Structure and Function of Biomolecules: Water, Macromolecules, Protein Folding and Six Model Proteins

This chapter answers Section XL1.1 of XL1 Biochemistry, one of the five optional sections of the Life Sciences (XL) paper: “Organization of life; Importance of water; Structure and function of biomolecules: Carbohydrates, lipids, proteins, and nucleic acids; Protein structure, folding/misfolding, and function; Myoglobin, hemoglobin, lysozyme, ribonuclease A, carboxypeptidase, and chymotrypsin.” XL0 Chemistry covers the chemistry of biomolecules; here the question is always how a structure does its job. It runs from biological organisation and water, through the four classes of biomolecule, to protein structure, folding and misfolding, and ends with six case-study proteins: two oxygen carriers and four enzymes. The numericals are fractional saturation of myoglobin and hemoglobin, oxygen delivery, the Hill coefficient, and base composition from Chargaff's rules.

1. Organisation of life and the importance of water

Life is organised in levels, and each level has properties its parts lack. A few elements (C, H, O, N, P, S) make up most of a cell's mass. They form small molecules (sugars, amino acids, nucleotides, fatty acids), which are joined into macromolecules by condensation, with loss of water, and taken apart by hydrolysis: polysaccharides, proteins and nucleic acids are polymers, while lipids are not polymers but assemble non-covalently into membranes. Macromolecules build supramolecular complexes (ribosomes, chromatin, membranes), then organelles, cells, tissues and organisms. A cell is an open system that stays far from equilibrium only by taking in free energy.

Water is the medium of all this, and its properties come from one fact: the bent (104.5°), polar molecule can donate two and accept two hydrogen bonds. Hence its high specific heat and heat of vaporisation (thermal buffering), its cohesion, and its high dielectric constant (about 78 at 25 °C), which weakens the attraction between ions so that salts and polar molecules dissolve and hydrate. Water also ionises slightly (Kw = 1.0 × 10⁻¹⁴ at 25 °C), so pH and the ionisation of acidic and basic groups set the charge of every protein and nucleic acid.

The hydrophobic effect is water's most important gift to structure. A nonpolar group cannot hydrogen-bond, so the water around it is forced into an ordered cage; when nonpolar groups cluster, that water is released and its entropy rises. This drives the burial of nonpolar side chains, the formation of micelles and bilayers, and base stacking in DNA.

⚠️ Exam trap
A frequent wrong answer says nonpolar groups “attract” each other in water; their van der Waals contacts are weak. The driving force is the entropy gain of the released water, while the chain itself loses entropy on folding.

2. Carbohydrates and lipids

Monosaccharides are aldoses or ketoses; D or L is decided by the chiral carbon farthest from the carbonyl, and natural sugars are mostly D. Pentoses and hexoses close into five-membered (furanose) or six-membered (pyranose) rings, which creates a new chiral anomeric carbon (α or β). The two anomers interconvert through the open chain (mutarotation); D-glucose in water is about one-third α and two-thirds β. A sugar with a free anomeric carbon is reducing. A glycosidic bond links the anomeric carbon to another sugar: maltose is Glc(α1→4)Glc and lactose is Gal(β1→4)Glc, both reducing, while sucrose, Glc(α1↔2β)Fru, joins both anomeric carbons and is non-reducing.

Polysaccharides: the linkage decides the job
PolymerUnit and linkageShape and role
AmyloseGlucose, α(1→4)Unbranched helix; plant storage
Amylopectinα(1→4) with α(1→6) branches every 24–30 residuesBranched; plant storage
Glycogenα(1→4) with α(1→6) branches every 8–12 residuesMany non-reducing ends for fast release; animal storage
CelluloseGlucose, β(1→4)Straight chains H-bonded into fibres; plant cell wall
ChitinN-acetylglucosamine, β(1→4)Arthropod exoskeleton, fungal wall
PeptidoglycanGlcNAc and MurNAc alternating, β(1→4), cross-linked by peptidesBacterial cell wall; lysozyme's substrate

Sugars also carry information: sulfated glycosaminoglycans form matrix proteoglycans, and oligosaccharides on glycoproteins and glycolipids act in recognition, as in the ABO blood groups. Lipids are defined by insolubility in water. Fatty acids are saturated (straight) or unsaturated (cis double bonds put kinks that lower the melting point). Triacylglycerols store energy compactly, being more reduced and stored without water. Glycerophospholipids and sphingolipids are amphipathic: two-tailed, cylinder-shaped molecules form bilayers, whereas one-tailed, cone-shaped ones form micelles. Cholesterol buffers membrane fluidity and is the precursor of steroid hormones, bile acids and vitamin D; arachidonate gives eicosanoids, and inositol phospholipids give second messengers.

3. Nucleic acids

A nucleotide is a base (purine A or G; pyrimidine C, T or U) on C1′ of a pentose, with phosphate. 3′→5′ phosphodiester bonds give each strand a 5′ to 3′ polarity. In B-DNA two antiparallel strands wind into a right-handed helix of about 10.5 base pairs per turn, 0.34 nm apart, about 2 nm across, with a major and a minor groove; proteins read the sequence mostly in the major groove. G·C pairs make three hydrogen bonds and A·T pairs two, but base stacking contributes most to stability, so the melting temperature rises with GC content, and melting raises absorbance at 260 nm (hyperchromicity). A-DNA is right-handed with about 11 bp per turn; Z-DNA is left-handed, about 12 bp per turn, favoured by alternating purine–pyrimidine sequences.

Chargaff's rules follow from pairing: in double-stranded DNA, A = T and G = C, so purines equal pyrimidines. Example: a 5,000 bp duplex with 30% A has 10,000 nucleotides, of which A = T = 3,000 and G = C = 2,000. RNA's 2′-OH changes everything: duplex RNA adopts the A-form, and in alkali the 2′-OH attacks the neighbouring phosphodiester to form a 2′,3′-cyclic phosphate, so RNA is alkali-labile while DNA is not. Nucleotides also serve as energy currency (ATP), coenzyme parts (NAD⁺, FAD, CoA) and signals (cAMP), and some RNAs are catalysts.

⚠️ Exam trap
Chargaff's equalities hold only for double-stranded DNA. A single-stranded viral genome need not have A = T, and a question that gives “% A” for single-stranded DNA cannot be solved for G.

4. Protein structure: four levels

Proteins are chains of the 20 standard L-α-amino acids. The peptide bond has partial double-bond character, so the six atoms around it are planar and almost always trans (cis occurs mainly before proline). Backbone freedom lies only in the rotations φ (N–Cα) and ψ (Cα–C); the Ramachandran plot shows the sterically allowed φ/ψ pairs, widest for glycine and narrowest for proline. Primary structure is the sequence; secondary structure is local backbone hydrogen bonding; tertiary structure is the full fold of one chain; quaternary structure is the arrangement of subunits, such as the α₂β₂ of hemoglobin.

Regular secondary structures
ElementGeometryHydrogen bonds and residues
α-helixRight-handed; 3.6 residues per turn; rise 0.15 nm per residue; pitch 0.54 nmC=O of residue n to N–H of residue n + 4, parallel to the axis; side chains point outward; proline breaks it, glycine destabilises it
β-sheetExtended strands side by side, parallel or antiparallel; pleatedBetween adjacent strands; side chains alternate above and below the sheet
β-turnFour residues reverse the chain directionC=O of residue 1 to N–H of residue 4; glycine and proline are common

The tertiary fold buries nonpolar side chains in a core and leaves polar ones on the surface; it is held by the hydrophobic effect, hydrogen bonds, ion pairs, van der Waals contacts and, in many extracellular proteins, disulfide bonds. Recurring motifs (β-α-β, helix-turn-helix) combine into domains that fold independently. Fibrous proteins serve mechanics: α-keratin is a coiled coil of α-helices; collagen is a triple helix of three left-handed chains with Gly-X-Y repeats (X often Pro, Y often 4-hydroxyproline), and prolyl hydroxylation needs ascorbate, so scurvy weakens connective tissue.

5. Folding and misfolding

Anfinsen's experiment with ribonuclease A showed that sequence determines structure. RNase A (124 residues, four disulfide bonds) was unfolded in 8 M urea with β-mercaptoethanol, which reduces the disulfides, and lost activity. Removing both agents together by dialysis restored nearly full activity, with the correct four pairings out of the 105 possible for eight cysteines (7 × 5 × 3 × 1). If the reduced protein was instead oxidised while still in urea and urea was removed afterwards, “scrambled” RNase with about 1% activity resulted; a trace of β-mercaptoethanol let the disulfides reshuffle and activity returned. The native structure is therefore the thermodynamically most stable one under physiological conditions.

A random search of all conformations would take longer than the age of the universe (Levinthal's paradox), so folding follows a funnel: hydrophobic collapse to a compact molten globule with much secondary structure, then slower locking of tertiary contacts. Slow steps are catalysed by protein disulfide isomerase and peptidyl-prolyl cis–trans isomerase. Molecular chaperones prevent aggregation but carry no folding information: Hsp70 binds exposed hydrophobic segments of new chains in an ATP-driven cycle, and chaperonins such as bacterial GroEL–GroES enclose a single chain in a chamber where it can fold alone.

Misfolding has clinical faces. Many proteins can convert to amyloid, protease-resistant fibrils with a cross-β structure: amyloid-β plaques in Alzheimer's disease and α-synuclein in Parkinson's disease. In prion diseases (Creutzfeldt–Jakob disease, bovine spongiform encephalopathy) the α-helix-rich normal PrP^C is converted by the β-rich PrP^Sc into more PrP^Sc with no change in sequence, so the misfolded form propagates itself. In cystic fibrosis the ΔF508 CFTR folds poorly and is destroyed by ER quality control before reaching the membrane.

⚠️ Exam trap
Sickle-cell hemoglobin is not a misfolded protein. HbS folds normally; the β6 Glu→Val change creates a sticky hydrophobic patch, so deoxygenated HbS tetramers polymerise into fibres that distort the red cell.

6. Myoglobin and hemoglobin: binding and cooperativity

Myoglobin is a single chain of 153 residues folded into eight helices (A–H) around a heme. The heme Fe²⁺ binds four pyrrole nitrogens and the proximal His (F8); O₂ binds the sixth position, where the distal His (E7) hydrogen-bonds it and sterically hinders CO, cutting CO's large advantage over O₂ seen with free heme. Oxidised Fe³⁺ (met-form) cannot bind O₂. With one site, binding is hyperbolic: θ = pO₂/(P50 + pO₂), where P50 is the pO₂ at half saturation, only a few torr for myoglobin. Example: with P50 = 2.8 torr, at 20 torr θ = 20/22.8 = 0.88, so myoglobin stays loaded as a muscle O₂ store.

Hemoglobin is an α₂β₂ tetramer (α 141, β 146 residues), each subunit with a myoglobin-like fold. It shifts between a low-affinity T state and a high-affinity R state. O₂ binding pulls the Fe into the heme plane, dragging the proximal His and helix F, which breaks ion pairs at the α₁β₂ interface and favours R in the remaining subunits: cooperativity gives a sigmoidal curve with P50 about 26 torr in red cells. The Hill equation, log[θ/(1 − θ)] = n log pO₂ − n log P50, gives a Hill coefficient n of about 3 for hemoglobin (less than the four sites, since cooperativity is not infinite) and 1 for myoglobin. The concerted (MWC) model moves all subunits together between T and R; the sequential (KNF) model lets each subunit change on binding.

Allosteric control of hemoglobin
EffectorHow it actsEffect on the curve
H⁺ (Bohr effect)Protonated His146 of β forms a salt bridge with Asp94 that stabilises TRight shift in acidic, active tissue
CO₂Forms carbamates at the N-termini, which stabilise T, and also lowers pHRight shift
2,3-BPGOne per tetramer, in the positively charged central cavity between the β subunits of TRaises P50; rises at high altitude
Fetal γ chainsSer at γ143 instead of His lowers 2,3-BPG bindingHigher O₂ affinity, so HbF draws O₂ from maternal blood
COBinds the heme more than 200-fold more tightly than O₂ and pushes the other sites towards RLeft shift; remaining O₂ is not unloaded
🧠 Memory trick
Delivery is the difference in θ between lungs and tissues. With n = 3 and P50 = 26 torr, θ is 0.98 at 100 torr and 0.31 at 20 torr, so 67% of the bound O₂ is released; a non-cooperative carrier with the same P50 would release only about 36%.

7. Four model enzymes: lysozyme, ribonuclease A, carboxypeptidase A and chymotrypsin

Each enzyme illustrates a different catalytic strategy
EnzymeBond cleavedKey groups and strategy
Lysozyme (hen egg white, 129 residues)β(1→4) bond between C1 of MurNAc and C4 of GlcNAc in peptidoglycanGeneral acid catalysis, substrate strain, covalent intermediate
Ribonuclease A (bovine pancreas, 124 residues)RNA phosphodiester on the 3′ side of pyrimidine nucleotidesConcerted general acid–base catalysis by two His; Lys41 stabilises the transition state
Carboxypeptidase A (pancreas)C-terminal peptide bond; prefers aromatic or bulky aliphatic C-terminal residues, not Arg, Lys or ProMetal-ion catalysis by Zn²⁺; Glu270 general base; induced fit
Chymotrypsin (pancreas)Peptide bond on the carboxyl side of Phe, Tyr, Trp (less so Leu, Met)Covalent and acid–base catalysis by a Ser–His–Asp triad; oxyanion hole

Lysozyme, in tears, saliva and egg white, lyses bacteria, especially Gram-positive ones with exposed peptidoglycan. Six sugars bind in a cleft (sites A–F) and the bond between sites D and E is cut. Glu35, in a nonpolar pocket and hence protonated, donates a proton to the glycosidic oxygen; the sugar in site D is distorted towards a half-chair; Asp52 forms a covalent glycosyl-enzyme intermediate, which water then hydrolyses with Glu35 acting as a base, retaining the β configuration. RNase A uses two histidines: His12 removes the proton from the 2′-OH, which attacks phosphorus while His119 protonates the leaving 5′-oxygen, giving a 2′,3′-cyclic phosphate; the roles reverse to hydrolyse it to a 3′-phosphate. Because one His must be basic and the other acidic, activity against pH is bell-shaped, and DNA, lacking a 2′-OH, is not cut.

Carboxypeptidase A is a zinc exopeptidase: Arg145 holds the free C-terminal carboxylate, so only the terminal residue is removed, and the residue's side chain must fit a hydrophobic pocket; Lipscomb's structures showed Tyr248 moving in on substrate binding, a classic case of induced fit. Chymotrypsin is the model serine protease. His57, oriented by Asp102, takes the proton from Ser195, whose oxygen attacks the carbonyl; the tetrahedral oxyanion is stabilised by backbone N–H groups of Gly193 and Ser195 (the oxyanion hole); the amine half leaves and an acyl-enzyme remains, which water then hydrolyses. With p-nitrophenyl acetate a fast burst of p-nitrophenolate, equal to the active sites, precedes a slow steady rate set by deacylation. The deep hydrophobic S1 pocket explains the specificity; trypsin's Asp189 at the bottom of S1 selects Lys and Arg. Chymotrypsinogen is activated when trypsin cuts Arg15–Ile16; the new N-terminal Ile16 pairs with Asp194 and forms the oxyanion hole and S1 pocket. DIFP inactivates by phosphorylating Ser195, and TPCK alkylates His57.

Key takeaways

  • Macromolecules form by condensation and break by hydrolysis; lipids assemble rather than polymerise. Water's high dielectric constant dissolves ions, and the hydrophobic effect, an entropy gain of released water, drives folding and bilayers.
  • Linkage decides function: α(1→4)/α(1→6) glucose stores fuel (glycogen branches every 8–12 residues), β(1→4) builds fibres; sucrose is non-reducing. In dsDNA A = T and G = C; RNA's 2′-OH makes it alkali-labile.
  • The α-helix has 3.6 residues per turn, 0.15 nm per residue and i→i+4 hydrogen bonds; β-sheets bond between strands. Anfinsen's RNase A experiment shows the native fold is the free-energy minimum set by sequence; chaperones only prevent aggregation.
  • Misfolding gives amyloid (Alzheimer's, Parkinson's), self-propagating prions and ER-degraded ΔF508 CFTR; sickle-cell disease is polymerisation of a normally folded HbS.
  • Myoglobin binds O₂ hyperbolically (n = 1); hemoglobin is cooperative (n ≈ 3, P50 ≈ 26 torr) through T→R switching. H⁺, CO₂ and 2,3-BPG stabilise T and shift the curve right; HbF and CO shift it left.
  • Lysozyme (Glu35, Asp52, strain), RNase A (His12, His119, 2′,3′-cyclic phosphate), carboxypeptidase A (Zn²⁺, Glu270, induced fit) and chymotrypsin (Ser195–His57–Asp102, oxyanion hole, acyl-enzyme) model acid–base, metal-ion and covalent catalysis.

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.

  1. Which class of biomolecule is NOT a polymer built by condensation of repeating monomer units, but instead assembles into larger structures through non-covalent interactions?

    1. Proteins, built from amino acids joined by peptide bonds
    2. Nucleic acids, built from nucleotides joined by phosphodiester bonds
    3. Polysaccharides, built from monosaccharides joined by glycosidic bonds
    4. Membrane lipids, which form bilayers through the hydrophobic effect
    Show answer

    Answer: D — Membrane lipids, which form bilayers through the hydrophobic effect

    Lipids are not polymers: phospholipids and cholesterol are separate molecules that associate non-covalently into bilayers, driven by the hydrophobic effect. Proteins, nucleic acids and polysaccharides are true polymers made by condensation (loss of water) and broken by hydrolysis. Polysaccharides are the tempting distractor because glycogen looks like a large aggregate, but its glucose units are covalently linked.

  2. When a protein folds, its nonpolar side chains become buried in the core. What is the main thermodynamic reason this burial is favourable in water?

    1. Water molecules ordered around exposed nonpolar groups are released, so the entropy of the water increases
    2. Nonpolar side chains attract one another strongly, so the enthalpy of their association is large and negative
    3. Nonpolar side chains form hydrogen bonds with backbone amide groups once they are inside the core
    4. The polypeptide chain gains conformational entropy as it collapses into a compact folded state
    Show answer

    Answer: A — Water molecules ordered around exposed nonpolar groups are released, so the entropy of the water increases

    This is the hydrophobic effect: nonpolar groups cannot hydrogen-bond, so water around them is held in ordered cages; clustering the groups releases that water and its entropy rises. Van der Waals attraction between nonpolar groups exists but is weak and is not the main driving force. The chain itself loses, not gains, conformational entropy on folding, which opposes folding; nonpolar side chains have no groups to hydrogen-bond with amides.

  3. A disaccharide gives a negative test with Benedict's reagent and shows no mutarotation in solution. Which disaccharide fits these observations?

    1. Maltose, Glc(α1→4)Glc
    2. Lactose, Gal(β1→4)Glc
    3. Sucrose, Glc(α1↔2β)Fru
    4. Cellobiose, Glc(β1→4)Glc
    Show answer

    Answer: C — Sucrose, Glc(α1↔2β)Fru

    In sucrose the glycosidic bond joins the anomeric C1 of glucose to the anomeric C2 of fructose, so neither ring can open: there is no free anomeric carbon to reduce Cu²⁺ or to interconvert between α and β. Maltose, lactose and cellobiose all use the anomeric carbon of only one residue; the other glucose keeps a free anomeric carbon, so they are reducing and mutarotate.

  4. Glycogen and amylose are both glucose polymers. Why is glycogen better suited than amylose to release glucose rapidly in muscle at the start of exercise?

    1. Its glucose units are joined by β(1→4) bonds, which are hydrolysed faster than α(1→4) bonds
    2. Its α(1→6) branches every 8–12 residues create many non-reducing ends at which enzymes act at once
    3. Its single reducing end is where glucose is released, and this end is more exposed than in amylose
    4. It is stored as a compact unbranched helix, so less water has to be removed before it is degraded
    Show answer

    Answer: B — Its α(1→6) branches every 8–12 residues create many non-reducing ends at which enzymes act at once

    Glycogen phosphorylase removes glucose from non-reducing ends. Branching every 8–12 residues gives glycogen a very large number of such ends, so many enzyme molecules work simultaneously; branching also makes the polymer more soluble. Glycogen's main bonds are α(1→4), not β(1→4); it has only one reducing end, which is not where degradation happens; and it is highly branched, not an unbranched helix (that describes amylose).

  5. Which of the following statements about nucleic acid structure are correct?

    1. A double-stranded DNA in which A makes up 30% of the bases must contain 30% G
    2. RNA, but not DNA, is cleaved by dilute alkali because its 2′-OH attacks the adjacent phosphodiester to form a 2′,3′-cyclic phosphate
    3. Z-DNA is a right-handed helix favoured by alternating purine–pyrimidine sequences
    4. In B-DNA there are about 10.5 base pairs per turn and adjacent base pairs are 0.34 nm apart
    Show answer

    Answer: B — RNA, but not DNA, is cleaved by dilute alkali because its 2′-OH attacks the adjacent phosphodiester to form a 2′,3′-cyclic phosphate; D — In B-DNA there are about 10.5 base pairs per turn and adjacent base pairs are 0.34 nm apart

    The 2′-OH makes RNA alkali-labile through a 2′,3′-cyclic phosphate, and B-DNA has about 10.5 bp per turn with a 0.34 nm rise. With 30% A, Chargaff's rules give T = 30% and G = C = 20%, not 30% G. Z-DNA is left-handed, not right-handed, although it is indeed favoured by alternating purine–pyrimidine sequences.

  6. A linear double-stranded DNA molecule of 5,000 base pairs contains 30% adenine (as a percentage of all its bases). How many guanine residues does the molecule contain in total, counting both strands?

    Numerical answer — type the value.

    Show answer

    Answer: 2000

    The molecule has 2 × 5,000 = 10,000 nucleotides. A = 30% = 3,000, and by Chargaff's rule T = A = 3,000. The remaining 10,000 − 6,000 = 4,000 are G + C, and G = C, so G = 2,000. Answering 3,000 (taking G = A) or 1,000 (counting one strand) are the usual slips.

  7. In a right-handed α-helix, the backbone C=O group of residue n forms a hydrogen bond with the backbone N–H group of which residue?

    1. Residue n + 3 of the same chain, giving a tighter helix with exactly three residues per turn
    2. Residue n + 2 of the same chain, so that every second residue closes a small ring
    3. Residue n + 4 of the same chain, with the bond running roughly parallel to the helix axis
    4. A residue lying on a neighbouring polypeptide strand, roughly perpendicular to the chain
    Show answer

    Answer: C — Residue n + 4 of the same chain, with the bond running roughly parallel to the helix axis

    In the α-helix each C=O(n) bonds to N–H(n + 4), within the same chain and nearly parallel to the axis, which gives 3.6 residues per turn. The n + 3 pattern describes the tighter 3₁₀ helix. Hydrogen bonds between neighbouring strands, perpendicular to the chains, define the β-sheet, not the helix.

  8. Ribonuclease A is fully unfolded in 8 M urea containing β-mercaptoethanol. In experiment X, urea and β-mercaptoethanol are removed together by dialysis and the protein is then exposed to air. In experiment Y, the reduced protein is first oxidised by air while still in 8 M urea, and only then is the urea removed. Which result and conclusion are correct?

    1. Both X and Y regain full activity, because once disulfide bonds form they alone specify the native fold
    2. X stays inactive while Y regains activity, because the native fold requires its disulfide bonds to form before the chain is allowed to fold up
    3. Neither regains activity, because the folding information is supplied by chaperones that are absent from the purified system
    4. X regains nearly full activity but Y gives scrambled, about 1% active protein: the native pairing is the most stable one, set by sequence
    Show answer

    Answer: D — X regains nearly full activity but Y gives scrambled, about 1% active protein: the native pairing is the most stable one, set by sequence

    In X the chain folds first under native conditions and the correct four disulfides then form, restoring nearly full activity. In Y the disulfides form while the chain is still unfolded, so they pair almost at random among the 105 possibilities, locking in wrong structures with about 1% activity (and a trace of thiol lets them reshuffle to the native form). Hence the native structure is the thermodynamically favoured one encoded by sequence. The view that disulfides alone specify the fold is contradicted by Y, and no chaperone was needed in X.

  9. Which of the following statements about protein folding and misfolding are correct?

    1. Chaperonins such as GroEL–GroES enclose a polypeptide in a chamber where it can fold without aggregating, using ATP
    2. Molecular chaperones carry the structural information that specifies the final three-dimensional fold of their substrates
    3. In prion disease the β-sheet-rich PrP^Sc converts α-helix-rich PrP^C into more PrP^Sc without any change in amino acid sequence
    4. Sickle-cell hemoglobin is a misfolded protein that is rapidly degraded by the proteasome inside red cells
    Show answer

    Answer: A — Chaperonins such as GroEL–GroES enclose a polypeptide in a chamber where it can fold without aggregating, using ATP; C — In prion disease the β-sheet-rich PrP^Sc converts α-helix-rich PrP^C into more PrP^Sc without any change in amino acid sequence

    Chaperonins provide an enclosed, ATP-driven folding chamber, and prions propagate by templated conversion of PrP^C to PrP^Sc with the same sequence. Chaperones only prevent aggregation and give the chain repeated chances; the information for the fold lies in the sequence, as Anfinsen showed. HbS folds normally; the β6 Glu→Val surface change makes deoxy-HbS polymerise into fibres, which is not misfolding followed by degradation.

  10. Fetal hemoglobin (α₂γ₂) has a higher O₂ affinity than adult hemoglobin (α₂β₂) inside red cells, which lets the fetus take O₂ from maternal blood. What is the molecular basis?

    1. The γ chains carry Ser instead of His at position 143, which weakens 2,3-BPG binding in the central cavity
    2. The γ chains lack the proximal histidine, so the heme iron is left free to bind O₂ more tightly
    3. Fetal red cells contain no 2,3-BPG at all, so hemoglobin stays locked in the high-affinity R state
    4. HbF dissociates into monomers like myoglobin, so its O₂ binding becomes non-cooperative and of high affinity
    Show answer

    Answer: A — The γ chains carry Ser instead of His at position 143, which weakens 2,3-BPG binding in the central cavity

    2,3-BPG binds in the positively charged cavity between the two β chains and stabilises the T state. In γ chains His143 is replaced by Ser, removing positive charges, so HbF binds 2,3-BPG weakly and has a lower P50. Fetal red cells do contain 2,3-BPG, so the claim that they lack it is the tempting but wrong distractor. γ chains keep the proximal His, and HbF remains a cooperative tetramer.

  11. Patient X has carbon monoxide occupying 50% of his heme sites; patient Y is anaemic, with half the normal hemoglobin and no CO. Both carry about the same amount of O₂ per litre of arterial blood, yet X is far more severely hypoxic in his tissues. Why?

    1. CO oxidises the heme iron to Fe³⁺ in the remaining subunits, so these subunits release their bound O₂ into plasma before the blood reaches the tissues
    2. CO binds 2,3-BPG and removes it from the central cavity, which shifts the O₂ curve of patient X to the right
    3. CO bound to some subunits shifts the tetramer towards the R state, raising the affinity of the other hemes so they hold their O₂ in tissues
    4. CO stops 2,3-BPG synthesis in red cells, so the hemoglobin of patient X becomes non-cooperative and releases O₂ too early
    Show answer

    Answer: C — CO bound to some subunits shifts the tetramer towards the R state, raising the affinity of the other hemes so they hold their O₂ in tissues

    CO acts like a bound ligand that favours the R state: in a partly CO-liganded tetramer the remaining hemes have raised O₂ affinity, so the curve shifts left and becomes less sigmoidal, and little O₂ is unloaded at tissue pO₂. In anaemia the remaining hemoglobin is normal and unloads normally. CO does not oxidise iron (that gives methemoglobin, which cannot bind O₂ at all), and a right shift would increase, not decrease, unloading.

  12. Myoglobin binds O₂ with a single site and a P50 of 2.8 torr. What fraction of myoglobin molecules carry O₂ at a pO₂ of 20 torr? Give the answer as a fraction to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.88

    For one site θ = pO₂/(P50 + pO₂) = 20/(2.8 + 20) = 20/22.8 = 0.877, which is 0.88. The curve is hyperbolic, so no Hill exponent enters.

  13. Hemoglobin binding is described by the Hill equation θ = pO₂ⁿ/(P50ⁿ + pO₂ⁿ) with n = 3.0 and P50 = 26 torr. Blood leaves the lungs at pO₂ = 100 torr and active muscle capillaries are at pO₂ = 20 torr. What percentage of the hemoglobin binding sites unload O₂ between lungs and muscle, i.e. 100 × (θ at 100 torr − θ at 20 torr)? Give the nearest integer.

    Numerical answer — type the value.

    Show answer

    Answer: 67

    26³ = 17,576. At 100 torr θ = 1,000,000/(17,576 + 1,000,000) = 0.983. At 20 torr θ = 8,000/(17,576 + 8,000) = 0.313. Difference = 0.670, i.e. 67%. Using n = 1 instead gives only about 36%, which is why cooperativity matters for delivery.

  14. An oxygen-binding protein is 20% saturated at pO₂ = 15 torr and 80% saturated at pO₂ = 40 torr. Assuming it obeys the Hill equation over this range, what is its Hill coefficient? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 2.83

    From log[θ/(1 − θ)] = n log pO₂ − n log P50, n = Δlog[θ/(1 − θ)]/Δlog pO₂. At 80%: log(0.8/0.2) = log 4 = 0.6021; at 20%: log(0.25) = −0.6021; difference 1.2041. log(40/15) = log 2.667 = 0.4260. n = 1.2041/0.4260 = 2.83, a cooperative protein like hemoglobin.

  15. In chymotrypsin, which residue's side chain acts as the nucleophile that attacks the carbonyl carbon of the scissile peptide bond?

    1. His57, which accepts a proton during the reaction
    2. Ser195, which forms the acyl-enzyme intermediate
    3. Asp102, which holds the triad in place
    4. Gly193, which lines the oxyanion hole
    Show answer

    Answer: B — Ser195, which forms the acyl-enzyme intermediate

    Ser195's hydroxyl oxygen, made strongly nucleophilic when His57 takes its proton, attacks the carbonyl carbon and ends up as the ester of the acyl-enzyme. His57 is the general base (and later the general acid), not the nucleophile; Asp102 orients His57; Gly193 contributes a backbone N–H to the oxyanion hole.

  16. Chymotrypsin is mixed with a large excess of p-nitrophenyl acetate. Yellow p-nitrophenolate appears in a rapid initial burst whose size equals the concentration of active enzyme, followed by a slower, linear release. What does this kinetic pattern show?

    1. The enzyme has two classes of active site, one fast and one slow, present in equal amounts
    2. The product p-nitrophenolate gradually inhibits the enzyme until a lower steady rate is reached
    3. Binding of the substrate is slow and rate-limiting, while all the chemical steps that follow it are very fast
    4. Acylation of the enzyme is fast and deacylation of a covalent acyl-enzyme is the slow, rate-limiting step
    Show answer

    Answer: D — Acylation of the enzyme is fast and deacylation of a covalent acyl-enzyme is the slow, rate-limiting step

    Each active site quickly releases one p-nitrophenolate as Ser195 is acetylated, giving a burst equal to the enzyme concentration; then each site must hydrolyse the acetyl-enzyme before it can react again, so the steady rate reports the slow deacylation. This is direct evidence for a covalent intermediate and a two-phase (ping-pong) mechanism. If binding were rate-limiting there would be no burst at all, and product inhibition would give a curve, not a burst stoichiometric with the enzyme.

  17. Which of the following statements about the model enzymes lysozyme, ribonuclease A, carboxypeptidase A and chymotrypsin are correct?

    1. Ribonuclease A cleaves DNA and RNA at similar rates, since it attacks the phosphodiester backbone common to both
    2. In carboxypeptidase A, Zn²⁺ polarises the scissile carbonyl and helps activate the attacking water molecule
    3. Lysozyme hydrolyses the β(1→4) bond between C1 of N-acetylmuramic acid and C4 of N-acetylglucosamine
    4. Chymotrypsinogen is activated when enteropeptidase cleaves it after a lysine near its C-terminus
    Show answer

    Answer: B — In carboxypeptidase A, Zn²⁺ polarises the scissile carbonyl and helps activate the attacking water molecule; C — Lysozyme hydrolyses the β(1→4) bond between C1 of N-acetylmuramic acid and C4 of N-acetylglucosamine

    Zn²⁺ in carboxypeptidase A polarises the carbonyl and lowers the pKa of bound water, and lysozyme cuts the MurNAc(C1)–GlcNAc(C4) β(1→4) bond of peptidoglycan. RNase A needs the 2′-OH of RNA to form its 2′,3′-cyclic intermediate, so DNA is not a substrate. Chymotrypsinogen is activated by trypsin cleaving the Arg15–Ile16 bond near the N-terminus; enteropeptidase activates trypsinogen.

  18. Which set of observations best supports the mechanism in which ribonuclease A uses His12 and His119 to cleave RNA through a 2′,3′-cyclic phosphate intermediate?

    1. Activity requires Zn²⁺ and is abolished by EDTA, and the products of the digestion all carry 5′-phosphate groups
    2. Activity rises steadily with pH up to pH 11, because both histidines must be deprotonated so that each can act as a base
    3. A 2′,3′-cyclic nucleotide appears first and later becomes a 3′-phosphate; DNA is not cut; the pH–activity curve is bell-shaped
    4. Diisopropyl fluorophosphate inactivates the enzyme by covalently labelling a single, unusually reactive serine in its active site
    Show answer

    Answer: C — A 2′,3′-cyclic nucleotide appears first and later becomes a 3′-phosphate; DNA is not cut; the pH–activity curve is bell-shaped

    The cyclic intermediate accumulating and then opening to a 3′-phosphate shows the two-step transphosphorylation and hydrolysis; the lack of DNA cleavage shows the 2′-OH is the attacking group; the bell-shaped pH profile shows one His must be basic (His12) and the other acidic (His119). RNase A is not a metalloenzyme and gives 3′-phosphate products; a profile rising to pH 11 would mean two bases, not an acid–base pair; DIFP labelling of serine is the signature of serine proteases such as chymotrypsin.