Amino acids, proteins, folding and structure

Protein structure is more than a formula

Proteins participate in catalysis, transport and structural support, among other cellular roles. Their interactions depend on the arrangement of chemical groups and on the environment. The same elemental composition is insufficient to describe a particular sequence, folded organisation or assembly of subunits.

For the examples here, glycine is H₂N–CH₂–COOH and alanine is H₂N–CH(CH₃)–COOH. These are chosen neutral NH₂/COOH representations of α-amino acids. Glycine’s central carbon has two hydrogen atoms; alanine has a methyl side chain. Alanine configuration is explicitly left unspecified in these inventory examples.

In water, protonated and deprotonated groups can be important. A net-neutral formula does not select between uncharged NH₂/COOH and internally charged NH₃⁺/COO⁻ ends. State the representation before interpreting a drawing or formal charge.

A peptide link removes a formal water inventory

Choose conventional α-linked Gly–Ala with uncapped NH₂ and COOH termini. Its condensed connectivity is H₂N–CH₂–C(=O)–NH–CH(CH₃)–C(=O)OH. The first residue is at the N terminus, and the second at the C terminus. The α-carboxyl carbon of glycine is linked to alanine’s α-amino nitrogen through an amide bond.

Adding glycine C₂H₅NO₂ and alanine C₃H₇NO₂ gives C₅H₁₂N₂O₄. Subtract one H₂O inventory for the selected peptide link to obtain C₅H₁₀N₂O₃. The formal net is glycine + alanine → Gly–Ala + H₂O; atoms, net charge and total electron count are conserved for these chosen states.

The reversed sequence Ala–Gly has the same molecular formula but different ordered connectivity. This ledger does not establish spontaneous reaction in water or describe cellular protein synthesis. An actual mechanism uses its own substrates, activation and additional participants; balancing the formal inventory does not determine it.

Residue counts do not specify sequence

Choose the conventional α-linked tripeptides Ala–Gly–Ala and Ala–Ala–Gly, each with uncapped neutral NH₂/COOH ends and unspecified alanine configurations. Both contain two alanine and one glycine inventories with two peptide links. Thus 2C₃H₇NO₂+C₂H₅NO₂−2H₂O gives C₈H₁₅N₃O₄.

Both sequences begin with Ala, but Ala–Gly–Ala ends with Ala while Ala–Ala–Gly ends with Gly. Their second and third residue positions differ. Formula equality does not imply equal spectra, equilibrium organisation or biological function.

Primary structure describes the ordered amino-acid sequence. Secondary structure includes local backbone arrangements such as α helices and β sheets. Tertiary organisation concerns a polypeptide’s three-dimensional arrangement, while quaternary organisation concerns assembled subunits. These descriptions require more information than a formula or a short residue count.

Mean charge is a population average

Consider a separate hypothetical ampholyte with three forms H₂A⁺, HA and A⁻ carrying formal charge numbers +1, 0 and −1. Assign pK_a1=2.00, pK_a2=9.00 and controlled pH, using ideal concentration relations. Relative weights are w₊=10^(pK_a1−pH), w₀=1 and w₋=10^(pH−pK_a2). Divide by W=w₊+1+w₋ to obtain fractions.

At pH=5.50 the charged-form weights are equal, both 10⁻³·⁵. Each charged fraction is about 0.000316028; the middle fraction is about 0.999367944. Mean charge z̄=f₊−f₋ is exactly zero in this assigned symmetric model. That does not mean every molecule has zero charge or that the middle form has no internally charged groups.

At pH=7.00, weights are 0.0000100, 1 and 0.0100. Fractions are approximately 0.000009900892, 0.990089207 and 0.009900892. Their sum is one, and mean charge is −0.009890991. The graph samples the same three-state model. It does not give a real protein’s isoelectric point: actual side chains, local environments and additional states require their own accounting.

A two-state free-energy model gives populations

For an explicitly assigned reversible two-macrostate equilibrium, label folded F and unfolded U. Let ΔG_UF=G_U−G_F include the stipulated ensemble’s contributions, so p_U/p_F=exp(−ΔG_UF/RT). These effective macrostate free energies include the state multiplicities; they are not merely energies of two individual drawings.

Invent ΔG_UF=+5.00 kJ mol⁻¹ at 300 K. Converting to 5000 J mol⁻¹ gives ratio 0.134722355. Since p_F+p_U=1, the folded fraction is 0.881272847 and the unfolded fraction 0.118727153. The positive difference favors F under the chosen conditions.

Those fractions do not specify a folding path, time constant or barrier. Nor do they prove that every protein has just two states or a single fixed rigid structure. Additional conformations, association, aggregation and changing conditions can require a different model.

An observed function needs the right molecular organisation

Peptide bonds provide covalent connectivity; noncovalent contacts and, in appropriate proteins, additional covalent links contribute to organisation. Local backbone hydrogen bonding in helices and sheets is not interchangeable with the peptide bond itself. An interaction site can also depend on the arrangement of several residues or subunits.

Relating structure to function requires evidence about the actual sequence, chemical modifications, partners and conditions. Our short peptides and assigned population models teach inventory and inference; they are not measured protein structures or tests of a cellular function. Keep composition, sequence, protonation, equilibrium populations and kinetic behavior distinct.

Worked examples

Try it yourself

1. Find the neutral formal formula of α-linked Gly–Gly–Ala with uncapped NH₂/COOH ends and unspecified alanine configuration.

   Show the answer and reasoning

   C₇H₁₃N₃O₄.

   2C₂H₅NO₂+C₃H₇NO₂−2H₂O conserves the chosen atom inventory. It does not describe a synthesis mechanism.

2. Compare Ala–Gly–Gly and Gly–Gly–Ala in the same α-link/end convention. Give the formula and both terminal residue identities.

   Show the answer and reasoning

   Both C₇H₁₃N₃O₄; first has N-terminal Ala/C-terminal Gly, second N-terminal Gly/C-terminal Ala.

   Composition is the same, while ordered connectivity and terminal residue identities differ.

3. At controlled pH=9.00 in the three-state model, find the fractions and mean charge.

   Show the answer and reasoning

   Fractions about 4.99999975×10⁻⁸, 0.499999975 and 0.499999975; z̄=−0.499999925.

   Weights are 10⁻⁷, 1 and 1. Retain the tiny positive-form population: the mean is close to −0.5, but not exactly −0.5.

4. For a different assigned two-state ensemble with G_U−G_F=−3.00 kJ mol⁻¹ at 300 K, find folded and unfolded fractions.

   Show the answer and reasoning

   p_F=0.230991066 and p_U=0.769008934.

   The ratio exp[+3000/(R×300)]=3.32917175 favors U; normalizing it gives the fractions. No kinetic conclusion follows.

What to carry forward

• Specify peptide sequence, α-linkage, termini and protonation separately.
• Neutral formula alone does not select uncharged or zwitterionic groups.
• Residue composition and ordered sequence can differ in information.
• Mean charge comes from normalized form populations.
• Assigned equilibrium free energies give populations, not pathways or rates.