1 Powers and logarithms
1.1 Definition and calculation rules for powers
Suppose we have a positive number \(b\) which we call the basis, and we want to multiply this number by itself for \(n\) times, where \(n\) is an integer number larger than 0. Then we say that we want to raise \(b\) to the power \(n\). The philosopher and mathematician René Descartes (Figure 1.1) introduced the notation for power raising in an appendix to the famous Discours de la méthode1. He wrote, for example, \(b^2\) instead of \(b \times b\), just as a convenient abbreviation. We still use this notation, and write in general:
\[ b^{n} = b \times \cdots \times b \tag{1.1}\]
where on the right hand side the number \(b\) is multiplied by itself \(n\) times. We call \(n\) the exponent of \(b\) in such an equation. As an example if \(b=2\) and \(n=3\) then
\[ b^{n} = 2^{3} = 2 \times 2 \times 2 = 8 \]
Using the definition Equation 1.1 for this notation, we can determine rules for the multiplication of two powers with the same basis \(b\) as well as for raising a power to a power. You should memorize these rules to be able to calculate with powers easily:
Theorem 1.1 (Adding exponents) \[ b^{n} \times b^{m} = b^{n + m} \]
Theorem 1.2 (Exponentiating powers) \[ b^{n^m} = {\left( b^{n} \right)}^m = {\left( b^{m} \right)}^n = b^{n m} \]
Proof. The proof2 of Theorem 1.1 is a matter of using the definition of \(b^{n}\) and \(b^{m}\) and putting the \(\times{}\) symbol between them:
\[ b^n \times b^m \equiv \underbrace{b \times b \times \cdots \times b}_\text{n times b} \times \underbrace{b \times b \times \cdots \times b}_\text{m times b} = \underbrace{b \times b \times \cdots b}_\text{n + m times b} \equiv b^{n + m} \]
By definition, the last series of \(n+m\) \(b\) is the same as \(b^{n + m}\).
Proving Theorem 1.2 goes in a similar fashion.
There is also a rule that applies to multiplications of numbers with a different basis, for example bases \(a\) and \(b\), but with the same exponent, \(n\).
Theorem 1.3 (Numbers with the same exponent) \[ a^n \times b^n = {(a b)}^n \]
Proof. The proof of this rule is again relatively simple, because you can re-arrange the sequence of \(a\)’s and \(b\)’s as follows: \[ a^n \times b^n = \underbrace{a \times a \times \cdots \times a}_\text{n times a} \times \underbrace{b \times b \times \cdots \times b}_\text{n times b} = \underbrace{ab \times ab \times \cdots ab}_\text{n times ab} = {(ab)}^n \]
Extending the definitions of power raising
We understand what \(b^n\) means if \(n\) is a positive integer number, larger than \(0\), because until now, the power was an instruction to write \(n\) times the basis \(b\) with a \(\times\) sign in between. The goal is now to extend this to bases \(b\) and exponents \(n\) from more general number sets, like the negative integers including 0 and the rational numbers \(\mathbb{Q}\) while applying the same rules as defined above. It is trivial to extend the definition to positive and negative integer bases \(b\), but we can also do this when \(b\) is a ratio. For, if we just follow the same rules then
\[ {\left ( \dfrac{p}{q} \right )}^n = \dfrac{p}{q} \times \dfrac{p}{q} \times \cdots \times \dfrac{p}{q}% \]
We can also write this as
\[ \dfrac{p \times p \times \cdots \times p}{q \times q \times \cdots \times q} = \dfrac{p^n}{q^n} = p^n \dfrac{1}{q^n} = p^n \dfrac{1^n}{q^n} = p^n {\left ( \dfrac{1}{q} \right )}^n% \]
Using this result, we can now create two definitions for \(b^n\) when \(n\) is a negative integer and when \(n=0\), that are consistent with the rule for addition of exponents, Theorem 1.1. Suppose, in the previous result, that \(p=q\). Then
\[ {\left( \dfrac{p}{q} \right)}^n = {\left( \dfrac{q}{q} \right)}^n = 1^n = 1% \] so
\[ q^n {\left ( \dfrac{1}{q} \right )}^n = 1% \]
If we now define \(q^{-n}\) as being \(q^{-n} \equiv {\left ( \dfrac{1}{q} \right )}^n = \dfrac{1}{q^n}\), and also define \(b^0\) for any \(b\) as \(b^0 \equiv 1\) then we can consistently apply rule Theorem 1.1, namely:
\[ {\left( \dfrac{q}{q} \right)}^n = q^n q^{-n} = q^{n + -n} = q^0 = 1% \]
So, we have added two definitions for power raising which we need to extend our rules to zero and negative powers:
Definition 1.1 (Negative integer power) \[ b^{-n} \equiv \dfrac{1}{b^n} \]
Definition 1.2 (Zero power) \[ b^{0} \equiv 1 \]
Answer
Because we can not prove these equations for any power \(n \in \mathbb{Z}\) since the original definition of power raising only considered powers in the set \(n \in \mathbb{N}\). Only consistency of applying the same rules for powers from \(\mathbb{N}\) to those from \(\mathbb{Z}\) requires these additional definitions.The following question is whether there is a way to interpret \(b^n\) when \(n\) is a ratio \(p/q\), where \(p,q \in \mathbb{N}\)! Applying Theorem 1.2 for exponentiation of exponents consistently on fractions we would have that
\[ b^{p/q} = b^{p \cdot 1/q} = {\left ( b^{1/q} \right )}^p \]
Suppose now that \(p=q\), then \(b^{p/q} = b^{1} = b\). Hence, for a consistent interpretation of \(n = p/q\) it should be true that
\[ {\left ( b^{1/q} \right )}^q = b% \]
In other words, we are looking for the number \(b^{1/q}\) which, if we raise it to the power \(q\) yields \(b\). There is an existing definition for such a number, namely \(\sqrt[q]{b}\). Therefore, the only consistent interpretation of \(b^{1/q}\) is:
\[ b^{1/q} \equiv \sqrt[q]{b}% \]
because, by the definition of the root, \({\left( \sqrt[q]{b} \right )}^q = b\). So we have a third definition for power raising, namely
Definition 1.3 (Rational power) \[ b^{1/n} \equiv \sqrt[n]{b} \]
To summarize, we can say that we know what \(b^n\) means for all positive and negative rational numbers \(b\) and all positive and negative rational numbers \(n\), including \(b=0\) and \(n=0\). Strictly speaking, we do not yet know what \(b^n\) means when \(b\) or \(n\) are irrational numbers3 (like \(b=\pi{}\) or \(n=\pi{}\)), but let’s assume that mathematicians have proven the consistency of similar interpretations for rational numbers for the irrational numbers too. In practice there is no problem anyway, since on paper or with a calculator we only calculate with rational numbers4.
Powers of units
When calculating with powers of quantities that have an associated physical unit the rules introduced above can be applied to the unit symbols and their interconversion factors as well.
For example, how many cubic micrometer fit into a liter? A cubic micrometer equals \((1 \; \mu\text{m})^3 = (10^{-6} \; \text{m})^3 = 10^{-18} \; \text{m}^3\). A liter equals \((1 \; \text{dm})^3 = (0.1 \; \text{m})^3 = 10^{-3} \; \text{m}^3\). The ratio of these two quantities is
\[ \frac{10^{-3} \; \text{m}^3}{10^{-18} \; \text{m}^3} = 10^{-3 + 18} = 10^{15} \] Hence there fit \(10^{15}\) cubic micrometers in one liter.
1.2 Definitions and calculation rules for logarithms
A logarithm is defined as the inverse function of power raising.
Definition 1.4 (Logarithm) The logarithm with basis \(b\) of a number \(x\), written as \(\log_b x\), is defined as the number \(n\) to which we should raise \(b\) in order to obtain \(x\).
In other words, if \(b^n = x\), then the logarithm of \(x\) with basis \(b\) is
\[ \log_b x = n \]
We can derive three basic formulas for calculating with logarithms. You should memorize these formulas to be able to easily calculate with logarithms.
Theorem 1.4 (Logarithm of a product) \[ \log x y = \log x + \log y \]
Theorem 1.5 (Logarithm of a quotient) \[ \log \dfrac{x}{y} = \log x - \log y \]
Theorem 1.6 (Logarithm of a power) \[ \log x^n = n \log x \]
Proof. All three formulas can be derived from the definition of the logarithm and from the previously discussed rules for calculating with exponents. Suppose we define \(p=\log_b x\) and \(q=\log_b y\), or equivalently, \(x=b^p\) and \(y=b^q\). Then it follows from from the rule for the addition of powers (Theorem 1.1) that
\[ x y = b^p b^q = b^{p+q} \]
Applying the definition just given it follows that
\[ \log_b x y = \log_b b^{p+q} \equiv p + q = \log_b x + \log_b y \]
with which we have proven Theorem 1.4.
Using the same definitions (\(p=\log_b x\) and \(q=\log_b y\)), it follows from the rule for the exponentiation of powers (Theorem 1.2) that
\[ x^n = {\left( b^p \right)}^n = b^{np} \]
from which we derive that
\[ \log_b x^n = \log_b b^{np} = n p = n \log_b x \]
With this we have proven Theorem 1.6.
Using the following formula we can map logarithms from one basis to another.
Theorem 1.7 (Changing the basis from \(a\) to \(b\)) \[ \log_b x = \dfrac{\log_a x}{\log_a b} \]
Proof. You can derive this formula from the rules for calculating with powers, Theorem 1.6 and Definition 1.4.
We want to recalculate \(x = b^p\) to a power \(q\) of \(a\), say \(x = b^p = a^q\). In that case, by definition, \(p = \log_b x\) and \(q = \log_a x\). From our assumption \(x = b^p = a^q\) it follows that
\[ \begin{align*} \log_a b^p &= \log_a a^q \\ p \log_a b &= \log_a a^q \\ p \log_a b &= q \end{align*} \] Or
\[ p = \dfrac{q}{\log_a b} \]
And therefore
\[ \log_b x = \dfrac{\log_a x}{\log_a b} \]
The unit of a logarithm
According to Definition 1.4 the logarithm of a quantity is a number that is a power to which a base number should be raised. We do not have the concept of the physical unit of such a power. Yet, we often take the logarithm of a quantity that has a unit. For example, when calculating the pH, we calculate the base-10 logarithm of the proton concentration: \(\text{pH} = -\log_{10} \left( [\ce{H+}] \right)\). Clearly, it makes a difference whether we express the concentration in molar or millimolar. How do we avoid a confusion about which units to use for \([\ce{H+}]\)?
The solution is to be more precise about the definition of pH: if we agree that logarithms have no physical dimension then we should make the number of which we take the logarithm dimensionless. Formally, we should say that the pH is the base-10 logarithm of the concentration divided by a reference concentration, and we agree upon a reference concentration of 1 Molar. The ratio then becomes a dimensionless number, a relative concentration. Therefore, the precise definition of pH is
\[ \text{pH} = -\log_{10} \left( \frac{[\ce{H+}]}{1 \; \text{M}} \right) \] If the number in brackets is to become dimensionless, then clearly we should express \([\ce{H+}]\) in molar. The same holds for any other physical quantity of which we take the logarithm: we take the logarithm of a dimensionsless ratio of the physical quantity relative to a reference quantity with the same dimension. Clearly, to avoid confusion, we must agree upon that reference quantity.
The exponential function and the natural logarithm
The exponential function exp is defined5 as
\[ \exp(x) = \lim_{n \to \infty} {\left( 1 + \dfrac{x}{n} \right)}^n \]
This function has a number of properties, which we accept without proof, that make it coincide with the notion of raising a number \(e\) to the power \(x\), or in other words, we could write \(\exp(x) = e^x\). The number \(e\), or \(\exp(1)\), is called Euler’s number and equals6
\[ e = \lim_{n \to \infty} {\left( 1 + \dfrac{1}{n} \right)}^n \]
The number \(e\) is used as the basis for the so-called natural logarithm. The natural logarithm of a number \(x\), also written as \(\ln x\), which is the same as \(\log_e x\). The number \(e\) is an irrational number. So, we cannot write it down, but we can approach it as close as we want by developing the convergent series
\[ e \approx 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \cdots = 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \ldots \]
where \(n!{}\) is the factorial of \(n\), which is defined as \(n! = 1 \times 2 \times 3 \times \cdots \times n\). The number \(e\) equals approximately \(2.718\).
What is so natural about this irrational number to call the logarithm with \(e\) as a base the natural logarithm? That has to do with the fact that exponents and logarithms with basis \(e\) occur, in contrast to those of other bases, often in a natural way in answers to mathematical problems. For example, an anti-derivative of \(1/x\) equals \(\ln x\), or in other words:
\[ \dd{\ln x}{x} = \dfrac{1}{x} \]
In contrast, the derivative of \(\log_2 x\) also contains a constant which happens to be equal to \(\ln 2 \approx 0.693\):
\[ \dd{\log_{2} x}{x} = \dfrac{1}{x \cdot \ln{2}} \]
Another example (see also ): for any basis \(a \neq e\)
\[ \dd{a^x}{x} = \ln a \cdot a^x \]
whereas
\[ \dd{e^x}{x} = e^x \]
We will see many uses of \(e^x\) and \(\ln x\) in our biological examples later on.
Understanding the definition of the function \(e^x\)
An understanding of the definition \(\exp(x) = e^x = \lim_{n \to \infty} {\left( 1 + \dfrac{x}{n} \right)}^n\) can be deduced from the following argument.
Suppose you have one euro on a bank account for which you obtain interest at a rate \(x\) per year. For example, if your bank pays 1% interest per year then \(x = 0.01\). If your bank pays interest on a yearly basis then after a year that euro has increased to an amount \(1 + x\). However, if your bank pays interest on a half-year basis but at the same yearly interest rate then the amount of money after half a year equals \(1 + \frac{x}{2}\), and after a year it equals \[ 1 + \frac{x}{2} + \frac{x}{2} + \left(\frac{x}{2}\right)^2 = 1 + 2 \frac{x}{2} + \left(\frac{x}{2}\right)^2 = \left(1 + \frac{x}{2}\right)^2 \]
This is \(\left( \frac{x}{2} \right)^2\) more than when interest is payed on a yearly basis because of the additional interest that you get on the interest in the previous half year. Continuing this argument, if your bank pays interest on a quarterly basis, the total amount of money after a year will be
\[ \left( 1 + \frac{x}{4} \right)^4 \]
If your bank would pay interest every \(n\)-th part of the year the amount of money after a year would be
\[ \left( 1 + \frac{x}{n} \right)^n \]
When interest would be payed on a continuous basis (every “split second”) then your account after a year would approach the already mentioned limit above for the function \(e^x\).
1.3 Exercises
Simplify the following expressions
Exercise 1.
\(5^{1/3} \cdot 200^{1/3}\)
Exercise 2.
\(\sqrt{1+x} \cdot (1+x)^{3/2}\)
Exercise 3.
\(\dfrac{a^{x-1}}{a^{x+2}}\)
Exercise 4.
\(\dfrac{1}{\sqrt{e^{-x}e^{7x}}}\)
Exercise 5.
\(\dfrac{\ln(a)}{\ln(b^x)+\ln(c^x)} e^{\ln(e^5)}\)
Exercise 6.
\(\ln{\left(x^2-2x+1\right) - \ln{\left(x - 1\right)}}\)
Solve the following expressions for \(x\) using factorization where possible
Exercise 7.
\(x^2-\sqrt{2}x-\frac{5}4 = 0\)
Exercise 8.
\(9x^2-12x+4= 0\)
Exercise 9.
\(5x^2-4x-1 = 0\)
Exercise 10.
\(x^2 + 3 x - 10 = 0\)
Solve the following expressions for \(x\)
Exercise 11.
\(\sqrt{{(2^3)}^x} = 64\)
Exercise 12.
\(e^{x^2-2x}=e^8\)
Exercise 13.
\(\ln(4-x) = \frac{1}2\)
Exercise 14.
\(\log_2 \left(5\right) + \log_2 \left(x\right)=\log_2 \left( 10 \right)\)
Exercise 15.
\(\ln(\sqrt{x}) = \sqrt{\ln x}\)
Exercise 16.
\(5 \ln x - 2 \ln x^2 = 4\)
Exercise 17.
\({\left( \dfrac{ x^2}{\sqrt{x}} \right)}^2 = 125\)
Exercise 18.
Which condition holds for \(a\) and \(b\) when \(\dfrac{a \ln{\left( {(x - 1)}^2 \right)}}{\ln{\left( {(x - 1)}^b \right)}} = 1\)? Which values can not be assumed by \(a\) and \(b\)?
Applications
Exercise 19. Free energy of a reaction
Suppose we have a reversible reaction (note: all chemical reactions are reversible) \(\ce{A <-> B}\). When the probability of the reaction \(\ce{A -> B}\) occurring in a vessel of molecules \(\ce{A}\) and \(\ce{B}\) equals that of the reverse reaction \(\ce{B -> A}\) then we have a chemical equilibrium. We then say that vessel contains no free energy anymore. The ratio \(\frac{[B]}{[A]}\) then equals the equilibrium constant \(K_{eq}\) of this reaction. When the reaction is not at equilibrium, the ratio \(\frac{[B]/[A]}{K_{eq}}\) is a measure of the free energy per mol of reacting molecules. If the ratio is smaller than 1 the reaction still contains free energy, and when it equals 1 the free energy equals 0. In fact, the Gibbs free energy of a reaction is defined as
\[ \Delta G = R T \ln{ \left( \frac{[B]/[A]}{K_{eq}} \right)} \]
where \(R\) is a proportionality constant (the gas constant, unit \(\text{J}\text{K}^{-1}\text{mol}^{-1}\)) and \(T\) is the absolute temperature (unit \(\text{K}\)) in the vessel.
a. The standard Gibbs free energy \(\Delta G^0\) of the reaction can be calculated from \(K_{eq}\) and is independent of the concentrations of \(\ce{A}\) and \(\ce{B}\). Work out the relation between \(\Delta G^0\) and \(K_{eq}\).
b. Can you think of a reason why the free energy per mol of reactant in a reaction is proportional to \(\ln{\left(\frac{[B]/[A]}{K_{eq}}\right)}\) rather than to \(\frac{[B]/[A]}{K_{eq}}\)?
Exercise 20. Information theory
Claude Shannon (1948) formulated a theory that quantified information present in messages. Such messages could be any sequence of symbols, including biological sequences! It quantifies the amount of information that we gain from a message if we only know in advance (before receiving the message) what the probability is of each of the individual symbols (in this concept, information is something relative, not absolute!).
If there are \(n\) different symbols \(s_i\), having probabilities \(p_i\) of occurring in a sequence, then the amount of information \(H\) present in one symbol of that message equals
\[ H = -\sum_{i=1}^n p_i \log{(p_i)} \]
In a message of length \(N\) symbols this equals \(N \times H\). If we use the base-2 logarithm then this information is measured in bits and if we use the natural logarithm then it is measured in nats.
a. If we only know that the symbols A, T, G and C occur at frequencies 0.2, 0.25, 0.35 and 0.2 in the coding strands of the DNA of an organism then what is the amount of information present (in bits) in a nucleotide of a coding strand of that organism?
b. What is the amount of information that we gain when receiving a message (a DNA sequence) if the frequencies were 0.1, 0.1, 0.4 and 0.4? Why is it less than in the previous case?
c. Based on the previous observations: guess at what probabilities for each symbol we gain maximal information and at what probabilities minimal information? What are the values of these minima and maxima?
Exercise 21. The genetic code
A codon is a triplet sequence of RNA nucleotides that dictates either a specific amino acid or signals the termination of translation. The four types of RNA nucleotides are Adenine (A), Uracil (U), Cytosine (C), and Guanine (G), and each codon comprises a sequence of three of these nucleotides.
a. Calculate the total number of possible codons.
b. Determine how many codons a minimal genetic code would require. Assess whether a system with 2-nucleotide codons would suffice.
c. Write a function describing the number of codons as a function of codon length. Identify the type of function and sketch its graph.
Exercise 22. The polymerase chain reaction
The polymerase chain reaction (PCR) is a sequencing method that allows for the amplification of DNA fragments, where the number of DNA fragments doubles with each cycle of the reaction. After \(n\) cycles, the number of DNA fragments can be modeled by the equation:
\[ N = N_0 2^n \]
where \(N_0\) is the initial number of DNA fragments.
a. If the initial number of DNA fragments is 50 and the PCR is run for 15 cycles, calculate the number of DNA fragments produced after the 15 cycles.
b. Determine how many cycles are required to produce at least 10 million DNA fragments from an initial number of 100 DNA fragments. Use logarithms and the change of base formula to solve for \(n\).
c. During the PCR, you observe that the error rate in DNA sequencing decreases exponentially with each cycle of error correction, following the equation: \(E = E_0 e^{-kt}\), where \(E_0\) is the initial error rate, \(k\) is a constant, and \(t\) is the number of cycles of error correction. If the initial error rate is 0.1 (10%) and the error rate reduces to 0.01 (1%) after 5 cycles, determine the constant \(k\).
Exercise 23. DNA substitution
The BLOSUM (Blocks Substitution Matrix) is used to evaluate the similarity between protein sequences through sequence alignment. The BLOSUM matrix contains log-odds ratios that quantify the likelihood of one amino acid being substituted for another in similar protein sequences. The log-odds ratio for a substitution of amino acid i with amino acid j are derived from observed relative frequencies compared to expected relative frequencies: \[ S(i,j) = \log_2 \left(\frac{p_{i,j}}{q_i \, q_j} \right) \] where \(p_{i,j}\) is the observed frequency of substitution between amino acid \(i\) and \(j\), and \(q_i\) and \(q_j\) are the expected frequencies of amino acids \(i\) and \(j\), respectively.
a. The log-odds ratio of substitution \(i\) to \(j\) is \(-2\). What does this indicate about the likelihood of the substitution compared to the expected frequencies?
b. Why is the base-2 logarithm often used in bioinformatics for calculating log-odds ratios? What is the impact of using different logarithm bases (e.g., natural logarithm) on the interpretation of results?
Exercise 24. Gene expression analysis
Gene expression data from an experiment can be analyzed by determining the expression levels of certain genes under two different conditions: Normal and Diseased. A volcano plot can be used to visualize differentially expressed genes between the two conditions, where the x-axis shows the \(\log_2\) fold change (\(\log_2 \left(\text{FC}\right)\)) and the y-axis shows the \(-\log_{10}(\text{p -value})\), an indication of the statistical significance of the expression change. The \(\log_2 \left(\text{FC}\right)\) is calculated using the formula:
\[ \log_2 \left(\text{FC}\right) = \log_2 \left( \frac{\text{Expression(Diseased)}}{\text{Expression(Control)}} \right) \]
a. If the expression counts for a gene are 500 in the Disease group and 125 in the Control group, what is the \(\log_2 \left(\text{FC}\right)\) for this gene?
b. A gene with a \(\log_2 \left(\text{FC}\right)\) of -2 and a p-value of 0.0005 has been identified. What does this negative \(\log_2 \left(\text{FC}\right)\) indicate about the expression of this gene in the Diseased group compared to the Control?
Exercise 25. Unit conversions
Carry out the magnitude conversions for the following units.
a. \(1\unit{m} = \ldots \unit{cm}\), \(1\unit{m}^2 = \ldots \unit{cm}^2\), and \(1\unit{m}^3 = \ldots \unit{cm}^3\) ?
b. \(1 \unit{dm}^2 = \ldots \unit{L}\), \(1 \unit{cm}^2 = \ldots \unit{L}\), and \(1 \unit{nm}^3 = \ldots \unit{L}\)
c. \(1 \unit{fL} = \ldots \unit{\mu L}\), \(1 \unit{mL} = \ldots \unit{L}\), and \(1 \unit{\mu L} = \ldots \unit{fL}\) ?
d. \(1 \unit{mol} = \ldots \unit{molecules}\), \(1 \unit{\mu mol} = \ldots \unit{molecules}\), and \(1 \unit{pmol} = \ldots \unit{molecules}\) ?
Carry out the magnitude conversions for the following composite units.
e. \(50 \unit{nM}\unit{min}^{-1} = \ldots \unit{M}\unit{s}^{-1}\)
f. \(10^{-12} \unit{cm}^2\unit{s}^{-1} = \ldots \unit{m}^2\unit{min}^{-1}\)
g. \(10^{16} \unit{cm}^{-3} = \ldots \unit{M}\)
Exercise 26. The rate of the DNA polymerase
The length of the DNA of E. coli, a bacterium, is \(1.5 \unit{mm}\) and consists of \(4558953 \unit{bp}\). Its DNA polymerase runs at a speed of \(800 \unit{bp}\unit{s^{-1}}\). How much time does it take for this enzyme to replicate E. coli’s DNA by 50%?
Exercise 27. Membrane proteins
Membranes contain membrane-proteins, many of which are transporters. Also, the periplasmic space, the space between the outer and inner membranes of Gram-negative bacteria, contains many proteins. This question is about these proteins.
a. Calculate how many membrane proteins fit in E. coli’s cell membrane assuming that their radius is \(5 \unit{nm}\). Assume a radius of \(1 \unit{\mu m}\) and a perfect spherical shape of an E. coli cell. We know that approximately 50% of the membrane surface is occupied by membrane protein, the rest being lipid surface.
b. How many proteins fit in its periplasm if this compartment is \(15 \unit{nm}\) thick (on top of the \(1 \unit{\mu m})\)?
c. What is the ratio of the protein numbers in the membrane and periplasm over the number of proteins in its cytoplasm?
This appendix, La géometrie Descartes (1886), published in 1637, has been very influential in the development of science. In it, Descartes also introduced the technique of mapping algebraic equations (functions, among others) on axes, using a coordinate system. Newton and Leibniz developed this technique in their research in mathematical physics, and we still use it extensively, also in this syllabus, to study properties of functions. In Descartes honor, this coordinate system is called the “Cartesian” system. Also see Burton (2010) for an account of the influence of this work.↩︎
Note that these proofs are informal. A formal way would be to use proof by induction, i.e. prove the property for a small natural number like \(n=1\) or \(n=2\) and then show that if it holds for any number \(n\) it also holds for \(n+1\).↩︎
Irrational numbers are numbers that can not be expressed as a ratio of integers \(p/q\), like \(\pi\) or \(\sqrt{2}\).↩︎
Because in the computer an irrational number like \(\pi\) is approximated with a finite number of digits, like \(3.1415\), which is equal to the ratio \(31415/10000\).↩︎
A definition that seems to drop out of the blue. In fact, the exponential function is derived from the natural logarithm, which itself is defined as the anti-derivative of \(1/x\).↩︎
See the box below for an explanation of this this formula.↩︎