Biomolecules • Topic 2 of 3

Proteins & Amino Acids

Proteins are the most abundant biomolecules of the living cell and are built from α-amino acids. An α-amino acid carries both an amino group (–NH2) and a carboxyl group (–COOH) attached to the same (α) carbon, with the general formula R–CH(NH2)–COOH. The side chain R distinguishes the twenty standard amino acids.

Classification of amino acids. By the R group: neutral (one –NH2, one –COOH, e.g. glycine, alanine), acidic (extra –COOH, e.g. glutamic acid, aspartic acid) and basic (extra –NH2, e.g. lysine, arginine). Nutritionally they are essential (must come from diet, e.g. valine, leucine, lysine) or non-essential (the body can synthesise them, e.g. glycine, alanine). Except glycine, all are chiral and natural proteins use the L-forms.

Zwitterion. Amino acids exist not as neutral R–CH(NH2)–COOH but as internal salts — the proton moves from –COOH to –NH2, giving R–CH(NH3+)–COO−, a zwitterion (dipolar ion). This explains their high melting points, water solubility and amphoteric (acid + base) behaviour. In acid the –COO− picks up H+ (cation); in base the –NH3+ loses H+ (anion).

Isoelectric point (pI). The pH at which the amino acid exists almost entirely as the zwitterion with zero net charge is the isoelectric point. At this pH it does not migrate in an electric field and its solubility is lowest. Each amino acid has a characteristic pI, which is the basis of separation by electrophoresis.

Peptide bond. The –COOH of one amino acid condenses with the –NH2 of the next, eliminating water and forming an amide (–CO–NH–) link called a peptide bond. Two amino acids give a dipeptide, many give a polypeptide; a protein is a polypeptide with a definite three-dimensional shape.

Structure of proteins is described at four levels. Primary structure is the exact sequence of amino acids in the chain — any change can alter function (e.g. sickle-cell haemoglobin). Secondary structure is the local folding held by hydrogen bonds between backbone C=O and N–H groups, giving the right-handed α-helix (intramolecular H-bonds, as in keratin) or the pleated β-sheet (intermolecular H-bonds between stretched chains, as in silk fibroin). Tertiary structure is the overall 3-D folding of the whole chain (fibrous, e.g. keratin/collagen, or globular, e.g. insulin/enzymes) stabilised by H-bonds, disulphide (–S–S–) bridges, ionic and van der Waals forces. Quaternary structure is the assembly of two or more polypeptide sub-units, e.g. haemoglobin's four chains.

Denaturation. Heat, strong acid/base, heavy-metal ions or alcohol break the H-bonds and other weak forces (but not peptide bonds). The secondary and tertiary structures unravel while the primary sequence stays intact, so the protein loses biological activity — e.g. the coagulation of egg white on boiling or the curdling of milk.

Enzymes are biological catalysts, almost all globular proteins, that speed up cellular reactions by lowering activation energy — e.g. maltase, urease, pepsin. They are remarkably specific: each enzyme acts on one substrate or one type of reaction, often described by the lock-and-key (and induced-fit) model. The substrate binds at the active site, the reaction occurs, and the product is released, regenerating the enzyme. Enzyme names usually end in -ase.

Zwitterion of an amino acid and the peptide bondZwitterion (dipolar ion)H3N+CHRCOO−proton shifts –COOH → –NHPeptide bond (amide)—CO–NH–C—− H2O
Solved Examples
1
Worked Example
Draw and explain the zwitterionic form of glycine. Why does it have a high melting point and good water solubility?
Solution
  1. Glycine is H2N–CH2–COOH.
  2. An internal proton transfer from –COOH to –NH2 gives the zwitterion +H3N–CH2–COO−.
  3. The molecule is now ionic (dipolar), so strong electrostatic forces hold the crystal together.

Answer: The zwitterion +H3N–CH2–COO− behaves like a salt — strong ionic forces raise the melting point, and its charges make it highly soluble in water.

2
Worked Example
Define the isoelectric point and explain its use in separating amino acids.
Solution
  1. At low pH the amino acid is a cation; at high pH it is an anion.
  2. At one intermediate pH the positive and negative charges balance — net charge zero (pure zwitterion).
  3. This pH is the isoelectric point (pI); here the species does not move in an electric field.
  4. Since each amino acid has its own pI, electrophoresis at a chosen pH makes them migrate at different rates and separates them.

Answer: The pI is the pH of zero net charge; because pI differs for each amino acid, electrophoresis separates them.

3
Worked Example
Distinguish between essential and non-essential amino acids with two examples each.
Solution
  1. The body can build some amino acids from other compounds — these are non-essential.
  2. Others cannot be made in the body and must come from food — these are essential.

Answer: Essential (diet-supplied): valine, leucine, lysine. Non-essential (body-synthesised): glycine, alanine.

4
Worked Example
Compare the α-helix and β-pleated sheet secondary structures.
Solution
  1. Both are held by H-bonds between backbone C=O and N–H groups.
  2. In the α-helix the chain coils as a right-handed spiral with intramolecular H-bonds (e.g. wool keratin).
  3. In the β-sheet the chains are stretched and lie side by side, held by intermolecular H-bonds, giving a pleated sheet (e.g. silk fibroin).

Answer: α-helix = coiled, intramolecular H-bonds (keratin); β-sheet = extended parallel chains, intermolecular H-bonds (silk).

5
Worked Example
What is denaturation of a protein? Which structural levels are affected and which is not?
Solution
  1. Heat, acids, bases, heavy-metal ions or alcohol disrupt the weak forces (H-bonds, ionic, disulphide) holding the protein's shape.
  2. The secondary and tertiary structures unfold (the globular protein uncoils).
  3. Peptide bonds are not broken, so the primary sequence is retained.
  4. The unfolded protein loses its biological activity (e.g. boiled egg white).

Answer: Denaturation destroys the secondary and tertiary structures (loss of function) while the primary structure remains intact; coagulation of egg white on boiling is an example.

6
Worked Example
Explain how an enzyme catalyses a reaction and why it is highly specific.
Solution
  1. An enzyme is a globular protein with an active site of definite shape and chemistry.
  2. Only a substrate that fits this site binds (lock-and-key / induced-fit model).
  3. Binding lowers the activation energy, so the reaction proceeds far faster.
  4. The product leaves and the enzyme is regenerated unchanged.

Answer: The substrate binds at a shape-matched active site, which lowers the activation energy; because only the matching substrate fits, the enzyme is highly specific.

Key Points

  • Proteins are polymers of α-amino acids R–CH(NH2)–COOH; classed as neutral/acidic/basic and as essential/non-essential. Natural proteins use L-amino acids.
  • Amino acids exist as zwitterions +H3N–CHR–COO−, are amphoteric, and have a characteristic isoelectric point (pI) of zero net charge.
  • A peptide (amide) bond –CO–NH– forms by condensation of –COOH with –NH2; many such bonds give a polypeptide/protein.
  • Protein structure has four levels: primary (sequence), secondary (α-helix / β-sheet, H-bonds), tertiary (overall 3-D fold), quaternary (assembly of sub-units, e.g. haemoglobin).
  • Denaturation breaks weak forces (not peptide bonds), unfolding 2°/3° structure and destroying activity. Enzymes are globular protein catalysts that lower activation energy and act with high specificity.
Quick Check — 4 MCQs
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Q1.In its zwitterionic form, an amino acid is best represented as:
Explanation: The proton transfers from –COOH to –NH2, giving the dipolar (net-neutral) ion +H3N–CHR–COO−.
Q2.The right-handed coiled secondary structure stabilised by intramolecular hydrogen bonds is the:
Explanation: The α-helix is a right-handed coil held by H-bonds within the same chain; the β-sheet uses intermolecular H-bonds.
Q3.Boiling an egg coagulates its protein. This is an example of:
Explanation: Heat breaks the weak forces holding the 2°/3° structure (denaturation); peptide bonds remain intact, so it is not hydrolysis.
Q4.Which level of protein structure is defined by the sequence of amino acids?
Explanation: The primary structure is the exact order of amino acids joined by peptide bonds in the polypeptide chain.
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