Carbohydrates are polyhydroxy aldehydes or ketones (or compounds that yield these on hydrolysis), broadly classified by their behaviour on hydrolysis.
| Category | Hydrolysis Behaviour | Example |
|---|---|---|
| Monosaccharides | Cannot be hydrolysed into smaller units | Glucose, fructose, ribose |
| Oligosaccharides | Hydrolyse to give 2-10 monosaccharide units | Sucrose, maltose, lactose (all disaccharides) |
| Polysaccharides | Hydrolyse to give many (hundreds/thousands of) monosaccharide units | Starch, cellulose, glycogen |
A reducing sugar can reduce Tollens'/Fehling's reagent, since it has a free (or potentially free, via ring-opening) aldehyde or α-hydroxy ketone group available. Glucose, fructose, maltose, and lactose are all reducing sugars. Sucrose is a notable non-reducing sugar, since both its anomeric carbons are involved in the glycosidic linkage, leaving no free reducing group available.
Explain why sucrose does not give a positive Tollens' test, while maltose does.
In sucrose, the glycosidic bond forms between the anomeric carbon of glucose (C1) and the anomeric carbon of fructose (C2) — both potential reducing (anomeric) centres are tied up in this linkage, leaving no free aldehyde or hemiketal group available to react with Tollens' reagent. In maltose, the glycosidic bond forms between the anomeric carbon of one glucose unit and a non-anomeric carbon (C4) of the second glucose unit — this leaves the second unit's anomeric carbon free, retaining its ability to open into a reactive aldehyde form and reduce Tollens' reagent.
Glucose has the open-chain structure $\text{CH}_2\text{OH-(CHOH)}_4\text{-CHO}$ — an aldohexose with 4 chiral (stereogenic) centres, giving it $2^4 = 16$ possible stereoisomers, of which D-glucose is the biologically predominant form.
In solution, glucose predominantly exists as a stable six-membered ring (pyranose form), formed when the C5 -OH group attacks the C1 aldehyde carbon intramolecularly, creating a new stereocentre at C1 — this new centre is called the anomeric carbon, and gives rise to two distinct forms: α-D-glucose and β-D-glucose (anomers), differing only in the spatial orientation of the -OH at C1.
Mutarotation is the spontaneous interconversion between the α and β anomeric forms (through the open-chain form as an intermediate) when either pure anomer is dissolved in water, until an equilibrium mixture of both forms (plus a small amount of open-chain form) is reached — observed experimentally as a gradual change in the solution's optical rotation over time until it stabilises at a constant value.
Pure α-D-glucose, freshly dissolved in water, shows a specific rotation of +112°, which gradually changes and stabilises at +52.5°. Explain this observation.
This is mutarotation: the pure α-anomer in solution slowly opens up to the open-chain form and re-closes randomly, forming an equilibrium mixture of α-D-glucose (specific rotation +112°) and β-D-glucose (specific rotation +18.7°) in a fixed ratio (roughly 36:64). The overall measured rotation of this equilibrium mixture settles at an intermediate, constant value (+52.5°), reflecting the weighted average of both anomers present.
| Disaccharide | Constituent Monosaccharides | Glycosidic Linkage | Reducing? |
|---|---|---|---|
| Sucrose | Glucose + Fructose | C1 of glucose — C2 of fructose (α,β-1,2) | No |
| Maltose | Glucose + Glucose | C1 — C4 (α-1,4) | Yes |
| Lactose | Glucose + Galactose | C1 — C4 (β-1,4) | Yes |
| Polysaccharide | Linkage Type | Structure & Function |
|---|---|---|
| Starch | α-1,4 (amylose, linear) and α-1,6 (amylopectin, branched) | Energy storage in plants |
| Cellulose | β-1,4 glycosidic linkage | Structural component of plant cell walls; not digestible by humans (lack the required enzyme) |
| Glycogen | α-1,4 and α-1,6 (highly branched) | Energy storage in animals ("animal starch") |
Explain why humans can digest starch but not cellulose, even though both are polymers of glucose.
Starch is held together by α-1,4-glycosidic linkages, which the human digestive enzyme amylase can specifically recognise and break down. Cellulose is held together by β-1,4-glycosidic linkages — a different spatial geometry that human digestive enzymes cannot recognise or hydrolyse, since we lack the enzyme (cellulase) needed to break this specific type of bond. This is purely a difference in bond geometry, since the constituent monomer (glucose) is identical in both polymers.
An α-amino acid has both an amino group ($-\text{NH}_2$) and a carboxyl group ($-\text{COOH}$) attached to the same (α) carbon. Amino acids are classified as essential (must be obtained through diet, since the body cannot synthesise them) or non-essential (the body can synthesise these itself).
In the solid state and in aqueous solution near neutral pH, an amino acid exists predominantly as a zwitterion (dipolar ion) — the carboxyl group loses its proton (becoming $-\text{COO}^-$) while the amino group gains that proton (becoming $-\text{NH}_3^+$), giving a molecule with both a positive and negative charge but an overall net charge of zero. This internal acid-base reaction is why amino acids have unusually high melting points and behave like ionic solids compared to simple organic molecules of similar size.
The specific pH at which an amino acid exists predominantly as the neutral zwitterion, carrying no net electrical charge overall — this is called the isoelectric point, and it's a characteristic value for each amino acid depending on its specific side chain (R group).
Explain why glycine has a much higher melting point than a simple carboxylic acid or amine of comparable molecular size.
Glycine exists predominantly as a zwitterion ($^+\text{H}_3\text{N-CH}_2\text{-COO}^-$) rather than as a simple neutral molecule. This dipolar, internally-charged structure allows for strong electrostatic (ionic) interactions between neighbouring molecules in the solid lattice, similar to an ionic salt — requiring considerably more energy to break apart and melt than the comparatively weak intermolecular forces (van der Waals, ordinary hydrogen bonding) present in typical small neutral organic molecules of similar molecular weight.
A peptide bond forms between the carboxyl group of one amino acid and the amino group of another, releasing a water molecule (a condensation reaction) — chemically, this is simply an amide linkage ($-\text{CO-NH}-$). Proteins are long polymers (polypeptides) built from many amino acids connected by peptide bonds.
| Structural Level | Description |
|---|---|
| Primary structure | The specific linear sequence of amino acids joined by peptide bonds |
| Secondary structure | Regular, repeating local folding patterns, stabilised by hydrogen bonds between backbone atoms — e.g., the α-helix and β-pleated sheet |
| Tertiary structure | The overall 3D folding of the entire polypeptide chain into a specific compact shape, stabilised by interactions between side chains (R groups) — hydrogen bonds, disulphide bridges, ionic and hydrophobic interactions |
| Quaternary structure | The spatial arrangement of two or more separate polypeptide subunits combined into one larger functional protein complex |
Explain the key structural difference between the α-helix and β-pleated sheet forms of protein secondary structure.
An α-helix is a right-handed, coiled/spiral structure held together by hydrogen bonds between the backbone C=O of one amino acid and the backbone N-H of an amino acid four residues further along the same chain. A β-pleated sheet is a more extended, zig-zag arrangement, where hydrogen bonds form between backbone atoms of adjacent, side-by-side polypeptide strands (or segments of the same strand folded back on itself), rather than within a single continuous coil.
Denaturation is the loss of a protein's specific 3D (secondary, tertiary, and/or quaternary) structure — and consequently its biological activity/function — caused by disruption of the weak, non-covalent interactions holding that structure together (hydrogen bonds, ionic interactions, hydrophobic interactions). Importantly, denaturation does not break the primary structure — the peptide bonds and amino acid sequence remain intact.
Heat, changes in pH, exposure to certain chemicals (heavy metal ions, urea), and mechanical agitation can all disrupt a protein's native folded structure, causing it to unfold ("denature") into a random, biologically inactive conformation. A classic everyday example: cooking an egg irreversibly denatures its proteins (the clear albumin visibly turns white and solid).
Explain, at a structural level, why a denatured protein typically loses its biological (e.g., enzymatic) activity even though its primary structure remains unchanged.
A protein's biological function (e.g., an enzyme's catalytic activity) depends critically on its precise 3D shape — specifically the exact spatial arrangement of its active site, formed by the correct tertiary (and sometimes quaternary) folding. Denaturation disrupts the weak non-covalent interactions responsible for this specific folding, causing the protein to unfold into a random, less organised shape — even though every peptide bond and the amino acid sequence (primary structure) remains completely intact, the loss of correct 3D shape destroys the precise geometric fit required for biological function.
Enzymes are biological catalysts, almost all of which are proteins, that dramatically speed up specific biochemical reactions by lowering the activation energy required. They act with high specificity, generally catalysing only one particular reaction or a closely related set of reactions, based on the precise 3D shape of their active site.
The substrate (reactant molecule) fits into the enzyme's active site much like a key fits a specific lock — either through a rigid, pre-formed complementary shape (lock-and-key model) or through a shape that adjusts slightly to better accommodate the substrate upon binding (induced fit model), forming an enzyme-substrate complex that lowers the reaction's activation energy and speeds up conversion to product.
Explain why a small change in pH or temperature can significantly reduce (or completely destroy) an enzyme's catalytic activity.
Since enzymes are proteins, their catalytic activity depends entirely on maintaining a precise 3D shape at the active site. Changes in pH or temperature beyond an enzyme's optimal range can disrupt the weak non-covalent interactions (hydrogen bonds, ionic interactions) holding this shape together — effectively causing partial or complete denaturation. Even a slightly altered active-site shape may no longer bind the substrate correctly, sharply reducing or eliminating the enzyme's catalytic efficiency.
Vitamins are classified by solubility: fat-soluble vitamins (A, D, E, K) are stored in the body's fatty tissue and liver, so excessive intake can potentially accumulate to harmful levels; water-soluble vitamins (B-complex, C) are not significantly stored in the body and are regularly excreted in urine, so they generally need more frequent dietary intake and rarely accumulate to toxic levels.
| Vitamin | Deficiency Disease |
|---|---|
| A | Night blindness (xerophthalmia) |
| B₁ (Thiamine) | Beriberi |
| C (Ascorbic acid) | Scurvy |
| D | Rickets (children) / osteomalacia (adults) |
| K | Impaired/increased blood clotting time |
Explain why excessive supplementation of vitamin A is more dangerous than excessive supplementation of vitamin C.
Vitamin A is fat-soluble and gets stored in the body's fatty tissues and liver rather than being readily excreted — over time, excessive intake can accumulate to toxic levels (hypervitaminosis A). Vitamin C is water-soluble, and any excess beyond what the body needs is simply excreted in urine rather than stored, making toxic accumulation from oral intake far less likely under normal circumstances.
Nucleic acids (DNA and RNA) are polymers built from repeating nucleotide units. Each nucleotide consists of three components: a nitrogenous base (purine or pyrimidine), a pentose sugar (deoxyribose in DNA, ribose in RNA), and a phosphate group.
A nucleoside is just the sugar plus the nitrogenous base (no phosphate). A nucleotide is a nucleoside with a phosphate group attached — nucleotides are the actual repeating monomer units that link together (via phosphodiester bonds between the sugar of one unit and the phosphate of the next) to form the nucleic acid polymer chain.
| Base Type | Bases | Ring Structure |
|---|---|---|
| Purines | Adenine (A), Guanine (G) | Fused double-ring structure |
| Pyrimidines | Cytosine (C), Thymine (T, DNA only), Uracil (U, RNA only) | Single-ring structure |
What are the three components of a single DNA nucleotide, and which pentose sugar does it contain?
A DNA nucleotide contains: (1) a nitrogenous base (one of adenine, guanine, cytosine, or thymine), (2) the pentose sugar deoxyribose (which lacks an -OH at the 2' carbon, compared to ribose), and (3) a phosphate group.
In double-stranded DNA, bases pair specifically: Adenine always pairs with Thymine (via 2 hydrogen bonds), and Guanine always pairs with Cytosine (via 3 hydrogen bonds) — this specific, complementary pairing is what allows DNA's two strands to wind together into a stable double helix, and is fundamental to accurate DNA replication and the genetic code.
| Feature | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Bases | A, T, G, C | A, U, G, C |
| Strands | Double-stranded helix | Usually single-stranded |
| Primary function | Long-term storage of genetic information | Protein synthesis, gene expression (mRNA, tRNA, rRNA) |
A double-stranded DNA molecule contains 30% adenine. Calculate the percentage of each of the other three bases.
Since $\text{A} = \text{T}$ (complementary base pairing), $\text{T} = 30\%$ as well.
Remaining percentage for G + C $= 100 - 30 - 30 = 40\%$, and since $\text{G} = \text{C}$, each is $40/2 = 20\%$.
So: A = 30%, T = 30%, G = 20%, C = 20%.
Hormones are chemical messengers produced by specific endocrine glands and secreted directly into the bloodstream, travelling to distant target tissues/organs to regulate physiological processes such as growth, metabolism, and reproduction — chemically, they may be proteins/peptides (e.g., insulin), steroids (e.g., testosterone, estrogen), or amino-acid derivatives (e.g., adrenaline).
Explain why insulin, a protein hormone, cannot be taken orally as a tablet and must instead be injected.
Insulin is a protein, and like any protein consumed orally, it would be broken down by digestive enzymes (proteases) in the stomach and intestines into individual amino acids before it could ever reach the bloodstream intact — destroying its specific 3D structure and biological activity in the process. Injecting insulin directly into the bloodstream (or subcutaneously) bypasses the digestive system entirely, allowing the intact, functional hormone to reach its target tissues.