10. Biomolecules

What you will learn

• Classify carbohydrates and interpret stereochemical drawing conventions.
• Connect open-chain evidence with cyclic sugar structures and anomers.
• Compare glycosidic linkages and reducing behaviour.
• Relate amino-acid sequence, peptide geometry and protein folding.
• Explain enzyme, vitamin and hormone roles without oversimplified universal rules.
• Read nucleic-acid components, direction and complementarity in both 2D and 3D.

Big idea: biological function depends on connections, direction and shape—not just a list of atoms.

Biomolecules reuse familiar functional groups in precisely organised structures. Changing one linkage or reversing one sequence can change behaviour while leaving much of the atom inventory the same. This chapter makes those hidden differences visible.

For each paired model, practise the same cycle: identify the labelled connections, rotate the spatial view, then redraw the exam convention with the correct orientation. State when a drawing is a projection, a connectivity shorthand or a schematic shape model.

Biological carbohydrates are classified by both their carbonyl chemistry and what hydrolysis can break apart.

Carbohydrates are polyhydroxy aldehydes or ketones, or substances that give such units on hydrolysis. The traditional “hydrate of carbon” formula pattern is not a reliable definition: some substances fit the ratio without being carbohydrates, and important carbohydrate derivatives depart from it.

A monosaccharide cannot be hydrolysed into a simpler carbohydrate unit. Oligosaccharides give a small number of units; disaccharides give two. Polysaccharides contain many linked units. Classifying by unit count is different from classifying a monosaccharide as an aldose or ketose.

An aldose has an aldehyde carbonyl in its open-chain form; a ketose has a ketone carbonyl. Triose, tetrose, pentose and hexose refer to carbon count. A ketohexose and an aldohexose can share a molecular formula while differing in carbonyl position and stereochemistry.

Glucose can be obtained through hydrolysis of suitable larger carbohydrates. Its chemistry supports an unbranched carbon skeleton with a carbonyl group and several hydroxyl groups. Carbonyl-derivative formation, controlled oxidation and acetylation each probe a different structural feature. An experiment is useful only if its conclusion is kept within what it actually measures.

Sucrose hydrolysis supplies glucose together with fructose, while sufficiently complete starch hydrolysis supplies glucose units. Glucose acetylation can account for five hydroxyl groups in its relevant derivative. The cyclic hemiacetal description explains why its behaviour is not identical to an always-open free aldehyde under every test or derivatisation condition.

Controlled bromine-water oxidation supports an aldehyde-derived end group by converting it to a carboxylic acid. Stronger oxidation can also alter the terminal primary-alcohol end. Formation of an acetate derivative shows the number of available hydroxyl groups. These observations together are more persuasive than any one reaction used alone.

For the open-chain D-glucose Fischer projection, place the aldehyde end at the top. The hydroxyl positions from C2 through C5 are right, left, right, right. In a Fischer convention, horizontal bonds point towards the viewer and vertical bonds point away. D/L describes configuration relative to a reference; it does not by itself state the sign of optical rotation.

The hydroxyl at C5 can add internally to the C1 carbonyl, forming a cyclic hemiacetal. A six-membered pyranose ring results. C1 becomes a new stereogenic anomeric centre, producing alpha and beta forms. In the conventional D-glucose Haworth drawing, the C5-linked CH₂OH is up; alpha has its anomeric OH on the opposite face, while beta has it on the same face.

A Haworth drawing is a useful flat convention, not a claim that the ring is physically flat. The real ring can pucker. In solution, anomers can interconvert through opening and reclosing, changing measured optical rotation towards an equilibrium value. This is mutarotation. A derivative that locks the anomeric group can prevent the same free hemiacetal behaviour.

Fructose is a ketohexose in its open-chain form. It can form cyclic hemiketals, including five-membered furanose representations used in common carbohydrate discussions. It is not correct to imply that every fructose molecule in every solution exists in only one ring form. Its anomeric carbon is C2, not C1.

A reducing sugar can access a form that reduces an appropriate test reagent under the test conditions. Ketoses can also give positive reducing tests through isomerisation in the alkaline test environment. A reducing test therefore does not prove that the only equilibrium form is an open-chain aldehyde.

A glycosidic linkage joins monosaccharide units through an anomeric-centre-derived connection. Sucrose joins the anomeric centres of glucose and fructose and is nonreducing in the ordinary comparison. Maltose and lactose retain a free anomeric hemiacetal site and can behave as reducing sugars.

Sucrose hydrolysis can reverse the sign of a mixture’s optical rotation because the product contributions differ. Inversion in this context describes that optical change. It is not the same as backside stereochemical inversion at one saturated carbon in substitution chemistry.

Starch is a glucose-based storage polysaccharide containing largely alpha-linked arrangements. Amylose is mainly unbranched; amylopectin contains branches. Glycogen is another branched glucose storage system. Cellulose has beta 1→4 glucose linkages supporting an extended structural arrangement. Similar monomers can therefore give very different materials because linkage and branching differ.

Carbohydrates provide energy-related, storage, structural and molecular-recognition roles. Their stereochemistry is chemically consequential: enzymes recognise particular configurations and links. A formula that counts only carbon, hydrogen and oxygen does not capture those differences.

Why does sucrose lack the same reducing end that maltose retains?

Locate each anomeric carbon. In sucrose both are tied into the glycosidic connection; in maltose one anomeric hemiacetal remains free.

Result: The difference is anomeric-centre availability, not simply whether the molecule contains oxygen or two sugar units.

1. Which carbon becomes the glucose anomeric centre on ordinary ring closure?
2. Which carbon is the fructose anomeric centre?
3. Why can cellulose and starch behave differently despite both being glucose polymers?

Answers

1. C1.
2. C2.
3. Their linkages and higher arrangements differ.

Proteins turn a sequence of amino-acid residues into a structure. Both the sequence and the folding matter.

An ordinary alpha-amino acid has amino and carboxyl-derived groups attached to the same alpha carbon, along with a side chain and usually hydrogen. Glycine has two hydrogens at that carbon and is achiral there. In a zwitterion, a molecule contains both a positively charged ammonium group and a negatively charged carboxylate group.

The side chain supplies many differences in polarity, charge, reactivity and packing. Most common protein amino-acid alpha centres use the L configuration; that label should not be confused with a universal optical-rotation sign. Proline has a side chain connected back to the backbone nitrogen, giving a distinctive ring constraint.

Acidic and basic side chains can change charge with pH. Polar and nonpolar classifications describe interaction tendencies rather than rigid behaviour in every environment. Essential amino acids are those an organism cannot synthesise sufficiently for its needs in the relevant physiological setting. Essentiality can depend on species, life stage and conditions; it is not a classification based only on molecular size.

A peptide bond joins the carboxyl-derived carbon of one residue to the amino nitrogen of another. Write a sequence from its amino terminus towards its carboxyl terminus unless another convention is stated. The reversed residue sequence is generally a different peptide.

$$\mathrm{C_{2}H_{5}NO_{2}} + \mathrm{C_{3}H_{7}NO_{2}} + \mathrm{C_{5}H_{11}NO_{2}} \longrightarrow \mathrm{C_{10}H_{19}N_{3}O_{4}} + 2 \mathrm{H_{2}O}$$

Form a linear tripeptide from glycine, alanine and valine in that order. Using rounded molar masses $75$, $89$, $117$ and $18\,\mathrm{g\,mol^{-1}}$ for water, find its molar mass.

A linear chain of three residues has two peptide links and loses two water molecules in the formal condensation count. Sequence information is Gly–Ala–Val; the atom-balanced formula alone does not distinguish its sequence isomers.

Result: $M=245\,\mathrm{g\,mol^{-1}}$ using the stipulated rounded data.

Primary structure is the covalent residue sequence. Secondary structure describes recurring local backbone arrangements, such as alpha helices and beta sheets, stabilised by backbone interactions. In the familiar alpha helix, a carbonyl oxygen hydrogen-bonds to a later backbone N–H group in a regular pattern. Beta sheets place extended strands beside one another, with hydrogen bonds between strands.

Tertiary structure is the overall three-dimensional fold of one polypeptide chain. Side-chain interactions, hydrogen bonds, ionic interactions, hydrophobic packing and sometimes disulfide links contribute. Quaternary structure concerns the arrangement of multiple chains into an assembly. Not every protein has all these levels in the same way.

Fibrous proteins commonly support extended structural roles; globular proteins commonly form compact folded shapes. These are broad structural descriptions. A protein’s function depends on its full architecture, not just its amino-acid inventory.

Heat, pH changes or chemical conditions can disrupt the interactions that maintain a protein’s native fold. This is denaturation. It does not normally mean every peptide bond has been hydrolysed. Loss of the native arrangement can remove biological activity even when the primary sequence remains largely intact. Some denaturation is reversible; some is not.

A protein loses activity after a change in pH but its peptide sequence remains intact. Which structural change is most directly implicated?

Activity may require a precise folded arrangement and correct charge states. Disrupting those interactions can alter shape without cutting all covalent backbone links.

Result: A change in conformation or interaction state, consistent with denaturation or related loss of the native functional structure.

1. Why is glycine not chiral at its alpha carbon?
2. How many peptide links are present in an ordinary linear chain of five residues?
3. What structural level specifies residue order?

Answers

1. That carbon has two identical hydrogen substituents.
2. $4$.
3. Primary structure.

An enzyme speeds a particular chemical route by binding participants in a productive environment.

Enzymes are biological catalysts. Most are proteins, while some RNA molecules also catalyse reactions. An active site provides a particular arrangement of groups that binds a substrate and supports its conversion. Specificity means that structure and chemical interactions restrict which substrates and reactions are favoured; it need not mean one enzyme has only one possible substrate in every context.

Binding forms an enzyme–substrate complex and can induce structural adjustments. A productive active site stabilises relevant transition-state features or offers a better route. Product release regenerates the catalyst. A rigid lock-and-key picture is a useful first sketch but does not capture all induced-fit behaviour.

$$E+S\rightleftharpoons ES\longrightarrow E+P$$

Changing temperature or pH affects activity through both reaction kinetics and enzyme structure. A rise in temperature may initially increase rate, but excessive heating can disrupt the active structure. The optimum is enzyme- and condition-dependent, not a universal temperature for every biological catalyst.

At sufficiently high substrate availability, a fixed enzyme population can become largely occupied. Adding still more substrate then gives a smaller rate increase. Increasing enzyme amount under otherwise suitable conditions can change the available catalytic capacity. These are distinct experimental changes.

Why does an enzyme not shift the equilibrium constant merely by lowering a barrier?

It supplies a route available in both directions. The same reactant and product states still determine the equilibrium condition at the same temperature.

Result: It changes the speed of approach to equilibrium, not the equilibrium constant for the same overall reaction.

When interpreting an activity graph, first identify what was changed: substrate concentration, enzyme amount, temperature, pH or an inhibitor. A decrease in activity alone does not uniquely identify the molecular cause.

1. What distinguishes an enzyme’s active site from an arbitrary surface patch?
2. Must every enzyme be a protein?

Answers

1. Its arranged groups support the relevant binding and catalytic pathway.
2. No. Catalytic RNA provides an important exception.

Vitamins are small-amount requirements with distinct chemical roles, not a single interchangeable class of energy-providing molecules.

Vitamins are organic substances needed in relatively small amounts for normal biological function. They have different structures and roles. Many support enzyme systems or specialised physiological processes. Calling something a vitamin does not mean it supplies energy like a bulk fuel molecule.

The familiar fat-soluble group comprises A, D, E and K. Vitamin C and the B group are water-soluble. Solubility helps explain transport and storage patterns but does not make a universal rule that every water-soluble vitamin is immediately lost. Vitamin B12 is an important exception with substantial body stores.

These are teaching associations, not one-symptom diagnostic rules. A deficiency can have several manifestations, and the same observation can have other causes. Excess intake of a substance is not automatically beneficial simply because deficiency is harmful.

Why is “all water-soluble vitamins are never stored” too strong?

Solubility is a useful organising property, but the actual transport and storage system matters. B12 has long-lasting stores despite being water-soluble.

Result: State the broad pattern and its important exception rather than replacing it with an absolute rule.

1. Which letters form the usual fat-soluble vitamin group?
2. Which water-soluble vitamin is an important storage exception?

Answers

1. A, D, E and K.
2. Vitamin B12.

Nucleic acids store sequence information in a directional backbone. Pairing links the sequences, while orientation tells you how to read them.

A nucleoside contains a nitrogenous base joined to a sugar. Adding phosphate gives a nucleotide. DNA uses deoxyribose, while RNA uses ribose. Adenine and guanine are purines; cytosine, thymine and uracil are pyrimidines. DNA normally uses thymine, while RNA normally uses uracil.

The sugar–phosphate backbone uses phosphodiester links. The prime marks distinguish sugar-carbon positions from base positions. A chain has direction, conventionally labelled 5′ to 3′. These are not temperature marks and should remain clear in an exam drawing.

In the familiar DNA double helix, two complementary strands are antiparallel. Adenine pairs with thymine and guanine with cytosine in the standard Watson–Crick pattern. Hydrogen bonds support specific pairing; stacking and the surrounding environment also contribute to helix stability. The backbone bonds within a strand are covalent, not the same as the interstrand base-pair hydrogen bonds.

One DNA strand is written $5′\text{-ACTGGA-}3′$. Write its aligned complement and then write that complement in its own 5′→3′ direction.

Pair A with T and C with G at each aligned position. The aligned strand must run in the opposite direction. Reverse its written order if you want to report that strand from its own 5′ end.

Result: Aligned: 3′-TGACCT-5′. In its own 5′→3′ direction: 5′-TCCAGT-3′.

A double-stranded DNA sample contains 1000 nucleotide residues in total, including 220 adenines. Find the other base counts and the standard-pair hydrogen-bond count.

In a fully paired double-stranded sample, thymine count equals adenine count and cytosine equals guanine. Divide the remaining residues equally between C and G. Count one base pair for each A or G on its respective pair type.

Result: T = 220, G = C = 280. The standard-pair hydrogen-bond count is $2(220)+3(280)=1280$. These equalities apply to the fully paired double-stranded sample, not automatically to one isolated strand.

DNA stores information that can be copied through complementary pairing. RNA has several roles in expression and regulation, including messenger, transfer and ribosomal functions. RNA often forms internal paired structures even though it is commonly described as single-stranded. The chemistry of a nucleotide is therefore connected with both sequence and three-dimensional folding.

The double-helical DNA model associated with Watson and Crick drew on chemical reasoning and crucial diffraction evidence, including work by Rosalind Franklin and Maurice Wilkins. The achievement connected base pairing with a plausible copying principle; it was not the discovery of DNA itself.

Har Gobind Khorana’s work on controlled nucleic-acid synthesis helped connect defined nucleotide sequences with the genetic code. He shared the 1968 physiology or medicine prize with Robert Holley and Marshall Nirenberg for work on interpreting the code and its role in protein synthesis.

DNA profiling compares selected variable sequence features rather than normally reading every base of a whole genome. Inherited variation makes profiles informative for identity-related or relationship questions. A profile must be interpreted with the sampling, laboratory method and statistical context; a pattern by itself is not a complete account of an event.

1. How does a nucleotide differ from a nucleoside?
2. Are base-pair hydrogen bonds the bonds that join adjacent nucleotides along a strand?
3. Do equal A/T counts have to hold in one isolated DNA strand?

Answers

1. It includes phosphate in addition to the base–sugar nucleoside unit.
2. No. Adjacent nucleotides are joined by covalent phosphodiester links.
3. No. The familiar equality is for an appropriately paired double-stranded sample.

A hormone is a chemical signal. Its effect depends on who can recognise it, not merely on where the molecule travels.

Endocrine hormones are chemical messengers released by specialised tissues and carried to responsive targets. A target responds through suitable receptors and its internal signalling machinery. Circulating through a tissue does not guarantee that every cell responds in the same way.

Hormones belong to different chemical classes. Peptide or protein hormones include insulin. Steroid hormones have a characteristic steroid-derived framework. Other hormones are amino-acid-derived, including thyroid-related and catecholamine examples. Shared signalling function does not imply shared molecular structure or mode of action.

Hormone production, release and response are regulated. More signal is not automatically better, and an observed biological change is not enough to infer a unique hormonal cause. For this chemistry chapter, focus on molecular class, recognition and the difference between a chemical signal and a bulk nutrient.

Why can one circulating hormone produce a response in one cell type but little response in another?

Receptor presence, receptor state and downstream cellular machinery determine how the signal is interpreted. The same external concentration does not give identical recognition systems.

Result: The target’s molecular recognition and signalling context control its response.

The chapter’s common theme is structure-specific function. Sugar linkages, peptide sequences, nucleic-acid direction and receptor recognition all show why a molecular formula is only the beginning of a biological explanation.

1. What makes a cell a responsive target for a hormone?
2. Must all hormones be proteins?

Answers

1. Suitable recognition and signalling machinery, including relevant receptors.
2. No. Steroid and amino-acid-derived hormones illustrate other chemical classes.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

10.1 Distinguish a monosaccharide from a disaccharide by hydrolysis.

10.2 What two pieces of information are encoded by “ketohexose”?

10.3 Why is D not a universal positive-rotation label?

10.4 Give the OH-side sequence for C2–C5 in the usual open-chain D-glucose Fischer projection.

10.5 Identify the bond-making event in glucose pyranose closure.

10.6 Distinguish alpha and beta D-glucopyranose in a conventional Haworth drawing.

10.7 Why is a Haworth hexagon not a measured flat molecular structure?

10.8 Why can fructose give a reducing test despite being classified as a ketose?

10.9 Explain sucrose’s ordinary nonreducing classification.

10.10 Which structural feature distinguishes cellulose from starch most fundamentally?

10.11 Why is glycine achiral at its alpha carbon?

10.12 What is a zwitterion?

10.13 How many peptide links form in an ordinary linear chain of seven residues?

10.14 Using rounded masses $75$, $89$ and water $18\,\mathrm{g\,mol^{-1}}$, find Gly–Ala dipeptide molar mass.

10.15 Does Ala–Gly necessarily equal Gly–Ala?

10.16 Compare primary and secondary protein structure.

10.17 What remains largely intact in ordinary denaturation but is broken in extensive hydrolysis?

10.18 Explain why enzyme activity can fall on excessive warming.

10.19 Does an enzyme change the equilibrium constant of its unchanged overall reaction at fixed temperature?

10.20 Name the usual fat-soluble vitamin group and one important exception to a simplistic water-soluble storage rule.

10.21 What is the chemical difference between a nucleoside and a nucleotide?

10.22 Distinguish the DNA backbone bond from the interstrand base-pair interaction.

10.23 Write the reverse complement of 5′-ACTGGA-3′ in its own 5′→3′ direction.

10.24 A fully paired double-stranded DNA sample has $18\%$ adenine. Find the other percentages.

10.25 Connect molecular structure to function using one example from each of carbohydrates, proteins and nucleic acids.

Answers

10.1 A monosaccharide does not hydrolyse to a simpler carbohydrate unit; a disaccharide gives two monosaccharide units.

10.2 A ketone-type open-chain carbonyl and a six-carbon skeleton.

10.3 D/L is a configurational relation, while rotation sign is an independently measured property.

10.4 Right, left, right, right.

10.5 The C5 hydroxyl adds internally to the C1 carbonyl, creating a cyclic hemiacetal.

10.6 The anomeric OH is opposite to the C5-linked CH₂OH face for alpha and on the same face for beta.

10.7 It is a drawing convention; real rings can pucker.

10.8 The alkaline test conditions can permit isomerisation to reducing forms.

10.9 Both anomeric centres are involved in its glycosidic linkage, leaving no free ordinary hemiacetal reducing end.

10.10 Their glycosidic linkage configurations and resulting higher arrangements differ.

10.11 It has two identical hydrogen substituents at that carbon.

10.12 A species with both positively and negatively charged groups in the same molecule, often net neutral overall.

10.13 $6$.

10.14 $146\,\mathrm{g\,mol^{-1}}$.

10.15 No. Sequence direction changes the structure even when the residue inventory matches.

10.16 Primary is residue sequence; secondary is recurring local backbone arrangement such as helices or sheets.

10.17 The peptide-linked primary sequence can remain largely intact in denaturation; hydrolysis breaks backbone covalent links.

10.18 A faster chemical rate can be overtaken by disruption of the active folded structure.

10.19 No. It changes the kinetic route and approach to equilibrium.

10.20 A, D, E and K; water-soluble B12 can have substantial long-lasting stores.

10.21 The nucleotide includes phosphate as well as the base–sugar unit.

10.22 The backbone uses covalent phosphodiester links; standard interstrand pairing includes hydrogen bonds between complementary bases.

10.23 5′-TCCAGT-3′.

10.24 T = 18%, C = G = 32%. The equality is for the paired sample.

10.25 Carbohydrate linkage controls polymer arrangement; protein sequence and folding control recognition or catalysis; nucleic-acid sequence and direction support pairing and information use. A molecular formula alone omits these features.