IB DP Biology: Carbohydrates and lipids: macromolecules, sugars, fats and phospholipids (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise 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. There is no higher level only content in B1.1, so every question suits SL and HL. Data values are supplied in the questions.

Learn the content first in the study guide and the revision notes. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

Questions

1. Outline two properties of carbon atoms that allow the formation of a very large number of different compounds. [2]

2. An electron micrograph shows a cell surface membrane. The membrane measures 28 mm across on the image and the magnification is ×4 000 000.

(a) Calculate the actual thickness of the membrane in nanometres. [2] (b) Express your answer to (a) in micrometres. [1]

3. Glucose and fructose (both C6H12O6, relative molecular mass 180.16) join to form sucrose, and the water released has a relative molecular mass of 18.02.

(a) Name the type of reaction and the bond formed. [2] (b) Calculate the relative molecular mass of sucrose. [2] (c) Outline what happens to the water molecule when sucrose is digested. [2]

4. Molecule X has a ring made of four carbon atoms and one oxygen atom, with a fifth carbon atom attached to the ring as a side group. Molecule Y has five carbons and an oxygen in its ring, plus a sixth carbon as a side group.

(a) Classify X and Y as pentose or hexose. [2] (b) State one example of each. [2]

5. Explain how four properties of glucose make it suitable as the main carbohydrate fuel carried in the blood. [4]

6. Explain how the structure of glycogen makes it suitable as an energy store in animals. [4]

7. Cellulose and starch are both polymers of glucose. Explain how the structure of cellulose suits its function in the plant cell wall. [4]

8. Red blood cells carry ABO antigens on their surface.

(a) State the type of molecule that carries these antigens. [1] (b) A patient with blood group A is mistakenly given group B blood. Explain why this is dangerous. [3] (c) Suggest why blood group O red cells can be given to patients of other groups in an emergency. [2]

9. The table shows melting points of four fatty acids, each with 18 carbon atoms. The values are supplied for this question.

Fatty acid	C=C bonds	Melting point / °C
Stearic	0	69
Oleic	1	13
Linoleic	2	−5
Linolenic	3	−11

(a) Describe the relationship between the number of C=C bonds and melting point. [2] (b) Explain the relationship. [3] (c) State whether oleic acid is solid or liquid at a human body temperature of 37 °C. [1] (d) Suggest why many plant seeds store energy as oils rich in unsaturated fatty acids, while mammals store fats richer in saturated fatty acids. [2]

10. A hibernating mammal stores 400 g of triglyceride. Its energy use is 150 kJ per day. Lipid yields 37 kJ per gram and carbohydrate 17 kJ per gram (data supplied).

(a) Calculate the energy stored in the fat. [1] (b) Calculate how many days the fat store lasts, to the nearest day. [2] (c) Calculate the mass of carbohydrate needed to store the same energy, to the nearest gram. [2] (d) Give two properties of triglycerides, other than energy density, that suit long-term storage. [2] (e) Explain why a thick layer of fat is more useful for an Arctic seal than for a tropical frog. [2]

11. Phospholipids and triglycerides are both made from glycerol and fatty acids.

(a) State the number of water molecules released when one triglyceride is formed from glycerol and fatty acids, and when one phospholipid is formed from glycerol, two fatty acids and a phosphate group. [2] (b) Explain, using the term amphipathic, why phospholipids form a bilayer in water. [3] (c) The table shows how easily three substances cross an artificial phospholipid bilayer (illustrative values, arbitrary units).

Substance	Relative rate of crossing
Testosterone	95
Glucose	0.2
Sodium ions	0.001

Explain the differences in rate. [3] (d) Testosterone is a steroid. State how a steroid can be recognised from its molecular diagram. [1]

Answers

1. A carbon atom forms up to four covalent bonds [1]. It can bond to other carbon atoms in chains and rings, and to other non-metals such as hydrogen, oxygen and nitrogen, with single or double bonds [1]. [2] Examiner insight: A bare “carbon can form four bonds” gives only one idea; the second point needs the bonding to other carbon atoms or the variety of skeletons.

2. (a) Actual size = image size ÷ magnification [1]. 28 mm ÷ 4 000 000 = 7 × 10⁻⁶ mm = 7 nm [1]. (b) 7 nm ÷ 1000 = 0.007 µm [1]. [3] Examiner insight: Show the rearranged formula and convert mm to nm explicitly; a correct final value is easier to credit when the working is visible.

3. (a) Condensation [1]; glycosidic bond [1]. (b) 180.16 + 180.16 − 18.02 [1] = 342.30 [1]. (c) A water molecule is split [1] into -H and -OH, which are added to the two monosaccharides (hydrolysis) [1]. [6] Examiner insight: Answers on hydrolysis gain credit when they say water is split and the -H and -OH are incorporated; “water is used” is incomplete.

4. (a) X is a pentose [1] (five carbons in total); Y is a hexose [1] (six carbons in total). (b) X: ribose or deoxyribose [1]. Y: glucose, fructose or galactose [1]. [4] Examiner insight: Count every carbon, including the one outside the ring; counting only ring carbons reverses both answers.

5. Soluble, because the many -OH groups form hydrogen bonds with water, so it dissolves in plasma [1]. Transportable, since it is dissolved and small, so it moves in the blood and into cells [1]. Chemically stable, so it is not altered while being carried [1]. Oxidation in respiration releases energy for ATP production [1]. [4] Examiner insight: Each property needs its link to the function; a list of four property names without the consequence tends to earn little.

6. Glycogen is a polymer of alpha-glucose that is coiled and highly branched, which makes it compact [1]. Its large molecular size makes it relatively insoluble, so it does not affect the water balance of the cell by osmosis [1]. Glucose units are added by condensation and removed by hydrolysis [1]. The many branch ends allow many units to be mobilised at once for rapid energy release [1]. [4] Examiner insight: Mention of branching earns most credit when tied to either compactness or the number of free ends.

7. Cellulose is made of beta-glucose, with alternate units inverted [1], giving straight, unbranched chains [1]. Chains lie side by side in bundles cross-linked by hydrogen bonds [1]. Many hydrogen bonds together give high tensile strength to the wall, which resists the cell bursting [1]. [4] Examiner insight: “Hydrogen bonds between chains” is the key link; stating that the bonds are individually weak but numerous shows understanding rather than recall.

8. (a) Glycoprotein [1]. (b) The patient has antibodies against B antigens [1]. These bind the B antigens on the donated cells [1] and cause clumping of the red blood cells, which can block blood vessels [1]. (c) Group O cells lack A and B antigens [1], so the recipient’s antibodies have nothing to bind to [1]. [6] Examiner insight: State who has the antibodies and what they bind; answers that blame “the wrong blood type” without naming antigens and antibodies often do not progress.

9. (a) As the number of C=C bonds increases, the melting point decreases [1]. Quote data, for example stearic acid with none melts at 69 °C but linolenic acid with three melts at −11 °C [1]. (b) C=C bonds put bends in the hydrocarbon chain [1], so molecules cannot pack closely [1] so the molecules separate with a smaller energy input [1]. (c) Liquid, because its melting point (13 °C) is below 37 °C [1]. (d) Plant oils stay liquid at ambient temperatures so can be mobilised in a seed [1]. Mammals’ body temperature is high enough for fats richer in saturated acids to remain firm in adipose tissue [1]. [8] Examiner insight: For a “describe” question, a trend plus data quoted from the table usually gains more than a trend alone.

10. (a) 400 × 37 = 14 800 kJ [1]. (b) 14 800 ÷ 150 [1] = 98.7, so 99 days [1]. (c) 14 800 ÷ 17 [1] = 871 g [1]. (d) Insoluble, so no osmotic effect [1]; stable over long periods [1]. (Compact storage in adipose tissue is also acceptable.) (e) Fat conducts heat poorly, so it reduces heat loss [1]. The seal is an endotherm in cold water and needs to retain body heat, while a frog in a warm habitat is an ectotherm that does not maintain a high body temperature, so insulation gives little benefit [1]. [9] Examiner insight: Include units with every calculated value and relate the insulation point to both body temperature and habitat, as the question asks you to compare.

11. (a) Triglyceride: 3 waters [1]. Phospholipid: 3 waters (two from the fatty acids, one from the phosphate) [1]. (b) Phospholipids are amphipathic, with a hydrophilic phosphate head and hydrophobic fatty acid tails [1]. In water the heads point towards the surrounding water while the tails turn inwards, away from it [1], so two layers form with tails inside, a bilayer [1]. (c) Testosterone is non-polar, so it dissolves in the hydrophobic core and crosses quickly [1]. Glucose is polar and crosses slowly [1]. Sodium ions are charged and are almost blocked by the core [1]. (d) Four fused carbon rings [1]. [9] Examiner insight: Use comparative language tied to the table values, and link each substance to polarity rather than only to size.

Where marks are usually lost

• Saying condensation “removes” or hydrolysis “adds” water without naming the -H and -OH groups.
• Counting only the ring carbons when classifying a sugar.
• Missing the unit, or converting between nm, µm and mm with the wrong factor.
• Describing starch as insoluble with no link to osmosis or compact storage.
• Giving cellulose strength as “tough” without alternating beta-glucose and hydrogen bonds.
• Confusing which antigens and which antibodies a person has.
• Describing unsaturated fatty acids without mentioning C=C bonds and chain bends.
• Calling the phosphate head hydrophobic, or the tails attracted to water.
• Giving insulation as “keeps warm” with no reference to endotherms or habitat.

Next steps

• Revisit the revision notes and the study guide.
• Tick off the topic on the printable checklist and return to the IB DP Biology hub.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

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