IB DP Biology: Carbohydrates and lipids: macromolecules, sugars, fats and phospholipids (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology topic B1.1 (Carbohydrates and lipids), statements B1.1.1 to B1.1.13, from the IB Biology guide, first assessment 2025. All of it is SL and HL. For full explanations read the study guide, then test yourself with the practice set.

Course pages: IB DP Biology hub, printable checklist, Theme B: Form and function and free 10-minute diagnostics.

Key definitions

Term	Meaning
Covalent bond	Two atoms held together by a common electron pair
Monomer / polymer	A small unit / a long chain of linked units
Condensation	Joining two molecules with the release of water
Hydrolysis	Breaking a bond by splitting water into -H and -OH
Monosaccharide	A single sugar unit, a pentose (5 C) or hexose (6 C)
Lipid	Dissolves in non-polar solvents, only sparingly in water
Amphipathic	Having both a hydrophilic and a hydrophobic region
Glycoprotein	A protein with attached carbohydrate chains

B1.1.1 Carbon

• Four covalent bonds per carbon atom: single bonds, or single plus double bonds.
• Partners: C, H, O, N, P, S.
• Skeletons: unbranched chain, branched chain, single ring, multiple rings.
• SI prefixes: kilo 10³, centi 10⁻², milli 10⁻³, micro 10⁻⁶, nano 10⁻⁹.

Method: convert units

1. Write the prefix as a power of ten.
2. Multiply or divide by the factor between the units.
3. Check the size makes sense (a bilayer of 7 nm is 0.007 µm, not 7000 µm).

B1.1.2 and B1.1.3 Condensation and hydrolysis

Polymer	Monomer	Bond
Polysaccharide	monosaccharide	glycosidic
Polypeptide	amino acid	peptide
Nucleic acid	nucleotide	phosphodiester

• An unbranched chain with n monomers contains n − 1 bonds: that many waters are released on building and used on breaking.
• Worked reminder: three glucose units joined in a line release 2 waters: 3 × 180.16 − 2 × 18.02 = 504.44.
• Digestion is hydrolysis: large polymers become monomers small enough to absorb.

B1.1.4 Monosaccharides

• Count all carbons, including the one outside the ring.
• Pentose examples: ribose, deoxyribose. Hexose examples: glucose, fructose, galactose.
• Glucose: soluble (many -OH groups), transportable, chemically stable, yields energy on oxidation.

B1.1.5 and B1.1.6 Polysaccharides

Feature	Starch	Glycogen	Cellulose
Found in	plants	animals	plant cell walls
Monomer	alpha-glucose	alpha-glucose	beta-glucose
Shape	coiled; amylopectin branched	coiled and highly branched	straight chains
Function	energy store	energy store, quickly mobilised	structural strength

Why storage polysaccharides suit storage

• Coiling and branching make them compact.
• Large molecules are relatively insoluble, so no osmotic effect.
• Units added or removed easily by condensation and hydrolysis.
• Branching gives many ends, so many units are released at once.

Why cellulose is strong

• Beta-glucose units alternate in orientation, giving straight chains.
• Chains lie in bundles and are cross-linked by hydrogen bonds.
• Many weak bonds add up to high tensile strength.

B1.1.7 Glycoproteins

• Carbohydrate chains face outwards from the plasma membrane and act as recognition markers.
• ABO: the antigen on red blood cells depends on the terminal sugar; group O cells lack A and B antigens.
• A person has antibodies against ABO antigens they do not carry.

B1.1.8 to B1.1.11 Lipids and fats

Compound	Built from	Water released
Triglyceride	glycerol + 3 fatty acids	3
Phospholipid	glycerol + 2 fatty acids + phosphate	3

Fatty acid type	C=C bonds	Effect
Saturated	none	straight, high melting point
Monounsaturated	one	bent, lower melting point
Polyunsaturated	two or more	more bends, lowest melting point

• Plants: storage oils, rich in unsaturated fatty acids, liquid.
• Endotherms: storage fats richer in saturated fatty acids, firmer at body temperature.
• Energy density (data given in questions): lipid about 37 kJ per gram, carbohydrate about 17 kJ per gram, ratio 2.18.
• Fat = long-term store. Glycogen = short-term store.
• Insulation: poor heat conductor; matters most for endotherms in cold habitats, such as marine mammals in cold water.

Method: compare two stores

1. Divide the energy required by the energy per gram of each store.
2. Compare masses (or divide the larger by the smaller).
3. Add the biological reason: insolubility, no osmotic effect, compact.

B1.1.12 and B1.1.13 Bilayers and steroids

• Heads hydrophilic (face water), tails hydrophobic (face inwards): the bilayer forms spontaneously.
• Core blocks ions and large polar molecules.
• Steroids have four fused rings (three of six carbons, one of five) and are non-polar.
• Oestradiol and testosterone cross the bilayer directly.

Worked reminders

Magnification. Actual size = image size ÷ magnification. A bilayer measured as 21 mm on an image at ×3 000 000 is 21 mm ÷ 3 000 000 = 7 × 10⁻⁶ mm = 7 nm.

Mass of a cellulose chain. Each glucose unit in a chain contributes C6H10O5, 162.14, because one water is lost per bond. A chain of 5000 units has a mass of about 811 000 (five thousand times 162.14, plus one water).

Fat reserve. A 70 kg person with 18% body fat holds 70 × 0.18 = 12.6 kg of fat. At 37 kJ per gram this is 12 600 × 37 = 466 200 kJ. At a use of 10 000 kJ per day it would last 46.6 days if it were the only fuel. A glycogen store of 400 g holds 400 × 17 = 6800 kJ, under one day of that use. This is the quantitative reason fat is the long-term store.

Melting point trend. Stearic acid (0 C=C) melts at about 69 °C and oleic acid (1 C=C) at about 13 °C: a fall of 56 °C for one double bond. State the trend first, then explain it with bent chains and weaker packing.

Command words in this topic

• State / Outline: a short answer or a brief account, with no explanation needed.
• Distinguish: give differences between two items, side by side, not separate descriptions.
• Explain: give a cause as well as a fact, for example “relatively insoluble because the molecule is large, so no osmotic effect”.
• Compare: give similarities and differences, such as starch and glycogen being alpha-glucose polymers but glycogen being more highly branched.
• Calculate: show the working and the unit.

Exam habits for B1.1

• Name the bond (glycosidic, peptide, phosphodiester, ester) when asked about a condensation reaction.
• When a question gives a table, quote paired values to support a trend.
• Link every structural feature to a function in the same sentence.
• Write the unit after each calculation and show the rearranged formula first.

Must-know distinctions

• Condensation vs hydrolysis: water released vs water used.
• Alpha vs beta glucose: -OH on carbon 1 down vs up; coiled stores vs straight fibres.
• Starch vs glycogen: plants, less branched vs animals, more branched.
• Saturated vs unsaturated: fully single-bonded chain vs chain with a double bond or more.
• Triglyceride vs phospholipid: three fatty acids vs two plus phosphate.
• Hydrophilic vs hydrophobic: attracted to water vs repelled from it.

Quick self-test

1. State how many covalent bonds a carbon atom can form.
2. Convert 2.5 mm to micrometres.
3. A polypeptide chain contains 120 amino acids in one chain. How many peptide bonds does it have?
4. State what is added to each end of a broken bond in hydrolysis.
5. Distinguish a pentose from a hexose.
6. State two reasons why glycogen is a good short-term energy store.
7. Explain why cellulose chains form strong fibres.
8. Name the carbohydrate-protein molecules that carry ABO antigens.
9. Calculate the mass of lipid that stores 5000 kJ, using 37 kJ per gram. Give the answer to 3 significant figures.
10. Explain why a fatty acid with two C=C bonds melts at a lower temperature than a saturated one.
11. Define amphipathic.
12. Explain why testosterone crosses a membrane without a channel.

Answers

1. Four.
2. 2.5 mm = 2500 µm.
4. -H on one end and -OH on the other, from a split water molecule.
5. Pentose has five carbons; hexose has six.
6. Compact because of coiling and branching; relatively insoluble so no osmotic effect; many ends allow fast release (any two).
7. Straight chains of beta-glucose lie in bundles, cross-linked by many hydrogen bonds.
8. Glycoproteins.
9. 5000 ÷ 37 = 135 g.
10. The C=C bonds bend the chain, so molecules pack less closely and need less energy to separate.
11. It has both a hydrophilic and a hydrophobic region.
12. It is a non-polar steroid and dissolves in the hydrophobic core of the bilayer.

Where marks are usually lost

• Saying water is “added” in condensation, or not stating that hydrolysis splits it.
• Using “dissolves” for starch without saying why large size makes it relatively insoluble.
• Claiming cellulose is “a different shape” without naming alternating beta-glucose orientation and hydrogen bonds.
• Leaving out units, or a prefix error such as nm to µm by a factor of 100.
• Describing phospholipid heads as hydrophobic.
• Stating “unsaturated” without referring to C=C bonds.
• Giving the ratio of energy density without the biological consequence.
• Treating ABO antibodies as made against the person’s own antigens.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Topic B1.1, Carbohydrates and lipids.