Biochemistry: Carbohydrates, Lipids & Proteins

Complete Study Guide

Topics Covered

1. Carbohydrates and Lipids (B1.1)
2. Proteins (B1.2)
3. HL Extension — Chromatography, Electrophoresis & Biochemical Tests
4. Exam Strategy & Common Mistakes
5. Mixed Practice — Exam Style

Aligned to IB Biology HL 2025 syllabus — B1.1, B1.2

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Section 1: Carbohydrates and Lipids (B1.1)

1.1 Water — The Solvent of Life

Although water ($\text{H}_2\text{O}$) is not a carbohydrate or lipid, its unique properties underpin all biochemistry. IB Biology frequently tests water’s properties and their biological significance.

1.2 Monosaccharides

Monosaccharides are the simplest carbohydrates — single sugar units with the general formula $(\text{CH}_2\text{O})_n$ where $n = 3$ to $7$.

$\alpha$-glucose vs $\beta$-glucose:

Both have the same molecular formula ($\text{C}_6\text{H}_{12}\text{O}_6$) but differ in the orientation of the hydroxyl group ($\text{-OH}$) on carbon 1:

• $\alpha$-glucose: $\text{-OH}$ on carbon 1 is below the ring plane (same side as the $\text{-CH}_2\text{OH}$ group on carbon 6 is NOT — it points down on the opposite side to the $\text{-CH}_2\text{OH}$)
• $\beta$-glucose: $\text{-OH}$ on carbon 1 is above the ring plane

1.3 Disaccharides

Disaccharides form when two monosaccharides join by a condensation reaction (also called a dehydration synthesis), releasing one molecule of water and forming a glycosidic bond.

$\text{monosaccharide} + \text{monosaccharide} \xrightarrow{\text{condensation}} \text{disaccharide} + \text{H}_2\text{O}$

The reverse reaction is hydrolysis — adding water to break the glycosidic bond.

1.4 Polysaccharides

Polysaccharides are polymers of many monosaccharide units linked by glycosidic bonds.

1.5 Lipids — Triglycerides

Lipids are a diverse group of hydrophobic (water-insoluble) molecules. The most important lipids for IB are triglycerides and phospholipids.

A triglyceride consists of one molecule of glycerol bonded to three fatty acid chains by ester bonds (formed via condensation reactions).

$\text{glycerol} + 3 \text{ fatty acids} \xrightarrow{\text{3 condensation reactions}} \text{triglyceride} + 3\text{H}_2\text{O}$

Functions of triglycerides:

• Energy storage — triglycerides store more than twice as much energy per gram as carbohydrates ($\approx 38 \text{ kJ g}^{-1}$ vs $\approx 17 \text{ kJ g}^{-1}$) because they have a higher proportion of C-H bonds
• Thermal insulation — subcutaneous fat reduces heat loss (e.g. blubber in marine mammals)
• Buoyancy — lipids are less dense than water
• Protection — fat pads around organs provide cushioning

1.6 Saturated vs Unsaturated Fatty Acids

1.7 Phospholipids

Phospholipids have the same basic structure as triglycerides, but one fatty acid is replaced by a phosphate group (which may be linked to an additional small molecule like choline).

This gives phospholipids an amphipathic structure:

• Hydrophilic head (phosphate group — polar, interacts with water)
• Hydrophobic tails (two fatty acid chains — non-polar, repelled by water)

In aqueous environments, phospholipids spontaneously form a bilayer — the basis of all biological membranes (the fluid mosaic model).

1.8 BMI and Health Risks (IB Application)

The Body Mass Index (BMI) is used as a simple indicator of whether body mass is within a healthy range:

$\text{BMI} = \frac{\text{mass (kg)}}{\text{height (m)}^2}$

BMI Range	Classification
Below 18.5	Underweight
18.5 — 24.9	Normal
25.0 — 29.9	Overweight
30.0 and above	Obese

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Section 2: Proteins (B1.2)

2.1 Amino Acid Structure

Proteins are polymers of amino acids. All amino acids share the same basic structure:

There are 20 different amino acids used in protein synthesis, each with a different R group (side chain). The R group determines the chemical properties (polar, non-polar, charged, etc.) of each amino acid.

2.2 Peptide Bonds

Amino acids join by condensation reactions to form peptide bonds between the amino group ($\text{-NH}_2$) of one amino acid and the carboxyl group ($\text{-COOH}$) of another, releasing water.

$\text{amino acid}_1 + \text{amino acid}_2 \xrightarrow{\text{condensation}} \text{dipeptide} + \text{H}_2\text{O}$

A chain of many amino acids linked by peptide bonds is called a polypeptide. A functional protein may consist of one or more polypeptide chains.

2.3 Levels of Protein Structure

2.4 R Group Interactions in Tertiary Structure

The tertiary structure is stabilised by interactions between the R groups (side chains) of amino acids:

2.5 Fibrous vs Globular Proteins

Collagen:

• Triple helix of three polypeptide chains wound around each other
• Every third amino acid is glycine (smallest R group, fits inside the helix)
• Hydrogen bonds between chains provide tensile strength
• Found in tendons, ligaments, bone, cartilage, skin

Haemoglobin:

• Quaternary structure: 4 polypeptide subunits (2 $\alpha$-chains, 2 $\beta$-chains)
• Each subunit contains a haem group (prosthetic group) with an iron ($\text{Fe}^{2+}$) ion that binds one $\text{O}_2$ molecule
• Exhibits cooperative binding — binding of $\text{O}_2$ to one subunit increases the affinity of the remaining subunits for $\text{O}_2$

2.6 Proteomics

Proteomics is the study of the complete set of proteins (the proteome) produced by an organism, tissue, or cell at a given time.

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Section 3: HL Extensions — Chromatography, Electrophoresis & Biochemical Tests

3.1 Paper Chromatography

Chromatography separates mixtures based on differential solubility in a mobile phase (solvent) vs a stationary phase (paper).

The $R_f$ value (retardation factor) identifies substances:

$R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent front}}$

3.2 Gel Electrophoresis

Gel electrophoresis separates proteins or DNA fragments by size (and, for proteins, charge) in an electric field through a gel matrix.

How it works:

1. Samples are loaded into wells in a gel (polyacrylamide for proteins; agarose for DNA)
2. An electric field is applied across the gel
3. Molecules migrate through the gel toward the electrode of opposite charge
4. Smaller molecules move faster and further through the gel pores
5. After separation, bands are visualised using staining (e.g. Coomassie blue for proteins, ethidium bromide or SYBR green for DNA)

3.3 Biochemical Tests for Macromolecules

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Section 4: Exam Strategy & Common Mistakes

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Mixed Practice — Exam Style

1. [Carbohydrates] Which of the following is a disaccharide formed from glucose and galactose?

   A. Maltose

   B. Sucrose

   C. Lactose

   D. Cellulose

2. [Protein Structure] The tertiary structure of a protein is primarily determined by:

   A. Peptide bonds between amino acids

   B. Hydrogen bonds between backbone $\text{C=O}$ and $\text{N-H}$ groups

   C. Interactions between R groups (side chains) of amino acids, including hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions

   D. Hydrogen bonds between separate polypeptide chains

3. [Lipids] Which statement correctly describes the difference between saturated and unsaturated fatty acids?

   A. Saturated fatty acids have double bonds between carbon atoms; unsaturated fatty acids have only single bonds

   B. Saturated fatty acids have only single bonds between carbon atoms, making them straight; unsaturated fatty acids have one or more double bonds, creating kinks

   C. Saturated fatty acids are always liquid at room temperature; unsaturated fatty acids are always solid

   D. Saturated and unsaturated fatty acids have identical melting points

4. [Water Properties — Distractor] A student states: “Water has a high specific heat capacity because it is a liquid.” Evaluate this claim:

   A. Correct — all liquids have high specific heat capacity

   B. Incorrect — water has a high specific heat capacity because hydrogen bonds between water molecules require significant energy to break, meaning more heat energy is needed to raise the temperature

   C. Correct — specific heat capacity is a property of the liquid state

   D. Incorrect — water actually has a low specific heat capacity

5. [Condensation and Hydrolysis] Which pair correctly matches the reaction type to the process?

   A. Condensation — digestion of starch to maltose

   B. Hydrolysis — formation of a peptide bond between two amino acids

   C. Condensation — synthesis of a polypeptide from amino acids (removing water)

   D. Hydrolysis — synthesis of a triglyceride from glycerol and fatty acids

6. [Protein Function] Collagen is a fibrous protein. Which structural feature contributes most to its high tensile strength?

   A. Globular shape allowing it to dissolve in blood plasma

   B. Triple-helix structure with hydrogen bonds between three polypeptide chains and staggered covalent cross-links

   C. Quaternary structure with four subunits each containing a haem group

   D. Single polypeptide chain folded into a compact sphere

7. [Biochemical Tests] A solution gives a negative result with Benedict’s test but a positive result after acid hydrolysis followed by Benedict’s test. The solution most likely contains:

   A. Glucose

   B. Starch

   C. Sucrose

   D. A protein

8. [Polysaccharides — Distractor] A student claims that glycogen and cellulose have the same structure because they are both made of glucose. Evaluate this claim:

   A. Correct — both are identical polymers of glucose

   B. Incorrect — glycogen is made of $\alpha$-glucose with $\alpha$-1,4 and $\alpha$-1,6 bonds (highly branched); cellulose is made of $\beta$-glucose with $\beta$-1,4 bonds (straight chains forming microfibrils); the different bond types give completely different structures and functions

   C. Correct — the only difference is that glycogen is found in animals and cellulose in plants

   D. Incorrect — glycogen is made of fructose, not glucose

9. [Gel Electrophoresis] In SDS-PAGE, proteins are separated primarily by:

   A. Their charge, because each protein has a unique charge at physiological pH

   B. Their size (molecular mass), because SDS gives all proteins a uniform negative charge, so smaller proteins migrate further through the gel

   C. Their colour, allowing visual separation without staining

   D. Their primary amino acid sequence, which determines migration speed directly

10. [Proteomics] The human proteome is larger than the human genome (in terms of number of distinct proteins vs number of genes). The best explanation is:

    A. Humans have more proteins than genes because each protein codes for multiple genes

    B. Alternative RNA splicing allows one gene to produce multiple different mRNA sequences and thus multiple different proteins; post-translational modifications further increase protein diversity

    C. Each gene produces exactly one protein, so the proteome and genome are the same size

    D. Mutations increase the number of proteins beyond the number of genes

Show Answers

1. C — Lactose. Lactose is a disaccharide composed of glucose and galactose, linked by a $\beta$-1,4 glycosidic bond. It is found in mammalian milk. Maltose (A) is glucose + glucose. Sucrose (B) is glucose + fructose. Cellulose (D) is a polysaccharide, not a disaccharide.

2. C — R group interactions. Tertiary structure is the overall 3D folding of a single polypeptide, determined by interactions between R groups: hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions, and van der Waals forces. A (peptide bonds) determines primary structure. B (backbone H-bonds) determines secondary structure. D (inter-chain bonds) relates to quaternary structure.

3. B — Saturated fatty acids have only $\text{C-C}$ single bonds, forming straight chains that pack closely (solid at room temperature). Unsaturated fatty acids have one or more $\text{C=C}$ double bonds that create kinks, preventing close packing (liquid at room temperature). A reverses the definitions.

4. B — Water’s high specific heat capacity is due to the extensive hydrogen bonding between water molecules. Breaking these bonds absorbs significant energy before temperature rises. Not all liquids have high specific heat capacity — it is a unique property of water arising from its molecular structure.

5. C — Condensation reactions build polymers by joining monomers and releasing water. Peptide bond formation (joining amino acids) and polysaccharide synthesis are condensation reactions. A and B are reversed — digestion (A) is hydrolysis; peptide bond formation (B) is condensation. D is also reversed.

6. B — Collagen’s tensile strength comes from its triple-helix structure: three polypeptide chains wound around each other, stabilised by hydrogen bonds between chains and staggered covalent cross-links between adjacent collagen molecules. A describes globular proteins. C describes haemoglobin. D describes a generic globular protein.

7. C — Sucrose. Sucrose is a non-reducing sugar (negative with Benedict’s). Acid hydrolysis breaks it into glucose and fructose (both reducing sugars), which then test positive with Benedict’s. Glucose (A) would test positive without hydrolysis. Starch (B) would test positive with iodine, and its hydrolysis products would also be positive with Benedict’s, but starch itself gives a negative Benedict’s result — the question specifically describes a non-reducing sugar behaviour. Proteins (D) are not detected by Benedict’s test.

8. B — Although both are glucose polymers, glycogen uses $\alpha$-glucose ($\alpha$-1,4 and $\alpha$-1,6 bonds) forming a highly branched structure for rapid glucose mobilisation, while cellulose uses $\beta$-glucose ($\beta$-1,4 bonds) forming straight chains with inter-chain hydrogen bonds that create strong microfibrils. The different isomers and bond types produce entirely different structures and functions.

9. B — In SDS-PAGE, SDS denatures proteins and coats them with a uniform negative charge proportional to their length. This means proteins are separated by size only — smaller proteins encounter less resistance in the gel matrix and migrate further. A would apply to native (non-denaturing) electrophoresis but not SDS-PAGE.

10. B — Alternative RNA splicing allows different exon combinations from the same pre-mRNA to produce different mature mRNAs (and thus different polypeptides). Post-translational modifications (phosphorylation, glycosylation, cleavage) further diversify the protein products. The human genome has approximately 20,000 protein-coding genes, but the proteome contains many more distinct proteins. A reverses the gene-protein relationship. C is incorrect because of splicing.

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May 2026 Prediction Questions

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Question 1 — Protein Structure and Function [8 marks]

(a) Distinguish between the primary, secondary, tertiary, and quaternary structures of proteins, including the bonds or interactions responsible for each level. [4 marks]

(b) Explain how the tertiary structure of an enzyme determines its specificity for its substrate. [2 marks]

(c) A student heats an enzyme to 80 degrees C and observes that activity falls to zero. Explain this result in terms of protein structure. [2 marks]

Model Answer:

(a) Primary structure: the sequence of amino acids in the polypeptide chain, held together by peptide bonds formed by condensation reactions (1). Secondary structure: local regular folding into alpha-helices or beta-pleated sheets, stabilised by hydrogen bonds between backbone C=O and N-H groups (1). Tertiary structure: overall 3D shape of a single polypeptide, determined by interactions between R groups — hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions (1). Quaternary structure: two or more polypeptide subunits held together by the same types of interactions as tertiary structure, e.g. haemoglobin (four subunits) (1).

(b) The tertiary structure creates a uniquely shaped active site (1); only a substrate with a complementary shape (and complementary charge distribution) can form an enzyme-substrate complex — this is the basis of enzyme specificity (induced fit model: the active site may adjust slightly on substrate binding) (1).

(c) At 80 degrees C, thermal energy breaks the hydrogen bonds and other weak interactions that maintain tertiary structure (1); the enzyme denatures — the active site changes shape and can no longer bind the substrate, so no enzyme-substrate complexes form and activity falls to zero (1).

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Question 2 — Polysaccharide Structure and Function [7 marks]

(a) Compare the structures of starch (amylose), glycogen, and cellulose in terms of the monomer, glycosidic bond type, and degree of branching. [3 marks]

(b) Relate the structural differences in (a) to the different functions of each polysaccharide. [3 marks]

(c) Iodine solution turns blue-black in the presence of starch but not cellulose. Suggest a reason for this. [1 mark]

Model Answer:

(a)

	Starch (amylose)	Glycogen	Cellulose
Monomer	$\alpha$-glucose	$\alpha$-glucose	$\beta$-glucose
Bond type	$\alpha$-1,4 glycosidic	$\alpha$-1,4 and $\alpha$-1,6	$\beta$-1,4 glycosidic
Branching	Unbranched (coiled helix)	Highly branched	Unbranched (straight chains)

Award 1 mark per correctly completed comparison row, max 3.

(b) Starch: stored in plant cells in a coiled, compact form that is insoluble and does not affect osmosis; gradual hydrolysis releases glucose for energy (1). Glycogen: highly branched structure in animal liver and muscle cells provides many chain ends for rapid glucose release during exercise or between meals (1). Cellulose: straight, unbranched chains form hydrogen bonds between parallel strands, creating strong microfibrils that provide structural support in plant cell walls (1).

(c) Iodine molecules are trapped inside the helical coil structure of amylose, forming a blue-black complex (1). Cellulose is straight and does not form a helix, so iodine cannot be trapped and no colour change occurs.

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Question 3 — The Role of Water in Living Systems [6 marks]

(a) Explain why water is described as a polar molecule and how this gives rise to hydrogen bonding between water molecules. [2 marks]

(b) Describe and explain three properties of water that are important to living organisms, linking each property to its biological significance. [4 marks]

Model Answer:

(a) Oxygen is more electronegative than hydrogen; electrons are pulled towards oxygen, creating a partial negative charge ($\delta^-$) on the oxygen and partial positive charges ($\delta^+$) on the hydrogen atoms — this makes the molecule polar (1). The $\delta^+$ hydrogen of one water molecule is attracted to the $\delta^-$ oxygen of a neighbouring molecule, forming a hydrogen bond; each water molecule can form up to four hydrogen bonds (1).

(b) Any three of the following, with correct explanation and biological significance (1 + 1 each, maximum 4 marks):

• High specific heat capacity: Many hydrogen bonds must be broken to raise temperature; water resists temperature changes (1), buffering aquatic organisms against rapid temperature fluctuations and helping mammals maintain a stable core body temperature (1).
• High latent heat of vaporisation: Significant energy is needed to convert liquid water to vapour (1); sweat evaporation absorbs large amounts of heat from the skin, providing effective cooling in endotherms (1).
• Cohesion and surface tension: Hydrogen bonds between water molecules create cohesive forces (1); this allows water to form a continuous column in xylem vessels (cohesion-tension mechanism) for transport in tall plants, and provides a surface for small organisms to walk on (1).
• Excellent solvent: The polar water molecule surrounds and separates ions and polar molecules (1); metabolites, hormones, and nutrients dissolve in water, allowing blood plasma and cytoplasm to transport substances throughout the organism (1).
• Maximum density at 4 degrees C / ice less dense than liquid water: Ice floats on water due to the open lattice structure of hydrogen bonds in ice (1); this insulates bodies of water in winter, allowing aquatic life to survive beneath the ice layer (1).

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IB Biology HL — Biochemistry: Carbohydrates, Lipids & Proteins — Complete Study Guide — 2025 Syllabus — Good luck!