 = 1 cm3 = 1 mm3 = 1 nm3
= 10-3m3 = 10-6m3 = 10-9m3 = 10-12m3
The following table summarises the conversion between few common units to their SI quivalents:
Unit °Angstrшm Inch Ounce Pound Centigrade degree millimeters mercury Atmosphere Calorie
Symbol A in. oz lb C mmHG atm cal
SI-equivalent 10-10m 0.0254m 28.3g 0.4536kg (tC + 273)K [(5/9)(F-32)]C 133.322Pa 101325Pa 4.186J
PRACTICE. Try these questions without a calculator.
1. The height of the author of these notes is 2.11m. Given that 1 feet (ft) is 0.3048m, what is his height in feet?
2. Travelling to other countries you may find this useful: A gallon of fuel costs Ј2.55. (A gallon is approximately 4.5 litres). What is the price per litre?
3. Convert 4570 milligrams to grams.
Skip back to Section 2!
27
Remark: We used чb to make clear that we divide by b, e.g., a/b or we used Ч to describe the task of multiplication, e.g. a Ч a to denote the process of multiplying a · a. Most symbols, like for example % are recognised regardless of nationality, culture and have not changed over time. However, communicating your experimental results with colleagues in Germany and France you should note the following differences. While in Britain we write 0.4 (nought point four) there it is written as 0, 4. Thousands are also the other ways round: in Britain we write 1, 000 while for example in Germany one thousand is written 1.000. Of course measurement units are another example where some countries adopt different conventions. The emphasis is on convention ­ there is no true or correct standard and the next time you drive with a German to a pub (on what for him is the wrong side of the road) remember that "miles" and "pints" aren't natural to him or her. 10 Concentrations In the following two sections we deal with two common tasks in laboratory work: calculating concentrations and giving approximate results. In your laboratory work, a solution is defined in terms of the amount of solute (material) and the amount of solvent (liquid). Concentrations are expressed most commonly in terms of weight by volume (w/v) in which a given amount of solid is dissolved in a solvent to a fixed total volume, including the solute. Less frequently solutions are expressed as weight/weight (w/w), in which case a fixed mass of solvent is added to dissolve the solute. In summary, the following cases can be distinguished [10]: 1. Percentage weight/volume: % (w/v) = weight in grams of solute per 100mL of solution. Example: 1g (100mL)-1 = 1% (w/v) solution. In some textbooks or journal papers you may see concentration measured in milligrams per cent (mg%). This is defined as the weight of solute in milligrams per 100mL of solution. 2. Percentage volume/volume: % (v/v) = volume in millilitres of solute per 100mL of solution. Example: 1mL methanol plus 99mL water = 1% (v/v) solution. 3. Percentage weight/weight: % (w/w) = the weight in grams of solute per 100g of solution. Example: 15g of salt plus 85g of water = 15% (w/w) solution. If the concentration is very low it is common for either parts per million (p.p.m.) or parts per billion (p.p.b.) to be used, the assumption being made that the density of water approximates to 1g cm-3. Gas mixture concentrations are expressed as p.p.m or p.p.b on a volume/total volume basis. Molarity (M) is another very common expression of concentration. It expresses the number of moles of a substance that are present in a given volume of solution. A mole of a substance in grams (the gram mole) is numerically equal to its molecular mass. For example, for sodium chloride, which has a molecular mass of 58.5 daltons (from atomic weights:
quantity concentration = volume Molarity, Mole
28
Na=23 daltons, Cl=35.5 daltons) one mole is 58.5g. A 1M solution of a substance contains one mole of the substance in 1 litre of solution. Various units of concentration are therefore possible and the conversion rules for units, introduced in the previous section, are important. Familiarise yourself with the following interconversions of mol, mmol and µmol in different volumes to give different concentrations:
M 1 mol dm-3 1 mmol cm-3 1 µmol mm-3
mM 1 mmol dm-3 1 µmol cm-3 1 nmol mm-3
µM 1 µmol dm-3 1 nmol cm-3 1 pmol mm-3
PRACTICE. Try the following questions.
1. Assuming that in 100mL of water are 3g of salt. How much is in 20mL?
2. 100mL of a 10% (w/v) solution is prepared. Assume that the salt is evenly distributed throughout the solution. 50mL of the solution is removed.
(a) What is the amount of salt present, in grams? (b) What is the concentration of the 50mL sample that is removed?
3. What are the following concentrations in % (w/v)?
(a) 5g of glucose in a final volume of 50mL. (b) 7.5g of glucose in a final volume of 75mL.
Skip back to page 4 and 10!
11 Accuracy In Section 2 the scientific or standard notation was introduced as a way to present very large and very small numbers. Handling experimental data it is often necessary to decide what level of accuracy is required. The required accuracy is expressed by the number of significant figures. To express 349 to two significant figures, only the first two digits are displayed while the remaining digits are set to zero. Since the third digit is larger than 5, the result is 350. In general, if the last significant figure is smaller than five you round down, while for the last digit being greater than five you round up. The purpose of this is to minimise the error in rounding. For example, expressing 19 732 to three significant figures, we notice that position four is less than 5 and hence we leave the last significant figure as it is. The answer is therefore 19 700. So far we have not mentioned what to do if the last significant figures is equal to 5. To minimise the error it is better to use the rule that if the last significant figure is odd, i.e., {1, 3, 5, 7, 9}, then it should be rounded up, but if the last figure is even it should be rounded down. For example, 365 to two significant figures: The number in position three is 5 and hence the rule is applied. Since the last significant figure, 6 is even, we round down. The answer is 360. Our previous examples only considered integer values. To express 2.342 to two decimal places, we notice that the third digit after the
Significant figures The even/odd rule Decimal places
29
decimal point is the value 2, which is less than five and hence the answer is 2.34. To compare both, significant figures and decimal places, consider the number 0.0457 and express it first to two significant figures: We count only non-zero values, so the first figure considered is 4. The value at position three is greater than 5; hence the answer is 0.046. To express 0.0457 to two decimal places, we notice that the value at position three is 5, so we round up and 0.0457 becomes 0.05 to two decimal places. PRACTICE. Try the following questions. 1. Express 7849 and 375 to two significant figures. Describe the error for rounding up and down. 2. Represent the following to three decimal places: (i) 45.09653 (ii) 0.464782 (iii) 0.00089 (iv) 1289.632 3. Represent the following to three significant figures: (i) 23.347893 (ii) 128904 (iii) 0.003429 (iv) 267491.954
12 Relations and Functions A set X is a collection of arbitrary objects (e.g. points in the plane, real numbers or symbols), called the elements of X. If x is an element of X, we write x X. Curly brackets are used to write a set as a list of elements, X = {x}. A relation is a set of ordered pairs; the set of first elements in each ordered pair is called the domain, and the set of second elements is called the range. A function is a relation for which each value in the domain corresponds to a unique value in the range. A function is also referred to as a mapping or map for short (See Figure 12.1). The term mapping suggests some form of `rule' that describes how elements in the domain are related to the elements in Y . For example, in Section 6, we discussed logarithms: assuming a base b, and given a number x, the logarithm determines the exponent y = logb(x) such that by = x. With base 10 and x = 1000 the logarithm is 3 since 103 = 1000. While we keep the base fixed, there is no reason why we shouldn't explore what happens for other numbers. In Section 6, we produced a table for some obvious candidates. Assuming a fixed base b, and replacing a particular number by the symbol x, we treat x and its logarithm y as variables. In the equation y = logb(x) the brackets are used to emphasise the dependency of y on x. y is then also called the dependent variable while x the independent variable. With this generalisation we obtain a logarithmic function. In another example, let the domain, denoted X, be a subset of all genes in a genome and let the range, Y , denote all known `functional classes'1. The relationship between genes and gene products is a relation not a function because more than one gene can code for the same product.
Relations are multi-valued while functions assign unique elements Skip back to Section 6! Independent and dependent variables.
1Most areas of research have evolved their individual way of communicating and representing knowledge. The mathematician's and the life scientist's vocabulary overlap occasionally. The term `function' is an example. In mathematics the definition of a function is unambiguous and we are going to use the alternative term `mapping' if there is a risk of confusion.
30
x1
f
y1
y2
x2
y3
X
Y
function f
x1
R
y1
x2
y2
X
Y
relation R
Figure 12.1: Examples of a relation (bottom) and a function (top). While a relation is multi-valued, one element in domain X can be related with more than one element in the range Y .
The concept of mathematical relations is frequently used by computer scientists (think of relational databases). However, the definition of a function, which is stricter than that of a relation, provides the basis for very powerful methodologies developed within the areas of applied mathematics, physics and engineering. There are several ways of representing functions but a plot in the Cartesian plane remains the most revealing form of presentation. In 1637, the French philosopher and mathematician Descartes published a book entitled La Gґeomґetrie which set out a new way of linking algebra and geometry. The system, which grew out of his work describes a point with reference to two perpendicular lines or axes. Figure 12.2 illustrates a relation in the Cartesian space, denoted X ЧY and formed by all points in the plane (ordered pairs (x, y)). For a function, the set of ordered pairs {(x, y)} is called its graph.
Y -axis
X ЧY
Cartesian coordinates
R
y
(x, y)
x (0, 0)-origin
X -axis
Figure 12.2: A relation R defined in the Cartesian plane X Ч Y is a subset of X Ч Y , i.e., a set of ordered pairs {(x, y)}. Let us consider a somewhat idealised situation in which you conducted an experiment in which you studied the growth of bacteria. Starting off
31
with one cell at an arbitrary time 0, you make four consecutive measurements after equal time intervals. Let us assume you obtained the following data: time 0 1 2 3 4 count 1 2 4 8 16
Visualised with their Cartesian coordinates, in Figure 12.3 (on the left), the experimental data suggest that we may be able to model the data and then simulate the experiment with a function (why not a relation?). In Figure 12.3 on the right we see the plot of the function y = 2x which shows a good approximation to the experimental data and if it is reasonable representation of the biology would allow us to make predictions.
bacteria growth cell count 20
plot of the function y=2x 20
15
15
count y=2x
10
10
5
5
0
0
0
1
2
3
4
5
0
1
2
3
4
5
time
x
Figure 12.3: Experimental data on the left and plot of exponential function y = 2x on the right. The function (model) allows us to make predictions.
The principal purpose of experimentation is to generalise from the special case obtained in our experiment to what one would expect to observe if the experiment is repeated and measurement errors are minimal. In other words, we try to recognise a pattern in the data. Ultimatively, we like to postulate some principle by which the natural system under consideration operates. The process of mathematical modelling is illustrated in Figure 12.4. Despite the enormous advances in applied mathematics and the very successful representation of natural systems in physics, the complexity of biological systems has meant that we often face a dilemma, Albert Einstein summarised as follows: So far as the laws of mathematics refer to reality, they are not certain. And so far as they are certain, they do not refer to reality. Despite the difficulties in building accurate mathematical models of genetic systems, mathematics and statistics play a major role in state-ofthe-art molecular biology, biochemistry and bioinformatics. The discovery of structure in data, the formulation of causal entailment in genetic systems will always be the natural scientist's job; mathematical modelling is a way of thinking that helps you in the discovery! The need for theory and mathematical modelling was nicely summarised by Henri
The purpose of mathematical modelling
32
Phenomenal World ambience NATURAL SYSTEM
decoding (prediction)
Mathematical World the self
Natural Law
FORMAL SYSTEM
causal entailment
encoding (observation, measurement) inferential entailment
Figure 12.4: The modelling relation between a natural system S and a formal or mathematical system M. If the modelling relation brings both systems into congruence by suitable modes of encoding and decoding, it describes a Natural Law . In this case M is a model of S, that is, S is a realisation of M.
Poincarґe: `Science is built up of facts, as a house is with stones. But a collection of facts is no more a science than a heap of stones is a house.' Figure 12.3 shows how we can generalise from our experimental data. Given y = 2x we can calculate the value y for any time in the future or in between the instances at which we made our measurements. For example, at x = 2.5, using the model y = 2x, we find y = 22.5 = 5.66. Despite the apparent success of our modelling approach there is scope for questions. The mathematical function y = 2x grows to infinity as x increases but if we had continued our measurements, would the number of cells increase indefinitely? With your cultures growing on a plate of limited size, this is unlikely. Also, how do we get any function y = f (x) from any set of training data? Admittedly, the data were deliberately chosen to fit to the exponential function and with more general situations and measurement errors we enter the area of statistics which is not part of this course. In statistics, specifically regression analysis, a number of very powerful methods are available to fit functions to experimental data. Most software packages, which handle data, have such functions built-in.
exponential functions 10
y=10x
y=ex y=2x
8
Statistics, regression analysis
6
y
4
2
0
-3 -2 -1
0
1
2
3
4
x
Figure 12.5: A selection of exponential functions.
33
logarithmic functions 3
2
y=log (x) 2
y=log (x)
1
e
y=log (x) 10 0
y
-1
-2
-3
0
1
2
3
4
5
6
7
x
Figure 12.6: A selection of logarithmic functions.
The irrational, if not bizarre, number e and its functions such as loge(x) and ex play an important role in applied mathematics2. Examples are:
1. In a system at thermal equilibrium at an absolute temperature T , the numbers n1 and n2 of molecules in two states with energies E1 and E2 respectively are related according to the equation
n1
E2 -E1 = e kT
n2
in which k is the Boltzmann constant (= 1.38 Ч 10-23JK-1 = R/N , where R = 8.31JK-1mol-1 is the gas constant and N = 6.02 Ч 1023mol-1 is the Avogadro constant). If we take the natural log- arithms on both sides, we can express the difference between the energies as
E2 - E1 = kT loge
n1 n2
2. In a first-order reaction with rate constant k, the extent of reaction after time t is proportional to 1 - e-kt. See Figure 12.7. More on this later!
3. The decay of a radio-isotope to a non-radioactive form can be described in the form `the instantaneous change in a quantity with time is equal to a constant times the current value'. Translated into a mathematical expression, we get the following differential equation
d m(t) = k · m(t) dt
(12.1)
2An important property of the function y = ekx is that it is the only function which is equal to its own derivative, i.e., if y = ekx, dy/dx = ekx.
34
In other words, the rate of decay of a radioactive substance ­ and the amount of radiation it emits ­ is at every moment proportional to its mass m. The solution of this differential equation is
m(t) = m0 · ekt
(12.2)
where m(t) is the number of radioactive particles in the sample at time t, and m0 is the amount in the starting material (the mass at t = 0) and k is the decay constant and is unique to each isotope. From the equation above, which is plotted in Figure 12.7, we can see that m will gradually approach 0 but never reach it. This explains why, years after disposal, radioactive waste can still be a hazard. The value of k determines the rate of decay of the substance and is usually measured by the half-life time, the time it takes a radioactive substance to decay to one-half of its initial mass. Different substances have vastly different half-life times. For example, the common isotope of uranium3, U 238, has a half-life of about five billion years, ordinary radium, Ra226, about sixteen hundred years, while Ra220 has a half-life of only twenty-three milliseconds. I am going to explain differentials and differential equations in Section 15; don't worry if you are unfamiliar with them!
4. If money is compounded continuously (that is, every instant) at an annual interest rate k, the balance grows exponentially in time.
5. When sound waves travel through air (or any other medium) their intensity is governed by a first-order differential equation where y denotes the intensity and x the distance travelled. A similar law, known as Lambert's law, holds for the absorption of light in a transparent medium.
Note that with numerous elements, such as fractions, in the exponent, it is sometimes typographically inconvenient to write function ex and instead we may write exp(x) which is identical.
y(t)=100 e-kt for k=-0.001 100
y(t)=1-e-kt for k=-0.05 1
80
0.8
60
0.6
y y
40
0.4
20
0.2
0
0
0
1000
2000
3000
4000
5000
0
time t
20
40
60
80
100
time t
Figure 12.7: Radioactive decay example (left) and example of firstorder reaction (right).
3The expression U 238 is a notation used in chemistry and is not the power of a number U !
35
exp(x) ex
Equation (12.2), describing the exponential decay of radioactive ma- terial, is interesting because there are a number of natural processes that can be modelled by this equation. We are going to see it again in Sec- tion 15 and Section 16 where we discuss a population model and rate equations, respectively. Another example that has led to its fame is ra- diocarbon dating. The method can be explained with the following example. If a fossilised bone contains 25% of the original amount of radioactive carbon C164, what is its age? The idea to answer this question was honored with the Nobel Prize in chemistry for W. Libby in 1960. He concluded that in the atmosphere, the ratio of radioactive carbon C164 and ordinary carbon C162 is constant, and that the same holds for living organisms. When an organism dies, the absorption of C164 by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the carbon ratio in the fossil with that in the atmosphere. The half-life of C164 is 5730 years. A result is obtained from the solution (12.2) to differential equation (12.1):
m(t) = m0 · ekt
Here, m0 is the initial amount of C164. By definition, the half-life (5730 years) is the time after which the amount of radioactive substance, C164, has decreased to half of its original value. Thus,
m0
·
ek·5730
=
1 2 m0,
e5730·k
=
1 ,
2
k = loge 0.5 = -0.000121 . 5730
The time after which 25% of the original amount of C164 is still present can now be calculated from
m0
·
e-0.000121·t
=
1 4 m0,
t
=
loge 1/4 -0.000121
=
11460
years
.
Hence the mathematical answer is that the bone has an age of 11460 years. Actually, the experimental determination of the half-life of C164 involves an error of about 40 years. Also a comparison with other methods shows that radiocarbon dating tends to give values that are too small, hence, 12000 or 13000 is probably a more realistic answer to our present problem. In Section 6, I mentioned that scientific data are often plotted using a logarithmic scale in graphical representations of the data. Considering the exponential function y = 2x, Figure 12.8 illustrates why it makes sense to visualise data in this way. Without a logarithmic scale, the y values increase exponentially and become very large, very quickly. Plotted on a log-scale, we obtain a simple straight line and it then becomes easier to read off values for a larger range of x. In Section 16, (see Figure 16.3) we are going to encounter a useful example for log-scales. Before we conclude this section, we have a look at two more curious functions: the logarithmic spiral and the `hanging chain'. Suspending a chain between two poles, its shape is closely modelled by the equation
a y=
e-
x a
+
e-
x a
.
2
Probably one of the most famous equations which has inspired mathematicians, biologists and artists alike is the logarithmic spiral:
r = a · eb
Radiocarbon dating More about the "half-life" on page 51 Log-scale plots
36
y log10 y
1200
y=2x
y=2x 104
1000
103
800 102 600 101 400 100 200
0
10-1
-4 -2
0
2
4
6
8
10
-4 -2
0
2
4
6
8
10
x
x
Figure 12.8: Representing data with log-scales. For both plots the same values of the function y = 2x are calculated. What is different, is the way the information is visualised.
where r is radius of a line with angle to the abscissa. The shape of the logarithmic spiral resembles the cross-section of nautilus shell but many other examples, which show this pattern have been found in the natural world. See the books [11, 12, 8] and in particular [1], which has many beautiful illustrations. Many other books on `chaos theory' and `fractals', are filled with examples with astonishingly good matches between mathematics and nature. Does this suggest that nature and natural laws are mathematical? Is in the sky? Do mathematicians discover mathematics or is it an invention that fits the world around us only so well because it is the creation of the human mind, which is part of the natural world? The book by Evelyn-Fox Keller [6] provides a fascinating discussion of the use of models in physics and the difference between the physical and biological sciences.
y y
150 180
Logarithmic spiral
90 8
120
60
6
4
2
4
30
3.5
3
0 2.5
Hanging chain y = a/2 (e x/ a+e- x/ a)
2
1.5
210
330
1
240
300
270
x
0.5
0
-2
-1
0
1
2
x
Figure 12.9: Logarithmic spiral (left), a = 1, b = 0.1 and 0 6. The `hanging chain', a = 1 (right).
A conclusion from this and the following sections should be that plots of functions are a powerful tool to analyse real-world problems and to visualise results. Central to all this is the concept of a mapping or function. For example, a mapping which we use when travelling by train is that between a set of names of stations and a set of times of day:
Conclusions
37
Manchester Stockport Wilmslow Watford Station Euston
10.00 10.09 10.17 12.24 12.44
As so often in mathematical modelling, this timetable makes a somewhat idealistic assumption ­ that the train from Manchester to London is fast and reliable... . Here is a mapping which might be of interest to parachutists. It gives the total distance fallen for various times after they jump, before the parachute opens, and neglecting air resistance:
Time in seconds:
Distance in metres:
0

0
1

4.9
2

19.6
3

44.1
4

78.4
5

122.5
A function was defined as any rule or method whereby, for any and every object in the original set, we can find a (unique) corresponding element in its range. Different ways of symbolising the same function helps us to understand, or to centre our attention on, different aspects of it. We used words to describe functions;, Venn diagrams with arrows as in Figure 12.1; algebraic equations; tables or sets of ordered pairs and graphs. The graph of a function is particularly useful and we will use this frequently in subsequent sections. You should not hesitate to make drawings, sketches to illustrate written ideas, tables and equations!
PRACTICE. Try the following questions.
1. If a bacterial culture contains N(t) bacteria at time t, then the growth of the population can be modelled by the equation:
N (t) = N0 · 2n
where N0 is the number of bacteria at the start and n is the number of generation times that have occurred. For example [10], a Bacillus subtilis bacterium divides approximately every 40min and a culture was found to contain 103 bacterial cells. How many cells are present after 10h?
2. For the decay of a radio-isotope as described above, the half-life for an isotope is the time taken for the amount of material to decrease by 50%. For 32P, the half-time t1/2 is 14.3 days. What is the value of the decay constant k?
3. A population increases at an annual rate of 4% to 360 000 over a period of ten years. Assuming exponential growth, what was the original size of the population?
38
Remark: The irrational number e = 2.718 . . . is indeed strange, and fascinating. The number e and exponential function e(·) are hugely important in science and engineering. Most of us get used to using e, directly or indirectly, without ever questioning it. If you are baffled by this number you are in good company. Most of your lecturers will not be able to tell you what e really is, what it means and why it appears in so many areas and disciplines. If you want to know more, I recommend the books by Maor [8] and Lakoff [7].
13 More on Functions
In the previous section, we pointed out that the aim of most scientific experiments is to try to identify relationship from data. In the Cartesian plane this means that we wish to quantify the relationship between variables x and y. We referred to y as the dependent variable, and the function f is to describe the process by which a value x is transformed to the `output' y. In mathematical notation, the statement "function f is a mapping from domain X to range Y " is written as f: X Y
and the statement "function f maps an argument x into the value y" is written as x y where y = f (x) .
In this representation of a function, f is deliberately not specified and we have numerous possibilities to consider. In previous sections we considered for example: x y = logb(x) , x y = 2x , x y = ex . One of the simplest (hence useful and important) functions is that of a straight line. Figure 13.1 shows the plot of the general straight line for which the equation is
y = a+bx .
(13.1)
In this equation, b and a are a coefficient and constant respectively. The straight line cuts the y-axis (or ordinate) at a distance a from the origin. Because of this relationship, a is known as the intercept on the y-axis. The intercept on the x-axis is found by putting y = 0, which gives a + bx = 0, and so x = -b/a. The meaning of b follows from consideration of how y changes when x changes. Suppose that y changes from y1 to y2 as x changes from x1 to x2. Then y1 = a + bx1 , y2 = a + bx2 If the first equation is subtracted from the second, the constant a disappears: y2 - y1 = a + bx2 - a - bx1 = bx2 - bx1 = b(x2 - x1) and if we divide both sides of the equation by (x2 - x1) and interchange the left- and right-hand sides we obtain the expression for b:
Straight line equation: y = a+bx Intercept a Gradient b
39
(ordinate) y y2 y1
-
a b
0
y = a + bx
y2 - y1
x2 - x1 a
b
=
y2 x2
- y1 - x1
x1
x2 (abscissa) x
Figure 13.1: Plot of the general straight line y = a + b x.
b = y2 - y1 x2 - x1
(13.2)
The coefficient b has a meaning very similar to that of the slope or gradient of a hill: the steeper the hill, the greater the gradient. It describes a rate of change: if we change x by one unit, y changes by b units. Suppose you need to find the equation of the line given only two points in the plane through which the line passes: (0, 3) and (3, 9). The gradient is given by:
b
=
y2 x2
- -
y1 x1
9-3
= 3-0
6 = 3
= 2.
The equation of a straight line is y = a + bx, so if we substitute for x, y and b the given values and the gradient from above:
9 = a + (2 · 3) a=9-6 =3.
The equation of the line is therefore given as y = 3 + 2x.
PRACTICE. Rearrange the following equations into `straight line' form. What are the gradients (slopes) and intercepts (on the x-axis)? 1. y - 3x = 1
"Rate of change" Determining the equation of a straight line
2. 2y - 3x - 1 = x
40
3. 0 = 4y + 5(2x - 3 + 10) 4. Assuming two variables x and y are linked by a linear relationship, find the equation of the line from the following data: (0, 2) and (2, 5) fall on the line. 5. The straight line has a y intercept 3 and includes the ordered pair (4, 4), what is the equation of the linear relationship? 6. Given the point (7, 3) lies on the line and that b = 4 find the equation for the straight line. 7. Sketch the following functions: y = 2x + 3, and -1 = y - x 41
y
11 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 x 11 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 x Figure 13.2: Plain grids for the exercise. Notice the aspect-ratio of the y and x axis. This is the usual representation from computer programmes - why? 42
y
14 Proportionality In this section we briefly look at a simple special case of the linear equation (13.1): y = a+bx . Consider the example y = 3x , where each time there is a change in x, there is a change in y which is two-fold greater than that of x. The equation can be rearranged so that x and y are represented as a proportion: y/x = 2 Any two quantities that can be represented as a proportion such that y/x = a constant are said to be proportional. The symbol for proportionality is `' and indicates that both sides of the equation are not equal but that they are related and can be represented as a proportion. In general, we can write A B implies A = kB where k is a constant terms the constant of proportionality. The example y = 3x is a special of equation (13.1): y = bx which shows that if two variables, here x and y, are related by this special linear equation, the constant of proportionality is equal to the gradient (y/x = 3) (Figure 14.1).
Proportion: A B implies A = kB
y
y=3x 12 9 6 3
0
1
2
3
4
5
x
Figure 14.1: Plot of x and y such that y = 3x.
43
15 Differential Equations
In Section 13, we introduced the concept of a function f : f: X Y x y = f (x) .
In the equation y = f (x), the variable y is referred to as the dependent variable while x is called the independent variable. We call this a general equation because function f is not explicitly specified. A particular example for the general case y = f (x) is the exponential equation we encountered previously: y(t) = y0 · ekt . In this equation y0 and k are coefficients, i.e., some fixed numbers, while t is used to replace x. The letter t is most commonly used to describe time. Therefore in case the equation above is intended to model some natural phenomena, time t is the independent variable and to emphasise the dependence of y on t we write y(t).
d distance
v
m s
2
v
=
2
m s
d2
1.5
1
v
=
1
m s
d1
0.5
v
=
0.5
m s
t
t
t1 t2 time
02468
Figure 15.1: Translation (horizontal movement) of a mass with constant velocity.
For many practical situations, the equation y = f (x) is unknown. However, through observation and experimentation we can often assess rates of changes of variable y. For example, watching a moving car, we observe its velocity as the rate of change of position1. As illustrated in Figure 15.1, we find that for a car with constant velocity,
v = d1 = d2 = d3 = · · · t1 t2 t3
and therefore for constant velocity, the quotient of the travelled distance
and the time required are independent of the length and position of the
time interval considered.
The derivative of a function y = f (x) is the rate at which the
quantity y = f (x) is changing with respect to the independent variable
x.
We
denote
the
derivative
as
dy dx
and
translating
the
previous
sentence
into mathematical jargon, we obtain the definition
1Speed is a measure of the distance travelled during a given time period, whereas velocity is a measure of the change in position during a given period of time.
Skip back to page 39 if the notation appears unfamiliar. Equation y = f (x): Dependent variable y. Independent variable x. Function f . Cf. page 40! d Derivative of function dx y = f (x).
44
dy
f (x + x) - f (x)
= lim
dx x0
x
(15.1)
The derivative is therefore defined as the limit of the change in which we imagine the increment x tends to 0. If y is dependent on only one variable, the derivative is equivalent to the gradient, introduced in Section 13 and illustrated in Figure 13.1. See Figure 15.2 for an illustration of the derivative of a general function. In Figure 15.2, we can see how the derivative is related to the limit of the gradient of the chord joining the points (x, f (x)) and (x+x, y+y).
y
y = f (x)
(x+x, y+y) Q
y (x, y) P x
0
x
Figure 15.2: Derivative of function f (x) at the argument x. dy/dx is the limit of y/x as Q approaches P .
An equation relating an unknown function and one or more of its derivatives is called a differential equation. Because a rate of change is about the difference between some quantity now and its value an instant into the future, equations of this kind are called differential equations. We find that many natural laws (principles by which nature appears to operate) are formulated as equations that relate not the bio-physical quantities of primary interest but the rates at which those quantities change with time, or the rates at which those rates change with time. In fact most molecular, genetic or natural processes are in fact dynamic processes. Although we ought to describe most processes by differential equations, this is often only feasible for simple systems. In this case one can often make assumptions which turn differential equations into simpler linear equations. Let us consider an example which was already briefly mentioned in Section 13. We denote the size of a population with the letter P and in order to emphasise the dependency on time we write P (t). The time rate of change of a population P (t) with constant birth and death rates is, in many simple cases, proportional to the size of the population. That is, a possible mathematical model is the differential equation
dP = k · P , dt
(15.2)
where k is the constant of proportionality. Even if the value of the constant k is known, the differential equation dP/dt = kP has infinitely many different solutions. The solutions are all of form
P (t) = C · ekt ,
(15.3)
Differential equations Population modelling
45
where any constant C defines one possible solution. To verify this assertion, we take the derivative of P (t):
dP (t) = Ckekt = k(Cekt) = kP (t) dt
for all numbers t.
Once more the number e and the exponential function appear and the reason is that it is the only function that is equal to its derivative. (See also page 34). Suppose that P (t) = Cekt is the population of bacteria at time t, that the population at time t = 0 (hours, h) was 1000, and that the population doubled after 1 h. This additional information about P (t) yields the following equations:
1000 = P (0) = Ce0 = C 2000 = P (1) = Cek .
It follows that C = 1000 and that ek = 2, so k = loge 2 0.693147. With this value of k the differential equation in (15.2) is
dP dt
= (loge 2)P
0.693147 · P
.
Substitution of k = loge 2 and C = 1000 in equation (15.3) yields the particular solution
P (t) = 1000e(loge 2)t = 1000· eloge 2 t = 1000·2t (because eloge 2 = 2)
that satisfies the given conditions. The "power" of mathematical modelling lies in the fact that we can use this particular solution to predict future populations of the bacteria colony. For instance, the predicted number of bacteria in the population after one and a half hours (when t = 1.5) is P (1.5) = 1000 · 23/2 2828 .
The condition P (0) = 1000 in this example is called an initial condition because we frequently write differential equations for which t = 0 is the "starting time". The population growth example illustrates the process of mathematical modelling (Figure 15.4):
1. Formulate your problem related to a biological system in mathematical terms, that is, construct the mathematical model.
2. Analyse, solve and/or simulate the mathematical model.
3. Interpret mathematical results in the context of the real process, try to answer the question you are investigating.
4. Validate the mathematical model with experimental data.
In the population example, the problem we investigate is to determine the population at some future time. The mathematical model consists of a set of variables, (P and t), coefficients, (k), together with one or more equations relating the variables, (dP/dt = kP , P (0) = P0), that are
Initial condition Mathematical modelling
46
4
C = 12
C=1
2
C = 0.5 0
P(t)
C = -0.5
-2
C = -12
C = -1
-4
-4 -3 -2 -1
0
1
2
3
t
Figure 15.3: Graphs of P (t) = Cekt with k = loge 2 and C = -12, -6, -3, -2, -0.5, 0.5, 1, 3, 6, 12.
known or assumed to hold. The mathematical analysis consists of solving these equations, (here, for P as a function of t). Mathematical results, predictions or simulations have to be interpreted with observations made for the real system. The best way to validate a model is to compare simulation results with experimental data. This leads us into the domain of statistical techniques which are also used to identify parameters (here k) from data.
Mathematical Model
Formulation
Mathematical Analysis
Natural System
Experimentation
Experimental Data Sets
Interpretation Mathematical Results
Statistical Analysis
Figure 15.4: The process of mathematical modelling. It is important to realise that the process of mathematical modelling makes assumptions and that it is quite possible that no one solution of the differential equation exists. The population model P (t) = Cekt does only approximately describe a bacteria colony growing on an agar plate with limited space. Our population function P (t) was also a continuous approximation to the actual population, which of course grows by integral increments. Mathematical modelling requires therefore a tradeoff between what is biologically, physically or chemically realistic and what is mathematically possible.
47
Remark: For some reason, unknown to mathematicians, differential
equations are perceived as difficult by most who have no degree in math-
ematics. These difficulties are quite reasonable because mathematicians
would have to admit that some of the implicit assumptions of the defini-
tion dy/dx are `suspicious' to say the least. In order to handle questions
about rates of changes in physics, Isaac Newton and Gottfried Leibniz
invented in the eighteenth century a new branch of mathematics, called
calculus. At the root of calculus is the metaphor that `instantaneous
change is average change over an infinitely small interval'. Thus, for
a function f (x) and an interval of length x, instantaneous change is
formulated as
f
(x+x)-f x
(x)
.
The instantaneous change in f (x) at x is
arrived at when x is "infinitely small". The limit metaphor we used
in (15.1) states that x approaches zero, without reaching it. Baffled?
There is no doubt that Newton's calculus and its applications to physics
led to a revolution in physics and changed our view of the world. How-
ever, despite the apparent rigour of mathematics, many of the greatest
ideas in math rely on `tricks'. After all, if dx describes an infinitely small
quantity, so does dy and dy/dx appear to be a meaningless fraction 0/0.
The full explanation is far from being trivial and for us and in our con-
text,
it
is
sufficient
to
view
d dx
as
an
operator
which
describes
a
rate
of
changes.
For a short but very good introduction to mathematical modelling and
calculus I strongly recommend the small book `Nature's Numbers' by
Ian Stewart [11]. Another book, ideally suited for those who thought
they would hate maths, is the beautiful book by Philip Ball on pattern
formation in nature [1].
16 Rate Equations In this section we summarise some of the insights gained and techniques learned in this course. You will find that virtually all ideas we have discussed so far, come together in the discussion of rate-equations as a model of kinetic processes. Throughout biomolecular sciences we meet many processes which exhibit growth or decay, and may be described by rate equations. Several general examples were already listed in Section 12 (see page 34) and Section 15 (the population model). Others are enzyme properties such as activation, activity, proteolysis as well as cellular properties such as the growth considered previously. The common element in all these, is that the rate of the process is proportional to the amount of the substrate itself. Take for example a protease E, which cleaves a specific peptide bond in a substrate protein S, and thereby activating it to yield the modified cleaved form S . The rate of cleavage (proteolysis) is proportional to the amount of inactive substrate S. We can write this as rate of proteolysis = -kp · amount of S where kp is a constant, referred to as the rate constant for this process. We write -kp because the process reduces the amount of S with increasing time. What I called "amounts", is more commonly written as concentrations. For concentrations, biochemists prefer to use following
Rates? See page 11 Proportionality? See page 43 Concentrations? See page 28
48
notational form1 [S]. The rate of proteolysis is therefore the change in substrate concentration over time:
d[S] dt
=
-kp
·
[S]
(16.1)
Where the notation d/dt represented an instantaneous rate of change in time and is called a derivative (See page 45). Let us consider an example of such a process with kp equal to 0.1 per second, and an initial concentration of 20mg of substrate. We have to give our rate constant units which relate how much is changed in a given unit of time. At time zero, t = 0, we have 20mg of substrate, and after one second, approximately 0.1mg · 20 are converted to the active form, giving 2mg. This leaves approximately 20 - 2 = 18mg still uncleaved (and inactive). After another second, 0.1mg·18 equal 1.8mg are converted, leaving 18 - 1.8 = 16.2mg, and so on. These calculations are only approximate since the differential equation assumes infinitesimally small intervals of time rather than our chosen one second. Nevertheless, consider the table of eight data points, along with the corresponding value for S .
t01 2
3
4
5
6
7
8
S 20 18 16.2 14.58 13.12 11.81 10.63 9.57 8.61
S' 0 2 3.8 5.42 6.88 8.19 9.37 10.43 11.39
The graphs of the data are plotted in Figure 16.1. Do you recognise the shape of the plots? Where have you seen them before?
20
S 15
Notation: concentrations in square brackets. Remember: "per second" is written /s or s-1
amounts (mg)
10
5 S
0
0
2
4
6
8
time (seconds)
Figure 16.1: Graphs of the amounts for inactive protein S and the modified cleaved form S . In fact, the data show an exponential relationship and are suitably modelled by the function y = e-x. Indeed, as we discussed in Section 15 on differential equations, the solution to our differential equation (16.1) is (See also Figure 15.3) [S] = [S]0 · e-kp·t
1Biochemical processes take place in a defined space (e.g. cell compartments). Assuming a compartment of known size, [A] for some metabolite A is equivalent to the number of molecules of A. We could then use refer to the concentration or the number of molecules interchangeably.
Remember: exp(x) ex
49
At t = 0, we have e0 = 1 and therefore [S] = [S]0, where [S]0 is the initial value, that is, the initial concentration or amount of the substrate protein. Let us rename S as product P , then given that [P ] = [S]0 - [S] we can substitute the equation for [S] to obtain an equation of P : [P ] = [S]0 - [S] = [S]0 - [S]0 · e-kp·t = [S]0 · 1 - e-kp·t Figure 16.2 shows the plot of the two mathematical equations for S and P. 20 P(t) 15
Initial value.
amounts (mg)
10
5 S(t)
0
0
5
10
15
20
time (seconds)
Figure 16.2: Graphs of the rate equations for S = S0 · e-kpt and P = S0 · (1 - e-kpt), S0 = 20mg, kp = 0.1s-1.
These are all first-order kinetic processes, which are very useful to describe a number of different systems. Very often, the amount of protein present is proportional to the activity of the protein, or some secondary measurement we can make regarding an enzyme or its properties. Typically, the rate-equations are used to represent irreversible processes, where "A" is "converted" or changed in some way to "B". This leads to the general equation
A = A0 · e-kt .
If we can express our measured quantity as a fractional measure of the initial amount or activity, this can help us obtain a value for k. Rewriting the equation above, we obtain
A = e-kt A0 Like substraction is the complementary or inverse operation to addition, or multiplication to division, so is the logarithm related to the exponential. Taking the natural logarithm, loge ln, on both sides of the equation:
loge
A A0
= -kt .
First-order kinetic processes Logarithms? See page 16
50
0
log (S/S ) e0
-0.5 -1 -0.5-(-1.5) =1
k = 1/(-10) = -0.1 p
-1.5 5-15=10
-2
0
5
10
15
20
time (seconds)
Figure 16.3: A value for the rate constant kp can be obtained from the gradient of loge(S/S0)(t).
We can now obtain a value for k by plotting loge(A/A0) versus t and determining the gradient of the straight line. For our previous example, this is illustrated in Figure 16.3. Another useful parameter we can discuss in the given context is that of the half-life of a process. This describes the time it takes for half of the initial quantity A to be converted into B. It does not matter how much A is there initially, the half-life, denoted t1/2 is still the same. Consider the general rate equation for first-order kinetics:
A = A0 · e-kt
At the half-life, t1/2, A = A0/2 = B, therefore,
A0 2
=
A0
· e-kt1/2
Rearranging gives
1 = e-kt1/2 2
or
loge
1 2
= -k · t1/2
Therefore
t1/2
=
loge 2 k
.
17 Michaelis-Menten Modelling
In Section 15 we introduced differential equations and as demonstrated in Section 16, we often observe rates of changes, dy/dx of functions, y, rather than the functions y themselves. In this Section we continue the theme, highlight problems with differential equations and introduce an important formula. Consider the reaction between two molecules A and B. Biochemists represent such an reaction using the following symbols and notation1: A+B P
1Who thought that only mathematicians are fond of symbols and notation?
Half-life of a process
51
As we have seen before, the rate of reaction can often be modelled as being proportional both to the concentration of A and to the concentration of B, i.e.,
d[P ] = k[A][B] dt
(17.1)
where [A], [B], and [P ] are the concentrations of A, B and P , respectively, at time t and k is a constant. This equation is not very useful as it stands; it would not tell us how much product has been formed after a given amount of time. We have to rearrange the equation, simplify it and then solve it to obtain an equation with only [P ] and t in it. The equation (17.1) reminds us of equation (15.2), the population model on page 45:
dP (t) = k · P (t) dt
(population model),
where k is the constant of proportionality and P (t) denotes the size of the population at time t. They are however not the same as there are four variables ([A], [B], [P ]), and t instead of only two, (P, t), in the population model. Taking account of the fact that every molecule of P that appears is the result of the disappearance of one molecule of A and one molecule of B; it follows that ([A]+[P ]) and ([B]+[P ]) are constants. Define [A]0 and [B]0 as the initial values of [A] and [B] at time t = 0 and when [P ] = 0. We can then replace [A] and [B] in equation (17.1) with ([A]0 - [P ]) and ([B]0 - [P ]), respectively:
d[P ] dt
=
k([A]0
- [P ])([B]0
- [P ])
.
(17.2)
This is then a `proper' differential equation with only one variable, [P ], depending on time t. Finding a solution to this differential equation is far more complicated than for the very similar population model. And yet the biochemical model is quite simple as it only models basic irreversible reactions. For a matter of completeness, and ignoring various steps of restructuring equation (17.2), we would obtain the following solution:
[A]0([B]0 - [P ]) = e([A]0-[B]0)kt [B]0([A]0 - [P ]) With this equation, given a time t and initial values for [A], [B], [P ] at t = 0, we could calculate a concentration [P ]. The moral of the story is that although we could, "in theory", find a solution for the model above, even simple kinetic models can lead to intractable mathematics. For example, the simplest model commonly discussed in enzyme catalysis is the Michaelis-Menten mechanism which leads to the following pair of simultaneous differential equations:
d[Y ] dt = k1
[E]0 - [Y ]
[A]0 - [Y ] - [P ]
- k-1y - k2y + k-2 [E]0 - [Y ] [P ]
d[P ] dt
=
k2[Y
]
-
k-2
[E]0 - [X]
[P ] .
52
Do not worry about the meaning of these equations, they serve to illus-
trate the complexity and what we can do about it. It is possible to remove
[Y ] from these equations: first, using the second equation to express [Y ]
in
terms
of
[P ],
and
differentiating
this
gives
an
expression
for
d[Y ] dt
in
terms of [P ] and its derivatives. These two can then be substituted into
the first equation to give a single differential equation containing only
two variables [P ] and t. However, this is as far as we can go, because the
resulting differential equation has no known analytical solution. (We can
solve and numerically simulate the equation for a specific set of initial
conditions. Mathematicians would however prefer to study properties of
general solutions - on paper.).
A common way out of this problem is to make assumptions that
simplify the equations. For the example above, one can for instance
assume (or adjust experimental condition such) that [A]0 is much larger than [E]0 and to consider only the time scale in which d[Y ]/dt can be regarded as negligible. This basic restriction leads to what is known as
steady-state kinetics.
A simple enyzme-catalysed reaction which converts substrate S to
product P is represented by biochemist as:
S + E ES P
where E represents the enzyme and ES the enzyme-substrate complex. The rate of reaction can be measured as the change in product concentration with time. We can define this as the velocity, v, of the reaction:
d[P ] v= dt
where [P ] is the concentration of product P . This rate of reaction is a change of product concentration over time and hence its unit is specified as mol · L-1 · min-1. The rate of reaction as a function of the substrate concentration leads to what is known as the Michaelis-Menten equa- tion:
v = Vmax · [S] Km + [S]
(17.3)
in which [S] is the concentrations of the substrate, respectively, at time t. Vmax is often called maximum velocity or limiting rate because v cannot exceed it under steady-state conditions. Km is called the Michaelis constant with unit mol · L-1. Since it is like v a velocity, it has the same units as v. The substrate concentration [S] is measured in `moles' per litre, mol · L-1. In Figure 17.1, the rate of reaction is plotted against substrate concentration. This plot is called the Michaelis-Menten plot. The Michalis-Menten plot in Figure 17.1 shows that it is difficult to estimate Vmax since the curve tends to Vmax but will never reach it under experimental conditions. To estimate Vmax it is therefore necessary to transform the curve defined by equation (17.3) into a straight-line equa- tion of the form (13.1). A way to do this was developed by Lineweaver
Steady-state kinetics Michaelis-Menten equation Michaelis-Menten plot Lineweaver-Burk plot
53
velocity of reaction
Michaelis-Menten plot 1.4 V max 1.2
1
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
substrate concentration
Figure 17.1: Michaelis-Menten plot.
and Burk:
v = Vmax · [S] Km + [S]
invert, ...
1 = Km + [S] v Vmax · [S]
separate Km and [S], ...
=
Km Vmax · [S]
+
[S] Vmax · [S]
that is,
1 =
Km
·
1
+
1
v Vmax [S] Vmax
This equation is now in the straight-line form y = a + bx, where
1 y= ;
1 x= ;
b = Km ;
a=
1
v
[S]
Vmax
Vmax
The graph of 1/v against 1/[S] will therefore produce a straight line and the y intercept is equal to 1/Vmax so Vmax can be found. The gradient is Km/Vmax and hence if Vmax is known, we can determine Km as shown in Figure 17.2.
54
1/ v 1/ v
Lineweaver-Burk plot 8 7 6
Lineweaver-Burk plot 1.05 1
5
0.95
4
3
0.9
2 0.85
1
Intercept = 1/ V = 0.833 max
0
0.8
0
0.2
0.4
0.6
0.8
1
0
0.005
0.01
0.015
0.02
0.025
1/[ S]
1/[ S]
Figure 17.2: Lineweaver-Burk plot as a way to estimate the limiting rate Vmax .
Remark: Please note that this section does not provide a full introduction to modelling enzyme kinetic reactions. The material of this Section demonstrated how the previously introduced basic mathematical concepts come together when we try to devise mathematical models of cellular processes. There are numerous (interesting) issues associated with the assumptions and applications to cellular dynamics (e.g. pathway modelling). For example, one might argue that if we have to make so many assumptions and simplifications to establish a mathematical model, how can the model then be useful? I would argue that the purpose of mathematical modelling is not only to provide accurate, quantitative predictions but the modelling process itself helps the life scientist to design experiments, generate and test hypotheses. If you don't believe me, Sir Paul Nurse, winner of The 2001 Nobel Prize in Physiology or Medicine wrote a year before his award in the journal Cell: "Dealing with these system properties, which ultimately must underlie our understanding of all cellular behavior, will require more abstract conceptualizations than biologists have been used to in the past. We might need to move into a strange more abstract world, more readily analyzable in terms of mathematics than our present imaginings of cells operating as a microcosm of our everyday world.". What he refers to is a change of thinking, away from molecular characterizations of the nuts and bolts in cell systems to an understanding of functional activity. Technological developments in the post-genome era make it increasingly difficult to make sense of experimental data without mathematical modelling and data analysis.
55
18 Supplementary Material The following books can be recommended for additional exercises and further reading (prices are estimated): [10] Introductory mathematics for the life sciences. 229 pages. ISBN 0 7484 0428 7. Ј16 [4] Basic mathematics for biochemists. 221 pages. ISBN 9 780198 502166. Ј14 [3] GCSE math designed for post-16 students. 586 pages. ISBN 0 7487 5510 1. Ј16 There are a number of excellent software packages for mathematics and data analysis. The packages are well established and for which numerous introductory books have been published. For some of these programmes student and campus licenses are available. · Matlab from MathWorks. · Mathematica from Wolfram Research. · Mathcad · Maple [5] The Internet provides of course a vast source of information, including a lot of material on mathematics. The following general reference may be useful. For example try searching the following sites for the keyword `logarithm' ! · Google search engine: http://www.google.com · Brittanica encyclopedia: http://www.britannica.com/ · Encarta from Microsoft: http://encarta.msn.com/ Other web-sites of interest are: Mathematics WWW Virtual Library: http://euclid.math.fsu.edu/Science/math.html Eric Weisstein's World of Mathematics: http://mathworld.wolfram.com/ 56
19 More Exercises Fractions Try to simply the following fractions. 12 3 1. + = 54
2. 2 ·
12 +
=
5 35
x 2x 3. + = y 3y
4. 5 ч 1 · 3 = 62 2
5. a - 2 · d = b c3
6.
a3 - b b-c
ч
c
a -
b
=
7.
d2 c
-
d2 c2
=
4x
8.
5y 10y2
=
57
Algebraic Expressions and Equations Simplify the following expressions by finding common factors. 1. a · b + a · c = 2. a2 · b + a · c = a2c a3d 3. b + a2 Rearrange the following equations. 1. Express b in terms of a, c and d, i.e., determine b =?): b · c + a · d = d2 - a · d 2. Express a in terms of c, b and d: a · d2 - b · a = c · a2 3. Express c in terms of a and b for the following expression: b-1 =a+b c+1 58
Variables and Constants In the following problems, rearrange the equations to give y in terms of x and any constants. Remember that in these examples, x and y are the only variables and that all other terms are constants or coefficients. For each example, also work out whether y increases or decreases when x increases. 1. A · y + C · x = 12 2. c · y2 - 17 = 18x 3. C = Z · x 1+y 59
Powers 1. Show that 10m = 10m-n 10n 2. Show that 10(a-b)10(b-a) = 1
3. Simplify 10a10a10a10a10a =?
4. Simplify (10a)3(10a)-4 =?
5.
Simplify
1 a2
(3a
-
4a3b)
=?
Using your knowledge of powers, especially with base 10, evaluate the following (without using your calculator ). Show your working. You can check your final answers with your calculator if you wish.
0.00006
1.
=
2000
2. 0.05 · 200 = 0.002
0.0009 1 4.9 · 105
3.
=
7000 30 0.1
4. (0.0005)2 =
Simplify each of the following: 60
1. 25 · 26 2. 34 · 37 3. 53 · 54 · 56 4. 3 · 32 · 35 75 5. 72 312 6. 34 28 7. 24 105 · 103 8. 104 37 · 36 9. 35 25 · 26 10. 22 · 27 11. 53 4 12. 3 · 54 2 13. 103 4 14. 2 · 32 · 53 4 33 15. 4 52 4 16. 73 61
SI Prefixes and Units
Work out the following in the correct SI units and using standard SI prefixes.
3 · 105m2
1.
=
3 · 102m
5 · 103ms-2 2. 5 · 10-2s-2 =
8 · 103ms-2 3. 2 · 103s-1 =
8 · 102mol · s-1 4. 2 · 108s-1 =
0.01mol2
5. 5
=
1000mol
62
Logarithms Use the rules for manipulating logarithms to simplify the following expressions. 1. log(ab2) - 2 log b = a2 2. log b2 + 2 log b = 3. log 32 - log 3 - log 18 = 4. Express x in terms of y. All other terms are constants. y = (1 + y2)ex 5. Express x in terms of y. All other terms are constants. y = -AB ln x 63
Functions Rearrange the following into straight line form. What are the gradients (slopes) and intercepts (on the y-axis) of these straight lines? Remember that gradients have signs. You don't need to plot these functions to obtain these values. 1. y - 3x = 1 2. 2y - 3x - 1 = x 3. 0 = 4y + 5(2x - 3) + 10 64
Graphs Sketch graphs of the following functions. The scales have been deliberately omitted from the axes. This has been done to make you think about the scales yourself. You should work out a few points by substituting values into the equations before plotting them. It is not critically important that your plots should be perfectly proportioned, but they should have the right shape. For each plot, try and say something about the way the gradient changes, if at all. Does y increase when x increase? Is the gradient always the same sign (positive or negative)? 1. y = x2 - 6x + 2 Y X
2.
y=
5 x
+7
Y
X
65
3.
y
=
20x x+5
.
How
does
y
vary
with
x?
What
is
the
maximum
value
y
can take?
Y
X
4. Plot these two functions on the axes below. You will certainly need your calculator in order to do this. Make sure you are using the natural logarithm function and not log10 y = ln x and y = 2.718x Y X What do you notice about these plots? Is there any relationship between them? From your graph, estimate the gradient at x = 1 for the function y = 2.718x. What do you notice about this value? (Hint: For the first function, use values of x from 0.1 to 1.5, and for the second function use values of x from -1.0 to 1.0). 66
20 More Exercises: Solutions
Fractions
Try to simply the following fractions.
1.
12 5
+
3 4
=
48 20
+
15 20
=
3
3 20
2. (Do the bracket first)
2 5
·
(
1 3
+
2 5
)
=
2 5
·
(
5 15
+
6 15
)
=
2 5
·
11 15
=
22 75
3.
x y
+
2x 3y
=
3x 3y
+
2x 3y
=
5x 3y
4.
(
5 6
ч
1 2
)
·
3 2
=
(
5 6
ч
2 1
)
·
3 2
=
10 6
·
3 2
=
5 2
5.
a b
-
(
2 c
·
d 3
)
=
a b
-
(
2d 3c
)
=
3ac 3bc
-
2bd 3bc
=
3ac-2bd 3bc
6. Note: (b - c) = -1 · (c - b) = -(c - b) then you can cancel these
terms out and simplify:
a3 -b b-c
ч
a c-b
=
a3 -b b-c
·
c-b a
=
a3 -b b-c
·
-(b-c) a
=
b-a3 a
7.
- = - = = d2
d2
cd2
d2
cd2 -d2
d2 (c-1)
c
c2
c
c2
c2
c2
4x
8.
5y 10y2
=
4x 5y
ч 10y2
=
4x 5y
·
1 10y2
=
4x 50y3
=
2x 25y3
Algebraic Expressions and Equations
Simplify the following expressions by finding common factors.
1. a · b + a · c = a(b + c)
2. a2 · b + a · c = a(ab + c)
3.
a2 c b
+
a3 d a2
=
a2(
c b
+
ad a2
)
=
a2(
c b
+
d a
)
Rearrange the following equations.
1. Express b in terms of c, b and d, i.e.,determine b =?
The simplest way to solve this one is to isolate the term in b on
the left hand side by subtracting ad from both sides. Then you can
factorise in a on the right hand side, before finally dividing both
sides by c.
bc + ad = d2 - ad
bc = d2 - ad - ad = d2 - 2ad = d(d - 2a)
b=
d(d-2a) c
or
d2 -2ad c
2. Express a in terms of c, b and d.
Factorise in a on the left hand side, and then divide both sides of
the equation by a. You could also do this second step first if you
wish (but be careful).
a · d2 - b · a = c · a2 a(d2 - b) = ca2
a(d2 -b) a
=
ca2 a
=
d2 - b
=
ca
a
=
d2 -b c
67
3. Express c in terms of a and b for the following expression.
(Treat the a+b term as a single term, and then you can divide both
sides of the equation by (a + b). Similarly, you can then multiply
both sides of the equation by (c + 1). Then it is simply a matter
of subtracting 1 from both sides.)
b-1 c+1
=
a
+
b
c+1
=
b-1 a+b
c
=
b-1 a+b
-1
Variables and Constants In the following problems, rearrange the equations to give y in terms of x and any constants. Remember that in these examples, x and y are the only variables and that all other terms are constants or coefficients. For each example, also work out whether y increases or decreases when x increases.
1. A · y + C · x = 12. Take Cx from both sides. Then divide both
sides by A.
Ay = 12 + Cx
y
=
12-C x A
y as x
2. c · y2 - 17 = 18x Add 17 to both sides. Divide both sides by c.
Then take square roots. y increases if we only take the positive
root.
cy2 - 17 = 18x
cy2 = 18x + 17
y2
=
18x+17 c
y=
18x+17 c
y as x
3.
C 1+y
= Z · x.
This is just like the earlier rearrangement.
You can
treat the (1 + y) as a single term, and multiply both sides of the
equation by (1 + y) and divide both sides by Zx. Then subtract 1
from both sides.
1
+
y
=
C Zx
y
=
C Zx
-
1
y as x
Simple Powers
1.
Show that
10m 10n
= 10m-n
10m ·
1 10n
=
10m10-n
=
10m-n
2. Show that 10(a-b)10(b-a) = 1
10(a-b)10(b-a) = 10(a-b)+(b-a) = 10a-a+-b = 100 = 1
or
10 10 = = = 1 (a-b) (b-a)
10(a-b) 10-(b-a)
10(a-b) 10(a-b)
3. Simplify 10a10a10a10a10a =?
= 10a+a+a+a+a = 105a
68
4. Simplify (10a)3(10a)-4 =?
=
(10a )3 (10a )4
=
1 10a
=
10-a
5.
Simplify
1 a2
(3a
-
4a3b)
=?
3a a2
-
4a3 b a2
=
3 a
- 4ab
or
3-4a2 b a
Using your knowledge of simple powers, especially with base 10, evaluate the following (without using your calculator ). Show your working. You can check your final answers with your calculator if you wish.
1.
0.00006 2000
=
6·10-5 2·103
=
6 2
· 10-8
= 3 · 10-8
2.
0.05·200 0.002
=
5·10-2 ·2·102 2·10-3
=
2·5·100 2·10-3
= 5 · 103
=
5000
3.
0.0009 1 4.9·105 7000 30 0.1
=
9·10-4 ·49·104 7·103 ·3·101 ·10-1
=
3·10-4 ·7·104 103
=
21 · 10-3
=
2.1 · 10-2
=
0.021
4. (0.0005)2 = (5 · 10-4)2 = 25 · 10-8 = 2.5 · 10-7
Simplify each of the following:
1. 25 · 26 = 211
2. 34 · 37 = 311
3. 53 · 54 · 56 = 513
4. 3 · 32 · 35 = 38
5.
75 72
= 73
6.
312 34
= 38
7.
28 24
= 24
8.
105 ·103 104
=
104
9.
37·36 35
= 38
10.
= 2 25·26
2
22·27
11. (53)4 = 512
12. (3 · 54)2 = 32 · 58
13. 103 4 = 1012
14. (2 · 32 · 53)4 = 24 · 38 · 512
15.
= 3 3 33
4
43
16.
= 52 4
58
73
712
69
SI Prefixes and Units Work out the following in the correct SI units and using standard SI prefixes.
1. Keep the values and units separate. Do values first, units last (or the other way around) but not together. Treat the units like other index values.
3·105 m2 3·102 m
= 103 · m2m-1
=
103m = 1000m
2.
5·103 ms-2 5·10-2 s-2
=
103 ms-2 10-2 s-2
= 103+2m = 105m
3. Keep the 8 divided by 2 separate too. Treat the powers of 10 together. Calculate the correct units independently. Rearrange (if necessary) as the final step.
8·103 ms-2 2·103 s-1
= 4 · 100ms-1
= 4ms-1
4.
8·102 mol·s-1 2·108 s-1
=
4 · 102-8mol · s-1s
=
4 · 10-6mol
5.
0.01mol2 1000mol
=
1·10-2 1·103
mol
=
10-5mol
Logarithms Use the rules for manipulating logs to simplify the following expressions. 1. Treat the log(ab2) term as log(ac) where c = b2. Then log(ac) = log a + log c (but c = b2) so... Write it now as = log a + log b2. Use 2 log b = log b2 to finish it off.
log(ab2) - 2 log b = log a + log b2 - log b2 = log a
2.
log(
a2 b2
)
+
2
log
b
=
log
a2
-
log
b2
+
logb2
=
log
a2
= 2 log a
3. log(32) - log 3 - log 18 = 2 log 3 - log 3 - log(6 · 3) = log 3 - log 6 - log 3 = - log 6
4. Express x in terms of y. All other terms are constants. Rearrange equation to isolate x term. Take natural logs of both sides. Rearrange the left-hand-side.
y
= (1 + y2)ex
leading
to
ex
=
y 1+y2
leading
to
x
=
ln(
y 1+y2
)
5. Express x in terms of y. All other terms are constants. Don't confuse exponentials and logarithms. Remember how to resolve y = ex. Do not forget ln e = 1.
y = -AB ln x
leading
to
ln
x
=
-
y AB
leading
to
x
=
e-
y Ab
70
Functions
Rearrange the following into straight line form. What are the gradients (slopes) and intercepts (on the y-axis) of these straight lines? Remember that gradients have signs. You do not need to plot these functions to obtain these values.
1. y - 3x = 1 y = 3x + 1 m = 3, c = 1
2. 2y-3x-1 = x 2y = x+3x+1 2y = 4x+1 y = 2x+1/2 m = 2, c = 0.5
3. 0 = 4y + 5(2x - 3) + 10 4y + 10x - 15 + 10 = 0 4y =
-10x + 5
y
=
-
10 4
x
+
5 4
=
-2.5x
+
1.25
m = -2.5, c = 1.25
Graphs 1. y = x2 - 6x + 2. The gradient starts out negative and becomes positive with a point of inflection at x = 3.
Y 30
25
20
15
10
5
-4
-2
-5
X
2
4
6
8
-10
2.
y=
5 x
+ 7.
The function
decreases
as x increases,
but
is
not
deter-
mined at x = 0. It converges to y = 7 at very low and very high x.
Y 15
12.5
10
7.5
5
2.5
-6 -4 -2 -2.5
X
2
4
6
8
71
3.
y
=
20x x+5
.
The function increases with x, is not determined at
x = -5, and slowly increases until a maximum value of 20.
Y 15 12.5 10 7.5 5 2.5 X 2.5 5 7.5 10 12.5 15
4. Plot these two functions on the axes below. y = ln x and y = 2.718x
They are inverse functions. e = 2.718, so gradient at x = 1 is e. This holds true at every point, so that the gradient of ex always equals ex.
Y 3
2
1
-2 -1.5 -1 -0.5 -1
X 0.5 1 1.5 2
-2
-3
72
Bibliography [1] Ball, P. (1999): The self-made tapestry: pattern formation in nature. Oxford University Press. [2] Bodanis, D. (2000): E = m · c2 A Biography of the World's Most Famous Equation. Walker Books. [3] Bowles, D. and Metcalf, P. (1997): GCSE Math: The modular course. Stanley Thornes (Publishers) Ltd, Cheltenham. [4] Cornish-Bowden, A. (1999): Basic Mathematics for Biochemists. 2nd edition. Oxford University Press [5] Dodson, C.T.J. and E.A.Gonzalez (1995): Experiments in Mathematics using Maple. Springer Verlag, Berlin-Heidelberg. [6] Keller, E.F.: Making Sense of Life. Harvard University Press, 2002 [7] Lakoff, G. and Nunez, R.E. (2000): Where mathematics comes from. Basic Books. [8] Maor, E. (1994): e: The story of a number. Princeton University Press, Princeton NJ. [9] Mason, J. and L. Burton, K. Stacy (1985): Thinking Mathematically. Addison-Wesley Publishing Company. [10] Phoenix, D. (1997): Introductory Mathematics for the Life Sciences. Taylor & Francis Ltd., London. [11] Stewart, I. (1995): Nature's Numbers: Discovering order and pattern in the universe. Reissued 2001. Phoenix Paperbacks. [12] Stewart, I. (1998): Life's Other Secret: The new mathematics of the living world. Penguin Books. [13] Suggate, J. and A. Davis, M. Goulding (1998): Mathematical Knowl- edge for Primary Teachers. David Fulton Publishers, London. [14] Wolkenhauer, O. (2001): Data Engineering: Fuzzy Mathematics in Systems and Data Analysis. John Wiley, New York. 73

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Title: Data Handling Skills
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Author: Olaf Wolkenhauer
Subject: mathematical analysis
Published: Tue Sep 24 09:16:27 2002
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