Biological molecules

Class 11 Biology, unit 2, book pages 43–82

Practise 72 MCQs Take the unit quiz Review 70 cards

This unit studies the chemistry of life. It starts with what biochemistry is and the bonds that hold biological molecules together, then shows how small monomers are joined into polymers by condensation and broken down by hydrolysis. Water is described as the medium of life. The four fundamental kinds of biological molecules follow in turn: carbohydrates (monosaccharides, disaccharides and polysaccharides), proteins (amino acids, peptide bonds, the importance of sequence, classification and roles), lipids (acylglycerols, phospholipids, terpenes, waxes, steroids and prostaglandins) and nucleic acids (nucleotides, the phosphodiester bond, ATP and NAD, the Watson and Crick model of DNA, the three types of RNA and the central dogma). The unit ends with conjugated molecules. Many of the book's figures are structures or processes that must be drawn and labelled in the exam.

Learning outcomes (30)

Define biochemistry/molecular biology.
Describe briefly the different types of bonds found in biology (hydrogen bonds, covalent bonds, interactions, ionic, hydrophobic and hydrophilic interactions etc.).
Distinguish carbohydrates, proteins, lipids and nucleic acids as the four fundamental kinds of biological molecules.
Describe and draw sketches of the condensation-synthesis and hydrolysis reactions for the making and breaking of macromolecule polymers.
State the properties of water (high polarity, hydrogen bonding, high specific heat, high heat of vaporization, cohesion, hydrophobic exclusion, ionization and lower density of ice) that allow it to be the medium of life.
Define carbohydrates and classify them.
Compare and contrast the properties and roles of monosaccharides and write their formula.
Compare the isomers and stereoisomers of glucose.
Distinguish the properties and roles of disaccharides.
Describe glycosidic bonds in disaccharides.
Describe the structure, properties and roles of polysaccharides starch, glycogen, cellulose and chitin.
Define protein, amino acid and recognise essential amino acid and structural formula of amino acid.
Outline the synthesis and breakage of peptide linkages.
Justify the significance of the sequence of amino acids through the example of sickle cell haemoglobin.
Classify proteins as globular and fibrous proteins.
List the roles of structural proteins and functional proteins with 3 examples.
Define lipids.
Describe the properties and roles of acylglycerols, phospholipids, terpenes and waxes.
Illustrate the molecular structure (making and breaking) of an acylglycerol, a phospholipid and a terpene.
Write the chemical structure of a single phospholipid (glycerol as a three carbon molecule, phosphate group, one unsaturated fatty acid tail and one saturated fatty acid tail).
Evaluate steroids and prostaglandins as important groups of lipids.
Describe nucleic acids and molecular structure of nucleotides.
Distinguish among the nitrogenous bases found in the nucleotides of nucleic acids.
Outline the examples of a mononucleotide (ATP) and a dinucleotide (NAD).
Illustrate the formation of phosphodiester bond.
Explain the double helical structure of DNA as proposed by Watson and Crick.
Explain the general structure of RNA.
Distinguish in terms of functions and roles, the three types of RNA.
Discuss the central dogma.
Define conjugated molecules and describe the roles of common conjugated molecules i.e. glycolipids, glycoproteins, lipoproteins and nucleoproteins.

2.1Biochemistry / molecular biology

Every living cell is made of chemical compounds and runs on chemical reactions. Biochemistry is the subject that studies these compounds and reactions, and it is the foundation of the whole chapter.

Biochemistry: The study of the composition and structure of the different chemical compounds found in living organisms, and of the chemical processes taking place within or related to living organisms.
Molecular biology: The specialised branch of biochemistry. The two names are sometimes used interchangeably.

Biochemistry is a sub-discipline of both biology and chemistry. By using the knowledge and techniques of chemistry, biochemistry solves the problems faced by living organisms.

2.2Different types of bonds and interactions found in biological moleculesOften examined

Chemical bonds, and the weaker forces between and within molecules, decide how stable a biological molecule is and what it can do. This section goes through the bonds and interactions the book names, and the biological molecules in which each one matters.

2.2.1 Chemical bonds

The two most prevalent chemical bonds in biological molecules are the covalent bond and the ionic bond.

Covalent bond: A bond formed by the mutual sharing of electrons between two atoms so that each gets a stable electronic configuration.
Ionic bond: A bond formed between two atoms when one atom donates one or more electrons and becomes a positively charged ion, while the other atom gains these electrons and becomes a negatively charged ion. The oppositely charged ions are held together by the strong electrostatic force of attraction between them.

2.2.1.1 Covalent bonds

Biomolecules are mainly made of non-metals such as carbon, hydrogen, oxygen, sulphur, nitrogen and phosphorus, so covalent bonds are the most prevalent chemical bond in them. The covalent bonds found in carbohydrates, proteins, lipids, nucleic acids and vitamins include:

carbon–carbon bonds and carbon–hydrogen bonds
peptide bonds (proteins)
glycosidic bonds (carbohydrates)
phosphodiester bonds (nucleic acids)
disulphide bonds
ester bonds (lipids)
thioester bonds

2.2.1.2 Ionic bonds

Ionic bonds are less common than covalent bonds in biological molecules, but they still play some important roles. Salts are an important part of biological systems, and ionic bonding is the most common type of bonding found in salts. Ionic bonds are important in the structure of biological molecules in these places:

between some amino acids in proteins
between the negatively charged phosphate groups of DNA and other molecules
between metal ions and the protein part of enzymes, and in enzyme-substrate interactions

Figure 2.1: Different types of bonds and interactions found in a protein molecule. A folded polypeptide is held by hydrogen bonds, ionic bonds, disulphide bridges and hydrophobic interactions (all coloured) between side chains and backbone groups.

2.2.2 Intermolecular and intramolecular interactions

Apart from chemical bonds, various weak interactive forces exist between molecules (intermolecular forces) and between different parts of the same molecule (intramolecular forces). They differ in type and strength and are very important in deciding the physical properties of substances. They play a crucial role in the structure, function and behaviour of biological molecules and in many cellular processes. Protein folding, the double helical structure of DNA, the structure of the cell membrane and molecular recognition would never be possible without them.

2.2.2.1 Hydrogen bonding

A hydrogen bond is a type of intermolecular force that occurs between a hydrogen atom bonded to a highly electronegative atom (oxygen, nitrogen or fluorine) and a lone pair of electrons present on a neighbouring electronegative atom. It may be intermolecular (between atoms of different molecules) or intramolecular (between atoms of the same molecule).

The double helical structure of DNA is due to hydrogen bonds between the nitrogenous bases: three hydrogen bonds between cytosine and guanine and two between adenine and thymine. Protein-DNA interactions and various other biological processes also involve hydrogen bonding.

2.2.2.2 Hydrophobic interactions

Hydrophobic interactions are non-covalent interactions between non-polar molecules, or non-polar regions of molecules, in the presence of water. "Hydrophobic" literally means "afraid of water". Water is polar, so hydrophobic molecules or regions minimise their contact with water by clustering and coming together. This hydrophobic effect has important biological implications such as protein folding, membrane formation, ligand binding and protein-protein interactions.

2.2.2.3 Hydrophilic interactions

"Hydrophilic" literally means "water-loving". Hydrophilic molecules are polar: they have partially positive and partially negative regions (poles) because the shared electrons are unevenly distributed between the bonded atoms.

Hydration is a hydrophilic process in which an ionic substance placed in water dissociates into positive and negative ions, which are then surrounded by water molecules. This helps the dissolution and stabilisation of substances in aqueous solutions. Hydrophilic amino acid residues on the surface of proteins interact with water and help to keep proteins soluble and stable in the cell.

Figure 2.2: Hydrophilic and hydrophobic interactions in the cell membrane. The hydrophilic heads (coloured) face the water on both sides, and the hydrophobic tails face each other away from water.

In the cell membrane the phospholipids are arranged so that their hydrophilic heads point outwards and interact with water, while the hydrophobic tails point towards each other and away from water. DNA and RNA have phosphate groups in their backbone, which makes them hydrophilic, so they dissolve and work effectively in the watery nucleoplasm and cytoplasm. Hydrophilic active sites on enzymes let them interact with water and take part in biochemical reactions.

Summary of the bonds and interactions in 2.2
Bond or interaction	What happens	Where the book places it
Covalent bond	Atoms share electrons	C–C, C–H, peptide, glycosidic, phosphodiester, disulphide, ester and thioester bonds
Ionic bond	Opposite ions attract	Between amino acids, phosphate groups of DNA, metal ions and enzyme proteins
Hydrogen bond	H on N, O or F attracted to a lone pair on a neighbouring atom	Base pairs of DNA (3 for C and G, 2 for A and T), protein folding, water
Hydrophobic interaction	Non-polar parts cluster together away from water	Protein folding, membrane formation, ligand binding
Hydrophilic interaction	Polar parts and ions are surrounded by water (hydration)	Phospholipid heads, protein surfaces, DNA and RNA phosphate groups

2.3Protoplasm and biological molecules

The living mass of a cell, and of an organism in general, is called protoplasm. All the biochemical reactions needed to keep a cell alive take place in protoplasm. It is chiefly composed of bioelements and biomolecules, which make and run the machinery of the cell, that is, the cellular organelles.

2.3.1 Chemical composition of protoplasm

Out of the 92 naturally occurring elements on earth, approximately 25 are found in living organisms and are called bioelements. The human body is made of only 16 of these. Only six bioelements make up about 99% of the protoplasm; they are the major bioelements. Those that make up less than 1% are the minor bioelements. The bioelements combine to form thousands of different biomolecules, which may be inorganic (water and minerals) or organic (carbohydrates, lipids, proteins and nucleic acids).

Figure 2.3: Proportions of the six major bioelements in the human body. Together they make up 99%.

Major and minor bioelements
Type	Elements and their proportion in the human body
Major bioelements (99%)	Oxygen 65%, carbon 18%, hydrogen 10%, nitrogen 3%, calcium 2%, phosphorus 1%
Minor bioelements (less than 1%)	Potassium 0.35%, sulphur 0.25%, chlorine 0.15%, sodium 0.15%, magnesium 0.05%

2.3.2 Fundamental biological molecules

The four biological molecules, carbohydrates, proteins, lipids and nucleic acids, are of fundamental nature in protoplasm because they are present in every cell, from the simplest prokaryotic cell to the most advanced mammalian cell. They are needed to maintain the structure of the cell and its specific metabolic activities.

The four fundamental biological molecules
Biomolecule	Where it is found and what it does
Carbohydrates	Present in the cytoplasm; act as fuel for the metabolic activities of the cell
Proteins	Present in the membranes, ribosomes, cytoskeleton and enzymes of the cell
Lipids	Present in the membranes and cytoplasm; provide a reserved energy source, give shape, and protect and insulate the cell
Nucleic acid DNA	Present in the chromosome; controls the cell activity
Nucleic acid RNA	Present in the nucleoplasm and cytoplasm; takes genetic information from DNA and plays a role in protein synthesis

Table 2.1: Proportions of various biomolecules in bacterial and mammalian cells
Biomolecule	Bacterial cell	Mammalian cell
Water	70%	70%
Protein	15%	18%
Carbohydrates	3%	4%
Lipids	2%	3%
DNA	1%	0.25%
RNA	6%	1.1%
Other organic molecules (enzymes, hormones, metabolites)	2%	2%
Inorganic ions (Na⁺, K⁺, Ca²⁺, Mg²⁺, Cl⁻, SO₄²⁻)	1%	1%

2.4Condensation and hydrolysisOften examined

Large biological molecules are built from small ones and broken down again by two opposite kinds of reaction, condensation and hydrolysis.

Macromolecule (polymer): A compound of high molecular weight that is made from many repeating units.
Micromolecule (monomer): An individual unit of a polymer.
Condensation (dehydration synthesis): A reaction in which two monomers join and a molecule of water is released. Water is removed (dehydration) and a bond is made (synthesis).
Hydrolysis: The breakdown of a dimer or polymer into its monomers by the addition of water.
Dimer: The product of the condensation of two monomers, for example maltose.

Condensation

When two monomers such as α-glucose join, a hydroxyl group (–OH) is removed from one monomer and a hydrogen (–H) is removed from the other. They are released as water and a bond forms between the monomers. The product of two α-glucose molecules is maltose, a dimer.

Figure 2.4: Condensation (dehydration synthesis) of two α-glucose monomers to form maltose (dimer). The –OH from carbon 1 of one glucose and the H from the –OH on carbon 4 of the other (coloured) leave as water, and the oxygen left behind joins the two rings.

If the same reaction is repeated several times, the product is a polymer. For example, many α-glucose molecules condense to form amylose and many water molecules are released. The number of water molecules released depends on the number of bonds formed: 19 water molecules are released when 20 monomers condense to form a polymer. Condensation does not take place unless the proper enzyme is present and the monomers are in an activated, energy-rich form.

Worked example: counting the water in condensation and hydrolysis

A polysaccharide chain is built from 50 glucose units,

C X_{6} H X_{12} O X_{6}

. (a) How many bonds are formed and how many water molecules are released? (b) What is the molecular formula of the chain? (c) How many water molecules are needed to hydrolyse it completely?

Each condensation joins two units with one bond and releases one water molecule. A chain of $n$ units has $n - 1$ bonds.
$bonds = water released = n - 1 = 50 - 1 = 49$
The chain has all the atoms of 50 glucose molecules minus the atoms of 49 water molecules.
$C X_{300} H X_{600} O X_{300} - 49 H X_{2} O = C X_{300} H X_{502} O X_{251}$
Hydrolysis is the reverse reaction. Breaking the 49 bonds returns the 49 water molecules, and 50 glucose units are set free.
$water needed = n - 1 = 49$

Units (n)	Bonds formed	Water released in condensation	Water used in hydrolysis
2 (dimer)	1	1	1
3	2	2	2
20 (book)	19	19	19
50	49	49	49
n	n − 1	n − 1	n − 1

(a) 49 bonds and 49 water; (b) C X_{300} H X_{502} O X_{251}; (c) 49 water

Figure 2.5: Condensation of α-glucose monomers to form maltose (dimer) and a polymer (amylose), and hydrolysis of the dimer and the polymer back to α-glucose monomers. Each hexagon is one glucose unit.

Hydrolysis is essentially the reverse of condensation, the breakdown of a dimer or polymer into its monomers by the addition of water. During hydrolysis an –OH group from water is attached to one monomer and an –H to the other. For example, the breakdown of amylose in the presence of the enzyme amylase, with splitting of water molecules, releases many α-glucose monomers. All digestion reactions are examples of hydrolysis, and they are controlled by enzymes such as carbohydrases, proteases, lipases and nucleases.

Condensation compared with hydrolysis
Point	Condensation	Hydrolysis
Direction	Joins monomers to build a dimer or polymer	Breaks a dimer or polymer into monomers
Water	Released (dehydration synthesis)	Added to the molecule
Bonds	A bond is made between the monomers	The bond between the monomers is broken
Example	α-Glucose + α-glucose → maltose + water	Amylose + water → many α-glucose, by amylase

2.5Water as medium of lifeOften examined

Water is one of the main constituents of the earth, and the origin, sustainability and continuity of life depend on it. More than two thirds of the earth is covered by water, and the distribution of life on earth is directly proportional to the availability of water: deserts have few living organisms, while rivers and oceans are full of life.

Approximately 70% of any organism is water. It is the most abundant component in any organism: the lowest, 20%, is in seeds and bones and the highest, 85–90%, is in brain cells. The jellyfish has an exceptionally large amount of water, 99%, which makes its body transparent.

2.5.1 Properties of water

Eight properties of water make it the medium of life: high polarity, hydrogen bonding, cohesion and adhesion, high specific heat capacity, high heat of vaporisation, hydrophobic exclusion, ionisation and the lower density of ice.

1. High polarity

A covalent bond is formed by the mutual sharing of electrons between two atoms. Usually the sharing is fairly equal and the bond is non-polar. In water the sharing of electrons between oxygen and hydrogen is not equal, so the covalent bond is polar: the shared electrons are pulled closer to the more electronegative atom, making it partially negative and the other atom partially positive. In $H X_{2} O$ the oxygen atom has a slight negative charge ( $δ -$ ) and each hydrogen has a slight positive charge ( $δ +$ ), although the molecule as a whole is neutral. Water is therefore a polar molecule: it has one slightly negative pole and two slightly positive ones.

Figure 2.6: Polarity of the water molecule. Oxygen carries a slight negative charge and each hydrogen a slight positive charge.

This polarity makes water an excellent, or universal, solvent for polar substances. Ionic compounds and electrolytes dissolve easily in water, and non-polar substances with charged groups in their molecules also dissolve. When they dissolve they dissociate into positive and negative ions, which are in a more favourable state to react with other molecules and ions. This is why all the chemical reactions in living beings occur in an aqueous medium.

2. Hydrogen bonding

The polarity of water molecules makes them interact with each other. The charged regions on each molecule are attracted to oppositely charged regions on neighbouring molecules, forming weak bonds. Since the positively charged region in this bond is always an H atom, the bond is called a hydrogen bond. It is often shown as a dotted line because it is easily broken. Because of hydrogen bonding, water is a liquid at temperatures suitable for life. The high cohesion and adhesion of water are also due to hydrogen bonds, which makes water a transport medium.

Figure 2.7: Hydrogen bonds between water molecules. Each dotted (coloured) line joins a slightly positive hydrogen of one molecule to the slightly negative oxygen of the next; each molecule can bond to several neighbours.

3. Cohesion and adhesion

Cohesion: The attraction among water molecules, which enables them to stick together. Water flows freely due to cohesion.
Adhesion: The attraction of water molecules to polar surfaces.

Both are due to hydrogen bonds. These properties let water circulate in living bodies and act as a transport medium.

4. High specific heat capacity

Heat capacity is the amount of heat required for the minimum increase in temperature of a substance. The specific heat capacity of water is the number of calories required to raise the temperature of 1 g of water by 1 °C, that is, 1 calorie (4.18 joules). Water has a very high heat capacity compared with any other substance because much of the heat it absorbs is used to break hydrogen bonds, so it does not show itself as a rise in temperature. A very large amount of heat therefore raises the temperature of water very little. This makes water a temperature stabiliser or regulator for organisms in a hot environment and protects living material against sudden thermal changes.

5. High heat of vaporisation

Heat of vaporisation is the amount of heat required to convert a unit mass of a liquid into gas. For water it is high, 574 calories per gram. A large amount of heat can therefore be lost with a minimal loss of water from the body. This gives animals an efficient way to release excess heat in a hot environment: when an animal sweats, body heat is used to vaporise the sweat and the animal is cooled. The evaporation of only 2 ml out of one litre of water lowers the temperature of the remaining 998 ml by 1 °C.

Worked example: why sweating cools the body

Use the numbers in the book to check the statement "evaporation of only 2 ml out of one litre of water lowers the temperature of the remaining 998 ml by 1 °C", and explain which two properties of water are involved.

Take 1 ml of water as 1 g. The heat of vaporisation is 574 calories per gram, so the heat used to evaporate 2 g is:
$2 \times 574 = 1148 calories$
The specific heat capacity is 1 calorie per gram per °C, so the heat given up when 998 g of water cools by 1 °C is:
$998 \times 1 \times 1 = 998 calories$
The two values are close (998 is a little less than 1148), so the book's statement is a good approximation: the heat taken away by 2 ml of evaporating water is enough to cool the other 998 ml by about 1 °C.
Explain. High heat of vaporisation: a lot of heat is needed to turn each gram of sweat into vapour, so evaporation removes a lot of heat from the body. High specific heat capacity: the rest of the body water changes temperature only slowly, so cooling is smooth and not sudden.

Quantity	Value
Heat of vaporisation of water	574 cal per g
Specific heat capacity of water	1 cal per g per °C
Heat to evaporate 2 g	1148 cal
Heat lost when 998 g cools by 1 °C	998 cal

1148 cal \approx 998 cal

6. Hydrophobic exclusion

Hydrophobic exclusion is the reduction of the contact area between water and hydrophobic substances placed in water. For example, a few drops of oil on the surface of a water solution tend to join into a single drop. Biologically, hydrophobic exclusion plays a key role in maintaining the integrity of lipid bilayer membranes.

Figure 2.8: Hydrophobic exclusion. Non-polar molecules in water come together so that less of their surface touches water.

7. Ionisation

The dissociation of a molecule into ions is called ionisation. When a water molecule ionises, it releases an equal number of positive hydrogen ions and negative hydroxyl ions. The reaction is reversible, but equilibrium is maintained at 25 °C.

H X_{2} O H X^{+} + OH X^{-}

Figure 2.9: Ionisation of water into a hydrogen ion and a hydroxyl ion.

The $H X^{+}$ and $OH X^{-}$ ions affect and take part in many of the reactions that occur in cells, for example, they help to maintain or change the pH of the medium.

8. Lower density of ice

Ice floats on water because it is less dense than water. Ice has a giant structure with the maximum number of hydrogen bonds among water molecules, so the molecules are arranged like a lattice. In freezing weather, ice forms on the surface of ponds and lakes and forms an insulating layer above the water below. This provides a living environment for some organisms until the ice melts, and organisms can live under the ice.

Figure 2.10: Lattice-like arrangement of water molecules in ice. Each dot is a water molecule joined to its neighbours by hydrogen bonds in an open, hexagonal network.

Summary: the properties of water that make it the medium of life
Property	What it means	Importance for life
High polarity	Unequal sharing of electrons: slightly negative O, slightly positive H	Universal solvent; reactions occur in aqueous medium
Hydrogen bonding	Weak bonds between neighbouring water molecules	Liquid at temperatures suitable for life; cohesion and adhesion
Cohesion and adhesion	Water sticks to water and to polar surfaces	Water flows and acts as a transport medium
High specific heat capacity	Much heat is used to break hydrogen bonds	Temperature stabiliser; protects against sudden thermal change
High heat of vaporisation	574 cal needed to vaporise 1 g	Sweating cools animals in a hot environment
Hydrophobic exclusion	Non-polar substances are pushed together by water	Keeps lipid bilayer membranes intact
Ionisation	$H X_{2} O H X^{+} + OH X^{-}$	H⁺ and OH⁻ take part in reactions and set the pH
Lower density of ice	Ice forms an open lattice and floats	An insulating layer; organisms survive under the ice

2.6CarbohydratesOften examined

Carbohydrates are compounds of carbon, hydrogen and oxygen. The word literally means "hydrates of carbon", that is, carbon associated with water. According to this classic concept, carbohydrates are organic biomolecules in which all the carbon atoms are hydrated, with the general formula $C X_{n} (H X_{2} O) X_{n}$ , where $n$ is the number of carbon atoms and the number of water molecules. For example, glucose has the molecular formula $C X_{6} (H X_{2} O) X_{6}$ or $C X_{6} H X_{12} O X_{6}$ .

This formula is not exactly applicable in some cases. Lactic acid ( $C X_{3} H X_{6} O X_{3}$ ) and acetic acid ( $C X_{2} H X_{4} O X_{2}$ ) fit the formula but are not carbohydrates, whereas rhamnose ( $C X_{6} H X_{12} O X_{5}$ ) and deoxyribose ( $C X_{5} H X_{10} O X_{4}$ ) are carbohydrates although they do not fit it. Hence carbohydrates are now defined chemically as follows.

Carbohydrates: Organic compounds that are either polyhydroxy aldehydes or polyhydroxy ketones (monosaccharides) or their complex derivative compounds (polysaccharides), which on hydrolysis produce either polyhydroxy aldehydes or polyhydroxy ketones or both.

2.6.1 Classification of carbohydrates

Carbohydrates are commonly known as sugars or saccharides, because most of the familiar carbohydrates are sweet in taste. The word saccharide comes from the Greek word "sakcharon", which means sugar. Carbohydrates are classified on the basis of the number of saccharide units into three groups: monosaccharides, oligosaccharides and polysaccharides.

Figure 2.11: Chemical nature of carbohydrates. Glucose is a polyhydroxy aldehyde and fructose is a polyhydroxy ketone; the carbonyl group (coloured) is on carbon 1 in glucose and on carbon 2 in fructose.

Table 2.2: Comparison of the characteristics of carbohydrates
Monosaccharides	Oligosaccharides	Polysaccharides
Consist of a single saccharide unit	Composed of 2 to 10 saccharide units	Composed of more than 10 saccharide units
Simplest carbohydrates, so they cannot be hydrolysed further	Less complex; on hydrolysis they yield at least 2 and at most 10 monosaccharides	Highly complex; on hydrolysis they yield at least 11 monosaccharides
Highly soluble in water	Less soluble in water	Generally insoluble in water
The sweetest of all carbohydrates	Less sweet in taste	Tasteless

2.7MonosaccharidesOften examined

Monosaccharides are the true carbohydrates: they are either polyhydroxy aldehydes or polyhydroxy ketones. The general molecular formula is $(CH X_{2} O) X_{n}$ , where $n$ is the number of carbon atoms, which ranges from 3 to 7. All the carbon atoms in a monosaccharide except one carry a hydroxyl group (–OH); the remaining carbon atom is part of the aldehyde or the ketone group.

Classification of monosaccharides

Monosaccharides are classified by their functional group and by their number of carbon atoms. On the basis of the functional group, those containing an aldehyde are called aldoses and those containing a ketone are called ketoses. On the basis of carbon number they fall into five groups: trioses (3C), tetroses (4C), pentoses (5C), hexoses (6C) and heptoses (7C).

Table 2.3: Examples and functions of monosaccharides
Class	Formula	Aldoses	Ketoses	Function
Trioses (3C)	$C X_{3} H X_{6} O X_{3}$	Glyceraldehyde	Dihydroxyacetone	Intermediates in photosynthesis and cellular respiration
Tetroses (4C)	$C X_{4} H X_{8} O X_{4}$	Erythrose	Erythrulose	Intermediates in bacterial photosynthesis
Pentoses (5C)	$C X_{5} H X_{10} O X_{5}$	Ribose, deoxyribose ( $C X_{5} H X_{10} O X_{4}$ )	Ribulose	Ribose and deoxyribose are components of RNA and DNA respectively; ribulose is an intermediate in photosynthesis
Hexoses (6C)	$C X_{6} H X_{12} O X_{6}$	Glucose, galactose	Fructose	Glucose is the respiratory fuel (initial substrate); fructose is an intermediate in respiration; galactose is a component of milk sugar
Heptoses (7C)	$C X_{7} H X_{14} O X_{7}$	Glucoheptose	Sedoheptulose	Intermediates in photosynthesis

Properties of monosaccharides

They are the simplest form of carbohydrates, so they cannot be hydrolysed into simpler compounds.
They are highly soluble in water. The many hydroxyl groups can form hydrogen bonds with water molecules.
They are called sugars because of their sweet taste. The specially arranged hydroxyl groups interact with and activate the sweet taste receptors on the tongue.
They are reducing sugars: they can reduce other compounds and are themselves oxidised. They have a free aldehyde or ketone group whose oxygen atom may donate electrons to the recipient compound.
They show stereo-isomerism, because the hydrogen and hydroxyl groups are oriented differently at the several asymmetric (chiral) carbons. An asymmetric carbon is one attached to four different groups.
Those with five or more carbons may exist in two structural forms: the open-chain (acyclic) form and the ring (cyclic) form.

Chemical structures of monosaccharides

Monosaccharides are usually found in the open-chain form in the crystalline state. When they are dissolved in water, most of them (pentoses and hexoses) change into the ring form. Two types of ring are found, furanose and pyranose.

Furanose: A five-cornered ring with one oxygen and four carbon atoms. C1 and C4 are joined through the oxygen. Pentoses and ketohexoses form the furanose ring.
Pyranose: A six-cornered ring with one oxygen and five carbon atoms. C1 and C5 are joined through the oxygen. Only aldohexoses form the pyranose ring.

Figure 2.12: Types of ring structure in monosaccharides: the six-membered pyranose and the five-membered furanose.

Take the aldohexose glucose ( $C X_{6} H X_{12} O X_{6}$ ) and the aldopentose ribose ( $C X_{5} H X_{10} O X_{5}$ ). Both can exist in the open-chain form in dried form, but in water they exist as rings. When they dissolve in water, the oxygen of the aldehyde group joins with the second-last carbon, which is C5 in glucose and C4 in ribose. In this way an oxygen link is made between C1 and C5 in glucose and between C1 and C4 in ribose. The ring of glucose is called glucopyranose and the ring of ribose is called ribofuranose.

Figure 2.13: Conversion of (A) open-chain glucose into the ring form glucopyranose and (B) open-chain ribose into the ring form ribofuranose. The oxygen link closes the ring at carbon 1 (coloured).

Ring forms of glucose and ribose
Point	Glucopyranose	Ribofuranose
Parent sugar	Glucose, an aldohexose ( $C X_{6} H X_{12} O X_{6}$ )	Ribose, an aldopentose ( $C X_{5} H X_{10} O X_{5}$ )
Ring closes between	C1 and C5	C1 and C4
Ring size	Six corners (5 C and 1 O)	Five corners (4 C and 1 O)
Ring type	Pyranose	Furanose

2.8Isomers and stereoisomers of glucoseOften examined

The phenomenon in which compounds have the same molecular formula but differ in structural arrangement or formula is called isomerism. For example, glucose and fructose are both hexoses with the same molecular formula $C X_{6} H X_{12} O X_{6}$ , but they have different structures of their functional groups. Glucose is an aldo sugar and fructose is a keto sugar, so they are functional group isomers of each other.

Stereoisomerism in glucose

Stereoisomers: Molecules that have the same molecular formula but differ only in the relative arrangement of –H and –OH groups in 3D space at the anomeric and asymmetric carbon atoms.
Anomeric carbon: The carbon in the ring structure of a monosaccharide that is derived from the carbonyl carbon of the aldehyde or ketone group of the open chain, for example C1 in glucose and ribose.
Asymmetric carbon: A carbon that makes bonds with four different atoms or groups. In glucose four carbons (C2, C3, C4 and C5) are asymmetric. The last asymmetric carbon, the second last in the carbon chain (C5 in glucose), is called the penultimate carbon.

Figure 2.14: The carbonyl carbon (coloured) and the asymmetric carbons (dots) in the open-chain structure of glucose.

The number of stereoisomers of a monosaccharide depends on the number of asymmetric carbons in its structure and is calculated by the formula ** $2^{n}$ **, where $n$ is the number of asymmetric carbon atoms. Glucose has four, so it has $2^{4} = 16$ stereoisomers. They are classified into the following groups.

(i) Anomers

Anomers are stereoisomers that differ in the position of the –H and –OH groups on C1 (the anomeric carbon) of a pentose (for example ribose) or a hexose (for example glucose). They are called α or β isomers. If the –OH on C1 is found downward the sugar is α; if it is upward it is β.

Figure 2.15: Anomers (α and β stereoisomers) of glucose. They differ only in the direction of the –OH group on the anomeric carbon, C1 (coloured).

(ii) Enantiomers

Enantiomers are stereoisomers that are non-superimposable mirror images of one another. An example is the D (dextro) and L (levo) isomers of glucose. D-isomers are also called right-handed forms and L-isomers left-handed forms. In D-glucose the penultimate carbon (C5) has its –OH group on the right side, whereas in L-glucose it has the –OH group on the left side. Out of the 16 stereoisomers of glucose, 8 are enantiomers of the other 8.

Figure 2.16: An example of enantiomers: D-glucose and L-glucose are mirror images. The circled –OH on the penultimate carbon (C5) decides D or L.

(iii) Diastereoisomers

Diastereoisomers are stereoisomers that have different positions of –H and –OH groups at more than one asymmetric carbon atom. Unlike enantiomers, they are not mirror images of each other. For example, D-glucose and D-altrose are diastereoisomers of each other.

(iv) Epimers

Epimers are stereoisomers that have a different arrangement of –H and –OH groups at only one asymmetric carbon atom, for example D-glucose and D-mannose.

Figure 2.17: Eight stereoisomers of D-glucose (the D-aldohexoses). The circled –OH groups at the penultimate carbon indicate the D-isomers. Glucose is boxed.

The four groups of stereoisomers of glucose
Type	What differs	Mirror images?	Example in the book
Anomers	–OH on C1 (down = α, up = β)	No	α-glucose and β-glucose
Enantiomers	Every asymmetric carbon is reversed (D and L)	Yes	D-glucose and L-glucose
Diastereoisomers	More than one asymmetric carbon	No	D-glucose and D-altrose
Epimers	Only one asymmetric carbon	No	D-glucose and D-mannose

Worked example: how many stereoisomers?

How many stereoisomers are possible for (a) glucose, (b) fructose, (c) ribose? Mark the asymmetric carbons of fructose. (d) How many of the stereoisomers of glucose are D-sugars?

Count the asymmetric carbons: those joined to four different groups. The carbonyl carbon and the carbons bearing two hydrogen atoms are not asymmetric.
Glucose (aldohexose): C2, C3, C4 and C5 are asymmetric, so $n = 4$ .
$2^{4} = 16$
Fructose (ketohexose): C1 and C6 are $CH X_{2} OH$ groups, C2 is the carbonyl carbon, so only C3, C4 and C5 are asymmetric ( $n = 3$ ), as the dots in the figure show.
$2^{3} = 8$
Ribose (aldopentose): C2, C3 and C4 are asymmetric ( $n = 3$ ).
$2^{3} = 8$
D-forms of glucose: every D-isomer has an L-enantiomer, so half of the 16 are D-sugars and 8 are L-sugars.
$16 \div 2 = 8$

Open-chain fructose. The three dots (coloured) mark the asymmetric carbons C3, C4 and C5.

Sugar	Type	Asymmetric carbons	n	Stereoisomers $2^{n}$
Glucose	Aldohexose	C2, C3, C4, C5	4	16
Fructose	Ketohexose	C3, C4, C5	3	8
Ribose	Aldopentose	C2, C3, C4	3	8

(a) 16; (b) 8; (c) 8; (d) 8

2.9Properties and roles of disaccharidesOften examined

Two monosaccharides combine to form a disaccharide, which is a kind of oligosaccharide. Carbohydrates that are derivatives of monosaccharides and yield 2 to 10 saccharide units on hydrolysis are called oligosaccharides. On the basis of the number of saccharide units they are classified into disaccharides, trisaccharides, tetrasaccharides and so on. The most common among these are the disaccharides.

Disaccharides are less sweet in taste and comparatively less soluble in water than monosaccharides.
They can be hydrolysed to give monosaccharides.
They are familiar energy sources for living organisms. Examples are maltose, lactose and sucrose.
The general formula of a disaccharide is \ce{C12H22O11}.

2.9.1 Sucrose

Sucrose is commonly known as cane sugar. It is widely used as a sweetener at homes for making sweet dishes. In plants sucrose is also called the transport disaccharide, because the food prepared in plants is transported in the form of sucrose. It is very soluble, so it can be moved efficiently in high concentration in plants, and it is relatively unreactive chemically. On hydrolysis sucrose yields one molecule of α-glucose and one molecule of β-fructose.

2.9.2 Maltose

Maltose is commonly known as malt sugar. It is an intermediate disaccharide produced during the breakdown of starch and glycogen and acts as an energy-producing molecule. Maltose is produced in the brewing industry by the malting process, in which barley starch is broken down using the enzyme amylase. Maltose is generally found in germinating seeds and sprouting potatoes, where it acts as the energy source for the growth and development of young seedlings.

2.9.3 Lactose

Lactose is commonly known as milk sugar, because it is found in the milk of mammals. The ratio of lactose in milk is variable: 3–4% in goat milk, 4–6% in cow milk and 5–8% in human milk. It is also produced as a by-product during cheese production. Lactose is used as a direct energy source by infants. It acts as a substrate for the synthesis of various macro-biomolecules involved in vital nervous and immunological processes.

The three common disaccharides
Disaccharide	Common name	Made of	Where it is found and its role
Sucrose	Cane sugar	α-glucose + β-fructose	Sweetener; the transport disaccharide of plants; very soluble and relatively unreactive
Maltose	Malt sugar	α-glucose + α-glucose	Intermediate in the breakdown of starch and glycogen; energy source in germinating seeds and sprouting potatoes; made in brewing
Lactose	Milk sugar	β-galactose + β-glucose	Milk of mammals (goat 3–4%, cow 4–6%, human 5–8%); energy source for infants

2.10Glycosidic bonds in disaccharidesOften examined

A glycoside is basically a ring-form hexose (6C) or pentose (5C) sugar molecule that is attached, usually, to another sugar molecule.

Glycosidic bond: The covalent chemical bond between two glycosides (ring sugars) that holds them together.

A glycosidic bond is formed by a condensation reaction between two sugar molecules, mostly hexoses. The –H of one ring sugar interacts with the –OH of another sugar and one water molecule is released. Glycosidic linkages are present in disaccharides, oligosaccharides and polysaccharides.

How a glycosidic bond forms

Two ring sugars come close: an –OH on the first sugar, at its carbon 1, and an –OH on a carbon of the second sugar.
The –OH of the first sugar and the –H of the –OH of the second sugar are removed together as one molecule of water.
The oxygen that remains forms a bridge, C–O–C, between the two sugars. This is the glycosidic bond.
The bond is named by the carbon numbers of the two linked carbons, and as α or β according to the first sugar.

Glycosidic bonds are named on the basis of the carbon numbers of the interacting carbons of both linked sugars. The linkage is also called alpha or beta on the basis of the alpha or beta type of the first sugar involved in the linkage. For example, an α-1,4 glycosidic linkage is formed when the first sugar is alpha and donates its –OH group from its C1, and the other sugar donates the H of its fourth carbon.

2.10.1 Glycosidic linkage in sucrose

Sucrose is formed by the condensation of α-glucose and β-fructose. The –OH group at C1 of α-glucose interacts with the H of the –OH group at C2 of β-fructose and a water molecule is released, forming an α-1,2-glycosidic linkage (also written α-1,β-2).

Figure 2.18: Formation of the glycosidic bond in sucrose. The –OH on C1 of α-glucose and the H of the –OH on C2 of β-fructose (coloured) leave as water.

2.10.2 Glycosidic linkage in maltose

Maltose is formed by the condensation of two molecules of α-glucose. The –OH group at C1 of the first α-glucose interacts with the H of the –OH group at C4 of the other α-glucose, releasing a water molecule and forming an α-1,4-glycosidic linkage (also written α-1,α-4).

Figure 2.19: Formation of the glycosidic bond in maltose.

2.10.3 Glycosidic linkage in lactose

Lactose is formed by the condensation of β-galactose and β-glucose. The –OH group at C1 of β-galactose interacts with the H of the –OH group at C4 of β-glucose, releasing a water molecule and forming a β-1,4-glycosidic linkage (also written β-1,β-4).

Figure 2.20: Formation of the glycosidic bond in lactose.

Glycosidic linkages in the three disaccharides
Disaccharide	Monomers	Carbons joined	Linkage
Sucrose	α-glucose + β-fructose	C1 of glucose and C2 of fructose	α-1,2 (α-1,β-2)
Maltose	α-glucose + α-glucose	C1 of the first glucose and C4 of the second	α-1,4
Lactose	β-galactose + β-glucose	C1 of galactose and C4 of glucose	β-1,4

Breakage. Each of these reactions is reversible. When water is added, under the action of an enzyme, the glycosidic bond is broken by hydrolysis: sucrose gives α-glucose and β-fructose, maltose gives two α-glucose molecules and lactose gives β-galactose and β-glucose.

2.11PolysaccharidesOften examined

Carbohydrates that yield more than ten monosaccharide units on hydrolysis are called polysaccharides. They are the largest group of carbohydrates.

Homopolysaccharides: Polysaccharides composed by the condensation of only one kind of monosaccharide, for example starch, glycogen, cellulose and chitin.
Heteropolysaccharides: Polysaccharides composed by the condensation of different kinds of monosaccharides, for example agar, pectin and peptidoglycan.

Polysaccharides function chiefly as food and energy stores (starch, glycogen) and as structural material (cellulose, chitin). They are convenient storage molecules for several reasons:

Their large size makes them more or less insoluble in water, so they exert no osmotic or chemical influence in the cell.
They fold into compact shapes.
They are easily converted to sugars by hydrolysis when required.

Starch, cellulose and chitin, together with glycogen, are discussed below.

2.11.1 Structure, properties and roles of starch

Starch is a homopolysaccharide formed by the condensation of hundreds of α-glucose monomers. It is the storage carbohydrate of plants, mainly stored in roots, stems and seeds. Cereal grains and potato tubers are rich sources of starch in the human diet. Starch is digested in the oral cavity and small intestine by the enzyme amylase. On hydrolysis it yields maltose first, and then maltose is digested by the enzyme maltase to glucose. The presence of starch in a sample can be confirmed by the iodine test: it gives a blue colour with iodine solution. There are two types of starch, amylose and amylopectin.

Figure 2.21: Structure of starches. Amylose is an unbranched chain; amylopectin has an α-1,6 branch (coloured link) on an α-1,4 chain.

The two types of starch
Point	Amylose	Amylopectin
Structure	Unbranched, a linear chain of glucose units	Branched: a linear chain with more chains as branches
Linkages	α-1,4-glycosidic linkages	α-1,4 linkages in the chains and α-1,6-glycosidic linkages at the branches
Solubility in water	Soluble in hot water only	Completely insoluble in water

2.11.2 Structure, properties and roles of glycogen

Like starch, glycogen is a homopolysaccharide composed of many α-glucose monomers. It is the storage carbohydrate of animals, mainly stored in the liver and muscles, so it is also known as animal's starch. Its digestion is quite similar to that of starch. The presence of glycogen in a sample can be confirmed by the iodine test: it gives a red colour with iodine solution. The structure of glycogen resembles that of amylopectin starch, but glycogen has much more branching than amylopectin.

Figure 2.22: Structure of glycogen. A highly branched molecule: every bead is a glucose unit and branches start from the main chain at α-1,6 branch points.

2.11.3 Structure, properties and roles of cellulose

Cellulose is the most abundant carbohydrate on earth. It is also a homopolysaccharide but, unlike starch and glycogen, it is formed by the condensation of hundreds of β-glucose units. It is the structural carbohydrate of plants and a major constituent of the plant cell wall. Cotton and paper are pure forms of cellulose. Cellulose shows no colour with iodine solution. Its structure resembles amylose in that it is unbranched, but it has β-1,4-glycosidic linkages between the glucose residues.

Figure 2.23: Structure of cellulose, with β-1,4-glycosidic linkages between β-glucose units.

2.11.4 Structure, properties and roles of chitin

Chitin is the second most abundant organic molecule on earth. It is also a homopolysaccharide. It is a structural carbohydrate found in the cell walls of fungi and in the exoskeleton of arthropods. Because it occurs in the fungal cell wall it is also known as fungal cellulose. Chitin is a derivative of N-acetyl glucosamine monomers, which are a modified form of glucose. It has an unbranched structure, and its monomers are linked together by β-1,4-glycosidic linkages.

Figure 2.24: Structure of chitin. The N-acetyl group (coloured) on carbon 2 distinguishes it from cellulose.

Starch, glycogen, cellulose and chitin compared
Polysaccharide	Monomer	Linkage	Branching	Colour with iodine	Where found and role
Starch	α-glucose	α-1,4 (amylose); α-1,4 and α-1,6 (amylopectin)	Amylose none; amylopectin branched	Blue	Storage in plants: roots, stems, seeds, cereal grains, potato tubers
Glycogen	α-glucose	α-1,4 and α-1,6	Highly branched	Red	Storage in animals: liver and muscles
Cellulose	β-glucose	β-1,4	Unbranched	No colour	Structure: plant cell wall; cotton and paper
Chitin	N-acetyl glucosamine	β-1,4	Unbranched	Not given in the book	Structure: fungal cell walls and arthropod exoskeleton

Worked example: starch and cellulose

Both starch and cellulose are polymers of glucose. Differentiate between them in three points.

Look first at the monomer. Both are made of glucose, but of different forms: starch of α-glucose and cellulose of β-glucose.
Look at the bond between the units. Starch has α-1,4 links (and α-1,6 links at the branches of amylopectin); cellulose has only β-1,4 links and is unbranched.
Look at the job and the iodine test. Starch is the storage carbohydrate of plants and turns blue with iodine; cellulose is the structural carbohydrate of the plant cell wall and gives no colour.
Finally, add a point on digestion if more is wanted: amylase digests starch in the human gut, but humans cannot digest cellulose, which is why it works as fibre.

Point	Starch	Cellulose
Monomer	α-glucose	β-glucose
Linkage	α-1,4 (and α-1,6 branches in amylopectin)	β-1,4 only; unbranched
Role	Storage carbohydrate of plants	Structural carbohydrate of the plant cell wall
Test with iodine	Blue	No colour
Digestion in humans	Digested by amylase	Not digested; works as roughage

Biochemical tests for carbohydrates (STEAM activity 2.1)

The presence or absence of starch, glycogen, cellulose and glucose in a sample can be confirmed by biochemical tests.

Biochemical tests for carbohydrates
Test	Method	Observation	Inference
Iodine test for starch in solution (A)	Take 5 ml of water in a test tube, add a small amount of starch and boil to make starch solution. Add a few drops of iodine solution to the clear starch solution.	Dark blue-black colour	Starch is confirmed in solution
Iodine test for starch in food (B)	Cut a small slice of a food material such as potato and add a few drops of iodine solution on it.	Dark blue-black colour on the slice	Starch is confirmed in the food
Iodine test for glycogen (C)	Take glycogen solution in a test tube and add a few drops of iodine solution.	Red colour	Glycogen is present in solution
Iodine test for cellulose (D)	Take cellulose solution in a test tube and add a few drops of iodine solution.	No change in the colour of iodine solution	Starch and glycogen are absent and cellulose may be present
Benedict's test for glucose	Take 5 ml of Benedict's solution in a test tube, add a few drops of the test solution (glucose solution) and boil for 2–3 minutes.	Dirty green, yellow or red precipitate	Glucose is present
Fehling's solution test for glucose	Take 3 ml of Fehling's solution in a test tube, add 3 ml of the test solution (glucose solution) and heat.	Red precipitate	Glucose is present
Molisch's test	Take 1 ml of the test solution (sucrose), add 4–5 drops of alcoholic alpha-naphthol solution, then slowly pour 5 ml of concentrated \ce{H2SO4} and do not mix.	Purple ring at the junction of the two solutions	The book infers "disaccharide sugar present"

2.12ProteinsOften examined

Proteins play most of the critical roles in cells and are required for the structure, functioning, maintenance and regulation of the body systems. They are the main structural components of the cell. All proteins contain C, H, O and N, some also contain P and S, and a few have Fe, I and Mg built into the molecule.

2.12.1 Structure of proteins

Protein: Chemically, a polymer of amino acids, or a polypeptide chain. A protein may consist of a single polypeptide or more than one polypeptide.

2.12.2 Amino acids

Amino acids are the building blocks of proteins. Many amino acids are known to occur, but only 20 are commonly found in proteins. They are built on a common plan. Each contains a carbon atom called the α (alpha) carbon. To this carbon are attached a hydrogen atom, an amino group (–NH₂), a carboxyl group (–COOH) and a variable group known as the R group. The R group has a different structure in each of the 20 biologically important amino acids and determines their individual chemical properties.

Figure 2.25: General structure of an amino acid. Every amino acid has the same amino group, carboxyl group, hydrogen and α carbon; only the R group (coloured) differs.

The two simplest amino acids are glycine (R is a single hydrogen atom) and alanine (R is a \ce{CH3} group); they are shown joined in Figure 2.26.

Glycine R = H

Alanine R = CH₃

2.12.3 Essential and non-essential amino acids

Amino acids are categorised into two groups.

Essential amino acids: Amino acids that cannot be produced by our bodies and must be acquired through the diet. They can be obtained from foods such as milk, egg, fish, chicken, meat, vegetables and lentils. They build and repair muscle and bones and act as precursor molecules for the synthesis of hormones and neurotransmitters. Deficiency from an irregular diet can cause many health problems.
Non-essential amino acids: Amino acids that can be synthesised in the human body by modifying other amino acids and other food components, so they need not be part of the diet. Deficiency is rare, but illness or starvation may cause it. They are involved in the removal or detoxification of toxins, are essential in RBC and WBC synthesis, and improve brain and cardiac functioning.

Essential and non-essential amino acids
Group	Number	Amino acids
Essential	9	Lysine, leucine, isoleucine, histidine, methionine, phenylalanine, threonine, valine, tryptophan
Non-essential	11	Alanine, arginine, asparagine, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, tyrosine

2.13Synthesis and breakage of peptide linkagesOften examined

When two or more amino acids bind with each other by condensation, a bond called a peptide bond is formed. On hydrolysis of proteins these peptide bonds are broken to release multiple amino acids. Peptide bonds are formed when animal and plant cells store extra amino acids as polypeptides or proteins for their structural and functional role. Every cell carries out this polymerisation of amino acids. Peptide bonds are vital for the maintenance of the primary structure of proteins.

Figure 2.26: Formation of a dipeptide and a peptide bond. The –OH of the carboxyl group of glycine and an H of the amino group of alanine (coloured) leave as water, and a C–N peptide bond joins the two.

How a peptide bond is made

Condensation of amino acids

Two amino acids come together on the ribosome, with the instructions carried by mRNA (this process is called translation).
The –OH from the carboxylic acid group of one amino acid and an –H from the amino group of the other amino acid are liberated and form a molecule of water.
A bond is established between the C of the carboxylic acid group and the N of the amino group of the two amino acids. This is the peptide bond, and the product of two amino acids is a dipeptide.
A new amino acid can be added to the chain from its –C terminal end in the same way. A tripeptide (three amino acids) is formed and another water molecule is released. When several amino acids are linked by many peptide bonds, a polypeptide chain is formed.

A dipeptide has two ends: one is called the amino or –N terminal end, the other the carboxylic acid or –C terminal end.

Breakage. Peptide bonds are broken during catabolism of proteins in the gut, by digestive enzymes that act on proteins and polypeptide chains, such as pepsin, trypsin and erepsin (written "erypsin" in the book). Peptide bonds may also be broken at cellular level by lysosomal enzymes, to release amino acids for new protein synthesis or for ATP synthesis.

Worked example: peptide bonds in haemoglobin

Normal adult haemoglobin (\ce{HbA}) has four polypeptide chains: two α-chains of 141 amino acids each and two β-chains of 146 amino acids each. Find (a) the total number of amino acids, (b) the number of peptide bonds, (c) the number of water molecules released when the protein was built, and (d) the number of nucleotides in the mRNA that codes for one β-chain.

Add the amino acids of all four chains.
2\times141+2\times146=282+292=574
In each chain, n amino acids are joined by n-1 peptide bonds. There are 4 chains, so 4 bonds are "missing" from a single chain of 574.
\text{bonds}=574-4=570
Each peptide bond is formed by condensation and releases one water molecule.
\text{water released}=570
Three nucleotides in mRNA (one codon) code for one amino acid, so one β-chain of 146 amino acids needs:
146\times3=438\ \text{nucleotides}

Chain	Amino acids	Peptide bonds (n-1)
α-chain 1	141	140
α-chain 2	141	140
β-chain 1	146	145
β-chain 2	146	145
Total	574	570

\text{(a) 574;}\quad\text{(b) 570;}\quad\text{(c) 570;}\quad\text{(d) 438 (at least)}

2.13.1 Structural conformations in proteins

A protein does not stay in its primary structure after synthesis, but changes into other conformations (particular forms, shapes or structures). There are four levels of protein organisation.

Figure 2.27: Structural conformations (levels of organisation) in proteins: primary, secondary, tertiary and quaternary structure. The second chain in the quaternary structure is coloured.

The four structures of proteins
Level	What it is	Held by
Primary structure	A linear polypeptide with a specific sequence and number of amino acids. It is shown by all proteins at the time of their synthesis on the ribosomal surface.	Peptide bonds
Secondary structure	A helical shape (α-helix) or flattened sheet-like shape (β-pleated sheet), established between groups bearing opposite charges of different amino acids.	Hydrogen bonds
Tertiary structure	In some proteins the linear polypeptide is changed into an α-helix, and the α-helix folds again and again to form a globular shape.	Ionic bonds and disulphide bridges
Quaternary structure	Some proteins exist in a very complex structure in which more than one globule is attached together.	Hydrophobic interaction

2.14Significance of amino acid sequenceOften examined

The sequence of amino acids in a polypeptide is the characteristic feature of the primary structure of a protein, and it is responsible for the proper functioning of the protein. It is determined by the sequence of nucleotides in DNA. Even a point mutation (a change of a single or few nucleotides in DNA) can disturb the amino acid sequence of a particular protein, and it can cause severe defects in the body, as happens in sickle cell anaemia, a hereditary disease.

Normal red blood cells are disc-shaped and look like doughnuts without holes in the centre. They move easily through blood vessels. Red blood cells contain an iron-rich protein called haemoglobin, which carries oxygen from the lungs to the rest of the body. **Normal haemoglobin (\ce{Hb^A})** contains four polypeptides: two α-chains of 141 amino acids each and two β-chains of 146 amino acids each.

Sickle cell anaemia is a serious disorder in which the body makes sickle or crescent shaped red blood cells. Sickle cells contain abnormal haemoglobin called **sickle haemoglobin (\ce{Hb^S})**, which makes the cells develop a sickle shape. Sickle cells are stiff and sticky. They tend to block blood flow in the blood vessels of the limbs and organs, and blocked blood flow can cause pain and organ damage.

Sickle cell anaemia is caused by a point mutation in the β-globin gene, in which only one nucleotide is replaced by another. This changes the amino acid sequence of the β-chain of haemoglobin. \ce{Hb^S} shows only one difference from \ce{Hb^A}: glutamic acid is replaced by valine at position number six of the β-chain.

Figure 2.28: Difference in the β-chain of \ce{Hb^A} and \ce{Hb^S}. One amino acid at position 6 (coloured) changes from glutamic acid (Glu) to valine (Val), and the red blood cells change shape.

Normal and sickle haemoglobin compared
Point	Normal haemoglobin (HbA)	Sickle haemoglobin (HbS)
β-chain, positions 4 to 7	Thr, Pro, Glu, Glu	Thr, Pro, Val, Glu
Amino acid at position 6	Glutamic acid	Valine
Cause	Normal β-globin gene	Point mutation in the β-globin gene (one nucleotide replaced)
Red blood cells	Disc-shaped, move easily through blood vessels	Sickle or crescent shaped, stiff and sticky; block blood flow, causing pain and organ damage

2.15Classification of proteins

Based on their structure and shape, proteins are classified into two groups, fibrous and globular proteins.

Fibrous proteins: Proteins with a fibre or filament-like shape. They exist in the secondary structure during their function. They are insoluble in aqueous medium, elastic in nature and cannot be crystallised. Examples: collagen, fibrinogen, actin, myosin and keratin.
Globular proteins: Proteins with a spherical or globule-like shape. They exist in the tertiary or quaternary structure during their function. They are soluble in aqueous medium, inelastic in nature and can be crystallised. Examples: enzymes, hormones, antibodies and channel proteins.

Fibrous proteins compared with globular proteins
Point	Fibrous proteins	Globular proteins
Shape	Fibre or filament like	Spherical or globule like
Structure during function	Secondary structure	Tertiary or quaternary structure
Solubility in aqueous medium	Insoluble	Soluble
Elasticity	Elastic	Inelastic
Crystallisation	Cannot be crystallised	Can be crystallised
Examples	Collagen, fibrinogen, actin, myosin, keratin	Enzymes, hormones, antibodies, channel proteins

2.16Role of proteinsOften examined

Proteins are very important molecules in our cells and are involved in virtually all cell functions. Each protein has a specific function. Some proteins are involved in the support or composition of body parts, which are the structural roles, while others are involved in physiological activities such as bodily movement or defence against germs, which are the functional roles.

Table 2.4: List of structural proteins
Type	Role of the protein
Collagen	It establishes the matrix of bone and cartilages and is a component of ligaments and tendons
Elastin	It provides support for connective tissues such as tendons and ligaments
Keratin	It strengthens protective coverings such as hair, nails, quills, feathers, horns and beaks
Histone	It arranges the DNA into the chromosome

Table 2.5: List of functional proteins
Type	Role of the protein
Enzymes	Most enzymes are proteins which control metabolism, that is, they speed up the biochemical reactions
Hormones	Some hormones are proteins; they regulate physiological activities such as the glucose level, calcium level, digestion and blood pressure
Antibodies	Produced by white blood cells (WBCs) in response to an antigen (a foreign particle) and provide immunity
Haemoglobin	Found in red blood cells; transports oxygen mainly and carbon dioxide to some extent
Fibrinogen	Found in blood plasma; involved in the blood clotting process
Ovalbumin and casein	Ovalbumin is found in egg white and casein is a milk protein; both are involved in the storage of amino acids

Biochemical tests for proteins (STEAM activity 2.2)

The presence or absence of protein in a given sample can be confirmed by biochemical tests.

Biochemical tests for proteins
Test	Method	Observation	Inference
Biuret test	Take 2 ml of the test solution (protein solution), add 2 ml of concentrated sodium hydroxide solution and two to three drops of 1% copper sulphate solution, then shake well.	Pink or violet colour	Protein is present in the solution
Millon's test	Take 1 ml of the test solution (protein), add 1 ml of Millon's reagent. Shake well and let it settle.	Initially a white precipitate, which turns red when heated	Protein is present in the solution
Ninhydrin test	Take 2 ml of the test solution (protein), add 3–4 drops of aqueous ninhydrin reagent, boil and let it cool.	The book gives: initially a white precipitate, which turns red when heated	Amino acids or protein are present in the solution

2.17LipidsOften examined

Lipid is the collective name for a variety of organic compounds such as fats, oils, waxes and fat-like substances (steroids) found in the body.

Lipids: A heterogeneous group of organic compounds that are insoluble in water (hydrophobic) but soluble in organic solvents such as acetone, alcohol and ether.

Lipids are composed of carbon, hydrogen and oxygen, but they have relatively less oxygen in proportion to carbon and hydrogen than carbohydrates do. For instance, tristearin is a simple lipid with the molecular formula \ce{C57H110O6}. Because of their high content of carbon and hydrogen, lipids contain double the amount of energy of carbohydrates.

In general, lipids:

are components of cell membranes (phospholipids and cholesterol);
act as energy stores (triglycerides);
include steroid hormones;
are involved in protection, waterproofing, insulation and buoyancy.

2.18Properties and roles of acylglycerols, phospholipids, terpenes and waxesOften examined

2.18.1 Properties of acylglycerol

Acylglycerols have a lower density than water.
They are insoluble in water but soluble in some organic solvents such as chloroform, ether and hot alcohol.
Acylglycerols having saturated and long chain fatty acids have a higher melting point than those with unsaturated and short chains. For example, tristearin melts at 71 °C, while triolein melts at −17 °C.
Acylglycerols have great emulsifying power.
Hydrolysis of acylglycerols releases fatty acids and glycerol when heated in an aqueous acidic medium.
Acylglycerols can undergo oxidation to produce odorous and flavoured compounds on decay.

2.18.2 Roles of acylglycerols

Acylglycerols make the major part of the lipids in the body.
Triacylglycerols are the major lipids in fat deposits and in food.
When their level is high they may cause various diseases such as obesity, diabetes and hyperlipoproteinaemia.
Hydrogenation of acylglycerols produces margarine and butter. Margarine lacks the vitamins that are present in butter.

2.18.3 Properties of phospholipids

Phospholipids have the unique property of being amphipathic: they have both polar and non-polar ends.
They fix the proteins within the lipid bilayer of cell membranes.
They are important components of bile and lipoproteins.

2.18.4 Functions of phospholipids

They regulate the permeability of the membrane.
They are involved in the absorption of fat from the intestine.
They help in the electron transport chain (ETC) in the mitochondria.
They inhibit the accumulation of fats in the liver.
They transport and remove cholesterol from the cells.
They are structural components of the cell membrane and are associated with proteins.
They act as surfactants in the respiratory system.
They help in the coagulation of blood.
They synthesise different lipoproteins, prostaglandins and so on.

2.18.5 Roles of terpenes

They are anti-inflammatory and analgesic.
They are antioxidant, anticonvulsive and depressive.
They are anticancer, antitumour and anti-mutagenic.
They are neuroprotective, anti-allergic and antibiotic.
They have anti-diabetic activities.

2.18.6 Roles of waxes

Wax is effective in polishing, water proofing and anti-fouling.
It is used to make cosmetics and crayons, candles and car waxes.
Waxes are used to produce paints, inks, adhesives and tyres.

Summary of properties and roles
Lipid	Main properties	Main roles
Acylglycerols	Less dense than water; insoluble in water; high melting point if saturated and long chain; emulsifying power	Major lipids in fat deposits and food; margarine and butter by hydrogenation
Phospholipids	Amphipathic (polar and non-polar ends); fix proteins in the bilayer	Permeability of membranes, fat absorption, cholesterol transport, surfactants in the respiratory system
Terpenes	(Not described in this section)	Anti-inflammatory, antioxidant, anticancer, neuroprotective, antidiabetic
Waxes	Highly hydrophobic	Polishing, water proofing, cosmetics, candles, paints, inks, adhesives, tyres

2.19Molecular structure of an acylglycerol, a phospholipid and a terpeneOften examined

Some common lipids are acylglycerols, waxes, phospholipids, terpenes, prostaglandins and steroids.

2.19.1 Acylglycerol

Acylglycerols are the most abundant lipids in living things. Chemically, acylglycerols can be defined as esters of glycerol and fatty acids.

Ester: The compound produced as the result of a chemical reaction of an alcohol with an acid, in which a water molecule is released.
Esterification: The reaction of an alcohol with an acid to form an ester and water.

Figure 2.29 shows esterification with a simple example: methanol (an alcohol) reacts with ethanoic acid (an acid) to form methyl ethanoate (an ester) and water.

Methanol

Ethanoic acid

Methyl ethanoate (an ester)

\ce{CH3OH + CH3COOH <=> CH3COOCH3 + H2O}

Figure 2.29: Esterification

Glycerol: A trihydroxy alcohol that contains three carbons, each bearing an –OH group.
Fatty acid: A type of organic acid containing one carboxylic acid group attached to a hydrocarbon. Fatty acids contain an even number of carbons, from 2 to 30. Each fatty acid is represented as R–COOH, where R is a hydrocarbon tail.

Glycerol

When a glycerol molecule combines chemically with one fatty acid, a monoacylglycerol (monoglyceride) is formed. With two fatty acids a diacylglycerol (diglyceride) is formed, and with three fatty acids a triacylglycerol (triglyceride) is formed. Triacylglycerols are also called neutral lipids, because all three –OH groups of glycerol are occupied by fatty acids and no charge-bearing –OH group is left.

Figure 2.30: Formation of a triacylglycerol (neutral lipid). Each of the three –OH groups of glycerol joins a fatty acid by an ester bond, and three water molecules are released (the –OH of each acid and the H of each glycerol –OH, coloured).

Mono-, di- and triacylglycerols
Compound	Fatty acids joined to glycerol	Water molecules released
Monoacylglycerol (monoglyceride)	1	1
Diacylglycerol (diglyceride)	2	2
Triacylglycerol (triglyceride, neutral lipid)	3	3

Properties and types of fatty acids

About 30 different fatty acids are found, and they vary in length. Acetic acid (2C) and butyric acid (4C) are the simplest fatty acids, whereas palmitic acid (16C) and stearic acid (18C) are the most common fatty acids. Some properties of fatty acids, such as melting point, solubility in organic solvents and hydrophobic nature, increase with an increase in the number of carbon atoms. Fatty acids are either saturated or unsaturated.

Saturated fatty acids: Fatty acids in which all of the internal carbon atoms carry hydrogen side groups, so they contain the maximum number of hydrogen atoms possible, for example palmitic acid. They tend to be solid at room temperature (higher melting point) and are more common in animal lipids (fats).
Unsaturated fatty acids: Fatty acids with one or more pairs of carbon atoms joined by a double bond, so they are not fully saturated with hydrogen, for example oleic acid. They are liquid at room temperature (lower melting point) and are more common in plant lipids (oils).

A saturated and an unsaturated fatty acid. The double bond (coloured) bends the chain of oleic acid.

Table 2.6: Common types of fatty acids
Name	Typical source	No. of carbons	Condensed formula	Melting point (°C)
Saturated
1. Lauric acid	Coconut oil	12	CH₃(CH₂)₁₀COOH	44
2. Myristic acid	Butter fat	14	CH₃(CH₂)₁₂COOH	58
3. Palmitic acid	Most fats and oils	16	CH₃(CH₂)₁₄COOH	63
4. Stearic acid	Most fats and oils	18	CH₃(CH₂)₁₆COOH	70
Unsaturated
5. Oleic acid	Olive oil	18	CH₃(CH₂)₇CH=CH(CH₂)₇COOH	4
6. Linoleic acid	Vegetable oils	18	CH₃(CH₂)₄CH=CHCH₂CH=CH(CH₂)₇COOH	−5
7. Linolenic acid	Soybeans and canola oils	18	CH₃CH₂CH=CHCH₂CH=CHCH₂CH=CH(CH₂)₇COOH	−11
8. Arachidonic acid	Poultry, fish, egg	20	CH₃(CH₂)₄CH=CHCH₂CH=CHCH₂CH=CHCH₂CH=CH(CH₂)₃COOH	−50

Triglycerides containing hydrocarbon chains melt at a low temperature when the chains are unsaturated; this is useful for living things.

Saturated and unsaturated fatty acids compared
Point	Saturated fatty acids	Unsaturated fatty acids
Double bonds	None between the carbons	One or more double bonds
Hydrogen	Maximum possible number of H atoms	Not fully saturated with hydrogen
At room temperature	Tend to be solid (higher melting point)	Liquid (lower melting point)
Found mostly in	Animal lipids (fats)	Plant lipids (oils)
Examples in Table 2.6	Lauric, myristic, palmitic, stearic acid	Oleic, linoleic, linolenic, arachidonic acid

2.19.2 Waxes

Waxes are highly hydrophobic compounds. There are two types of waxes.

Natural waxes are simple lipids. They are typically esters of long chain fatty acids and long chain alcohols, such as bee's wax (found in honeycomb) and cutin (on the leaf surfaces of plants). They are chemically inert and resistant to atmospheric oxidation. Waxes have protective functions in plants and animals.
Synthetic waxes are generally derived from petroleum or polyethylene, for example paraffin wax, which is used to make candles.

2.19.3 Phospholipids

Phospholipids are derived from phosphatidic acid. A phosphatidic acid molecule is most similar to a diglyceride: it contains a glycerol, two fatty acids esterified with the first and second –OH groups of glycerol, and a phosphate group esterified with the third –OH group of glycerol. A phospholipid is formed when phosphatidic acid combines with one of four organic compounds: choline (a nitrogenous base), ethanolamine (an amino alcohol), inositol (an amino alcohol) and serine (an amino acid).

The most common type of phospholipid is phosphatidylcholine, also called lecithin, in which choline is attached to the phosphate group of phosphatidic acid. One end of the molecule, containing the phosphate group and the additional compound, is hydrophilic (polar and readily soluble in water). The other end, containing the fatty acid side chains, is hydrophobic (non-polar and insoluble in water). These phospholipids are major constituents of the lipid bilayer of the cell membrane.

Figure 2.31: Phosphatidylcholine (lecithin). A polar head (choline, phosphate and glycerol) carries two non-polar fatty acid tails, one saturated and one unsaturated with a cis double bond (coloured).

Phosphatidylcholine Glycerol with a saturated and an unsaturated fatty acid, a phosphate and choline

Building a phospholipid (as in Figure 2.31)

Start with glycerol, a three-carbon molecule with an –OH group on each carbon.
Join a saturated fatty acid to the first –OH group and an unsaturated fatty acid (with one cis double bond) to the second –OH group by ester bonds. This is a diacylglycerol; the two fatty acid chains are the hydrophobic tails.
Join a phosphate group to the third –OH group. This molecule is phosphatidic acid.
Attach choline to the phosphate group. Glycerol, phosphate and choline together form the polar, hydrophilic head.

Triacylglycerol compared with phospholipid
Point	Triacylglycerol	Phospholipid
Parts	Glycerol and three fatty acids	Glycerol, two fatty acids, a phosphate group and another compound such as choline
Polarity	Neutral lipid: non-polar	Amphipathic: polar head and non-polar tails
Role	Major lipid of fat deposits and food	Major constituent of the lipid bilayer of the cell membrane

2.19.4 Terpenes

All the terpenes are synthesised from a five-carbon building block known as the isoprene unit. This unit condenses in different ways to form many compounds. Two isoprene units form a monoterpene, for example menthol; four form a diterpene, for example vitamin A and phytol (the chlorophyll tail); and six form a triterpene, for example ambrein. Natural rubber is a polyterpene.

Isoprene unit Figure 2.32: the five-carbon building block of terpenes

The isoprene units that make up monoterpenes, diterpenes, triterpenes and polyterpenes.

2.20Steroids and prostaglandins

2.20.1 Steroids

Steroids are lipids of high molecular weight, which can be crystalline. A steroid nucleus consists of 17 carbon atoms arranged in four attached rings: three of the rings contain six carbon atoms and the fourth contains five. The length of the side chains that extend from these rings distinguishes one steroid from another. Steroids are synthesised from isoprene units.

Figure 2.33: Steroid nucleus: three six-carbon rings (A, B and C) and one five-carbon ring (D), with the 17 ring carbons numbered and two methyl groups (18 and 19).

Cholesterol is a structural component of the cell membrane. It is also the precursor of a large number of equally important steroids, which include bile acids, the male sex hormone testosterone and the female sex hormones progesterone and oestrogen (estrogen). Bile salts, which emulsify fats, and vitamin D, which helps to regulate calcium metabolism, are also steroids.

2.20.2 Prostaglandins

Prostaglandins exist in virtually every mammalian tissue, acting as local hormones. They are derived from arachidonic acid (a fatty acid in Table 2.6). Their functions vary widely depending on the tissue: some reduce blood pressure, whereas others raise it. In the immune system, various prostaglandins help to induce fever and inflammation and also intensify the sensation of pain. They also help to regulate the aggregation of platelets, an early step in the formation of blood clots. In fact, the ability of aspirin to reduce fever and decrease pain depends on the inhibition of prostaglandin synthesis.

Steroids and prostaglandins
Group	What it is	Examples and roles in the book
Steroids	Lipids with a nucleus of 17 carbons in four rings (three of six carbons, one of five)	Cholesterol (membrane component and precursor); bile acids and bile salts; testosterone, progesterone, oestrogen; vitamin D
Prostaglandins	Local hormones derived from arachidonic acid	Raise or lower blood pressure; induce fever and inflammation; intensify pain; platelet aggregation in clotting

Biochemical tests for lipids (STEAM activity 2.3)

The presence or absence of lipids in a sample can be confirmed by biochemical tests.

Biochemical tests for lipids
Test	Method	Observation	Inference
A. Spot test	Put a drop of the test solution (lipid solution) on a filter paper and allow it to dry.	A clear or translucent greasy spot appears on the paper	Lipid is present in the solution
B. Solubility in water	Take a small amount of fat (butter or ghee) in a test tube and half fill the tube with distilled water.	Fat and water remain separate	Fat is insoluble in water
C. Heating	Heat the water and fat in the test tube.	Fat melts on heating but does not mix with water and makes a clear upper layer over the lower water layer	Fat is insoluble in water and does not mix even in melted form
D. Emulsion test	Take 2 ml of the test material (lipid), add 2 ml of absolute alcohol and shake well. Now add 4 ml of cold water, shake well again, then allow to stand.	A cloudy white suspension (emulsion) is formed after shaking	Lipid is confirmed
E. Sudan III test	Take 2 ml of the test material (lipid), add 2 ml of water, then add 4–5 drops of Sudan III solution. Shake well and let it settle.	A dark red coloured ring or layer forms at the surface of the mixture when it settles	Lipids are present in the solution

2.21Nucleic acidOften examined

Nucleic acids were first reported in 1869 by a Swiss physician who isolated a new compound from the nuclei of pus cells (white blood cells). This compound was neither a protein nor a lipid nor a carbohydrate, so it was a novel type of biological molecule. He named it nuclein because it was located in the nucleus. Its basic structure and chemical nature were determined in 1920, and it was renamed nucleic acid because of its acidic nature.

2.21.1 Chemical structure of nucleic acids

Nucleic acids are of two types, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are linear unbranched polymers. The monomers of nucleic acids are called nucleotides.

Composition of a nucleotide

The nucleotides of DNA are called deoxyribonucleotides and those of RNA are called ribonucleotides. Each nucleotide consists of three parts: a pentose sugar, a phosphate and a nitrogen-containing ring structure called a base.

The pentose sugar is deoxyribose in deoxyribonucleotides and ribose in ribonucleotides.
Phosphoric acid is a common component of both kinds of nucleotide. It provides the acidic properties of DNA and RNA.
The nitrogen-containing ring structures are called bases because of the unshared pair of electrons on their nitrogen atoms, which can thus acquire a proton.

Figure 2.34: Structure of a nucleotide. A phosphate group is attached to carbon 5′ of the pentose sugar and a nitrogenous base to carbon 1′. The group on carbon 2′ is OH in RNA and H in DNA.

During the formation of a nucleotide, first the nitrogenous base is linked with the 1′ carbon of the pentose sugar. This combination is called a nucleoside. When phosphoric acid is linked with the 5′ carbon of the pentose sugar of a nucleoside, the nucleotide is formed.

Nucleoside: Pentose sugar + nitrogenous base (the base joined to carbon 1′ of the sugar).
Nucleotide: Nucleoside + phosphoric acid (the phosphate joined to carbon 5′ of the sugar).
Nucleoside monophosphate: A nucleotide with one phosphoric acid.
Nucleoside diphosphate: A nucleotide with two phosphoric acids.
Nucleoside triphosphate: A nucleotide with three phosphoric acids.

Nucleoside and nucleotides compared
Part	Joined to	Name of the product
Base + pentose sugar	Base on carbon 1′ of the sugar	Nucleoside
Nucleoside + 1 phosphate	Phosphate on carbon 5′	Nucleoside monophosphate (a nucleotide)
Nucleoside + 2 phosphates	Phosphates on carbon 5′	Nucleoside diphosphate
Nucleoside + 3 phosphates	Phosphates on carbon 5′	Nucleoside triphosphate

2.22Nitrogenous bases in the nucleotideOften examined

There are two major classes of nitrogenous bases, the single ring pyrimidines and the double ring purines.

Pyrimidines are of three types: cytosine (C), thymine (T) and uracil (U). Thymine is found only in DNA, while uracil is found only in RNA.
Purines are of two types: adenine (A) and guanine (G).

Pyrimidines and purines compared
Class	Rings	Bases	Found in
Pyrimidines	Single ring	Cytosine (C), thymine (T), uracil (U)	C in DNA and RNA; T only in DNA; U only in RNA
Purines	Double ring	Adenine (A), guanine (G)	Both DNA and RNA

Figure 2.35: Components of nucleotides. The two pentose sugars (below), then the phosphoric acid and the bases drawn as structures.

Phosphoric acid

Pyrimidine

Cytosine (C)

Uracil (U), found in RNA

Thymine (T), found in DNA

Purine

Adenine (A)

Guanine (G)

The nucleotides which take part in the formation of DNA or RNA must contain three phosphates, but during their incorporation into the DNA or RNA polymer each nucleotide loses its two terminal phosphates. The different terms used for nucleosides and nucleotides are given in Table 2.7.

Table 2.7: Different types of nucleosides and nucleotides of RNA and DNA
Nitrogenous base	Ribonucleosides (ribose + base)	Ribonucleotides (ribose + base + phosphate)	Deoxyribonucleosides (deoxyribose + base)	Deoxyribonucleotides (deoxyribose + base + phosphate)
Adenine	Adenosine	AMP, ADP, ATP	d-Adenosine	dAMP, dADP, dATP
Guanine	Guanosine	GMP, GDP, GTP	d-Guanosine	dGMP, dGDP, dGTP
Cytosine	Cytidine	CMP, CDP, CTP	d-Cytidine	dCMP, dCDP, dCTP
Uracil / Thymine	Uridine	UMP, UDP, UTP	d-Thymidine	dTMP, dTDP, dTTP

2.23Formation of phosphodiester bond (polymerisation of nucleotides)Often examined

Nucleotides are joined together by a condensation reaction, like other biomolecules. Unlike proteins, carbohydrates and lipids, however, the molecule released is not water but pyrophosphate (two phosphate groups bound together). The mechanism of polymerisation of nucleotides can be described in the following steps.

The phosphate group at 5′ of the precursor nucleotide triphosphate (NTP) is held close to the free OH group at 3′ of the last nucleotide of the growing polynucleotide chain, and forms a phosphodiester bond between the two nucleotides.
Meanwhile, the bond between the alpha and beta (second) phosphate groups of the NTP breaks and a pyrophosphate group is liberated.
The pyrophosphate group is hydrolysed (split by the addition of water), releasing a great deal of energy which drives the polymerisation reaction forward to completion.

Figure 2.36: Polymerisation of nucleotides. The 5′ phosphate of the incoming nucleotide triphosphate joins the 3′ OH of the chain (coloured bond); pyrophosphate is released and hydrolysed.

In this way nucleotides begin to link by phosphodiester bonds and a polymer of nucleotides (polynucleotide) is formed. Polynucleotides have a free 5′ phosphate group at one end and a free 3′ hydroxyl group at the other end. By convention, sequences are named from 5′ to 3′.

Phosphodiester bond: The bond that joins two nucleotides: the phosphate on carbon 5′ of one nucleotide is joined to the OH on carbon 3′ of the next, forming a phosphate bridge to two sugars.
Pyrophosphate: Two phosphate groups bound together; it is the molecule released when two nucleotides condense.

Condensation of nucleotides compared with other biomolecules
Point	Condensation of sugars, amino acids and fatty acids	Condensation of nucleotides
Molecule released	Water	Pyrophosphate (two phosphates)
Bond formed	Glycosidic, peptide or ester bond	Phosphodiester bond
Energy	Supplied separately	Released by hydrolysis of pyrophosphate; drives polymerisation to completion

2.24Chemical nature and role of ATP and NADOften examined

Adenosine triphosphate (ATP) is a mononucleotide. As shown in Figure 2.37, ATP has three parts connected by covalent bonds: (a) adenine, a nitrogen base, (b) ribose, a five-carbon sugar, and (c) three phosphates. The two covalent bonds linking the three phosphates together are called high-energy bonds. ATP can be converted to ADP and inorganic phosphate (\text{P}_\text{i}) by hydrolysis. ATP is known as the energy currency of cells.

\ce{ATP + H2O -> ADP + P_i}

Hydrolysis of ATP

Figure 2.37: Structure of ATP. Adenine + ribose = adenosine; adding one, two or three phosphates gives AMP, ADP and ATP. The two bonds between the phosphates (coloured) are the high-energy bonds.

ATP

The adenosine family
Name	Parts	Number of phosphates
Adenosine	Adenine + ribose	0
AMP (adenosine monophosphate)	Adenosine + 1 phosphate	1
ADP (adenosine diphosphate)	Adenosine + 2 phosphates	2
ATP (adenosine triphosphate)	Adenosine + 3 phosphates	3

Nicotinamide adenine dinucleotide (NAD) consists of two nucleotides. One nucleotide consists of nicotinamide, sugar and phosphate. The other nucleotide consists of adenine, sugar and phosphate. The two nucleotides are joined by their phosphate groups, forming a dinucleotide. NAD is a coenzyme.

Figure 2.38: Structure of NAD, drawn as blocks. Two nucleotides (nicotinamide + sugar + phosphate, and adenine + sugar + phosphate) joined through their phosphate groups.

NAD

ATP and NAD compared
Point	ATP	NAD
Kind of molecule	Mononucleotide	Dinucleotide (two nucleotides)
Parts	Adenine, ribose, three phosphates	Nicotinamide + sugar + phosphate, and adenine + sugar + phosphate
Role (as the book states)	Energy currency of cells	Coenzyme

2.25Watson and Crick model of DNAOften examined

In 1951, Erwin Chargaff found that the nitrogenous bases in DNA show specific ratios. He observed that the amount of adenine is always equal to the amount of thymine and the amount of guanine is always equal to the amount of cytosine in DNA. This implies that the total purines and total pyrimidines are in a 1:1 ratio in any DNA. This conclusion is known as Chargaff's rule.

\text{A}=\text{T},\qquad \text{G}=\text{C},\qquad \dfrac{\text{A}+\text{G}}{\text{T}+\text{C}}=\dfrac{\text{purines}}{\text{pyrimidines}}=1

Chargaff's rule

In those days the X-ray diffraction analysis of DNA by Maurice Wilkins and Rosalind Franklin was published. They claimed for the first time that DNA is a duplex (double helix) molecule. The width of the duplex is 2 nm while the length of each turn is 3.4 nm. In 1953, on the basis of these observations, a graduate student Francis Crick and a research fellow James Watson of Cambridge University proposed a physical model of DNA, now called the Watson and Crick model of DNA.

Figure 2.39: Watson and Crick model of DNA. (a) Key features of the double helix. (b) Partial chemical structure: sugar-phosphate backbones run in opposite directions (5′ to 3′ and 3′ to 5′) and the bases are paired by hydrogen bonds (dotted, coloured).

According to this model:

A DNA molecule is made up of two polynucleotide chains which are attached together by base pairs.
To make base pairing possible, the two chains are opposite in direction: one chain runs from 5′ to 3′ downward and the other from 5′ to 3′ upward. The chains are antiparallel to each other.
Both chains show a constant width of 2 nm.
The base pairing is very specific: adenine pairs with thymine and guanine pairs with cytosine.
The base pairs are held together by hydrogen bonds: three hydrogen bonds between guanine and cytosine and two between adenine and thymine.
Each turn of the duplex consists of 10 base pairs.
Both polynucleotide chains are complementary to each other.
There is no restriction on the sequence of nucleotides along the length of a DNA strand. The sequence can vary in countless ways, and it is specific for different species, organisms and even individuals.

Base pairing in DNA
Base pair	Bases	Hydrogen bonds	Kind of pair
A with T	Purine with pyrimidine	2	Weaker pair
G with C	Purine with pyrimidine	3	Stronger pair

Worked example: base pairing in a DNA segment

One strand of a DNA segment reads 5′-ATGCCATG-3′. (a) Write the complementary strand with its ends marked. (b) How many hydrogen bonds hold the two strands together? (c) In another DNA sample 20% of the bases are adenine. Find the percentage of each of the other bases.

Pair each base with its partner: A with T, T with A, G with C, C with G. The second strand runs in the opposite direction, so it is written 3′ to 5′ under the first strand.
\begin{aligned}5′&\text{-A T G C C A T G-}3′\\3′&\text{-T A C G G T A C-}5′\end{aligned}
Count the pairs. The segment has 4 A-T or T-A pairs (A, T, A, T) and 4 G-C or C-G pairs (G, C, C, G).
\text{4 pairs with 2 bonds}+\text{4 pairs with 3 bonds}
Add up the hydrogen bonds.
4\times 2+4\times 3=8+12=20
By Chargaff's rule A = T, so T is also 20%. That leaves 100 − 40 = 60% for G and C together, and G = C, so each is 30%.
\text{A}=\text{T}=20\%,\quad \text{G}=\text{C}=\dfrac{100-40}{2}=30\%

Base composition for part (c)
Base	A	T	G	C	Purines (A + G)	Pyrimidines (T + C)
Percentage	20	20	30	30	50	50

\text{(a) 3′-TACGGTAC-5′;}\quad\text{(b) 20 hydrogen bonds;}\quad\text{(c) T 20\%, G 30\%, C 30\%}

2.26Ribonucleic acid (RNA)

RNA is also a polymer of nucleotides. Its detailed chemical nature has already been discussed in the previous topics (Sections 2.21 to 2.23). Unlike DNA, RNA is generally single stranded and does not form a double helix like DNA. However, some regions of RNA show a secondary double-stranded structure in their complementary regions.

There are three major classes of RNA, each with a special function in protein synthesis. These RNAs are transcribed from a DNA template.

DNA and RNA compared (from Sections 2.21 to 2.26)
Point	DNA	RNA
Sugar	Deoxyribose	Ribose
Base found only in it	Thymine (T)	Uracil (U)
Strands	Two strands, double helix	Generally single stranded (some regions double stranded)
Monomers	Deoxyribonucleotides	Ribonucleotides

2.27Roles and types of RNAOften examined

2.27.1 Messenger RNA (mRNA)

mRNA consists of a single strand of variable length. Its length depends upon the size of the gene as well as the protein for which it is taking the message. For example, for a protein molecule of 100 amino acids, the mRNA will have a length of 300 nucleotides. Every three nucleotides in mRNA encode a specific amino acid, and such triplets of nucleotides along the length of mRNA are called codons of the genetic code. mRNA is about 3 to 4% of the total RNA in the cell. mRNA takes the genetic message from the nucleus to the ribosome in the cytoplasm to form a particular protein. This process is known as translation.

2.27.2 Ribosomal RNA (rRNA)

A ribosome consists of rRNA and protein. rRNA is transcribed by the genes present on the DNA of several chromosomes. It is called rRNA because it eventually becomes part of the ribosome. The rRNA is packaged with a variety of proteins into ribosomal subunits. The base sequence of rRNA is similar from bacteria to higher plants and animals. rRNA has the largest size among the RNAs. Approximately 80% of the total RNA of a cell is rRNA. It is a part of the ribosome where protein synthesis takes place. In other words, rRNA provides a platform for protein synthesis.

2.27.3 Transfer RNA (tRNA)

tRNA is the smallest of the RNA molecules and consists of 75 to 90 nucleotides. A tRNA is a single stranded molecule, but it shows a duplex appearance in some regions where complementary bases are bonded to one another. It shows a flat cloverleaf shape in two dimensional views. Its 5′ end always terminates in a guanine base while the 3′ end is always terminated with the base sequence CCA. The amino acid is attached to tRNA at this end. The nucleotide sequence of the rest of the molecule is variable.

tRNA has three loops.

The middle loop in all tRNAs is composed of 7 bases, the middle three of which form the anticodon. The anticodon is complementary to a specific codon of mRNA.
The D loop recognises the activation enzyme.
The theta (θ) loop recognises the specific place on the ribosome for binding during protein synthesis.

There is at least one tRNA molecule for each of the 20 amino acids found in proteins. Sixty tRNAs have been identified. However, human cells contain about 45 different kinds of tRNA molecules, each transporting a specific amino acid from the cytoplasm to the surface of the ribosome for protein synthesis.

Figure 2.40: Cloverleaf model of tRNA. The 3′ end carries the amino acid; the anticodon (coloured) pairs with the codon of mRNA.

The three major classes of RNA
Point	mRNA	rRNA	tRNA
Share of total cellular RNA	About 3 to 4%	About 80%	Not stated in the book
Size	Variable length, depends on the gene	Largest of the RNAs	Smallest: 75 to 90 nucleotides
Shape	Single strand	Packaged with proteins into ribosomal subunits	Cloverleaf with three loops
Function	Carries the genetic message (codons) from the nucleus to the ribosome	Part of the ribosome; platform for protein synthesis	Carries a specific amino acid to the ribosome; the anticodon pairs with the mRNA codon

2.28Central dogma of lifeOften examined

Genes contain the information needed for the synthesis of protein molecules. The flow of biological information from DNA to RNA and from RNA to protein is known as the central dogma of life. Protein synthesis takes place in two major steps, transcription and translation. Together, transcription and translation are known as gene expression.

\text{DNA}\xrightarrow{\text{transcription}}\text{RNA}\xrightarrow{\text{translation}}\text{protein}

Central dogma

Figure 2.41: Central dogma of molecular biology. In the nucleus, DNA is transcribed into mRNA; the mRNA moves to the cytoplasm where, at the ribosome, it is translated into a growing amino acid chain.

2.28.1 Transcription

Transcription is the first step of gene expression. When a new protein is needed by a cell, the two DNA strands of the gene that codes for the protein unwind from each other. A single strand of messenger RNA (mRNA) is then synthesised by pairing up complementary mRNA nucleotides (adenine, cytosine, guanine and uracil) with the exposed DNA (non-coding) strand. After the segment is copied as mRNA, both strands of DNA are "zipped" back together into their normal shape. Uracil is paired in mRNA as complementary to adenine instead of thymine.

Steps of transcription

The two DNA strands of the gene unwind from each other, exposing the bases of the template (non-coding) strand.
Free mRNA nucleotides (A, C, G and U) pair with the exposed bases by complementary base pairing: G with C, C with G, T with A and A with U.
The mRNA strand grows as nucleotides are added, and the segment of DNA that has been copied is zipped back into its double helix.
The finished single-stranded mRNA is released and goes to the ribosome.

Transcription: the DNA strands unwind, free mRNA nucleotides pair with the exposed template strand (U instead of T), and the DNA zips back as the mRNA is released.

2.28.2 Translation

Translation is the second step of gene expression. The mRNA synthesised in the nucleus of the cell moves into the cytoplasm, where it binds to the ribosome. The mRNA codes are translated by ribosomes with the help of tRNAs of opposite anticodon, thus synthesising a protein with a specific amino acid sequence. In this process each set of 3 mRNA nucleotides called a codon specifies an amino acid in the polypeptide. Each cell has many different types of tRNAs for transferring each type of amino acid to ribosomes when needed. During protein synthesis by the ribosome, amino acids are linked together by peptide bonds to form a polypeptide chain or protein.

Steps of translation

The mRNA moves from the nucleus into the cytoplasm and binds to a ribosome.
A tRNA whose anticodon is complementary to the codon on the mRNA brings its specific amino acid to the ribosome.
The ribosome joins the new amino acid to the growing chain by a peptide bond, and the empty tRNA leaves.
The ribosome moves along the mRNA, codon by codon, until the polypeptide is complete and the chain folds to form the protein.

Translation: the ribosome reads the mRNA codon by codon. Each tRNA pairs its anticodon with a codon and brings one amino acid, which is joined to the growing chain by a peptide bond (coloured).

Transcription compared with translation
Point	Transcription	Translation
Step of gene expression	First	Second
Where it starts from	DNA (the exposed non-coding strand)	mRNA (bound to a ribosome)
Product	mRNA, single stranded	Polypeptide chain (protein)
Pairing	mRNA nucleotides pair with DNA bases; U pairs with A	tRNA anticodon pairs with mRNA codon
Where (book figure)	In the nucleus	In the cytoplasm, at the ribosome

2.29Conjugated molecules

Molecules joined to other kinds of molecules are called conjugated molecules. The examples are glycolipids, glycoproteins, lipoproteins and nucleoproteins.

Glycolipids: Complex lipids containing one or more simple sugars in connection with long fatty acids or alcohol. They are present in the white matter of the brain, the myelin sheath of nerve fibres and the chloroplast membrane.
Glycoproteins: Formed when proteins are covalently attached to carbohydrates. They are widely distributed in the cells and function as hormones, transport proteins, structural proteins and receptors. The blood group antigens contain glycoproteins, which also play an important role in blood grouping.
Lipoproteins: Formed by the combination of protein with phospholipids. Phospholipid-protein complexes are widely distributed in plant and animal material. They occur in milk, blood, cell nucleus, egg yolk, membranes and chloroplasts of plants.
Nucleoproteins: Consist of a simple basic protein and nucleic acid. They are found in chromosomes and ribosomes.

Conjugated molecules
Conjugated molecule	Made of	Where found or what it does
Glycolipid	Lipid + sugar	White matter of brain, myelin sheath of nerve fibres, chloroplast membrane
Glycoprotein	Protein + carbohydrate (covalent)	Hormones, transport proteins, structural proteins, receptors; blood group antigens
Lipoprotein	Protein + phospholipid	Milk, blood, cell nucleus, egg yolk, membranes, chloroplasts
Nucleoprotein	Basic protein + nucleic acid	Chromosomes and ribosomes

STEAM activity 2.4

Performing Benedict's test for reducing sugars and confirmation of the presence of starch through the iodine test.
Confirmation of the presence of proteins through the Biuret test.
Confirmation of the presence of lipids through the emulsion test.
Demonstration of the presence of nucleic acids in biological materials, for example onion.

Careers in biochemistry

Science, Technology and Society Connections (book): "List the career opportunities in the field of biochemistry." Biochemistry, the study of chemical processes that take place in living organisms, is a broad field that offers a wide range of career options. Biochemists can pursue stem cell or genetic research that has the potential to result in dramatic medical or scientific breakthroughs. Some biochemists study the body's immune response to germs and allergens or the effectiveness of drugs in treating a wide array of afflictions. Other biochemists work in the commercial food or agricultural field looking for ways to improve products and crops. The many and diverse applications of biochemistry include pharmacology, genetics, immunology, bioinformatics, environmental science, forensics, toxicological studies and food science. The career options are nearly endless, and still unfolding.

Formula sheet

General formula of a carbohydrate

\ce{C_n(H2O)_n}

Carbohydrates have hydrogen and oxygen in the ratio of water; a monosaccharide is (\ce{CH2O})_n with n=3 to 7, and glucose is \ce{C6H12O6}.

no unit

Number of stereoisomers

\text{number of stereoisomers}=2^n

n is the number of asymmetric carbon atoms; glucose has n=4, so 2^4=16 stereoisomers (8 D and 8 L forms).

no unit

Disaccharide formation

\ce{C6H12O6 + C6H12O6 -> C12H22O11 + H2O}

Two monosaccharides join by a glycosidic bond; one water molecule is lost.

no unit

Water released or used in a chain

\text{bonds}=\text{water molecules}=n-1

n is the number of monomers (sugar, amino acid) in the chain; condensation releases and hydrolysis uses n-1 water molecules.

no unit

Peptide bonds in a polypeptide

\text{peptide bonds}=n-1

n is the number of amino acids in the polypeptide chain.

no unit

Esterification

\ce{R-OH + R'-COOH <=> R'-COO-R + H2O}

An alcohol and an acid form an ester and water; the same reaction joins glycerol to fatty acids.

no unit

Chargaff's rule

\text{A}=\text{T},\quad \text{G}=\text{C},\quad \dfrac{\text{A}+\text{G}}{\text{T}+\text{C}}=1

Holds for the bases of double-stranded DNA: total purines equal total pyrimidines.

no unit

Hydrogen bonds in a DNA base pair

\text{A-T}:\ 2\qquad \text{G-C}:\ 3

The number of hydrogen bonds between the two bases of a pair.

no unit

Nucleotides in an mRNA

\text{nucleotides}=3\times\text{number of amino acids}

Each codon of three nucleotides codes for one amino acid; 100 amino acids need 300 nucleotides.

no unit

Hydrolysis of ATP

\ce{ATP + H2O -> ADP + P_i}

Energy is released when the high-energy phosphate bond is broken.

no unit

Key terms

Biochemistry: The study of the composition and structure of the chemical compounds found in living organisms and the chemical processes taking place within or related to them.
Molecular biology: The specialised branch of biochemistry; the two names are sometimes used interchangeably.
Covalent bond: A bond formed by the mutual sharing of electrons between two atoms to get a stable electronic configuration.
Ionic bond: A bond formed when one atom donates electrons to become a positive ion and another atom gains them to become a negative ion.
Macromolecule: A compound of high molecular weight made from many repeating units (a polymer).
Monomer: An individual unit of a polymer (a micromolecule).
Condensation: A reaction in which two monomers join and a molecule of water is released (dehydration synthesis).
Hydrolysis: The breakdown of a dimer or polymer into its monomers by the addition of water.
Cohesion: The attraction among water molecules, which enables them to stick together.
Adhesion: The attraction of water molecules to polar surfaces.
Carbohydrates: Organic compounds that are polyhydroxy aldehydes or polyhydroxy ketones, or give these on hydrolysis.
Furanose: A five-cornered sugar ring with one oxygen and four carbon atoms.
Pyranose: A six-cornered sugar ring with one oxygen and five carbon atoms.
Stereoisomers: Molecules with the same molecular formula that differ only in the relative arrangement of their atoms in space.
Anomeric carbon: The ring carbon of a monosaccharide derived from the carbonyl carbon; it gives the alpha and beta forms.
Asymmetric carbon: A carbon bonded to four different atoms or groups.
Glycosidic bond: The covalent bond that holds two ring sugars together in a disaccharide or polysaccharide.
Homopolysaccharide: A polysaccharide made of only one kind of monosaccharide, such as starch, glycogen and cellulose.
Heteropolysaccharide: A polysaccharide made of different kinds of monosaccharides.
Protein: A polymer of amino acids, or a polypeptide chain.
Peptide bond: The covalent bond between the carboxyl group of one amino acid and the amino group of the next, formed with loss of water.
Essential amino acids: Amino acids that cannot be made by the body and must be taken in the diet.
Non-essential amino acids: Amino acids that the body can make by modifying other amino acids.
Point mutation: A change of a single or a few nucleotides in DNA, which can change the amino acid sequence of a protein.
Fibrous proteins: Fibre-like, insoluble, elastic proteins that act in the secondary structure, such as collagen and keratin.
Globular proteins: Spherical, soluble, inelastic proteins that act in the tertiary or quaternary structure, such as enzymes and antibodies.
Lipids: A heterogeneous group of organic compounds that are insoluble in water but soluble in organic solvents.
Ester: The compound produced when an alcohol reacts with an acid.
Esterification: The reaction of an alcohol with an acid to form an ester and water.
Glycerol: A trihydroxy alcohol with three carbons, each bearing an OH group.
Fatty acid: An organic acid with one carboxyl group attached to a hydrocarbon chain.
Saturated fatty acids: Fatty acids with no double bonds between carbons; every carbon carries hydrogen.
Unsaturated fatty acids: Fatty acids with one or more double bonds between carbon atoms.
Phospholipid: A lipid derived from phosphatidic acid, with a polar head and two nonpolar fatty acid tails.
Terpenes: Lipids built from five-carbon isoprene units.
Steroids: Lipids with a nucleus of 17 carbon atoms in four fused rings, such as cholesterol.
Prostaglandins: Local hormones derived from arachidonic acid.
Nucleic acid: A linear unbranched polymer of nucleotides; DNA or RNA.
Nucleotide: The monomer of nucleic acids: pentose sugar, phosphate and nitrogenous base.
Nucleoside: A pentose sugar joined to a nitrogenous base.
Pyrimidines: Single-ring nitrogenous bases: cytosine, thymine and uracil.
Purines: Double-ring nitrogenous bases: adenine and guanine.
Phosphodiester bond: The bond joining the 5′ phosphate of one nucleotide to the 3′ OH of the next in a polynucleotide.
Pyrophosphate: Two phosphate groups bound together; released when nucleotides condense.
ATP: Adenosine triphosphate, a mononucleotide of adenine, ribose and three phosphates; the energy currency of cells.
NAD: Nicotinamide adenine dinucleotide, a dinucleotide coenzyme.
Chargaff's rule: In DNA, adenine equals thymine and guanine equals cytosine, so purines and pyrimidines are in a 1:1 ratio.
Double helix: The two antiparallel polynucleotide chains of DNA coiled together, 2 nm wide with 3.4 nm per turn.
Antiparallel: Running in opposite directions, as the two strands of DNA do.
RNA: Generally single-stranded polymer of ribonucleotides; the three major classes are mRNA, rRNA and tRNA.
mRNA: Messenger RNA; carries the genetic message from the nucleus to the ribosome.
rRNA: Ribosomal RNA; the largest RNA, part of the ribosome and a platform for protein synthesis.
tRNA: Transfer RNA; the smallest RNA, with a cloverleaf shape, carrying a specific amino acid to the ribosome.
Codon: A triplet of nucleotides on mRNA that specifies one amino acid.
Anticodon: The three bases in the middle of the tRNA loop that are complementary to a codon of mRNA.
Central dogma of life: The flow of biological information from DNA to RNA and from RNA to protein.
Transcription: The first step of gene expression: synthesis of mRNA on an exposed DNA strand.
Translation: The second step of gene expression: synthesis of a polypeptide at the ribosome from the mRNA codons.
Gene expression: Transcription and translation together.
Conjugated molecules: Molecules joined to other kinds of molecules: glycolipids, glycoproteins, lipoproteins and nucleoproteins.

Multiple choice questionsp. 79–80

Select the correct answer.

An amino acid molecule has the structure shown (groups numbered 1 to 4). Which two of the groups combine to form a peptide link between two amino acids?

Structure of an amino acid. Groups 1 (amino), 2 (H), 3 (carboxyl) and 4 (R) are numbered as in the question.

A1 and 2
B1 and 3Answer
C2 and 3
D2 and 4

Why

The peptide link is the \text{C–N} bond formed when the -\text{COOH} group (3) of one amino acid joins the -\text{NH}_2 group (1) of another, releasing water. Groups 1 and 3 are therefore the answer.

Group 2 is the hydrogen on the \alpha-carbon and does not take part; group 4 is the variable R group. Option C (2 and 3) and option D (2 and 4) pick the hydrogen or the R group, which stay on the \alpha-carbon.

Which class of molecule is the major component of the cell membrane?

AphospholipidAnswer
Bcellulose
Cwax
Dtriglyceride

Why

The cell membrane is a lipid bilayer of phospholipids: each molecule has a polar head and two non-polar fatty acid tails (book, section 2.18.3).

Cellulose (B) is a polysaccharide of plant cell walls, wax (C) is a protective coating, and triglyceride (D) is a storage fat. None of them forms the membrane bilayer.

Glycerol is the backbone molecule for:

AATP
Bterpenes
Cneutral lipidsAnswer
Dsteroids

Why

Glycerol (a trihydroxy alcohol) is the backbone of acylglycerols. When all three OH groups are esterified with fatty acids, the molecule is a triacylglycerol, which the book calls a neutral lipid because no charged OH group is left (section 2.19.1).

ATP (A) has ribose, not glycerol, as its sugar. Terpenes (B) are built from isoprene units and steroids (D) from four fused rings, so neither has a glycerol backbone.

A fatty acid is unsaturated if it:

Acontains hydrogen
Bcontains double bondsAnswer
Ccontains an acid group
Dall of them

Why

An unsaturated fatty acid has one or more pairs of carbon atoms joined by a double bond, so it does not hold the maximum number of hydrogen atoms (section 2.19.1, Table 2.6).

All fatty acids contain hydrogen (A) and a carboxylic acid group (C), so these do not distinguish saturated from unsaturated acids. Option D is wrong because only option B is true.

In RNA the nitrogen base that takes the place of thymine is:

Aadenine
Bcytosine
Cguanine
DuracilAnswer

Why

Thymine occurs in DNA. In RNA the complementary base to adenine is uracil, which the book says is found only in RNA (section 2.22).

Adenine (A) and cytosine (B) are found in both DNA and RNA, and guanine (C) is also in both, so none of them replaces thymine.

The ending -ose means a substance is a:

AsugarAnswer
Blipid
Cprotein
Dnucleic acid

Why

Names of monosaccharides and disaccharides end in -ose (glucose, fructose, sucrose, maltose), and these are sugars (sections 2.7 and 2.9).

The ending does not indicate a lipid (B), protein (C) or nucleic acid (D); those names are based on other words such as -in, -ids or -ic acid.

Note. The book does not state the -ose rule in words; the answer follows from the names of its sugars.

Glycolipids and lipoprotein are important components of:

Acellular membraneAnswer
Bcell wall
Cboth of them
Dnone of them

Why

Glycolipids are found in the membranes of brain and nerve myelin sheath and chloroplast membrane, and lipoproteins contain phospholipid–protein complexes that form part of membranes (section 2.29). Both are components of cellular membranes, so option A is the answer.

Option C (both of them) is tempting but is not the intended answer: lipoproteins are also found in milk and blood, which are not membranes. Cell walls (B) are made of cellulose or chitin, not these molecules.

Note. The book prints no key. Option C is arguable because lipoproteins also occur in membranes (egg yolk membrane, chloroplasts); A is the answer the book's text supports best.

When two amino acids are linked to form a peptide linkage, ........... is removed.

Ahydroxyl
BwaterAnswer
Ccarbon
Dnitrogen

Why

A peptide bond forms when the –OH from the carboxyl group of one amino acid and the –H from the amino group of the other are removed, and they combine to form water (section 2.13).

Hydroxyl (A) is only half of the water molecule and is not released alone. Carbon (C) and nitrogen (D) remain in the peptide bond itself.

What is the theoretical number of chemically different dipeptides that may be assembled from two amino acids?

Aone
BtwoAnswer
Cthree
Dfour

Why

With two different amino acids, X and Y, two dipeptides are possible: X–Y and Y–X. The order matters because each dipeptide has a free amino (N) end and a free carboxyl (C) end (section 2.13).

One (A) would be correct only if both amino acids were the same; three or four (C, D) are impossible with two residues.

Note. The question assumes two different amino acids. With identical amino acids only one dipeptide is possible.

A polar molecule is ........... in water.

AsolubleAnswer
Binsoluble
Creactive
Dinnert

Why

Polar molecules have partly charged ends that interact with water's polar molecules and form hydrogen bonds, so they dissolve: a polar molecule is soluble in water (section 2.5).

Insoluble (B) describes non-polar molecules such as fats, which water repels. Reactive (C) and inert (D) are not properties linked to polarity in water.

Which statement correctly describes a property of water?

Aa relatively large amount of heat is needed to increase its temperatureAnswer
Bat normal room temperature, its molecules are bound together by ionic bonds
Cthe highest density of water occurs below its freezing point
Dwater acts as solvent for nonpolar molecules

Why

Water has a high specific heat capacity: a large amount of heat is needed to raise its temperature by 1 °C (section 2.5.1, property 4). Option A states this correctly.

B is wrong because water molecules are held together by hydrogen bonds, not ionic bonds. C is wrong because ice, not liquid water near its freezing point, is the least dense form. D is wrong because water dissolves polar substances; non-polar molecules are repelled.

Estrogen, vitamin-D and cholesterol are all examples of:

Aglycolipids
Blipoproteins
Cterpenes
DsteroidsAnswer

Why

All three are steroids. The book lists cholesterol, bile acids, testosterone, progesterone, estrogen and vitamin D among the steroids, all built on the four-ring steroid nucleus (section 2.20.1).

They are lipids but not glycolipids (A), lipoproteins (B) or terpenes (C), which are built from isoprene units.

Which term includes all others?

AcarbohydrateAnswer
Bstarch
Cmonosaccharide
Dpolysaccharide

Why

Carbohydrate is the broadest term: monosaccharides, oligosaccharides (including disaccharides) and polysaccharides (including starch) are all carbohydrates (section 2.6.1).

Starch (B), monosaccharide (C) and polysaccharide (D) are each only one part of the carbohydrate group, so none of them includes all the others.

Choose the pair of terms that correctly completes this sentence: Nucleotides are to ...... as ...... are to proteins.

Anucleic acids; amino acidsAnswer
Bamino acids; polypeptides
Cglycosidic linkages; polypeptide linkages
Dpolymers; polypeptides

Why

Nucleotides are the building blocks of nucleic acids, and amino acids are the building blocks of proteins. The correct pair is therefore nucleic acids; amino acids (option A).

Option D (polymers; polypeptides) is tempting, but a polymer is a class, not the molecule that nucleotides build; the analogy needs the molecule itself. B and C put the wrong molecules in the pattern.

The enantiomer of D-glucose is:

AD-galactose
BL-galactose
Cboth of them
Dnone of themAnswer

Why

An enantiomer is the non-superimposable mirror image of a sugar. The mirror image of D-glucose is L-glucose (book, section 2.8, Fig. 2.16). Neither D-galactose nor L-galactose is the mirror image of glucose: galactose differs from glucose at C-4.

L-galactose (B) is tempting because it is a left-handed sugar, but it is the enantiomer of D-galactose. Since neither listed option is L-glucose, option D is the answer.

Note. The book prints no key. The answer is D because L-glucose is not offered. If an official key says B, that key is wrong: L-galactose is not the mirror image of D-glucose.

Short answer questionsp. 80–81

13 marks

How would you describe biochemistry?

Solution

Biochemistry is the study of the composition and structure of the chemical compounds found in living organisms, and of the chemical processes that take place in, or relate to, living organisms (section 2.1).

It applies the knowledge and techniques of chemistry to the problems faced by living organisms.
It explains life processes such as photosynthesis, respiration, digestion, inheritance and muscle contraction in chemical terms.
Molecular biology is the specialised branch of biochemistry, although the two terms are often used interchangeably.

23 marks

What are bioelements?

Solution

Bioelements are the chemical elements found in living organisms. About 25 of the 92 naturally occurring elements are bioelements, and the human body is built from 16 of them.

Major bioelements make up about 99% of protoplasm: oxygen (65%), carbon (18%), hydrogen (10%), nitrogen (3%), calcium (2%) and phosphorus (1%).
Minor bioelements are each less than 1% of protoplasm, for example potassium (0.35%), sulphur (0.25%), chlorine, sodium (0.15% each) and magnesium (0.05%).
The bioelements combine to form the thousands of biomolecules, both inorganic (water, minerals) and organic (carbohydrates, lipids, proteins, nucleic acids).

33 marks

Describe the chemical composition of protoplasm.

Solution

Protoplasm is the living mass of a cell. Water is its largest component, and the rest is made of the four biomolecules together with ions and other organic molecules (section 2.3, Table 2.1).

Biomolecule	Bacterial cell	Mammalian cell
Water	70%	70%
Protein	15%	18%
Carbohydrates	3%	4%
Lipids	2%	3%
DNA	1%	0.25%
RNA	6%	1.1%
Other organic molecules (enzymes, hormones, metabolites)	2%	2%
Inorganic ions (Na⁺, K⁺, Ca²⁺, Mg²⁺, Cl⁻, SO₄²⁻)	1%	1%

The 16 bioelements of the body, mainly C, H, O and N, combine to form these molecules; the proportions are much the same in bacteria and mammals.

43 marks

What are the four fundamental kinds of biological molecules? Explain.

Solution

The four fundamental kinds of biological molecules are carbohydrates, proteins, lipids and nucleic acids. They are present in every cell, from the simplest prokaryote to a mammalian cell (section 2.3.2).

Molecule	Monomer (building block)	Main role in the cell
Carbohydrates	Monosaccharides (e.g. glucose)	Fuel for metabolism; cell structure in plants (cellulose) and fungi (chitin)
Proteins	Amino acids	Membranes, ribosomes, cytoskeleton and enzymes
Lipids	Fatty acids and glycerol (and other lipids)	Energy reserve; membranes (phospholipids); steroid hormones; insulation
Nucleic acids	Nucleotides	DNA controls cell activity; RNA takes genetic information from DNA and takes part in protein synthesis

53 marks

Why is the covalent bond in water polar?

Solution

In a water molecule each hydrogen shares a pair of electrons with oxygen. Oxygen is more electronegative than hydrogen, so it pulls the shared electrons closer to itself.

Oxygen becomes slightly negative (δ−) and each hydrogen slightly positive (δ+), although the molecule as a whole is neutral.
Because the bond is unequally shared, it is a polar covalent bond, not a non-polar one as in C–H bonds.
The bent shape of the molecule means the charges do not cancel, so water is a polar molecule with one negative pole and two positive poles.

62025 Annual I, Set B, Q.2 (iii)3 marks

Why is water regarded as the universal solvent?

Solution

Water is a polar molecule, so it can dissolve almost any polar or ionic substance. This is why water is called the universal solvent for polar substances (section 2.5.1).

Ionic compounds (electrolytes) dissolve readily: water molecules surround the ions and separate them into positive and negative ions.
Non-polar molecules that carry charged groups can also dissolve, as their charged regions attract water.
Dissolved substances are in a more favourable state to react, which is why all chemical reactions in living things take place in an aqueous medium.

73 marks

What is the importance of hydrogen bonding?

Solution

A hydrogen bond forms between the partly positive H of one molecule and the partly negative O or N of a neighbouring molecule. Each bond is weak but many together are important (section 2.2.2.1).

In water, hydrogen bonds give cohesion and adhesion, so water is a liquid at suitable temperatures and can flow and act as a transport medium.
They make water resist temperature change (high specific heat) and high heat of vaporisation, which protect living things.
In biomolecules they hold DNA together (two bonds between A and T, three between G and C) and maintain the folding of proteins.

83 marks

Why can a very large amount of heat increase the temperature of water very little?

Solution

Water has a high specific heat capacity: 1 calorie raises the temperature of 1 g of water by 1 °C (section 2.5.1, property 4).

Much of the heat absorbed by water is used to break the hydrogen bonds between its molecules, not to make the molecules move faster.
Because the energy is spent on bonds, the kinetic energy of the molecules, which is what temperature measures, rises only a little.
So a very large amount of heat produces only a small rise in temperature.

93 marks

How does water protect living things against sudden thermal change?

Solution

Water acts as a temperature stabiliser or regulator because of its high specific heat capacity, and its large volume in the body keeps the temperature steady (section 2.5.1).

A large amount of heat can be absorbed with little change in temperature, so organisms in a hot environment are protected from sudden rises.
Water releases heat slowly as it cools, which keeps the living material from sudden drops in temperature.
Ice floats on water because it is less dense, forming an insulating layer on ponds and lakes, so organisms can live under the ice.

103 marks

What is the importance of the high heat of vaporisation of water to animals?

Solution

The heat of vaporisation of water is 574 calories per gram. It is the heat needed to change a unit mass of liquid into gas (section 2.5.1, property 5).

Animals lose a large amount of heat with very little loss of water when they sweat, so they cool down efficiently in a hot environment.
When sweat evaporates, the body heat used to vaporise it cools the animal, so body temperature is kept in a safe range.
Evaporating only 2 ml of water lowers the temperature of the remaining 998 ml by 1 °C, which shows how effective this cooling is.

113 marks

Describe the classification of carbohydrates.

Solution

Carbohydrates are classified by the number of saccharide (sugar) units they yield on hydrolysis (section 2.6.1, Table 2.2).

Class	Number of saccharide units	Examples and features
Monosaccharides	One unit; cannot be hydrolysed further	Glucose, fructose; highly soluble and the sweetest
Oligosaccharides	2 to 10 units	Disaccharides such as sucrose, maltose and lactose; less soluble and less sweet
Polysaccharides	More than 10 units (at least 11)	Starch, glycogen, cellulose and chitin; insoluble and tasteless

123 marks

Describe the classification of monosaccharides.

Solution

Monosaccharides are classified in two ways: by their functional group and by the number of carbon atoms (section 2.7, Table 2.3).

By functional group: aldoses contain an aldehyde group (e.g. glucose) and ketoses contain a ketone group (e.g. fructose).
By carbon number: trioses (3 C), tetroses (4 C), pentoses (5 C), hexoses (6 C) and heptoses (7 C).
Examples: glyceraldehyde and dihydroxyacetone are trioses; ribose is an aldopentose; glucose and galactose are aldohexoses.

133 marks

Describe the conversion of open chain of ribose into ring chain.

Solution

In the open chain, ribose has an aldehyde group at C-1 and a hydroxyl group on C-4. In water these two groups react with each other, so the molecule closes into a ring (section 2.7, Fig. 2.13).

The –OH on C-4 attacks the aldehyde carbon C-1. This reaction is intramolecular, so the ring forms within a single molecule.
The C=O double bond opens. The oxygen of the C-4 hydroxyl becomes the ring oxygen linking C-1 and C-4, and the old aldehyde oxygen becomes the OH on C-1.
A five-membered ring forms: ribose becomes ribofuranose. C-1 is now the anomeric carbon, which can have the OH either below (α) or above (β) the ring.

Pentoses and ketohexoses form furanose rings. Aldohexoses such as glucose form six-membered pyranose rings instead.

143 marks

Draw and label the ring forms of alpha and beta glucose.

Solution

Glucose in water forms a six-membered pyranose ring. The ring oxygen links C-1 and C-5, and C-1 is the anomeric carbon. The two forms differ only in the position of the OH on C-1 (section 2.8).

Ring forms of glucose (Haworth projection). Left: α-glucose, OH on C1 below the ring. Right: β-glucose, OH on C1 above the ring. The accent marks the anomeric carbon.

The α and β forms are anomers. In solution they interconvert through the open chain.

153 marks

Justify that the laboratory-manufactured sweeteners are “left-handed” sugars and cannot be metabolised by the “right-handed” enzymes.

Solution

Artificial sweeteners such as tagatose and sucralose are L-sugars, the mirror images of the D-sugars that living organisms use (section 2.8, Science Titbits).

Enzymes, receptors and proteins are shaped to react only with particular enzymes or substrates. The active site of an enzyme fits a right-handed (D) sugar in the way a right hand fits a right-handed glove.
A left-handed sugar cannot fit the active site, so the enzyme cannot bind or split it, and the sugar cannot be metabolised for energy.
This is why the sweeteners can be used without providing energy to the body.

163 marks

Illustrate the formation and breakage of (a) sucrose (b) maltose (c) lactose.

Solution

Each disaccharide is formed by condensation of two monosaccharides, which releases a water molecule and makes a glycosidic bond. Breakage is hydrolysis, which adds a water molecule (sections 2.9 and 2.10).

Formation (arrow to the right, H₂O removed) and breakage (arrow to the left, H₂O added) of (a) sucrose, (b) maltose and (c) lactose.

Disaccharide	Monosaccharides	Glycosidic bond
(a) Sucrose	α-glucose + β-fructose	α-1,2
(b) Maltose	α-glucose + α-glucose	α-1,4
(c) Lactose	β-galactose + β-glucose	β-1,4

173 marks

Draw the structural formula of amino acid.

Solution

All amino acids share a common plan: a central α-carbon bonded to a hydrogen atom, an amino group (–NH₂), a carboxyl group (–COOH) and a variable R group (section 2.12.2, Fig. 2.25).

General structure of an amino acid. The α-carbon carries the amino group, the carboxyl group, H and the R group.

The R group differs in each of the 20 common amino acids and gives each its properties. Glycine has R = H and alanine has R = CH₃.

183 marks

Describe the synthesis of peptide bond.

Solution

A peptide bond is a covalent C–N bond between the carboxyl carbon of one amino acid and the amino nitrogen of the next (section 2.13, Fig. 2.26).

The –OH of the carboxyl group of one amino acid and the –H of the amino group of the other are removed.
Together they form water. The carbon and nitrogen left behind join to make the C–N peptide bond.
Repeated condensation gives a dipeptide, then a polypeptide. On the ribosome during translation, each new amino acid is added to the free carboxyl end of the chain.

Condensation of glycine and alanine. The –OH and –H leave as water and the new CO–NH link (accent) is the peptide bond.

193 marks

Describe the four types of structure of proteins.

Solution

A protein has four levels of structure, each stabilised by different bonds (section 2.13.1, Fig. 2.27).

Level	Description	Bonds holding it
Primary	Linear sequence of amino acids joined by peptide bonds	Peptide (covalent) bonds
Secondary	α-helix or β-pleated sheet formed by coiling or folding of the chain	Hydrogen bonds between peptide groups
Tertiary	A single chain folded into a globular shape	Ionic bonds, disulphide bridges, hydrogen and hydrophobic interactions
Quaternary	Two or more polypeptide chains (globules) held together	Hydrophobic interactions

203 marks

Describe (a) globular proteins (b) fibrous proteins.

Solution

Proteins are classified by their shape, which depends on their structure (section 2.15).

Feature	(a) Globular proteins	(b) Fibrous proteins
Shape	Spherical or globular	Fibre or filament like
Structure in function	Tertiary or quaternary	Secondary
Solubility	Soluble in aqueous medium	Insoluble in aqueous medium
Examples	Enzymes, hormones, antibodies, channel proteins	Collagen, fibrinogen, actin, myosin, keratin

213 marks

Describe the classification of lipids.

Solution

The book does not give a formal classification of lipids. It defines lipids as a heterogeneous group of organic compounds that are insoluble in water but soluble in organic solvents, and lists the common groups (sections 2.17 and 2.19).

Acylglycerols (fats and oils): esters of glycerol with fatty acids; mono-, di- and triacylglycerols.
Phospholipids and waxes: phospholipids are the main membrane lipids; waxes are esters of long-chain fatty acids and long-chain alcohols.
Terpenes, steroids and prostaglandins: terpenes are built from isoprene units; steroids have a four-ring nucleus; prostaglandins are derived from arachidonic acid.

Note. The book lists these groups without a single classification scheme; the grouping above follows its headings.

223 marks

What role do lipids play in living organisms?

Solution

Lipids have several roles, from energy storage to structure (section 2.17).

Energy: they are a reserved energy source. Lipids contain more carbon and hydrogen than carbohydrates, so they store about twice as much energy.
Structure: phospholipids and cholesterol form the membrane, and lipids give shape and protection to cells.
Regulation and protection: steroid hormones and vitamin D regulate body processes; lipids also insulate, waterproof and provide buoyancy.

233 marks

Why do phospholipids form a thin layer on the surface of an aqueous solution?

Solution

Phospholipids are amphipathic: one end is polar and hydrophilic (the head with phosphate and choline), and the other end is non-polar and hydrophobic (the two fatty acid tails) (section 2.18.3).

The polar head group is attracted to water and dissolves in it, so the head stays in the water.
The non-polar tails are repelled by water, so they point away from it, out of the solution and into the air.
The molecules therefore spread as a single layer at the surface, the same arrangement that gives the phospholipid bilayer of cell membranes.

243 marks

What is isoprene unit? Explain.

Solution

The isoprene unit is a five-carbon building block from which all terpenes are synthesised (section 2.19.4, Fig. 2.32).

Isoprene unit (2-methylbuta-1,3-diene) Isoprene unit: five carbon atoms with two double bonds and a methyl group on the second carbon.

Terpenes form when isoprene units condense in different ways. Two units give a monoterpene (e.g. menthol), four give a diterpene (e.g. vitamin A and the phytol tail of chlorophyll), and six give a triterpene (e.g. ambrein).
Natural rubber is a polyterpene made from many isoprene units.
Steroids are also synthesised from isoprene units.

253 marks

Describe a steroid nucleus.

Solution

A steroid nucleus is a basic skeleton of 17 carbon atoms arranged in four fused rings (section 2.20.1, Fig. 2.33).

Rings A, B and C each have six carbon atoms; ring D has five carbon atoms.
The different side chains attached to these rings distinguish one steroid from another, for example cholesterol, testosterone, progesterone, estrogen, bile acids and vitamin D.
The nucleus is synthesised from isoprene units, and steroids are lipids of high molecular weight that are often crystalline.

263 marks

How might an error in the DNA of an organism affect protein function?

Solution

An error in DNA is usually a point mutation, in which one or a few nucleotides are changed. The sequence of the gene then changes and so does the protein it codes for (section 2.14).

The sequence of amino acids (the primary structure) is determined by the sequence of nucleotides in DNA. A changed nucleotide can change one amino acid.
A changed amino acid alters how the chain folds, and so the shape of the protein and its ability to function.
Sickle cell anaemia is an example: one amino acid change (glutamic acid to valine) in the β-chain of haemoglobin makes the red cells sickle-shaped and blocks blood flow.

273 marks

Define gene is a sequence of nucleotides as part of DNA, which codes for the formation of a polypeptide.

Solution

A gene is a region of DNA, a sequence of nucleotides, that carries the information needed to build one specific polypeptide, and so one specific protein (section 2.25, Science Titbits).

Information is written as triplets of bases. Each triplet (codon) in mRNA specifies one amino acid.
In transcription the gene is copied into mRNA, which leaves the nucleus.
In translation the ribosome reads the mRNA codons, tRNA brings the matching amino acids, and peptide bonds join them into a polypeptide.

Write the differences between:

Solution

Each pair is compared on the feature that the book uses to distinguish it.

(a) major and minor bioelements

Major bioelements (O, C, H, N, Ca, P) each make up more than 1% of the protoplasm and about 99% in total. Minor bioelements (K, S, Cl, Na, Mg) are each less than 1%.

(b) dimer and polymer

A dimer is a molecule of two monomers joined together (e.g. maltose from two glucose units). A polymer is a macromolecule of many monomers (e.g. amylose, with hundreds of glucose units).

In a polar covalent bond electrons are shared unequally, so one atom is slightly negative (e.g. O–H in water). In a non-polar bond they are shared equally (e.g. C–H).

(d) polyhydroxy aldehyde and polyhydroxy ketone

A polyhydroxy aldehyde (aldose) has the carbonyl group at the end of the chain (e.g. glucose). A polyhydroxy ketone (ketose) has it inside the chain (e.g. fructose).

(e) alpha and beta glucose

They are anomers. In α-glucose the OH on C-1 is below the ring; in β-glucose it is above the ring.

(f) D-glucose and L-glucose

They are enantiomers, non-superimposable mirror images. In D-glucose the OH on the penultimate carbon (C-5) is on the right; in L-glucose it is on the left. Only D-sugars are used by living organisms.

(g) amylase and amylopectin

The question probably means amylose; amylase is an enzyme. Amylose is unbranched (α-1,4 linkages) and soluble in hot water, while amylopectin is branched (α-1,6 linkages) and insoluble in water.

(h) amylopectin and glycogen

Both are branched polymers of α-glucose with α-1,4 and α-1,6 linkages. Glycogen (animal starch, stored in liver and muscle) is much more branched than amylopectin (plant starch).

(i) primary and secondary structure of proteins

Primary structure is the linear sequence of amino acids joined by peptide bonds. Secondary structure is the α-helix or β-pleated sheet formed by hydrogen bonds between the chain's groups.

(j) tertiary and quaternary structure of proteins

Tertiary structure is one polypeptide folded into a globule by ionic bonds, disulphide bridges and hydrogen bonds. Quaternary structure is a protein made of more than one polypeptide, held together by hydrophobic interactions.

(k) purine and pyrimidine

Purines (adenine, guanine) have a double ring. Pyrimidines (cytosine, thymine, uracil) have a single six-membered ring.

(l) saturated and unsaturated fatty acids

Saturated fatty acids have no double bonds, so they hold the maximum hydrogen and are solid at room temperature (fats). Unsaturated fatty acids have one or more C=C double bonds and are liquid at room temperature (oils).

(m) DNA and RNA

DNA is double-stranded, contains deoxyribose and the bases A, T, G, C, and is found in the chromosomes. RNA is usually single-stranded, contains ribose and the bases A, U, G, C, and takes part in protein synthesis.

Extensive answer questionsp. 81–82

18 marks

Describe the chemical composition of protoplasm.

Solution

Definition

Protoplasm is the living mass of a cell, and of organisms generally. All the biochemical reactions and activities a cell needs to survive take place in it (section 2.3).

Bioelements

Of the 92 naturally occurring elements, about 25 are found in living organisms and are called bioelements. The human body is made of 16 of them. Six major bioelements make up about 99% of protoplasm, and the rest are minor bioelements that are each less than 1%.

Type	Elements	Share of protoplasm
Major bioelements	Oxygen 65%, carbon 18%, hydrogen 10%, nitrogen 3%, calcium 2%, phosphorus 1%	about 99% in total
Minor bioelements	Potassium 0.35%, sulphur 0.25%, chlorine 0.15%, sodium 0.15%, magnesium 0.05%	less than 1% each

Biomolecules

Protoplasm is mostly water. The remaining 30% is made of biomolecules, which the bioelements form by combining with each other. The proportions in a bacterial cell and a mammalian cell are compared below.

Table 2.1: proportions of biomolecules in bacterial and mammalian cells
Biomolecule	Bacterial cell	Mammalian cell
Water	70%	70%
Protein	15%	18%
Carbohydrates	3%	4%
Lipids	2%	3%
DNA	1%	0.25%
RNA	6%	1.1%
Other organic molecules	2%	2%
Inorganic ions	1%	1%

Summary

Protoplasm is about 70% water. Its dry matter is built from carbon, hydrogen, oxygen and nitrogen, with calcium and phosphorus as the main minor partners, forming the four fundamental biomolecules, together with a small share of ions and other organic molecules.

28 marks

Distinguish carbohydrates, proteins, lipids and nucleic acids as the four fundamental kinds of biological molecules.

Solution

Comparison

Feature	Carbohydrates	Proteins	Lipids	Nucleic acids
Elements	C, H, O	C, H, O, N; some contain P, S	C, H, O (less O than carbohydrates)	C, H, O, N, P
Monomers	Monosaccharides	Amino acids	Fatty acids and glycerol (not true polymers)	Nucleotides
Bond joining monomers	Glycosidic bonds	Peptide bonds	Ester bonds	Phosphodiester bonds
Solubility in water	Monosaccharides soluble; polysaccharides insoluble	Varies by type	Insoluble	Soluble (phosphate groups)
Examples	Glucose, starch, cellulose	Enzymes, haemoglobin, collagen	Fats, oils, phospholipids, steroids	DNA, RNA, ATP, NAD
Main roles	Fuel and structure	Structure, enzymes, transport, defence	Energy store, membranes, hormones	Store and transfer genetic information
Location	Cytoplasm; cell walls	Membranes, ribosomes, enzymes	Membranes, cytoplasm	DNA in chromosomes; RNA in nucleoplasm and cytoplasm

Key differences

Carbohydrates are the cell's fuel and are the only group that is mainly built from sugar units with the formula (CH₂O)n.
Proteins are the most varied: their function depends on the sequence and folding of 20 amino acids.
Lipids are the only group that is largely insoluble in water, and they store the most energy per gram.
Nucleic acids carry the genetic code; their sequence of bases, not their shape, is what matters.

Summary

All four are polymers or contain monomers, but they differ in the monomer, the bond between monomers and the function they serve in the cell.

38 marks

Describe and draw sketches of dehydration synthesis and hydrolysis reactions for making and breaking of macromolecule polymers.

Solution

Definitions

A macromolecule (polymer) is a large molecule made of many repeating units, the monomers. Condensation (dehydration synthesis) joins monomers and releases water. Hydrolysis breaks the bond between monomers by adding water.

Dehydration synthesis (condensation)

Steps

A –OH group is removed from one monomer and an –H from the other.
The two groups combine to give water (H₂O).
The monomers are joined by a new covalent bond, here an α-1,4 glycosidic bond that makes maltose from two α-glucose units. Each further monomer adds another bond and releases another water molecule.

Hydrolysis

Steps

Water is added across the bond between the two monomers.
The –OH from the water attaches to one monomer and the –H to the other, which splits the bond.
In the body, hydrolysis is carried out by enzymes such as amylase, which breaks amylose into maltose and glucose.

Dehydration synthesis (top) joins two α-glucose units to form maltose, releasing water; hydrolysis (bottom) reverses it by adding water.

Summary

Condensation makes polymers and releases water; hydrolysis breaks polymers into monomers by adding water. Digestion is an example of hydrolysis, so the two processes are the reverse of each other.

48 marks

How do the properties of water make it the cradle of life?

Solution

Introduction

About 70% of an organism is water. Water is the medium in which all the chemical reactions of life take place, and its properties come mainly from its polar covalent bonds and hydrogen bonding (section 2.5).

Polarity of water. Oxygen is slightly negative and the hydrogens slightly positive, so neighbouring molecules attract each other by hydrogen bonds.

Properties and their role in life

Property	Why it matters for life
High polarity	Dissolves ionic and polar substances, so chemical reactions occur in aqueous solution
Hydrogen bonding	Keeps water liquid at temperatures suitable for life and gives it cohesion and adhesion
Cohesion and adhesion	Water flows freely and moves through living bodies as a transport medium
High specific heat	Acts as a temperature stabiliser, protecting organisms from sudden temperature change
High heat of vaporisation (574 cal per g)	Sweating cools animals efficiently with little water loss
Hydrophobic exclusion	Keeps lipid bilayer membranes intact and proteins folded
Ionisation (H⁺ and OH⁻)	These ions help maintain or change the pH of the medium
Lower density of ice	Ice floats, forming an insulating layer under which organisms can live in winter

Summary

Water's polarity makes it a universal solvent, its hydrogen bonds hold it liquid and give it high heat capacity and cohesion, and its ionisation and density provide stable conditions. Together these properties make it the medium in which life can begin and continue.

58 marks

Distinguish the properties and role of monosaccharides.

Solution

Definition and formula

Monosaccharides are the simplest carbohydrates. They are polyhydroxy aldehydes or ketones with the general formula (CH₂O)n, where n is 3 to 7 (section 2.7).

Properties

They cannot be hydrolysed into simpler compounds.
They are highly soluble in water, because their many hydroxyl groups form hydrogen bonds with water.
They taste sweet, which is why they are called sugars.
They are reducing sugars: the aldehyde or ketone group can reduce other compounds and is oxidised itself.
They show stereoisomerism because of their asymmetric carbons.
Pentoses and hexoses with five or more carbons exist in an open-chain form and a ring form.

Examples and roles

Class	Formula	Examples	Role
Trioses (3 C)	C₃H₆O₃	Glyceraldehyde, dihydroxyacetone	Intermediates in photosynthesis and respiration
Tetroses (4 C)	C₄H₈O₄	Erythrose, erythrulose	Intermediates in bacterial photosynthesis
Pentoses (5 C)	C₅H₁₀O₅	Ribose, deoxyribose, ribulose	Ribose and deoxyribose form RNA and DNA; ribulose is an intermediate in photosynthesis
Hexoses (6 C)	C₆H₁₂O₆	Glucose, galactose, fructose	Glucose is the respiratory fuel; fructose is a respiration intermediate; galactose is part of milk sugar
Heptoses (7 C)	C₇H₁₄O₇	Glucoheptose, sedoheptulose	Intermediates in photosynthesis

Summary

Monosaccharides are small, soluble, sweet and reducing sugars. Their role is mainly as an immediate fuel (glucose) and as intermediates in photosynthesis and respiration, and as the building blocks of nucleic acids and of polysaccharides.

68 marks

Write the empirical formula of monosaccharides and classify them.

Solution

Empirical formula

The general formula of monosaccharides is (\text{CH}_2\text{O})_n, where n is the number of carbon atoms. For example, glucose is \text{C}_6\text{H}_{12}\text{O}_6, which is (\text{CH}_2\text{O})_6.

Each carbon except one carries a hydroxyl group, and the remaining carbon is part of an aldehyde or a ketone group.

Classification

Monosaccharides are classified in two ways (section 2.7, Table 2.3).

By functional group: aldoses (contain an aldehyde group, e.g. glucose) and ketoses (contain a ketone group, e.g. fructose).
By number of carbon atoms: trioses (3 C), tetroses (4 C), pentoses (5 C), hexoses (6 C) and heptoses (7 C).

Class	Empirical formula	Aldose	Ketose
Triose (3 C)	C₃H₆O₃	Glyceraldehyde	Dihydroxyacetone
Tetrose (4 C)	C₄H₈O₄	Erythrose	Erythrulose
Pentose (5 C)	C₅H₁₀O₅	Ribose, deoxyribose	Ribulose
Hexose (6 C)	C₆H₁₂O₆	Glucose, galactose	Fructose
Heptose (7 C)	C₇H₁₄O₇	Glucoheptose	Sedoheptulose

Summary

The empirical formula is (\text{CH}_2\text{O})_n with n = 3 to 7. Monosaccharides are named by their carbon number and whether they are aldoses or ketoses.

78 marks

Compare the stereoisomers of glucose.

Solution

Stereoisomers of glucose

Stereoisomers have the same molecular formula but differ only in the arrangement of –H and –OH groups in three-dimensional space. Glucose has four asymmetric carbons (C-2 to C-5), so it has 2^4 = 16 stereoisomers: 8 belong to the D series and 8 to the L series (section 2.8).

α-glucose and β-glucose are anomers. They differ only in the OH on C1 (the anomeric carbon).

Types of stereoisomers

Type	Definition	Example
Anomers	Differ in the –OH on C-1 of the ring (anomeric carbon)	α-glucose and β-glucose
Enantiomers	Non-superimposable mirror images; differ at every asymmetric carbon	D-glucose and L-glucose
Diastereomers	Not mirror images; differ at more than one asymmetric carbon	D-glucose and D-allose
Epimers	Differ in the arrangement at only one asymmetric carbon	D-glucose and D-mannose

Summary

Of the 16 stereoisomers of glucose, the 8 D-sugars are the ones used by living organisms. α and β forms interconvert in solution, while enantiomers are fixed mirror images.

88 marks

Distinguish the properties and role of disaccharides.

Solution

General properties

A disaccharide is two monosaccharides combined by a condensation reaction. The general formula is \text{C}_{12}\text{H}_{22}\text{O}_{11}.
They are less sweet and less soluble in water than monosaccharides.
They can be hydrolysed back to monosaccharides and are a familiar energy source for living organisms.

Comparison of the common disaccharides

Feature	Sucrose	Maltose	Lactose
Common name	Cane sugar	Malt sugar	Milk sugar
Monosaccharides	α-glucose + β-fructose	α-glucose + α-glucose	β-galactose + β-glucose
Glycosidic bond	α-1,2	α-1,4	β-1,4
Reducing sugar?	No (non-reducing)	Yes	Yes
Source and role	Transport sugar of plants; sweetener	Produced from starch and glycogen; energy in germinating seeds and sprouting potatoes; brewing	Milk of mammals; direct energy source for infants; substrate for other biomolecules

Summary

Sucrose is the form in which plants transport sugar, maltose is an intermediate of starch digestion, and lactose supplies energy to infants. Sucrose is non-reducing because both of its anomeric carbons are tied up in the glycosidic bond.

98 marks

Describe glycoside bond in the transport of disaccharides.

Solution

What a glycosidic bond is

A glycoside is a ring sugar that is attached to another sugar, and the covalent bond that holds two sugars together is a glycosidic bond. It forms by condensation: the –H of one sugar combines with the –OH of the other, and one water molecule is released (section 2.10).

Naming

The bond is named by the carbon numbers of the two sugars it joins and by the α or β form of the first sugar. In sucrose, the α-glucose C-1 is joined to the β-fructose C-2, which is an α-1,2 glycosidic bond.

Formation of sucrose (top) and its hydrolysis (bottom). The glycosidic bond joins C1 of α-glucose to C2 of β-fructose.

Role in transport

Sucrose is the transport disaccharide of plants: glucose made in photosynthesis is converted into sucrose and moved through the plant.
Because the anomeric carbons of both sugars are used in the α-1,2 bond, sucrose is non-reducing and chemically unreactive, so it can be carried in high concentration without reacting.
It is very soluble, so it can be moved efficiently through the plant in solution.
When the plant needs energy, the glycosidic bond is hydrolysed and the glucose and fructose are used.

Summary

The glycosidic bond is a stable condensation link. Its stability makes sucrose a safe, soluble form for transporting energy through a plant.

108 marks

Distinguish the properties and roles of polysaccharides.

Solution

Definition and types

Polysaccharides yield more than ten monosaccharide units on hydrolysis. Homopolysaccharides are made from one kind of monosaccharide (starch, glycogen, cellulose, chitin); heteropolysaccharides are made from different monosaccharides (agar, pectin, peptidoglycan) (section 2.11).

Properties

They are large, so they are more or less insoluble in water and exert no osmotic effect in the cell.
They fold into compact shapes, which makes them convenient storage molecules.
They are easily converted into sugars by hydrolysis when the cell needs energy.
They are tasteless, unlike monosaccharides and disaccharides.

Roles

Role	Polysaccharides	What they do
Food and energy store	Starch (plants), glycogen (animals)	Stored as compact granules and broken down to glucose when energy is needed
Structural material	Cellulose (plant cell wall), chitin (fungal cell wall and arthropod exoskeleton)	Give rigidity and protection to cells and bodies

Summary

Polysaccharides are large, insoluble and compact. Their role is storage (starch and glycogen) or structure (cellulose and chitin), and they are the largest group of carbohydrates.

118 marks

Describe the properties and roles of starch, glycogen, cellulose and chitin.

Solution

Comparison

Feature	Starch	Glycogen	Cellulose	Chitin
Monomer	α-glucose	α-glucose	β-glucose	N-acetylglucosamine (modified glucose)
Linkage	α-1,4 and α-1,6 (branches)	α-1,4 and α-1,6, with much more branching	β-1,4 (unbranched)	β-1,4 (unbranched)
Iodine test	Blue-black	Red	No colour	Not tested in the book
Solubility	Amylose soluble in hot water; amylopectin insoluble	Insoluble	Insoluble	Insoluble
Where found	Roots, stems and seeds; cereal grains and potato tubers	Liver and muscle of animals	Plant cell walls; cotton and paper	Fungal cell walls; arthropod exoskeleton
Role	Plant energy store	Animal energy store	Structural carbohydrate of plants; indigestible fibre	Structural carbohydrate of fungi and arthropods

Structure in brief

Starch is a mixture of amylose (linear) and amylopectin (branched), both of α-glucose.
Glycogen resembles amylopectin but has even more branches.
Cellulose is formed by hundreds of β-glucose units, and its β-linkages make straight chains that hydrogen-bond side by side.
Chitin is made of N-acetylglucosamine units with β-1,4 linkages, and is also called fungal cellulose.

Summary

Starch and glycogen store energy because they can be broken down to glucose; cellulose and chitin provide structure because their β-linkages make strong, unbranched fibres.

122026 Annual I, Set B, Q.2 (iv)2026 Annual I, Set D, Q.2 (iv)8 marks

Justify the significance of the sequence of amino acids through the example of sickle cell haemoglobin.

Solution

The primary structure decides function

The sequence of amino acids in a polypeptide is its primary structure. It is set by the sequence of nucleotides in DNA, and it determines how the protein folds and works (section 2.14).

Normal and sickle haemoglobin

Normal haemoglobin (HbA) contains four polypeptides: two α-chains of 141 amino acids and two β-chains of 146 amino acids. In sickle cell anaemia the gene for the β-chain has a point mutation, in which one nucleotide is replaced by another.

Feature	Normal haemoglobin (HbA)	Sickle haemoglobin (HbS)
β-chain, position 6	Glutamic acid	Valine
Amino acid sequence	…Thr–Pro–Glu–Glu…	…Thr–Pro–Val–Glu…
Shape of red blood cell	Disc-shaped, like a doughnut without a hole	Sickle or crescent shaped
Properties of cell	Flexible, moves easily through vessels	Stiff and sticky; blocks blood flow in limbs and organs

Why one change matters

A single nucleotide change alters the codon, so a different amino acid is placed at position 6 of the β-chain.
Glutamic acid (acidic, hydrophilic) is replaced by valine (non-polar, hydrophobic). The new hydrophobic patch changes how the haemoglobin molecules fit together.
The abnormal haemoglobin molecules stick together, distorting the red cells into sickle shapes.
Blocked blood flow causes pain and damage to organs.

Summary

A change of one amino acid among hundreds alters the protein's shape and function, so the sequence of amino acids is significant for the health of the whole organism.

138 marks

List examples and the roles of structural and functional proteins.

Solution

Structural proteins

Structural proteins give support and shape to the body. They are mostly fibrous (section 2.16, Table 2.4).

Protein	Role
Collagen	Establishes the matrix of bone and cartilage; forms ligaments and tendons
Elastin	Gives support to connective tissues such as tendons and ligaments
Keratin	Strengthens protective coverings: hair, nails, quills, feathers, horns and beaks
Histone	Arranges DNA into chromosomes

Functional proteins

Functional proteins take part in physiological activity, such as movement, defence and metabolism (Table 2.5).

Protein	Role
Enzymes	Control metabolism by speeding up biochemical reactions
Hormones (protein)	Regulate glucose level, calcium level, digestion and blood pressure
Antibodies	Produced by white blood cells in response to antigens; provide immunity
Haemoglobin	Found in red blood cells; transports oxygen and some carbon dioxide
Fibrinogen	Found in plasma; takes part in blood clotting
Ovalbumin and casein	Storage proteins in egg white and milk that store amino acids

Summary

Structural proteins build and support the body; functional proteins carry out its activities. Each protein has a specific function that depends on its structure.

148 marks

Describe the properties and roles of: (a) acylglycerol (b) phospholipids (c) terpenes (d) waxes.

Solution

(a) Acylglycerol

Properties:

Low density (lighter than water), insoluble in water but soluble in organic solvents such as chloroform, ether and hot alcohol.
Saturated acylglycerols with long chains melt at higher temperatures than unsaturated ones with short chains (tristearin melts at 71 °C, triolein at −17 °C).
They have great emulsifying power and are hydrolysed into fatty acids and glycerol when heated in an acidic aqueous medium.
They can be oxidised to odorous, flavoured compounds when they decay.

Roles:

They make up most of the body's lipids and are the main lipids in fat deposits and food.
High levels can cause obesity, diabetes and hyperlipoproteinaemia.
Hydrogenation of acylglycerols produces margarine and butter-like fats.

(b) Phospholipids

Properties: phospholipids are amphipathic. The polar head (phosphate with choline, ethanolamine, inositol or serine) is hydrophilic, and the two fatty acid tails are non-polar and hydrophobic.

Phosphatidylcholine (lecithin): a polar head of choline, phosphate and glycerol, and two non-polar fatty acid tails. The second tail has a cis double bond, which gives the kink.

Roles: they regulate membrane permeability and form the lipid bilayer of the cell membrane; they help absorb fat from the intestine, take part in the electron transport chain in mitochondria, transport cholesterol, form part of bile and lipoproteins, help blood coagulation, and act as surfactants in the respiratory system.

(c) Terpenes

Properties: all terpenes are synthesised from the five-carbon isoprene unit. Two units make a monoterpene (e.g. menthol), four a diterpene (e.g. vitamin A, phytol), six a triterpene (e.g. ambrein), and natural rubber is a polyterpene.

Roles: terpenes are anti-inflammatory, analgesic, antioxidant, anticonvulsive and antidepressive. They are also anticancer, antitumour, anti-mutagenic, neuroprotective, anti-allergic, antibiotic and anti-diabetic.

(d) Waxes

Properties: waxes are highly hydrophobic and chemically inert, and resist atmospheric oxidation. Natural waxes (e.g. bee's wax, cutin on leaves) are esters of long-chain fatty acids and long-chain alcohols; synthetic waxes such as paraffin come from petroleum or polyethylene.

Roles: they protect plants and animals, make water-proof and anti-fouling coatings, and are used in polishing, cosmetics, crayons, candles, paints, inks, adhesives and tyres.

Summary

Acylglycerols store energy, phospholipids build membranes, terpenes provide a range of medicinal and protective compounds, and waxes waterproof and protect surfaces.

Note. The book lists the terpene role as “depressive”; this is probably a misprint for antidepressive.

152025 Annual I, Set B, Q.5 OR8 marks

Evaluate the role of the following as important groups of lipids and describe their roles in living organisms: (a) steroid (b) prostaglandins.

Solution

(a) Steroids

Steroids are lipids of high molecular weight with a four-ring nucleus of 17 carbon atoms: three six-membered rings (A, B, C) and one five-membered ring (D). Different side chains give different steroids (section 2.20.1).

Steroid nucleus: rings A, B and C are six-membered and ring D is five-membered. Side chains attached to these rings distinguish one steroid from another.

Roles:

Cholesterol is a structural component of the cell membrane and the precursor of the other steroids.
Bile acids and bile salts emulsify fats so that they can be digested.
Sex hormones: testosterone (male), progesterone and estrogen (female).
Vitamin D helps to regulate calcium metabolism.

Evaluation: steroids are essential. Cholesterol keeps membranes stable, and the hormones and vitamin D control reproduction and calcium levels, so the body cannot survive without them.

(b) Prostaglandins

Prostaglandins exist in virtually every mammalian tissue, where they act as local hormones. They are derived from arachidonic acid, and their effect depends on the tissue (section 2.20.2).

Some reduce blood pressure and others raise it.
In the immune system they help induce fever and inflammation, and they intensify the sensation of pain.
They help regulate the aggregation of platelets, an early step in forming blood clots.
Aspirin reduces fever and pain because it inhibits prostaglandin synthesis.

Evaluation: prostaglandins are important local regulators. Their effects are useful when they protect and heal, but they also cause fever, pain and inflammation, which is why drugs that block them are used.

Summary

Steroids are structural molecules and carriers of control signals (hormones, vitamin D, bile). Prostaglandins are short-acting local regulators of blood flow, pain and clotting.

168 marks

Describe the molecular level structure of nucleotides.

Solution

Components

A nucleotide is the monomer of nucleic acids. It is made of three parts: a pentose sugar, a phosphate group and a nitrogenous base (section 2.21.1, Fig. 2.34).

Structure of a nucleotide: pentose sugar with the base on C1′ (N-glycosidic bond), phosphate on C5′ and a free 3′ OH.

Steps in forming a nucleotide

The nitrogenous base is linked to C-1′ of the pentose sugar. The product is a nucleoside (for example adenosine).
A phosphate group is linked to C-5′ of the nucleoside, which gives a nucleotide: nucleoside monophosphate.
Adding a second or third phosphate gives a nucleoside diphosphate or triphosphate (for example ATP).

Sugar and base

Feature	DNA	RNA
Sugar	Deoxyribose (H at C-2′)	Ribose (OH at C-2′)
Bases	A, G, C, T	A, G, C, U
Nucleotides (monophosphates)	dAMP, dGMP, dCMP, dTMP	AMP, GMP, CMP, UMP

Summary

Each nucleotide is a pentose sugar with a base on C-1′, a phosphate on C-5′ and a free 3′ OH. These parts are joined into polynucleotides by phosphodiester bonds.

178 marks

Distinguish among the nitrogenous bases found in the nucleotides of nucleic acids.

Solution

Two classes of bases

Nitrogenous bases are of two classes. Pyrimidines have a single six-membered ring and include cytosine (C), thymine (T) and uracil (U). Purines have a double ring and include adenine (A) and guanine (G) (section 2.22).

Base	Class	Rings	Found in	Pairs with	Hydrogen bonds
Adenine (A)	Purine	Double	DNA and RNA	Thymine (DNA) or uracil (RNA)	2
Guanine (G)	Purine	Double	DNA and RNA	Cytosine	3
Cytosine (C)	Pyrimidine	Single	DNA and RNA	Guanine	3
Thymine (T)	Pyrimidine	Single	DNA only	Adenine	2
Uracil (U)	Pyrimidine	Single	RNA only	Adenine	2

Chargaff's rule

In DNA the amount of A equals T and the amount of G equals C, so the total purines equal the total pyrimidines (1:1).

Summary

Purines are larger and pyrimidines smaller. Thymine is the DNA base and uracil the RNA base, both pairing with adenine; guanine and cytosine always pair with each other through three hydrogen bonds.

Note. Section 2.2.2.1 of the book says “thiamine” where thymine is meant; the answer uses thymine.

188 marks

Describe the structure of a mononucleotide (ATP) and a dinucleotide (NAD).

Solution

ATP: a mononucleotide

Adenosine triphosphate (ATP) has three parts: adenine (a nitrogenous base), ribose (a five-carbon sugar) and three phosphate groups. The two covalent bonds between the three phosphates are high-energy bonds. ATP is converted to ADP and inorganic phosphate (Pi) by hydrolysis, and it is called the energy currency of the cell (section 2.24).

NAD: a dinucleotide

Nicotinamide adenine dinucleotide (NAD) consists of two nucleotides joined through their phosphate groups. One nucleotide has nicotinamide, ribose and phosphate; the other has adenine, ribose and phosphate. NAD is a coenzyme.

Structure of ATP (top, a mononucleotide with three phosphates) and NAD (bottom, a dinucleotide joined through its phosphates).

Comparison

Feature	ATP	NAD
Type	Mononucleotide	Dinucleotide
Bases	Adenine	Nicotinamide and adenine
Phosphates	Three, in a chain	Two, linking the two nucleotides
Role	Energy currency of the cell	Coenzyme in biochemical reactions

Summary

ATP is a single nucleotide with three phosphates, whose high-energy bonds make it the energy currency of the cell. NAD is two nucleotides joined by their phosphate groups and acts as a coenzyme.

198 marks

Explain the formation of phosphodiester bond.

Solution

What is formed

Nucleotides are joined by condensation, but the molecule released is not water. It is pyrophosphate, two phosphate groups joined together. The new link between the 3′ carbon of one sugar and the 5′ phosphate of the next is a phosphodiester bond (section 2.23).

Steps

The phosphate group at 5′ of the incoming nucleotide triphosphate (NTP) is held close to the free 3′ OH of the last nucleotide of the growing chain.
The 3′ OH attacks the phosphate and a phosphodiester bond forms between the two nucleotides. Meanwhile the bond between the alpha and beta phosphates of the NTP breaks, and a pyrophosphate group is released.
The pyrophosphate is hydrolysed (split by water) into two phosphates. The energy released drives the polymerisation to completion.

Formation of a phosphodiester bond. The 3′ OH of the growing chain attacks the 5′ phosphate of the incoming nucleotide, and pyrophosphate is released.

Result

Polynucleotides have a free 5′ phosphate at one end and a free 3′ OH at the other. By convention the sequence is read from 5′ to 3′.

Summary

The phosphodiester bond joins the 3′ OH of one nucleotide to the 5′ phosphate of the next. Hydrolysis of pyrophosphate supplies the energy for the reaction.

208 marks

Explain the double helical structure of DNA as proposed by Watson and Crick.

Solution

Background

Chargaff found in 1951 that A equals T and G equals C. X-ray diffraction work by Wilkins and Franklin showed that DNA is a duplex with a width of 2 nm and a turn of 3.4 nm. Watson and Crick combined these data in 1953 to propose their double helix model (section 2.25).

Features of the model

DNA is made of two polynucleotide chains twisted around a common axis.
The chains are antiparallel: one runs 5′ to 3′ and the other runs 3′ to 5′.
The sugar and phosphate groups form the outside of the helix, and the bases point inwards.
The width of the helix is 2 nm; each turn is 3.4 nm long and contains 10 base pairs, so each base pair rises 0.34 nm.
The bases pair specifically: adenine with thymine (two hydrogen bonds) and guanine with cytosine (three hydrogen bonds).

The double helix unwound into a flat ladder. The two chains run in opposite directions. Base pairs are held by hydrogen bonds (dotted lines).

Base pairing

Pair	Hydrogen bonds	Why
A–T	2	A purine always pairs with a pyrimidine, so each pair spans the same 2 nm width
G–C	3	The same width, with an extra hydrogen bond

Summary

The Watson and Crick model shows that DNA is a stable, antiparallel double helix in which specific base pairing allows the sequence to be copied exactly.

218 marks

What is a gene? How does a gene code for the formation of a polypeptide?

Solution

Gene

A gene is a region of DNA made of a sequence of nucleotides. It is the physical and functional unit of heredity, and it carries the information needed to build one specific polypeptide (section 2.25 and 2.28).

Central dogma

The flow of information goes from DNA to RNA to protein. This is the central dogma of life. Transcription and translation together are called gene expression (section 2.28).

The central dogma: the gene is transcribed into mRNA in the nucleus, and the mRNA is translated on the ribosome in the cytoplasm into a polypeptide.

Transcription

The two DNA strands of the gene unwind.
Free RNA nucleotides pair with the exposed DNA strand: A pairs with U, G with C, C with G, and T with A (uracil replaces thymine).
The mRNA is made and the DNA zips back together.

Translation

The mRNA leaves the nucleus and binds to a ribosome in the cytoplasm.
Each triplet of mRNA bases is a codon that specifies one amino acid. tRNA molecules, each carrying a specific amino acid and bearing an anticodon, pair with the codons in turn.
Peptide bonds join the amino acids into a polypeptide chain, which then folds into the protein.

Summary

A gene is a DNA sequence that is transcribed into mRNA and translated, codon by codon, into a polypeptide.

228 marks

Explain general structure of RNA.

Solution

Chemical nature

RNA is a polymer of ribonucleotides. Each one has ribose, a phosphate and one of the bases A, G, C or U. The chain is linked by phosphodiester bonds and has a free 5′ phosphate and a free 3′ OH (section 2.26).

Shape

Unlike DNA, RNA is generally single-stranded and does not form a double helix. Some regions, however, have complementary bases that pair with each other and form short double-stranded stems, so the single strand folds back on itself.

General structure of RNA: a single strand whose complementary regions pair (A with U, G with C) to form a stem, with an unpaired loop.

Comparison with DNA

Feature	DNA	RNA
Strands	Double helix	Usually single strand
Sugar	Deoxyribose	Ribose
Bases	A, G, C, T	A, G, C, U
Location	Nucleus (chromosomes)	Nucleoplasm and cytoplasm
Function	Stores genetic information	Carries and uses the information in protein synthesis

Summary

RNA is a single-stranded polynucleotide of ribonucleotides that folds back on itself where its bases are complementary.

238 marks

Explain the structure and role of three types of RNA.

Solution

mRNA: messenger RNA

mRNA is a single strand of variable length, depending on the gene and the protein it codes for. Each three nucleotides form a codon. It carries the message from the nucleus to the ribosome, and it is about 3 to 4% of the total RNA.

rRNA: ribosomal RNA

rRNA is the largest type of RNA and makes up about 80% of the RNA of a cell. It is transcribed from genes on the chromosomes and packaged with proteins into the two ribosomal subunits, where it provides the platform for protein synthesis.

tRNA: transfer RNA

tRNA is the smallest RNA, with 75 to 90 nucleotides. It is a single strand that folds into a flat clover-leaf shape, with a 3′ CCA end to which the amino acid is attached. Its anticodon pairs with the mRNA codon.

Clover-leaf model of tRNA: the amino acid attaches at the 3′ CCA end, the anticodon loop pairs with an mRNA codon, and the D loop and θ loop bind the enzyme and ribosome.

Summary

Type	Structure	Role	Share of RNA
mRNA	Single strand of variable length; codons	Carries the message from the nucleus to the ribosome	3 to 4%
rRNA	Largest RNA; combined with proteins in ribosomal subunits	Platform for protein synthesis	about 80%
tRNA	Clover-leaf, 75 to 90 nucleotides; anticodon; amino acid at 3′ end	Carries the specific amino acid to the ribosome	smallest molecules

248 marks

Describe the roles of the following conjugated molecules: (a) glycolipids (b) glycoproteins (c) lipoproteins (d) nucleoproteins.

Solution

Conjugated molecules are formed when proteins, lipids or nucleic acids are joined to other kinds of molecules (section 2.29).

Molecule	Made of	Where found	Roles
(a) Glycolipids	Lipid joined to one or more simple sugars	White matter of brain, myelin sheath of nerve fibres, chloroplast membrane	Components of nerve insulation and membranes
(b) Glycoproteins	Protein covalently attached to carbohydrate	Widely distributed in cells	Hormones, transport proteins, structural proteins and receptors; blood group antigens
(c) Lipoproteins	Protein combined with phospholipid	Milk, blood, cell nucleus, egg yolk membrane, chloroplasts	Lipid–protein complexes that form part of membranes and carry lipids in blood and milk
(d) Nucleoproteins	Simple basic protein with nucleic acid	Chromosomes and ribosomes	Package DNA into chromosomes; ribosomes carry out protein synthesis

Summary

Conjugated molecules combine the functions of two groups: glycolipids and glycoproteins take part in cell recognition and membranes, lipoproteins carry lipids, and nucleoproteins store genetic information and build ribosomes.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 71 mark

The protein present in tendons and ligaments is:

ACasein
BKeratin
CElastinAnswer
DFibrinogen

Why

Elastin provides support for connective tissues such as tendons and ligaments (collagen is also a component of them). Keratin (B), the tempting option because it is also a fibrous structural protein, strengthens hair, nails, feathers, horns and beaks. Casein (A) is the milk protein and fibrinogen (D) is a blood plasma protein used in clotting.

2026 Annual I, Set B, MCQ 81 mark

A triglyceride molecule is formed by:

AOne fatty acid and one glycerol
BTwo fatty acids and one phosphate
CThree fatty acids and one glycerolAnswer
DAmino acids and glycerol

Why

A triacylglycerol (triglyceride) is the ester of one glycerol and three fatty acids: each of the three –OH groups of glycerol joins a fatty acid and three water molecules are released. Option A describes a monoacylglycerol (only one fatty acid). Option B is tempting but a phosphate group belongs to a phospholipid, which has glycerol, two fatty acids and a phosphate. Amino acids (D) are the units of proteins, not of lipids.

2025 Annual I, Set B, Q.2 (iii)3 marks

Justify the statement "Water is a universal solvent".

Solution

Polarity of water (1 mark)

In a water molecule the shared electrons are pulled towards oxygen, so oxygen is slightly negative (\delta-) and each hydrogen slightly positive (\delta+). Water is a polar molecule.

Dissolving ionic and polar substances (1 mark)

The charged ends of water molecules attract the ions of ionic compounds (electrolytes) and the polar groups of other molecules, surround them and separate them. Non-polar molecules that carry charged groups also dissolve because their charged regions attract water.

Importance in living things (1 mark)

On dissolving, substances dissociate into positive and negative ions, which are in a more favourable state to react. This is why all the chemical reactions of living organisms occur in an aqueous medium. Water is therefore called the universal solvent (for polar substances).

2026 Annual I, Set B, Q.2 (iv)3 marks

Describe three effects of amino acid sequence change in haemoglobin.

Solution

The sequence of amino acids is the primary structure of a protein and it decides how the protein works. In haemoglobin the change of one amino acid out of 146 in the β-chain (HbA to HbS) has three effects.

Effect 1: the haemoglobin molecule itself changes (1 mark)

Glutamic acid at position 6 of the β-chain is replaced by valine, so normal haemoglobin (HbA) becomes abnormal sickle haemoglobin (HbS).

Effect 2: the red blood cells change shape (1 mark)

Normal red blood cells are discs, like doughnuts without holes, and move easily through vessels. Cells with HbS become sickle or crescent shaped, stiff and sticky.

Effect 3: blood flow is blocked and the disease sickle cell anaemia results (1 mark)

Sickle cells block blood flow in the vessels of the limbs and organs, which causes pain and organ damage.

How one nucleotide leads to the disease

A point mutation in the β-globin gene replaces one nucleotide by another, so the codon changes.
One amino acid of the β-chain changes: glutamic acid is replaced by valine at position 6 (Thr-Pro-Glu-Glu becomes Thr-Pro-Val-Glu).
The abnormal sickle haemoglobin (HbS) makes the red blood cells sickle or crescent shaped, stiff and sticky.
The sickle cells block blood flow in the vessels of the limbs and organs, causing pain and organ damage.

2025 Annual I, Set B, Q.2 (xiii) OR1x3 = 3 marks

How is maltose different from sucrose in glycosidic linkage, monosaccharide units and specific role?

Solution

Both are disaccharides (\ce{C12H22O11}) formed by a condensation reaction that releases one water molecule, but they differ as follows.

Point	Maltose	Sucrose
Glycosidic linkage	α-1,4: the –OH on C1 of the first glucose joins the H of the –OH on C4 of the second glucose	α-1,2 (α-1,β-2): the –OH on C1 of α-glucose joins the H of the –OH on C2 of β-fructose
Monosaccharide units	Two α-glucose molecules	One α-glucose and one β-fructose
Specific role	Malt sugar: an intermediate in the breakdown of starch and glycogen and an energy source in germinating seeds and sprouting potatoes	Cane sugar: the transport disaccharide of plants (very soluble and relatively unreactive), and the common household sweetener

2025 Annual I, Set B, Q.5 OR3+3 = 6 marks

Elaborate the structures and roles of steroids and prostaglandins as important group of lipids.

Solution

Steroids (3 marks)

Structure. Steroids are lipids of high molecular weight which can be crystalline. The steroid nucleus has 17 carbon atoms arranged in four attached rings: three rings (A, B and C) of six carbons and a fourth ring (D) of five carbons. The side chains that extend from these rings distinguish one steroid from another. Steroids are synthesised from isoprene units.

Steroid nucleus: rings A, B and C are six-membered and ring D is five-membered. Side chains attached to these rings distinguish one steroid from another.

Roles. Cholesterol is a structural component of the cell membrane and the precursor of other steroids: bile acids and bile salts (they emulsify fats), the male sex hormone testosterone and the female sex hormones progesterone and oestrogen. Vitamin D, which helps to regulate calcium metabolism, is also a steroid.

Prostaglandins (3 marks)

Structure. Prostaglandins are lipids derived from arachidonic acid, a fatty acid. They occur in virtually every mammalian tissue and act as local hormones.

Prostaglandins are made from arachidonic acid. Aspirin reduces fever and pain by inhibiting their synthesis.

Roles. Their functions depend on the tissue: some lower blood pressure while others raise it; they help to induce fever and inflammation and intensify pain; and they regulate the aggregation of platelets, an early step in blood clotting. Aspirin reduces fever and pain because it inhibits prostaglandin synthesis.

Group	Structure	Roles
Steroids	Nucleus of 17 carbons in four rings (three of six carbons, one of five)	Cholesterol in membranes and as precursor; bile salts; testosterone, progesterone, oestrogen; vitamin D
Prostaglandins	Derived from arachidonic acid	Local hormones: blood pressure, fever, inflammation, pain, platelet aggregation

2026 Annual I, Set D, MCQ 21 mark

Which polysaccharide is storage carbohydrate in animals?

AStarch
BGlycogenAnswer
CCellulose
DChitin

Why

Glycogen is the storage carbohydrate of animals, kept mainly in the liver and muscles, so it is called animal's starch. Starch (A) is the tempting wrong answer: it is also a storage polysaccharide of α-glucose, but it is stored in plants. Cellulose (C) and chitin (D) are structural, not storage, polysaccharides.

2025 Annual I, Set D, MCQ 21 mark

Prostaglandins are derivatives of:

AAmino acids
BSteroids
CArchidonic acidAnswer
DNucleic acids

Why

Prostaglandins are derived from arachidonic acid, a fatty acid (spelled "Archidonic" in the paper). Steroids (B) are another group of lipids, but they are synthesised from isoprene units, not the source of prostaglandins. Amino acids and nucleic acids are the units of proteins and the nucleic acids, not lipid precursors.

2026 Annual I, Set D, MCQ 81 mark

A phospholipid molecule contains:

AThree fatty acids only
BGlycerol, two fatty acids a phosphate groupAnswer
COnly glucose units
DOnly amino acids

Why

A phospholipid is built from glycerol, two fatty acids and a phosphate group (with a compound such as choline attached to the phosphate). Option A, three fatty acids only, describes a triacylglycerol, which has no phosphate. Glucose units only (C) describes a polysaccharide and amino acids only (D) a protein.

2026 Annual I, Set D, Q.2 (iv)1x3 = 3 marks

Compare sickle cell hemoglobin and normal hemoglobin. (Any three points)

Solution

Both are haemoglobin, the iron-rich protein of red blood cells that carries oxygen; they differ in one amino acid of the β-chain.

Point	Normal haemoglobin (HbA)	Sickle haemoglobin (HbS)
Amino acid at position 6 of the β-chain	Glutamic acid (sequence Thr-Pro-Glu-Glu)	Valine (sequence Thr-Pro-Val-Glu)
Cause	Normal β-globin gene	Point mutation in the β-globin gene: one nucleotide replaced by another
Shape of the red blood cells	Disc-shaped, like doughnuts without holes	Sickle or crescent shaped
Effect on blood flow	Cells move easily through blood vessels	Cells are stiff and sticky and block blood flow, causing pain and organ damage

2025 Annual I, Set D, Q.2 (viii)2+1 = 3 marks

The diagram shows molecule of Lecithin: (a) Identify parts A, B, C and D. (b) Mention the polarity of E and F.

Lecithin as given in the paper: three stacked parts (A, B, C) form the head E, and three tails form F; D is the bent tail.

Solution

Lecithin (phosphatidylcholine) is the commonest phospholipid. It is built from glycerol, two fatty acids (one saturated, one unsaturated) and a phosphate group with choline attached.

(a) Identify parts A, B, C and D. (2 marks)

Part	Name
A	Choline (the nitrogenous base)
B	Phosphate group
C	Glycerol (a three-carbon molecule)
D	Unsaturated fatty acid: the tail with a cis double bond that bends the chain

Lecithin (phosphatidylcholine) labelled: choline, phosphate and glycerol form the polar head; the two fatty acids are the non-polar tails.

(b) Mention the polarity of E and F. (1 mark)

E (choline, phosphate and glycerol, the head) is polar or hydrophilic: it is readily soluble in water. F (the fatty acid tails) is non-polar or hydrophobic: it is insoluble in water. A molecule with both a polar and a non-polar end is amphipathic.

Note. In the paper's drawing the right-hand tail has a bend, which stands for the cis double bond of an unsaturated fatty acid.

2025 Annual I, Set D, Q.2 (xi)3 marks

How are Starch and Glycogen different from each other although both are polysaccharides?

Solution

Both are homopolysaccharides made of many α-glucose units, and both store energy. They differ in the following points.

Point	Starch	Glycogen
Where stored	Storage carbohydrate of plants: roots, stems and seeds (cereals, potato tubers)	Storage carbohydrate of animals: liver and muscles (animal's starch)
Branching	Two types: amylose (unbranched, α-1,4) and amylopectin (branched with α-1,6 links)	Resembles amylopectin but is much more highly branched
Colour with iodine	Blue (blue-black)	Red

2025 Annual I, Set D, Q.5 OR2x3 = 6 marks

Identify and explain the properties of water which enable it to: (a) Act as temperature stabilizer (b) Help organisms to live in warm environment (c) Help organisms to live under ice

Solution

Part	Property of water	Reason in one line
(a) Temperature stabiliser	High specific heat capacity	Hydrogen bonds absorb heat first, so the temperature rises slowly
(b) Warm environment	High heat of vaporisation (574 cal per g)	Sweat evaporating carries away much heat from the body
(c) Under ice	Lower density of ice	Ice floats and insulates the liquid water beneath

(a) Act as temperature stabilizer (2 marks)

The property is the high specific heat capacity of water: 1 calorie (4.18 joules) is needed to raise the temperature of 1 g of water by 1 °C, far more than for most substances.

Explanation. Much of the heat that water absorbs is used first to break hydrogen bonds between its molecules, so it does not appear as a rise in temperature. A very large amount of heat therefore raises the temperature of water very little. Water protects living material against sudden thermal change and acts as a temperature stabiliser or regulator, for example for organisms in a hot environment.

(b) Help organisms to live in warm environment (2 marks)

The property is the high heat of vaporisation: 574 calories are needed to convert 1 g of liquid water into gas.

Explanation. A large amount of heat can be lost with a very small loss of water. When an animal sweats, body heat is used to vaporise the sweat and the animal is cooled. The evaporation of only 2 ml out of one litre of water lowers the temperature of the remaining 998 ml by 1 °C. This is an efficient way to release excess heat in a hot environment.

The property is the lower density of ice than liquid water.

Explanation. Ice has a giant lattice-like structure with the maximum number of hydrogen bonds between water molecules, so it is less dense and floats. In freezing weather ice forms on the surface of ponds and lakes and acts as an insulating layer above the water below. The water under it stays liquid, so organisms can live there until the ice melts.

A frozen pond: floating ice insulates the liquid water below, where organisms survive.

Exam focus

Explain how a change in amino acid sequence affects haemoglobin: glutamic acid to valine at position 6 of the β-chain in sickle cell anaemia. (short)
Asked in both 2026 sets (Set B Q.2 (iv) as three effects, Set D Q.2 (iv) as a compare table); the book's long question 12 is the same topic.
Go to answer
Properties of water: temperature stabiliser, warm environment, life under ice, and universal solvent. (long)
2025 Set D Q.5 OR (6 marks, three properties) and 2025 Set B Q.2 (iii) (universal solvent); the book's long question 4 and short questions 6 to 10 cover all eight properties.
Go to answer
Structure of lecithin (phospholipid): identify choline, phosphate, glycerol and fatty acids and give the polarity of head and tails. (short)
2025 Set D Q.2 (viii) is a labelled-diagram question; 2026 Set D MCQ 8 asks the parts of a phospholipid. Learn to draw Figure 2.31.
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Structures and roles of steroids and prostaglandins. (long)
2025 Set B Q.5 OR (6 marks) and 2025 Set D MCQ 2 (prostaglandins from arachidonic acid); book long question 15.
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Differentiate starch and glycogen; know which polysaccharide is stored where and which is structural. (short)
2025 Set D Q.2 (xi) (3 marks) and 2026 Set D MCQ 2 (glycogen); 2026 Set B MCQ 2 (cellulose) tests the same table.
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Compare maltose and sucrose (or lactose): glycosidic linkage, monomers and role. (short)
2025 Set B Q.2 (xiii) OR (3 marks, one mark per point); book short question 16 draws their formation.
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Name structural proteins and their roles (collagen, elastin, keratin) and functional proteins. (long)
2025 Set B MCQ 7 (protein of tendons and ligaments) and the book's long question 13 (three examples of each).
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Describe the Watson and Crick model of DNA and the structure of RNA types. (long)
No unit 2 paper question in 2025 or 2026 (the paper items on mutation and nucleic acids sit in unit 13), but it is a unit SLO and book long questions 20 to 23.
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