Higher Orders of Structure — Q-Bank
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Unit 3 Question Bank

Four orders · α-helix · β-sheet · domains · folding · prions · collagen
25 MCQ · five options9 Definitions3 Written answersHarper's verified
Format note: the TMU Biochemistry paper gives five suggested answers (A–E), not four — these MCQs match that. Items tagged TMU 2019 or TMU 2020/21 come from the real papers. Answers are verified against Harper's Illustrated Biochemistry; the "marking schemes" in the source folder are other students' answer sheets, not official, so they are never used as the authority.
0 / 25 answered
1Which type of bond or interaction is MOST important in determining the secondary structure of a protein?
A. Hydrophobic interactions
B. Disulfide bonds
C. Hydrogen bonds
D. Salt bridges
E. van der Waals interactions
Answer: C
Hydrogen bonds between the backbone carbonyl oxygen and amide hydrogen create and stabilise both the α-helix and the β-sheet. Hydrophobic interactions dominate tertiary and quaternary structure; disulfide bonds are covalent cross-links, not the defining force of secondary structure.Harper's ch.5, p.47 · TMU Lecture 3 Slide 35 · TMU 2020/21 paper Section IV
2Which force is MOST important in stabilising the tertiary and quaternary structure of proteins?
A. Hydrogen bonds
B. Peptide bonds
C. Disulfide bonds
D. Hydrophobic interactions
E. Coordinate bonds to metal ions
Answer: D
Hydrophobic interactions. The driving force in folding is the segregation of non-polar side chains into the protein's interior, away from solvent — this is what forms the molten globule and ultimately the native conformation. Peptide bonds stabilise primary structure; hydrogen bonds dominate secondary structure.TMU Lecture 3 Slide 35 · Harper's ch.5, p.47
3Non-covalent interactions that stabilise secondary, tertiary and quaternary structure do NOT include ( ).
A. hydrophobic forces
B. electrostatic bonds
C. van der Waals forces
D. hydrogen bonds
E. disulfide bonds
Answer: E
Disulfide bonds are covalent — they link the sulfhydryl groups of two cysteinyl residues. They do stabilise structure, but they are the exception to the “higher orders are stabilised by non-covalent forces” rule. This exact question closes the TMU lecture deck.TMU Lecture 3 Slide 49
4An α-helix contains, on average, how many amino acid residues per complete turn?
A. 3.6
B. 2.0
C. 3.3
D. 4.5
E. 5.4
Answer: A
3.6 residues per turn, with a pitch (rise per turn) of 0.54 nm. Do not confuse this with the collagen triple helix, which has 3.3 residues per turn — that similar-looking number is the standard distractor.Harper's ch.5, p.38 · TMU Lecture 3 Slide 12
5The pitch of an α-helix — the distance it rises per complete turn — is ( ).
A. 0.15 nm
B. 0.54 nm
C. 0.34 nm
D. 1.08 nm
E. 3.60 nm
Answer: B
0.54 nm. (0.15 nm is the rise per residue, and 0.34 nm is the rise per base pair in DNA — a distractor drawn from Unit 23.)Harper's ch.5, p.38, Figure 5–2
6The stability of an α-helix arises primarily from hydrogen bonds formed between the carbonyl oxygen of one residue and the amide hydrogen of ( ).
A. the immediately adjacent residue
B. a residue on an adjacent polypeptide chain
C. the fourth residue further along the same chain
D. the second residue further along the same chain
E. the C-terminal residue
Answer: C
The fourth residue down the chain, with the bonds running parallel to the helix axis. Hydrogen bonding to an adjacent segment of chain is the β-sheet, which is the commonest wrong answer here.Harper's ch.5, p.38 · TMU Lecture 3 Slide 14
7In an α-helix, the R groups of the amino acid residues ( ).
A. point into the centre of the helix
B. alternate between inside and outside
C. form hydrogen bonds with each other
D. project outward from the helix
E. lie in the plane of the peptide bond
Answer: D
They face outward. There is almost no free space inside an α-helix — the core is tightly packed backbone stabilised by van der Waals interactions. Side chains take no part in forming secondary structure, though they influence its stability.Harper's ch.5, p.38, Figure 5–3 · TMU Lecture 3 Slide 9
8Only right-handed α-helices occur in proteins because ( ).
A. hydrogen bonds cannot form in a left-handed arrangement
B. ribosomes can only synthesise right-handed structures
C. left-handed helices would place all R groups inside
D. the peptide bond can only rotate in a clockwise sense
E. proteins contain only L-amino acids, favouring the right-handed form
Answer: E
It follows directly from Unit 1: proteins are built exclusively from L-α-amino acids, and for L-residues the right-handed helix is by far the more stable arrangement. This is a good example of a Unit 1 fact being tested inside a Unit 3 question.Harper's ch.5, p.38
9Proline disrupts an α-helix because ( ).
A. its nitrogen has no hydrogen to donate to a hydrogen bond
B. its side chain is too large to fit inside the helix
C. it cannot form a peptide bond with the next residue
D. it carries a positive charge at physiological pH
E. it is the only D-amino acid found in proteins
Answer: A
Proline's nitrogen is part of its ring and lacks a hydrogen, so it cannot donate the hydrogen bond an α-helix depends on. It can only be stably accommodated within the first turn; elsewhere it produces a bend. Glycine also induces bends, but for a different reason — its R group is so small the residue is too flexible.Harper's ch.5, p.38
10Compared with the α-helix, the polypeptide backbone of a β-sheet is ( ).
A. more tightly coiled
B. highly extended, giving a pleated zigzag appearance edge-on
C. identical but left-handed
D. held together by disulfide rather than hydrogen bonds
E. devoid of hydrogen bonding
Answer: B
The β-sheet backbone is highly extended; viewed edge-on the residues form a zigzag or pleated pattern, with the R groups of adjacent residues pointing in opposite directions. Like the α-helix it depends on hydrogen bonds — but formed with adjacent segments of chain.Harper's ch.5, p.39 · TMU Lecture 3 Slide 15
11In an ANTIPARALLEL β-sheet, the adjacent segments of polypeptide chain ( ).
A. run in the same direction, amino to carboxyl
B. are joined by disulfide bonds
C. run in opposite directions, amino to carboxyl
D. belong to different proteins
E. are separated by an α-helix
Answer: C
Opposite directions. The hydrogen bonds differ accordingly: in the antiparallel sheet pairs of bonds alternate close together and wide apart and lie approximately perpendicular to the backbone; in the parallel sheet they are evenly spaced but slanted.Harper's ch.5, p.39, Figure 5–5
12Which two amino acids are often present in β-turns?
A. Leucine and isoleucine
B. Aspartate and glutamate
C. Lysine and arginine
D. Proline and glycine
E. Phenylalanine and tyrosine
Answer: D
Proline and glycine — the same two residues that break α-helices. Proline's rigid ring forces a change of direction and glycine's flexibility permits one, so between them they make the tight bend a turn requires. A β-turn typically spans four residues, hydrogen bonded between the first and the fourth.Harper's ch.5, p.39 · TMU Lecture 3 Slides 18–19
13A domain is best defined as ( ).
A. a short conserved sequence critical to a protein's function
B. a combination of secondary structures 10–40 residues long
C. the spatial arrangement of subunits in an oligomeric protein
D. a region of a protein that lacks any ordered structure
E. a region that folds independently and performs a distinct function
Answer: E
A domain folds independently and is sufficient to perform a particular chemical or physical task. Option A describes a motif; option B a supersecondary structure; option C quaternary structure. Harper's example: protein kinases have two domains — a β-rich N-terminal domain binding ATP and an α-rich C-terminal domain binding the substrate.Harper's ch.5, pp.40–41 · TMU Lecture 3 Slide 25
14Quaternary structure refers to ( ).
A. the number, type and spatial arrangement of the subunits
B. the complete three-dimensional conformation of a single polypeptide
C. the folding of short contiguous segments into ordered units
D. the sequence of amino acids in a polypeptide chain
E. the covalent modifications added after translation
Answer: A
The key words are number, type and spatial arrangement of subunits. Note that a monomeric protein has no quaternary structure at all — myoglobin, one chain, has none; haemoglobin, α₂β₂, does.Harper's ch.5, p.41 · TMU Lecture 3 Slides 28–29
15The notation α₂β₂γ describes a protein with ( ).
A. three subunits of two different types
B. five subunits of three different types
C. two subunits of two different types
D. five identical subunits
E. two domains and one motif
Answer: B
Greek letters distinguish different subunit types and subscripts give the number of each: 2 + 2 + 1 = five subunits of three different types. By the same convention α₄ is a homotetramer and α₂β₂ — adult haemoglobin — is a heterotetramer.Harper's ch.5, p.41 · TMU Lecture 3 Slide 28
16Interconversion between two CONFORMATIONS of a molecule ( ).
A. requires the breaking and reforming of covalent bonds
B. always changes the molecule from the L- to the D-form
C. occurs without breaking covalent bonds, by bond rotation
D. requires the action of a specific protease enzyme
E. is only possible at temperatures above 60 C
Answer: C
Conformation concerns the spatial relationship of every atom, and conformers interconvert by rotation about single bonds, with configuration retained. Changing configuration — for example L- to D-amino acid — is what requires breaking covalent bonds.Harper's ch.5, p.36 · TMU Lecture 3 Slide 3
17On a Ramachandran plot, the φ and ψ angles characteristic of the α-helix and the β-sheet fall respectively in the ( ).
A. upper and lower right-hand quadrants
B. upper and lower left-hand quadrants
C. lower and upper right-hand quadrants
D. lower and upper left-hand quadrants
E. centre and upper right of the plot
Answer: D
α-helix in the lower left quadrant, β-sheet in the upper left quadrant — this order is stated explicitly in both the Harper's summary and the TMU slide. The plot's dots mark allowable φ/ψ combinations; most combinations are prohibited by steric hindrance for all residues except glycine.Harper's ch.5, pp.37–38, 47 · TMU Lecture 3 Slide 7
18The “molten globule” in protein folding is ( ).
A. an insoluble aggregate of irreversibly misfolded protein chains
B. a completely unfolded random coil with no secondary structure
C. the fully mature, functional conformation of the protein
D. a chaperone protein that binds and assists nascent chains
E. a partially folded intermediate whose modules rearrange to the native form
Answer: E
It is the second stage of folding: forces driving hydrophobic regions into the interior away from solvent collapse the chain into a partially folded state, within which the pre-formed modules of secondary structure rearrange until the mature conformation is attained. The process is orderly but not rigid.Harper's ch.5, pp.44–45 · TMU Lecture 3 Slide 37
19Chaperone proteins assist folding by ( ).
A. binding nascent chains and rescuing those already misfolded
B. catalysing the exchange of incorrectly paired disulfide bonds
C. isomerising X-Pro peptide bonds between the cis and trans forms
D. cleaving proproteins into their mature active form
E. hydroxylating prolyl and lysyl residues in collagen
Answer: A
Those are the two chaperone roles. Option B is protein disulfide isomerase; option C is proline-cis,trans-isomerase; option E is prolyl/lysyl hydroxylase in collagen synthesis. Chaperones participate in the folding of over half of mammalian proteins.Harper's ch.5, p.45 · TMU Lecture 3 Slide 38
20Approximately what percentage of the X-Pro peptide bonds of mature proteins are in the CIS configuration?
A. 0%
B. 6%
C. 25%
D. 50%
E. 94%
Answer: B
All X-Pro bonds are synthesised trans, but about 6% of those in mature proteins are cis — particularly common in β-turns. The trans → cis isomerisation is catalysed by proline-cis,trans-isomerase (the cyclophilins).Harper's ch.5, p.45 · TMU Lecture 3 Slide 39
21In prion disease, the pathological protein PrPˢᶜ differs from the normal PrPᶜ in that it is ( ).
A. rich in α-helix and remains soluble and monomeric
B. encoded by a viral gene acquired during infection
C. rich in β-sheet and forms insoluble, protease-resistant aggregates
D. lacking several amino acids because of a deletion mutation
E. unable to bind the copper ions required for its function
Answer: C
Normal PrPᶜ is monomeric and α-helix rich; pathological PrPˢᶜ is β-sheet rich, with hydrophobic side chains exposed to solvent, so molecules associate into insoluble protease-resistant aggregates. Crucially, the amino acid sequence is the same — only the conformation differs, which is why these are called protein conformation diseases.Harper's ch.5, pp.45–46 · TMU Lecture 3 Slide 42
22Prion diseases are transmitted by ( ).
A. a slow virus that integrates into the neuronal genome
B. a mutation in the gene encoding PrPᶜ in every case
C. transfer of prion messenger RNA between neurons
D. PrPˢᶜ templating the conversion of host PrPᶜ into the disease form
E. a bacterial toxin that denatures neuronal proteins
Answer: D
Prions are protein particles that lack nucleic acid. No viral or bacterial gene encoding them could ever be identified. Transmission occurs because PrPˢᶜ serves as a template for the conformational conversion of normal PrPᶜ — a shape is propagated, not a sequence. Prion disease can manifest as infectious, genetic or sporadic.Harper's ch.5, pp.45–46 · TMU Lecture 3 Slide 41
23In the collagen triple helix, every third amino acid residue is glycine because ( ).
A. glycine is the only residue that can form hydrogen bonds between chains
B. glycine is required for hydroxylation by prolyl hydroxylase
C. glycine carries the covalent cross-links that strengthen the fibre
D. glycine is the only amino acid that can adopt a left-handed conformation
E. the R groups of the three strands pack so closely that one must be a hydrogen atom
Answer: E
Pure packing. Where the three strands come into contact the side chains are so closely apposed that one of the three must be a hydrogen — and glycine is the only amino acid whose R group is H. This gives the repeating Gly-X-Y pattern, in which Y is generally proline or hydroxyproline. A mutation replacing one glycine causes osteogenesis imperfecta.Harper's ch.5, p.47
24Scurvy impairs collagen maturation because vitamin C is required by ( ).
A. prolyl hydroxylase and lysyl hydroxylase
B. lysyl oxidase
C. procollagen N-peptidase
D. glucosyl and galactosyl transferases
E. protein disulfide isomerase
Answer: A
Ascorbic acid is the required cofactor for prolyl and lysyl hydroxylase. Without it there are fewer hydroxyproline and hydroxylysine residues, so fewer stabilising hydrogen bonds, and the fibre is conformationally unstable — hence bleeding gums, swollen joints, poor wound healing and ultimately death. Lysyl oxidase is the copper-requiring enzyme, deficient in Menkes syndrome.Harper's ch.5, pp.46–47 · TMU Lecture 3 Slide 44
25Which technique determines protein three-dimensional structure by measuring the absorbance of radiofrequency electromagnetic energy by atomic nuclei?
A. X-ray crystallography of protein crystals
B. Nuclear magnetic resonance spectroscopy
C. Gel filtration chromatography on Sephadex
D. Mass spectrometry of tryptic peptides
E. Isoelectric focusing in a pH gradient
Answer: B
NMR spectroscopy, whose “NMR-active” isotopes are ¹H, ¹³C, ¹⁵N and ³¹P. Its advantage over X-ray crystallography is that it analyses proteins in aqueous solution rather than requiring ordered crystals that diffract X-rays.Harper's ch.5, pp.43–44 · TMU Lecture 3 Slide 36
1 Secondary structure — 3′ · classic Section I term+
The folding of short (3- to 30-residue), contiguous segments of polypeptide backbone into geometrically ordered units.

It arises when a series of consecutive aminoacyl residues adopt similar φ and ψ angles. The two commonest forms are the α-helix and the β-pleated sheet, together with bends, turns and loops. It is stabilised principally by hydrogen bonds between backbone carbonyl oxygens and amide hydrogens. Side chains are not involved in forming secondary structure, though they help determine its stability and type.Harper's ch.5, p.36 · TMU Lecture 3 Slide 9
2 The α-helix — 3′+
A regular secondary structure in which the polypeptide backbone is twisted by an equal amount about each α-carbon, with φ ≈ −57° and ψ ≈ −47°.

A complete turn contains an average of 3.6 residues and rises 0.54 nm (the pitch). The R groups face outward. Because proteins contain only L-amino acids, only right-handed α-helices occur. Stability arises from hydrogen bonds parallel to the helix axis, between the carbonyl oxygen of one peptide bond and the amide hydrogen of the fourth residue down the chain.Harper's ch.5, p.38
3 Domain — 3′ · very likely Section I term+
A section of protein structure that folds independently into a stable conformation, sufficient to perform a particular chemical or physical task — such as binding a substrate or interacting with a regulatory molecule.

Different domains work together to provide the complete function of the protein. A small polypeptide such as triose phosphate isomerase or myoglobin may consist of a single domain; protein kinases contain two — a β-sheet-rich amino-terminal domain that binds ATP and an α-helix-rich carboxyl-terminal domain that binds the substrate.Harper's ch.5, pp.40–41
4 Motif — 2′+
A short, conserved region of a protein, frequently the most conserved part of a domain and critical to that domain's function — in an enzyme it may contain the active site.

Examples include the zinc finger (about 30 residues forming an elongated loop held at its base by a single Zn²⁺ ion coordinated to four residues — four Cys, or two Cys and two His) and nuclear localisation sequences.TMU Lecture 3 Slides 20, 22
5 Supersecondary structure — 2′+
Combinations of secondary structure, 10–40 residues in length, found recurrently in numerous proteins — structurally intermediate between secondary and tertiary structure.

The commonest are combinations of α-helix and β-sheet. The helix-loop-helix motif, which provides the oligonucleotide-binding portion of DNA-binding proteins such as repressors and transcription factors, is the standard example.Harper's ch.5, p.39 · TMU Lecture 3 Slide 20
6 Quaternary structure — 3′+
The number and types of polypeptide subunits (protomers) of an oligomeric protein and their spatial arrangement.

Monomeric proteins consist of one chain and have no quaternary structure. Greek letters distinguish subunit types and subscripts their number: α₄ is a homotetramer, α₂β₂ (adult haemoglobin) a heterotetramer. It is stabilised by the same non-covalent forces as tertiary structure, and sometimes by interchain disulfide bonds.Harper's ch.5, p.41
7 Molten globule — 3′+
A partially folded polypeptide formed during the second stage of protein folding, when the forces driving hydrophobic regions into the interior away from solvent collapse the chain.

Within it, the modules of secondary structure rearrange until the mature, native conformation is attained. The process is orderly but not rigid: considerable flexibility exists in the order in which elements may be rearranged.Harper's ch.5, pp.44–45
8 Prion — 3′+
A protein particle that lacks nucleic acid and causes fatal transmissible neurodegenerative disease — Creutzfeldt-Jakob disease in humans, scrapie in sheep, bovine spongiform encephalopathy in cattle.

Prion diseases are protein conformation diseases: the pathological isoform PrPˢᶜ acts as a template that converts the host's normal PrPᶜ — monomeric and α-helix rich — into the β-sheet-rich PrPˢᶜ, which aggregates into insoluble, protease-resistant deposits. The amino acid sequence is unchanged; only the conformation differs.Harper's ch.5, pp.45–46
9 Conformation (vs configuration) — 2′+
Conformation is the spatial relationship of every atom in a molecule. Interconversion between conformers occurs without rupture of covalent bonds, with retention of configuration, typically by rotation about single bonds.

Configuration is the geometric relationship between a given set of atoms — for example that which distinguishes L- from D-amino acids — and its interconversion requires breaking covalent bonds.Harper's ch.5, p.36
1 Describe the primary, secondary, tertiary and quaternary structure of a protein. 8′ — TMU study question, past paper

Framing the answer

Protein structure is described in four orders. The primary structure is the basic structure, held together by covalent bonds; the secondary, tertiary and quaternary structures together constitute the spatial structure, or conformation, and are stabilised principally by non-covalent forces. Conformation is dictated by the primary sequence.

Primary structure

The sequence of amino acids in the polypeptide chain, read from the N-terminus to the C-terminus, joined by covalent peptide bonds. It is gene-encoded and determines all higher orders of structure.

Secondary structure

The folding of short (3–30 residue), contiguous segments of the polypeptide backbone into geometrically ordered units, arising when a series of consecutive residues adopt similar φ and ψ angles. Side chains are not involved, though they influence stability and type. Stabilised by hydrogen bonds between backbone carbonyl oxygens and amide hydrogens. The principal forms are:

FormFeatures
α-helix3.6 residues per turn, pitch 0.54 nm, φ ≈ −57° and ψ ≈ −47°; right-handed only; R groups face outward; hydrogen bonds parallel to the axis, to the fourth residue down the chain
β-pleated sheetHighly extended backbone with a pleated zigzag appearance; R groups of adjacent residues point in opposite directions; hydrogen bonds formed with adjacent segments of chain; may be parallel or antiparallel
Turns, bends and loopsConnect adjacent regions of secondary structure; proline and glycine are common in β-turns; loops contain residues beyond the minimum needed, and often lie on the surface as epitopes

Recurring combinations of these, 10–40 residues long, are termed supersecondary structures (e.g. helix-loop-helix), intermediate between secondary and tertiary structure.

Tertiary structure

The entire three-dimensional conformation of a polypeptide — how the secondary structural features (helices, sheets, bends, turns and loops) assemble into domains, and how those domains relate spatially to one another. A domain is a section of structure that folds independently into a stable conformation and is sufficient to perform a particular task; protein kinases, for example, have two, one binding ATP and one the substrate. Tertiary structure is stabilised mainly by hydrophobic interactions, with contributions from hydrogen bonds, salt bridges, van der Waals interactions and intrachain disulfide bonds.

Quaternary structure

The number and types of polypeptide subunits of an oligomeric protein and their spatial arrangement. Present only in proteins of more than one chain: monomeric proteins have none. Homodimers contain two identical chains, heterodimers two different ones; Greek letters and subscripts denote composition, so α₄ is a homotetramer and α₂β₂ — adult haemoglobin — a heterotetramer. Stabilised by the same non-covalent forces, and in some proteins by interchain disulfide bonds.

Marking guide: four orders named with correct definitions 3 · primary stabilised by covalent peptide bonds 0.5 · α-helix described with at least two quantitative features 1.5 · β-sheet described with extended backbone and adjacent-chain H-bonding 1 · domain defined within tertiary structure 1 · quaternary defined as number, type and arrangement of subunits 1.
2 What are the major forces that stabilise protein structure? 5′ — TMU study question, past paper

The general principle

Primary structure is stabilised by covalent peptide bonds. The higher orders of structure are stabilised primarily — and often exclusively — by non-covalent interactions. Each individual interaction is weak; their number is what makes the folded conformation stable.

The non-covalent forces

ForceNature and importance
Hydrogen bondsBetween the carbonyl oxygen and the amide hydrogen of peptide bonds. The dominant force in secondary structure — parallel to the axis in the α-helix, between adjacent segments in the β-sheet. Also stabilise loops and tertiary contacts
Hydrophobic interactionsNon-polar side chains segregate into the interior of the protein, away from solvent. The dominant force in tertiary and quaternary structure, and the driving force of the molten globule stage of folding
Salt bridges (electrostatic or ionic bonds)Between oppositely charged side chains, for example a lysine ammonium group with an aspartate carboxylate
van der Waals interactionsWeak and short-range, but numerous in the tightly packed core of a folded protein

The covalent exception

Some proteins contain disulfide (–S–S–) bonds linking the sulfhydryl groups of two cysteinyl residues. These are covalent, and therefore the exception to the rule above.

  • Intrachain disulfide bonds form within a single polypeptide and further enhance the stability of its folded tertiary conformation.
  • Interchain disulfide bonds link different polypeptides and stabilise the quaternary structure of certain multimeric proteins.

Summary statement

Primary structure — covalent peptide bonds. Secondary structure — mainly hydrogen bonds. Tertiary and quaternary structure — mainly hydrophobic interactions, supported by hydrogen bonds, salt bridges and van der Waals forces, and in some proteins reinforced by covalent disulfide bonds.

Marking guide: distinction between covalent primary and non-covalent higher orders 1 · four non-covalent forces named 2 · hydrogen bonds identified with secondary structure 0.5 · hydrophobic interactions identified with tertiary/quaternary structure 1 · disulfide bonds noted as the covalent exception, ideally with intra/interchain distinction 0.5.
3 Elucidate the process of protein folding and the consequences when it fails. 8′ — 'elucidate'

The problem

A typical polypeptide can adopt ≥10⁵⁰ distinct conformations. If a chain searched them at random, folding would take billions of years; in reality proteins fold in milliseconds. Folding cannot therefore be a random search.

Thermodynamics provides the direction

The biologically relevant, or native, conformation is generally the one that is most energetically favoured. Knowledge of the native conformation is therefore already specified in the primary sequence — conformation is dictated by primary structure.

Folding is modular — two stages

  1. Local order. As the newly synthesised polypeptide emerges from the ribosome, short segments fold into secondary structural units that provide local regions of organised structure. The problem is thereby reduced from an astronomical search to the selection of an appropriate arrangement of a relatively small number of pre-formed elements.
  2. The molten globule. Forces driving hydrophobic regions into the interior, away from solvent, collapse the partially folded chain into a molten globule, within which the modules of secondary structure rearrange until the mature conformation is reached. The process is orderly but not rigid. For oligomeric proteins, individual protomers tend to fold before associating with other subunits.

Auxiliary proteins assist folding

ProteinFunction
ChaperonesBind polypeptides before synthesis is complete, preventing premature folding into an incorrect conformation; and rescue proteins thermodynamically trapped in a misfolded dead end by unfolding hydrophobic regions and providing a second chance. They participate in the folding of over half of mammalian proteins
Protein disulfide isomeraseDisulfide bond formation is non-specific; by catalysing disulfide exchange — rupture of an S–S bond and its reformation with a different partner cysteine — the enzyme drives the protein toward its native pairings
Proline-cis,trans-isomeraseAll X-Pro bonds are synthesised trans, but about 6% of those in mature proteins are cis, particularly in β-turns; this enzyme catalyses the isomerisation

Note the contrast with the laboratory: many denatured proteins refold spontaneously in vitro, but far more slowly, and some fail entirely, forming insoluble aggregates of unfolded or partly folded chains held together by hydrophobic interactions.

Consequences of failure — protein conformation diseases

Prion diseases. Prions are protein particles that lack nucleic acid, causing fatal transmissible spongiform encephalopathies — Creutzfeldt-Jakob disease, scrapie, bovine spongiform encephalopathy. Normal human PrPᶜ is monomeric and rich in α-helix; the pathological PrPˢᶜ is rich in β-sheet, with hydrophobic side chains exposed to solvent, so the molecules associate into insoluble, protease-resistant aggregates. Transmission occurs because PrPˢᶜ serves as a template for the conformational conversion of PrPᶜ: the amino acid sequence is unchanged, and only the conformation is propagated.

Alzheimer's disease. Misfolding of β-amyloid, a 4.3-kDa polypeptide produced by proteolytic cleavage of amyloid precursor protein, is a prominent feature.

Nutritional failure of maturation. Scurvy illustrates the same principle from a different direction: vitamin C deficiency impairs prolyl and lysyl hydroxylase, so collagen fibres never acquire conformational stability.

Marking guide: the 10⁵⁰ conformations problem and the speed of real folding 1 · native conformation thermodynamically favoured and specified by primary sequence 1 · two-stage modular folding with molten globule defined 2 · the three auxiliary proteins named with functions 1.5 · prion mechanism as templated conformational conversion 1.5 · one further example — Alzheimer's or scurvy 1.