Chapter 6: Biomolecules and Related Techniques

In the 1920s, most chemists were certain of one thing: no one would ever hold an enzyme in their hand. Enzymes were thought to be mysterious “vital forces” — something living tissue possessed, not a substance that could be weighed out on a balance. The leading opinion of the day held that enzymes were not ordinary chemicals at all.

James Sumner, working in a small laboratory in the United States, disagreed. He chose jack beans, a rich source of the enzyme urease, which splits urea. Year after year he ground the beans, extracted them, and tried to coax the active material out of solution. Colleagues told him he was wasting a career on an impossible problem. One arm had been amputated in a boyhood accident, and much of the delicate bench work had to be managed one-handed.

In 1926 he finally saw tiny crystals settle out of his extract. He dissolved them, tested them, and found they still split urea. He tested them further and found something more important: the crystals were protein. The enzyme was not a vital force. It was a molecule — one that could be purified, crystallised, and studied like any other chemical. Sumner was awarded the Nobel Prize for this work twenty years later.

Diagram

Fig. 6.1 – An opening illustration showing a jack bean pod beside a flask of extract, with an arrow to needle-shaped protein crystals under a hand lens; label “urease crystals — the first enzyme shown to be a protein”.

Curiosity Corner

Q1. If an enzyme is only a protein, why does boiling an egg — or boiling an enzyme — destroy it forever? Q2. Why does a piece of plain chapati begin to taste sweet if you chew it long enough? Q3. Cows and termites live on grass and wood. Why can they digest cellulose when we cannot, even though our starch is also made of glucose? Q4. Why does a doctor give a dehydrated patient glucose rather than a spoonful of starch? Q5. Butter is solid and mustard oil is liquid at the same room temperature. What difference in their molecules decides this?

This chapter will let you answer each of these questions for yourself.

6.1 Introduction to Biomolecules

Every living thing — a bacterium in curd, a mango tree, a human being — is built and run by chemicals. If you were to take any living cell apart and analyse it, you would find a limited set of carbon-based compounds doing all the work. These compounds are called biomolecules, and they are often described as the “chemicals of life”.

Definition

A biomolecule is any organic compound that is produced by a living organism and takes part in its structure, growth, maintenance or metabolic activity.

Biomolecules are organic compounds, which means that carbon forms their backbone. Along with carbon, they are built mainly from hydrogen and oxygen, with nitrogen present in proteins and nucleic acids, and phosphorus and sulphur appearing in particular groups. Only a handful of elements, joined in different arrangements, account for the enormous variety of living matter.

Small Molecules and Large Molecules

Not all biomolecules are the same size. Some are small and simple — a single sugar such as glucose, a single amino acid, a fatty acid, a nucleotide. These are called micromolecules, and each has a molecular weight of a few hundred at most.

Living cells then join these small units together, one after another, into very long chains. A large molecule built by linking many small repeating units is called a macromolecule, and the small repeating unit it is built from is called a monomer. The long chain itself is a polymer.

Definition

A macromolecule is a very large biomolecule formed by the covalent linking of many smaller repeating units called monomers.

This one idea — small units joined into long chains — explains most of the chemistry in this chapter. Many glucose units joined together give starch or cellulose. Many amino acids joined together give a protein. Many nucleotides joined together give DNA or RNA. The monomers are few and familiar; the polymers they build are what make one organism different from another.

Compare Table

The Major Types of Biomolecules

For the purposes of this chapter, biomolecules are grouped into a small number of classes, each with its own chemistry and its own job in the cell. The classification is summarised in FC 6.1.

Flowchart

FC 6.1 – Types of biomolecules: Biomolecules → (i) Carbohydrates → mono-, di- and polysaccharides; (ii) Proteins → built from amino acids; (iii) Lipids → simple, conjugated and derived; (iv) Vitamins → fat-soluble and water-soluble; (v) Enzymes → protein biocatalysts

Carbohydrates are the cell’s most readily available fuel, and in plants and bacteria they also form the structural framework of the cell wall. Proteins are the most versatile class: they build tissues, transport oxygen, defend the body and, as enzymes, carry out almost every chemical reaction in a cell. Lipids store energy in a compact form and build the membranes that separate a cell from its surroundings. Vitamins are needed only in tiny amounts, but without them many enzymes cannot work at all. Enzymes are treated as a separate group in this chapter because of their central importance, although chemically almost all of them are proteins.

A word of caution about a group you will meet elsewhere. Hormones are often listed alongside these classes, but hormones are not a separate chemical type — insulin is a protein, while testosterone is a lipid. Hormones are classified by what they do, not by what they are made of.

Why This Chapter Matters for Biotechnology

Biotechnology is, at bottom, the business of handling biomolecules. A fermenter is a vessel in which microorganisms convert one biomolecule into another. Insulin manufactured by recombinant bacteria is a protein that must be purified from a soup of thousands of other proteins. Every diagnostic test — the blood-sugar strip, the pregnancy test, the food-adulteration check — works by detecting one specific biomolecule and ignoring the rest.

That is why this chapter ends with laboratory techniques rather than with more chemistry. Knowing what a molecule is has limited value unless you can also separate it, identify it and measure it. In the ISC practical course you will do exactly this: test foods for carbohydrates, proteins and lipids, estimate protein by Bradford’s assay, and separate leaf pigments by paper chromatography.

Try Yourself

Take any three items from your kitchen — for example, sugar, groundnut oil and a boiled egg. For each one, write down which class of biomolecule it mainly represents, and whether that biomolecule is a micromolecule or a macromolecule.

6.2 Carbohydrates

Carbohydrates are the most abundant naturally occurring organic compounds on earth. They are found in animals, plants and microorganisms alike, and they do two very different jobs. They are the cell’s first and fastest source of energy, and they are also the material out of which plants, fungi and bacteria build their outer walls. Energy and architecture — both come from the same family of molecules.

Why the Name “Carbohydrate”

The name was coined when chemists noticed that these compounds contain only carbon, hydrogen and oxygen, and that the hydrogen and oxygen are present in the same 2 : 1 ratio as in water. Glucose, for instance, has the molecular formula \(C_6H_{12}O_6\), which can be rewritten as \(C_6(H_2O)_6\) — six carbons apparently “hydrated” with six waters. Sucrose, \(C_{12}H_{22}O_{11}\), can likewise be written as \(C_{12}(H_2O)_{11}\). This suggested a general formula:

\[C_x(H_2O)_y\]

where \(x\) and \(y\) are whole numbers that may or may not be equal.

The formula turned out to be a rough guide rather than a rule. As more carbohydrates were discovered, exceptions appeared on both sides. Deoxyribose (\(C_5H_{10}O_4\)) and rhamnose (\(C_6H_{12}O_5\)) are genuine carbohydrates that do not fit the formula, because they have lost an oxygen. Conversely, acetic acid (\(C_2H_4O_2\)) and lactic acid (\(C_3H_6O_3\)) fit the formula perfectly but are not carbohydrates at all.

A definition based on chemical structure was therefore needed, and this is the one now used.

Definition

Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones, and their derivatives.

Read that definition carefully, because it does two things at once. “Polyhydroxy” means the molecule carries several \(-\text{OH}\) groups, which is why sugars dissolve so readily in water. “Aldehyde or ketone” means it also carries one carbonyl group, and almost all the chemical behaviour of sugars — the colour tests you will perform in the laboratory — traces back to that single carbonyl group.

6.2.1 Classification of Carbohydrates

Carbohydrates are grouped according to how many simple sugar units they contain, and whether those units can be released by hydrolysis.

Flowchart

FC 6.2 – Classification of carbohydrates: Carbohydrates → Monosaccharides (cannot be hydrolysed further; e.g. glucose, fructose) → Disaccharides (2 units; e.g. maltose, sucrose, lactose) → Oligosaccharides (2–9 units) → Polysaccharides (many units; homopolysaccharides e.g. glycogen, cellulose, chitin, starch / heteropolysaccharides e.g. peptidoglycan)

Oligosaccharides sit between disaccharides and polysaccharides: they contain a small number of monosaccharide units, generally two to nine, and most of them still taste sweet. Polysaccharides contain ten or more units, usually many hundreds or thousands, and do not taste sweet at all.

6.2.2 Monosaccharides

Monosaccharides are the simple sugars. They are the monomers of the whole carbohydrate family, and by definition they cannot be hydrolysed into anything simpler that is still a carbohydrate.

Monosaccharides are named on two criteria taken together:

Length of the carbon chain — triose (3C), tetrose (4C), pentose (5C), hexose (6C) and heptose (7C).
Nature of the functional group — an aldose carries an aldehyde group, a ketose carries a ketone group.

The two criteria are combined into a single name. Glucose has six carbons and an aldehyde group, so it is an aldohexose. Fructose also has six carbons but carries a ketone group, so it is a ketohexose.

S. No.	Monosaccharide	Aldose	Ketose
1.	Triose (\(C_3H_6O_3\))	Glyceraldehyde	Dihydroxyacetone
2.	Tetrose (\(C_4H_8O_4\))	Erythrose	Erythrulose
3.	Pentose (\(C_5H_{10}O_5\))	Ribose, deoxyribose	Ribulose
4.	Hexose (\(C_6H_{12}O_6\))	Glucose, galactose	Fructose
5.	Heptose (\(C_7H_{14}O_7\))	Sedoheptose	Sedoheptulose

Open-Chain and Ring Structures

A monosaccharide can exist in two structural forms, and both matter. In the open-chain form, the carbon atoms are drawn in a straight line with the carbonyl group at or near one end. In solution, however, the molecule prefers to curl up on itself.

What happens is this. One of the alcohol (\(-\text{OH}\)) groups on the chain attacks the carbonyl carbon of the same molecule. If the carbonyl is an aldehyde, the product is a hemiacetal; if it is a ketone, the product is a hemiketal. Either way, a new bond forms between two carbons of the same chain through an oxygen atom, and the molecule closes into a ring. The sequence is shown in Fig. 6.2.

Diagram

Fig. 6.2 – Conversion of open-chain glucose into the hemiacetal and then the ring structure; label the aldehyde group at C-1, the \(-\text{OH}\) at C-5 that attacks it, and the ring oxygen formed.

The ring can be either six-membered, called the pyranose form, or five-membered, called the furanose form. The pyranose form is the more stable of the two. This is because ring closure between C-1 and C-5 produces a more relaxed, less strained ring than closure between C-1 and C-3 or C-1 and C-7 would.

When the open chain is rewritten as a ring, two conventions are followed:

Groups written on the right side of the open chain are drawn below the plane of the ring, and those on the left side are drawn above it.
For the carbon involved in hemiacetal or hemiketal formation, the opposite rule applies. So in D-sugars, \(-\text{CH}_2\text{OH}\) is written above the plane and \(-\text{H}\) below.

Ring closure has one further consequence that is easy to miss. The carbonyl carbon, which was symmetrical in the open chain, now has four different groups around it — it becomes a new centre of asymmetry. This new asymmetric carbon is C-1 in aldoses and C-2 in ketoses. Depending on whether the \(-\text{OH}\) at this carbon points one way or the other, the sugar is called an α-sugar or a β-sugar.

This is not a small technical point. As you will see shortly, whether glucose units are joined in the α form or the β form decides whether the resulting polymer is digestible food or indigestible fibre.

Glucose

Glucose is a polyhydroxy aldehydic monosaccharide with six carbons, \(C_6H_{12}O_6\). It is the single most important sugar in biology, because it is the immediate source of energy for a living cell and the raw material fed into respiration.

Glucose is widely available in nature — in cereals, fruits and vegetables — and it is the unit from which many larger carbohydrates are constructed. It forms disaccharides such as maltose, sucrose and lactose, and polysaccharides such as glycogen, starch and cellulose. Its ring structure is shown in Fig. 6.3.

Diagram

Fig. 6.3 – Ring structure of glucose in the pyranose form with all six carbons numbered, showing the \(-\text{CH}_2\text{OH}\) group at C-6 and the \(-\text{OH}\) at C-1.

Curiosity Corner

Q4. Why does a doctor give a dehydrated patient glucose rather than a spoonful of starch?

A. Glucose is a monosaccharide — it needs no digestion at all. It can pass directly through the intestinal wall, or be delivered straight into the bloodstream in a saline drip, and enter respiration within minutes. Starch is a polysaccharide of many hundreds of glucose units; before a single molecule of it can be used, enzymes must hydrolyse the chain step by step into glucose. When a patient is weak and needs energy immediately, that delay matters.

Fructose

Fructose is a polyhydroxy ketonic monosaccharide, also with the formula \(C_6H_{12}O_6\). It is commonly called fruit sugar, and it is found mainly in fruits such as apple, mango, grapes and banana, in honey, and in smaller amounts in vegetables such as onion, carrot and sweet potato. It is also added to processed foods in the form of high-fructose corn syrup.

Fructose and glucose are a useful pair to compare. They have exactly the same molecular formula, yet they behave differently, taste differently and give different results in laboratory tests. The reason is the position and type of the functional group: an aldehyde at C-1 in glucose, a ketone at C-2 in fructose. Molecules with the same formula but different structures are called isomers.

Diagram

Fig. 6.4 – Ring structures of α-D-fructose (furanose form) alongside glucose (pyranose form) for comparison; label the five-membered and six-membered rings.

Ribose and Deoxyribose

Two pentose sugars deserve special mention because they carry the genetic material of every organism.

Ribose is a pentose sugar, \(C_5H_{10}O_5\). It carries a hydroxyl group at the 2′ carbon, and this \(-\text{OH}\) makes the molecule chemically reactive. Ribose is the sugar of RNA, and its 2′ hydroxyl is the reason RNA is more easily broken down than DNA.

Deoxyribose is a pentose sugar, \(C_5H_{10}O_4\). As the name suggests, it is ribose minus one oxygen: the hydroxyl at the 2′ carbon has been replaced by a hydrogen. Losing that reactive group makes deoxyribose, and therefore DNA, distinctly more stable. This stability is precisely why DNA — and not RNA — serves as the genetic material in all organisms except some viruses.

Diagram

Fig. 6.5 – Ribose and deoxyribose side by side, with the 2′ carbon circled in both to show \(-\text{OH}\) in ribose and \(-\text{H}\) in deoxyribose.

6.2.3 Disaccharides

A disaccharide is formed when two monosaccharide units — the same or different — are joined together. The bond that joins them is called a glycosidic linkage.

Definition

A disaccharide is a carbohydrate composed of two monosaccharide units joined by a glycosidic linkage.

The linkage is formed by condensation: the \(-\text{OH}\) group of one sugar reacts with the \(-\text{OH}\) group of the other, and one molecule of water is eliminated. This is shown in Fig. 6.6.

Diagram

Fig. 6.6 – Formation of a glycosidic bond between two glucose molecules to give maltose; show the two \(-\text{OH}\) groups involved, the loss of \(-\text{H}_2\text{O}\), and label the resulting 1,4-glycosidic bond.

The linkage is described by two pieces of information: whether it is α or β, and which carbon numbers are joined. So “α, 1→4” means the bond runs from carbon 1 of the first sugar, in the α configuration, to carbon 4 of the second.

Side Note

Glycosides are compounds formed by condensation between a monosaccharide and the hydroxyl group of a second compound, which may or may not itself be another monosaccharide.

S. No.	Disaccharide	Monosaccharide units	Type of glycosidic linkage
1.	Lactose (milk sugar)	β-D-galactose + α-D-glucose	β, 1→4
2.	Maltose (malt sugar)	α-D-glucose + α-D-glucose	α, 1→4
3.	Sucrose (cane sugar)	α-D-glucose + β-D-fructose	α-1 → β-2

Maltose is formed by two molecules of glucose linked by an α-(1→4) glycosidic bond. It is known as malt sugar and is found in germinating seeds such as sprouting malt seeds. It is also produced during the digestion of starch by amylase. Maltose is a reducing sugar, because one of its two carbonyl groups is left free.

Sucrose is formed by an α-(1→2) glycosidic bond between glucose and fructose. Popularly known as table sugar, it occurs in sugarcane and sugarbeet. Sucrose is a non-reducing sugar. The reason lies in the bond itself: it is formed between C-1 of glucose and C-2 of fructose, which are the carbonyl carbons of both sugars. With both carbonyl groups locked into the linkage, neither is free to react.

Lactose is formed by a β-(1→4) glycosidic bond between glucose and galactose. Called milk sugar, it is found in milk and dairy products. Like maltose, it is a reducing sugar because it retains a free carbonyl group.

Diagram

Fig. 6.7 – Structures of maltose, sucrose and lactose, with the glycosidic bond marked in each and the free carbonyl group indicated in maltose and lactose.

6.2.4 Polysaccharides

Polysaccharides are high molecular weight carbohydrates that yield many monosaccharide molecules on hydrolysis. This is by far the most abundant form of carbohydrate in nature — most of the carbohydrate on earth, by weight, is polysaccharide.

They are classified on the basis of the units they contain:

Homopolysaccharides are built from only one type of monosaccharide.
Heteropolysaccharides are built from more than one type of monosaccharide.

Glycogen

Glycogen is the main reserve food of animals, and for that reason it is often called animal starch. It is stored in the liver and in muscle, and makes up roughly 7% of the wet weight of the liver.

Structurally it resembles amylopectin, but with two differences that matter: it contains more glucose units per molecule (molecular weight \(1\)–\(2 \times 10^7\)) and it branches far more frequently, with chains of only about 10–20 glucose units between branch points. The main chain is held by α-(1→4) glycosidic bonds, and the branch points are α-(1→6) glycosidic bonds. Glycogen gives a red colour with iodine.

Why should a storage molecule be so heavily branched? Because enzymes attack a glycogen molecule only from the free ends of its chains. Every branch creates another free end. A highly branched molecule therefore has hundreds of points at which glucose can be released simultaneously — exactly what a muscle needs when energy demand rises suddenly.

Diagram

Fig. 6.8 – Branched structure of glycogen, showing α-1,4 linkages along the chains and an α-1,6 linkage at a branch point, with free ends marked.

Cellulose

Cellulose is the most abundant biomolecule on earth. It forms the main constituent of the plant cell wall and is made up of about 25,000–30,000 β-D-glucose residues linked together by β-1,4-glycosidic linkages.

That single letter — β instead of α — has enormous consequences. Animals whose main carbohydrate-digesting enzyme is α-amylase cannot break a β-glycosidic bond, because the enzyme’s active site simply does not fit a bond of that shape. Cellulose therefore passes through the human digestive system unchanged. In cattle it is digested, but not by the animal itself: symbiotic bacteria and ciliates living in the gut produce the necessary enzymes.

In humans, cellulose still serves a purpose. It adds bulk to the diet, acts as roughage and prevents constipation. Commercially it is used in the manufacture of paper, cloth (cotton, rayon), explosives, celluloids and collodion.

Curiosity Corner

Q3. Cows and termites live on grass and wood. Why can they digest cellulose when we cannot, even though our starch is also made of glucose?

A. Starch and cellulose are both polymers of glucose, but the glucose units are joined differently — α-1,4 in starch, β-1,4 in cellulose. Our digestive enzymes are shaped to fit the α linkage only. Cattle and termites do not possess a cellulose-digesting enzyme either; what they possess is a gut full of symbiotic microorganisms that do. The animal supplies the shelter, the microbes supply the enzyme.

Compare Table

Chitin

Chitin is the second most abundant polysaccharide after cellulose. It occurs in the cell wall of fungi and in the exoskeleton of arthropods — the hard outer covering of insects, prawns and crabs.

Structurally it is closely related to cellulose, with one substitution: the \(-\text{OH}\) group at C-2 of each glucose residue is replaced by an N-acetylamino group. Chitin is therefore a polymer of N-acetylglucosamine units joined together by β-1,4-glycosidic linkages. Because the linkage is β, chitin shares cellulose’s great mechanical strength and resistance to digestion.

Starch

Starch is the storage polysaccharide of plants and the main carbohydrate in the human diet. It is not a single compound but a mixture of two:

α-Amylose (about 15–20%) consists of long, unbranched chains of 200–1000 glucose molecules linked by α-1,4 glycosidic linkage. Its molecular weight ranges between 10,000 and 50,000.
Amylopectin (about 80–85%) is the branched component, with a molecular weight of 50,000–100,000 and about 300 to 5,500 glucose units per molecule. Its straight chains use α-1,4 linkages as in maltose, but it branches through α-1,6 linkages, with an average chain length of about 24 glucose units.

Starch is insoluble in water at normal temperature. On heating it breaks into large fragments called dextrins, and on continued hydrolysis it is gradually converted into glucose. The progress of this hydrolysis can actually be watched, because the colour given with iodine solution changes as the chains get shorter.

Flowchart

FC 6.3 – Colour change during starch hydrolysis with iodine: Starch (blue) → dextrins (purple) → shorter fragments (red) → maltose and glucose (no colour)

Technique

Iodine test for starch Principle: Iodine molecules slip inside the helical coil of the amylose chain, forming a coloured complex; the colour depends on how long the coil is. Steps: Take the food sample in a test tube → add a few drops of iodine solution → observe the colour. Used for: Detecting starch in a food sample, which gives a blue-black colour; this is ISC practical 9(i), and the same test tracks the progress of starch digestion by salivary amylase.

Peptidoglycan

Peptidoglycan is a heteropolysaccharide, and it is the molecule that gives the bacterial cell wall its strength. It is made up of chains of alternating molecules of N-acetyl glucosamine (NAG) and N-acetyl muramic acid (NAM), linked by β-1,4-glycosidic bonds. These sugar chains are then cross-linked to one another by short peptide chains — hence the name peptido-glycan.

This structure is of direct importance in biotechnology and medicine. The enzyme lysozyme, present in tears, saliva and egg white, attacks the glycosidic bond between NAM and NAG, cutting the chain and bursting the cell. Several antibiotics work by preventing the peptide cross-links from forming, so that the wall cannot be completed as the bacterium grows.

Diagram

Fig. 6.9 – Structure of peptidoglycan showing alternating NAG and NAM units in the glycan chains, joined by short oligopeptide cross-links; mark the NAM–NAG bond attacked by lysozyme.

Side Note

Teichoic acids, found in the cell wall of Gram-positive bacteria, are chains of glycerol phosphate or ribitol phosphate connected to the peptidoglycan wall by phosphate ester bridges.

6.2.5 Physical Properties of Carbohydrates

Sugars are sweet, crystalline, water-soluble and odourless. Their solubility in water follows directly from the many \(-\text{OH}\) groups on the molecule, which form hydrogen bonds with water.
The glycosidic \(-\text{OH}\) group of mutarotating sugars reacts with alcohols to form α- and β-glycosides or acetals. Glucose forms glucosides and fructose forms fructosides.

6.2.6 Chemical Properties of Carbohydrates

Almost every chemical property of a sugar comes back to one question: is its carbonyl group free, or is it tied up in a bond?

Reduction

Monosaccharides are reduced to their corresponding alcohols by mild reducing agents such as sodium amalgam. Glucose yields sorbitol, while fructose yields two isomeric forms — sorbitol and mannitol. With strong reducing agents, glucose is reduced to levulinic acid.

Why does fructose give two products where glucose gives one? Because reduction of the ketone group at C-2 of fructose can produce a new asymmetric carbon in either of two orientations, whereas reduction of the terminal aldehyde of glucose can produce only one.

Oxidation

With a mild oxidising agent such as HOBr, glucose is oxidised to gluconic acid. Ketoses do not respond to this reaction, and so this reaction is used to distinguish aldoses from ketoses. With strong oxidising agents such as \(\text{HNO}_3\), glucose is oxidised to saccharic acid (glucaric acid), while ketoses are converted to furfural.

Metal hydroxides such as \(\text{Cu(OH)}_2\) and AgOH oxidise the free aldehyde or ketone group of mono- or disaccharides and are themselves reduced in the process. These form the active ingredients of Fehling’s solution and Benedict’s solution.

Reducing and Non-reducing Sugars

Definition

A reducing sugar is a sugar whose carbonyl group is free — that is, not involved in hemiacetal or glycosidic bond formation — and which therefore gives a positive test with Fehling’s and Benedict’s reagents.

Glucose, fructose, lactose and maltose are reducing sugars. Sucrose is a non-reducing sugar, because both its carbonyl groups are consumed in forming the glycosidic linkage.

Technique

Benedict’s / Fehling’s test for reducing sugars Principle: A free carbonyl group reduces the blue cupric ion (\(\text{Cu}^{2+}\)) of the reagent to the red cuprous ion (\(\text{Cu}^{+}\)), which appears as a brick-red precipitate of cuprous oxide. Steps: Take the sugar solution in a test tube → add Benedict’s or Fehling’s solution → heat in a water bath → observe the colour change from blue to green, yellow, orange and finally brick red. Used for: Detecting reducing sugars; in ISC practical 9(i) it distinguishes reducing from non-reducing sugars, and clinically the same chemistry underlies traditional tests for glucose in urine.

Exam Corner

Q. A sugar solution gives no colour with Benedict’s reagent. The solution is then boiled with a little dilute acid, cooled, and tested again — and now it gives a brick-red precipitate. What was the sugar?

A. It was sucrose, or another non-reducing sugar. Boiling with acid hydrolyses the glycosidic bond and releases the two monosaccharides, freeing their carbonyl groups. Those free carbonyls then reduce the reagent. This is a standard way of confirming that a sugar is non-reducing rather than simply absent.

Barfoed’s reagent contains cupric acetate and is used to distinguish monosaccharides from disaccharides. On reaction with monosaccharides it is converted to the cuprous ion, and it does so faster than with disaccharides, showing that monosaccharides are the more active reducing sugars.

Tests that Distinguish Glucose from Fructose

Since glucose and fructose share the same formula, tests that separate them must depend on the difference between an aldehyde and a ketone group.

Pinoff’s test — glucose reacts with ammonium molybdate and a few drops of acetic acid to give a light blue colour, whereas fructose gives a deep blue to bluish green colour.
Selivanoff’s test — the sugar is warmed with resorcinol and dilute HCl. Ketoses such as fructose give a rapid cherry-red colour, while aldoses such as glucose react much more slowly and give little or no colour in the same time.

Other Reactions of Sugars

Reaction with phenylhydrazine: When monosaccharides are treated with excess phenyl hydrazine, yellow coloured crystals of osazone are formed. Each sugar’s osazone has a characteristic crystal shape and melting point, so the reaction can be used to identify the type of sugar.

Reaction with amino acids: Glucose reacts with the amino acid alanine to produce Schiff’s base. This reaction is responsible for the browning of bread when it is heated — the same chemistry that browns a roti on the tawa and gives roasted food its colour and aroma.

Reaction with methyl alcohol: When glucose in solution is treated with methyl alcohol and HCl, the corresponding glucoside is formed.

Numerical 6.1

A cellulose molecule from a plant cell wall is found to contain 27,000 glucose residues. Given that each glucose residue in the chain contributes a mass of 162 u (the mass of glucose, 180 u, less the mass of one water molecule lost during condensation), calculate the approximate molecular weight of this cellulose molecule.

Try Yourself

Chew a small piece of plain chapati or bread slowly for two minutes without swallowing, and note how the taste changes. Then write down which carbohydrate you started with, which one you are tasting at the end, and what kind of bond had to be broken in between.

6.3 Proteins and Amino Acids

If carbohydrates are the fuel of the cell, proteins are its workforce. They are the most important and abundant constituent of protoplasm, making up about 12–13% of its weight. Nothing else in the cell does so many different jobs: proteins build tissue, carry oxygen, defend the body against infection, carry messages between organs and, as enzymes, run almost every chemical reaction that keeps a cell alive.

The name protein comes from a Greek word meaning “of first importance”, and it was proposed by Berzelius in the 1830s. The chemistry behind the name is remarkably simple.

Proteins usually contain carbon, hydrogen, oxygen and nitrogen. Sulphur and phosphorus are also present in some of them. Nitrogen is the element to watch: carbohydrates and lipids contain none, so the presence of nitrogen is often the first clue that a biomolecule is a protein.

Definition

Proteins are condensation polymers of α-amino acids, joined together by peptide bonds.

6.3.1 Amino Acids: The Building Blocks

An amino acid is an organic compound containing carbon, hydrogen, oxygen and nitrogen. Every amino acid has at least one amino group (\(-\text{NH}_2\)) and one carboxyl group (\(-\text{COOH}\)) attached to the same central carbon atom, called the α-carbon.

The remaining two valencies of that central carbon are satisfied by one hydrogen atom and one variable group, written as \(R\) and called the side chain. The general structure is shown in Fig. 6.10.

Diagram

Fig. 6.10 – General structure of an α-amino acid: central α-carbon bearing \(-\text{NH}_2\), \(-\text{COOH}\), \(-\text{H}\) and the variable side chain \(R\); label each of the four groups.

Because four different groups are attached to the four bonds of the α-carbon, the molecule is asymmetrical, or chiral. The single exception is glycine, where the side chain is simply a hydrogen atom, so two of the four groups are identical.

Everything that makes one amino acid different from another lies in that \(R\) group. The backbone is identical in all of them.

How Many Amino Acids?

More than 300 types of amino acids occur in nature, but only 20 of them are found in proteins. Every protein of plant or animal origin is built by combining these same 20 units in different numbers and different sequences — which is why they are often called the magic 20.

Consider what that means. Proteins differ not only in how many amino acids they contain but, far more importantly, in the order in which those amino acids are arranged in the chain. Twenty different units arranged in a chain a few hundred long can produce more distinct sequences than there are atoms in the observable universe. This is the source of the enormous variety of proteins in living things.

Side Note

Two further amino acids have since been added to the list of standard amino acids — selenocysteine, the 21st, and pyrrolysine, the 22nd.

D and L, d and l

Two pairs of symbols are used with amino acids, and they mean different things.

The small letters \(d\) (dextrorotatory) and \(l\) (levorotatory) describe how the compound rotates plane polarised light — a property discussed in Section 6.7. The capital letters D and L describe stereo configuration, that is, the spatial arrangement of groups around the α-carbon. An amino acid with the \(-\text{NH}_2\) group written on the right is a D-amino acid; one with it on the left is an L-amino acid.

All the amino acids found in proteins are of the L-configuration. D-amino acids are not entirely absent from biology, however: they have been reported in some antibiotics and in bacterial cell walls.

Types of Amino Acids

Amino acids are classified according to how they behave in solution, which depends on how many amino and carboxyl groups each one carries.

Acidic amino acids possess one amino group and two carboxyl groups (mono amino–dicarboxylic amino acids), for example aspartic acid and glutamic acid.
Basic amino acids possess two amino groups and one carboxyl group (diamino–monocarboxylic amino acids), for example lysine, arginine and histidine.
Neutral amino acids possess one amino group and one carboxyl group (mono amino–monocarboxylic amino acids), for example glycine, alanine, serine, leucine, isoleucine, threonine and valine.

Side Note

Amino acids are also grouped by whether the body can make them. Adults cannot synthesise nine of them — lysine, histidine, methionine, phenylalanine, tryptophan, threonine, valine, isoleucine and leucine — so these essential amino acids must be supplied in the diet. The rest, such as glycine, alanine, serine, aspartic acid, glutamic acid, proline, tyrosine, cysteine, asparagine and glutamine, are non-essential and can be made from the corresponding keto acids. Arginine is semi-essential: adults make enough of it in the ornithine cycle, but infants and growing children cannot make enough to support both urea synthesis and protein synthesis, and so need a dietary supply.

6.3.2 Physical and Chemical Properties of Amino Acids

Zwitter Ions and Amphoteric Behaviour

At the normal physiological pH of about 7.4, an amino acid does not exist as the neutral molecule you would draw on paper. Its carboxyl group dissociates to form a negatively charged carboxylate ion (\(-\text{COO}^-\)), while its amino group picks up a proton to become positively charged (\(-\text{NH}_3^+\)).

The molecule therefore carries a positive charge at one end and a negative charge at the other, at the same time. Such dipolar ions are called zwitter ions.

Because an amino acid can donate a proton like an acid and accept one like a base, it is described as amphoteric, and such substances are referred to as ampholytes, that is, amphoteric electrolytes.

The Isoelectric Point

The zwitter-ionic form is the iso-electric form of the amino acid. In this form the molecule carries a net charge of zero — the positive and negative charges are equal and cancel out.

Definition

The iso-electric pH (pI) is the pH at which an amino acid exists in its zwitter-ionic form and carries no net charge, so that in an electric field it moves neither towards the anode nor towards the cathode.

The isoelectric point can be calculated from the dissociation constants of the two ionisable groups:

\[pI = \frac{1}{2}\left(pK_1 + pK_2\right)\]

where \(pK_1\) is the dissociation constant of the \(-\text{COOH}\) group and \(pK_2\) that of the \(-\text{NH}_2\) group. Both are dimensionless quantities, as is \(pI\) itself.

Follow what happens to alanine, which has two ionisable groups, as base is added and the pH is raised. This is shown in Fig. 6.11.

Diagram

Fig. 6.11 – The three ionic forms of alanine: Form I (net charge +1, pH < 2) → Form II, the zwitter ion (net charge 0, pH ≈ 6) → Form III (net charge −1, pH > 10); show \(-\text{OH}^-\) above each arrow.

At low pH, alanine exists as Form I with a net charge of +1. As base is added, the \(-\text{COOH}\) group donates a proton to the medium and becomes \(-\text{COO}^-\), giving Form II. This form has an overall charge of zero — it is the zwitter ion, and the pH at which it occurs is the \(pI\) of alanine. The dissociation constant for this first step is \(pK_1\).

Raise the pH further and a proton is released from the protonated amino group of Form II, giving Form III with a net charge of −1. The dissociation constant for this second step is \(pK_2\).

Numerical 6.2

For glycine, the dissociation constant of the carboxyl group is \(pK_1 = 2.34\) and that of the amino group is \(pK_2 = 9.60\). Calculate the isoelectric pH of glycine, and state whether glycine will carry a net positive or net negative charge in a buffer of pH 8.0.

Absorption of Light

Amino acids do not absorb visible light, which is why an aqueous solution of them is colourless. The aromatic amino acids — tryptophan, tyrosine, phenylalanine and histidine — are the exception in the ultraviolet region: they absorb UV light of wavelength above 240 nm, while the others do not. Most of the UV absorption of a protein above 240 nm is therefore due to its tryptophan content.

This is more than a curiosity. It is the basis on which protein concentration is estimated in a spectrophotometer without adding any reagent at all.

The Side Chain Decides Everything

Inside a protein, the carboxyl and amino groups of most amino acids are locked into peptide linkages and are no longer free to react. What remains free, and therefore what decides how the amino acid behaves within a protein, is the side chain.

Side chains fall into two broad camps. Non-polar side chains are hydrophobic: they do not liberate or accept protons and do not form hydrogen or ionic bonds, and they are insoluble in water. Polar side chains are hydrophilic and dissolve readily.

In an aqueous solution of a protein, this difference sorts the molecule out spatially. The non-polar side chains fold inward and cluster together in the interior of the molecule, away from water, while the polar ones remain on the surface in contact with it. That single tendency is what gives a protein its three-dimensional shape — a point we return to in Section 6.3.4.

6.3.3 The Peptide Bond

Amino acids are joined into chains by peptide bonds. A peptide bond is an amide linkage formed between the carboxyl group (\(-\text{COOH}\)) of one amino acid and the α-amino group (\(-\text{NH}_2\)) of the next, with the loss of one molecule of water. The process is called condensation, and it is shown in Fig. 6.12.

Diagram

Fig. 6.12 – Formation of a peptide bond between two amino acids: show the \(-\text{COOH}\) of the first and the \(-\text{NH}_2\) of the second, the elimination of \(-\text{H}_2\text{O}\), and the resulting \(-\text{CO}-\text{NH}-\) linkage.

Condensation of several amino acids in this way builds a polypeptide, and polypeptides build proteins.

Peptide bonds are covalent bonds, and this has a practical consequence: they are not easily broken by ordinary means such as heating. When you boil an egg the protein changes completely in appearance and texture, yet not one peptide bond has been broken. Something else has happened — and understanding what that something is requires the next subsection.

6.3.4 Structure of Proteins

Proteins show four levels of structural organisation: primary, secondary, tertiary and quaternary. Each level builds on the one below it.

Diagram

Fig. 6.13 – The four levels of protein structure shown as a single sequence: a linear chain of amino acids (primary) → coiled α-helix and folded β-pleated sheet (secondary) → a compact folded three-dimensional shape (tertiary) → several such units associated together (quaternary).

Primary Structure

The primary structure refers to the number and sequence of amino acids in a protein chain, together with the location of disulphide bonds, if any are present.

This is the most important level of all, and for one reason: the primary structure determines everything above it. A chain of amino acids folds into a particular three-dimensional shape because of the order in which its amino acids are arranged. Change one amino acid in the sequence and the folding may change, and with it the protein’s properties. Insulin is the classic example of a protein whose function is dictated by its precise sequence.

Secondary Structure

The polypeptide chain does not remain a straight line. It folds in regular, repeating ways to give the secondary structure, and these folds are held in place by hydrogen bonds formed between the oxygen atom of one amino acid and the hydrogen atom of an adjacent one.

Note the contrast with primary structure. Primary structure is held by strong covalent peptide bonds; secondary structure is held by weak hydrogen bonds. That difference explains almost everything about how proteins respond to heat.

A protein chain may assume any of the following secondary forms.

(a) Alpha-helix. This is the most common form of coiling, a right-handed spiral with a tightly packed, coiled polypeptide backbone at its core. The side chains of the constituent amino acids project outward from the central axis, which prevents them from getting in each other’s way.

The hydrogen bonds run along the length of the helix rather than across it: each is formed between the carbonyl oxygen of one amino acid and the amide (\(-\text{NH}\)) hydrogen of the fourth amino acid ahead of it. Each turn of an α-helix contains 3.6 amino acid residues, with a pitch of 5.4 Å, and the rise along the central axis is 1.5 Å per amino acid residue.

The helical arrangement produces stiff bundles or rods of fibres. Keratin, myosin, tropomyosin, fibrin and haemoglobin all contain α-helical regions, which shows that both fibrous and globular proteins can adopt this structure.

Diagram

Fig. 6.14 – The α-helix, showing the right-handed coil, side chains projecting outward, and hydrogen bonds drawn between every first and fourth residue; mark the pitch of 5.4 Å and 3.6 residues per turn.

Certain amino acids interfere with the formation of an α-helix and are therefore called helix breakers. These include proline, glycine, the charged amino acids (aspartic acid, glutamic acid, arginine, lysine and histidine) and amino acids with bulky side chains such as tryptophan.

The secondary structure of proteins was first studied by Astbury in the 1930s using X-ray diffraction analysis. The structure of the right-handed α-helix was proposed by Linus Pauling and Robert Corey, work for which Pauling received the Nobel Prize in Chemistry.

(b) Beta-pleated sheet. The term β was applied by Pauling and Corey because this structure was described after the α-helix. The word pleated indicates that it is a sheet-like structure, and the important contrast with the helix is that here the polypeptide chain is fully stretched rather than coiled. The amino acids are placed at a distance of 3.5 Å from each other.

The sheet forms when two or more polypeptide chains become interconnected by hydrogen bonds, or when the same chain folds back on itself so that its own amino acids can establish hydrogen bonds among them. The chains may be placed parallel to each other, as in β-keratin of feathers, or anti-parallel, as in fibroin of silk. The antiparallel arrangement allows maximum hydrogen bonding, because the donor and acceptor groups line up directly opposite one another.

This structure is mostly shown by fibrous proteins, although some globular proteins such as lysozyme and carboxypeptidase also contain it.

Diagram

Fig. 6.15 – β-pleated sheet showing extended polypeptide chains side by side in antiparallel arrangement, with hydrogen bonds drawn between adjacent chains.

(c) Random structures. When the pattern of folding of a polypeptide chain is not orderly, it gives rise to a randomly arranged coil. The name is slightly misleading: although the arrangement has no regular repeating pattern, it is highly specific and reproducible for a particular protein chain, and it forms in the same way every time that chain is made.

Compare Table

Tertiary Structure

When a long polypeptide chain, with or without helical regions, is coiled and variously folded upon itself, it acquires a highly specific three-dimensional configuration. This is the tertiary structure, and myoglobin is a standard example.

Tertiary structure is maintained mainly by four types of interaction.

(a) Disulphide bonds. These are covalent linkages formed from the sulphydryl (\(-\text{SH}\)) groups of two cysteine residues, which together produce a cystine residue. The two cysteines may be separated by many amino acids along the chain, or may even lie in two different polypeptide chains. Being covalent, disulphide bonds are the strongest of the four and act as permanent staples holding distant parts of the fold together.

(b) Hydrophobic interactions. Amino acids with non-polar side chains tend to be located in the interior of the polypeptide molecule, associated with other hydrophobic amino acids, while polar amino acids are located on the surface. In a membrane, however, the arrangement is exactly reversed — the hydrophilic amino acids are tucked into the interior of the chain and the hydrophobic ones are found on the surface, which is the face exposed to the non-polar lipid molecules of the membrane. The rule is not “polar inside, non-polar outside” but rather “whichever face fits the surroundings”.

(c) Hydrogen bonds. Amino acids having oxygen-bound hydrogen in their side chains can form hydrogen bonds with the oxygen of a carboxyl group or with the carbonyl groups of peptide bonds. The presence of hydrogen bonds in a protein chain increases its solubility.

(d) Ionic interactions. A negatively charged group such as \(-\text{COO}^-\) in the side chain of one amino acid can interact with a positively charged group such as \(-\text{NH}_3^+\) in the side chain of another.

Side Note

Chaperones are special proteins required for the proper folding of a protein chain during protein synthesis. They do not decide the final shape — the sequence does that — but they prevent a half-made chain from folding wrongly or clumping with its neighbours before it is finished.

Quaternary Structure

A protein is said to possess quaternary structure if it consists of two or more polypeptide chains united by forces other than covalent bonds — that is, not by peptide or disulphide bonds. The forces that stabilise these aggregates are hydrogen bonds and electrostatic or salt bonds formed between residues on the surfaces of the polypeptide chains.

Such proteins are termed oligomeric, and the individual polypeptide chains of which they are composed are called protomers, monomers or sub-units. The same forces that fold a single chain into its tertiary structure act here to link different chains together.

Haemoglobin is the standard example. It has four polypeptide chains — two α and two β sub-units — held together in one functional molecule.

6.3.5 Protein Sequencing by MALDI-MS

Knowing that primary structure determines everything raises an obvious practical question: how do you find out what the sequence actually is? For a modern laboratory, the answer is mass spectrometry, and the technique named in your syllabus is MALDI-MS.

MALDI-MS stands for Matrix Assisted Laser Desorption Ionization – Mass Spectrometry. The difficulty it solves is a real one. A mass spectrometer measures the mass of ions flying through a vacuum, but a protein is a large, fragile molecule — fire a laser directly at it and it simply burns. The trick, and the reason for the word matrix in the name, is to protect the protein while still delivering the energy needed to lift it into the gas phase.

Technique

MALDI-MS (Matrix Assisted Laser Desorption Ionization – Mass Spectrometry) Principle: The protein is embedded in a chemical matrix that absorbs laser energy on its behalf; the matrix vaporises and carries the intact protein or peptide with it as a charged ion, whose mass-to-charge ratio is then measured, and the measured masses are used to read off the amino acid sequence. Steps: 1. The protein is cut into smaller peptides using a specific enzyme such as trypsin. 2. The peptides are mixed with the matrix and allowed to dry on a metal plate. 3. A short laser pulse strikes the spot; the matrix absorbs the energy, vaporises, and ionises the peptides without destroying them. 4. The ions are accelerated in an electric field and travel down a flight tube; lighter ions arrive at the detector sooner than heavier ones. 5. The masses obtained are compared with one another, and the difference in mass between successive peaks identifies the amino acid that lies between them. Used for: Determining the amino acid sequence of a protein, identifying an unknown protein by matching its peptide masses against a database, and checking that a recombinant protein such as insulin produced in a fermenter has exactly the intended sequence.

Flowchart

FC 6.4 – Protein sequencing by MALDI-MS: purified protein → enzymatic digestion into peptides → mix with matrix and dry on plate → laser pulse desorbs and ionises → ions accelerated down flight tube → detector records mass-to-charge ratios → mass differences between peaks read as amino acid sequence

6.3.6 Properties of Proteins

Physical Properties

Proteins are colourless and tasteless in general, and are homogenous and crystalline.
They may be spherical or fibrillar in shape, and are accordingly called globular or fibrillar proteins.
The molecular weight of proteins ranges from \(5 \times 10^3\) to \(1 \times 10^6\).
Proteins show colloidal nature. Because of their huge size they diffuse at a very slow rate, and they show the Tyndall effect — they scatter light passing through the solution and make it appear visibly turbid.
Proteins are amphoteric, for the same reason amino acids are: they carry a carboxyl group at one end and an amino group at the other. Each protein therefore has its own isoelectric point (pI), the pH at which the number of anions equals the number of cations and the net charge is zero. Placed in an electric field, a protein migrates in a direction that depends on the net charge it carries, and therefore on the pH of the medium.
Proteins are least soluble at their isoelectric point, and their solubility increases as the medium is made either more acidic or more alkaline. Some proteins that are poorly soluble in water dissolve when neutral salts such as NaCl are added — an effect called salting-in. Conversely, adding a high concentration of neutral salt precipitates proteins out of aqueous solution, which is called salting-out. Milk protein (casein) is precipitated easily at or near its isoelectric point.
Hair remains stretched in dry conditions because of a reduction in the number of intra-chain hydrogen bonds in keratin. When moistened with water, hydrogen bond formation between amino acids is promoted, and this makes the hair flexible and smooth — which is why shaving and hair-setting are easier after wetting.

Denaturation

Denaturation is the best known property of proteins, and it is usually followed by coagulation. It occurs when a protein is exposed to conditions different from those it normally works in.

Physical agents that cause denaturation include shaking, heat treatment, cooling, freezing, rubbing, high pressure and UV rays. Chemical agents include organic solvents such as alcohols and acetone, aromatic anions, and anionic detergents such as sodium dodecyl sulphate (SDS).

What is actually destroyed in denaturation is the shape, not the chain. The weak hydrogen bonds, hydrophobic interactions and ionic bonds that held the secondary and tertiary structure together are disrupted, while the covalent peptide bonds of the primary structure remain intact. The sequence survives; the folding does not.

Some proteins cannot regain their original form once denatured. Many others can be brought back to their original state if normal conditions are restored, and this reversible process is called renaturation or refolding.

Curiosity Corner

Q1. If an enzyme is only a protein, why does boiling an egg — or boiling an enzyme — destroy it forever?

A. Because a protein’s function depends on its precise three-dimensional shape, and that shape is held by weak bonds. Heat supplies enough energy to break hydrogen bonds, hydrophobic interactions and ionic bonds, so the fold collapses even though the peptide backbone is untouched. Egg white turns solid and opaque as the unfolded chains tangle and coagulate. An enzyme treated the same way loses the exact geometry of its active site, and with it the ability to recognise its substrate. In many proteins the tangling is irreversible, which is why a boiled egg never becomes raw again.

Chemical Properties

Hydrolysis: Proteins can be hydrolysed to their constituent amino acids by treating them with concentrated acids, bases or proteolytic enzymes.
Ninhydrin reaction: All α-amino acids are oxidised by ninhydrin (triketo-hydrindene) to form a blue coloured compound. This reaction is used for the quantitative estimation of amino acids. Proline and hydroxyproline are exceptions and produce a yellow coloured complex.
Biuret reaction: Compounds having peptide linkages produce a characteristic pinkish violet colour when treated with an alkaline solution of copper sulphate (Biuret reagent). The reaction is given by all proteins except dipeptides, because at least two peptide bonds are needed for the copper ion to be held.
Xanthoproteic test: When proteins are boiled with concentrated nitric acid, they yield a yellow coloured product which changes to orange on the addition of alkali. This reaction is given due to the presence of the aromatic amino acids tyrosine, tryptophan and phenylalanine in the protein.
Reaction with Millon’s reagent: Millon’s reagent is a solution of mercuric and mercurous nitrates in nitric acid. When proteins are heated with it, a red coloured compound is formed. The reaction is specific to tyrosine, and tryptophan also responds to it.

Technique

Biuret test for proteins Principle: Cupric ions in an alkaline medium form a coloured coordination complex with the nitrogen atoms of peptide bonds; the more peptide bonds present, the deeper the colour. Steps: Take the sample solution in a test tube → add the alkaline copper sulphate reagent → mix and let it stand → observe the deep blue-violet colour. Used for: Detecting proteins in a sample and estimating them roughly by depth of colour. It is ISC practical 9(ii), and because the colour depends on peptide bonds rather than on any particular amino acid, it responds to almost every protein.

Exam Corner

Q. A student tests two food samples. Sample A gives a violet colour with Biuret reagent but no yellow colour with concentrated nitric acid. Sample B gives both. What can be concluded?

A. Both samples contain protein, since both give a positive Biuret test. Only Sample B contains aromatic amino acids such as tyrosine, tryptophan or phenylalanine, because only the aromatic ring is nitrated to give the yellow xanthoproteic product. The two tests answer different questions — Biuret asks “is protein present?”, xanthoproteic asks “which amino acids are in it?”

6.3.7 Classification of Proteins

Proteins are classified on the basis of their chemical composition into three groups.

Flowchart

FC 6.5 – Classification of proteins by chemical composition: Proteins → Simple (only amino acids) / Complex or conjugated (amino acids + prosthetic group → glycoproteins, lipoproteins, metalloproteins, chromoproteins, phosphoproteins, nucleoproteins) / Derived (from native proteins by heat, enzymes or chemicals → primary derived and secondary derived)

1. Simple Proteins

These are made up of only amino acids. They may be fibrous, as in keratin, or globular, as in albumin, globulin and histones.

2. Complex or Conjugated Proteins

In addition to amino acids, these proteins contain a non-proteinaceous substance called a prosthetic group. They are classified further on the basis of the nature of that group.

Glycoproteins: They contain carbohydrates as their prosthetic group. If the amount of carbohydrate is small (less than 4%) they are called glycoproteins; if it is more than 4% they are called mucoproteins. Examples are albumin and globulins among glycoproteins, and mucin of saliva among mucoproteins.
Lipoproteins: They form complexes with lipids, for example lipovitelline from egg yolk.
Metalloproteins: These proteins are linked with various metals. The metal may be strongly bound, as in the binding of heavy metals such as Ag, Zn, Hg and Cu with collagen, albumin or casein, or weakly bound, as in the binding of calcium with calmodulin.
Chromoproteins: The prosthetic group of these proteins is usually a coloured compound, for example haemoglobin, hemocyanin and cytochrome.
Phosphoproteins: These proteins are linked with inorganic phosphate (\(\text{H}_3\text{PO}_4\)), for example casein of milk and vitelline of egg yolk.
Nucleoproteins: These proteins are conjugated with nucleic acids, for example histones and protamines. It is the association of histones with DNA that packages the genome inside the nucleus.

3. Derived Proteins

These are derived from native proteins by the action of heat, enzymes or chemical agents. The group also includes artificially produced synthetic polypeptides. Derived proteins are further classified into two groups.

(a) Primary derived proteins. Here the size of the protein molecule is not altered — the chain is the same length, but its state has changed.

Coagulated proteins are produced by the action of heat or alcohol on proteins, for example coagulated egg white.
Metaproteins are produced by further action of acid or alkali at about 30–60°C, for example acid and alkali metaproteins.
Proteans appear as the first products of acid hydrolysis or enzyme hydrolysis of native proteins, for example edestan and myosan.

(b) Secondary derived proteins. These are derivatives of proteins in which hydrolysis has occurred, and they are, as a rule, smaller than the native proteins.

Proteoses are produced when hydrolysis proceeds beyond the level of metaproteins.
Peptones are formed as hydrolysis continues further.
Oligopeptides, containing a small number of amino acid units.
Polypeptides, the smallest products still recognisable as chains.

Notice that (i) to (iv) form a sequence of decreasing size. If you can remember that hydrolysis chops a protein into steadily smaller pieces, you do not need to memorise the four names separately — proteose, peptone, oligopeptide, polypeptide is simply the order in which the fragments get shorter.

Compare Table

6.3.8 Functions of Proteins

The functions of proteins are best appreciated all at once, because the range is what is remarkable about them.

S. No.	Function	Example
1.	Catalytic	Enzymes
2.	Contraction	Actin, myosin
3.	Protection	Fibrin, antibodies, interferons
4.	Regulatory — gene regulation	Histones, repressor proteins
5.	Regulatory — hormonal regulation	ACTH, insulin
6.	Regulatory — intracellular functions	Calmodulin
7.	Structural	Collagen, elastin, keratin
8.	Transport	Albumin (fatty acids, bilirubin), haemoglobin and myoglobin (\(\text{O}_2\)/\(\text{CO}_2\)), transferrin (iron)
9.	Storage and nutrition	Casein, ovalbumin, glutelin
10.	Carriers and receptors	Present on the membranes of different cells

Look down the list and one point stands out. Interferons, insulin and antibodies all appear here — and all three are manufactured commercially today by recombinant bacteria and cultured cells. Protein chemistry is not merely descriptive background to biotechnology; the industry’s most valuable products are proteins.

Try Yourself

A protein is heated to 90°C and then cooled back to room temperature, but it does not recover its activity. Write down, in order, which bonds were broken and which survived, and use that to explain why cooling alone was not enough to restore the protein.

6.4 Lipids

Ask a chemist to define a lipid and you will get an unusual answer. Carbohydrates and proteins are defined by what they are made of; lipids are defined largely by how they behave. They are a rather heterogeneous class of compounds whose common feature is that they dissolve in organic solvents and do not dissolve in water. Mixed with water, they do not form a solution at all — they form an emulsion, which is why oil floats in droplets on the surface of dal and why milk is opaque.

The term lipid was introduced by Bloor. Chemically, lipids are compounds of carbon, hydrogen and oxygen in which the ratio of hydrogen to oxygen is very high — much higher than in a carbohydrate. That high proportion of hydrogen is exactly why fats yield more than twice as much energy per gram as sugars do.

Definition

Lipids are esters of fatty acids with various alcohols, which are insoluble in water and soluble in organic solvents.

According to Bloor, lipids can be classified into three categories: simple lipids, conjugated (compound) lipids and derived lipids.

Flowchart

FC 6.6 – Classification of lipids: Lipids → Simple lipids (esters of fatty acids with alcohols → fats and oils, waxes) → Conjugated or compound lipids (phospholipids, glycolipids, sulpholipids, lipoproteins) → Derived lipids (products of hydrolysis → fatty acids, glycerol, sterols, steroids)

6.4.1 Fatty Acids

Since every lipid is an ester of a fatty acid, it makes sense to begin with the acid itself.

Fatty acids are characteristic constituents of almost all the lipids. Each one has a long hydrocarbon chain and a terminal carboxyl group. The shortest fatty acid commonly listed among those found in living organisms is butyric acid, with 4 carbon atoms, while the most abundant fatty acids are 16 or 18 carbons long.

The general formula of fatty acids is written as \(\text{R}-\text{COOH}\), or more informatively as:

\[\text{CH}_3(\text{CH}_2)_n\text{COOH}\]

where \(\text{R}\) is the hydrocarbon chain and \(n\) is the number of \(-\text{CH}_2-\) groups in it. The value of \(n\) may vary from 2, for butyric acid, to 22, for lignoceric acid.

Why the Carbon Number is Always Even

Look down any list of natural fatty acids and you will notice something odd: the carbon numbers are 4, 6, 8, 10 and so on — never 5, 7 or 9. This is not a coincidence.

Living cells synthesise fatty acids by joining acetyl CoA units end to end, and acetyl CoA contributes two carbons at a time. A chain built two carbons at a time can only ever have an even number of carbons.

Why Fatty Acids Behave in Two Ways at Once

The \(-\text{COOH}\) group of a fatty acid is strongly polar. At physiological pH it ionises by losing a proton, and it has a tendency to attract water — it is hydrophilic. The hydrocarbon chain at the other end is non-polar and therefore hydrophobic.

So a single fatty acid molecule has a water-loving head and a water-hating tail. Molecules that possess both hydrophilic and hydrophobic properties in this way are called amphipathic.

Definition

An amphipathic molecule is one that contains both a hydrophilic (water-attracting) region and a hydrophobic (water-repelling) region within the same molecule.

This one property is the reason cell membranes exist. Placed in water, amphipathic molecules arrange themselves automatically so that their heads face the water and their tails hide from it — and that spontaneous arrangement is the foundation of every membrane in every cell.

Saturated and Unsaturated Fatty Acids

The hydrocarbon chain of a fatty acid may be saturated or unsaturated.

A saturated fatty acid does not contain any double bond (\(-\text{C}=\text{C}-\)) or triple bond (\(-\text{C}\equiv\text{C}-\)) in its hydrocarbon chain. It therefore carries the maximum possible number of attached hydrogens — it is “saturated” with hydrogen. The general formula of a saturated fatty acid is:

\[\text{C}_n\text{H}_{2n+1}\text{COOH}\]

An unsaturated fatty acid contains one or more double bonds, and so carries fewer hydrogens than the maximum.

The shape consequence is what matters. A saturated chain is straight, so saturated molecules pack closely together and hold each other firmly — the substance is solid. Each double bond puts a permanent kink in the chain, so unsaturated molecules cannot pack neatly, slide past one another easily, and the substance stays liquid.

Diagram

Fig. 6.16 – A straight saturated fatty acid chain drawn beside a kinked unsaturated chain with one \(\text{C}=\text{C}\) double bond; show saturated molecules packing closely in a block and unsaturated molecules packing loosely.

Curiosity Corner

Q5. Butter is solid and mustard oil is liquid at the same room temperature. What difference in their molecules decides this?

A. The difference is the proportion of saturated to unsaturated fatty acids. Butter is rich in saturated fatty acids, whose straight chains stack together tightly and hold firm at room temperature, giving a solid. Mustard oil is rich in unsaturated fatty acids, whose double bonds kink the chains so that they cannot stack neatly; the molecules slip past one another and the substance flows. A simple lipid that is solid at room temperature is called a fat, and one that is liquid at room temperature is called an oil — but chemically they are the same kind of molecule, differing only in the shape of their chains.

Examples of saturated fatty acids and their sources are summarised below.

S. No.	Fatty acid	No. of C atoms	Source
1.	Butyric acid	4-C	Butter
2.	Valeric acid	5-C	Butter
3.	Caproic acid	6-C	Butter
4.	Caprylic acid	8-C	Fats of plant origin, butter
5.	Capric acid	10-C	Fats of plant origin, butter
6.	Lauric acid	12-C	Laurels, coconut oil, cinnamon, palm kernel, spermaceti
7.	Myristic acid	14-C	Myrtles, coconut oil, nutmeg, palm kernel
8.	Palmitic acid	16-C	Common in all plant and animal fats
9.	Stearic acid	18-C	Common in all plant and animal fats
10.	Arachidic acid	20-C	Groundnut oil
11.	Behenic acid	22-C	Seeds
12.	Lignoceric acid	24-C	Groundnut oil, cerebrosides

6.4.2 Simple Lipids

Simple lipids are the esters of fatty acids with alcohols. They include fats, oils and waxes.

Fats and Oils

These are the esters of fatty acids with glycerol, an alcohol carrying three hydroxyl groups. Because glycerol has three \(-\text{OH}\) groups, it can be esterified in stages:

When one hydroxyl group of glycerol is esterified with a fatty acid, a monoacylglycerol (monoglyceride) is formed.
When two hydroxyl groups are esterified, a diacylglycerol (diglyceride) is formed.
When all three hydroxyl groups are esterified, a triacylglycerol or triglyceride is formed. This is also called a neutral fat.

When all three fatty acids of a fat are identical, it is said to be a simple fat. If they are not identical, it is called a mixed triglyceride. Most naturally occurring fats are mixed triglycerides.

The standard example of a simple fat is tripalmitin, formed from palmitic acid and glycerol, which is abundant in palm oil. Its chemical formula is \(\text{C}_{51}\text{H}_{98}\text{O}_{6}\).

Diagram

Fig. 6.17 – Structure of tripalmitin: a glycerol backbone with three ester linkages, each to a \(-(\text{CH}_2)_{14}\text{CH}_3\) palmitate chain; label the glycerol backbone and one ester bond.

Compare Table

Waxes

If the fatty acid is esterified with a monohydric alcohol of high molecular weight rather than with glycerol, a wax is formed. The fatty acids of waxes usually have 14 to 34 carbon atoms, and the alcohols have 16 to 36 carbon atoms.

Because both components have fully reduced hydrocarbon chains, waxes are insoluble in water and chemically inert. That inertness is precisely what makes them useful: they serve as barriers against water loss in insects, birds and furred animals. In many animals the epidermis secretes true waxes to reduce wetting or water loss. Lanolin, secreted by the skin of most fur-bearing animals, consists of palmitic acid, oleic acid or stearic acid together with cholesterol. Bees wax, secreted by honey bees, consists of palmitic acid esterified with either hexacosanol or triacontanol. Spermaceti wax is found in the head of the sperm whale.

Waxes also serve as a protective covering on fruits and leaves in the form of cuticle. The waxy bloom you can rub off the skin of a mango or a brinjal is exactly this — a water barrier the plant has laid down for itself.

6.4.3 Conjugated or Compound Lipids

These are the esters of fatty acids containing some other group in addition to an alcohol and a fatty acid. On the basis of the nature of that additional group, they are classified into four groups.

Phospholipids. In addition to a fatty acid and an alcohol, these contain a phosphoric acid residue, and they may also contain nitrogen-containing bases. Phospholipids are always present in animal cell membranes, where they maintain selective permeability. They differ from diglycerides in having one hydrophilic polar head and two hydrophobic non-polar tails, which makes them amphipathic; for this reason they are also called polar lipids. The commonest example is lecithin (phosphatidyl choline), present in both plants and animals, which contains glycerol, fatty acids, phosphoric acid and the nitrogenous base choline.
Glycolipids. These are compounds of fatty acids with a carbohydrate, containing nitrogen but no phosphoric acid. Cerebrosides, found in the myelin sheath of nerve fibres and in many other tissues, are the standard example.
Sulpholipids. These are similar to cerebrosides except that, in place of the sugar, sulphuric acid is present in the molecule.
Lipoproteins. These are composed of lipid materials bound with proteins. They are an important component of the cell membrane, and their special structure is responsible for the active permeability properties of the cell. On the basis of their electrophoretic mobility, plasma lipoproteins have been divided into a-, b- and pre-b lipoproteins and chylomicrons; the last are very large molecules which do not show electrophoretic mobility.

Diagram

Fig. 6.18 – A phospholipid molecule drawn as a polar head with two non-polar tails, beside a diagram of phospholipids arranged as a bilayer in a membrane with heads facing the water on both sides.

Exam Corner

Q. Why does a cell membrane form spontaneously when phospholipids are shaken in water, without any enzyme directing the process?

A. Because the arrangement is driven by the molecules themselves. Water repels the hydrophobic tails and attracts the hydrophilic heads. The only arrangement that satisfies both demands at once is a double sheet with the tails tucked inward and the heads facing outward on both sides. No instruction is needed — it is the lowest-energy arrangement available, and the molecules fall into it on their own.

6.4.4 Derived Lipids

Substances derived from the hydrolysis of simple and complex lipids are called derived lipids. They include fatty acids (both saturated and unsaturated), glycerol, steroids, sterols, ketone bodies and prostaglandins.

Sterols are steroids with a hydroxy group at C-3 and a side chain of about 10 carbon atoms at C-17. The most important example is cholesterol, which is found in the cell membrane of many organisms and in blood and bile. It is the precursor of the steroid hormones — cortisone, testosterone, estrogen and progesterone.

Diagram

Fig. 6.19 – Structure of cholesterol showing the four fused rings, the \(-\text{OH}\) group at C-3 and the hydrocarbon side chain at C-17.

Cholesterol has an unfortunate public reputation, but the cell cannot manage without it. Wedged between the phospholipids of a membrane, it stops the membrane from becoming either too rigid in the cold or too fluid in the heat. Ergosterol, which occurs in plants and yeast, is important as a precursor of vitamin D.

6.4.5 Properties of Lipids

Fats and oils are triesters of glycerol with saturated or unsaturated fatty acids. Their chemical reactions are therefore of two kinds: reactions of the ester group, repeated three times over, and reactions of the carbon–carbon double bonds.

1. Enzymatic hydrolysis. Lipases in the alimentary canal of animals act as efficient catalysts for the hydrolysis of the ester bonds of triacylglycerols, releasing three molecules of fatty acid and one mole of glycerol. In animals, however, mono- and di-acyl glycerols are absorbed very rapidly, and therefore very little free glycerol is actually formed in the alimentary canal.

2. Saponification. When boiled with sodium or potassium hydroxide solution, fats are hydrolysed into sodium or potassium salts of long chain fatty acids. These salts are what we call soaps, and the hydrolysis itself is referred to as saponification.

The soaps of \(\text{Na}^+\) and \(\text{K}^+\) are water soluble and are used for cleansing purposes, while lead (Pb) soaps are called grease and are used as lubricants.

Definition

Saponification number is the number of milligrams of KOH (potassium hydroxide) required to saponify 1 gram of fat.

A soap molecule works precisely because it is amphipathic, in the way described earlier. Its hydrocarbon tail dissolves in the grease on a dirty plate while its ionic head stays in the water — so the grease is dragged into the water rather than left behind. This reaction is demonstrated in ISC practical 16.

3. Hydrogenation. On catalytic hydrogenation at low pressures, hydrogen adds across the carbon–carbon double bonds of the acid components of triacylglycerols. This converts unsaturated chains into saturated ones, and results in the formation of saturated glycerides which are solid fats at room temperature. The process is called hardening, and it is how liquid vegetable oil is converted industrially into solid vanaspati.

4. Rancidification. When fats and oils are left exposed to warm, moist air, they develop an offensive odour and off-taste. They are said to have become rancid. Rancidification is caused by two types of reaction.

In hydrolytic rancidification, the ester linkages of a glyceride are hydrolysed to give back the original fatty acid. For example, when left uncovered, the butter fat tributyrin is hydrolysed by microorganisms in the air, producing butyric acid, which has an unpleasant odour.
In oxidative rancidification, atmospheric oxygen attacks the carbon–carbon double bonds in the unsaturated side chains of the glyceride. Short chain volatile carboxylic acids and aldehydes are produced, and these have extremely bad odours.

Notice which fats spoil faster. Since oxidative rancidity attacks double bonds, highly unsaturated oils go rancid more readily than saturated fats — which is why cooking oil is sold in dark bottles and why airtight storage matters.

5. Dehydration. When fat is heated with \(\text{NaHSO}_4\) or \(\text{KHSO}_4\), the glycerol in it is dehydrated to an unsaturated aldehyde called acrolein. Acrolein has a sharp, pungent odour, and this test is therefore used to confirm the presence of glycerol in a fat.

Technique

Acrolein test for fats Principle: Potassium or sodium bisulphate acts as a dehydrating agent; it removes water from the glycerol part of a fat, converting it into the unsaturated aldehyde acrolein, which is recognised by its pungent smell. Steps: Take the sample in a dry test tube → add a little potassium bisulphate (\(\text{KHSO}_4\)) or sodium bisulphate (\(\text{NaHSO}_4\)) → heat strongly → detect the pungent, irritating odour of acrolein. Used for: Confirming the presence of glycerol, and therefore of fat, in a sample. This is ISC practical 9(iii), where it is performed alongside the Sudan III test — Sudan III being a red fat-soluble dye that stains lipids, triglycerides and lipoproteins red.

6. Halogenation. When unsaturated fatty acids of lipids and their esters react with halogens such as bromine and iodine at room temperature in acetic acid or methanol solution, the iodine or bromine adds across the unsaturated bonds.

This reaction gives the iodine number, which is a measure of the degree of unsaturation of the fatty acids in a fat.

Definition

Iodine number is the number of grams of iodine absorbed by 100 grams of fat.

Since each double bond consumes iodine, a higher iodine number means a more unsaturated fat. For example, the iodine number of cotton seed oil is 109, of corn oil is 127, and of soybean oil is 133.

Numerical 6.3

Three oil samples A, B and C are tested and found to have iodine numbers of 88, 127 and 141 respectively. (a) Arrange the three samples in increasing order of unsaturation. (b) Which of the three is most likely to be solid at room temperature, and which is most likely to turn rancid fastest on standing in air? Give a reason for each answer.

Try Yourself

Look at the nutrition label on any packet of biscuits or namkeen in your house and find the entries for “saturated fat” and “trans fat”. Using what you have learnt about chain shape and hydrogenation, explain what industrial process is likely to have produced the fat listed there, and why the manufacturer used it instead of a liquid oil.

6.5 Vitamins

Vitamins are unlike every other biomolecule in this chapter. Carbohydrates, proteins and lipids are needed by the gram; vitamins are needed by the milligram or even the microgram. Yet a diet lacking a few milligrams of one vitamin can be fatal, while a diet short of a few grams of carbohydrate is merely a light meal.

The term vitamin was coined by Funk, who thought them to be vital amines. The name stuck even after it became clear that many vitamins are not amines at all.

Definition

Vitamins are organic micronutrients which cannot be synthesised by the body, and which must therefore be supplied in the diet in their natural form or as precursors.

Two words in that definition carry the weight. Micronutrient explains the tiny quantities involved. Cannot be synthesised explains why they must come from food at all — the body has the chemistry to build almost everything else it needs, but not these.

Vitamins have a regulatory role, and most of them act as cofactors of vital enzymes of the body. Their deficiency is recognised by the development of characteristic deficiency disorders, collectively called hypovitaminosis.

Why a Deficiency Shows Up So Quickly

This is worth pausing on, because it explains the whole subject. If a vitamin is the cofactor of an enzyme, then without that vitamin the enzyme cannot work — no matter how much of the enzyme protein the cell has made. One missing milligram of vitamin can therefore shut down an entire metabolic pathway.

That is why vitamin deficiency diseases are specific rather than general. Scurvy is not “poor nutrition in general”; it is the precise consequence of collagen that cannot be built properly. Beri-beri is the precise consequence of pyruvate that cannot be processed.

6.5.1 Types of Vitamins

Vitamins are a heterogeneous group of chemicals with no common structure. They are therefore classified not by what they are made of but by what they dissolve in.

Flowchart

FC 6.7 – Types of vitamins: Vitamins → Fat soluble vitamins (A, D, E, K) → Water soluble vitamins (B complex and C)

This division has real consequences for health. Fat-soluble vitamins are absorbed along with dietary fat and can be stored in the liver and fatty tissue, so a deficiency takes months to appear — but an excess cannot be flushed out easily and can accumulate to harmful levels. Water-soluble vitamins are not stored to any great extent; any surplus is excreted in urine, so they must be supplied regularly, and a deficiency appears comparatively soon.

Compare Table

Exam Corner

Q. A person eats a diet rich in vitamin C for one day and then stops entirely. Would they be protected from scurvy for the next several months?

A. No. Vitamin C is water soluble and is not stored in the body in any useful quantity — the surplus taken in on that one day is simply excreted in the urine within a day or two. Water-soluble vitamins must be supplied regularly. Had the vitamin been A or D, the answer would have been different, because those are stored in the liver.

6.5.2 Fat Soluble Vitamins

Name of vitamin	Chemical nature	Functions	Hypovitaminosis	Sources
Vitamin ‘A’	Retinol	Maintenance of normal growth, repair of epithelial tissues, the function of the retina (a constituent of visual purple – rhodopsin) and reproductive activity	Night blindness, xerophthalmia	Milk, butter, cheese, liver, carrot, tomato and leafy vegetables
Vitamin ‘D’	Calciferol	Regulates Ca and \(\text{PO}_4\) metabolism	Rickets in children, osteomalacia in adults	Can be formed in skin under sun light, fish liver oil, butter, egg
Vitamin ‘E’	Tocopherol	Anti-oxidant, required for normal functioning of muscular and reproductive systems	Sterility	Wheat germ, butter, egg, meat, liver, fish, chicken, fruits, green leaves
Vitamin ‘K’	Phylloquinone	Forms prothrombin, maintains liver function	Haemorrhage, irregular growth	Colon bacteria, vegetables, liver, egg, yolk, cheese and green leaves

Two entries in this table are worth reading twice.

Vitamin A is a constituent of rhodopsin, the visual pigment of the retina. When it runs short, the eye can no longer regenerate this pigment fast enough in dim light — which is exactly what night blindness is. The link between the chemistry and the symptom is direct.

Vitamin K forms prothrombin, a protein needed for blood to clot. Without it, a small injury does not stop bleeding, and haemorrhage results. Note also its unusual source: colon bacteria. Some of your vitamin K is manufactured for you by the bacteria living in your own intestine — a natural example of the same partnership that biotechnology exploits deliberately in a fermenter.

6.5.3 Water Soluble Vitamins

1. Vitamin B₁

Chemical nature: Thiamine.
Sources: Unpolished cereal grains, yeast, egg, yolk, liver, green vegetables.
Deficiency disease: Beri-beri.
Functions: Regulates carbohydrate metabolism as the co-enzyme TPP (thiamine pyrophosphate) of the decarboxylase enzyme for pyruvate metabolism; promotes appetite; necessary for normal functioning of the digestive system, nervous system, heart and muscles.

Note the source: unpolished cereal grains. Thiamine sits in the outer layer of the rice grain, and polishing removes it. Beri-beri became widespread in populations that switched to polished white rice — a deficiency disease created not by lack of food but by a change in how food was processed.

2. Vitamin B₂

Chemical nature: Riboflavin.
Sources: Liver, cheese, milk, egg, white meat, green leaves.
Deficiency diseases: Cataract, loss of hair, inflamed tongue of magenta colour, scaly dermatitis.
Functions: Cellular respiration and growth, carbohydrate metabolism; constituent of the co-enzymes FAD (flavin adenine dinucleotide) and FMN (flavin mononucleotide).

3. Vitamin B₃

Chemical nature: Nicotinic acid (niacin).
Sources: Milk, liver, legumes, coffee, tea, vegetables, egg.
Deficiency disease: Pellagra.
Functions: Part of the co-enzymes NAD and NADP; controls cellular respiration.

4. Vitamin B₅

Chemical nature: Pantothenic acid.
Sources: Yeast, liver, peas, egg, whole grains.
Deficiency symptoms: Gut malfunctioning, dermatitis, greying of hair.
Functions: Part of co-enzyme A; regulates carbohydrate and lipid metabolism.

5. Vitamin B₆

Chemical nature: Pyridoxine, and its related forms pyridoxal and pyridoxamine.
Sources: Yeast, cereals, grains, egg, yolk; also synthesised by intestinal bacteria.
Deficiency diseases: Dermatitis, hypochromic microcytic anaemia.
Functions: Important for protein metabolism; essential for the synthesis of haem for haemoglobin and cytochromes.

6. Vitamin B₇ (Biotin)

Sources: Egg yolk, milk, liver, intestinal bacteria, chocolate.
Deficiency symptoms: Dermatitis, paralysis, muscle pain (myalgia).
Functions: Co-enzyme in \(\text{CO}_2\) fixation (carboxylase) and in deamination reactions.

7. Vitamin B₉ (Folic acid)

Chemical nature: Pteroylglutamic acid.
Sources: Yeast, liver, green vegetables, soyabean.
Deficiency diseases: Macrocytic anaemia, leucopenia, sprue.
Functions: Part of the co-enzyme of nucleic acid metabolism; necessary for maturation of RBCs.

8. Vitamin B₁₂

Chemical nature: Cyanocobalamine.
Sources: Milk, liver, kidney, meat, egg and cheese.
Deficiency disease: Pernicious anaemia.
Functions: Necessary for maturation of RBCs and for nucleic acid metabolism.

9. Vitamin C

Chemical nature: Ascorbic acid.
Sources: Nearly all fruits, especially citrus fruits, and vegetables. It is heat labile.
Deficiency disease: Scurvy, marked by bleeding from the gums.
Functions: Maintains the intercellular material of cartilage, dentine and bone, and the capillary walls; necessary for collagen synthesis; important for wound healing and repair; regulates protein metabolism and helps in the formation of blood.

The phrase “heat labile” in the sources line is a practical warning, not a technicality. Vitamin C is destroyed by prolonged cooking, which is why citrus fruits and salads are better sources than boiled vegetables.

Scurvy also illustrates neatly how a vitamin deficiency works through a protein. Vitamin C is required for the enzymatic step that stabilises collagen. Without it, collagen is made but is structurally weak — so gums bleed, wounds refuse to heal and old scars reopen.

6.5.4 Co-enzyme Forms of Water Soluble Vitamins

Most water-soluble vitamins do not act on their own. They are first converted into a related compound which then attaches to an enzyme and does the actual chemical work. That compound is the vitamin’s co-enzyme form, and this is the reason vitamins are needed at all.

To follow this, three terms are needed.

Many enzymes consist of a protein part and a non-protein part. The protein part is called the apoenzyme, and the non-protein part is called the cofactor. Together they form the complete, working enzyme, called the holoenzyme. Neither part can function independently — the apoenzyme has the specificity, the cofactor has the chemical capability, and the reaction needs both.

Definition

A co-enzyme is an organic, non-protein cofactor which remains loosely attached to the apoenzyme and acts as an acceptor or donor of an atom or a group that is removed from, or added to, the substrate.

Because a co-enzyme carries chemical groups from one reaction to another in this way, co-enzymes are also called second substrates. NAD, FAD, FMN and CoQ act as hydrogen carriers, ATP donates a phosphate group, and CoA donates the acyl group.

Diagram

Fig. 6.20 – An apoenzyme (protein) shown with a co-enzyme fitting into it to form the holoenzyme, with the substrate binding at the active site; label apoenzyme, co-enzyme and holoenzyme.

Compare Table

Most of the co-enzymes are derived from the B group of vitamins, which is precisely why a shortage of a B vitamin brings so many different reactions to a halt at once.

Vitamin	Co-enzyme form	Type of reaction catalysed
Vitamin B₁ (Thiamine)	TPP (thiamine pyrophosphate)	Carboxylation reaction
Vitamin B₂ (Riboflavin)	FMN (flavin mononucleotide), FAD (flavin adenine dinucleotide)	Redox (oxidation–reduction) reaction
Vitamin B₃ (Niacin)	NAD (nicotinamide adenine dinucleotide)	Dehydrogenation reaction
Vitamin B₅ (Pantothenic acid)	Coenzyme A	Acetyl transfer reaction
Vitamin B₆ (Pyridoxine)	Pyridoxal phosphate	Transaminase reaction
Vitamin B₇ (Biotin)	Biocytin (biotin–lysine complex)	Carboxylase reaction
Vitamin B₉ (Folic acid)	Tetrahydrofolic acid	Transfer of one carbon
Vitamin B₁₂ (Cyanocobalamine)	Deoxyadenosyl cobalamine	Molecular rearrangement reaction

Try Yourself

Look at the table of co-enzyme forms above and at the list of enzyme classes you will meet in Section 6.6. For any two vitamins in the table, write down which class of enzyme its co-enzyme form is most likely to serve, and justify your choice from the type of reaction catalysed.

6.6 Enzymes

A chemical reaction that takes hours in a laboratory flask, at high temperature and in strong acid, may be completed inside a living cell in a fraction of a second, at body temperature and at near-neutral pH. The cell manages this with enzymes.

Definition

Enzymes are biocatalysts — substances produced by living cells which greatly increase the rate of a specific biochemical reaction without themselves being consumed in it.

The word biocatalyst is exact. In the presence of an enzyme certain chemical changes go on very rapidly; in its absence the same changes go on very slowly, or scarcely at all. Because of enzymes, chemical changes that would otherwise require harsh conditions can proceed in living cells at relatively low temperature.

How the Subject Began

The role of enzymes in digestion and fermentation was suspected long before anyone knew what an enzyme was. In the seventeenth century Van Helmont suggested that the digestion of food involved certain “ferments”. In 1812 Kirchoff reported that starch can be hydrolysed not only by dilute acids but also by a ferment. Payen and Persoz later extracted the enzyme diastase from malt extract.

Pasteur demonstrated that living yeast cells were responsible for alcoholic fermentation, and the term enzyme itself was proposed by Kuhne in 1878. Pasteur’s conclusion, however, carried an assumption with it — that fermentation required a living cell. Testing that assumption produced one of the decisive experiments in the history of biotechnology.

Real Incident / Discovery

In 1897 Buchner was grinding yeast cells with sand to break them open, and pressed out the juice. The cells in it were unquestionably dead. He added sugar to the extract, expecting nothing — and the liquid began to ferment, producing alcohol and carbon dioxide exactly as living yeast would have done. Buchner had shown that fermentation does not need a living cell at all; it needs only the substances the cell contains. He named the active extract zymase, now recognised as a group of enzymes involved in the alcoholic fermentation of sugars. Every industrial process in which an isolated enzyme is used in place of a whole organism rests on the point Buchner established that day.

The question of what enzymes actually were chemically was settled a generation later, by the crystallisation of urease described at the start of this chapter, and by Northrop, who went on to crystallise pepsin and trypsin.

6.6.1 Chemical Nature of Enzymes

Enzymes are soluble substances produced in living cells, and are almost always of the nature of proteins. This is why everything you learnt in Section 6.3 about protein structure applies directly here: an enzyme has a primary sequence, folds into a precise three-dimensional shape, and loses its activity when that shape is destroyed by heat or extremes of pH.

There is one exception. Certain RNA molecules have been found which can cut, splice and assemble themselves. These catalytic RNAs are now considered enzymes and are called ribozymes. T. R. Cech and S. Altman were awarded the Nobel Prize in 1989 for their work on ribozymes.

Ribozymes force a careful statement of the rule:

Definition

All enzymes are proteins, but all proteins are not enzymes — with ribozymes as the recognised exception on the first half of that statement.

6.6.2 Characteristics of Enzymes

The chief characters of enzymes may be summarised as follows.

They are always produced by the activity of the living protoplasm.
They are effective in very small quantities.
They are sensitive to heat, and most enzymes are rendered useless at high temperature.
They are generally very specific in their action — one enzyme brings about only one type of chemical change, and that too involving only one type of substrate.
Most, but not all, enzyme-catalysed reactions are reversible.
They exhibit all the properties of colloids.
Their activity can be inhibited by certain narcotics, such as phenylurethane, or by inorganic poisons.
They require a specific pH for their optimum activity.

Points (ii) and (iv) taken together explain why enzymes matter so much in biotechnology. A tiny quantity does an enormous amount of work, and it does exactly one job — so a manufacturer can add one enzyme to a mixture of thousands of compounds and change only the one bond intended.

Curiosity Corner

Q2. Why does a piece of plain chapati begin to taste sweet if you chew it long enough?

A. Because saliva contains the enzyme salivary amylase. Chapati is largely starch, which is tasteless because its molecules are far too large to fit the sweet receptors of the tongue. Amylase hydrolyses the α-1,4 glycosidic bonds of starch, breaking the long chains into maltose and shorter fragments. These smaller sugars can reach the receptors, and sweetness appears. Notice the specificity at work: the same amylase cannot touch the β-1,4 bonds of the cellulose in the same meal, which is why chewing a piece of paper never tastes sweet however long you persist.

6.6.3 Nomenclature and Classification of Enzymes

Enzymes are named in several different ways.

Some enzymes carry a traditional name, for example ptyalin, pepsin, rennin and trypsin.
Usually the suffix -ase is added to the name of the substrate. Urease attacks urea, lipase digests lipids, and amylase hydrolyses amylum (starch).
They may be named after the type of reaction they catalyse, for example dehydrogenase, oxidase and decarboxylase.
The suffix -lytic denotes enzyme activity on a chemical class of substrate — proteolytic for protein-digesting, amylolytic for carbohydrate-digesting and lipolytic for fat-digesting enzymes.
Some proteinases of plants and microorganisms are named from the species of origin, for example papain from papaya, ficin from Ficus and subtilisin from Bacillus subtilis.

A rational system of classification was proposed by the International Union of Biochemistry (IUB), based on reaction types and their mechanism. Under it, all enzymes are grouped into six major classes, each divided into subclasses, and every known enzyme bears a characteristic code number. Hexokinase, for example, is coded as E.C. 2.7.1.1.

Class	Reaction catalysed	Examples
Oxidoreductases	Oxidation–reduction reactions	Dehydrogenases, oxidases, catalase, peroxidase
Transferases	Transfer of a functional group (aldehyde, ketone, phosphate, sulphur) from one substrate to another	Hexokinase, amino transferases (transaminases)
Hydrolases	Hydrolysis of bonds by the addition of water; all digestive enzymes fall here	Amylase, trypsin, lipase
Lyases	Removal of groups from substrates by a mechanism other than hydrolysis, forming double bonds in the process	Aldolase, fumarase, carboxylase
Isomerases	Intramolecular rearrangement resulting in the formation of isomers	Phosphohexoisomerase, epimerase, alanine racemase
Ligases or synthetases	Synthesis by condensation of two groups, requiring ATP or a similar triphosphate	RNA synthetase, DNA ligase, glutamine synthetase

The last entry deserves a note. DNA ligase is a ligase in exactly the sense given here — it joins two pieces of DNA using energy from ATP. It is the enzyme that makes gene cloning possible, and you will meet it again in Class XII as one of the basic tools of recombinant DNA technology.

6.6.4 Mechanism of Enzyme Action

Activation Energy

An enzyme does not take direct part in the reaction it catalyses. It merely provides a platform or template on which the reacting molecules are brought into contact far more often, and far more favourably, than chance collisions at that temperature would allow.

Its principal effect is therefore to lower the activation energy of the reaction — the energy barrier that reactant molecules must climb before they can be converted into product.

Definition

Activation energy is the minimum energy that reacting molecules must possess in order to be converted into products; an enzyme accelerates a reaction by lowering this energy requirement.

The scale of the effect is best seen in a number. The decomposition of hydrogen peroxide, \(\text{H}_2\text{O}_2\), without a catalyst requires 18,000 calories per mole. In the presence of catalase, the same reaction requires only 2,000 calories per mole.

Diagram

Fig. 6.21 – Activation energy requirement of non-catalysed and enzyme-catalysed reactions: free energy on the y-axis and time on the x-axis; show reactant A, product B, the higher energy barrier without enzyme and the lower barrier with enzyme, and mark the free energy difference between A and B as unchanged.

Read that graph carefully, because one feature of it is frequently misunderstood. The enzyme lowers the height of the barrier, but it does not change the free energy difference between A and B. An enzyme makes a reaction faster; it does not make an impossible reaction possible, and it does not change how much energy the reaction finally releases.

Side Note

The Commission on Enzymes of the International Union of Biochemists recommended a standard unit for all enzymes, the International Unit (IU). One unit of any enzyme is that amount which catalyses the transformation of 1 micromole of substrate per minute under defined conditions of pH, temperature, substrate concentration and cofactors.

The Active Site

In 1913 Michaelis and Menten proposed that for a catalytic reaction to occur, the enzyme and substrate must first bind together to form an enzyme–substrate complex. The specific site on the enzyme where the substrate binds is called the active site.

The active site is not a separate part bolted onto the enzyme. It is formed by the three-dimensional folding of the protein molecule, which brings certain specific amino acids very close to one another even though they may lie far apart in the linear sequence. In lysozyme, for instance, the catalytic or active site is formed by amino acids numbered 35, 52, 62, 63 and 101 in a linear sequence of 129 amino acids.

This is the deepest reason why denaturation destroys enzyme activity. Unfolding does not break the chain, but it scatters those five residues to different parts of the molecule — and an active site whose parts are no longer adjacent is no active site at all.

1. Lock and Key Model

This model was proposed by Emil Fischer in 1890. According to it, just as only a specially shaped key fits into a particular lock, a specific catalytic site will accept only a substrate having a complementary shape.

The model accounts very well for the specificity of enzymes. In it, the active site is presumed to be pre-shaped to fit the substrate — rigid, and waiting.

Diagram

Fig. 6.22 – Lock and key model: enzyme molecule with a pre-shaped active site + succinic acid → enzyme–substrate complex → enzyme + fumaric acid (product); label the active site.

2. Induced Fit Model

This model was proposed by Koshland in 1960. According to it, the shape of the catalytic site of an enzyme is not pre-determined. Rather, it is the binding of the substrate that induces a conformational change in the enzyme.

That change brings the required amino acids into the correct spatial orientation for proper catalytic activity. The active site, on this view, is flexible — it closes around the substrate the way a hand closes around an object rather than the way a lock receives a key.

The model is supported by X-ray crystallographic studies, which show a definite change in the shape of the active site after binding with an appropriate substrate.

Diagram

Fig. 6.23 – Mechanism of enzyme action based on Koshland’s induced fit hypothesis, in three stages: (A) enzyme with active site and approaching substrate, (B) enzyme–substrate intermediate complex with the active site changed in shape, (C) enzyme releasing the products and returning to its original shape.

Compare Table

Exam Corner

Q. If the lock and key model already explained enzyme specificity, why was a second model needed at all?

A. Because a rigid lock cannot explain everything that enzymes actually do. Some enzymes act on several closely related substrates, which a fixed shape would exclude. More tellingly, X-ray studies showed that the active site physically changes shape when the substrate arrives — something a rigid site could not do. The induced fit model keeps the useful part of Fischer’s idea, that shape governs specificity, and drops the part that the evidence contradicted.

6.6.5 Enzyme Inhibition

The activity of enzymes can be reduced by substances that combine with the enzyme but do not serve as substrates. These substances are called inhibitors.

1. Competitive or Surface Analog Inhibition

Sometimes a substance is structurally similar — analogous — to the substrate molecule. Being similar in shape, it binds to the active site of the enzyme and thereby blocks the enzyme’s action. In other words, such substances compete with the normal substrate for the active sites on the enzyme.

This type of inhibition is reversible, and it can be removed by increasing the concentration of the real substrate. The reason follows directly from the mechanism: substrate and inhibitor are competing for the same site, so flooding the system with substrate tips the odds back in the substrate’s favour.

The classic example is succinic dehydrogenase, which is inhibited by malonic acid — a molecule with a structure very similar to that of its real substrate, succinic acid. Compounds such as oxalic acid and glutaric acid also inhibit succinic dehydrogenase. Very often the product of an enzymatic reaction acts as a competitive inhibitor of the substrate, because of the similarity in structure between the substrate and the end product.

Diagram

Fig. 6.24 – Competitive inhibition: (A) normal reaction, enzyme + succinic acid → enzyme–substrate complex → enzyme + fumaric acid; (B) competitive inhibition, enzyme + malonic acid → complex → no reaction.

2. Non-competitive Inhibition

In this type of inhibition the inhibitor is covalently linked to the enzyme and binds so tightly that its dissociation from the enzyme is very low. Such inhibitors are either metabolic poisons or heavy metal ions. They do not compete for the active site; instead they destroy the structure of the enzyme. This type of inhibition is irreversible in nature.

An example is the inhibition of acetylcholinesterase by nerve gas (organophosphorus compounds). Because the enzyme is not merely occupied but permanently damaged, adding more substrate is of no help whatever — which is precisely what makes such compounds so dangerous.

Diagram

Fig. 6.25 – Non-competitive inhibition: the enzyme with its active site and the substrate above it, and a non-competitive inhibitor bound at a site away from the active site, distorting the enzyme so the substrate can no longer fit.

Compare Table

6.6.6 Factors Affecting Enzyme Activity

1. Temperature

Between the range of 0°C and 45°C the activity of enzymes increases with an increase in temperature. For every rise of 10°C the rate of reaction is roughly doubled, that is, the temperature coefficient \(Q_{10}\) is 2. The optimum temperature for most enzymes lies between 25°C and 45°C.

At lower temperatures the enzyme becomes inactivated — but not destroyed, which is why refrigeration preserves food rather than ruining it, and why cooling restores activity. At higher temperatures the enzymes are destroyed, and on lowering the temperature back to the optimum the catalytic activity is not restored. The asymmetry is the whole point: cold pauses an enzyme, heat denatures it.

Some enzymes are exceptions. Bacteria living in hot springs possess enzymes that remain fully active at temperatures which would denature most proteins within seconds, an adaptation that biotechnology has since put to considerable use.

Diagram

Fig. 6.26 – Effect of temperature on enzyme activity: rate of reaction on the y-axis against temperature on the x-axis, rising to a peak at the optimum temperature and then falling sharply; mark the optimum and the region of denaturation.

2. pH

Each enzyme is catalytically active only within a limited range of hydrogen ion concentration. In other words, every enzyme has its own specific pH optimum, and its activity declines gradually on either side of that optimum.

The optimum pH for pepsin is between 1.5 and 2.5, whereas for trypsin it is between 8 and 11. These two figures tell a story about where each enzyme works — pepsin in the strongly acidic stomach, trypsin in the alkaline small intestine. An enzyme’s pH optimum matches the compartment it evolved to work in.

The underlying reason is charge. Changing the pH alters the ionisation of the amino acid side chains that form the active site, and a site whose charges have changed can no longer bind the substrate correctly.

Diagram

Fig. 6.27 – Effect of pH on enzyme activity: bell-shaped curves for pepsin (optimum about pH 2) and trypsin (optimum about pH 8–9) plotted on the same axes.

3. Enzyme Concentration

The rate of a chemical reaction increases proportionately with increasing concentration of the enzyme, provided that substrate is present in excess and the end products are continuously removed from the site of reaction. Both conditions matter: if substrate runs short, adding more enzyme achieves nothing, since the extra enzyme molecules have nothing to act upon.

4. Substrate Concentration

Substrate is the substance which is acted upon by the enzyme to produce the product.

The catalytic activity of an enzyme is expressed in terms of its turnover number.

Definition

Turnover number is the number of substrate molecules converted into product per unit time by one enzyme molecule, provided the substrate is present in excess.

Turnover numbers for most enzymes vary between 1 and 104 per second, but the range across enzymes is enormous.

S. No.	Enzyme	Turnover number
1.	Catalase	40,000,000
2.	Carbonic anhydrase	400,000
3.	Acetyl cholinesterase	14,000
4.	Urease	10,000
5.	LDH	1,000
6.	Chymotrypsin	100
7.	DNA polymerase-I	15
8.	Lysozyme	0.5

Now consider what happens as substrate concentration is raised. In the absence of substrate the enzyme shows no activity at all. As substrate concentration is increased gradually, the activity of the enzyme increases. At a sufficiently high substrate concentration the maximum velocity (\(v_{max}\)) of the reaction is achieved. If substrate concentration is increased beyond this point, the rate of reaction shows no appreciable further increase, and may even decline.

Why should it level off? Because at \(v_{max}\) every enzyme molecule is already occupied — all of them exist in the form of ES complex. Adding more substrate cannot help, since there is no free enzyme left to bind it. The enzyme, not the substrate, has become the limiting factor.

The substrate concentration that produces half-maximal velocity is termed the \(K_m\) value, or Michaelis constant.

Definition

The Michaelis constant (\(K_m\)) is the substrate concentration at which the reaction proceeds at half its maximum velocity, and it is expressed in units of concentration such as mol L⁻¹.

Diagram

Fig. 6.28 – Effect of substrate concentration on the velocity of an enzymatic reaction: velocity \(v\) on the y-axis against substrate concentration [S] on the x-axis; mark \(v_{max}\), the level \(\frac{1}{2}v_{max}\), and \(K_m\) read off the x-axis at that level.

The relationship plotted in Fig. 6.28 is expressed by the Michaelis–Menten equation:

\[v = \frac{v_{max}[S]}{K_m + [S]}\]

where \(v\) is the velocity of the reaction, \(v_{max}\) the maximum velocity, \([S]\) the substrate concentration and \(K_m\) the Michaelis constant.

The Michaelis constant is useful in several ways:

It helps in finding out the concentration of substrate required for maximum velocity.
It helps in determining whether two enzymes obtained from different sources are identical.
It indicates the degree of affinity of an enzyme for different substrates, where the enzyme can act on more than one. The smaller the \(K_m\), the higher is the affinity of the enzyme for that substrate.
It determines the significance of a reaction in cellular metabolism.

Point (iii) is worth reasoning through rather than memorising. A small \(K_m\) means that only a small amount of substrate is needed to drive the reaction to half speed — which can only be true if the enzyme grips that substrate tightly. A large \(K_m\) means a great deal of substrate must be supplied before the enzyme responds, which indicates weak binding.

Technique

Determination of \(K_m\) by varying substrate concentration Principle: At a fixed enzyme concentration, reaction velocity rises with substrate concentration until all enzyme molecules are saturated; plotting velocity against substrate concentration gives a curve from which \(v_{max}\) and hence \(K_m\) can be read. Steps: 1. Prepare a series of starch solutions of increasing concentration. 2. Add the same volume of diluted saliva (salivary amylase) to each and incubate for a fixed time. 3. Measure the amount of starch remaining or reducing sugar formed in each tube. 4. Plot reaction velocity against substrate concentration. 5. Locate \(v_{max}\) from the plateau, mark \(\frac{1}{2}v_{max}\), and read the corresponding substrate concentration on the x-axis as \(K_m\). Used for: Comparing the affinity of an enzyme for different substrates, and checking whether enzymes from two sources are the same. This is ISC practical 8, in which the action of enzymes on starch is studied under variable temperature and variable substrate concentration, and \(K_m\) is plotted by graph.

Numerical 6.4

In an experiment on salivary amylase, the reaction velocity was measured at increasing substrate concentrations and the following readings were obtained.

Substrate concentration (mmol L⁻¹)	1	2	4	8	16	32
Velocity (arbitrary units)	12	20	30	38	43	45

(a) Plot velocity against substrate concentration. (b) Estimate \(v_{max}\) from the graph. (c) Determine the \(K_m\) of the enzyme. (d) A second enzyme acting on the same substrate is found to have a \(K_m\) of 2 mmol L⁻¹. Which of the two enzymes binds the substrate more tightly, and why?

Try Yourself

Take three test tubes of starch solution with a little saliva added to each, and keep one in ice, one at room temperature and one in boiling water for ten minutes. Then test all three with iodine. Predict, before you begin, which tube will still show a blue-black colour and which will not — and explain each prediction in terms of what has happened to the enzyme molecule.

6.7 Optical Activity of Biomolecules

Two molecules can contain the same atoms, joined in the same order, by the same bonds — and still not be the same molecule. They can be mirror images of each other, like your left and right hands. No matter how you turn a left hand, it will not lie exactly over a right one.

This property is called chirality, and biomolecules are full of it. It is not a chemical curiosity. It is one of the deepest facts about living matter, and it has consequences for medicine, industry and every enzyme you studied in the previous section.

6.7.1 The Asymmetric Carbon

The source of chirality in a biomolecule is almost always a single carbon atom carrying four different groups on its four bonds. Such a carbon is called an asymmetric carbon or a chiral centre.

A carbon with four different groups can be arranged in space in two ways, and the two arrangements are mirror images that cannot be superimposed on each other.

Diagram

Fig. 6.29 – An asymmetric carbon atom bearing four different groups, drawn twice as non-superimposable mirror images on either side of a vertical mirror line, with a pair of hands shown alongside for comparison.

You have already met asymmetric carbons twice in this chapter. Every amino acid except glycine has one at the α-carbon, because four different groups — \(-\text{NH}_2\), \(-\text{COOH}\), \(-\text{H}\) and the side chain \(R\) — are attached to it. Glycine is the exception because its side chain is a hydrogen atom, so two of the four groups are identical and the molecule is not chiral. In sugars, ring closure creates a fresh asymmetric carbon at C-1 in aldoses and C-2 in ketoses, which is why α- and β-forms exist.

How Many Isomers?

If one asymmetric carbon gives two mirror-image forms, a molecule with several asymmetric carbons gives many. The number of possible optical isomers is:

\[N = 2^n\]

where \(N\) is the number of optical isomers and \(n\) is the number of asymmetric carbon atoms in the molecule. Glucose has four asymmetric carbons, and therefore \(2^4 = 16\) possible optical isomers — of which the cell uses essentially one.

Numerical 6.5

(a) A ketohexose such as fructose contains three asymmetric carbon atoms. Calculate the number of optical isomers possible for it. (b) A triose sugar contains one asymmetric carbon. How many optical isomers can it have? (c) If a laboratory synthesises a hexose from scratch by ordinary chemistry, and all possible isomers are formed in equal amounts, what fraction of the product will be the one an enzyme can actually use?

6.7.2 Rotation of Plane Polarised Light

Ordinary light vibrates in every plane at right angles to its direction of travel. If it is passed through a suitable filter, only the vibrations in one plane are allowed through, and the result is called plane polarised light.

Now pass that plane polarised light through a solution of a chiral compound. The plane of vibration comes out tilted — it has been rotated through a measurable angle. A compound that does this is said to be optically active, and the instrument used to measure the angle is a polarimeter.

Definition

Optical activity is the ability of a compound to rotate the plane of plane polarised light passing through its solution.

The direction of rotation gives the compound its name.

Definition

A compound that rotates the plane of polarised light to the right, that is clockwise, is dextrorotatory, and is indicated by \(d\) or by a plus sign \((+)\).

Definition

A compound that rotates the plane of polarised light to the left, that is anticlockwise, is laevorotatory, and is indicated by \(l\) or by a minus sign \((-)\).

Diagram

Fig. 6.30 – A polarimeter: light source → polariser producing plane polarised light → sample tube containing the solution → analyser showing the plane rotated through an angle α; label the angle of rotation.

The size of the rotation depends on the compound, on how much of it is present and on how far the light travels through the solution. To compare compounds fairly, chemists use the specific rotation:

\[[\alpha] = \frac{\alpha}{l \times c}\]

where \(\alpha\) is the observed angle of rotation in degrees, \(l\) is the length of the sample tube in decimetres (dm), and \(c\) is the concentration of the solution in grams per millilitre (g mL⁻¹).

Turn that equation around and its industrial use becomes obvious. If \([\alpha]\) is known for a pure sugar, then measuring \(\alpha\) tells you \(c\) — the concentration. This is how sugar content is measured routinely in the sugar industry, without any chemical reaction at all.

Compare Table

6.7.3 Configuration and Rotation are Two Different Things

This is the point at which most confusion arises, and it is worth stating flatly.

The capital letters D and L describe configuration — the spatial arrangement of groups around the reference carbon, as explained in Section 6.3. The small letters \(d\) and \(l\), or the signs \((+)\) and \((-)\), describe observed rotation — which way the compound actually turns polarised light in a polarimeter.

Configuration is determined by structure; rotation is determined by experiment. The two do not have to agree. A D-sugar may be laevorotatory, and an L-amino acid may be dextrorotatory. You cannot predict the sign of rotation from the letter D or L, and you cannot deduce the configuration from a polarimeter reading.

Fructose settles the point. It is D-fructose by configuration, yet it is strongly laevorotatory — so much so that its older name is levulose. Glucose, by contrast, is D-glucose and dextrorotatory, which is why it is commonly sold as dextrose.

Exam Corner

Q. A student writes that “all D-sugars are dextrorotatory because D stands for dextro”. Where has the reasoning gone wrong?

A. The letter D refers to configuration — where the reference \(-\text{OH}\) group sits in the structural formula — not to the direction in which the compound rotates polarised light. D-fructose is the standing counter-example: its configuration is D, but it rotates light strongly to the left. The two pieces of information are obtained in completely different ways, one from structure and one from a polarimeter, and neither can be deduced from the other.

6.7.4 Mutarotation and Inversion

When a crystalline sugar is first dissolved in water, its specific rotation is not constant. It changes gradually and then settles at a steady value. This gradual change is called mutarotation.

Freshly dissolved α-D-glucose has a specific rotation of about \(+112°\), and β-D-glucose about \(+18.7°\). Left standing, either solution drifts to the same final value of about \(+52.7°\). The explanation lies in the ring chemistry of Section 6.2: in solution the ring opens and closes continually, so α and β forms interconvert through the open-chain form until a fixed equilibrium mixture is reached.

A related and more dramatic change occurs when sucrose is hydrolysed. Sucrose solution is dextrorotatory, at about \(+66.5°\). Hydrolysing it gives an equimolar mixture of glucose (\(+52.7°\)) and fructose (\(-92°\)). Since fructose rotates light much more strongly to the left than glucose does to the right, the mixture as a whole becomes laevorotatory — the sign of rotation has actually reversed.

For this reason the hydrolysis of sucrose is called inversion, the product is called invert sugar, and the enzyme that catalyses it is called invertase. Honey is largely invert sugar, which is one reason it tastes sweeter than an equal weight of table sugar: the fructose released is sweeter than the sucrose it came from.

6.7.5 Why Chirality Matters in Biology

Here is the striking fact. Of the sixteen possible optical isomers of glucose, living cells use one. Of the two mirror-image forms of every amino acid, proteins use only the L-configuration. Life, on this planet, is built almost entirely one-handed.

The reason lies in the enzymes. An active site is a three-dimensional pocket, and a pocket that fits a left-handed molecule will not fit its right-handed mirror image — for exactly the reason a left glove will not fit a right hand. Enzymes are therefore stereospecific: they recognise one optical isomer and ignore the other.

This has immediate practical consequences.

A chemically synthesised sugar or amino acid is usually produced as an equal mixture of both mirror forms, called a racemic mixture, which shows no net optical rotation because the two rotations cancel out. Only half of such a mixture is biologically useful. A product made by a microorganism in a fermenter, on the other hand, comes out as the single correct isomer, because the enzymes that made it could produce nothing else. This is one of the standing advantages of biological manufacture over chemical synthesis, and you will meet it again in the industrial biotechnology of Class XII.

Real Incident / Discovery

In the late 1950s a sedative was marketed widely, including to pregnant women for morning sickness. It was sold, as most drugs then were, as a mixture of both mirror-image forms of the molecule, on the assumption that the two forms would behave alike. They did not. The drug was withdrawn in the early 1960s after it was found to cause severe birth defects in thousands of children. The episode changed pharmaceutical regulation permanently: drug authorities now require that where a medicine is chiral, the biological activity and safety of each mirror form be established separately rather than assumed to be identical. A distinction that had looked like a fine point of stereochemistry turned out to be a matter of life and disability.

Exam Corner

Q. Bacterial cell walls contain D-amino acids, although proteins contain only L-amino acids. Why might this be an advantage to the bacterium?

A. Because the enzymes that break down proteins are stereospecific for L-amino acids. A wall built partly from D-amino acids is therefore not recognised by ordinary proteolytic enzymes, and resists being digested. The bacterium has, in effect, built its armour out of a material its attackers’ tools do not fit — the same principle that makes β-linked cellulose and chitin so difficult to digest.

Try Yourself

Take a solution of table sugar and add a little dilute acid, then warm it gently for a few minutes. Without using a polarimeter, describe one chemical test from Section 6.2 that would tell you the hydrolysis had taken place, and state what the sign of optical rotation would have done over the same period.

6.8 Supramolecular Assembly

So far this chapter has moved steadily upward in size. Small units were joined by covalent bonds into large ones: glucose into cellulose, amino acids into proteins, fatty acids and glycerol into fats. That process has a limit, however. A cell cannot build a ribosome or a membrane by simply making a longer chain.

Above the level of the macromolecule, nature changes method. Instead of joining molecules covalently, it lets finished macromolecules associate with one another to form larger working structures. These structures are called supramolecular assemblies.

6.8.1 The Hierarchy of Organisation

The build-up runs in clear stages.

Simple precursors are first converted into small organic compounds that serve as intermediates in biosynthesis. These become the micromolecules — amino acids, sugars, nucleotides, fatty acids and glycerol. By covalent linkage of these micromolecules, the macromolecules are constructed: proteins, polysaccharides, polynucleotides (DNA and RNA) and lipids.

Interactions among macromolecules then lead to the next level of structural organisation, the supramolecular complex. Here one or more types of macromolecule come together to form specific assemblies that serve important subcellular functions.

Flowchart

FC 6.8 – Hierarchy of biological organisation: precursors → simple organic intermediates → micromolecules (amino acids, sugars, nucleotides, fatty acids, glycerol) → [covalent bonds] → macromolecules (proteins, polysaccharides, nucleic acids, lipids) → [non-covalent forces] → supramolecular assemblies (membranes, ribosomes, chromosomes, enzyme complexes) → organelles → cell

Definition

A supramolecular assembly is an organised structure formed when two or more macromolecules come together through non-covalent forces to serve a specific subcellular function.

Examples of supramolecular assemblies include multifunctional enzyme complexes, ribosomes, chromosomes, cytoskeletal elements and the cell membrane itself.

6.8.2 Held Together Without Covalent Bonds

The defining feature of a supramolecular assembly is what holds it together. These structures are maintained by non-covalent forces, not by covalent bonds.

The forces involved are the ones already familiar from protein structure:

Hydrogen bonds
Ionic attractions
Van der Waals’ forces
Hydrophobic interactions between macromolecules

Individually, each of these is weak. Collectively, when hundreds or thousands of them act between two large surfaces, they hold the assembly in a highly ordered functional state.

Compare Table

Exam Corner

Q. Non-covalent forces are weak. Would it not be safer for the cell to hold a ribosome together with covalent bonds?

A. Safer, perhaps, but useless. A structure held by covalent bonds could not be taken apart and reassembled without chemical surgery. Ribosomes must associate and dissociate around a messenger RNA repeatedly; membranes must let components move within them; chromatin must open at one gene and stay closed at another. Weak, reversible forces are what make an assembly dynamic. The cell has traded permanence for the ability to build, dismantle and rebuild on demand — and the trade is a good one, since large numbers of weak bonds acting at once are quite strong enough for ordinary conditions.

6.8.3 Some Common Supramolecular Assemblies

Different combinations of macromolecules give assemblies with quite different functions.

Proteins + carbohydrates → glycoproteins, which function largely in cell recognition and cell communication, and sit on the outer face of the cell membrane.
Lipids + proteins → lipoproteins, the basic fabric of the plasma membrane, which combines the lipid bilayer with proteins embedded in it.
Nucleic acids + proteins → chromatin, formed when DNA associates with histone proteins; this is the genetic component of the cell.
Nucleic acids + proteins → ribosomes, formed when ribosomal RNA associates with ribosomal proteins; these are the sites of protein synthesis.

Diagram

Fig. 6.31 – A single integrating chart for the whole chapter: BIOMOLECULES at the centre, branching to nucleotides → polynucleotides → DNA and RNA; amino acids → proteins → primary, secondary, tertiary and quaternary structure → supramolecular assemblies; monosaccharides → disaccharides → oligosaccharides → polysaccharides → cell wall; fatty acids (saturated and unsaturated) and lipids → phospholipids, glycolipids, steroids, waxes, triglycerides → cell membrane; with the non-covalent interactions (hydrogen bonding, electrostatic, van der Waals’, hydrophobic) shown feeding into the assembly level.

Notice something about entries (iii) and (iv). Both are combinations of nucleic acid with protein, yet one stores information for decades while the other reads that information into protein in seconds. The class of macromolecules used does not by itself determine the function; the arrangement does. This is the central idea of supramolecular organisation.

Real Incident / Discovery

In the mid-1950s Fraenkel-Conrat and Williams, working on tobacco mosaic virus, separated the virus completely into its two components — a strand of RNA and its coat protein. Neither fraction alone could infect a tobacco plant. They then mixed the purified RNA and the purified protein together in a test tube, with no cell, no enzyme and no energy source supplied. Rod-shaped virus particles reassembled on their own, and these reconstituted particles infected tobacco plants normally. The experiment demonstrated that the information needed to build the assembly is contained in the shapes of the macromolecules themselves. Given the right conditions, the parts find each other and fit together without being told how.

6.8.4 Why the Idea Matters

Self-assembly is not a laboratory trick; it is how much of a cell is built. A membrane forms because phospholipids arrange themselves; a ribosome forms because rRNA and protein recognise one another’s surfaces. The cell supplies the components and the conditions, and the structure appears.

For biotechnology this has two practical consequences.

First, it explains a common industrial problem. A recombinant protein produced in a bacterial fermenter may have the correct amino acid sequence and still be useless, because it has failed to fold and assemble correctly in an unfamiliar cellular environment. Getting the sequence right is only half the task.

Second, it identifies targets. Several important antibiotics work by binding to the bacterial ribosome and jamming this assembly, so that the bacterium can no longer make protein. A drug does not need to break a covalent bond to be effective — interfering with a non-covalent assembly is enough.

Try Yourself

Using Fig. 6.31, trace a single path from a small precursor molecule all the way up to the plasma membrane, naming the micromolecule, the macromolecule and the assembly at each stage. Then state clearly at which step covalent bonding stops and non-covalent association begins.

6.9 Techniques Used for Separation of Biomolecules

A cell extract is a mixture of thousands of different molecules. Knowing the chemistry of proteins or sugars is of little practical use unless a particular one can be pulled out of that mixture, identified and measured. This is the work of separation techniques, and it is where the biotechnology laboratory actually begins.

The most widely used family of separation methods is chromatography. Every form of chromatography works on the same underlying idea.

The mixture is made to travel across or through a fixed material called the stationary phase, carried by a moving liquid called the mobile phase. Each component of the mixture is pulled two ways at once — held back by the stationary phase, urged forward by the mobile phase. Components that are held more strongly travel slowly; those held weakly travel fast. Given enough distance, they separate.

Definition

Chromatography is a technique for separating the components of a mixture on the basis of their differing distribution between a stationary phase and a mobile phase.

What changes from one type of chromatography to another is simply the property on which that distribution depends. In ion exchange chromatography it is electric charge. In paper chromatography it is solubility and polarity.

6.9.1 Ion Exchange Chromatography

Ion exchange chromatography is a process used to separate ions and polar molecules on the basis of their charge.

The stationary phase is a solid support, prepared from synthetic resins, that carries fixed ions — charged groups permanently bonded to the resin. The mobile phase is usually a buffer solution containing some anions or cations.

The principle follows directly. A charged molecule in the sample is attracted to a resin of opposite charge and sticks to it, while molecules of the same charge as the resin, and uncharged molecules, are not held and simply flow out with the buffer. The bound molecules are recovered afterwards in a separate step called elution.

There are two types of ion-exchange chromatography, depending on the charge carried by the resin.

(a) Cation-exchange Chromatography

In cation-exchange chromatography, positively charged molecules behaving as cations are adsorbed to a negatively charged solid support or matrix, for example carboxy-methyl cellulose.

They are later separated, or eluted, from the matrix by using a more positively charged ion in the mobile phase, which displaces them from the matrix and binds in their place.

Follow a concrete case. If a solution containing positively charged proteins is allowed to flow through a matrix carrying negative charge, the proteins bind to the matrix. Other uncharged or negatively charged molecules flow straight out. The matrix-bound protein can then be eluted by increasing the concentration of sodium chloride in the eluting buffer: the \(\text{Na}^+\) ions compete with the positively charged proteins for binding to the matrix, and take their place. Proteins bearing less positive charge are dislodged first, followed by those with more positive charge.

That last sentence is the reason the method separates rather than merely captures. Proteins do not all come off at once; they come off in order of how tightly they were held.

(b) Anion-exchange Chromatography

Anion-exchange chromatography is used for the separation of negatively charged molecules. Negatively charged proteins are allowed to adsorb on a positively charged matrix, for example DEAE-cellulose (diethyl aminoethyl cellulose).

These are later eluted from the matrix by using a buffer containing more negatively charged ions, which displace them.

Factors other than net charge, such as affinity for the supporting matrix, also influence the rate of elution of substances in an ion-exchange column.

Diagram

Fig. 6.32 – Separation of solutes in ion-exchange chromatography in four stages: (i) mixture of solutes loaded onto the fixed-ion charged resin with mobile counter ions; (ii) charged and neutral solutes separated, unbound solutes flowing out; (iii) bound solutes released by buffer of increasing pH and ionic strength; (iv) column washed with starting buffer to give the regenerated ion exchanger.

Diagram

Fig. 6.33 – Ion-exchange chromatography compared: (a) cation exchanger, with sample cations displacing mobile counterions on a negatively charged resin while anions and neutral solutes flow out; (b) anion exchanger, with sample anions displacing counterions on a positively charged resin while cations and neutral solutes flow out.

Flowchart

FC 6.9 – Ion exchange chromatography workflow: equilibrate column with starting buffer → load sample mixture → oppositely charged molecules bind to resin, others flow through → wash column → elute bound molecules with buffer of increasing salt concentration or changing pH → collect fractions in order of binding strength → regenerate column with starting buffer

Technique

Ion exchange chromatography Principle: Molecules carrying a charge opposite to that of a fixed-ion resin bind to it reversibly, and can be released in order of binding strength by raising the salt concentration or changing the pH of the eluting buffer. Steps: Equilibrate the charged resin column with the starting buffer → load the sample → allow unbound and like-charged molecules to flow out → wash the column → elute the bound molecules with a buffer of increasing ionic strength or altered pH → collect the fractions separately. Used for: Separating and purifying proteins, amino acids and other charged biomolecules. It is one of the standard purification steps used industrially to recover a recombinant protein such as human insulin from the mixture of bacterial proteins it is produced in.

Exam Corner

Q. Two proteins, P and Q, have isoelectric points of 5.0 and 8.5. A mixture of them is loaded onto an anion-exchange column equilibrated at pH 7.0. Which one binds?

A. Protein P binds. At pH 7.0, which is above P’s isoelectric point of 5.0, P carries a net negative charge, and so it is attracted to the positively charged anion-exchange matrix. Q, whose isoelectric point of 8.5 lies above the buffer pH, carries a net positive charge at pH 7.0 and is repelled by the matrix, so it flows straight through. This is why the buffer pH is chosen before the run rather than after it — the choice decides which protein is captured and which is discarded.

Notice how much work the isoelectric point does here. A protein’s charge is not a fixed property of the molecule; it is set by the pH of the surrounding buffer. Change the buffer and the very same protein can be made to bind or not to bind. That is what makes ion exchange chromatography so adaptable.

6.9.2 Paper Chromatography

Paper chromatography works on a completely different property. It is a form of partition chromatography, which is based on differences in the solubility of a substance in two different solvents.

In paper chromatography, cellulose paper acts as the stationary phase and a non-polar solvent acts as the mobile phase. There is a subtlety in how the paper works: cellulose paper holds a thin layer of water on its surface by adsorption, and it is this bound water that makes the stationary phase polar.

So the mixture is really being offered a choice between two liquids — the water held on the paper, and the solvent moving up it. When a mixture is applied to the paper, its components move at different rates depending on their solubility in the solvent and their attraction to the paper. A component that dissolves readily in the moving solvent travels far; one that prefers the water on the paper lags behind.

The \(R_f\) Value

Since the solvent front itself moves a different distance in every run, the raw distance travelled by a spot is not a reliable identity. What is reliable is the ratio of the two distances.

Definition

The \(R_f\) value, or retention factor, is the ratio of the distance travelled by the substance to the distance travelled by the solvent, on a given stationary phase.

\[R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent}}\]

Because the numerator is always smaller than the denominator, the \(R_f\) value of all substances is always less than 1, and being a ratio of two lengths it is dimensionless.

Under fixed conditions of paper and solvent, each compound has its own characteristic \(R_f\). This is what allows an unknown spot to be identified — its \(R_f\) is compared with that of a known standard run under the same conditions.

Separation of Plant Pigments

The standard demonstration is the separation of photosynthetic pigments from a leaf extract. Four pigments separate out at different heights, in this order from the top: carotene, xanthophyll, chlorophyll-a and chlorophyll-b.

Diagram

Fig. 6.34 – Paper chromatogram of leaf pigments in a jar: the solvent front at the top, with bands for carotene, xanthophyll, chlorophyll-a and chlorophyll-b below it; mark the distance travelled by the solvent as D and the distances travelled by the four pigments as a, b, c and d respectively.

If the distance travelled by the solvent is \(D\) and the distances travelled by the four pigments are \(a\), \(b\), \(c\) and \(d\), then the \(R_f\) values are \(a/D\) for carotene, \(b/D\) for xanthophyll, \(c/D\) for chlorophyll-a and \(d/D\) for chlorophyll-b.

Read the chromatogram and it tells you about polarity as well as identity. Carotene travels furthest, so it is the least polar of the four and the most at home in the non-polar moving solvent. Chlorophyll-b travels least, so it is the most polar and clings hardest to the water held on the paper.

Technique

Paper chromatography Principle: Components of a mixture partition themselves between the water held on cellulose paper and a moving non-polar solvent, and therefore travel different distances along the paper according to their relative solubility in the two. Steps: Extract the sample in a suitable solvent → load a small spot of the extract near one end of the paper strip → place the paper in a jar with the rising solvent, keeping the loading point above the solvent level → allow the solvent to rise for the required time → mark the solvent front, dry the paper and locate the separated spots → measure the distances and calculate \(R_f\) for each. Used for: Separating and identifying the components of a mixture. In the ISC practical course it is used to separate photosynthetic pigments from a leaf, and to detect adulteration in spices — for example, distinguishing the permitted colour of turmeric from the added dye metanil yellow, which travels a different distance and so gives a different \(R_f\).

One point of technique is worth stressing because it decides whether the experiment works at all. The loading spot must always be kept above the level of the rising solvent. If the spot is dipped into the solvent, the sample simply dissolves into the reservoir and washes away, and no separation occurs.

Compare Table

Numerical 6.6

In a paper chromatography run, the solvent front travelled 12.0 cm from the loading point. Four pigment spots were located at 10.8 cm, 8.4 cm, 5.4 cm and 3.6 cm from the loading point. (a) Calculate the \(R_f\) value of each spot. (b) Identify each pigment, given the usual order of separation of leaf pigments. (c) In a repeat run the solvent was allowed to rise only 8.0 cm. Predict the distance at which the carotene spot would now be found, and state why the \(R_f\) value is a more useful figure to report than the distance itself.

Try Yourself

Draw a spot of ordinary black ink about 2 cm from the end of a strip of filter paper, and stand the strip in a little water so that the spot stays above the water level. Watch what happens over ten minutes. Write down how many different coloured components separate out, which one has the highest \(R_f\), and what that tells you about its polarity relative to the others.

6.10 Summary

Biomolecules are organic compounds produced by living organisms that take part in their structure, growth, maintenance and metabolic activity. They are built mainly from carbon, hydrogen and oxygen, with nitrogen in proteins and nucleic acids, and phosphorus and sulphur in particular groups.
Small, simple biomolecules such as glucose, amino acids, fatty acids and nucleotides are called micromolecules. When many such monomers are linked covalently, the result is a macromolecule or polymer — starch, protein, DNA.
The major types of biomolecules are carbohydrates, proteins, lipids, vitamins and enzymes. Hormones are not a separate chemical class; they are classified by function, and may be proteins or lipids.
Carbohydrates were originally named for the 2 : 1 hydrogen-to-oxygen ratio expressed by the formula \(C_x(H_2O)_y\), but exceptions on both sides made a structural definition necessary: carbohydrates are polyhydroxy aldehydes or ketones and their derivatives.
Carbohydrates are classified as monosaccharides, disaccharides, oligosaccharides (2–9 units) and polysaccharides (10 or more units).
Monosaccharides are named by the length of the carbon chain (triose to heptose) together with the functional group present — aldose for an aldehyde group, ketose for a ketone group. Glucose is an aldohexose; fructose is a ketohexose.
In solution, a monosaccharide closes into a ring by hemiacetal or hemiketal formation. The six-membered pyranose form is more stable than the five-membered furanose form. Ring closure creates a new asymmetric carbon, giving the α- and β-forms of the sugar.
Glucose is the immediate source of energy for the cell and the unit from which maltose, sucrose, lactose, starch, glycogen and cellulose are built. Fructose, the fruit sugar, has the same formula but a ketone group, and is therefore an isomer of glucose.
Ribose carries a hydroxyl group at the 2′ carbon; deoxyribose does not. The absence of that reactive group makes deoxyribose, and therefore DNA, more stable — which is why DNA serves as the genetic material.
A disaccharide is formed when two monosaccharides join by a glycosidic linkage, with the loss of one water molecule. Maltose is glucose + glucose (α, 1→4), lactose is galactose + glucose (β, 1→4) and sucrose is glucose + fructose (α-1 → β-2).
Maltose and lactose are reducing sugars because a free carbonyl group remains. Sucrose is non-reducing because both carbonyl groups are used up in forming the linkage.
Homopolysaccharides contain one type of monosaccharide; heteropolysaccharides contain more than one.
Glycogen is the reserve food of animals, stored in liver and muscle, with α-1,4 bonds in the main chain and α-1,6 bonds at branch points. Heavy branching gives many free ends, so glucose can be released from many points at once.
Cellulose is the most abundant biomolecule on earth and forms the plant cell wall, with about 25,000–30,000 β-D-glucose units joined by β-1,4 linkages. Animals with α-amylase cannot digest a β linkage, so cellulose acts as roughage in humans and is digested in cattle only by symbiotic gut microorganisms.
Chitin occurs in fungal cell walls and arthropod exoskeletons. It is a polymer of N-acetylglucosamine joined by β-1,4 linkages — cellulose with the \(-\text{OH}\) at C-2 replaced by an N-acetylamino group.
Starch is a mixture of unbranched α-amylose (15–20%) and branched amylopectin (80–85%). On hydrolysis it passes through dextrins to glucose, and the iodine colour changes from blue through purple and red to none.
Peptidoglycan is the heteropolysaccharide of the bacterial cell wall, made of alternating NAG and NAM units joined by β-1,4 bonds and cross-linked by short peptide chains. Lysozyme cuts the bond between NAM and NAG.
Sugars are sweet, crystalline, water-soluble and odourless. Their chemical behaviour — reduction, oxidation, and the Benedict’s, Fehling’s and Barfoed’s tests — depends on whether the carbonyl group is free or bonded. Selivanoff’s and Pinoff’s tests distinguish glucose from fructose, and phenylhydrazine forms osazone crystals characteristic of each sugar.
Proteins are condensation polymers of α-amino acids joined by peptide bonds, and are the most abundant constituent of protoplasm. The presence of nitrogen distinguishes them from carbohydrates and lipids.
Every amino acid has an amino group, a carboxyl group, a hydrogen atom and a variable side chain \(R\) on the same α-carbon. Because these four groups differ, the α-carbon is asymmetric in every amino acid except glycine.
Only 20 amino acids occur in proteins, and all of them are of the L-configuration. Proteins differ from one another not only in the number of amino acids but, more importantly, in their sequence.
Amino acids are classified by their reaction in solution as acidic (one amino, two carboxyl groups), basic (two amino, one carboxyl group) and neutral (one of each).
At physiological pH an amino acid exists as a zwitter ion, with \(-\text{COO}^-\) at one end and \(-\text{NH}_3^+\) at the other. Because it can both donate and accept protons it is amphoteric. The pH at which it carries no net charge is its isoelectric pH, \(pI = \frac{1}{2}(pK_1 + pK_2)\).
Proteins show four levels of structure. Primary structure is the number and sequence of amino acids, held by covalent peptide bonds, and it determines all the levels above it.
Secondary structure is held by hydrogen bonds and takes the form of the α-helix (3.6 residues per turn, pitch 5.4 Å, rise 1.5 Å per residue), the β-pleated sheet (chains fully stretched, 3.5 Å apart, parallel or antiparallel) or random structures.
Tertiary structure is the specific three-dimensional fold of a chain, maintained by disulphide bonds, hydrophobic interactions, hydrogen bonds and ionic interactions. Quaternary structure arises when two or more chains associate by non-covalent forces, as in haemoglobin’s four sub-units.
MALDI-MS — Matrix Assisted Laser Desorption Ionization – Mass Spectrometry — is used to sequence proteins. The protein is digested into peptides, embedded in a matrix that absorbs the laser energy on its behalf, ionised without being destroyed, and the mass differences between successive peaks are read as the amino acid sequence.
Proteins are colloidal, show the Tyndall effect and are amphoteric, and are least soluble at their isoelectric point. Adding neutral salt may dissolve them (salting-in) or precipitate them (salting-out).
Denaturation destroys the shape of a protein, not its chain: heat, extremes of pH, organic solvents and detergents break the weak hydrogen, hydrophobic and ionic bonds while the covalent peptide bonds survive. Where the change is reversible, the recovery is called renaturation.
Proteins are detected by the Biuret test (violet, given by all proteins except dipeptides), the xanthoproteic test (yellow, due to aromatic amino acids) and Millon’s reagent (red, specific to tyrosine); ninhydrin gives a blue colour with α-amino acids.
Proteins are classified as simple (only amino acids), complex or conjugated (with a prosthetic group — glycoproteins, lipoproteins, metalloproteins, chromoproteins, phosphoproteins, nucleoproteins) and derived (primary derived and secondary derived, produced by heat, enzymes or chemicals).
Proteins are catalytic, contractile, protective, regulatory, structural, transport, storage and receptor molecules — and several of them, such as insulin, interferons and antibodies, are today’s most valuable biotechnology products.
Lipids are esters of fatty acids with various alcohols; they are insoluble in water and soluble in organic solvents. They are classified as simple, conjugated and derived lipids.
A fatty acid has a hydrophilic \(-\text{COOH}\) head and a hydrophobic hydrocarbon tail, and is therefore amphipathic. Natural fatty acids have an even number of carbons because they are built two carbons at a time from acetyl CoA.
Saturated fatty acids have no double bond, general formula \(\text{C}_n\text{H}_{2n+1}\text{COOH}\), and straight chains that pack closely to give solids. Unsaturated fatty acids have kinked chains that pack loosely and stay liquid — which is why fats are solid and oils are liquid.
Simple lipids include fats and oils, which are esters of fatty acids with glycerol (mono-, di- and triacylglycerols, the last being neutral fat, e.g. tripalmitin), and waxes, which are esters with high molecular weight monohydric alcohols, e.g. bees wax and lanolin.
Conjugated lipids contain an additional group and comprise phospholipids (e.g. lecithin), glycolipids (e.g. cerebrosides), sulpholipids and lipoproteins. Phospholipids, with one polar head and two non-polar tails, form the cell membrane.
Derived lipids are products of hydrolysis and include fatty acids, glycerol, sterols and steroids. Cholesterol is the chief sterol and the precursor of the steroid hormones.
Lipids undergo enzymatic hydrolysis by lipases, saponification with alkali to give soaps (saponification number = mg KOH needed for 1 g fat), hydrogenation or hardening of double bonds, rancidification by hydrolysis or oxidation, dehydration to acrolein, and halogenation measured by the iodine number (g iodine absorbed by 100 g fat).
Vitamins are organic micronutrients that the body cannot synthesise and that must be supplied in the diet. Most act as cofactors of enzymes, which is why the deficiency of a few milligrams can halt an entire pathway and produce a specific disease.
Fat-soluble vitamins A, D, E and K are absorbed with dietary fat and stored in the body; their deficiency appears slowly and causes night blindness and xerophthalmia, rickets and osteomalacia, sterility and haemorrhage respectively.
Water-soluble vitamins are the B complex and vitamin C. They are not stored, so deficiency appears sooner: beri-beri (B₁), pellagra (B₃), macrocytic anaemia (B₉), pernicious anaemia (B₁₂) and scurvy (C).
Many enzymes consist of a protein apoenzyme plus a non-protein cofactor, together forming the holoenzyme. Water-soluble vitamins supply the co-enzyme forms: TPP from B₁, FMN and FAD from B₂, NAD from B₃, coenzyme A from B₅, pyridoxal phosphate from B₆, biocytin from B₇, tetrahydrofolic acid from B₉ and deoxyadenosyl cobalamine from B₁₂.
Enzymes are biocatalysts, almost all of them proteins; the catalytic RNAs called ribozymes are the recognised exception. They are effective in small quantities, highly specific, heat-sensitive and pH-dependent.
An enzyme works by lowering the activation energy of a reaction — the decomposition of \(\text{H}_2\text{O}_2\) needs 18,000 calories per mole uncatalysed and only 2,000 in the presence of catalase — without changing the energy difference between reactant and product.
The substrate binds at the active site, a pocket formed by the three-dimensional folding of the protein. In the lock and key model (Fischer) the site is rigid and pre-shaped; in the induced fit model (Koshland) the site is flexible and changes shape when the substrate binds.
Enzyme activity is affected by temperature (optimum 25–45°C; cold inactivates reversibly, heat denatures irreversibly), pH (each enzyme has its own optimum, pepsin 1.5–2.5, trypsin 8–11), enzyme concentration and substrate concentration.
As substrate concentration rises, velocity increases until \(v_{max}\) is reached and every enzyme molecule is bound as ES complex. The substrate concentration giving half-maximal velocity is the Michaelis constant \(K_m\); a smaller \(K_m\) means a higher affinity of the enzyme for that substrate.
Competitive inhibitors resemble the substrate, bind reversibly at the active site and can be displaced by adding more substrate (malonic acid on succinic dehydrogenase). Non-competitive inhibitors bind tightly away from the active site, destroy the enzyme’s structure and act irreversibly (nerve gas on acetylcholinesterase).
A carbon carrying four different groups is asymmetric, and gives two non-superimposable mirror-image forms; a molecule with \(n\) such carbons has \(2^n\) optical isomers.
A compound that rotates the plane of polarised light clockwise is dextrorotatory, \(d\) or \((+)\); one that rotates it anticlockwise is laevorotatory, \(l\) or \((-)\). Capital D and L describe configuration and cannot be used to predict the direction of rotation — D-fructose is laevorotatory.
Mutarotation is the gradual change in specific rotation as α- and β-forms of a sugar interconvert to equilibrium. Hydrolysis of sucrose reverses the sign of rotation, giving invert sugar.
Enzymes are stereospecific and use only one optical isomer — D-sugars and L-amino acids — which is why biological manufacture yields a single isomer while chemical synthesis yields a racemic mixture with no net rotation.
Above the macromolecule lies the supramolecular assembly, formed when finished macromolecules associate through non-covalent forces — hydrogen bonds, ionic attractions, van der Waals’ forces and hydrophobic interactions. Examples include glycoproteins, lipoproteins and the plasma membrane, chromatin, ribosomes and multienzyme complexes.
Chromatography separates the components of a mixture by their differing distribution between a stationary phase and a mobile phase.
Ion exchange chromatography separates molecules by charge. Cation exchangers such as carboxy-methyl cellulose carry a negative matrix and bind positively charged molecules; anion exchangers such as DEAE-cellulose carry a positive matrix and bind negatively charged ones. Bound molecules are eluted in order of binding strength by raising the salt concentration or changing the pH.
Paper chromatography separates by solubility and polarity, using cellulose paper holding a thin layer of water as the stationary phase and a non-polar solvent as the mobile phase. Each substance has a characteristic retention factor, \(R_f = \dfrac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent}}\), which is always less than 1 and dimensionless.

Flowchart

FC 6.10 – The chapter at a glance: micromolecules (monosaccharides, amino acids, fatty acids + glycerol) → [covalent bonds] → macromolecules (polysaccharides, proteins, lipids) → [folding and non-covalent forces] → functional forms (enzymes, membranes, supramolecular assemblies) → [separation techniques: ion exchange and paper chromatography] → purified, identified biomolecule ready for use in biotechnology

6.11 Key Terms

Biomolecule any organic compound produced by a living organism that takes part in its structure, growth, maintenance or metabolic activity.
Micromolecule a small, simple biomolecule such as a monosaccharide, amino acid, fatty acid or nucleotide.
Macromolecule a very large biomolecule formed by the covalent linking of many smaller repeating units.
Monomer the small repeating unit from which a macromolecule is built.
Polymer the long chain formed when many monomers are joined together.
Carbohydrate a polyhydroxy aldehyde or ketone, or a derivative of one.
Monosaccharide a simple sugar that cannot be hydrolysed into any simpler carbohydrate.
Disaccharide a carbohydrate composed of two monosaccharide units joined by a glycosidic linkage.
Oligosaccharide a carbohydrate made up of a small number of monosaccharide units, generally two to nine, which usually tastes sweet.
Polysaccharide a high molecular weight carbohydrate of ten or more monosaccharide units that yields many monosaccharides on hydrolysis.
Aldose a monosaccharide carrying an aldehyde functional group.
Ketose a monosaccharide carrying a ketone functional group.
Hemiacetal the product formed when an alcohol group of a sugar attacks its own aldehyde carbon, leading to ring closure.
Pyranose form the six-membered ring form of a sugar, which is more stable than the furanose form.
Furanose form the five-membered ring form of a sugar.
Anomers (α- and β-forms) the two forms of a ring sugar that differ only in the orientation of the \(-\text{OH}\) group at the new asymmetric carbon created by ring closure.
Isomers compounds having the same molecular formula but different structures, such as glucose and fructose.
Glycosidic linkage the bond joining two monosaccharide units, formed by condensation with the loss of one water molecule.
Condensation a reaction in which two molecules join together with the elimination of a molecule of water.
Homopolysaccharide a polysaccharide made up of only one type of monosaccharide.
Heteropolysaccharide a polysaccharide made up of more than one type of monosaccharide.
Glycogen the highly branched storage polysaccharide of animals, with α-1,4 bonds in the main chain and α-1,6 bonds at branch points.
Cellulose the most abundant biomolecule on earth, a β-1,4 linked polymer of glucose forming the plant cell wall.
Chitin a β-1,4 linked polymer of N-acetylglucosamine forming the fungal cell wall and the arthropod exoskeleton.
Amylose the long unbranched α-1,4 linked component of starch, forming about 15–20% of it.
Amylopectin the branched component of starch, forming about 80–85% of it, with α-1,6 linkages at the branch points.
Dextrins the large fragments produced when starch is broken down on heating, before complete hydrolysis to glucose.
Peptidoglycan the heteropolysaccharide of the bacterial cell wall, made of alternating NAG and NAM units cross-linked by short peptide chains.
NAG N-acetyl glucosamine, one of the two alternating sugar units of peptidoglycan.
NAM N-acetyl muramic acid, the other alternating sugar unit of peptidoglycan.
Lysozyme an enzyme present in tears, saliva and egg white that cuts the glycosidic bond between NAM and NAG in the bacterial cell wall.
Reducing sugar a sugar whose carbonyl group is free and which therefore gives a positive test with Benedict’s and Fehling’s reagents.
Non-reducing sugar a sugar whose carbonyl groups are involved in bond formation and which therefore fails to reduce Benedict’s and Fehling’s reagents, for example sucrose.
Benedict’s and Fehling’s solutions copper-containing reagents in which the blue cupric ion is reduced to the red cuprous ion by a free carbonyl group.
Protein a condensation polymer of α-amino acids joined by peptide bonds.
Amino acid an organic compound carrying an amino group, a carboxyl group, a hydrogen atom and a variable side chain on the same α-carbon.
Side chain (R group) the variable group attached to the α-carbon which distinguishes one amino acid from another.
Acidic amino acid an amino acid with one amino group and two carboxyl groups, for example glutamic acid.
Basic amino acid an amino acid with two amino groups and one carboxyl group, for example lysine.
Neutral amino acid an amino acid with one amino group and one carboxyl group, for example alanine.
Essential amino acid an amino acid that the adult body cannot synthesise and which must therefore be supplied in the diet.
Semi-essential amino acid an amino acid such as arginine which adults can make in sufficient quantity but growing children cannot.
Zwitter ion the dipolar form of an amino acid carrying \(-\text{COO}^-\) at one end and \(-\text{NH}_3^+\) at the other, with a net charge of zero.
Amphoteric able to act both as an acid and as a base.
Ampholyte an amphoteric electrolyte, such as an amino acid.
Isoelectric pH (pI) the pH at which a molecule carries no net charge and does not migrate in an electric field.
Peptide bond the covalent amide linkage formed between the carboxyl group of one amino acid and the amino group of the next, with the loss of water.
Primary structure the number and sequence of amino acids in a protein chain, together with the location of disulphide bonds.
Secondary structure the regular folding of a polypeptide chain, held in place by hydrogen bonds.
α-helix a right-handed coiled secondary structure with 3.6 amino acid residues per turn, a pitch of 5.4 Å and a rise of 1.5 Å per residue.
β-pleated sheet a sheet-like secondary structure in which the chains are fully stretched and placed 3.5 Å apart, arranged parallel or antiparallel.
Random structure a secondary arrangement without a regular repeating pattern, though still specific to a particular chain.
Helix breaker an amino acid such as proline, glycine or tryptophan that interferes with the formation of an α-helix.
Tertiary structure the specific three-dimensional configuration of a folded polypeptide chain.
Disulphide bond a covalent linkage formed between the sulphydryl groups of two cysteine residues.
Hydrophobic interaction the tendency of non-polar side chains to cluster away from water, which drives protein folding.
Chaperone a protein that assists the correct folding of another protein chain during protein synthesis.
Quaternary structure the association of two or more polypeptide chains by non-covalent forces into a single functional protein, as in haemoglobin.
MALDI-MS Matrix Assisted Laser Desorption Ionization – Mass Spectrometry, a technique for sequencing and identifying proteins from the masses of their peptide fragments.
Tyndall effect the scattering of light by a colloidal solution, which makes it appear visibly turbid.
Denaturation the loss of the secondary, tertiary and quaternary structure of a protein by heat, pH change or chemicals, without breaking its peptide bonds.
Renaturation the recovery of a denatured protein’s original folded structure when normal conditions are restored.
Salting-in the increase in solubility of a poorly soluble protein when a neutral salt is added.
Salting-out the precipitation of a protein from solution by the addition of a high concentration of neutral salt.
Biuret reaction the pinkish violet colour given by peptide linkages with an alkaline solution of copper sulphate.
Xanthoproteic test the yellow colour, turning orange with alkali, given by aromatic amino acids on boiling with concentrated nitric acid.
Simple protein a protein made up of amino acids only.
Conjugated protein a protein containing a non-proteinaceous prosthetic group in addition to amino acids.
Prosthetic group the non-protein component attached to a conjugated protein.
Derived protein a protein produced from a native protein by the action of heat, enzymes or chemical agents.
Globular protein a roughly spherical, compactly folded and generally soluble protein with a metabolic function.
Fibrous protein a long, rod-shaped, generally insoluble protein with a structural function.
Lipid an ester of a fatty acid with an alcohol, insoluble in water and soluble in organic solvents.
Fatty acid an organic acid with a long hydrocarbon chain and a terminal carboxyl group, of general formula \(\text{CH}_3(\text{CH}_2)_n\text{COOH}\).
Amphipathic possessing both a hydrophilic and a hydrophobic region within the same molecule.
Saturated fatty acid a fatty acid with no double or triple bond in its hydrocarbon chain, of general formula \(\text{C}_n\text{H}_{2n+1}\text{COOH}\).
Unsaturated fatty acid a fatty acid containing one or more double bonds, which kink the chain and keep the lipid liquid.
Simple lipid an ester of a fatty acid with an alcohol, including fats, oils and waxes.
Triglyceride a neutral fat formed when all three hydroxyl groups of glycerol are esterified with fatty acids.
Wax an ester of a fatty acid with a monohydric alcohol of high molecular weight, which is chemically inert and water-repellent.
Conjugated lipid a lipid containing an additional group besides an alcohol and a fatty acid, such as a phospholipid or glycolipid.
Phospholipid a conjugated lipid containing a phosphoric acid residue, with one polar head and two non-polar tails, forming the cell membrane.
Derived lipid a substance obtained from the hydrolysis of simple or complex lipids, such as glycerol, sterols or free fatty acids.
Cholesterol the chief sterol of animal cell membranes and the precursor of the steroid hormones.
Saponification the hydrolysis of a fat by alkali to give the sodium or potassium salts of fatty acids, that is, soaps.
Saponification number the number of milligrams of KOH required to saponify 1 gram of fat.
Hardening the catalytic hydrogenation of the double bonds of an oil to convert it into a solid fat.
Rancidification the development of an offensive odour and taste in fats and oils, caused by hydrolysis or by oxidation of double bonds.
Acrolein the pungent unsaturated aldehyde formed when the glycerol of a fat is dehydrated by bisulphate, used to confirm the presence of fat.
Iodine number the number of grams of iodine absorbed by 100 grams of fat, which measures the degree of unsaturation.
Vitamin an organic micronutrient that the body cannot synthesise and which must be supplied in the diet in small quantities.
Hypovitaminosis the deficiency disorder produced by an inadequate supply of a vitamin.
Apoenzyme the protein part of an enzyme, which cannot function on its own.
Cofactor the non-protein part of an enzyme, which may be a prosthetic group or a co-enzyme.
Co-enzyme an organic cofactor, loosely attached to the apoenzyme, that acts as an acceptor or donor of an atom or group.
Holoenzyme the complete, functional enzyme formed by an apoenzyme together with its cofactor.
TPP thiamine pyrophosphate, the co-enzyme form of vitamin B₁, involved in carboxylation reactions.
FMN and FAD flavin mononucleotide and flavin adenine dinucleotide, the co-enzyme forms of vitamin B₂, involved in oxidation–reduction reactions.
NAD and NADP nicotinamide adenine dinucleotide and its phosphate, the co-enzyme forms of vitamin B₃, involved in dehydrogenation reactions.
Enzyme a biocatalyst produced by living cells that greatly increases the rate of a specific biochemical reaction without being consumed in it.
Ribozyme a catalytic RNA molecule, the recognised exception to the rule that all enzymes are proteins.
Activation energy the minimum energy that reacting molecules must possess to be converted into products, which an enzyme lowers.
Active site the specific region of an enzyme, formed by its three-dimensional folding, at which the substrate binds.
Enzyme–substrate complex the temporary combination formed when a substrate binds to the active site of an enzyme.
International Unit (IU) the amount of an enzyme that catalyses the transformation of 1 micromole of substrate per minute under defined conditions.
Lock and key model Fischer’s model in which the active site is rigid and pre-shaped to fit only a complementary substrate.
Induced fit model Koshland’s model in which the active site is flexible and changes shape when the substrate binds.
Inhibitor a substance that combines with an enzyme and reduces its activity without acting as a substrate.
Competitive inhibition reversible inhibition in which a substrate-like molecule occupies the active site and can be displaced by excess substrate.
Non-competitive inhibition irreversible inhibition in which the inhibitor binds tightly away from the active site and destroys the structure of the enzyme.
Turnover number the number of substrate molecules converted into product per unit time by one enzyme molecule when substrate is in excess.
\(v_{max}\) the maximum velocity of an enzyme-catalysed reaction, reached when all enzyme molecules are bound as ES complex.
\(K_m\) (Michaelis constant) the substrate concentration at which the reaction proceeds at half its maximum velocity; a smaller value indicates greater affinity.
Asymmetric carbon a carbon atom bearing four different groups, which gives rise to two non-superimposable mirror-image forms.
Optical activity the ability of a compound to rotate the plane of plane polarised light passing through its solution.
Dextrorotatory rotating the plane of polarised light clockwise, indicated by \(d\) or \((+)\).
Laevorotatory rotating the plane of polarised light anticlockwise, indicated by \(l\) or \((-)\).
Polarimeter the instrument used to measure the angle through which a solution rotates plane polarised light.
Specific rotation the rotation produced by a solution under standard conditions of concentration and path length, used to identify a compound or measure its concentration.
Mutarotation the gradual change in the specific rotation of a freshly dissolved sugar as its α- and β-forms interconvert to equilibrium.
Invert sugar the laevorotatory mixture of glucose and fructose obtained by hydrolysing dextrorotatory sucrose.
Racemic mixture an equal mixture of both mirror-image forms of a compound, which shows no net optical rotation.
Stereospecific acting on only one optical isomer of a substrate, as enzymes do.
Supramolecular assembly an organised structure formed when two or more macromolecules associate through non-covalent forces to serve a specific subcellular function.
Chromatography a technique that separates the components of a mixture by their differing distribution between a stationary phase and a mobile phase.
Stationary phase the fixed material across or through which the mixture travels during chromatography.
Mobile phase the moving liquid that carries the mixture through the stationary phase.
Ion exchange chromatography a technique that separates ions and polar molecules on the basis of their charge, using a resin carrying fixed ions.
Cation-exchange chromatography ion exchange on a negatively charged matrix, which binds positively charged molecules.
Anion-exchange chromatography ion exchange on a positively charged matrix, which binds negatively charged molecules.
Elution the release of bound molecules from a chromatography matrix by changing the salt concentration or pH of the buffer.
Paper chromatography a partition technique using cellulose paper holding a thin layer of water as the stationary phase and a non-polar solvent as the mobile phase.
\(R_f\) value the ratio of the distance travelled by a substance to the distance travelled by the solvent, always less than 1 and dimensionless.

6.12 Points to Ponder

Life on earth uses only L-amino acids and only D-sugars, out of all the mirror-image forms available. Does this tell us something about how life began, or is it simply an accident that was locked in early and never reversed? What kind of evidence could settle the question?
A chemically synthesised drug is usually a racemic mixture, of which only half may be biologically useful. Is it reasonable to require pharmaceutical companies to test each mirror form separately, given the cost that adds to every new medicine? Consider both the patient’s position and the manufacturer’s.
Cellulose is the most abundant biomolecule on earth, and humans cannot digest a gram of it. Given the scale of world hunger, would it be a good idea to engineer a cellulose-digesting enzyme into the human gut, or into food crops? What could go wrong?
Vanaspati is made by hydrogenating liquid vegetable oil into a solid fat, which extends shelf life and lowers cost but changes the chemistry of what people eat. Where should the balance lie between industrial convenience, affordability and public health, and who should decide?
Enzymes are described as being highly specific, and this is usually presented as an advantage. Can you think of situations in which an enzyme’s narrow specificity is a disadvantage — for the organism, or for someone trying to use that enzyme industrially?
A person with lactose intolerance lacks a single enzyme. Milk is nevertheless treated as an essential food in dietary guidelines and school nutrition schemes across much of the world. How should public nutrition policy handle a deficiency that is normal in a large fraction of the adult population rather than rare?
Sumner spent years being told that crystallising an enzyme was impossible, and the prevailing theory of the day said he was wasting his time. How should a scientific community balance respect for established theory against the possibility that a lone dissenter is right? What would you have advised him?
Buchner showed that fermentation does not require a living cell. Today, industries choose between using a whole organism and using its isolated enzymes. What considerations — cost, control, contamination, waste, regulation — would you weigh in making that choice for a new product?
Beri-beri became widespread not because people ran out of food, but because rice began to be polished. Can you think of other instances in which a technological improvement in food processing created a nutritional problem? What does this suggest about how new food technologies should be assessed?
Vitamins are added to many packaged foods, and some governments mandate fortification of staples such as salt, oil or flour. Since fat-soluble vitamins accumulate in the body while water-soluble ones do not, does the distinction between the two types have any bearing on how fortification policy should be framed?
A recombinant protein may have exactly the intended amino acid sequence and still be useless because it has folded wrongly. What does this say about the limits of thinking of a gene as simply “the information for a protein”?
Antibiotics that jam the bacterial ribosome work by interfering with a non-covalent assembly rather than by breaking any chemical bond. What might this suggest about where else in a cell a drug could usefully act, and why such targets might be harder for bacteria to become resistant to — or easier?
Denaturation destroys a protein’s shape while leaving its sequence intact. If the sequence really determines the fold, why can so many proteins not simply refold on their own once normal conditions return?
Nearly every test in this chapter — Benedict’s, Biuret, iodine, acrolein, Sudan III — is over a century old, cheap, and needs no electricity. Modern laboratories have far more sensitive instruments. Is there still a case for teaching and using these classical tests, or should the practical course move on?
The same \(R_f\) value that identifies a leaf pigment can expose a dye illegally added to turmeric. What responsibilities does a person trained in such techniques have when they encounter evidence of adulteration in a product being sold to the public?

6.13 Objective Questions

A molecule is described as a macromolecule. This means it is:

A single unit of low molecular weight
Built by covalently linking many monomers
Held together only by hydrogen bonds
Always inorganic in nature

Which of the following is a heteropolysaccharide?

Glycogen
Cellulose
Peptidoglycan
Chitin

The two alternating sugar units of peptidoglycan are:

NAG and NAM
Glucose and galactose
Ribose and deoxyribose
Glucose and fructose

Lysozyme causes a bacterial cell to burst because it:

Denatures the membrane proteins
Cuts the glycosidic bond between NAM and NAG
Removes the peptide cross-links only
Oxidises the cell wall lipids

The monomer unit of chitin is:

β-D-glucose
N-acetylglucosamine
N-acetyl muramic acid
α-D-galactose

The pyranose form of a sugar is more stable than the furanose form because:

It contains more oxygen atoms
Ring closure between C-1 and C-5 produces a less strained ring
It has no asymmetric carbon
It cannot open into the chain form

Amylopectin differs from amylose in having:

β-1,4 linkages
α-1,6 linkages at branch points
No glycosidic bonds
A lower molecular weight

During the hydrolysis of starch, the colour given with iodine changes in the order:

Red → purple → blue → none
Blue → purple → red → none
None → blue → red → purple
Purple → none → blue → red

Barfoed’s reagent is used to distinguish:

Aldoses from ketoses
Monosaccharides from disaccharides
Proteins from lipids
Starch from glycogen

A student adds resorcinol and dilute HCl to a sugar solution and obtains a rapid cherry-red colour. The sugar is most likely:

Glucose
Fructose
Maltose
Cellulose

The only amino acid found in proteins that is not optically active is:

Alanine
Glycine
Serine
Proline

An amino acid with two carboxyl groups and one amino group will make a solution:

Acidic
Basic
Neutral
Non-conducting

The isoelectric pH of an amino acid is calculated as:

\(pK_1 - pK_2\)
\(\frac{1}{2}(pK_1 + pK_2)\)
\(\frac{1}{2}(pK_1 - pK_2)\)
\(pK_1 \times pK_2\)

In an α-helix, a hydrogen bond is formed between the carbonyl oxygen of one amino acid and the amide hydrogen of the:

Next amino acid
Second amino acid ahead
Fourth amino acid ahead
Last amino acid in the chain

Each turn of an α-helix contains:

2.0 amino acid residues
3.6 amino acid residues
5.4 amino acid residues
10 amino acid residues

Which of the following is a helix breaker?

Alanine
Proline
Valine
Leucine

Which bond is NOT involved in maintaining the tertiary structure of a protein?

Disulphide bond
Hydrogen bond
Peptide bond
Ionic interaction

In MALDI-MS, the function of the matrix is to:

Cut the protein into peptides
Absorb the laser energy and protect the protein
Separate peptides by charge
Stain the peptides for detection

Casein is precipitated most easily at:

Its isoelectric point
Very high pH
Very low salt concentration
Its optimum temperature

A protein that is conjugated with a nucleic acid is called a:

Glycoprotein
Chromoprotein
Nucleoprotein
Phosphoprotein

Natural fatty acids contain an even number of carbon atoms because they are synthesised from:

Glycerol
Acetyl CoA
Pyruvate
Amino acids

The number of milligrams of KOH required to saponify one gram of fat is called the:

Iodine number
Saponification number
Turnover number
Retention factor

Three oils have iodine numbers of 95, 118 and 140. The oil most likely to turn rancid fastest on exposure to air is the one with iodine number:

95
118
140
All will be equally affected

A wax differs from a fat in that the alcohol involved is:

Glycerol
A monohydric alcohol of high molecular weight
Cholesterol
Choline

Cholesterol is best described as a:

Simple lipid
Phospholipid
Derived lipid
Glycolipid

Which of the following is the coenzyme form of vitamin B₁?

FAD
TPP
NAD
Coenzyme A

Deficiency of nicotinic acid causes:

Beri-beri
Pellagra
Scurvy
Pernicious anaemia

The complete functional enzyme formed by an apoenzyme together with its cofactor is called the:

Holoenzyme
Proenzyme
Isoenzyme
Ribozyme

A vitamin deficiency cannot be corrected by eating more protein because:

Proteins are digested too slowly
The apoenzyme is only half the enzyme and the cofactor is still missing
Proteins are stored in the liver
Vitamins are made from lipids

Ribozymes are an exception to the statement that:

All proteins are enzymes
All enzymes are proteins
Enzymes lower activation energy
Enzymes are specific

An enzyme lowers the activation energy of a reaction but does not change:

The rate of the reaction
The free energy difference between reactant and product
The number of collisions between molecules
The formation of the ES complex

The active site of lysozyme is formed by amino acids that:

Lie next to one another in the linear sequence
Are brought close together by folding of the chain
Are attached after the protein is synthesised
Belong to a separate polypeptide chain

Two enzymes act on the same substrate. Enzyme P has a \(K_m\) of 0.5 mmol L⁻¹ and enzyme Q has a \(K_m\) of 5 mmol L⁻¹. It follows that:

P has a higher affinity for the substrate
Q has a higher affinity for the substrate
Both have the same affinity
Affinity cannot be compared using \(K_m\)

Increasing the substrate concentration will overcome the effect of:

A competitive inhibitor
A non-competitive inhibitor
Denaturation by heat
A change in pH

Malonic acid inhibits succinic dehydrogenase because it:

Binds covalently away from the active site
Resembles succinic acid in structure
Denatures the enzyme
Removes the cofactor

A molecule containing three asymmetric carbon atoms can have how many optical isomers?

D-fructose is laevorotatory. This shows that:

The D prefix always indicates left rotation
Configuration and direction of rotation are independent
Fructose has no asymmetric carbon
Fructose is a racemic mixture

The gradual change in the specific rotation of a freshly dissolved sugar until it reaches a steady value is called:

Inversion
Mutarotation
Racemisation
Denaturation

A racemic mixture shows no net optical rotation because:

It contains no asymmetric carbon
The rotations of the two mirror forms cancel out
It is insoluble in water
It has been denatured

Supramolecular assemblies are held together mainly by:

Peptide bonds
Glycosidic bonds
Non-covalent forces
Disulphide bonds

A ribosome is an assembly of:

Lipids and carbohydrates
Nucleic acids and proteins
Carbohydrates and proteins
Lipids and proteins

DEAE-cellulose is used as the matrix in:

Cation-exchange chromatography
Anion-exchange chromatography
Paper chromatography
Gel filtration

In ion exchange chromatography, bound proteins are eluted by:

Heating the column
Increasing the salt concentration of the buffer
Adding a reducing agent
Lowering the temperature

In paper chromatography, the stationary phase is:

The organic solvent
The water held on the cellulose paper
The silica resin
The buffer

The \(R_f\) value of a substance is always:

Greater than 1
Equal to 1
Less than 1
Expressed in centimetres

In a chromatogram of leaf pigments, the pigment that travels furthest from the loading point is:

Chlorophyll-b
Chlorophyll-a
Xanthophyll
Carotene

Assertion (A): Cellulose cannot be digested by humans although it is a polymer of glucose. Reason (R): Human digestive enzymes cannot hydrolyse β-1,4 glycosidic linkages.
Assertion (A): Glycogen is more heavily branched than amylose. Reason (R): Branching increases the number of free ends from which glucose can be released simultaneously.
Assertion (A): Sucrose does not give a positive Benedict’s test. Reason (R): Sucrose contains only non-reducing monosaccharide units.
Assertion (A): Boiling destroys the activity of an enzyme permanently in many cases. Reason (R): Heat breaks the covalent peptide bonds of the primary structure.
Assertion (A): The primary structure of a protein determines its three-dimensional shape. Reason (R): The order of amino acids decides which side chains are brought together on folding.
Assertion (A): The antiparallel arrangement of a β-pleated sheet allows maximum hydrogen bonding. Reason (R): In the antiparallel arrangement the donor and acceptor groups lie directly opposite one another.
Assertion (A): Fat-soluble vitamins can accumulate to harmful levels in the body. Reason (R): Fat-soluble vitamins are stored in the liver and adipose tissue and are poorly excreted.
Assertion (A): Vitamin C is a better source when citrus fruits are eaten raw than when vegetables are boiled. Reason (R): Ascorbic acid is heat labile.
Assertion (A): A non-competitive inhibitor cannot be displaced by adding excess substrate. Reason (R): A non-competitive inhibitor binds at the active site itself.
Assertion (A): Enzymes act on only one optical isomer of a substrate. Reason (R): The active site is a three-dimensional pocket that fits only one mirror-image form.
Assertion (A): Phospholipids form a bilayer spontaneously when shaken in water. Reason (R): Phospholipids are amphipathic, having a hydrophilic head and hydrophobic tails.
Assertion (A): The pH of the buffer is chosen before an ion exchange run rather than after it. Reason (R): The buffer pH decides the net charge a protein carries, and therefore whether it binds the matrix.
Assertion (A): The \(R_f\) value is reported instead of the distance travelled by a spot. Reason (R): The distance travelled by the solvent front differs from one run to another.
Assertion (A): A supramolecular assembly can be taken apart and rebuilt more easily than a macromolecule. Reason (R): Supramolecular assemblies are held by many weak non-covalent forces rather than by covalent bonds.
A sugar is used in yeast fermentation. Which sugar is most commonly used?

Lactose
Sucrose
Glucose
Cellulose

A student detects a pentose sugar in RNA. What is it?

Deoxyribose
Ribose
Fructose
Galactose

A disaccharide found in germinating seeds is:

Lactose
Maltose
Sucrose
Cellulose

A carbohydrate present in animal liver for storage is:

Starch
Glycogen
Cellulose
Chitin

A structural carbohydrate in plants is:

Glycogen
Cellulose
Starch
Inulin

A muscle contraction experiment shows the involvement of a fibrous protein. Identify it.

Actin
Albumin
Insulin
Casein

A nutrient insoluble in water but soluble in organic solvents is detected. It is:

Protein
Lipid
Vitamin C
Enzyme

A catalyst extracted from living cells is used in a reaction. It is:

Hormone
Enzyme
Vitamin
Lipid

A vitamin essential for vision is lacking. Identify it.

Vitamin A
Vitamin B
Vitamin C
Vitamin K

A protein composed of amino acids linked by peptide bonds is called:

Polysaccharide
Polypeptide
Lipoprotein
Steroid

A fat that remains solid at room temperature is called:

Oil
Wax
Fat
Sterol

The pH at which a protein has no net charge is called:

Neutral point
Buffer point
Isoelectric point
End point

A sugar gives a positive Benedict’s test due to the presence of:

Glycosidic bond
Free aldehyde or ketone group
Hydroxyl group
Carboxyl group

A carbohydrate is classified as a hexose. This means it has:

5 carbons
6 carbons
7 carbons
4 carbons

Ring formation in glucose occurs due to:

Peptide bond
Hydrogen bond
Hemiacetal formation
Ester bond

A polysaccharide is insoluble because:

It is non-polar
It has a large molecular size
It lacks oxygen
It has no bonds

A sugar is non-reducing because:

It lacks an OH group
Its carbonyl group is involved in bond formation
It has fewer carbons
It is insoluble

A protein loses its shape due to heat. This process is:

Hydrolysis
Denaturation
Oxidation
Reduction

Lipids are hydrophobic because they:

Contain nitrogen
Are non-polar
Contain peptide bonds
Are soluble in water

Enzymes are highly specific because:

They are lipids
They have active sites
They contain vitamins
They are carbohydrates

A vitamin acts as a cofactor because it:

Provides energy
Helps enzyme activity
Forms structure
Stores fat

Proteins are amphoteric because they contain:

Only \(\text{NH}_2\) group
Only COOH group
Both \(\text{NH}_2\) and COOH groups
Only OH group

Saturated fats differ from unsaturated fats by:

Presence of double bonds
Absence of double bonds
Presence of nitrogen
Presence of phosphate

A patient needs rapid energy during dehydration. Which is most suitable?

Starch
Glycogen
Glucose
Cellulose

A student performs the iodine test and gets a blue-black colour. The sample contains:

Glucose
Protein
Starch
Lipid

During digestion, starch is first broken into:

Glucose
Maltose
Lactose
Fructose

A sugar solution fails Benedict’s test but gives a positive result after hydrolysis. It is:

Glucose
Lactose
Sucrose
Fructose

A patient shows slow wound healing. Which vitamin deficiency is likely?

Vitamin A
Vitamin C
Vitamin D
Vitamin K

A student heats egg white and it solidifies. This shows:

Hydrolysis
Denaturation
Fermentation
Oxidation

A person consumes butter. The main type of lipid in it is:

Phospholipid
Triglyceride
Steroid
Glycolipid

An enzyme fails to work at high temperature. The reason is:

Increased activity
Denaturation
Activation
Synthesis

A person lacks the lactase enzyme. What happens?

Protein digestion stops
Lactose is not digested
Lipids are not digested
Vitamins are not absorbed

A diet lacks essential amino acids. What will be affected?

Lipid storage
Protein synthesis
Vitamin absorption
Enzyme activity

A protein with pI = 6 will stop moving in an electric field at:

pH 4
pH 6
pH 8
pH 10

At a pH lower than its pI, a protein will be:

Negatively charged
Neutral
Positively charged
Denatured

Two sugars have the same formula but different structures. These are:

Isomers
Polymers
Isotopes
Enzymes

A carbohydrate is highly branched for rapid energy release. It is:

Cellulose
Starch
Glycogen
Chitin

A sugar is reducing if it has:

Free carbonyl group
Glycosidic bond
Peptide bond
Phosphate group

DNA is more stable than RNA because it contains:

Ribose
Deoxyribose
Glucose
Fructose

A carbohydrate with a β-1,4 linkage is:

Starch
Glycogen
Cellulose
Sucrose

One protein has a globular shape and another is fibrous. Which is more soluble?

Fibrous
Globular
Both (a) and (b)
None of these

Oils differ from fats because they:

Are saturated
Are unsaturated
Contain proteins
Contain nitrogen

Enzyme activity decreases beyond the optimum pH. Why?

Substrate increases
Active site changes
Product increases
Energy decreases

A vitamin is fat-soluble. It will be:

Vitamin C
Vitamin B
Vitamin A
Vitamin B₁₂

A protein has quaternary structure. It contains:

One chain
Two or more chains
Only lipids
Only sugars

A lipid forms cell membrane structure. It is:

Triglyceride
Phospholipid
Wax
Steroid

Which sugar is least suitable for immediate energy?

Glucose
Fructose
Starch
Honey

Which sugar is most likely non-reducing?

Glucose
Maltose
Lactose
Sucrose

Which provides the fastest energy?

Polysaccharide
Disaccharide
Monosaccharide
Cellulose

Which nutrient is best for long-term energy storage?

Protein
Lipid
Vitamin
Enzyme

Which vitamin deficiency causes rickets?

Vitamin A
Vitamin B
Vitamin C
Vitamin D

Which enzyme property is most important for metabolism?

Colour
Specificity
Size
Shape

Which protein type is best for structural support?

Globular
Fibrous
Enzyme
Hormonal

Which lipid is best for waterproofing?

Wax
Fat
Oil
Sterol

Which vitamin is the best antioxidant?

Vitamin A
Vitamin E
Vitamin D
Vitamin C

To identify a reducing sugar, which test is best?

Iodine test
Benedict’s test
Biuret test
Xanthoproteic test

To differentiate glucose and fructose, which test is used?

Biuret
Selivanoff’s
Iodine
Sudan

To detect protein contamination in a sugar sample, use:

Benedict’s test
Biuret test
Iodine test
Acrolein test

A student wants to test for the presence of lipid in food. Which test is used?

Benedict’s
Sudan III
Iodine
Biuret

A student wants to test for proteins in food. Which test is used?

Benedict’s
Biuret
Iodine
Sudan

To test for lipids in a sample, use:

Biuret
Sudan III
Benedict’s
Iodine

To study enzyme activity, which factor should be controlled?

Colour
pH
Shape
Size

A student designs an experiment on enzyme activity versus temperature. What should be measured?

Colour change
Rate of reaction
Weight
Volume

To confirm protein denaturation, which condition is applied?

Cooling
Heating
Dilution
Filtration

For the following questions, choose from the options below:

Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
Assertion is true, but Reason is false.
Both Assertion and Reason are false.

Assertion (A): Carbohydrates are considered immediate sources of energy in living organisms. Reason (R): Glucose is easily oxidised during cellular respiration to release energy.
Assertion (A): Some carbohydrates do not follow the general formula \(C_x(H_2O)_y\). Reason (R): Compounds like deoxyribose and rhamnose deviate from this formula.
Assertion (A): Proteins show amphoteric nature. Reason (R): Proteins contain both amino (\(-\text{NH}_2\)) and carboxyl (\(-\text{COOH}\)) groups.
Assertion (A): Denaturation of proteins results in loss of biological activity. Reason (R): Denaturation disrupts the secondary and tertiary structures of proteins.
Assertion (A): Lipids are insoluble in water. Reason (R): Lipids consist mainly of non-polar hydrocarbon chains.
Assertion (A): Enzymes act as biological catalysts. Reason (R): Enzymes lower the activation energy of biochemical reactions.
Assertion (A): Vitamins are essential in small quantities for normal metabolism. Reason (R): Many vitamins function as cofactors or coenzymes in enzymatic reactions.
Assertion (A): Saturated fatty acids are usually solid at room temperature. Reason (R): Saturated fatty acids lack double bonds in their hydrocarbon chains.

Carbohydrates are also called “______ of life”.
Glucose is a ______ monosaccharide.
A protein is made up of repeating units called ______.
Lipids are soluble in ______ solvents but not in water.
The pH at which a protein has no net charge is called its ______.
Monosaccharides are classified based on the number of ______ atoms.
Fructose is a ______ sugar because it contains a ketone group.
When a protein loses its structure due to heat, the process is called ______.
Enzymes increase the rate of reaction by lowering ______ energy.
During digestion of starch, ______ sugar is formed.
The sugar present in milk is called ______.
A patient with a bleeding disorder may be deficient in vitamin ______.
A person unable to digest lactose lacks the enzyme ______.
A protein with pI = 6.5 will stop migrating in a region where the pH is ______.
If a protein is placed in a region where pH < pI, it will carry a ______ charge.
Ribose differs from deoxyribose by the presence of a ______ group at the 2′ carbon.
Sucrose is formed by the combination of ______ and ______.
Oils differ from fats because they contain more ______ fatty acids.
Enzyme specificity is due to the presence of a specific ______ site.

6.14 Check Your Concepts

Water makes up about 70% of a cell, yet it is not classified as a biomolecule. Justify this exclusion using the definition of a biomolecule.
Arrange the following in increasing order of structural complexity and state at which step covalent bonding gives way to non-covalent association: macromolecules, supramolecular assemblies, micromolecules, simple organic precursors.
Hormones are often listed as a separate class of biomolecules. Explain why this classification is not chemically sound, using two named examples.
Acetic acid fits the general formula \(C_x(H_2O)_y\) but is not a carbohydrate, while deoxyribose does not fit the formula but is a carbohydrate. Explain what this pair of exceptions tells us about the need for a structural definition.
Why does starch have no taste although it is built entirely from sweet-tasting glucose units?
Explain how ring closure in glucose creates a new centre of asymmetry that did not exist in the open-chain form, and name the two forms that result.
Glucose and fructose have the same molecular formula. Explain, in terms of functional groups, why one gives a positive result with a mild oxidising agent and the other does not.
Compare and contrast starch, glycogen and cellulose under the headings: monomer, type of linkage, branching, occurrence and digestibility in humans.
A single letter, α or β, separates a food store from a structural fibre. Explain this statement with reference to starch and cellulose.
Why is a heavily branched storage polysaccharide better suited to a working muscle than an unbranched one of the same total mass?
What happens if the bond between NAM and NAG in a bacterial cell wall is cut? Name the enzyme responsible and explain the consequence for the cell.
Chitin is described as “cellulose with one substitution”. State the substitution and explain why chitin retains cellulose’s mechanical strength.
Sucrose is non-reducing while maltose and lactose are reducing, although all three are disaccharides. Account for this difference in terms of the carbons involved in the glycosidic linkage.
A sugar solution gives no colour with Benedict’s reagent. After boiling with a little dilute acid and cooling, it gives a brick-red precipitate. Explain what has happened and identify the likely sugar.
Arrange the colours observed with iodine during the progressive hydrolysis of starch in the correct sequence, and state what each colour indicates about chain length.
Match the following:

Column A	Column B
a. Glycogen	i. Fungal cell wall
b. Chitin	ii. Bacterial cell wall
c. Peptidoglycan	iii. Plant cell wall
d. Cellulose	iv. Liver and muscle

Study Fig. 6.2. Label the aldehyde group of the open chain, the hydroxyl group that attacks it, the ring oxygen and the new asymmetric carbon, and state which structure is the hemiacetal.
Explain why the presence of nitrogen in a food sample is a useful first clue that it contains protein rather than carbohydrate or fat.
Twenty amino acids give rise to the enormous variety of proteins in nature. Explain how this is possible, and state which is more important — the number of amino acids in a chain or their order.
Why is glycine the only amino acid found in proteins that is not chiral?
Describe what happens to alanine as the pH of its solution is raised from below 2 to above 10, naming the form present at each stage and stating its net charge.
Study Fig. 6.11 and label the three ionic forms of alanine with their net charges, marking the form that exists at the isoelectric pH.
Explain why the side chain, and not the amino or carboxyl group, decides how an amino acid behaves once it is part of a protein.
Compare the α-helix and the β-pleated sheet under the headings: shape of the chain, position of the hydrogen bonds, spacing measurement and one named example.
Why are proline and tryptophan called helix breakers, although for different reasons?
Arrange the four levels of protein structure in order and state, for each, the type of bond that principally holds it together.
In an aqueous protein, non-polar side chains fold inward; in a membrane protein, the arrangement is reversed. Explain why the same molecule follows opposite rules in the two situations.
A protein is heated and loses all activity, yet chemical analysis shows that its amino acid sequence is unchanged. Explain this observation.
What happens if a protein is placed in a buffer whose pH equals its isoelectric point? Comment on both its net charge and its solubility.
Arrange the steps of protein sequencing by MALDI-MS in the correct order: ions accelerated down a flight tube; protein digested into peptides; mass differences read as sequence; peptides mixed with matrix and dried; laser pulse applied.
Explain why a mass spectrometer cannot simply be aimed at an intact protein, and how the matrix in MALDI-MS solves this problem.
Sample A gives a violet colour with Biuret reagent but no yellow colour with concentrated nitric acid. Sample B gives both. What can be concluded about each sample, and what different question does each test answer?
A student performs the Biuret test on a solution of a dipeptide and obtains no violet colour, and concludes that the sample contains no amino acids. Identify the error in this conclusion and explain the correct interpretation.
Distinguish between salting-in and salting-out, and explain why casein separates most readily at pH 4.6.
Compare globular and fibrous proteins under the headings: shape, solubility, structural level usually reached, function and one example of each.
The secondary derived proteins are listed as proteoses, peptones, oligopeptides and polypeptides. Explain the single principle that fixes this order, so that the list need not be memorised.
Explain why a fatty acid is described as amphipathic, and how this single property accounts for both the formation of a cell membrane and the cleaning action of soap.
Why do naturally occurring fatty acids almost always contain an even number of carbon atoms?
Two lipids are examined at room temperature; one is solid and the other liquid. Explain the difference in terms of chain shape and molecular packing, and state which is likely to have the higher iodine number.
What happens if an unsaturated oil is subjected to catalytic hydrogenation? Name the process and state the change in physical state, giving one industrial example.
Distinguish between hydrolytic and oxidative rancidification, and explain why highly unsaturated oils are more prone to the latter.
Match the following:

Column A	Column B
a. Lecithin	i. Derived lipid
b. Bees wax	ii. Phospholipid
c. Cholesterol	iii. Glycolipid
d. Cerebroside	iv. Simple lipid

Explain why the acrolein test detects fat, stating clearly which part of the fat molecule is actually being tested.
Why does a deficiency of a few milligrams of a vitamin bring an entire metabolic pathway to a halt, while a shortage of several grams of carbohydrate does not?
A person takes a very large dose of vitamin C on one day and none for the next month, and expects to be protected from scurvy. Explain why this reasoning fails, and state how the answer would differ for vitamin A.
Compare fat-soluble and water-soluble vitamins under the headings: storage in the body, excretion of excess, speed of appearance of deficiency and role as coenzymes.
Beri-beri spread in populations that adopted polished rice. Explain the connection between the processing step and the deficiency.
Explain, using the apoenzyme–cofactor relationship, why a cell well supplied with enzyme protein still cannot carry out a dehydrogenation reaction if the diet has lacked riboflavin for several weeks.
Match the following:

Column A	Column B
a. Vitamin B₁	i. Coenzyme A
b. Vitamin B₂	ii. Tetrahydrofolic acid
c. Vitamin B₅	iii. TPP
d. Vitamin B₉	iv. FMN and FAD

Explain why the statement “all enzymes are proteins” requires qualification, and name the class of molecule responsible for the exception.
Study Fig. 6.21. Label the reactant, the product, the activation energy with and without the enzyme, and the free energy difference between A and B. State which of these the enzyme changes and which it does not.
An enzyme cannot make an energetically impossible reaction happen. Explain this statement in terms of what an enzyme does and does not alter.
The active site of lysozyme is formed by amino acids at positions 35, 52, 62, 63 and 101 in a chain of 129. Use this fact to explain why denaturation abolishes enzyme activity without breaking a single peptide bond.
Compare the lock and key model with the induced fit model, and state the experimental evidence that made the second model necessary.
Cooling an enzyme pauses it, but heating destroys it. Explain the asymmetry, and relate it to why refrigeration preserves food.
Pepsin works best around pH 2 and trypsin around pH 8. Explain both why each enzyme has an optimum and why these two optima are so far apart.
What happens if substrate concentration is increased well beyond the point at which \(v_{max}\) has been reached? Explain your answer in terms of the ES complex.
Study Fig. 6.28. Label \(v_{max}\), half of \(v_{max}\) and \(K_m\), and explain in your own words why a smaller \(K_m\) indicates a higher affinity of the enzyme for its substrate.
A patient is poisoned by an organophosphorus compound. Explain why administering more of the natural substrate would not restore enzyme activity, whereas it would in a case of malonic acid inhibition.
A student sets up an experiment to measure the effect of substrate concentration on amylase activity, but changes the temperature of the water bath between readings. Identify the flaw in the protocol and state how it should be corrected.
Why does a molecule with four asymmetric carbon atoms have sixteen possible optical isomers, and what does it suggest about biological synthesis that a cell uses essentially one of them?
A student writes that “D-sugars are dextrorotatory because D stands for dextro”. Identify the error and support your answer with a named counter-example.
Explain why the hydrolysis of sucrose is called inversion, and why the resulting mixture is laevorotatory although one of its two products is dextrorotatory.
Explain why a compound made by a microorganism in a fermenter is usually a single optical isomer, whereas the same compound made by ordinary chemical synthesis is a racemic mixture.
Bacterial cell walls contain D-amino acids although proteins contain only L-amino acids. Suggest how this could be an advantage to the bacterium.
Explain why a cell holds a ribosome together with weak non-covalent forces rather than with covalent bonds, given that covalent bonds would be far stronger.
Chromatin and ribosomes are both assemblies of nucleic acid with protein, yet one stores information for decades and the other reads it in seconds. What does this tell us about the relationship between composition and function?
A recombinant protein produced in a bacterial fermenter has the correct amino acid sequence but no biological activity. Suggest what has gone wrong and explain why getting the gene right is not by itself sufficient.
Explain the common principle underlying all forms of chromatography, and state the specific property on which separation depends in ion exchange chromatography and in paper chromatography respectively.
Two proteins have isoelectric points of 4.5 and 9.0. A mixture is loaded onto a cation-exchange column equilibrated at pH 7.0. Predict which protein binds and which flows through, giving reasons.
Explain why proteins bound to an ion exchange column come off in a definite order rather than all at once when the salt concentration of the eluting buffer is raised.
Arrange the steps of an ion exchange chromatography run in the correct order: elute with buffer of increasing salt concentration; load the sample; equilibrate the column with starting buffer; regenerate the column; wash away unbound molecules.
A student sets up a paper chromatography experiment but dips the loading spot below the level of the solvent. Predict the result and explain the error.
Explain why the \(R_f\) value, rather than the distance travelled by a spot, is used to identify a compound, and state why \(R_f\) can never exceed 1.
In a chromatogram of leaf pigments, carotene travels furthest and chlorophyll-b the least. What does this tell you about the relative polarity of the two pigments, and why?
A student identifies a sugar with the formula \(C_6H_{12}O_6\) used for instant energy. Name it.
A sugar found in fruits is tested. What is it commonly called?
A carbohydrate cannot be hydrolysed further. What type is it?
A laboratory sample shows a biomolecule made of amino acids linked by peptide bonds. Identify the biomolecule and state one of its functions.
A food sample is insoluble in water but dissolves in ether. Identify the biomolecule and mention one example.
What does pI stand for?
A sugar shows an aldehyde group in its structure. What type of sugar is it?
A carbohydrate is composed of two monosaccharides. What is it called?
A student observes ring formation in glucose. What type of reaction causes it?
A pentose sugar is more reactive due to an \(-\text{OH}\) group at the 2′ carbon. Name that sugar.
What happens to a protein when the pH is higher than its pI?
During germination of seeds, which disaccharide is formed from starch?
A doctor tests blood sugar level. Which molecule is being measured?
In which direction will a positively charged protein migrate in an electric field?
At what pH will a protein with pI = 5 stop moving?
DNA is more stable than RNA. Which sugar difference explains this?
Identify a disaccharide that does not reduce Fehling’s solution.
Which polysaccharide gives a blue-black colour with iodine?
A patient is advised quick energy intake. Should they take glucose or starch? Give reasons.
A food is labelled “non-reducing sugar” and is tested. Which sugar is most likely present?
A student finds a sugar with the formula \(C_5H_{10}O_5\) in a nucleic acid sample. Identify the sugar and the nucleic acid.
A disaccharide known as “malt sugar” is detected. Name its components and bond type.
A plant cell wall sample is tested. Identify the major polysaccharide present.
A reserve food in the livers of many animals is analysed. Name this carbohydrate.
A sugar present in ATP is identified. Name the sugar.
What is meant by isoelectric point (pI)? How is it related to protein charge?
A student gets confused between an aldose and a ketose sugar. Explain the difference with one example each.
A sugar forms a five-membered ring. Explain what this indicates about its structure.
Explain why polysaccharides are generally insoluble in water.
A student observes that starch gives a blue-black colour with iodine. Explain the reason.
Explain why glycosidic bonds are important in carbohydrates.
A patient is given saline with glucose in hospital. Explain why glucose is preferred over polysaccharides.
A biologist studies an insect exoskeleton. Identify the polysaccharide present and its function.
Compare the behaviour of proteins in regions where the pH is less than, and greater than, their pI.
Explain why fructose and glucose have the same formula but different properties.
A carbohydrate shows branching and rapid energy release. Identify it and explain.
A sports coach suggests a glucose drink instead of fruit juice during a match. Evaluate the reason.
A lab test shows high glycogen in the liver. Evaluate its significance.
A student prefers honey over table sugar. Evaluate which provides faster energy, and why.
Suggest a way to differentiate reducing and non-reducing sugars experimentally.
Make a diagram to show ring formation in glucose and fructose.
In a classroom discussion, learners like Aditi, Nishhtha, Jasleen and Raj compare carbohydrates and lipids. They observe that carbohydrates are quick sources of energy, whereas lipids store energy for long-term use. Explain the structural and functional differences between carbohydrates and lipids.
A group of students including Faiza, Harbhajan, Anthony and Lepcha studies vitamins and enzymes. They learn that vitamins are required in small quantities and often act as cofactors, while enzymes speed up biochemical reactions. Explain the roles of vitamins and enzymes in metabolism.
A healthcare trainee group including Sudesh, Fardeen and Anita Joseph studies dietary deficiency disorders. They observe that lack of certain vitamins leads to diseases, while lack of proteins affects growth. Apply your knowledge to explain how deficiency of vitamins and proteins affects the human body.
A food science group including Dr. Iyer, Zubaida and Harpal studies digestion. They observe that carbohydrates, lipids and proteins are broken down by specific enzymes such as amylase, lipase and protease. Explain how enzyme specificity helps in the digestion of different biomolecules.
During a metabolic study, a group including Ananya, Rizwan and Bhartia observes that glucose provides immediate energy while glycogen serves as stored energy. Analyse the difference between monosaccharides and polysaccharides in terms of structure and function.
A research team led by Rohit, Noor and Oliver plans to design a balanced diet chart. Create a diet plan ensuring proper inclusion of carbohydrates, proteins, lipids, vitamins and enzymes, and justify your choices.
Identify the four amino acids shown in Fig. 6.36.

Diagram

Fig. 6.36 – Four amino acid structures labelled (A) to (D), each drawn with the amino group, carboxyl group and side chain shown: (A) side chain \(-\text{H}\), (B) side chain \(-\text{CH}_3\), (C) side chain \(-\text{CH}_2\text{OH}\), (D) side chain \(-\text{CH}_2\text{CH}_2\text{S}\text{CH}_3\).

6.15 Case-Based Questions

A food testing laboratory receives an unlabelled powdered sample said to be a nutritional supplement. A trainee analyst runs four tests. With iodine solution the sample turns blue-black. With Benedict’s reagent, warmed, it gives no colour change; but after boiling a fresh portion with dilute acid and cooling, the same test gives a brick-red precipitate. With Biuret reagent the sample gives a distinct violet colour. With Sudan III no red staining is seen.

Name the biomolecules present in the sample and the biomolecule that is absent, giving the evidence for each conclusion.
Explain why the Benedict’s test was negative at first but positive after treatment with acid.
Suggest one further test the analyst could perform to establish whether the protein in the sample contains aromatic amino acids.
The analyst reports the sample as “starch plus protein”. A senior scientist points out that the report is incomplete. What has been left out?

A nine-year-old child is brought to a clinic with bleeding gums, wounds on the legs that have not healed for several weeks, and an old scar that has reopened. The family reports that the child’s diet consists largely of boiled vegetables and rice, with almost no fresh fruit. The doctor notes that the child’s overall food intake is adequate in quantity.

Identify the vitamin most likely to be deficient and name the deficiency disease.
Explain how a deficiency of this vitamin produces the specific symptoms described, referring to the protein involved.
The family points out that the child does eat vegetables daily. Explain why boiling may not deliver this vitamin effectively.
Would the same reasoning apply if the deficient vitamin had been vitamin A? Justify your answer with reference to storage in the body.

A detergent manufacturer adds a protease enzyme to a washing powder to remove protein stains. Field reports arrive from two regions. In the first, users wash clothes in cold water and complain that stains are not removed. In the second, users boil their clothes with the detergent and report that the enzyme “stops working after the first wash”.

Explain, in terms of enzyme activity, why the cold-water users see poor stain removal.
Explain why boiling produces a different and more permanent problem, naming the process involved.
State which of the two situations can be reversed simply by changing the water temperature, and why the other cannot.
Suggest what property the manufacturer should look for when sourcing an enzyme for use in hot-water washing, and where in nature such an enzyme might be found.

A biotechnology company produces a recombinant human protein in bacteria. The cells are broken open, giving a mixture containing the target protein, whose isoelectric point is 5.2, along with two major bacterial contaminants with isoelectric points of 4.8 and 8.7. The downstream processing team plans to use ion exchange chromatography as the first purification step, and proposes running the column in a buffer at pH 7.0.

State the net charge carried by each of the three proteins in a buffer at pH 7.0, with reasons.
Which type of ion exchanger — cation or anion — should the team choose to capture the target protein? Justify your choice.
Explain which of the two contaminants will be removed easily in this step and which will not, and why.
Suggest how the bound proteins would be released from the column, and explain why they do not all come off at the same moment.

A food safety inspector collects a sample of turmeric powder from a market where adulteration with a synthetic dye has been reported. In the laboratory, an extract of the sample is spotted on a strip of chromatography paper alongside an extract of certified pure turmeric, and the strip is developed in a suitable solvent. The pure turmeric gives a single spot. The market sample gives two spots.

Name the technique being used and identify the stationary phase and the mobile phase.
Explain how the two spots in the market sample can be distinguished from one another quantitatively.
State one precaution that must be taken while placing the paper in the solvent, and explain what would happen if it were ignored.
The inspector must decide whether to report the trader. Discuss what responsibilities a person trained in this technique carries when the finding affects both a small trader’s livelihood and public health.

An edible oil company converts liquid sunflower oil into a solid cooking medium for sale in a low-income market, where the solid product is cheaper, keeps longer without refrigeration and is preferred by consumers for frying. Analysis shows that the iodine number of the raw oil is 132, while that of the finished product is 68. A public health body raises concerns about the process.

Name the industrial process used and explain the chemical change it brings about in the fatty acid chains.
Explain why the iodine number falls, and what this figure measures.
Explain why the finished product is solid at room temperature although the starting material was liquid.
Set out the competing considerations that the company, the consumer and the regulator each have to weigh. Do not attempt to settle the question; present the arguments fairly on each side.

A researcher studying antibiotic action isolates a bacterial strain and treats it in three ways. In treatment I, the culture is exposed to lysozyme purified from egg white. In treatment II, it is grown in the presence of an antibiotic that prevents peptide cross-links from forming. In treatment III, it is exposed to lysozyme but the bacteria have first been grown so that they are not dividing.

Name the macromolecule of the bacterial cell wall that is the target in all three treatments, and state its two alternating sugar units.
Explain the mechanism by which lysozyme damages the cell wall.
Explain why the antibiotic in treatment II is expected to affect growing cells much more than resting ones.
Predict, with reasons, whether treatment III will be more or less effective than treatment I.

Two students investigate amylase obtained from two different sources, A and B, using starch as substrate. Both enzymes are assayed at the same temperature and pH. Enzyme A reaches half its maximum velocity at a substrate concentration of 1.2 mmol L⁻¹, while enzyme B reaches half its maximum velocity at 6.0 mmol L⁻¹. Both eventually reach the same maximum velocity.

Name the constant that has been measured, and state its units.
Which enzyme has the greater affinity for starch? Justify your answer.
Both enzymes reach the same maximum velocity. Explain what limits the reaction rate at that point.
The students conclude that enzymes A and B must be the same enzyme because their maximum velocities match. Comment on the soundness of this conclusion.

A quality control laboratory receives two batches of a recombinant therapeutic protein from a fermenter. Both batches show the expected molecular weight and both give a positive Biuret test. However, batch 2 shows no biological activity when tested. The laboratory digests samples from both batches with trypsin and analyses the peptides by MALDI-MS. The peptide masses obtained from the two batches are identical.

Explain what MALDI-MS establishes about batch 2, and what it does not establish.
Suggest the most likely reason for the loss of activity in batch 2, given that the sequence is correct.
Which level or levels of protein structure would you suspect to be at fault? Give reasons.
Explain why the phrase “a gene contains the information for a protein” is an incomplete description of what is required to obtain a working protein.

A sports physiologist advises three athletes preparing for an endurance event. Athlete X is given a glucose drink immediately before the race. Athlete Y is advised to eat a starch-rich meal several hours earlier. Athlete Z prefers honey and takes a spoonful at the start line. A fourth athlete asks whether eating more roughage-rich salad at the start line would help.

Explain why glucose is effective for athlete X, referring to the digestion each carbohydrate requires.
Explain the purpose of athlete Y’s earlier starch-rich meal, and name the storage carbohydrate involved.
Identify the sugars present in honey and comment on athlete Z’s choice.
Explain why the fourth athlete’s suggestion would not provide usable energy, referring to the type of linkage involved.

A pharmaceutical company develops a new medicine whose molecule contains one asymmetric carbon. The compound is manufactured by chemical synthesis and comes out as an equal mixture of both mirror-image forms, which is cheaper than producing a single form. Preliminary tests show that the therapeutic effect resides in only one of the two forms. Regulators require the company to establish the biological activity and safety of each form separately, which will substantially increase the development cost and delay the launch.

Name the type of mixture the compound is produced as, and state what its net optical rotation would be.
Explain why an enzyme or receptor in the body may respond to one mirror form and not the other.
Suggest one manufacturing route that would yield predominantly a single form, and explain why it does so.
Present the case for and the case against the regulator’s requirement, taking account of patient safety, cost and access to medicines.

A school laboratory group is asked to determine which of four unlabelled bottles contains a fat, and to confirm the presence of glycerol in it. The group heats a portion of each sample strongly with potassium bisulphate and notes any odour produced. They also stain fresh portions with a red fat-soluble dye. One student proposes instead to boil each sample with sodium hydroxide and observe whether a soap forms.

Name the test that produces a pungent odour, and name the compound responsible for it.
Name the red fat-soluble dye and state what a positive result looks like.
Name the process the student’s alternative proposal depends on, and write what is formed from the fat.
Explain why the pungent-odour test specifically confirms glycerol, whereas the dye test only confirms that a lipid is present.

A student studies a sugar present in DNA.

Name the sugar.
Give its molecular formula.
State one structural feature.

A carbohydrate sample is classified as a pentose.

What does pentose mean?
Give one example.
State its biological role.

A student learns about malt sugar.

Identify the sugar.
Mention its components.
State where it is found.

A carbohydrate is found in a fungal cell wall.

Identify it.
State its monomer unit.
Mention its importance.

A student observes glucose forming a ring.

Explain why ring formation occurs.
Name the type of bond formed.
State which form is more stable.

A sugar is classified as a hexose.

Explain the term.
Give one example.
State its importance.

A disaccharide is formed by condensation.

What is a condensation reaction?
What is released during this process?
Name the bond formed.

A student studies reducing sugars.

Define a reducing sugar.
Give one example.
Explain why it gives a positive test.

A carbohydrate shows branching.

Explain how branching occurs.
Name the bond involved.
State its advantage.

A student studies cellulose.

Explain its structure.
Name the linkage present.
State why it is indigestible.

A patient requires quick energy.

Suggest a suitable carbohydrate.
Explain why it is effective.
State its source.

A student tests food with iodine.

What does a blue colour indicate?
Identify the substance.
Explain the result.

A lab test is done for reducing sugar.

Name the test.
State the expected observation.
Give one example of such a sugar.

Two sugars have the same formula but different structures.

Name the sugars.
Explain the difference.
State the term for such compounds.

A student compares starch and glycogen.

State one similarity.
State two differences.
Relate structure to function.

A sugar does not reduce Fehling’s solution.

Identify the sugar type.
Explain why it fails the test.
Give one example.

A nucleic acid contains ribose.

Identify the nucleic acid.
Compare with the DNA sugar.
State one implication of this difference.

A carbohydrate is insoluble but stores energy.

Identify it.
Explain why it is insoluble.
State the advantage of this property.

A doctor prefers glucose over starch for emergency care.

Justify the choice.
Compare the digestion rate.
State one practical advantage.

A scientist selects DNA for data storage.

Evaluate the role of the sugar.
Compare with RNA.
Conclude on suitability.

A student prefers honey for quick energy.

Evaluate the choice.
Identify the sugars present.
State the reason.

During a nutrition awareness session, students such as Zeenat, Jaya and Joe study the different biomolecules present in food. They learn that carbohydrates provide energy, proteins help in growth, lipids store energy, vitamins regulate functions, and enzymes act as catalysts. The instructor emphasises that each biomolecule has a unique structure and function.

Define biomolecules.
What are carbohydrates? Give one example.
What is the basic unit of proteins?
Why are enzymes called biological catalysts?

In a school laboratory, students such as Keerthi, Imtiaz, Fernanes and Norbu perform an experiment on enzyme activity. They observe that increasing temperature initially increases enzyme activity, but beyond a certain point the activity decreases sharply. They relate this to the structure of proteins.

What type of biomolecule are enzymes?
Why does enzyme activity increase with temperature initially?
What happens to enzymes at high temperature?
What is denaturation of proteins?

In a research lab, scientists such as Lakshmi Menon, Danish, Jaspal and Mark D’Silva analyse a protein sample exposed to extreme pH conditions. They observe that the protein loses its structure and enzyme activity is completely lost. The team studies the relationship between structure and function.

What level of protein structure is affected during denaturation?
How does pH affect enzyme activity?
Why does an enzyme lose its function after denaturation?
Analyse the importance of protein structure in biological systems.

You are part of a research team including Dr. Kapoor, Dr. Irfan, Dr. Chandani and Dr. Norbu. The team is designing an experiment to study the role of biomolecules in human nutrition. You are asked to create a detailed plan involving carbohydrates, proteins, lipids, vitamins and enzymes.

Suggest one test to identify carbohydrates.
Which test can be used to detect proteins?
How will you demonstrate enzyme activity?
Suggest one method to study the role of vitamins.

Study the structure of the biomolecule given in Fig. 6.35 and answer the questions that follow.

Identify the molecule.
What is its importance with respect to the production of an alcoholic beverage?
Which type of bond is found in this molecule?

Diagram

Fig. 6.35 – Two glucose ring units joined through an oxygen bridge between C-1 of the first and C-4 of the second, with all carbons numbered and the linking oxygen clearly marked, for identification as maltose.

The structure given in Fig. 6.37 is a hydrophobic compound.

Identify this compound.
What is the importance of this compound in human life?

Diagram

Fig. 6.37 – The four fused rings of a sterol with a hydroxyl group at C-3 and a branched hydrocarbon side chain at C-17, drawn for identification as cholesterol.

6.16 Practice With Data

The activity of an enzyme was measured at a series of temperatures, keeping pH and substrate concentration constant.

Temperature (°C)	0	10	20	30	40	50	60	70
Relative activity (%)	4	15	33	68	100	71	22	0

Plot relative activity against temperature and mark the optimum temperature.
Calculate the approximate factor by which activity rises between 20°C and 30°C, and comment on how this compares with the expected value of \(Q_{10}\).
Account for the shape of the curve on either side of the optimum, naming the process responsible for the fall.
The tube held at 0°C is warmed to 40°C, and the tube held at 70°C is cooled to 40°C. Predict the activity in each case and justify your prediction.

Two digestive enzymes, P and Q, were assayed over a range of pH values at constant temperature.

pH	1	2	3	5	7	8	9	11
Activity of P (%)	62	100	55	8	0	0	0	0
Activity of Q (%)	0	0	0	12	60	96	100	48

State the optimum pH of each enzyme.
Identify P and Q from the enzymes named in this chapter, and state the region of the digestive tract in which each acts.
Explain, in terms of the active site, why activity falls on both sides of the optimum.
Predict what would happen to enzyme P if the contents of the stomach were carried into the small intestine, and give a reason.

The velocity of an enzyme-catalysed reaction was measured at increasing substrate concentrations.

\([S]\) (mmol L⁻¹)	0.5	1.0	2.0	5.0	10.0	20.0
Velocity (µmol min⁻¹)	8	14	22	32	38	41

Plot velocity against substrate concentration.
Estimate \(v_{max}\) from the graph, stating its units.
Determine \(K_m\), showing on the graph how you obtained it.
Explain why the curve flattens at high substrate concentration rather than continuing to rise.

The experiment in B-Q150 was repeated in the presence of two different inhibitors, X and Y. The results obtained were:

Condition	\(v_{max}\) (µmol min⁻¹)	\(K_m\) (mmol L⁻¹)
No inhibitor	44	2.0
With inhibitor X	44	8.0
With inhibitor Y	17	2.0

Identify the type of inhibition produced by X and by Y.
Explain why \(v_{max}\) is unchanged in one case but reduced in the other.
State, with a reason, which inhibition could be overcome by adding a large excess of substrate.
Suggest one named example of each type of inhibitor from the chapter.

A mixture of amino acids was separated by paper chromatography. The solvent front travelled 15.0 cm from the loading point, and three spots were located at 12.0 cm, 9.6 cm and 6.0 cm.

Calculate the \(R_f\) value of each spot.
Arrange the three amino acids in order of increasing polarity, giving your reasoning.
In a repeat run the solvent was allowed to travel only 10.0 cm. Calculate the distance at which the first spot would now be expected.
Explain why \(R_f\) rather than distance travelled is quoted when a substance is being identified.

The dissociation constants of two amino acids are given below.

Amino acid	\(pK_1\) (–COOH)	\(pK_2\) (–NH₂)
Alanine	2.34	9.69
Leucine	2.36	9.60

Calculate the isoelectric pH of each amino acid.
State the net charge carried by alanine in a buffer of pH 3.0 and in a buffer of pH 11.0.
Towards which electrode will leucine migrate in a buffer of pH 8.0? Give a reason.
A mixture of the two amino acids is placed in a buffer at pH 6.0. Explain why they would be difficult to separate on the basis of charge alone.

Four proteins were found to have the isoelectric points shown below. Each was placed in a buffer at pH 7.0.

Protein	pI
A	4.2
B	5.9
C	7.0
D	9.4

State the net charge carried by each protein at pH 7.0.
Which protein will not migrate at all in an electric field at this pH, and why?
The mixture is loaded onto an anion-exchange column equilibrated at pH 7.0. Predict which proteins bind and which flow through.
Of the proteins that bind, state which would be eluted first as the salt concentration is raised, giving a reason.

A 2.5 g sample of an oil was found to absorb 3.2 g of iodine. A separate 1.5 g sample of the same oil required 285 mg of KOH for complete saponification.

Calculate the iodine number of the oil.
Calculate the saponification number of the oil.
Comment on the degree of unsaturation of this oil, comparing your answer with the iodine numbers of cotton seed oil (109) and soybean oil (133).
Predict how the iodine number would change if the oil were hydrogenated, and state the effect on its physical state.

In a Bradford’s assay, standard solutions of bovine serum albumin (BSA) gave the following optical density (OD) readings at a fixed wavelength.

BSA concentration (mg mL⁻¹)	0.2	0.4	0.6	0.8	1.0
Optical density	0.12	0.25	0.36	0.49	0.61

An unknown protein sample treated identically gave an OD of 0.42.

Plot the standard curve of OD against concentration.
Determine the protein concentration of the unknown sample from the graph.
The unknown sample had been diluted five times before the assay. Calculate the protein concentration of the original sample.
Explain why a set of standards must be run alongside the unknown in every assay, rather than using a curve prepared on an earlier day.

A solution of glucose was examined in a polarimeter. The observed angle of rotation was \(+13.2°\), the sample tube was 2.0 dm long, and the concentration of the solution was 0.125 g mL⁻¹.

Calculate the specific rotation of the sample, stating the formula used and what each symbol represents.
State whether the sample is dextrorotatory or laevorotatory, and how you can tell.
If the same solution were placed in a tube of 1.0 dm length, predict the observed angle of rotation.
Explain how this measurement could be used in a sugar factory to determine the concentration of an unknown sugar solution.

Freshly dissolved α-D-glucose has a specific rotation of \(+112°\) and freshly dissolved β-D-glucose has a specific rotation of \(+18.7°\). On standing, both solutions reach the same final value of \(+52.7°\).

Name the phenomenon described.
Explain, in terms of ring opening and closing, why both solutions arrive at the same value.
Taking the equilibrium mixture to contain a fraction \(x\) of the α-form and \((1-x)\) of the β-form, set up an equation for the final specific rotation and calculate the percentage of each form present at equilibrium.
State what would be observed if a solution of the equilibrium mixture were left standing for a further day.

Sucrose solution has a specific rotation of \(+66.5°\). On complete hydrolysis it yields an equimolar mixture of glucose (\(+52.7°\)) and fructose (\(-92°\)).

Calculate the approximate specific rotation of the equimolar product mixture.
Explain why the sign of rotation changes during the reaction, and name the process and the product.
Name the enzyme that catalyses this reaction and state one natural product rich in the resulting mixture.
Suggest one chemical test, other than polarimetry, by which the completion of this hydrolysis could be confirmed.

Two polysaccharide samples of equal mass were analysed. Sample I was an unbranched amylose chain of 12,000 glucose units. Sample II was a glycogen molecule of 12,000 glucose units with a branch point on average every 12 units.

State the number of non-reducing free ends available to an enzyme in Sample I.
Estimate the number of free ends in Sample II, showing your working.
Using your answers, explain why glycogen is better suited than amylose to the sudden energy demand of a working muscle.
Name the two types of glycosidic linkage present in Sample II and state where each occurs.

A 100 g sample of starch was analysed and found to contain 18% α-amylose and 82% amylopectin. The average molecular weight of α-amylose in the sample was 40,000 and that of amylopectin was 80,000. Each glucose residue in a chain contributes a mass of 162 u.

Calculate the mass of each component present in the 100 g sample.
Calculate the approximate number of glucose residues in one average molecule of α-amylose in this sample.
Amylopectin has an average chain length of about 24 glucose units between branch points. Estimate the number of branch points in one average amylopectin molecule.
Predict the colour this starch sample would give with iodine, and state how that colour would change as hydrolysis proceeded.