module 01:

module 02:
Periodic Table / Elements

module 03:
Chemical Bonds

module 04:

module 05:
Structures of Matter

module 06:
Substances of the environment

module 07:
Chemical Reactions

module 08:
Reactions in Equilibrium

module 09:

module 10:
Redox reactions

module 11:
Carbon chemistry

module 12:

module 13:
Qualitative analysis

module 14:
Quantitative analysis

module 15:
Chemical Industry

module 16:
Reactions in the environment



In this module you find basic data about the chemical processes in living organisms.
Much about (bio)chemical structures can be found in module 04, but also in this module, where needed, you will find info about the structures of biochemical substances (again).
The whole and the interrelations of all those reactions in living creatures is also called: the metabolism.
Mainly we talk about making ànd decomposing of all those substances needed or excreted by men, animal or plants.
In biochemistry perform lots of 'organic' compounds, from simpel tot complicated.
Striking is that about 95% of all substances in human body is built up from the elements Hydrogen, Carbon, Oxygen and Nitrogen.
Also striking is that the same body is composed of more than 50% water.
The more specific topics are: sugars (sacharides), proteins, fats (lipids), nucleic acids (DNA and RNA).

Of course you can find lots of websites about all kind of biochemical topics. example:

In principle we meet in biochemistry the same reactions as in general chemistry, certain in carbon chemistry. You wil meet the acid-base reactions, the redox, hydrolysis, condansation, polymerisation and much more.

Just start with a couple of questions that help you to get in the subject.

Question 1
Salicylic acid reacts with a solution of ammonia. Present its reaction equation.

Question 2
Try to read and understand the text below. Then you could make some calculations to justify the advise of the docter.
Comes a lady at the docters:
"I don't feel well, docter. very tired for many days."
The docter send her to the lab for blood analysis. Come back in ten days.
"O doc, do you have the lab results, please? I am still so tired."
"Certainly lady, sais doc, I have a list with data showing that the pH of your blood is 7,1 (too low). And your CO2-concentration is 1,12 mmol/l (also not quite normal. We need more investigation in the academic hospital. In the meantime I advise you to take tablets of bicarbonate daily. Probably your blood lacks some HCO3--ions.
Come back in ten days.
The patient goes home, did not understand a word, but will swallow those

Question 3
Explain the concepts in the next text:
Iron comes in the human body in the form of ions, through nutricion. The Fe3+ can change in Fe2+ in acid environment and with help of a reductor, for example ascorbate. The Fe2+ enters the pancreas at a pH of 7 and there Fe2+ becomes again Fe3+. Transferrine transports Fe3+; Ferritine stores Iron ions. Heme contains Fe2+.

Question 4
Is the next affirmation true of false?
"In the metabolisme many polymeres suffer hydrolysis, becoming monomers."


1. Proteins

1.1 Amino acids; how they react

1.2 Production of proteins

1.3 The Ureum cycle

2. Carbohydrates / Saccharides /
/ Sugars

2.1 Structures of carbohydrates

2.2 Energy in biochemistry

2.3 Photosynthesis and Respiration

2.4 The Krebs cycle (cycle of citric acid)

3. Fats / Lipids

3.1 Fatty acids

3.2 (un)saturated

4. Nucleic acids

4.1 The basis of nucleic acids / the building bricks of life

4.2 Trasport of information; genetics

5. Enzyms

5a. Introduction

5.1 Specificity of enzyms

5.2 Structure of enzyms

5.3 Optimal activity and denaturation of enzymes

5.4 Regulation of the enzym activity

5.5 Michaelis Menten

5.6 Just another explanation about Michaelis Menten

1. Proteins

Proteins are the most important component of the tissue of our body (skin, meat, nails, hair, and more). They are absolutely necessary and are produced by the body itself.
But we need for that building stones: amino acids. They must come from outside via nutrition, at least most of them. Some amino acids can be made by the body itself.
To obtain enough amino acids, the human body needs food like meat, beans, fish, nuts, gelatina and other products.
Att.: apart from the tissues, the proteins also are absolutely necessary in another form: enzyms.

1.1 How amino acids react

Question 5
True or fals (T / F)? about amino acids
amino acids serve to produce fats; T / F
There are ten for human essential amino acids; T / F
All amino acids have an amphoteric character. T / F

acid / base

Amino acids contain an amino group, mostly NH2, and a carboxylic group, -COOH.
Thats why an amino acid can react with bases ánd with acids. The amino acid is an ampholyte.
When the amino acid has, per molecule, 1 amino group and 1 carboxylic group, then the pH of the solution of this amino acid will be about neutral (6 - 8). Any amino acid with extra carboxylic groups will cause a lower pH and an amino acid with extra amino groups will cause a higher pH value.

Question 6
Find in tables the structure of Lysine and explain which pH a solution of Lysine will have.
Goto answer 12-06

There is a real option that, in a solution, one amino acid molecule will donate the H+ of its carboxylic group to the amino group of a similiar molecule.
And this amino group can belong to another amino acid molecule, but could also belong to the same molecule. In that last case, we talk about internal proton transfer.
On the one side, this molecule became --COO- and on the other side ---NH3+. A same molecule has so a positive side as well as e negative side; we call that particle a 'double ion'.
In short: an amino acid can - dependent on the environment - be neutral in two ways, be positive and be negative.

Question 7
What configuration will have a simple amino acid in an environment with pH = 7?
and if the pH = 2?
and with pH = 10?
Explain your answers.

Condensation and Hydrolysis

Apart from participatin in acid base reactions, amino acids do very well in another reaction type: condensation. There the amino group of one molecule is linked to the OH-group of another molecule, losing a water molecule. In this way two amino acids are linked to each other through a so called peptide bond.

Question 8
Find in a table the structures of Valine and Alanine; show how these amino acids can link to each other via condensation in two different ways (peptides are formed).

1.2 Production of proteins

In chemical terms we say: the proteins are made in a polycondensation process, where the amino acids are the 'monomeres'.

You may consider proteins as a co-polymerisation product becaus normally different amino acids participate (there are about 20 different amino acids) in the process.
The whole synthesis can also be considered in another way: as a process that is controled by DNA/RNA. That topic will com furtheron.

In the next scheme you can see the participating amino acids as double ions. (this appearance does not influence at all the condensation process.

At the two ends of such a dipeptide bond you still can see: an amino group and a carboxylic group. In other words: the possibility exists to continue the process.
Again these groups can 'condensate' with other amino acids and form tri, tetra- oligo- or polypeptides.

A protein can consist of 50 tot 2000 joined amino acids; the molecular mass varies between 5000 and 20000.

Question 9
  1. Name the essential amino acids.
  2. Why are they called like that: essential?

Optical isomery

In nature there is only one kind of amino acids, the α-amino acids. This has all to do with the subject 'optical isomery'.
All amino acids in nature have an amino group in the so called α-position regarding to the carboxylic group.

In the images we see Alanine. At the left side α-alanine (two times; the two structures are equal and they are not the mirror image of each other).
On the other side you see α-alanine and β-alanina, and these two are optical isomeres (they are not identic in threedimensional way, but the two are each others mirror image).

Question 10
Give of Alanine the four different structures (with and without charges)
Goto answer 12-10

Below you see the structure of penicillinum (twice).

Note the important functional groups: peptide and carboxylic, and the thiazol (where S substitutes an O).

Question 11
True or false?
  1. The formation of proteins under controle of DNA is a polymerisation reaction.     V/F
  2. The formation of proteins under control of DNA is polyhydrolysis reaction.      .V/F

There is an amino acid (Cysteíne) that can make another kind of bonding, namely a Sulphur bridge (S-S).
If a lot of amino acids make a polypeptide, than the long chain will start to fold / roll on into certain structures.
In the first place a so called Helix is formed, it will screw up and we name that the secondary structure of the proteins.
The primary structure is simply the sequence of the amino acids.
The Helix, in its turn, can be folded again and form a tridimensional structure that we call the tertiary protein structure. (see figure)

Question 12
Have a good look at the image and try to see / to indicate the secondary and the tertiary structures, and the active spot (the co-enzyme) of the enzyme.

The L-amino acids are the building stones of the proteins; they form the peptide bonds and are responsible for the Helices.

The primary structure of the proteins

So the sequence of amino acids in the polypeptides, in the proteins, we call the primary structure. It is the linear sequence of the amino acids.
This sequence is extremely important for the functioning of that particular protein. Substitution of one amino acid can be enough to change completely the activity of that protein in the organism, maybe with very negative consequences.

figure: albumine, The sequence of the amino acids and the positions of the S-bridges.

A protein molecule can be composed out of one or more polypeptides.

The albumine (main component of the white part of an egg) has one single folded polypeptide structure. Insuline however, the hormone that is made in the pancreas and controlls the amount of glucose in the blood, is composed of two polypeptide structures (see figure):

Question 13
Imagine a protein structure with 1000 monomeres of the amino acid Valine (see tables).
Calculate the molecular mass.
Goto answer 12-13

The amino acid Proline, if it appears somewhere in a polypeptide chain, forbids or disturbs the formation of α-helices. The consequence is: no Hydrogen bridges can be formed (because of lack of the necessary polarity).


Sickle cell anemia is inherited; something is missing. Reason is the one certain amino acid, somewhere in the plypeptide chain of Hemoglobine is substituted by something else. The right amino acid lacks.
The primary structure of the protein is wrong in the case of sickle cell anemia.

The secondary structure of proteins

The polypeptide chains will curl as a screw: a helix. This is the secondary structure.

Tertiary structure of proteins

A long polypeptide chain with a helix, still can be very long and fold in a threedimensional way: the tertiary structure.

Quaternary structure of proteins

The very large protein molecules are composed of more tertiary spheres: quaternary structure.

In the case of hemoglobine of the blood, four polypeptide chains, all of them already with a tertiary structure, are joined to form one big entity: the quaternary structure.

The forces that make the formation of secondary and tertiary structures possible are:
  1. Hydrogen bridges
  2. vanderWaals forces
  3. dipolar forces
  4. Sulfur bridges (Cisteíne)

Question 14

Have a close look at the above diagram of hemoglobine and answer the followint questions:
  1. what means 'nm'?
  2. Where in the enzyme can we find the helix?
  3. which forces stabelize the helix (indicated with dots . . . . . . . )?
  4. Indicat the co-enzyme. Where is that?
  5. Does Hb somewhere has an inorganic part?
  6. What kind of forces stabelize the tertiary structure?

Question 15
True or false?
  1. Proteins are polymeres;
  2. Proteins have a primary structure, being the sequence of the amino acids;
  3. A secundary structure connects various primary structures with each other;
  4. The tertiary structure is the helix screw structure;
  5. The secondary and tertiary structure owe their stability to Hydrogen- and Sulfur bridges/bonds.
  6. Proteins can be denaturated (revesible or irriversible), what means that they can be destroied completely with alcolho, with acid or by heating;
  7. The primary structure remains intact during denaturation. This structure only can be destroied in a chemical process (hydrolysis amino acids)
  8. There are 'structure proteins (like albumine) and 'enzym proteins' (like oxidase)

Classification of proteins

You can divide proteins according to the form: you have globular proteins and fibrilar proteins.
  1. Fibre proteins - at least 10 x longer than broad. Example: nails, skin: keratine.
  2. Globular proteins - spherical in form, so length and broadness do not differ very much. examples: enzymes, antibodies, membrane proteins, hemoglobine, chlorophyl, etcetera.
This division has directly to do with the two most important functions of proteins:
  1. production of tissue (muscles, skin, bone, colagene, etc.)
  2. catalysing (controle of the reactions, transmission of nerve impulses, immuno protection)

The study of proteins is basic for understanding genetics.
In the lab at least is needed for such studies: purification and separation of proteins with techniques like chromatography and electrophoresis.

Considering the composition of protein molecules, we can divide proteins in: simple and complex.
  1. Simple proteins or homoproteins: - they are only composed of amino acids and nothing else. Examples: insuline, albumine, keratine, fibrinogene (occuring in blood plasm)
  2. Complex proteins or heteroproteins: - they are connected with non-protein structures, like the prostetic group.

Question 16
How can we classify the proteins in meat? (they belong to what kind of proteins?)

Examples of group B:
  1. Chlorophyl, of who the prostetic group is Mg
  2. Hemoglobine, of who Heme + Fe is the prostetic group
  3. casein (in milk), of who the prostetic group is phosphate
  4. nucleic proteins: - mostly in the ribosomes, and of who the prostetic group is a ribonubleic acid (RNA) or (DNA)
  5. glycoproteins, of who the prostetic group is a saccharide; lipoproteins with lipides as a prostetic group.
  1. Complex proteins are divided according to their prostetic group.
  2. If the prostetic group has a certain color, we also can talk about chromoproteins.

About their fundamental function, the proteins are classified in structure proteins and enzymes:
  1. Structure proteins:
    belong to the cell structure and are building materials
  2. Enzymes:
    controle practically all chemical reactions in the cells.

Question 17
Hemoglobine owes his red color to the heme group.
Try to classify hemoglobine, and what is characteristic for that heme group?

1.3 The urea cycle

Main purpose of this cycle is the production of urea, the end product that is excreted as waist from the body. It is important that you understand that via this cycle the body - bit by bit - can lose superfluous nitrogen. Normally, Nitrogen enters the body via amino acids.
This cycle is connected to the Krebs cycle (combustion of sugar in many steps) and this cooperation creates a metabolism (a strong catabolism).

The urea cycle simultaneously serves the katabolism of old amino acids.
The waist leaves the body via urine.

2. Carbohydrate, saccharides, sugars

2.1 Structures of the saccharides

Question 18
Is the following statement true of false? Explain.
"the name 'Carbohydrates' is based on a misconception, on an old scientific misunderstanding.'

A simple division of the saccharides:
  1. Monosaccharides: glucose, fructose, galactose(C6H12O6)
  2. Disaccharides: saccharose (sugar), lactose, maltose (C12H22O11)
  3. Polysacchariden: amide (starch), amilose, cellulose( (C6H10O5)n)


The monosaccharides can occur in a linear or in a cyclic configuration (above you only see cyclic forms).

Below you can see how glucose can have the linear and the cyclic form:

The linear structure can easily be oxydised, with a weak oxydator; often we use for that:
Ag+ ammonia-silver-solution = reagets of Tollens. Another one, Fehlings' reagent, also acts very fine (Cu2+)

The cyclic configuration is not easily oxydised (there is no =O bond available). In practice the two structures, the linear and the cyclic, in a solution, are in an equilibrium with each other, so the linear structure is than always present and always ready to suffer oxydation.
Than the linear structure disappears in that process, the equilibrium will shift from the cyclic form tot the linear form (that continues to suffer oxydation), and at the end, all the monosaccharide is oxydised.

Question 19
Study well the next two images: glucose and fructose appear in two ways, the linear and the cyclic. The two configurations are - in aqueous environment - always in equilibrium with each other.
And as said, only the linear structure has the right group to oxydise (the reductor group).
Explain with your own words why during the redox process not only the linear form, but also the cyclic form will disappear.


Saccharose cannot be oxydised with Ag+ or with Cu2+.
Maltose and lactose can be oxydised with Ag+ or with Cu2+.

Question 20
The oxydisability of a cyclic structure depends on the position of the O (oxygen atom) in the ring.
If such a C - O bond in the ring can easily be opened (and the linear structure is made), than oxydation is possible. (see also module 11)
Analyse, with help of the structures, why some rings are and some are not easy to open.


polysacharide (6K)
Amide, starch, cannot be oxydised with Ag+ or with Cu2+.

Amide has a helix structure when in natural form
Than it also creates with I3--ions a dark blue complex.

Question 21
Iodine is a rather non polar substance and does normally not dissolve in water. But it does if that water also contains iodide ions, for example from KI.
Explain that.

Fehlings' Reagent

CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

Cu(OH)2(s) Cu2+ + 2OH-( = equilibrium 1)
Cu2+ + e- Cu+

Adding of Sodium tartrate: the tartrate ions attrect and crab the Cu2+-ions from equilibrium 1; therefore the precipitate Cu(OH)2 will disappear; at the same time appears a transparant solution with dark blue colour = Fehlings' Reagent (FR).
FR in basic environment has dissolved Cu2+-ions capable to oxydise a reductor (for example the aldehyde group), capturing an electron and creating an ion Cu+.
These ions Cu+, in a secondary reaction with OH- will form the orange red precipitate Cu2O(s.
This recognition reaction can be used as proval of the presence of monosaccharides.

Question 22
True or not true?
To investigate the presence of glucose in urine, a more specific mehtode than FR is the enzymatic method with glucose-oxidase.
Goto answer 12-22

Aspartame is an artificial, non-saccharide sweetener, used by people that try to loose weight.
They want to prevent that they become fat by eating too much sugar (or they have other reasons).
Fact is that sugars contain a lot of energy. In stead of sugar in the coffee, we can take substitutes like aspartame; but there are others.
Many sugar containing products (for example jam and coca cola) have also their diet products, i.e. without sugar, but with the substituting sweeteners.

The secret of aspartame is that it leaved the body as it entered it, without participating in the metabolisme, without any energy impact, without any fat making.
But a warning is needed: some sugar substitutes can be carcinogeneous, not healthy.

Question 23
Consider the aspartame structure again and show which groups you recognize.

2.2 Energy in biochemistry

In principle, all rules for exothermic and endothermic reactions are the same in biochemistry, including the processes in living creatures. There also we can find chemical reactions that deliver or cost energy.
But in living creatures, energy often is transferred via specific substances like ATP (Adenosine Tri Phosphaat), a molecule that often appears in reactions with energy transport.
Most living items must breath, where the main purpose is collecting Oxygen.
Green plants have also another option: the photosynthesis that is about the opposite of respiration. In the photosynthesis, Oxygen is produced in stead of used.
Respiration and photosynthesis are not the only biochemical processes with a real energy effect; in fact all biochemical reactions are exothermic or endothermic, but we wan't deal with that now.

2.3 Photosynthesis and respiration

Two important processes in nature, one the opposit of the other:
  1. the respiration: C6H12O6 + 6O2 6CO2 + 6H2O     exothermic(ΔH < 0)
  2. the photosynthesis: 6 CO2 + 6H2O C6H12O6 + 6O2       endothermic (ΔH > 0)
Not only men and animals, but also plants "breath"!
Contrarily to the photosynthesis, respiration needs chemical energy from outside the organism. (for example in the form of sugar).
The photosynthesis however produces chemical energy. (for example ATP and the sugars that can be consumed in respiration).
Plants have both possibilities; by day the photosynthesis dominates (under influence of light) and by night dominates the respiration.

Question 12-24
Imagine that somewhere a human being has succeeded (i.g. by mutation of its genes) to acquire the possibility to realise photosynhtesis somewhere in its body.
What could be the consequence?
Choose answer 12-24

2.4 The Krebs cycle, cytric acid cycle.

Just as the urea cycle, the Krebs cycle takes place at the mythogondria (the energy factories in cells).
The oxydator (Oxygen, O2) helps with the decarboxylation of the citric acid during the course of al those reactions in the cycle.
On complete round of the cycle includes one reduction of the two Carbon atoms that transform to 2CO2.
The connected Hydrogen atoms connect with energy carriers: FAD and NAD, that immediately after that will form ATP. ATP can we consider as a kind of fuel in living organisms.
The ATP molecules are full of chemical energy.

The complet Krebs cycle is built up of nine steps, with the total reaction:

Citrate oxalic acetate
C6H5O72- + H2O C4H2O52- + 2CO2 + 5H·

FAD- and NAD-molecules pick up the H-radicals (neutral and single H-atoms), and acquire thus a lot of energy.
Thereafter these H atoms react with Oxygen and form water H2O (a very exothermic reaction).
Water and Carbondioxyde are the end products.
The main function of that cycle is the acquirement and storage of chemical energy.

Question 25
What can you say about the electron transfer in the above processes?

3. Fats / lipides

In Nature we find fats mainly in animal fat and in vegetable oil. Those two can very well serve as food for men and animal.
The are made in a condensation process of glycerol with fatty acids. Glycero has three OH-groups and if all three participate in the condensation process, the products are 'triglycerides'.

3.1 Fatty acids / lipids

Fatty acids have a chemical structure with: of course a carboxylic group (they are indeed acids) + a carbon chain of about 15 to 20 C atoms.
Mostly the fatty acids occur in the compounds with glycerine, but now and then they occur freely (for example, stearic acid = candle, C17H35COOH).

In the figure you see the structures of three fatty acids.
One of them in stearic acid (candle); the other has a double bond, so is unsaturated; the third has several double bonds, so is plurally unsaturated.
The next image shows four models: three models of unsaturated fatty acid + one mode of glycerine.
The three fatty acids react, each of them with one OH-group of the glycerine, under simultaneously formation of three molecules of water.

And in the next image you can see the products: a fatty acid + the three produced water molecules:


In nature the lipids (fats)are the mono, di, and triglycerides of fatty acids, or also: the esters of glycerol and fatty acids.

"Fat" normally is saturated.
Do we use unsaturated fatty acids: than the fatte character decreases and the oily character increases.
Oily substances, normally, are more unsaturated than fats.

Unsaturated substances have one or more double bondings in the chain, and can be saturated, for example with Bromine (Br2)

A layer with unsaturated oil (yellow color) layer with fat (diffuse)
A layer with Br2(aq) (yellow color) a layer of water without Bromine (colorless)

Question 26
True or false?:
The function of bladder can be compared with the function of soap.

Question 27
Explain all concepts in the following text:
"Normally, margarine is made of vegetable oil. That oil is build up of unsaturated molecules.
The problem is dath oil is a liquid and how to spread this on your slice of bread?
We can resolve that problem (we did that in the old days) by saturating the double bonds with Hydrogen (catalyst = Zincum). More saturated means: harder, so it got a harder constitution.
But... in the same process, this substance is losing its healthy character (unsaturated).
The new way to change a liquid into a solid is the application of emulgators.

4. Nucleic acids

4.1 The bases, the building stones of the nucleic acids.

There are 4 + 1 bases, building stones for the nucleic acids. They compose the Desoxy Ribo Nucleic Acid (DNA) and the Oxy Ribo Nucleic Acid (RNA)
  • DNA: is built up and composed of Adenine, Guanine, Thimine and Cytosine (A,G,T,C)
  • RNA: Adenine, Guanine, Uracil and Cytosine (A,G,U,C)
  • The difference between Thymine and Uracil is not more than one methyl group.

The double helix of DNA, schematically and in model:
In such a (super)macro molecule of DNA, the two very long helix chains go toghether (see figure). There is always that attraction between G and C, or A and T.


Question 28
Inside the double helix, the molecules A, G, T, C e U participate in the bondings between the two strings of that double helix. (see figure).
Every bridge must deal with the same distance.
Explain why A and G always combine with T, C and U

4.2 Genetic information

We are not going to treat the biological process, but yes, in short, some chemical characteristics of the production of proteins in living creatures.
A summary:
The information stream goes from DNA to RNA (transcription).
The RNA, in its turn, translates her codes to the right protein (translation)

DNA (transcription) RNA (translation) proteins

The base sequence of a gene (a part of DNA or RNA) goes paralel to the amino acid sequence of the product that this gene is going to make (a polypeptide / protein).
The gene code is fixed in the sequence of the bases (A, C, T en U) in such a way that every three sequencing bases determine one amino acid.
So, e bit simplified: if a gene has 600 bases, a protein of 200 amino acids can be produced. The word 'determined' in fact, is better here.
You must understand the great importance of starting the whole process at the right base. There are even start (triplet)codes and stop codes (codons).

So there are four bases in the genetic material, in the genes, and with them you can determine 43 different amino acids. But you know that in living creatures about 20 different amino acids are used, not 64.
Three are reserved for stopping and starting the production, but there are stil 41 left more than necessary.
In practice several codes can determine one and the same amino acid. Such codons are called: synonym codes.

You know that the protein molecule (also an enzym) is formed out of amino acids in a polycondensation process.
The function of proteins, in particular that of enzyms, is extremely important in living creatures, also for the metabolism. Any single mistake in such an enzym can cause a stagnation in the metabolism, with the consequence of serious (inherited) abnormalities or even the end of life.
So the controll mechanism in the body, to secure the production of right proteins/enzyms is very good developed. The information and the controll comes from the DNA that stays in the cell nucleus.
The DNA itself does not go to the cell plasma (to the ribosomes) to do that controll itself. It sends copies op itself (RNA, in a transcription process) outside, to the ribosomes, where also the amino acids arrive, to be connected exactly in the right way.

The next scheme shows more or less such a happening:

During transcription parts of DNA (genes) are copied (with almost the same four types nucleic acids).

During translatie, at the ribosomes, the amino acids (20 types) are connected to make proteins.

Question 29
In the figure of the double helix of a DNA molecule, in the lower part we can see a sequence of bases: - C - T - G - A -  (from bottom to top).
T, after transcription of DNA to RNA, has suffered that small change to U, or, The RNA will have the sequence: - C - U - G - A -
Which amino acids can be formed here, via that part of RNA? Use a gene code table from biology.

5. Enzyms

5.2 The specificity of enzyms

Two general characteristics of enzyms:
  1. The structural part of enzyms is a protein. Att.: Every protein structure can be denaturated, in particular under influence of pH and temperature.
  2. Enzyms are (bio)catalysts and extremely specific and efficient; the holo-enzym = apo-enzym + co-enzym/prostetic group.

The activity of biocatalysts depends strongly of the molecular shape, in particular of the surface of that molecule.

For example: Imagine that certain molecules of the substrate (the reactant) want to react and make any product. This will only be possible if those molecules do contact each other, and more: that they contact in the right way. (Tow Train compartments also only connect in the right way).
If those molecules have to wait until - by accident - they collide in the right way, they can wait very long.
But look, there come the catalysts, the enzyms, to help out. They attract those molecules, fit them in their surface and thus these molecules are put toghether exactly in the right way, so they can react easy and fast.
When the molecules are toghether and connected, the enzym will lose them and the new molecule is a fact.

Again: a substrate wants to react.It fits exactly in the surface of the enzym, close to the action centre. There arrived, the substrate can react easily (with help of the position a, b and c).

The reaction equation:
E + S ES E+P

S = substrate, P = product

The functioning of the enzyms has all to do with the surface of their molecules.
No matter what is changing that surface, the enzym activity can be terribly disturbed.

Every enzym has its own temperatur and pH where the activity is maximal: the optimal pH and temperature.

Question 30
In the following scheme you can see another type of enzym catalysis.
Study the image and explain how the specificity of this enzym is guaranteed.

Question 31
Could the last reaction (S1+ S2 P) be a condensation reaction?
Explain your answer.

Mistakes in the DNA (by mutation for example) can create confusion in the (human) body. Maybe wrong enzyms will be produces, so with a wrong threedimensional shape.
Such deficient enzyms make the normal biochemical reaction in the metabolism difficult or even impossible. Such a person suffers one or another illness.

Question 32
Hemoglobine (Hb) can be considered a kind of enzym with an activator. What activator has Hb?

Question 33
Show in an energy diagram what 'activation energy' does with a chemical reaction.
Goto answer 12-33

5.3 The structure of enzyms; summery

The two substrates S1 and S2 connect very specifically to the enzym. Other molecules don't fit there (in a way of shapes)
Regularly but not in all cases, in the action centre of the enzym is a co-enzym (weakly bonded) or a prostetic group (strong bonded)
Chages in the shape of the action centre can strongly influence the enzym activity, even stop it.
The centre often is situated in a kind of non polar clove.

Co-enzyms and prostetic groups ( ) improve the enzym activity.

The complete enzym, including the cofactors and prostetic groups, are called the holo-enzym,
The enzym without cofactors and prostetic groups are called the apo-enzym.

Regularly the co-enzym is a vitamine, and the same co-enzym can serve different enzyms.
Some enzyms need an inorganic part like a metal ion (examples: Ca2+, Mg2+ of Zn2+).
These inorganic component is an 'activator'. In a functional way the activator equals a co-enzym, but the inorganisc components are not called: co-enzymes.

The enzym is - in a structural view - just a normal protein with:
  • a primary structure, made under control of DNA = the sequence of the amino acids
  • a secundary structure, = the  α-helix (screw) of the primary structure
  • a tertiary structure, = the result if the α-helix is folded into a threedimensional shape that determines the specificity of the enzym, as also the way the enzym can be inhibited.
  • a quaternary structure (not always present) is the joining of (four) equal tertiary structures.

different kinds of enzyms / nomenclature of enzyms:

Normally the name of an enzym start with the substrate name + immediately thereafter the activity of that enzym + suffix -ase.
Oxyreductases catalyse redox reactions (for example: glucose-oxidase)
Transferases transport functional groups of a donator to a receiver
Hydrolases split molecules (for example: peptidases, proteases)
Liases distract or add: certain functional groups (for example: decarboxylase)
Isomerases change isomeres (mutase)
Ligases connect different molecules with each other (piruvaate carboxylase)

Question 34
The following enzyms belong to what kind?
  1. The enzym that can transform D-glucose into L-glucose?
  2. The enzym that can transform phosphate into phosphite?
  3. The enzym that can change saccharose into glucose and fructose?

5.4 Optimal functioning and denaturating of enzyms

Proteins, and certainly the enzyms, are very sensitive for the environment in which they are, in particular the pH and the temperature. They have big influence on the threedimensional structure.
Denaturation for example, is desastrous: the enzym (a protein) loses its tertiary structure and with that, its shape. That enzym doesn't work any more.
So, deviation of the optimal pH and temperature will cause immediately some influence on the activity of the enzym.

A couple of possibilities to change the environment:
  • Heating gives more movement to the atoms, on particles in general and on bondings. The increased movement can disturb the vanderWaals and polar attraction. The covalent bondings do not brake easily at heating.
    If the denaturation was caused by temperature increase, that change is irreversible. Such an enzym can be thrown away.
  • Adding acid or alcohol can brake the Hydrogen bridges. Than the tertiary structure, and maybe the secondary structure can be lost.
    Its also possible to change the pH or to dilute or to add alcohol or salts.
    Such denaturation may be (partly) undone; there may be some reversibility.
In both cases, the tertiary and maybe the secondary structure can be lost.

Question 35

The figures show the sensibility of the enzym activity for temperature and pH. (the enzym activity is found op the y-ax)
Make in that figure a possible scale divistion of the x-ax and the y-ax (an estimation)

Question 36
Explain what will happen at cooking an egg during some minutes.

High temperatures are desastrous for enzyms. This cooking is irreversible denaturation, unrecoverable.
Cautious and modest decrease of temperature only will decrease the enzym activity, without any denaturation.
From that lower temperature going back to the normal temperature of 30 - 40°C, the activity will recover. (just an idea: just before dying you freeze your body, hoping that in a hundred years it can be recovered)

The big part of the enzyms functions the best at pH values of about 7 (so neutral). But there are a couple of exceptions like pepsine. That one has an optimal pH value of 1 - 2.
Strong changes in pH can also denaturate an enzym. This can be reversible, but is not always.

Pepsine 1,5 in the stomach
Amilase 6,6 in saliva
Lipase 8,0 resp. 7,0 in the pancreas and in the bowels
Saccharase 7,0 in the stomach

Question 37
What is the pH in the mouth? Make your answer acceptable.

You know that the enzym activity depends on the enzym surface. Denaturation includes that the shape, so also the surface of the enzym changes.

Question 38
In the human digestion system, different enzimes are working. Each of them has its specific pH-optimum (the optimal pH value where the enzime works the best).
place enzimes pH-optimum
In saliva amilase and maltase 6,6
In the stomach peptase, rennase 1,5 - 4
In the pancreas amilase, maltase, lipase, tryptase, polipeptidase 6,6 - 9
In the bowels maltase, saccharase, lactase, ereptase 6,6 – 8,5
  1. Explain the variations in pH-values in the four places
  2. Why does maltase only appears at the end of the digestion system?
Goto answer 12-38

5.5 Regulation of the enzym activitity:

Non catalised reactions, needing for example hours of cooking, even in the presence of strong acid or base, sometimes can occor in a couple of seconds at a decent temperature if the right enzym is present, and at the optimal pH and temperature.
The enzyms perform as catalysts by lowering strongly the needed activation energy of this specific biochemical reaction.
The lowered activating energy allows the process to occur at body temperature and at high speed.

There are different ways to regulate the enzym acitivity:
Different enzyms are not immediately ready to function, but have a "pre-enzym phase", or: a zymogene phase. This can be necessary to protect a vulnerable enzym.
It is also possible that an enzym is only needed in the case of accidents.
In the stomach we have 'pepsinogene', dat can produce pepsine enzym at a low pH. If it is produced, the pepsine cannot go back. The pepsinogene loses a part of amino acid chain of 44 amino acids.

The product itself can (if it has reached high concentrations) react back and thus forbid an enzyme to continue its activity.

If we know the enzym, specific substances can be added that can influence the enzym activity (inhibitors = molecules with a structure more or less similar to the normal substrate, but not exactly the same).
Such an inhibitor can make the enzym (temporarily) inactive. Another way of regulation.
In fact we talk here about drugs, that - for example - are used in chemo therapy, forbidding the metabolism of tumors.

Some enzyms have, apart from an action centre, another place where certain groups can connect: a so called allosteric centre.
Substances that fit in such an allosteric centre, can change the threedimensional shape of the enzym, including the action centre, and so change the specificity of this enzym, and with that, its activity. This can have a positive influence (an more active enzym) but also a negative influence.

Question 39
Are the next statements (a en b) true or false? Explain your answer.
  1. En enzym is a (bio)catalyst that influences the reaction rate.
  2. A catalyst will speed up (or slow down) the rate V of a chemical reaction: S P
  3. Give an explanation of the letters S, P and V in your own words.

Most biochemical reactions are also reactions in equilibrium. That means that the formation of the product, after some time, will reach a maximum rate at the moment that the equilibrium has been reached.
Without catalyst (see module 7), this can take an awful time: hours of even days or never.
In the presence of a catalyst (an enzym) the equilibrium is reached much faster. This does not mean that you will obtain more product; but yes, the product was formed faster, maybe within seconds.

The enzym, the biocatalyst, mostly speeds up the chemical proces with a factor of 106 or more.

The substrate forms a complex with an enzym. This complex is called: the transition state. The substrate changes into a product.

After that, the product and the enzym separate.

S + E  ES P  +  E
                       a           b

The formation of the comples (a) also is an equilibrium; the last reaction (b) is not.
The enzym speeds up the realisation of the equilibrium, but does not change the equilibrium itself. Only the formation rate of the product.

5.6 Michaelis Menten

(there is another paragraph about the same topic, an extra explanation from literature in 5.7)
Every reaction occurs with a certain reaction rate V and with its own reaction constant k
k1                   k3


Starting with three presupposition supposing ideal / optimal conditions, you may state that the rate, at its maximum, is Vmax

The three presuppositions are:
  1. a steady stat wherein the concentration of the intermediate ES does not change; the production of the ES complex takes place with the same rate as its disappearance;
  2. Every present enzyme participates;
  3. The system is in an optimal environment (saturated with S, the best pH, the best temperature);
In this way the reaction rate V has ist maximal value Vmax

about the rate of producint the intermediate ES:

about teh rate of the disappearance (to the right and to the left) of that intermediate ES:

KM is the constant of Michaelis

Question 40
Explain how you can find that expression, starting with the presupposition 1.
Goto answer 12-40

KM shows the affinity of the enzym for the substrate; a low KM means a lot of ES, or, a lot of substrate S,
and that means: a low KM and a bigger rate.

The relation between the reaction rate and the KM can be seen in the Michaelis Menten equation:

The values of KM    can vary between 0 and thousands.

Question 41
Read the following practical task and try to understand this description, and add your comments:
Divide a neutral extract of salive (pH=7) in two equal parts (A en B).
Part A is made acid with HCl until the pH = 2. This mixture does not show any more activity.
Part B is mixed with an equal volume of water. This mixture does not show any more activity.
After that we treat both solutions A and B with base until pH = 8.5. No more enzym activity is noticed.
At last, the two parts with pH 8.5 are treated with acid until pH = 2. Only part B shows activity again.

Question 42
Try to understand the following description of practice, using graphics:
"Comparing the enzyms, we determine the parameters Vmax and Km, and we measure the reaction rate (V) as a function of the substrate concentration [S]."

For a certain standard amount of hexokinase or glycokinase, we measure the production rate (V) of glucose phosphate (The product G6P) as a function of the glucose concentration [S].
The results show that Vmax of the two enzyms are equal (±100 nmol G6P are produced per minute), but that the Km-values differ considerably:
Km of hexokinase  = 0.1 mM, Km of glucokinase = 10 nM (factor 100)."

Have again a good look at the following equilibrium with the intermediate SE:
k1                k3


The constant of Michaelis shows the affinity of the enzym to the substrate.

The equation of Michaelis-Menten shows the relation between KM and the reaction rate.

We know various types of enzym inhibitors:
  1. product inhibitors
  2. drugs inhibitors
  3. allosteric inhibitors

You can find more explanation about Michaelis Menten in

Goto 5.7

5.6 An article about Michaelis Menten

Enzyme kinetics and the Michaelis-Menten equation


Study of the impact made on the rate of an enzyme-catalysed reaction by changes in experimental conditions is known as enzyme kinetics. Knowledge of kinetics can be a very useful tool in understanding the mechanism by which an enzyme carries out its catalytic activity.

The effect of substrate concentration on the initial rate of an enzyme-catalysed reaction is a central concept in enzyme kinetics. When data are generated from experiments of this type and the results plotted as a graph of initial rate (v, y-axis) against substrate concentration ([S] *, x-axis), many enzymes exhibit a rectangular hyperbolic curve like the one shown in the diagram below.
*Note: the use of square brackets, as for [S] above is short-hand notation for "concentration of S", a convention that will be used extensively in the derivation below.

Observations of this type set Leonor Michaelis and Maud Menten thinking about the underlying reasons why a curve should follow this shape and led them to derive an algebraic equation that now bears their names.  There are several modern ways to explain the way in which the Michaelis-Menten equation is derived, and one is spelt out below. Deriving the Michaelis-Menten Equation
Start with the generalised scheme for enzyme-catalysed production of a product (P) from substrate (S). The enzyme (E) does not magically convert S into P, it must first come into physical contact with it, i.e. E binds S to form an enzyme-substrate complex (ES).
Michaelis and Menten therefore set out the following scheme:

The terms k1, k-1 and k2 are rate constants for, respectively, the association of substrate and enzyme, the dissociation of unaltered substrate from the enzyme and the dissociation of product (= altered substrate) from the enzyme.  Note that there is the theoretical possibility of a reverse reaction, with ES complex forming from E and P, but this can be ignored because we are considering initial rates of reaction, i.e. when the enzyme is first provided with substrate, so there should not be any product available to combine with enzyme.
The overall rate of the reaction (v) is limited by the step ES to E + P, and this will depend on two factors - the rate of that step (i.e. k2) and the concentration of enzyme that has substrate bound, i.e. [ES].  This can be written as:
(Equation 1)

At this point it is important to draw your attention to two assumptions that are made in this scheme.  The first is the availability of a vast excess of substrate, so that [S]>>[E].  Secondly, it is assumed that the system is in steady-state, i.e. that the ES complex is being formed and broken down at the same rate, so that overall [ES] is constant.  The formation of ES will depend on the rate constant K1 and the availability of enzyme and substrate, i.e. [E] and [S].  The breakdown of [ES] can occur in two ways, either the conversion of substrate to product or the non-reactive dissociation of substrate from the complex.  In both instances the [ES] will be significant.  Thus, at steady state we can write:

The next couple of steps are rearrangements of this equation.  First of all we can collect together the rate constants on the right-hand side because they are both multiplied by [ES], this gives us:

Then dividing both sides by (k-1 + k2), this becomes:

Notice that the three rate constants are now on the same side of the equation.  As the name implies, these terms are constants, so we can actually combine them into one term.  This new constant is termed the Michaelis constant and is written KM.

Notice that the three rate constants in the definition of KM are actually inverted (the other way up) compared with our previous equation.  This is a 'trick' that makes for easier calculation at a later stage.  Substituting this definition of KM into our previous equation now gives us:
(Equation 2)

The total amount of enzyme in the system must be the same throughout the experiment, but it can either be free (unbound) E or in complex with substrate, ES.  If we term the total enzyme E0, this relationship can be written out:

This can be rearranged (by subtracting [ES] from each side) to give:

So, the [E] free in solution is equal to the total amount of enzyme minus the amount that has substrate bound.  Substituting this definition of [E] back into equation 2 gives us:

This can now be rearranged in several steps.  First of all, open the bracket so that the terms [E0] and [ES] are separately multiplied by [S]

Next, multiply each side by KM, this gives us:

Then collect the two [ES] terms together on the same side (you can either think of this as adding [ES][S] to both sides or as 'carry over and change the sign' - your preference will probably be an indication of how long ago you went to school). This gives:

Then because both terms on the right-hand side are multiplied by [ES] we can collect them together into a bracket:

Dividing both sides by (KM + [S]) now gives us:

Substituting this left-hand side into Equation 1 in place of [ES] results in:

The maximum rate, which we can call Vmax, would be achieved when all of the enzyme molecules have substrate bound.  Under conditions when [S] is much greater than [E], it is fair to assume that all E will be in the form ES.  Therefore [E0] = [ES].  Thinking again about Equation 1, we could substitute the term Vmax for v and [E0] for [ES].  This would give us:

Notice that k2[E0] was present in our previous equation, so we can replace this with Vmax, giving a final equation:

This final equation is actually called the Michaelis-Menten equation.

So what?
Perhaps this derivation still leaves you puzzled about the importance of the Michaelis-Menten equation.  The significance becomes clearer when you consider the case when the rate of reaction (v) is exactly half of the maximal reaction rate (Vmax).  Under those circumstances, the Michaelis-Menten equation could be written: 

On dividing both sides by Vmax this becomes:

Multiplying both sides by (KM + [S]) gives:

And then multiplying both sides by 2 further resolves the equation to:

2[S] on the right-hand side is the same as [S] + [S], so we can take away one [S] from each side.  Thus when the rate of the reaction is half of the maximum rate:

The KM of an enzyme is therefore the substrate concentration at which the reaction occurs at half of the maximum rate.  If we now reconsider the graph that came at the start of this tutorial it could be written:

What does this all mean in physical terms?  KM is an indicator of the affinity that an enzyme has for a given substrate, and hence the stability of the enzyme-substrate complex.

Look at the shape of the graph. At low [S], it is the availability of substrate that is the limiting factor. Therefore as more substrate is added there is a rapid increase in the initial rate of the reaction - any substrate is rapidly mopped up and converted to product.  At the KM, 50% of active sites have substrate bound. At higher [S] a point is reached (at least theoretically) where all of the enzyme has substrate bound and is working flat out. Adding more substrate will not increase the rate of the reaction, hence the levelling out observed in the graph.

There are limitations in the quantitative (i.e. numerical) interpretation of this type of graph, known as a Michaelis plot.  The Vmax is never really reached and therefore Vmax and hence KM values calculated from this graph are somewhat approximate.  A more accurate way to determine Vmax and KM (though still not perfect) is to convert the data into a linear Lineweaver-Burk plot.  See separate page:

"Experimental calculation of Vmax and KM "
If, incidentally, we consider that the rate constant k2 for the conversion of ES to E + P in the initial scheme is the step determining the overall rate of production of P (as we have in deriving the Michaelis-Menten equation) then k2 is actually the same term as kcat, the turnover number.