synthetic biology and protein engineering

how does protein folding occur?

valine

hydrophobic

tyrosine

aromatic

Arginine

basic

• + charge

• neutral pH

serine

can form hydrogen bond

glutamate

has negative charge

do proteins self assamble?

yes

what causes the dentauration of proteins?

• heat

• beta-mercaptoehtanol

• urea

how does heat denature protein?

how is the protein structure removed?

through dentauration

how is denaturation undone?

Why must trace β-mercaptoethanol be present when refolding denatured ribonuclease?

To keep cysteines reduced and allow reshuffling until the correct, native disulfide pairings form. Without it, incorrect “scrambled” disulfides lock the protein in a non-foldable state; correct disulfide formation is thus a post-folding rescue step, not part of the folding process itself.

What is protein folding and why is it indispensable for life?

The conformational rearrangement of a nascent, unstructured polypeptide into its precise 3-D functional form. Without it the protein has no defined architecture, no activity—and life cannot exist.

Front (Question)

What are the key structural features of an amino acid and how do they link into a polypeptide?

Back (Answer)

Central chiral α-carbon bonded to amino group, carboxyl group, H-atom, and variable R side-chain (L-isomer only). Condensation between the amino group of one residue and the carboxyl of the next forms a peptide bond, releasing water and extending the chain directionally from N-terminus to C-terminus.

Front (Question)

Why does the peptide bond limit how a protein chain can bend?

Back (Answer)

Resonance gives the peptide bond partial double-bond character, locking it into a rigid, planar trans conformation that prevents rotation around the C–N axis; this constraint sharply reduces the backbone’s freedom of movement, while neighboring φ and ψ angles remain rotatable.

Front (Question)

Why does a polypeptide chain have chemical “bulk” and variability in three dimensions?

Back (Answer)

Because every residue’s R-group side chain projects out of the plane of the rigid peptide-bond backbone, creating a 3-D surface with diverse chemical functionality and spatial bulk that determines interactions and overall shape.

Front (Question)

Why is the peptide bond preceding proline special in terms of cis/trans isomerism?

Back (Answer)

With most residues the trans form is ~1000× more stable, but proline’s cyclic side chain lowers the energy difference to only ~4×, allowing measurable populations of both cis and trans Xaa-Pro bonds; the slow inter-conversion can therefore serve as a kinetic folding switch or rate-limiting step in protein folding.

Front (Question)

Which two backbone dihedral angles define the path of a polypeptide chain, and what constrains their possible values?

Back (Answer)

Phi (φ, rotation about Cα–N) and psi (ψ, rotation about Cα–C) are the only freely rotatable backbone angles; their paired values (plotted on the Ramachandran diagram) determine the chain’s conformation, while steric clashes between atoms disallow most angle combinations.

Front (Question)

How does proline introduce a kink or rigid joint in the polypeptide backbone?

Back (Answer)

Its side chain forms a five-membered ring that bonds back to the α-nitrogen, locking the φ angle near –60° and eliminating rotation about the Cα–N bond; this lost degree of freedom creates a rigid pivot that kinks the chain and is therefore favored at turns, loops, and surface sites where flexibility must be restricted.

Front (Question)

What repeating feature of the protein backbone makes α-helices and β-sheets possible?

Back (Answer)

Every residue presents a backbone carbonyl oxygen (H-bond acceptor) and an amide N–H (H-bond donor); their regular spacing creates a linear array that can form cooperative hydrogen-bond networks, giving rise to α-helices, β-sheets, and overall fold stability.

Front (Question)

Why do α-helices, β-sheets and 310-helices cluster into small, specific regions on a Ramachandran plot?

Back (Answer)

Their repeating hydrogen-bond patterns lock the backbone into regular geometries, allowing only narrow ranges of the φ and ψ dihedral angles; any angle outside these ranges breaks the H-bond network and creates steric clashes, so only the clustered conformations are physically allowed.

Front (Question)

What are the defining geometric and H-bond parameters of the classic α-helix?

Back (Answer)

3.6 residues per turn, 5.4 Å pitch, φ ≈ –60°, ψ ≈ –45°; consecutive i → i+4 C=O···H–N hydrogen bonds along the backbone generate a stable right-handed helical rod (left-handed form is extremely rare).

Front (Question)

How does the 3₁₀-helix differ in geometry and H-bonding from the classic α-helix?

Back (Answer)

Tighter: 3.0 residues per turn, φ ≈ –50°, ψ ≈ –25°, connected by i → i+3 C=O···H–N bonds that close 10-atom rings (vs 13-atom rings in the α-helix); usually occurs as short stretches, not extended segments.

Front (Question)

What are the backbone angles and architectural features of an ideal β-strand, and why must strands pair into sheets?

Back (Answer)

φ ≈ –135°, ψ ≈ +135°, 3.5 Å axial rise per residue, with side chains alternating up/down; isolated strands are unstable, so lateral i → i±1 H-bonding between antiparallel or parallel strands is required to form a rigid, stable β-sheet.

Front (Question)

What structural role do β-turns and Ω-loops play in proteins?

Back (Answer)

They reverse the backbone direction, linking adjacent secondary-structure elements; 4-residue β-turns use an i→i+3 H-bond, exist in defined types, lack periodic backbone geometry, yet are structurally conserved and often essential for biological function.

Front (Question)

How does TEM β-lactamase illustrate the relationship between secondary-structure content and tertiary function?

Back (Answer)

The enzyme is 44 % α-helix, 17 % β-strand, 39 % loops/turns; its catalytic triad (Ser70, Glu166, Lys73) sits at loop/helix interfaces, showing that the linear sequence folds into a cohesive 3-D active site where secondary elements and loops cooperate for antibiotic-hydrolyzing function.

Front (Question)

Why are side chains (R groups) indispensable for a functional protein?

Back (Answer)

They provide the protein’s unique “chemical signature”; without them (e.g., poly-Gly) the chain would be an extended, chemically inert string. The specific identity and sequence of R-groups encode all folding instructions, determine 3-D structure, and enable biological function.

Front (Question)

Which amino acids are the strongest promoters of α-helix, β-strand, and β-turns according to Chou–Fasman propensities?

Back (Answer)

α-helix: Glu, Met, Ala; β-strand: Val, Ile, Tyr; turns: Asn, Gly, Pro, Asp, Ser. These empirical values are used to predict secondary structure from sequence alone.

Front (Question)

Why can Val, Ile, Pro, Gly, Ser, or Thr disfavor α-helix formation?

Back (Answer)

Branched Cβ (Val, Ile) creates steric clashes; Pro’s rigid ring breaks helical geometry; Gly is overly flexible; Ser/Thr side chains can H-bond to the backbone, competing with helical i→i+4 bonds—so their favorable placement depends on the surrounding tertiary context.

Front (Question)

Which amino acids top the second Chou–Fasman table as the strongest “winners” for each secondary structure, and why is this listing important?

Back (Answer)

α-helix: Glu, Met, Ala; β-strand: Val, Ile, Tyr; turns: Asn, Gly, Pro. These empirical propensities provided the quantitative scores used by the earliest algorithms to predict secondary structure directly from protein sequence."

Here are the same 10 facts re-cast as short Q&A flash-cards.

The answer to each question is a single, self-contained sentence that still makes sense when you read it alone.

Card 1

Q: What two sets of contacts must both be satisfied for the native state to win energetically?

A: Local contacts (like i→i+4 H-bonds inside an α-helix) and long-range contacts (distant residues that meet in 3-D to build the hydrophobic core or catalytic site).

Card 2

Q: Which class of interactions actually drives protein folding?

A: Thousands of weak non-covalent interactions—H-bonds, van der Waals, ionic and hydrophobic—cooperatively tip the free-energy balance toward the compact native state.

Card 3

Q: What physically is an H-bond in a protein?

A: A 2.5–3.5 Å dipole–dipole attraction between a hydrogen covalently bound to N or O (donor) and a lone pair on another N or O (acceptor).

Card 4

Q: Why can the peptide backbone form continuous H-bond ladders?

A: The repeating –CO–NH– unit presents a regular array of carbonyl oxygens (acceptors) and amide protons (donors) every 3.4 Å along the chain.

Card 5

Q: How does the energy of a salt bridge depend on its surroundings?

A: Coulombic energy E = kq₁q₂/Dr weakens ~80-fold in bulk water (D ≈ 80) compared with the low-dielectric protein interior (D ≈ 2–4).

Card 6

Q: Why are van der Waals forces important despite being weak individually?

A: Summed over thousands of atoms in a packed core, these ~0.3–1 kcal mol⁻¹ pairwise attractions provide major stabilising energy and enforce native-like packing.

Card 7

Q: What entropic benefit makes the hydrophobic effect the folding engine?

A: Sequestering non-polar side chains releases ordered water molecules, giving a large favourable –TΔS that drives chain collapse and core formation.

Card 8

Q: What extra stabilisation do aromatic side chains often contribute?

A: Face-to-face or edge-to-face π–π stacking between Tyr, Phe or Trp rings adds ~1–2 kcal mol⁻¹ per pair through delocalised electron clouds plus van der Waals contacts.

Card 9

Q: How does subtilisin turn an ordinary serine into a powerful nucleophile?

A: The Asp–His–Ser triad forms a charge-relay network in which Asp H-bonds to His, which polarises Ser-OH and lowers its pKa to generate the reactive alkoxide.

Card 10

Q: What principle governs both subtilisin–substrate and subtilisin–inhibitor recognition?

A: Complementary surface shapes plus extensive non-covalent contacts—H-bonds, hydrophobic patches and salt bridges—create tight, specific intermolecular interfaces.

Below are 10 compact Q&A notes—one per slide—each expanded with a little extra context I’ve added from scanning the visuals.

1. Ribonuclease refolding (slide 1)

Q: What does the urea-reduction/removal experiment prove?

A: That the amino-acid sequence alone encodes all the information needed for the protein to refold spontaneously into its native, catalytically active conformation—no cellular machinery required.

2. Ramachandran plot (slide 2)

Q: Why do dark-green clusters appear on a φ/ψ map?

A: They mark sterically allowed, highly favourable backbone angles; white space shows disallowed regions where atomic overlaps forbid the chain to exist—except for Gly (no β-carbon) and Pro (ring constraints) which can sample extra angles.

3. Non-covalent interaction strengths (slide 3)

Q: Rank H-bond, ionic, and van der Waals interactions by typical energy.

A: Ionic ≈ 10–20 kJ mol⁻¹ (strongest), H-bond 4–10 kJ mol⁻¹, van der Waals 2–4 kJ mol⁻¹ per contact. Although individually weak, hundreds of them summed together stabilise the folded state.

4. Thermodynamic driving force (slide 4)

Q: What simple thermodynamic requirement must be met for folding to occur?

A: ΔG = G_folded – G_unfolded must be negative; the native state must sit at a free-energy minimum relative to the ensemble of random coils.

5. Thermodynamic hypothesis reprised (slide 5)

Q: Why is Ribonuclease A considered the classic proof of the “thermodynamic hypothesis”?

A: Because complete denaturation (urea + reductant) abolishes structure and activity, yet simple removal of denaturants lets the chain refold in vitro to full activity—demonstrating the folded state is the global free-energy minimum.

6. Native vs denatured ensembles (slide 6)

Q: How do the structural and dynamic properties differ between native and denatured states?

A: Native: single, compact, well-defined 3-D structure, low dynamics, active. Denatured: highly dynamic, extended or molten-globule-like, no unique structure, inactive.

7. Conformational limits aid folding (slide 7)

Q: How does peptide-unit rigidity help the chain find its native shape?

A: The planar, trans-locked peptide bond and restricted φ/ψ angles drastically reduce the number of possible conformations, letting residues far apart in sequence approach each other often enough to form stabilising non-covalent contacts.

8. Gibbs free-energy definition (slide 8)

Q: What does ΔG quantify in protein folding?

A: The useful work (enthalpy – T·entropy) released when the chain moves from the unfolded ensemble to the single native state; a negative ΔG means the reaction is spontaneous under given conditions.

9. Magnitude of ΔG and stability (slide 9)

Q: Why is a larger negative ΔG_fold associated with greater protein stability?

A: Because ΔG_fold represents the depth of the energy well separating the native state from the unfolded ensemble; a deeper well means more energy is required to denature the protein, hence higher thermodynamic stability.

10. (Implicit summary slide)

Q: What overarching principle links slides 1–9?

A: Protein folding is governed by thermodynamics: the sequence biases the chain toward a set of favourable non-covalent interactions whose net result is a negative ΔG, making the native state the lowest free-energy conformation achievable without external help.

Here are 8 concise Q&A notes—one per slide—each expanded with extra interpretive points I’ve added after scanning the visuals.

1. (slide 74) What sets the “margin of safety” for a protein’s 3-D structure?

The net ΔG_fold is only –20 to –60 kJ mol⁻¹, equivalent to just a few hydrogen bonds. Because individual H-bonds contribute 10–20 kJ mol⁻¹, losing even two of them can erase stability and trigger unfolding—explaining why mutations or temperature jumps easily denature proteins.

2. (slide 75) How is the sign of ΔG ultimately decided?

ΔG = ΔH – TΔS. Folding is spontaneous only if the enthalpic gain (negative ΔH from new non-covalent bonds) outweighs the entropic penalty (negative ΔS from restricting chain freedom). Temperature tunes the balance: raising T makes the –TΔS term more punitive, which is why heat denatures proteins.

3. (slide 76) Where does the enthalpy bonus come from if peptide bonds stay intact?

Most covalent bonds are unchanged; the driving enthalpy arrives from newly formed persistent non-covalent contacts—H-bonds, salt bridges, van der Waals packing—that are absent or transient in the random-coil ensemble, yielding a sizable negative ΔH.

4. (slide 77) Why isn’t –8 MJ mol⁻¹ enough to lock a protein forever?

The –250 kJ from H-bonds, –100 kJ from ionic pairs and –8 000 kJ from thousands of vdW contacts look enormous, but the unfolded chain is already solvated and makes many solvent-mediated contacts; the incremental gain is therefore far smaller than the raw sum, leaving only a modest net ΔH.

5. (slide 78) What hidden energetic cost does water impose?

In water the unfolded state is stabilised by protein–water H-bonds and ionic solvation; breaking these and ordering water into “icebergs” around exposed hydrophobics carries an enthalpic price. Thus the apparent ΔH from intraprotein bonds must also pay for desolvation, further trimming net stability.

6. (slide 79) Does the Second Law forbid biological order?

No—local entropy can decrease (protein folds) only when the total entropy of the universe increases. Water molecules released from the hydrophobic surface gain motional freedom; their positive ΔS_system + ΔS_surroundings > 0 compensates for the negative conformational entropy of the polypeptide.

7. (slide 80) Why is conformational entropy an uphill battle?

An unfolded chain can sample ~10^N conformations for N residues, giving a large positive configurational entropy. Upon folding, this freedom collapses to essentially one compact state, costing –TΔS_conf ≈ +5–15 kJ mol⁻¹ per residue—an unfavourable term that must be offset by enthalpy and solvent entropy.

8. (slide 81) How do proteins still reach the native state despite the entropy penalty?

The hydrophobic effect (release of ordered water), extensive vdW packing, and cooperative H-bond/salt-bridge networks collectively provide enough favourable ΔH and ΔS_solvent to outweigh the large unfavourable ΔS_conf, driving ΔG negative and making the folded state the global free-energy minimum.

Here are 10 concise Q&A notes—one per slide—each expanded with extra interpretive points I’ve added after scanning the visuals.

1. (slide 82) How can burying hydrophobics ever be “entropically favourable”?

Water cages around exposed hydrophobics are highly ordered (low entropy). Bringing two hydrophobic surfaces together releases these caged waters into bulk, increasing solvent disorder; the positive ΔS_water outweighs the negative ΔS_chain, so ΔG becomes negative and association is spontaneous.

2. (slide 83) Why is the hydrophobic effect described as a two-part bargain?

Enthalpy: protein–protein contacts replace weaker protein–water contacts.

Entropy: ordered water shells are liberated.

Both terms cooperate to give the overall favourable free-energy change that drives collapse.

3. (slide 84) If non-covalent bonds give such a large ΔH, why is ΔG_fold still small?

The –TΔS_conf penalty for locking the chain is large (≈ +5–15 kJ mol⁻¹ per residue). Added to desolvation costs, it almost cancels the big enthalpy win, leaving only a slightly negative ΔG—hence marginal stability.

4. (slide 85) What is the microscopic signature of the native state?

A compact ensemble with many persistent intra-molecular H-bonds, ion pairs and vdW contacts, balanced against the entropy loss of constraining a formerly dynamic random coil.

5. (slide 86) Match the macroscopic property to the state:

- Extended, highly dynamic, exposed hydrophobics → unfolded.

- Compact, restricted motion, buried hydrophobics, stable secondary structure → native/folded.

6. (slide 87) How do disulfides act as “entropic bolts”?

They covalently link two chain segments after folding, reducing the number of conformations the unfolded state can sample. Lower ΔS_unfold means the native state is entropically favoured, giving extra thermodynamic stability—especially useful outside the reducing cytosol.

7. (slide 88) Why are disulfides rare in cytosolic proteins?

The intracellular environment is strongly reducing (high [GSH], low [GSSG]), so Cys-SH remains reduced; only in the oxidising extracellular milieu does disulfide formation become thermodynamically favourable.

8. (slide 89) What happens if Cys residues pair incorrectly before folding?

Scrambled non-native disulfides lock the chain in topologies that cannot reach the native state. Trace β-mercaptoethanol provides a reversible reducing environment, allowing thiols to shuffle until the correct S–S pairs form—an essential post-folding quality-control step.

9. (slide 90) Why does unfolding follow an “all-or-none” transition?

Protein folding is cooperative: loss of a few key interactions raises both enthalpy (fewer bonds) and entropy (more chain freedom). The ΔG of the denatured state quickly becomes lower, so intermediate, partly folded species are scarcely populated—giving the sharp sigmoidal transition seen in denaturant curves.

10. (slide 91) How does a single-residue mutation sometimes destabilise an entire domain?

Because interactions are networked; weakening one contact alters main-chain angles, reshuffles side-chain packing, and can break neighbouring H-bonds or vdW contacts. The cooperative network amplifies the local defect, often leading to global unfolding or functional loss.

Here are 7 concise Q&A notes—one per slide—each expanded with extra interpretive points I’ve added after scanning the visuals.

1. (slides 1-4, repeated) What is meant by “cooperativity” in protein structure?

A: Interactions are networked—perturbing one residue alters neighbours, which alters their neighbours—so the entire molecule responds as a unit. This reinforcement gives an all-or-none folding transition and means local damage can propagate globally.

2. (slide 5) What molecular species is actually present at the midpoint ([D] = [N]) of a denaturant titration?

A: An equilibrium mixture where K_D-N = 1 and ΔG = 0; individual molecules are either fully folded or fully unfolded—there is no stable “half-folded” intermediate because cooperativity suppresses partly folded states.

3. (slide 6) Why is cooperative folding advantageous in a fluctuating environment?

A: The sharp sigmoidal response gives a digital switch: small changes in temperature, pH or ligand concentration flip the whole population between functional (folded) and non-functional (unfolded), preventing the accumulation of potentially harmful, semi-folded species.

4. (slide 7, summary) List the four overarching principles that allow proteins to reach their native state.

A:

- Sequence encodes all necessary information (Anfinsen principle).

- Main-chain geometry (φ/ψ, peptide planarity) and side-chain chemistry create regular secondary structure and diverse non-covalent contacts.

- Thermodynamics balances favourable enthalpy (non-covalent bonds) against unfavourable conformational entropy, with water entropy (hydrophobic effect) tipping the scale.

- Cooperativity of interactions ensures a sharp, reversible transition to the lowest free-energy minimum, providing both stability and sensitivity to environmental change.

5. (extra from slide 7) Why are proteins described as “at the margin of stability”?

A: The net ΔG_fold is tiny (–20 to –60 kJ mol⁻¹) because the large enthalpic gain from thousands of non-covalent contacts is almost cancelled by the large entropic cost of ordering the chain and restricting its dynamics; this delicate balance makes them exquisitely responsive to mutation, temperature or solvent changes.

6. (extra from slide 7) How does the hydrophobic effect integrate into the thermodynamic ledger?

A: Sequestering non-polar side chains releases ordered water molecules, giving a positive ΔS_solvent that partially offsets the negative ΔS_chain, thereby making the overall ΔG negative without requiring an impossibly large enthalpy bonus.

7. (extra from slide 7) What practical consequence does cooperativity have for cellular proteostasis?

A: Because partially folded states are transient and lowly populated, chaperones can act early—binding to small, exposed hydrophobic patches—before toxic aggregation occurs; if folding were gradual, many more intermediates would need separate surveillance mechanisms.