Biomolecules Section 116 of 123

Peptides and proteins

the why on every wrong step

16 min readfree

Take alanine and serine, two of the twenty amino acids from the Amino acids section. Join the carboxyl group of alanine to the amino group of serine, and one molecule of water comes out. The product is a single molecule built from both, called a dipeptide. A protein is built by repeating the same reaction hundreds of times.

The peptide bond is an amide

The new bond joins alanine’s carbonyl carbon to serine’s nitrogen. A carbonyl carbon bonded to a nitrogen is an amide, the functional group from Esters & amides. When an amide links two amino acids, it is called a peptide bond. The name is new; the chemistry is that of any amide.

HCH₃OH₂NCCOH+HHCH₂OHOHNCCOHleave as wateralanineserine− H₂OHCH₃OHHCH₂OHOH₂NCCNCCOH+H₂Opeptide bond (an amide)N-terminusC-terminusAla-Ser: alanine keeps the free NH₂, serine keeps the free COOH.
Follow the coral atoms: alanine’s OH and one H from serine’s nitrogen leave as water, and the new C–N bond (highlighted) is the peptide bond.

Each amino acid unit in a chain is called a residue, because it is what remains of the amino acid once the water has gone. Two residues make a dipeptide and three make a tripeptide. A chain of a few dozen residues or more is a polypeptide, and a polypeptide that folds into a working shape is a protein.

The drawing shows only the overall change. Two amino acids mixed in a flask do not join this way; “Making a peptide on purpose”, below, shows how chemists do it.

Every chain has two different ends

Look again at Ala-Ser. Alanine gave its carboxyl group to the new bond, so its amino group is still a free NH₂. Serine gave its nitrogen, so its carboxyl is still a free COOH. Every peptide chain has these two ends. The end with the free amino group is the N-terminus, and the end with the free carboxyl group is the C-terminus.

Peptides are always written, and their residues numbered, from the N-terminus to the C-terminus. In Ala-Ser, alanine is residue 1 and holds the N-terminus. The full name follows the same order: every residue except the last takes the ending -yl, so Ala-Ser is alanylserine and Gly-Ala is glycylalanine.

Reversing the order gives a different compound. Gly-Ala and Ala-Gly have the same atoms and the same formula, but in Gly-Ala glycine carries the free NH₂, and in Ala-Gly alanine does.

Gly-AlaHHOHHCH₃H₂NCCNCCOOHglycinealanineN-terminusC-terminusAla-GlyHCH₃OHHHH₂NCCNCCOOHalanineglycineN-terminusC-terminus
Same two residues, opposite order. Look at which residue sits on the left, where the free NH₂ is: that residue is named first.
Because order matters, the number of possible sequences grows very fast. With twenty amino acids to choose from at each position, there are 20² = 400 dipeptides and 20³ = 8,000 tripeptides. A chain of n residues has 20n possible sequences, so even a small protein’s sequence is one out of an enormous number.

Amide resonance makes the peptide bond flat

The nitrogen of a peptide bond has a lone pair, and it sits right next to a C=O. As in every amide, resonance shares that lone pair with the carbonyl. Move the lone pair into the C–N bond and move the C=O π electrons onto the oxygen. The second structure has a C=N double bond, a positive charge on nitrogen and a negative charge on oxygen.

CαHCαONCCαHCαO−N+Clone pair on N, C=O doubleC=N double, + on N, − on O
Follow the two curved arrows from the left structure to the right one, then compare the charges.

Neither drawing is the real molecule. The real peptide bond is a blend of the two, called the resonance hybrid. The second structure contributes enough to make the C–N bond part-way double. It is 1.33 Å long, against 1.47 Å for an ordinary C–N single bond. The oxygen carries a partial negative charge and the nitrogen a partial positive one.

CαHCαOδ−Nδ+CC–N: 1.33 Åshaded: one flat plane, six atomsa plain C–N single bond is 1.47 Å
Each solid-plus-dashed pair is a partial double bond.

Three properties of the peptide bond follow from this one resonance.

  • The bond cannot rotate freely. A double bond resists twisting, and a partial double bond does too. So six atoms are held in one plane: the carbonyl C and O, the N and its H, and the α carbon on each side (the carbon that carries a residue’s side chain).
  • The nitrogen is not basic. Its lone pair is already shared with the carbonyl, so it is not free to take a proton the way an amine’s lone pair is. When strong acid does protonate an amide, the proton goes on the oxygen instead.
  • The amide is the least reactive acyl derivative, among the family R–C(=O)–X of carboxylic-acid relatives from the Carboxylic Acids & Derivatives chapter. The same donation that ties up the nitrogen’s lone pair also leaves the carbonyl carbon less δ+. So hydrolyzing a peptide bond needs hours in hot, strong acid, or an enzyme built for the job. That is why proteins last in water.

In almost every peptide bond, the two α carbons sit on opposite sides of the C–N bond, where they are farthest apart. This is the trans arrangement, and the drawings in this section show it. Bonds to proline are the main exception: they are cis far more often than other peptide bonds.

Now look at the whole chain. Each peptide unit is a flat, rigid plate, and neighboring plates meet at an α carbon. Only the two single bonds at each α carbon can rotate, so the backbone bends at those points and nowhere else. That restriction is what lets a chain settle into a few regular shapes.

OOHHRHRHHRCαCαCαCCNNrotation only at each Cαshaded: rigid, flat amide plates
Each shaded plate is one amide unit. The curved arrows sit on the two bonds at the middle α carbon.

Four levels of structure

A folded protein is described at four levels. Covalent peptide bonds set the first. The other three rest on weaker forces, with one covalent exception, the disulfide bond, described below.

  • Primary structure is the sequence of residues, joined by covalent peptide bonds.
  • Secondary structure is a local, repeating shape held by hydrogen bonds between backbone N–H and C=O groups. The two standard shapes are the α-helix and the β-sheet. The side chains take no part in them.
  • Tertiary structure is the three-dimensional fold of the whole chain. The side chains hold it, through the hydrophobic effect, hydrogen bonds, salt bridges and disulfide bonds (each explained below).
  • Quaternary structure is how two or more folded chains fit together. Hemoglobin, with its four chains, is the standard example. A protein made of a single chain has no quaternary structure.

Secondary structure: the α-helix and the β-sheet

Almost every residue contributes one N–H and one C=O to the backbone (proline, below, has no N–H). An N–H on one peptide unit can form a hydrogen bond to the C=O of another: the δ+ hydrogen is attracted to a lone pair on the δ− oxygen. One such bond is weak, but hundreds of them lined up in a regular pattern hold a shape.

In an α-helix, the backbone winds into a coil with 3.6 residues per turn. That spacing brings the C=O of each residue close to the N–H of the residue four places further along. So the C=O of residue 1 bonds to the N–H of residue 5, residue 2 bonds to residue 6, and so on up the coil. These hydrogen bonds all run roughly parallel to the axis, and the side chains point outward.

RRRRRRR1234567891011residue 1: N-terminal endC=O of residue 1 ··· H–N of residue 53.6 residues per turn
Numbered discs are residues and each dashed line is one hydrogen bond. The highlighted one joins residue 1 to residue 5. Side chains on the front and back of the coil are not drawn.

In a β-sheet, two or more stretches of chain, called strands, lie side by side in an almost fully stretched form. The hydrogen bonds run across, from an N–H on one strand to a C=O on the next. Neighboring strands can run in opposite directions, which makes an antiparallel sheet (drawn below), or in the same direction, which makes a parallel sheet.

HOHOHONCHRCNCHRCNCHRCOHOHOHCCHRNCCHRNCCHRNN → CN → C
The arrows give each strand’s N-to-C direction, and the dashed lines are hydrogen bonds. CHR is an α carbon with its H and side chain.

The flat amide plates of a strand meet at an angle at each α carbon, so the sheet is folded like a paper fan. That fold is where the full name, β-pleated sheet, comes from. The side chains point alternately above and below the sheet.

Proline fits poorly into either pattern. In a chain, its nitrogen is part of a ring and carries no H, so it cannot donate a hydrogen bond, and the ring limits how its backbone turns. It often ends a helix, which is why it is called a helix breaker.

Side chains are not part of either pattern, but they still affect which shape a stretch of chain takes up, as proline shows.

Why hydrophobic side chains end up inside

In a folded protein in water, the side chains of leucine, valine and phenylalanine are usually buried inside, while those of aspartate, lysine and serine sit on the surface. The rule behind it is simple: nonpolar side chains go inside, and polar or charged side chains stay outside, in contact with water.

LeuValPheIleAsp⁻H₂OLys⁺H₂OSerH₂OGlu⁻H₂OArg⁺H₂OThrH₂Oinside: nonpolar side chainssurface: polar and charged, facing water
Water surrounds the protein; the coral patch is its core.

The usual explanation, that nonpolar groups “avoid water”, is close enough to be useful but slightly misleading. The main effect is on the water. Water next to a nonpolar surface cannot hydrogen bond to it, so it is forced into a more ordered arrangement. Burying the nonpolar side chains frees that water.

Freed water gains entropy, a measure of how many ways the molecules can arrange themselves, and that gain drives the fold. This drive is called the hydrophobic effect, and it is the largest single force holding tertiary structure together.

Salt bridges and disulfide bonds

Side chains that can hydrogen bond, such as serine’s OH, do so with each other and with the backbone. Charged side chains form salt bridges. At pH 7, aspartate’s side chain is a carboxylate, COO⁻, and lysine’s is an ammonium, NH₃⁺. When the two sit close together in the fold, their opposite charges attract.

The one covalent cross-link is the disulfide bond. Two cysteine side chains each end in a thiol, S–H. Oxidation removes both hydrogens and joins the two sulfurs with an S–S bond. Reduction adds the hydrogens back and breaks it.

ONHCHCOCCHNHCH₂CH₂SHSHONHCHCOCCHNHCH₂CH₂SSoxidation (−2 H)reduction (+2 H)two cysteine thiols, S–Hone disulfide bond, S–S
Compare the sulfurs: two S–H groups on the left, one S–S bond (highlighted) on the right. The two cysteines can be far apart in the sequence.

Heating a protein breaks up its hydrogen bonds, salt bridges and hydrophobic packing, so the chain loses its folded shape. This is called denaturation. The peptide bonds survive it: a boiled egg white still holds intact protein chains, only tangled.

A disulfide bond is covalent, so it survives heating too. To unfold a protein completely, a lab adds a reducing agent to break the disulfides as well.

Worked example — reading a sequence for structure

A short peptide reads Cys-Leu-Asp-Lys-Val-Cys. What can you say about it from the sequence alone?

Two cysteines. They can form a disulfide bond, which would close the peptide into a loop. That covalent link limits its shape, so notice it first.

Asp and Lys. One side chain is acidic and the other basic, so at pH 7 one is negative and the other positive. Together they can form a salt bridge.

Leu and Val. Both are nonpolar, so they pack against each other, or against other nonpolar side chains, rather than facing the water.

SSdisulfide closes the loopH₃N⁺COO⁻COO⁻NH₃⁺salt bridgeCysLeuAspLysValCysLeu and Val:nonpolar
The six residues in order, with the disulfide, the salt bridge and the nonpolar pair marked. The loop is a sketch of which side chains meet, not a real shape.

The net charge at pH 7. Count every group that can carry a charge. The N-terminal NH₃⁺ gives +1, the C-terminal COO⁻ gives −1, aspartate gives −1 and lysine gives +1, so the net charge is 0, and the peptide would not move in an electric field. The amide nitrogens inside the chain add nothing, because their lone pairs are already shared with the carbonyls. Only the side chains and the two ends can carry a charge.

None of this needs you to know what the peptide does. Most questions at this level ask you to read a sequence for the kinds of side chain in it.

Making a peptide on purpose

Worked example — how many dipeptides from glycine and alanine?

Mix glycine and alanine with a reagent that removes water, and count what comes out.

Each molecule has both groups, an amine and a carboxyl. So either molecule can supply the carboxyl, and either can supply the amine.

Two choices for the first residue and two for the second give four dipeptides: Gly-Gly, Ala-Ala, Gly-Ala and Ala-Gly. As the Gly-Ala and Ala-Gly drawings above show, the last two are different compounds.

It does not stop there. Every dipeptide still has a free NH₂ at one end and a free COOH at the other, so the chains keep growing into a random mixture.

That mixture is the problem. Nothing tells either molecule which of its two groups should react. The solution is a protecting group, a temporary cap that stops one group from reacting and comes off later, as in Protecting groups. It takes three moves.

  • Protect. Cap the amino group of the residue that will be the N-terminus. The usual cap is Boc (tert-butoxycarbonyl), put on with Boc2O. It turns the amine into a carbamate, N–C(=O)–O–, whose nitrogen lone pair is tied up just as an amide’s is, so it no longer reacts as a nucleophile. Acid, usually trifluoroacetic acid (TFA), takes it off later. A second common cap, Fmoc, comes off with a mild base instead. Then cap the carboxyl of the other residue as a methyl ester. Each residue now has exactly one free group.
  • Couple. Join the one free carboxyl to the one free amine with a coupling reagent, usually DCC (below).
  • Deprotect the N-terminus and repeat. Each new residue joins at the free N-terminal end, so the chemist sets the sequence by choosing which Boc-amino acid to add in each cycle. At the end, remove the ester.
BocGly–COOH+H₂N–AlaOCH₃DCCBocGlyAlaOCH₃only COOH freeonly NH₂ freeTFA (acid) removes BocH₂N–GlyAlaOCH₃For a longer chain: couple the next Boc amino acidto this free NH₂ with DCC, then remove Boc again.aqueous base, then acid,removes the esterH₂N–GlyAla–COOHthe dipeptide Gly-Ala
Coral blocks are the protecting groups. Each coupling joins the one free COOH to the one free NH₂, so only one product can form.

The coupling step needs help. An amine and a carboxylic acid do not form an amide when mixed. The amine is a base, so it takes the acid’s proton instead, and the carboxylate that forms is too poor an electrophile to be attacked.

DCC, dicyclohexylcarbodiimide (Cy–N=C=N–Cy, where Cy is a cyclohexyl ring), solves this. The acid’s oxygen adds to the central carbon of DCC, giving an O-acylisourea. That turns the acid’s OH into a good leaving group. The amine then does ordinary nucleophilic acyl substitution: it adds to the C=O to give a tetrahedral intermediate, and the leaving group then departs as dicyclohexylurea.

1 · DCC ACTIVATES THE ACIDORO−CCyCyHNN+Ccarboxylateprotonated DCCORCyNCyHNOCCO-acylisourea: the activated acidafter the acid’s H⁺ moves to DCC2 · THE AMINE ADDS TO THE C=OOROC(=NCy)NHCyHHR′NCO−RHR′HN+OC(=NCy)NHCyCtetrahedral intermediate3 · THE C=O REFORMS AND THE LEAVING GROUP GOESO−RHR′HN+OC(=NCy)NHCyCORR′HNCthe amide (after N loses H⁺)+ CyNH–C(=O)–NHCydicyclohexylurea(after it gains H⁺)
Cy is cyclohexyl, C₆H₁₁. Follow the coral atoms from panel to panel. The highlighted bond in panel 1 is the new bond from the acid’s oxygen to DCC. Proton transfers are not drawn.
Solid-phase (Merrifield) synthesis makes the repetition practical. The first residue is anchored by its carboxyl to an insoluble polymer bead, and the chain grows from there toward its N-terminus. A large excess of reagent can drive every coupling to completion, and purifying after each step is just a wash, not a slow separation on a column. That is why a machine can run the whole cycle. At the end, strong acid cuts the finished chain off the bead.

Hydrolysis, and how sequences are determined

Heat a protein in 6 M HCl at about 110 °C for a day, and every peptide bond is hydrolyzed. This is the amide hydrolysis from the Carboxylic Acids & Derivatives chapter, run to completion, and it returns the mixture of free amino acids.

Those conditions are harsh. They destroy tryptophan, and they turn asparagine and glutamine into aspartic and glutamic acid, because the amide in each side chain is hydrolyzed too. So the result gives the protein’s composition, after a known correction. But hydrolysis destroys the sequence, which is why sequencing uses a method that removes one residue at a time from one end.

That method is the Edman degradation. It runs in three stages.

  • Coupling. Phenyl isothiocyanate, PhN=C=S (Ph is phenyl, C6H5), reacts with the free NH₂ at the N-terminus under mildly basic conditions. Its central carbon is double-bonded to both N and S, which leaves it δ+, and the amine adds to it. The product is a phenylthiourea. A thiourea is a urea with sulfur in place of oxygen: a C=S carbon bonded to two nitrogens.
  • Cleavage. In acid with no water present, the thiourea’s sulfur attacks the carbonyl carbon of the first peptide bond and closes a five-membered ring. The C–N bond to the rest of the chain then breaks. The first residue leaves inside the ring as an anilinothiazolinone (ATZ), and the rest of the peptide, one residue shorter, now has a free NH₂ of its own.
  • Conversion. The ATZ is unstable. Warmed in aqueous acid, it rearranges to an isomer, the phenylthiohydantoin (PTH): a five-membered ring holding both nitrogens, a C=S and a C=O. Each amino acid gives a different PTH, which is identified by chromatography (a separation method) against standards.
1 · COUPLING (MILDLY BASIC): THE AMINE ADDS TO PhN=C=SPhSNCR₁OHH₂NCHCNpeptidePh–NH–C(=S)–NH–CH(R₁)–C(=O)–NH–peptidea phenylthiourea (after one proton moves)2 · IN ACID: SULFUR ATTACKS THE FIRST C=O, THEN THE CHAIN LEAVESPhNHHR₁HpeptideOSNNCHCCPhNHHR₁HpeptideO−S+NNCHCCthe thiourea, foldedring closed; now the C=O reforms3 · THE FIRST RESIDUE LEAVES AS AN ATZ, WHICH BECOMES THE PTHPhNHR₁OSNCHCCATZ (anilinothiazolinone)+ H₂N–peptideone residue shorteraqueous acidOR₁HSPhCCHNCNPTH (phenylthiohydantoin)
Ph is phenyl, C₆H₅. Follow the coral carbon in each panel: first the carbon of PhN=C=S, then the carbonyl carbon that sulfur attacks. Proton transfers are not drawn.

Two features make the Edman degradation useful for reading a sequence. It works inward from one end, the N-terminus, and it removes exactly one residue per cycle, so the shortened peptide can go straight back through.

Small errors build up with each cycle, so an Edman run reads roughly thirty residues, far short of most proteins. A longer chain is first cut into shorter pieces by enzymes that cut only next to particular residues. A second sample is cut with a different enzyme, and the overlaps between the two sets of pieces fix the order.

Worked example — putting fragments in order

An unknown hexapeptide is cut two ways, and Edman sequencing reads each fragment. Trypsin cuts on the carboxyl side of lysine and arginine, and gives Ser, Ala-Phe-Arg and Gly-Lys. Chymotrypsin cuts on the carboxyl side of the aromatic residues phenylalanine, tyrosine and tryptophan, and gives Arg-Ser and Gly-Lys-Ala-Phe. What is the sequence?

Find the overlap. The trypsin fragment Ala-Phe-Arg starts inside Gly-Lys-Ala-Phe and ends with the Arg that begins Arg-Ser. So Gly-Lys-Ala-Phe comes first and Arg-Ser second.

Place the other trypsin pieces. Gly-Lys is the start of Gly-Lys-Ala-Phe. Ser is the end of Arg-Ser, and it is the only trypsin piece that does not end in Lys or Arg, so it must be the C-terminus. The sequence is Gly-Lys-Ala-Phe-Arg-Ser.

trypsincuts after Lys, ArgGlyLysAlaPheArgSerchymotrypsincuts after Phe, Tyr, TrpGlyLysAlaPheArgSerAla-Phe-Arg spans the chymotrypsin cutthe sequenceGlyLysAlaPheArgSer
Each dashed line is a cut. The green residues are the trypsin fragment Ala-Phe-Arg, marked in both enzyme rows.
Two trypsin cuts gave three fragments. In general, a chain with n cut sites gives n + 1 fragments, as long as none of the sites is the last residue. A lysine at the C-terminus has nothing after it to cut off. Trypsin also skips a lysine or arginine that is followed by proline.

What carries forward

Everything structural here rests on two earlier facts. An amide is resonance-stabilized, which makes it flat and unreactive. A hydrogen bond is weak on its own but decisive in large numbers. The biology applies both, plus one new idea, the hydrophobic effect.