Biomolecules
Carbohydrates (monosaccharides, stereoisomerism, mutarotation, glycosidic bond, disaccharides, polysaccharides), lipids (fatty acids, triglycerides, saponification, phospholipids, steroids), proteins (amino acids, peptide bond, primary/secondary/tertiary/quaternary structure, enzymes), nucleic acids (nucleotides, DNA, RNA, double helix structure, replication, transcription).
Complete Theory
5Carbohydrates (saccharides) are polyhydroxy compounds with general formula Cₙ(H₂O)ₘ. They are classified into monosaccharides (simple sugars), disaccharides (2 units), oligosaccharides (3–10), and polysaccharides (>10).
Monosaccharides: they are aldehydes or ketones with at least 2 —OH groups. They have the formula CₙH₂ₙOₙ.
Cyclization and mutarotation: aldoses with 4+ C form cyclic hemiacetals by intramolecular addition of OH to C=O. D-glucose in water exists as a mixture of:
Real-world applications: glucose is the main cellular energy source (glycolysis). Ribose is the sugar in RNA, deoxyribose in DNA. Artificial sweeteners (saccharin, aspartame) mimic the sweet taste of sugars.
Monosaccharides: they are aldehydes or ketones with at least 2 —OH groups. They have the formula CₙH₂ₙOₙ.
- Aldoses: aldehyde carbonyl group (—CHO) at position 1. Examples: D-glucose (aldohexose), D-ribose (aldopentose).
- Ketoses: ketone carbonyl (>C=O) at position 2. Example: D-fructose (ketose).
Cyclization and mutarotation: aldoses with 4+ C form cyclic hemiacetals by intramolecular addition of OH to C=O. D-glucose in water exists as a mixture of:
- α-D-glucopyranose (36%): anomeric OH in axial position (down in Haworth representation)
- β-D-glucopyranose (64%): anomeric OH in equatorial position (up)
- Open chain form (trace)
Real-world applications: glucose is the main cellular energy source (glycolysis). Ribose is the sugar in RNA, deoxyribose in DNA. Artificial sweeteners (saccharin, aspartame) mimic the sweet taste of sugars.
The glycosidic bond forms by condensation between the anomeric OH (C1) of one monosaccharide and an —OH (or —NH) group of another molecule, with elimination of H₂O. The bond can be α or β depending on the anomeric configuration.
Disaccharides:
Disaccharides:
- Sucrose (table sugar): α-D-glucopyranosyl-(1→2)-β-D-fructofuranoside. α(1→2) bond between glucose and fructose. No free anomeric OH → non-reducing.
- Lactose (milk sugar): β-D-galactopyranosyl-(1→4)-D-glucopyranose. β(1→4) bond. Has a free anomeric OH on glucose → reducing. Lactose-intolerant individuals lack the enzyme lactase.
- Maltose (malt sugar): α-D-glucopyranosyl-(1→4)-D-glucopyranose. α(1→4) bond. Reducing.
- Starch: mixture of amylose (linear, α(1→4) bonds) and amylopectin (branched, α(1→4) and α(1→6) bonds). It is the storage polymer of plants.
- Glycogen: highly branched amylopectin. Storage polymer of animals (liver, muscles).
- Cellulose: linear polymer of β-D-glucose with β(1→4) bonds. The β bonds give a rigid ribbon structure. Main component of plant cell wall. Mammals lack cellulase (enzyme that hydrolyzes β bonds) → cellulose is indigestible fiber.
- Chitin: polymer of N-acetylglucosamine with β(1→4) bonds. Component of arthropod exoskeleton.
Lipids are hydrophobic biological molecules (soluble in organic solvents, insoluble in water). Main classes are:
Fatty acids: long-chain carboxylic acids (C12–C24). They are divided into:
Phospholipids: similar to triglycerides but with a phosphate group (and an amine/alcohol) at position 3 of glycerol. Phospholipids constitute the cell membrane bilayer: hydrophobic tails inward, hydrophilic heads outward (toward the aqueous environment).
Steroids: compounds with a 4-ring fused structure (3 six-membered + 1 five-membered). Cholesterol is the precursor of steroid hormones (testosterone, estrogens, cortisol) and vitamin D.
Real-world applications: excess saturated fat increases LDL (bad) cholesterol, cardiovascular risk. Unsaturated fats (olive oil) increase HDL (good) cholesterol. Trans fatty acids (industrial hydrogenated fats) are particularly harmful. Saponification is the soap-making process.
Fatty acids: long-chain carboxylic acids (C12–C24). They are divided into:
- Saturated: no double bonds (e.g., palmitic acid C₁₆H₃₂O₂, stearic acid C₁₈H₃₆O₂). Solid at room temperature.
- Unsaturated: 1+ double bond (e.g., oleic acid C₁₈H₃₄O₂, cis-9). The cis double bond introduces a kink that prevents ordered packing → liquids at RT (oils).
- Essential: the body cannot synthesize them (linoleic and linolenic acids, omega-6 and omega-3).
Phospholipids: similar to triglycerides but with a phosphate group (and an amine/alcohol) at position 3 of glycerol. Phospholipids constitute the cell membrane bilayer: hydrophobic tails inward, hydrophilic heads outward (toward the aqueous environment).
Steroids: compounds with a 4-ring fused structure (3 six-membered + 1 five-membered). Cholesterol is the precursor of steroid hormones (testosterone, estrogens, cortisol) and vitamin D.
Real-world applications: excess saturated fat increases LDL (bad) cholesterol, cardiovascular risk. Unsaturated fats (olive oil) increase HDL (good) cholesterol. Trans fatty acids (industrial hydrogenated fats) are particularly harmful. Saponification is the soap-making process.
Proteins are polymers of α-amino acids (20 common, all L). Each amino acid has the general formula H₂N—CH(R)—COOH, with R (side chain) variable. The α carbon is chiral (except Glycine, R=H).
Amino acid classes:
Protein structure:
Real-world applications: protein denaturation (heat, pH, salts) destroys 2°–4° structures (e.g., cooking egg white). Drugs like aspirin inhibit enzymes (COX-1/COX-2) by blocking the active site. Recombinant proteins (insulin, monoclonal antibodies) are biotechnological products.
Amino acid classes:
- Polar charged +: Lysine (Lys), Arginine (Arg), Histidine (His) — basic
- Polar charged −: Aspartic acid (Asp), Glutamic acid (Glu) — acidic
- Polar neutral: Serine (Ser), Threonine (Thr), Asparagine (Asn), Glutamine (Gln)
- Nonpolar: Alanine (Ala), Valine (Val), Leucine (Leu), Isoleucine (Ile), Proline (Pro), Phenylalanine (Phe), Tryptophan (Trp), Methionine (Met)
- Special: Cysteine (Cys, forms disulfide bridges —S—S—)
Protein structure:
- Primary (1°): linear sequence of amino acids, genetically determined.
- Secondary (2°): regular local arrangement, stabilized by H-bonds between C=O and N—H of the backbone. α-helix: 3.6 residues/turn, H-bonds between residue n and n+4. β-sheet: aligned parallel/antiparallel chains, H-bonds between adjacent chains.
- Tertiary (3°): global folding of the polypeptide chain, stabilized by hydrophobic interactions, H-bonds, ionic bonds, disulfide bridges (Cys—Cys).
- Quaternary (4°): association of multiple subunits (e.g., hemoglobin: 4 subunits, 2α + 2β).
Real-world applications: protein denaturation (heat, pH, salts) destroys 2°–4° structures (e.g., cooking egg white). Drugs like aspirin inhibit enzymes (COX-1/COX-2) by blocking the active site. Recombinant proteins (insulin, monoclonal antibodies) are biotechnological products.
Nucleic acids (DNA and RNA) are polymers of nucleotides. Each nucleotide is composed of:
DNA (deoxyribonucleic acid): double helix structure (Watson & Crick, 1953):
DNA replication: semi-conservative process. Each strand serves as a template for synthesis of the new complementary strand. DNA polymerase adds nucleotides in the 5'→3' direction, using dNTP (deoxynucleoside triphosphates).
Transcription and translation: transcription copies a gene into mRNA (RNA polymerase). Translation (protein synthesis) occurs on ribosomes: each codon (base triplet of mRNA) specifies an amino acid (via tRNA with complementary anticodon). The genetic code is degenerate (18 of the 20 amino acids are encoded by 2+ codons).
- Nitrogenous base: purine (adenine A, guanine G) or pyrimidine (cytosine C, thymine T in DNA, uracil U in RNA)
- Sugar: D-ribose (RNA) or 2-deoxy-D-ribose (DNA)
- Phosphate group: bonded to C5' of the sugar via a phosphoester bond
DNA (deoxyribonucleic acid): double helix structure (Watson & Crick, 1953):
- Two antiparallel polynucleotide chains (5'→3' and 3'→5') helically wound.
- Nitrogenous bases inside, paired via hydrogen bonds: A=T (2 H bonds), G≡C (3 H bonds). Complementary base pairing is specific and geometrically explained (purine + pyrimidine = constant width).
- The backbone is sugar-phosphate, hydrophilic, on the outside.
- One complete turn: 10.5 base pairs, pitch 34 Å, diameter 20 Å.
DNA replication: semi-conservative process. Each strand serves as a template for synthesis of the new complementary strand. DNA polymerase adds nucleotides in the 5'→3' direction, using dNTP (deoxynucleoside triphosphates).
Transcription and translation: transcription copies a gene into mRNA (RNA polymerase). Translation (protein synthesis) occurs on ribosomes: each codon (base triplet of mRNA) specifies an amino acid (via tRNA with complementary anticodon). The genetic code is degenerate (18 of the 20 amino acids are encoded by 2+ codons).
Worked Examples
2Example 1Mutarotation of D-glucose
Given
Freshly dissolved D-glucose in water has optical rotation [α]ᴅ = +112°.
At equilibrium [α]ᴅ = +52.7°.
Pure β-D-glucopyranose has [α]ᴅ = +18.7°.
Pure α-D-glucopyranose has [α]ᴅ = +112°.
Find
Equilibrium composition (% α and % β)
Step-by-step solution
1Mutarotation is the equilibrium between α and β forms through the open-chain form (trace). The equilibrium value is the weighted average of the two forms.
2If x = fraction of α, then 1−x = fraction of β: 112x + 18.7(1−x) = 52.7.
3Solve: 112x + 18.7 − 18.7x = 52.7 → 93.3x = 34.0 → x = 0.364 (36.4% α).
4Thus 63.6% β + 36.4% α + traces of open-chain form.
✓ Final result: 36% α, 64% β (at equilibrium)
Example 2Saponification: soap yield calculation
Given
100 g of tristearin (C₅₇H₁₁₀O₆, MW 890 g/mol) are saponified with excess NaOH.
Find
Grams of sodium soap (sodium stearate, C₁₈H₃₅O₂Na, MW 306 g/mol) produced
Step-by-step solution
1Reaction: tristearin + 3NaOH → glycerol + 3 sodium stearate. 1 mol of tristearin gives 3 mol of soap.
2Moles of tristearin: 100 g / 890 g/mol = 0.1124 mol.
3Moles of sodium stearate: 0.1124 × 3 = 0.3372 mol.
4Mass: 0.3372 mol × 306 g/mol = 103.2 g of soap.
✓ Final result: 103.2 g of sodium stearate
Exercises with Solutions
3Exercise 1CarbohydratesMedium
Problem to solve
Is sucrose a reducing sugar? Explain.
Given data
Sucrose: glucose + fructose, α(1→2) bond
Step-by-step solution
1The glycosidic bond in sucrose is between C1 of glucose and C2 of fructose.
2Both anomeric carbons (C1 of glucose, C2 of fructose) are involved in the bond.
3Neither monosaccharide has a free anomeric OH → sucrose cannot open to the linear form with a free C=O.
4Consequence: sucrose does not reduce Tollens reagent (silver mirror) nor Fehling reagent (Cu²⁺ → Cu₂O).
✓ Final answer: No, it is non-reducing (both anomeric OH are engaged in the glycosidic bond)
Exercise 2ProteinsHard
Problem to solve
Describe the bonds/interactions that stabilize the tertiary structure of proteins and give an example of each.
Given data
Tertiary structure: global folding of a polypeptide chain
Step-by-step solution
1Hydrophobic interactions: nonpolar side chains (Leu, Val, Ile, Phe) tend to aggregate inside the protein, away from water. Example: the hydrophobic core of myoglobin.
2Hydrogen bonds: between polar side chains (Ser—OH with Asp—COO⁻ or between backbone). Example: in the active site of enzymes.
3Ionic bonds (salt bridges): between charged side chains (e.g., Lys⁺ with Asp⁻). Example: in thrombin.
4Disulfide bridges (—S—S—): covalent bond between two cysteine residues. Example: stabilization of insulin (chains A and B linked by 2 disulfide bridges).
✓ Final answer: Hydrophobic, H-bonds, ionic, disulfide bridges (examples: myoglobin/insulin)
Exercise 3DNAVery Hard
Problem to solve
Given the DNA sequence 5'-AGCTTAG-3', write the complementary sequence and explain why the two strands are antiparallel. Indicate the total number of H-bonds between the two sequences.
Given data
A=T (2 H), G≡C (3 H)
Step-by-step solution
1The complementary sequence is 3'-TCGAATC-5' (written 5'→3': 5'-CTAAGCT-3'). A pairs with T (2 H each), G with C (3 H each).
2The strands are antiparallel because the phosphodiester bonds run in opposite directions (5'→3' vs 3'→5'), necessitated by the orientation of H-bonds and helix geometry.
3H-bonds: A=T: 3 pairs × 2 H = 6 H (A1-T6, T2-A5, T6-A1). G≡C: 3 pairs × 3 H = 9 H (C4-G3, G3-C4). TOTAL = 15 H-bonds for the 7 base pairs.
4GC content determines the thermal stability of DNA (3 H vs 2 H).
✓ Final answer: 5'-CTAAGCT-3', 15 H-bonds (3×2 + 3×3)
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