Lesson 16: Biochemistry — Carbohydrates, Lipids, Proteins, and Nucleic Acids

Life runs on four families of large molecules, all built from the organic chemistry of Lesson 9. Carbohydrates store and supply energy; lipids store it long-term and build membranes; proteins do the work — enzymes, transport, structure, defence; nucleic acids store the instructions. This lesson treats each family the same way: monomers, the links that join them, the numbers that describe them, and the enzyme kinetics that govern how the protein machines perform.

Carbohydrates and lipids

Carbohydrates are built from monosaccharides such as glucose, C₆H₁₂O₆, joined by condensation into disaccharides (maltose: glucose + glucose − H₂O → C₁₂H₂₂O₁₁) and polysaccharides. The link's geometry decides digestibility: starch's α-links are hydrolysed by our enzymes; cellulose's β-links are not, so cellulose passes through as fibre while chemically similar starch fuels us. Hydrolysis reverses condensation — digestion is controlled hydrolysis. Lipids are esters: a triglyceride forms when glycerol condenses with three fatty acids, losing 3H₂O. Saturated fatty acids (no C=C) pack tightly — solid fats (butter); unsaturated ones are kinked at their double bonds, packing poorly — liquid oils. Partial hydrogenation of oils straightens some kinks and creates trans fats, whose health effects rewrote food law: structure is destiny, again.

Proteins: condensation at scale

Proteins are chains of amino acids joined by peptide bonds — each bond a condensation ejecting water. Twenty standard amino acids, one variable side chain (R group) each, combine into sequences whose folded 3D shapes create function at four levels: primary (sequence), secondary (α-helices and β-sheets, hydrogen-bonded), tertiary (overall fold, held by hydrogen bonds, ionic attractions, hydrophobic packing and covalent disulfide bridges), and quaternary (multi-chain assemblies such as haemoglobin). Heat or pH extremes denature a protein — disrupting the weak interactions of levels 2–4 while leaving the sequence intact — which is why egg white sets and why fevers are dangerous.

Worked example 1: the mass of a small protein

A protein consists of 50 amino acid residues. Take the average amino acid mass as 110 g/mol. Building the chain forms 49 peptide bonds and loses 49 water molecules. Mass = (50 × 110) − (49 × 18.02) = 5500 − 883 = 4617 g/mol (about 4.6 kDa). The method — monomer sum minus condensation losses — is exactly Lesson 14's nylon arithmetic wearing a lab coat. It also explains a standard exam trap: a protein of 100 residues does not weigh 100 × its average amino acid mass; the waters are gone.

Nucleic acids and the energy currency

DNA and RNA are polymers of nucleotides (base + sugar + phosphate). DNA's two strands run antiparallel, joined by hydrogen bonds between complementary bases: A pairs T (2 hydrogen bonds), C pairs G (3) — the pairing rules that make copying possible (Lesson 9's hydrogen bonding doing information work). Cellular energy transactions use ATP: hydrolysis ATP → ADP + phosphate releases about 30.5 kJ/mol under cellular conditions, coupling exergonic steps to endergonic ones throughout metabolism.

Worked example 2: energy bookkeeping

(a) Glucose oxidation, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, releases 2805 kJ/mol. Per gram: 2805 ÷ 180.16 = 15.6 kJ/g (about 3.7 kcal per gram — the nutrition label's "carbohydrate ≈ 4 kcal/g" comes from exactly this division, adjusted for digestibility). (b) If respiration captures energy in ATP at roughly 30 ATP per glucose, the captured energy is 30 × 30.5 = 915 kJ per mole of glucose — an efficiency of 915 ÷ 2805 = 33%. The rest leaves as heat, which is not waste for a warm-blooded animal: it is the central heating.

Enzymes: catalysis with a shape

Enzymes are protein catalysts whose active site binds a specific substrate (the lock-and-key picture, refined by induced fit). Rate rises with temperature then collapses as the enzyme denatures — an optimum curve; likewise each enzyme has a pH optimum. Kinetics follow saturation: at low substrate the rate rises proportionally; at high substrate every active site is busy and the rate plateaus at Vmax.

In the lab: catalase and the gas syringe

Liver or yeast catalase decomposes hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. Students vary temperature or pH and collect O₂ in a gas syringe. Measuring the *initial* rate from the tangent at t = 0 (Lesson 12) is essential — the substrate depletes, so later slopes sag. The uncertainty story has three layers. The syringe reads ±0.5 mL on perhaps 40 mL (about 1%) — not the problem. Timing the tangent is subjective, worth a few percent. The dominant error is temperature control: a water bath nominally at 40 °C may cycle ±2 °C, and with rates doubling per 10 °C that is a ±15% swing; worse, if the enzyme equilibrates slowly, the first run at each temperature is systematically the least reliable. Equilibration time, replicate runs and discarding the first reading at each condition are the working answers — the same professional triage as Lesson 12's kinetics lab, now with a denaturation cliff punishing anyone who pre-warms the enzyme too long.

Chemistry in the real world

Lactose intolerance is one missing enzyme (lactase) — milk sugar passes undigested to gut bacteria, whose fermentation products cause the symptoms. Industrial enzymes (proteases in biological washing powders, engineered to survive alkaline, warm wash water) are protein chemistry as consumer product; their denaturation limits are written on the box as "wash at 40 °C".

Common mistakes

• "Denaturing breaks peptide bonds." Denaturation disrupts secondary/tertiary interactions; the sequence survives (a boiled egg's proteins are unfolded, not hydrolysed).
• Forgetting condensation losses in biopolymer masses (Worked example 1).
• Saturated = solid because of stronger bonds. The C−C bonds are identical; it is the *packing* of straight versus kinked chains that changes melting behaviour.
• Base-pair swaps. A–T and C–G, with 2 and 3 hydrogen bonds respectively; C–G-rich DNA melts at higher temperature for exactly this reason.

Exam Experience

1. The bond linking two amino acids in a protein is: A) a glycosidic bond B) an ester bond C) a peptide (amide) bond D) a hydrogen bond Answer: C. Peptide bonds are amide links formed by condensation between the carboxyl of one amino acid and the amino group of the next. A joins sugars; B joins lipids; D holds secondary structure but is not the chain link.

2. Which change denatures an enzyme without breaking its primary structure? A) Hydrolysis B) Heating to 80 °C C) Adding a protease D) Changing the gene sequence Answer: B. Heat disrupts the weak interactions maintaining the fold. A and C cleave peptide bonds (primary structure); D changes the sequence itself.

3. A triglyceride is formed from glycerol and three fatty acids. How many water molecules are released, and what bond type forms? A) 1 water; peptide B) 3 waters; ester C) 3 waters; glycosidic D) No water; addition Answer: B. Each of the three −OH groups of glycerol esterifies one fatty acid carboxyl, ejecting one H₂O each: 3 ester bonds, 3 waters.

Constructed response (5 marks). A DNA segment contains 300 bases in total across both strands, of which 90 are adenine (A). (a) State how many thymine (T), guanine (G) and cytosine (C) bases the segment contains, explaining your reasoning. (b) The segment is heated until the strands separate; a second segment of the same length has 120 adenines. State which segment needs the higher temperature, and explain why in terms of bonding. Scoring guide: (a) 1 mark: A pairs T, so T = 90; 1 mark: remaining bases 300 − 180 = 120 split equally between G and C (G = C by pairing), so G = 60 and C = 60. (b) 1 mark: the second segment has 120 A → 120 T, leaving only 60 bases for G + C (30 each), so it is A–T richer; 1 mark: A–T pairs have 2 hydrogen bonds versus 3 for C–G, so the A–T-rich segment separates more easily / needs the lower temperature; 1 mark: correctly identifying the first segment as needing the higher temperature. Award the explanation marks only if hydrogen-bond counts are explicitly invoked.

Fats versus carbohydrates: the energy ledger

Lesson 8's energy-density logic explains the body's storage choices. Fatty acids are far more reduced (rich in C−H bonds, poor in oxygen) than sugars, so oxidising them releases more energy per gram: triglycerides yield about 37 kJ/g, carbohydrates about 17 kJ/g — migratory birds and hibernating bears bank fat for exactly this factor of two. The comparison runs through Lesson 15's table: glucose already carries six oxygens (it is, loosely, partly pre-burned), while a fatty acid like palmitic acid (C₁₆H₃₂O₂) carries just two. Worked check for palmitic acid: combustion ΔH ≈ −9977 kJ/mol over Mr = 256.43 gives 38.9 kJ/g, consistent with the dietary figure once water of metabolism and digestibility are accounted. ATP bookkeeping says the same in the body's currency: one palmitate yields roughly 106 ATP against glucose's ~30 — the molecular reason a gram of fat outlasts a gram of glycogen on any long journey, and why glycogen, stored wet, costs endurance athletes dearly per kilojoule carried.

The genome in numbers

DNA repays quantitative reading. The human genome is about 3 × 10⁹ base pairs; at roughly 660 g/mol per base pair, one cell's DNA weighs about 3 × 10⁹ × 660 ÷ 6.022 × 10²³ = 3.3 × 10⁻¹² g — three picograms encoding you. Copying it makes about one error per 10⁹ bases after proofreading and repair, so each cell division still introduces a handful of mutations across the genome — evolution's raw material and oncology's origin story, in one statistic. The same arithmetic explains PCR's power in forensics (Lesson 17): each thermal cycle doubles the target DNA, so 30 cycles multiply a single molecule by 2³⁰ ≈ 10⁹ — exponential growth deployed as an analytical amplifier, turning one cell's picograms into a testable sample in an afternoon.