A stretch of 18 consecutive residues of a protein is folded as one regular α-helix. Approximately how long is this stretch along the helix axis?

2.7 nm, because each residue advances along the axis by 0.15 nm

6.1 nm, because each residue advances along the axis by 0.34 nm

9.7 nm, because each residue advances along the axis by 0.54 nm

2.7 nm, because each residue advances along the axis by 0.15 nm

5.0 nm, because the 18 residues make five turns of 1.0 nm each

*E. coli* grown for many generations in ¹⁵N medium is transferred to ¹⁴N medium and allowed exactly three rounds of replication. If replication is semiconservative, CsCl density-gradient centrifugation of its DNA will show hybrid (¹⁵N/¹⁴N) and light (¹⁴N/¹⁴N) duplexes in the ratio

1 hybrid : 3 light, because the two parental strands lie in two of eight

1 hybrid : 7 light, because only one duplex in eight still carries old DNA

1 hybrid : 3 light, because the two parental strands lie in two of eight

1 hybrid : 1 light, because each round halves the hybrid fraction once more

1 heavy : 7 light with no hybrid, because the parental duplex stays intact

Which of the following statements about DNA replication in an animal cell are correct?

Okazaki fragments are made on the lagging strand and joined by DNA ligase after their RNA primers are replaced with DNA

Telomerase extends chromosome ends by reverse-transcribing an RNA template that is part of the enzyme

DNA polymerase δ can start a new chain on a bare single-stranded template without any primer

Okazaki fragments are made on the lagging strand and joined by DNA ligase after their RNA primers are replaced with DNA

Each origin may fire several times in one S phase, which is how large genomes finish replication on time

Telomerase extends chromosome ends by reverse-transcribing an RNA template that is part of the enzyme

Cultured mammalian cells are given a radioactive amino acid together with cycloheximide at a dose that completely blocks translation on cytosolic ribosomes. A small amount of protein still becomes labelled. These labelled proteins are most likely

Respiratory-complex subunits encoded by mitochondrial DNA and made on mitochondrial ribosomes

Nuclear-encoded mitochondrial proteins, which are made after being imported into the organelle

Secretory proteins, because ribosomes bound to the rough ER are resistant to cycloheximide

Histones, because their mRNAs lack poly(A) tails and are translated without the 80S ribosome

Respiratory-complex subunits encoded by mitochondrial DNA and made on mitochondrial ribosomes

Red blood cells are incubated with glucose and sodium fluoride, which inhibits enolase. Which glycolytic intermediate builds up immediately upstream of the blocked step?

2-Phosphoglycerate, the substrate that enolase dehydrates

Phosphoenolpyruvate, the high-energy product of enolase

Fructose-1,6-bisphosphate, the product of the committed step

Glyceraldehyde-3-phosphate, the substrate of the dehydrogenase

Which of the following statements about the Krebs cycle in animal mitochondria are correct?

Succinate dehydrogenase is bound to the inner mitochondrial membrane and passes electrons to ubiquinone as complex II

α-Ketoglutarate dehydrogenase, like pyruvate dehydrogenase, needs thiamine pyrophosphate and lipoate

Succinate dehydrogenase is bound to the inner mitochondrial membrane and passes electrons to ubiquinone as complex II

The two carbons brought in by an acetyl-CoA leave as the two CO₂ released in that same turn of the cycle

Malonate inhibits fumarase irreversibly, so that fumarate accumulates in poisoned mitochondria

α-Ketoglutarate dehydrogenase, like pyruvate dehydrogenase, needs thiamine pyrophosphate and lipoate

In the presence of 1 mM of a reversible inhibitor, an enzyme's apparent Km rises from 2 mM to 6 mM while its Vmax stays at 50 µmol min⁻¹. For the inhibition type involved, the affected constant changes by the factor 1 + [I]/Ki. Which conclusion is correct?

Competitive inhibition with Ki = 0.5 mM

Uncompetitive inhibition with Ki = 0.5 mM

Competitive inhibition with Ki = 3.0 mM

Competitive inhibition with Ki = 0.5 mM

Non-competitive inhibition with Ki = 1.5 mM

In chymotrypsin, the nucleophile that attacks the carbonyl carbon of the scissile peptide bond and forms the acyl-enzyme intermediate is the side chain of

Serine, made reactive by a histidine that is held in place by aspartate

Cysteine, made reactive by a neighbouring histidine as a thiolate ion

Serine, made reactive by a histidine that is held in place by aspartate

Histidine, which is itself acylated first and is later hydrolysed by water

Aspartate, which activates a water molecule as in the acid proteases

A hormone produces its full effect on target cells only after several hours, and the effect is abolished if actinomycin D, an inhibitor of transcription, is given together with the hormone. The hormone is most likely

Cortisol, acting through an intracellular receptor that binds hormone response elements in DNA

Adrenaline, acting through β-adrenergic receptors that raise cAMP and activate protein kinase A

Insulin, acting by moving pre-formed GLUT4 transporters from vesicles to the plasma membrane

Glucagon, acting through a G-protein-coupled receptor that activates glycogen phosphorylase

A cell carries both a Gs-coupled and a Gi-coupled receptor controlling the same adenylyl cyclase, and the hormones for both receptors are present. Which of the following treatments would raise the cell's cAMP level further?

Cholera toxin, which ADP-ribosylates Gsα and blocks its GTPase activity

Pertussis toxin, which ADP-ribosylates Giα and prevents its activation by the receptor

Caffeine, which inhibits cAMP phosphodiesterase

Cholera toxin, which ADP-ribosylates Gsα and blocks its GTPase activity

A drug that blocks phospholipase C-β and so prevents IP₃ formation

Pertussis toxin, which ADP-ribosylates Giα and prevents its activation by the receptor

Caffeine, which inhibits cAMP phosphodiesterase

A thiamine-deficient animal on a carbohydrate-rich diet shows raised blood pyruvate and lactate. The enzyme whose loss of activity best explains this is

Pyruvate dehydrogenase, whose first component needs thiamine pyrophosphate

Pyruvate carboxylase, which needs biotin to add CO₂ to pyruvate in mitochondria

Lactate dehydrogenase, which needs NAD⁺ made from dietary niacin

Pyruvate kinase, which needs pyridoxal phosphate as its cofactor

Pyruvate dehydrogenase, whose first component needs thiamine pyrophosphate

Vitamin K is required as a cofactor for

γ-Carboxylation of glutamate residues in prothrombin and other clotting factors

Hydroxylation of proline and lysine residues in procollagen

Hydroxylation of 25-hydroxycholecalciferol to calcitriol in the kidney

γ-Carboxylation of glutamate residues in prothrombin and other clotting factors

Isomerisation of all-trans-retinal to 11-cis-retinal to regenerate rhodopsin in rod cells

Which of the following pairings of a trace element with a protein or vitamin that contains it are correct?

Zinc: carbonic anhydrase, which hydrates CO₂ in red blood cells

Cobalt: cytochrome c oxidase, the terminal enzyme of the respiratory chain

Zinc: carbonic anhydrase, which hydrates CO₂ in red blood cells

Molybdenum: the corrin ring of vitamin B₁₂

Biochemistry and Molecular Biology: Biomolecules, the Central Dogma, Glycolysis and the Krebs Cycle, Enzymes, Hormones, Vitamins and Minerals

This chapter answers Section XL4.4 of XL4 Zoology, one of the five optional sections of the Life Sciences (XL) paper: “Nucleic acids, proteins, lipids and carbohydrates; replication, transcription and translation, Krebs cycle, glycolysis, enzyme catalysis, hormones and their actions, roles of vitamins and minerals.” It teaches the four classes of biomolecule and why their structure suits their job, how a cell copies DNA and turns it into protein, how glucose yields ATP, how enzymes work and are regulated, how water-soluble and lipid-soluble hormones act, and what each vitamin and mineral does. The numericals are Chargaff counts, melting temperatures, ATP yields under a stated P/O convention, Michaelis–Menten kinetics and replication time.

1. Carbohydrates and lipids: fuel, storage and structure

Carbohydrates are aldoses or ketoses of formula (CH₂O)ₙ. In water glucose and fructose close into rings, and the new chiral carbon (C1 of glucose, the anomeric carbon) gives α and β forms. A glycosidic bond joins the anomeric carbon of one sugar to a hydroxyl of the next, and its geometry decides its biology. Lactose, the sugar of mammalian milk, is Gal(β1→4)Glc and keeps a free anomeric carbon, so it is reducing; sucrose, Glc(α1↔β2)Fru, links both anomeric carbons and is non-reducing. Animals store glucose as glycogen: α(1→4) chains branched by α(1→6) links every 8 to 12 residues, whose many ends let glycogen phosphorylase release glucose quickly. Cellulose is β(1→4) glucose; its straight chains form fibres that no vertebrate enzyme cuts, so herbivores rely on gut microbes. Chitin, β(1→4) N-acetylglucosamine, forms the arthropod exoskeleton, and glycosaminoglycans such as hyaluronate and chondroitin sulfate hydrate cartilage matrix.

Lipids are defined by insolubility in water, not by one structure. Fatty acids are long-chain carboxylic acids; a cis double bond kinks the chain and lowers the melting point. Mammals cannot insert double bonds beyond C-9, so linoleic (ω-6) and α-linolenic (ω-3) acids are essential in the diet; arachidonate made from linoleate is the precursor of the eicosanoids (prostaglandins, thromboxanes, leukotrienes). Triacylglycerols store energy anhydrous and highly reduced, about 37 kJ g⁻¹ against about 17 kJ g⁻¹ for carbohydrate or protein. Phospholipids and sphingolipids are amphipathic and form bilayers; sphingomyelin is abundant in myelin. Cholesterol buffers membrane fluidity and is the precursor of steroid hormones, bile acids and vitamin D.

2. Proteins: from sequence to shape

Proteins are polymers of the 20 standard L-α-amino acids joined by peptide bonds. The peptide bond has partial double-bond character, so it is planar and almost always trans; only the φ and ψ angles at each α-carbon rotate. Primary structure is the sequence; secondary structure is local, hydrogen-bonded folding of the backbone; tertiary structure is the full fold of one chain, driven mainly by burial of hydrophobic side chains; quaternary structure is the assembly of chains, as in haemoglobin (α₂β₂), whose cooperative subunits give a sigmoid O₂-binding curve where single-chain myoglobin gives a hyperbola. Anfinsen showed that denatured ribonuclease refolds into active enzyme once urea and the reducing agent are removed: the sequence carries the information for the fold.

Regular structures of fibrous animal proteins
Structure	Geometry	Animal example
α-helix	Right-handed; 3.6 residues per turn, 0.15 nm rise per residue, 0.54 nm pitch; C=O of residue n bonds N–H of residue n + 4	α-keratin of hair, horn and nails
β-sheet	Extended strands (about 0.33–0.35 nm per residue) hydrogen-bonded side by side, parallel or antiparallel	Silk fibroin; β-keratin of feathers and reptile scales
Collagen triple helix	Three chains with glycine at every third residue (Gly–X–Y), rich in proline and hydroxyproline	Tendon, skin, bone matrix

⚠️ Exam trap

Proline and glycine break helices. Do not confuse the 0.54 nm pitch per helix turn with the 0.15 nm rise per residue, or with the 0.34 nm rise per base pair of B-DNA.

3. Nucleic acids, Chargaff's rules and melting

A nucleotide is a nitrogenous base, a pentose and phosphate. The purines are adenine and guanine; the pyrimidines are cytosine, thymine (DNA) and uracil (RNA). 3′→5′ phosphodiester bonds give the backbone a direction. B-DNA, the form in cells, is a right-handed double helix of two antiparallel strands with about 10.5 bp per turn and 0.34 nm rise per base pair; proteins read the sequence through its major and minor grooves. A pairs with T by two hydrogen bonds and G with C by three. RNA's 2′-OH lets it fold into catalytic and structural shapes (tRNA, rRNA, ribozymes) but makes it labile in alkali.

Chargaff's rules follow from base pairing: in double-stranded DNA A = T and G = C, so purines equal pyrimidines, while (A + T)/(G + C) varies from species to species. They do not hold for single-stranded genomes or for RNA. Worked example: a 4,000 bp duplex has 28% T. Then A = 28%, G + C = 44%, so G = 22% of 8,000 nucleotides = 1,760.

Heating separates the strands; stacked bases absorb less ultraviolet light than unstacked ones, so absorbance at 260 nm rises (the hyperchromic effect). The midpoint of this transition is the melting temperature (Tm). Tm rises with G + C content (three hydrogen bonds and stronger stacking) and with salt (cations screen the repelling phosphates), and falls with formamide or urea. For short oligonucleotides the Wallace rule, Tm ≈ 2(A + T) + 4(G + C) °C, is a working estimate: an 18-mer with 10 G + C has Tm ≈ 2 × 8 + 4 × 10 = 56 °C.

4. Replication, transcription and translation

DNA replication is semiconservative: Meselson and Stahl (1958) grew E. coli in ¹⁵N, shifted it to ¹⁴N and found by CsCl density-gradient centrifugation one hybrid band after one generation and equal hybrid and light bands after two. DNA polymerases add nucleotides only to a free 3′-OH, so they need an RNA primer (made by primase) and synthesise only 5′→3′. At each fork the leading strand is made continuously and the lagging strand as Okazaki fragments, whose primers are replaced and nicks sealed by DNA ligase; helicase unwinds, topoisomerases relieve supercoiling and a 3′→5′ exonuclease proofreads. Eukaryotic chromosomes use many origins: MCM helicases are loaded in G1 and activated once in S, so no stretch is copied twice; Pol ε extends mainly the leading strand and Pol δ the lagging one. The last lagging-strand primer at a linear end cannot be replaced, so chromosomes shorten unless telomerase, a reverse transcriptase carrying its own RNA template, extends them, as in germ cells and stem cells.

🎯 Why animals need many origins

A 100 Mb animal chromosome copied from one origin by two forks at about 50 nt s⁻¹ would need 50 × 10⁶/50 = 10⁶ s, more than ten days. With origins every 100 kb each fork copies only 50 kb, in about 1,000 s: S-phase length is set by origin spacing, not chromosome length.

Transcription copies the template strand into RNA. RNA polymerase needs no primer, reads the template 3′→5′ and builds RNA 5′→3′, so the RNA has the sequence of the coding strand (U for T). Eukaryotes divide the work: RNA Pol I makes the large pre-rRNA (28S, 18S, 5.8S), Pol II makes pre-mRNA and most snRNAs and miRNAs, and Pol III makes tRNA, 5S rRNA and U6 snRNA. In animal cells α-amanitin blocks Pol II at low concentration, Pol III only at high concentration and Pol I not at all.

Translation decodes mRNA in non-overlapping triplets. Of 64 codons, 61 specify amino acids and 3 (UAA, UAG, UGA) stop; AUG encodes methionine and starts. The code is degenerate (wobble at the third codon base lets one tRNA read several codons) and nearly universal: vertebrate mitochondria read UGA as tryptophan. Each aminoacyl-tRNA synthetase charges its tRNA in two steps (amino acid + ATP → aminoacyl-AMP + PPi, then transfer to the 3′-CCA end), and this sets fidelity. The eukaryotic 80S ribosome (40S + 60S) binds at the 5′ cap and scans to the first AUG; eEF1A·GTP delivers each aminoacyl-tRNA, the 28S rRNA forms the peptide bond (the ribosome is a ribozyme) and eEF2·GTP translocates. Counting two phosphoanhydride bonds for charging and one GTP each for delivery and translocation, a peptide bond costs four.

ℹ️ Note

Inhibitors sort the ribosomes: cycloheximide blocks cytosolic 80S ribosomes, chloramphenicol blocks bacterial and mitochondrial ones, puromycin mimics an aminoacyl-tRNA and ends chains on both, and diphtheria toxin ADP-ribosylates eEF2.

5. Glycolysis and the Krebs cycle

Glycolysis runs in the cytosol of every animal cell and splits glucose into two pyruvate in ten steps. The investment phase spends two ATP (hexokinase, phosphofructokinase-1) to make fructose-1,6-bisphosphate, which aldolase splits into two trioses. Then glyceraldehyde-3-phosphate dehydrogenase makes NADH, and phosphoglycerate kinase and pyruvate kinase make four ATP by substrate-level phosphorylation. Net: 2 ATP, 2 NADH, 2 pyruvate per glucose. PFK-1 is the committed, rate-setting step: ATP and citrate inhibit it, AMP and fructose-2,6-bisphosphate activate it. Without oxygen NAD⁺ must be regenerated: vertebrate muscle reduces pyruvate to lactate, which the liver turns back into glucose (Cori cycle); yeast makes ethanol. Fluoride inhibits enolase, and arsenate replaces phosphate at the dehydrogenase step so that the ATP of phosphoglycerate kinase is lost.

In the mitochondrial matrix the pyruvate dehydrogenase complex (cofactors TPP, lipoate, CoA, FAD, NAD⁺) converts pyruvate to acetyl-CoA, CO₂ and NADH. The step is irreversible and vertebrates lack the glyoxylate cycle, so they cannot make net glucose from fatty acids. The Krebs cycle (Krebs, 1937) condenses acetyl-CoA with oxaloacetate to citrate, then: citrate → isocitrate (aconitase) → α-ketoglutarate (isocitrate dehydrogenase; NADH, CO₂) → succinyl-CoA (α-ketoglutarate dehydrogenase; NADH, CO₂) → succinate (succinyl-CoA synthetase; GTP or ATP) → fumarate (succinate dehydrogenase; FADH₂) → malate (fumarase) → oxaloacetate (malate dehydrogenase; NADH). Per acetyl-CoA: 3 NADH, 1 FADH₂, 1 GTP or ATP and 2 CO₂. Succinate dehydrogenase, the only membrane-bound enzyme of the cycle, is also complex II. Intermediates drawn off for biosynthesis are replenished chiefly by biotin-dependent pyruvate carboxylase. Malonate competitively inhibits succinate dehydrogenase; fluoroacetate becomes fluorocitrate, which blocks aconitase.

ATP yield of glucose oxidation, taking 2.5 ATP per NADH and 1.5 per FADH₂
Stage (per glucose)	ATP or GTP made directly	NADH	FADH₂	ATP equivalents
Glycolysis	2	2 (cytosolic)	0	7 (malate–aspartate shuttle) or 5 (glycerol-3-phosphate shuttle)
Pyruvate dehydrogenase, ×2	0	2	0	5
Krebs cycle, ×2	2	6	2	20
Total	4	10	2	32 or 30

⚠️ Exam trap

With the older P/O values of 3 and 2 the same table gives 38 or 36. Use the convention the question states. The glycerol-3-phosphate shuttle, used in skeletal muscle and brain, hands cytosolic electrons to FAD and so yields one ATP less per cytosolic NADH.

6. Enzyme catalysis and kinetics

Enzymes speed reactions by lowering the activation free energy (ΔG‡); they change neither ΔG nor the equilibrium constant. The active site binds the transition state more tightly than the substrate, which is why transition-state analogues are potent inhibitors. Four tactics recur: general acid–base catalysis (proton transfer by side chains such as His); covalent catalysis, as in the acyl-enzyme of chymotrypsin, whose Ser–His–Asp catalytic triad makes Ser195 a strong nucleophile; metal-ion catalysis, as Zn²⁺ in carbonic anhydrase turns bound water into hydroxide; and proximity and orientation. Substrate binding often reshapes the site (induced fit). The seven EC classes are oxidoreductases, transferases, hydrolases, lyases, isomerases, ligases and translocases.

For one substrate at steady state the Michaelis–Menten equation is v = Vmax[S]/(Km + [S]). Km is the [S] at which v = Vmax/2; kcat = Vmax/[E]total is the turnover number; kcat/Km measures catalytic efficiency and approaches the diffusion limit, about 10⁸ to 10⁹ M⁻¹ s⁻¹, in the best enzymes. The Lineweaver–Burk plot of 1/v against 1/[S] is a line with y-intercept 1/Vmax, x-intercept −1/Km and slope Km/Vmax. Worked example: with Km = 4 mM and Vmax = 100 µmol min⁻¹, at [S] = 12 mM, v = 100 × 12/16 = 75 µmol min⁻¹. v reaches 90% of Vmax only at [S] = 9 Km.

Inhibition	Inhibitor binds	Apparent Km and Vmax	Lineweaver–Burk lines
Competitive	Free enzyme, at the active site	Km rises; Vmax unchanged; overcome by high [S]	Meet on the y-axis
Uncompetitive	ES complex only	Km and Vmax fall by the same factor	Parallel
Pure non-competitive	E and ES equally, away from the active site	Vmax falls; Km unchanged	Meet on the x-axis
Irreversible	Covalently, e.g. DIFP on serine hydrolases, aspirin on cyclooxygenase	Active enzyme removed, so Vmax falls	Not a reversible equilibrium

ℹ️ Note

Enzymes are also regulated by allostery (sigmoid kinetics, as for PFK-1; feedback inhibition of the committed step), covalent modification (phosphorylation activates glycogen phosphorylase) and zymogen activation (enteropeptidase turns trypsinogen into trypsin).

7. Hormones and how they act

A hormone is secreted in small amounts, travels in the blood and acts only on cells that carry its receptor; its chemistry decides where that receptor is. Peptides and catecholamines are water-soluble and cannot cross the membrane, so they bind cell-surface receptors and act within seconds to minutes through second messengers, mostly by modifying existing proteins. Steroids and thyroid hormones are lipophilic, travel on carrier proteins, enter the cell and bind intracellular receptors of the nuclear-receptor family, which bind hormone response elements in DNA and alter transcription; the effect takes hours and is blocked by inhibitors of RNA or protein synthesis.

Hormone chemistry decides the receptor
Chemical class	Examples	Receptor
Peptides and proteins	Insulin, glucagon, ADH, oxytocin, growth hormone	Cell surface: G-protein-coupled, tyrosine kinase or cytokine receptors
Amino-acid derivatives	Adrenaline (from tyrosine); melatonin (from tryptophan); T₃ and T₄ (tyrosine + iodine)	Catecholamines: G-protein-coupled; thyroid hormones: nuclear
Steroids (from cholesterol)	Cortisol, aldosterone, oestradiol, testosterone; insect 20-hydroxyecdysone	Intracellular nuclear receptors (ecdysone: EcR with Ultraspiracle)
Eicosanoids (from arachidonate)	Prostaglandins, thromboxanes	Cell surface, acting locally

G-protein-coupled receptors (seven transmembrane helices) activate heterotrimeric G proteins by GDP–GTP exchange on Gα. Gs stimulates adenylyl cyclase, and cAMP activates protein kinase A, which in liver drives glycogen breakdown in response to glucagon or adrenaline; Gi inhibits the cyclase. Gq activates phospholipase C-β, which splits PIP₂ into IP₃ (releases Ca²⁺ from the ER) and diacylglycerol (activates protein kinase C). The signal ends when Gα hydrolyses its GTP and phosphodiesterase destroys cAMP; every step of the cascade amplifies. The insulin receptor is a receptor tyrosine kinase: it autophosphorylates, recruits IRS proteins and, through PI3-kinase and Akt, moves GLUT4 transporters to the surface of muscle and fat cells. Atrial natriuretic peptide and nitric oxide act through guanylyl cyclases that make cGMP.

⚠️ Exam trap

Cholera toxin ADP-ribosylates Gsα so that it cannot hydrolyse GTP; pertussis toxin ADP-ribosylates Giα so that receptors cannot activate it. Both raise cAMP. Caffeine raises cAMP by inhibiting phosphodiesterase.

8. Vitamins and minerals

Vitamins are organic compounds an animal needs in small amounts but cannot make in sufficient quantity. Most water-soluble ones are coenzyme precursors. The fat-soluble vitamins A, D, E and K are stored in liver and fat and can be toxic in excess. Whether a compound is a vitamin depends on the species: most mammals make ascorbate, but humans, other haplorhine primates and guinea pigs lack L-gulonolactone oxidase and need it in food.

What each vitamin does
Vitamin	Active form and role	Deficiency
B₁ thiamine	TPP: pyruvate and α-ketoglutarate dehydrogenases, transketolase	Beriberi
B₂ riboflavin	FAD and FMN of flavoproteins	Cracked lips, sore tongue
B₃ niacin	NAD⁺, NADP⁺; partly made from tryptophan	Pellagra: dermatitis, diarrhoea, dementia
B₆ pyridoxine	Pyridoxal phosphate: transamination, amino-acid decarboxylation	Dermatitis, neuropathy, convulsions
B₇ biotin	Carboxylases (pyruvate, acetyl-CoA); raw egg-white avidin binds it	Dermatitis
B₉ folate	Tetrahydrofolate: one-carbon transfers for nucleotides	Megaloblastic anaemia; neural tube defects
B₁₂ cobalamin (Co)	Methylmalonyl-CoA mutase, methionine synthase; absorption needs intrinsic factor	Megaloblastic (pernicious) anaemia, neuropathy
C ascorbate	Reductant for prolyl and lysyl hydroxylases of collagen; antioxidant	Scurvy
A retinol	11-cis-retinal of rhodopsin; retinoic acid regulates genes	Night blindness, xerophthalmia
D cholecalciferol	Calcitriol (25-OH added in liver, 1α-OH in kidney) raises Ca²⁺ absorption	Rickets, osteomalacia
E tocopherol	Lipid-soluble antioxidant of membranes	Haemolysis, nerve damage
K	γ-Carboxylation of glutamate in clotting factors II, VII, IX, X	Bleeding; warfarin antagonises it

Minerals serve as structure, electrolytes or cofactors. Ca²⁺ builds bone hydroxyapatite, triggers contraction via troponin C and signals via calmodulin; Na⁺, K⁺ and Cl⁻ set osmotic balance and membrane potentials. Among trace elements, iron sits in haem (haemoglobin, myoglobin, cytochromes) and iron–sulfur clusters; copper in cytochrome c oxidase, Cu/Zn superoxide dismutase and haemocyanin, the oxygen carrier of many molluscs and arthropods; zinc in carbonic anhydrase and zinc-finger transcription factors; magnesium forms Mg-ATP, the true substrate of kinases; iodine is part of T₃ and T₄ (deficiency causes goitre); selenium enters glutathione peroxidase as selenocysteine; cobalt is the metal of vitamin B₁₂; molybdenum is a cofactor of xanthine oxidase.

Key takeaways

Glycosidic geometry decides function: glycogen (α1→4, α1→6 branches) is fuel, cellulose (β1→4) and chitin are structure. Linoleic and α-linolenic acids are essential for mammals.
In dsDNA A = T and G = C; Tm rises with G + C and salt, and 2(A + T) + 4(G + C) estimates it for short oligonucleotides. α-helix: 3.6 residues and 0.54 nm per turn, 0.15 nm per residue; B-DNA: 0.34 nm per base pair.
Replication is semiconservative and runs 5′→3′ from primers, with Okazaki fragments on the lagging strand. Pol I, II and III make rRNA, mRNA and tRNA/5S rRNA; 61 codons are sense and 3 stop.
Glycolysis nets 2 ATP and 2 NADH per glucose (control at PFK-1); each acetyl-CoA gives 3 NADH, 1 FADH₂, 1 GTP and 2 CO₂. At 2.5/1.5 ATP per NADH/FADH₂, glucose yields 32 ATP (30 with the glycerol-3-phosphate shuttle).
v = Vmax[S]/(Km + [S]); competitive inhibitors raise Km only, uncompetitive ones lower both, pure non-competitive ones lower Vmax only. Water-soluble hormones act via surface receptors and second messengers in minutes; steroid and thyroid hormones act on transcription via nuclear receptors over hours.
B vitamins are coenzyme precursors (TPP, FAD, NAD⁺, PLP, biotin, THF, cobalamin); A, D, E and K are fat-soluble. Fe, Cu, Zn, Se, Mo and Co are cofactors of named proteins.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A stretch of 18 consecutive residues of a protein is folded as one regular α-helix. Approximately how long is this stretch along the helix axis?
1. 6.1 nm, because each residue advances along the axis by 0.34 nm
2. 9.7 nm, because each residue advances along the axis by 0.54 nm
3. 2.7 nm, because each residue advances along the axis by 0.15 nm
4. 5.0 nm, because the 18 residues make five turns of 1.0 nm each
Show answer
Answer: C — 2.7 nm, because each residue advances along the axis by 0.15 nm
Each residue of an α-helix rises 0.15 nm, and one turn of 3.6 residues rises 0.54 nm. So 18 residues make 18/3.6 = 5 turns × 0.54 nm = 2.7 nm, the same as 18 × 0.15 nm. The 6.1 nm value uses 0.34 nm, the rise per base pair of B-DNA, the commonest confusion; 9.7 nm wrongly treats the 0.54 nm pitch per turn as a rise per residue; and a turn is 0.54 nm, not 1.0 nm.
A linear double-stranded DNA molecule of 5,000 bp contains 30% adenine, expressed as a percentage of all its bases. The total number of guanine residues in the molecule (both strands together) is ____.
Numerical answer — type the value.
Show answer
Answer: 2000
By Chargaff's rules A = T = 30%, so G + C = 40% and G = 20%. The molecule has 2 × 5,000 = 10,000 nucleotides, so G = 0.20 × 10,000 = 2,000. Taking 5,000 nucleotides instead of 10,000 gives 1,000, the usual slip.
Using the Wallace rule, Tm (°C) = 2(A + T) + 4(G + C), where A, T, G and C are the numbers of each base in the oligonucleotide, estimate the melting temperature of the duplex formed by 5′-GACCTGATCGAAGCTTGCAC-3′ with its complementary strand. Tm = ____ °C.
Numerical answer — type the value.
Show answer
Answer: 62
The 20-mer has A = 5, T = 4, G = 5 and C = 6. So A + T = 9 and G + C = 11, giving Tm = 2 × 9 + 4 × 11 = 18 + 44 = 62 °C. Swapping the weights (4 for A + T, 2 for G + C) gives 58 °C; the G + C pairs carry the larger weight because they have three hydrogen bonds and stack more strongly.
E. coli grown for many generations in ¹⁵N medium is transferred to ¹⁴N medium and allowed exactly three rounds of replication. If replication is semiconservative, CsCl density-gradient centrifugation of its DNA will show hybrid (¹⁵N/¹⁴N) and light (¹⁴N/¹⁴N) duplexes in the ratio
1. 1 hybrid : 7 light, because only one duplex in eight still carries old DNA
2. 1 hybrid : 3 light, because the two parental strands lie in two of eight
3. 1 hybrid : 1 light, because each round halves the hybrid fraction once more
4. 1 heavy : 7 light with no hybrid, because the parental duplex stays intact
Show answer
Answer: B — 1 hybrid : 3 light, because the two parental strands lie in two of eight
One heavy duplex gives 2³ = 8 duplexes after three rounds. Its two ¹⁵N strands are never broken up, so they end up in exactly two duplexes, each paired with a new ¹⁴N strand: 2 hybrid and 6 light, a 1 : 3 ratio. The 1 : 1 ratio is the result after two rounds, not three; 1 hybrid : 7 light wrongly counts one hybrid duplex instead of two; and heavy plus light with no hybrid is the conservative prediction that Meselson and Stahl ruled out.
Which of the following statements about DNA replication in an animal cell are correct?
1. DNA polymerase δ can start a new chain on a bare single-stranded template without any primer
2. Okazaki fragments are made on the lagging strand and joined by DNA ligase after their RNA primers are replaced with DNA
3. Each origin may fire several times in one S phase, which is how large genomes finish replication on time
4. Telomerase extends chromosome ends by reverse-transcribing an RNA template that is part of the enzyme
Show answer
Answer: B — Okazaki fragments are made on the lagging strand and joined by DNA ligase after their RNA primers are replaced with DNA; D — Telomerase extends chromosome ends by reverse-transcribing an RNA template that is part of the enzyme
Okazaki fragments are the discontinuous lagging-strand products, sealed by ligase once their primers are replaced, and telomerase is a reverse transcriptase that carries its own RNA template. No DNA polymerase can start a chain de novo: it needs a primer with a free 3′-OH, which primase provides. Origins are licensed in G1 and fire only once per S phase; firing twice would re-replicate DNA, and large genomes finish on time because they have many origins.
Cultured mammalian cells are given a radioactive amino acid together with cycloheximide at a dose that completely blocks translation on cytosolic ribosomes. A small amount of protein still becomes labelled. These labelled proteins are most likely
1. Nuclear-encoded mitochondrial proteins, which are made after being imported into the organelle
2. Secretory proteins, because ribosomes bound to the rough ER are resistant to cycloheximide
3. Histones, because their mRNAs lack poly(A) tails and are translated without the 80S ribosome
4. Respiratory-complex subunits encoded by mitochondrial DNA and made on mitochondrial ribosomes
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Answer: D — Respiratory-complex subunits encoded by mitochondrial DNA and made on mitochondrial ribosomes
Cycloheximide inhibits only the cytosolic 80S ribosome. Mitochondria keep their own bacteria-like ribosomes, which are insensitive to it (they are blocked by chloramphenicol instead) and translate the few polypeptides encoded by mtDNA, all subunits of the respiratory complexes and ATP synthase. Nuclear-encoded mitochondrial proteins, the tempting choice, are made on cytosolic ribosomes before import, so they are blocked. ER-bound ribosomes are ordinary 80S ribosomes, and histone mRNAs, though they lack poly(A), are still translated on 80S ribosomes.
An animal chromosome of 90 Mb has replication origins spaced uniformly every 150 kb. All origins fire at the start of S phase, each sends out two forks in opposite directions, and every fork moves at 25 nucleotides per second until it meets the fork from the neighbouring origin. Ignoring end effects, the time needed to replicate the chromosome is ____ minutes.
Numerical answer — type the value.
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Answer: 50
Forks from adjacent origins meet halfway, so each fork copies 150/2 = 75 kb = 75,000 nt. Time = 75,000/25 = 3,000 s = 50 min. The chromosome length (90 Mb, that is 600 replicons) does not enter, because all replicons are copied in parallel. Forgetting that replication is bidirectional gives 150,000/25 = 6,000 s = 100 min.
Red blood cells are incubated with glucose and sodium fluoride, which inhibits enolase. Which glycolytic intermediate builds up immediately upstream of the blocked step?
1. 2-Phosphoglycerate, the substrate that enolase dehydrates
2. Phosphoenolpyruvate, the high-energy product of enolase
3. Fructose-1,6-bisphosphate, the product of the committed step
4. Glyceraldehyde-3-phosphate, the substrate of the dehydrogenase
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Answer: A — 2-Phosphoglycerate, the substrate that enolase dehydrates
Enolase dehydrates 2-phosphoglycerate to phosphoenolpyruvate. Blocking it makes its substrate, 2-phosphoglycerate (in equilibrium with 3-phosphoglycerate), accumulate, while phosphoenolpyruvate, which lies downstream, is depleted, so the cells stop making pyruvate, lactate and the ATP of pyruvate kinase. Fructose-1,6-bisphosphate and glyceraldehyde-3-phosphate lie several steps earlier and are not the immediate substrate. This is why fluoride is added to blood samples meant for glucose estimation.
Heart muscle oxidises lactate completely to CO₂ and water. Assume 2.5 ATP per NADH and 1.5 ATP per FADH₂, that cytosolic NADH enters the mitochondrion through the malate–aspartate shuttle without loss, and that GTP counts as one ATP. The number of ATP formed per molecule of lactate is ____.
Numerical answer — type the value.
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Answer: 15
Lactate → pyruvate (lactate dehydrogenase, cytosol): 1 NADH = 2.5. Pyruvate → acetyl-CoA (pyruvate dehydrogenase): 1 NADH = 2.5. One turn of the Krebs cycle: 3 NADH = 7.5, 1 FADH₂ = 1.5 and 1 GTP = 1, total 10. Sum = 2.5 + 2.5 + 10 = 15. Forgetting the NADH made when lactate is oxidised to pyruvate gives 12.5.
Which of the following statements about the Krebs cycle in animal mitochondria are correct?
1. Succinate dehydrogenase is bound to the inner mitochondrial membrane and passes electrons to ubiquinone as complex II
2. The two carbons brought in by an acetyl-CoA leave as the two CO₂ released in that same turn of the cycle
3. Malonate inhibits fumarase irreversibly, so that fumarate accumulates in poisoned mitochondria
4. α-Ketoglutarate dehydrogenase, like pyruvate dehydrogenase, needs thiamine pyrophosphate and lipoate
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Answer: A — Succinate dehydrogenase is bound to the inner mitochondrial membrane and passes electrons to ubiquinone as complex II; D — α-Ketoglutarate dehydrogenase, like pyruvate dehydrogenase, needs thiamine pyrophosphate and lipoate
Succinate dehydrogenase is the cycle's only membrane-bound enzyme and is complex II. α-Ketoglutarate dehydrogenase is a multienzyme complex built like pyruvate dehydrogenase and uses TPP, lipoate, CoA, FAD and NAD⁺. The two CO₂ of a turn come from the oxaloacetate portion; the acetyl carbons leave only in later turns, as isotope labelling shows. Malonate is a competitive inhibitor of succinate dehydrogenase, so succinate, not fumarate, accumulates.
An enzyme that obeys Michaelis–Menten kinetics gives an initial rate of 40 µmol min⁻¹ at [S] = 2 mM and 60 µmol min⁻¹ at [S] = 6 mM, with the same amount of enzyme in both assays. Its Vmax is ____ µmol min⁻¹.
Numerical answer — type the value.
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Answer: 80
From v = Vmax[S]/(Km + [S]): 40(Km + 2) = 2Vmax gives Vmax = 20Km + 40, and 60(Km + 6) = 6Vmax gives Vmax = 10Km + 60. Equating, 10Km = 20, so Km = 2 mM and Vmax = 80 µmol min⁻¹. Check: 80 × 2/4 = 40 and 80 × 6/8 = 60. Reading 60 as Vmax ignores that the rate is still rising with [S].
In the presence of 1 mM of a reversible inhibitor, an enzyme's apparent Km rises from 2 mM to 6 mM while its Vmax stays at 50 µmol min⁻¹. For the inhibition type involved, the affected constant changes by the factor 1 + [I]/Ki. Which conclusion is correct?
1. Uncompetitive inhibition with Ki = 0.5 mM
2. Competitive inhibition with Ki = 3.0 mM
3. Competitive inhibition with Ki = 0.5 mM
4. Non-competitive inhibition with Ki = 1.5 mM
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Answer: C — Competitive inhibition with Ki = 0.5 mM
A rise in Km with Vmax unchanged is the signature of competitive inhibition. The factor is 6/2 = 3 = 1 + [I]/Ki, so [I]/Ki = 2 and Ki = 1/2 = 0.5 mM. Ki = 3.0 mM confuses the factor itself with Ki. Uncompetitive inhibition would lower both Km and Vmax, and pure non-competitive inhibition would lower Vmax and leave Km unchanged, the opposite of what was observed.
In chymotrypsin, the nucleophile that attacks the carbonyl carbon of the scissile peptide bond and forms the acyl-enzyme intermediate is the side chain of
1. Cysteine, made reactive by a neighbouring histidine as a thiolate ion
2. Serine, made reactive by a histidine that is held in place by aspartate
3. Histidine, which is itself acylated first and is later hydrolysed by water
4. Aspartate, which activates a water molecule as in the acid proteases
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Answer: B — Serine, made reactive by a histidine that is held in place by aspartate
Chymotrypsin is a serine protease: Ser195 is the nucleophile, His57 acts as a general base that takes its proton, and Asp102 orients and stabilises the histidine. The serine is acylated and the acyl-enzyme is then hydrolysed. A cysteine nucleophile belongs to papain-like proteases, and water activated by aspartates belongs to aspartic proteases such as pepsin; histidine is the base, not the acylated residue.
A hormone produces its full effect on target cells only after several hours, and the effect is abolished if actinomycin D, an inhibitor of transcription, is given together with the hormone. The hormone is most likely
1. Cortisol, acting through an intracellular receptor that binds hormone response elements in DNA
2. Adrenaline, acting through β-adrenergic receptors that raise cAMP and activate protein kinase A
3. Insulin, acting by moving pre-formed GLUT4 transporters from vesicles to the plasma membrane
4. Glucagon, acting through a G-protein-coupled receptor that activates glycogen phosphorylase
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Answer: A — Cortisol, acting through an intracellular receptor that binds hormone response elements in DNA
A slow effect that needs new transcription is the mark of a lipophilic hormone working through a nuclear receptor: cortisol binds the glucocorticoid receptor, which moves to the nucleus and binds hormone response elements. The other three act within seconds to minutes by modifying proteins that already exist (PKA phosphorylation, GLUT4 translocation, phosphorylase activation), so blocking transcription would not abolish them. Insulin is the tempting choice because it also has slower effects on genes, but the mechanism given here does not need transcription.
A cell carries both a Gs-coupled and a Gi-coupled receptor controlling the same adenylyl cyclase, and the hormones for both receptors are present. Which of the following treatments would raise the cell's cAMP level further?
1. Cholera toxin, which ADP-ribosylates Gsα and blocks its GTPase activity
2. A drug that blocks phospholipase C-β and so prevents IP₃ formation
3. Pertussis toxin, which ADP-ribosylates Giα and prevents its activation by the receptor
4. Caffeine, which inhibits cAMP phosphodiesterase
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Answer: A — Cholera toxin, which ADP-ribosylates Gsα and blocks its GTPase activity; C — Pertussis toxin, which ADP-ribosylates Giα and prevents its activation by the receptor; D — Caffeine, which inhibits cAMP phosphodiesterase
Cholera toxin locks Gsα in its GTP-bound, active state, so the cyclase is driven continuously. Pertussis toxin stops the Gi receptor from activating Giα, removing the brake on the cyclase. Caffeine slows the destruction of cAMP. Blocking phospholipase C-β affects the IP₃–DAG–Ca²⁺ branch used by Gq-coupled receptors and does not act on adenylyl cyclase or phosphodiesterase.
A thiamine-deficient animal on a carbohydrate-rich diet shows raised blood pyruvate and lactate. The enzyme whose loss of activity best explains this is
1. Pyruvate carboxylase, which needs biotin to add CO₂ to pyruvate in mitochondria
2. Lactate dehydrogenase, which needs NAD⁺ made from dietary niacin
3. Pyruvate kinase, which needs pyridoxal phosphate as its cofactor
4. Pyruvate dehydrogenase, whose first component needs thiamine pyrophosphate
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Answer: D — Pyruvate dehydrogenase, whose first component needs thiamine pyrophosphate
Thiamine pyrophosphate is the cofactor of the E1 component of pyruvate dehydrogenase. Without it pyruvate cannot enter the Krebs cycle, so it accumulates and is diverted to lactate, which is why carbohydrate loads worsen beriberi. Pyruvate carboxylase uses biotin, not thiamine, and lactate dehydrogenase uses NAD⁺; neither depends on thiamine. Pyruvate kinase needs K⁺ and Mg²⁺, not pyridoxal phosphate, and it lies upstream of pyruvate.
Vitamin K is required as a cofactor for
1. Hydroxylation of proline and lysine residues in procollagen
2. Hydroxylation of 25-hydroxycholecalciferol to calcitriol in the kidney
3. γ-Carboxylation of glutamate residues in prothrombin and other clotting factors
4. Isomerisation of all-trans-retinal to 11-cis-retinal to regenerate rhodopsin in rod cells
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Answer: C — γ-Carboxylation of glutamate residues in prothrombin and other clotting factors
Vitamin K is the cofactor of γ-glutamyl carboxylase, which converts glutamate to γ-carboxyglutamate in factors II, VII, IX and X; the extra carboxyl groups bind Ca²⁺ and anchor the factors to membranes. Warfarin blocks the recycling of vitamin K and so acts as an anticoagulant. Collagen hydroxylation needs vitamin C, the renal step that makes calcitriol belongs to vitamin D metabolism, and the retinal isomerisation belongs to the vitamin A cycle of the retina.
Which of the following pairings of a trace element with a protein or vitamin that contains it are correct?
1. Copper: haemocyanin, the oxygen carrier of many molluscs and arthropods
2. Cobalt: cytochrome c oxidase, the terminal enzyme of the respiratory chain
3. Zinc: carbonic anhydrase, which hydrates CO₂ in red blood cells
4. Molybdenum: the corrin ring of vitamin B₁₂
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Answer: A — Copper: haemocyanin, the oxygen carrier of many molluscs and arthropods; C — Zinc: carbonic anhydrase, which hydrates CO₂ in red blood cells
Haemocyanin binds O₂ between two copper ions, and carbonic anhydrase uses Zn²⁺ to turn bound water into a hydroxide that attacks CO₂. Cytochrome c oxidase contains copper and haem iron, not cobalt. Cobalt, not molybdenum, sits in the corrin ring of vitamin B₁₂; molybdenum is a cofactor of enzymes such as xanthine oxidase.