Exercises — Peptide bond; primary, secondary, tertiary, quaternary protein structure

WHAT: A peptide bond is an amide group, written $-\text{CO}-\text{NH}-$. HOW it forms: the $-COOH$ (carboxylic acid) of one amino acid reacts with the $-N{H}_2$ (amine) of the next, losing one water molecule (amide formation). Notice the link contains a nitrogen ($N$), not a second oxygen — that is what makes it an amide and not an ester.

Mnemonic from the parent: "Please Stop Touching Quietly."

Rule: $n$ pieces joined in a line need $n-1$ joins, and each join expels one $H_{2}O$.

• Peptide bonds $=n-1=8-1=7$.
• Water molecules $=n-1=7$. Why $n-1$? Think of a chain of 8 beads: the gaps between beads number one fewer than the beads themselves.

Water released $=n-1$, so $n-1=19\Rightarrow n=20$ amino acids. Peptide bonds $=n-1=19$ (same as the number of waters — one water per bond).

Peptide bonds only form within a continuous chain, never across the gap between two separate chains.

• Chain A: $6-1=5$.
• Chain B: $10-1=9$.
• Total $=5+9=14$ peptide bonds. Why not $16-1=15$? Because the two chains are not joined end-to-end by a peptide bond; they are held together (quaternary) by weaker forces, so you count each chain separately.

WHAT causes the flatness: the lone pair on nitrogen delocalises into the $C=O$ π system (resonance). This gives the $C-N$ bond partial double-bond character. WHY that fixes the geometry: a double bond cannot rotate freely, so the six atoms of the peptide unit ($C_{\alpha },C,O,N,H,C_{\alpha }$) are locked into one plane. WHY it helps folding: because each flat plate can only pivot at the $C_{\alpha }$ hinges, the backbone has very few ways to satisfy every $N-H$ and $C=O$ with a hydrogen bond. The regular right-handed coil (α-helix) is one such neat solution — the rigidity forces order instead of a random floppy tangle.

Heat supplies energy that shakes apart weak interactions:

• Secondary (backbone H-bonds), tertiary (R-group H-bonds, ionic, hydrophobic, and even some $-S-S-$ can scramble), and quaternary (chain–chain forces) are all lost — this is denaturation.
• Primary is preserved: peptide bonds are strong covalent bonds and are not broken by cooking heat. The sequence (spelling) is unchanged; only the folding is gone. Why the egg looks so different: the unfolded chains tangle and trap water, scattering light → opaque solid.

Directly: the primary structure changes — one letter of the sequence is swapped. Knock-on effect: valine is hydrophobic, so it creates a new "sticky" patch on the surface. This changes which R-groups attract/repel, altering the tertiary/quaternary folding and packing. The mutated haemoglobin molecules clump into fibres, deforming the cell. Lesson: primary sequence dictates all higher levels — change the spelling and the fold can follow.

(a) Read N → C and change every residue except the last to the "-yl" ending: Ala-Gly-Ser → Alanylglycylserine. (b) $n-1=3-1=2$ peptide bonds. (c) The N-terminus is the end with the free $-N{H}_2$, i.e. the Alanine end (where we started reading). Why direction matters: Ala-Gly-Ser and Ser-Gly-Ala are different molecules — like "cat" vs "tac."

Note (ii): a disulphide is covalent but it is not a peptide bond — it links side-chains, so it shapes tertiary structure, not primary sequence.

Globular enzyme: it folds so that hydrophobic (water-fearing) R-groups tuck inside, while polar/charged groups face the water outside. Result: a compact, water-friendly ball → soluble, and its folded pocket acts as an active site (globular enzymes). Fibrous keratin: long parallel chains packed by many H-bonds and $-S-S-$ links, with hydrophobic surfaces exposed along the fibre → water-insoluble, strong, structural. One principle, two outcomes: hiding hydrophobic groups drives a compact ball; exposing them along aligned chains gives a tough thread.

$n-1=1-1=0$ peptide bonds and $0$ water released — there is no second piece to join to, so no condensation happens. It has a primary structure (a sequence of length one) but no secondary, tertiary, or quaternary structure, since those require a chain long enough to fold or multiple chains to assemble. Why this matters: it confirms the formula $n-1$ behaves correctly at its smallest input — a single bead has zero gaps.

Myoglobin lacks quaternary structure: quaternary structure only exists when there are ≥2 chains assembled. With one chain, the highest level present is tertiary. Haemoglobin (2α + 2β) does have quaternary structure (haemoglobin). The edge-case lesson: quaternary structure is optional; a fully functional protein (myoglobin) can top out at tertiary. Do not force a quaternary answer where there is only one chain.

The final linear chain of 6 residues has $6-1=5$ peptide bonds, hence 5 water molecules total — regardless of route, because bonds released = bonds present in the product.

• (a) Sequential: $5$ additions → $5$ waters.
• (b) Two tripeptides: each is $3-1=2$ bonds → $2+2=4$ waters; joining them adds 1 more bond → 1 more water; total $4+1=5$. Both give 5. Why they must match: the number of waters equals the number of peptide bonds in the finished molecule — a conserved bookkeeping quantity independent of assembly order.

Shared idea: both form by condensation (loss of $H_{2}O$) — that part is right. The difference:

• A glycosidic bond links two sugars via an ether-like $-O-$ (C–O–C) oxygen bridge.
• A peptide bond links two amino acids via an amide $-CO-NH-$ with nitrogen in the link. So they are analogous polymerisation reactions but chemically different linkages (oxygen bridge vs nitrogen amide). Same strategy, different joint.

Peptide bonds in an $n$-residue single chain ::: $n-1$ Water released making that chain ::: $n-1$ Total peptide bonds in $k$ separate chains ::: (total residues) $-k$ Levels lost on denaturation ::: secondary, tertiary, quaternary (primary survives) Level requiring ≥2 chains ::: quaternary Bond type of a peptide link ::: amide, $-CO-NH-$ (nitrogen in the link)