Biology  by Bradford
IB Biology 2025 · Theme B · Form and function

B1.1 Carbohydrates and lipids. Practice questions with markscheme.

46 original IB-style questions on B1.1, written from the 2025 guide: 19 multiple-choice, 17 short-answer, 5 data-based, 3 extended-response part, 2 drawing. Below is a 20-mark standard-level practice paper built from them, ready to hand out as a class quiz or homework, or to sit yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

13 statements at SL and HL.

  1. B1.1.1SL / HL Chemical properties of a carbon atom allowing for the formation of diverse compounds upon which life is based
  2. B1.1.2SL / HL Production of macromolecules by condensation reactions that link monomers to form a polymer
  3. B1.1.3SL / HL Digestion of polymers into monomers by hydrolysis reactions
  4. B1.1.4SL / HL Form and function of monosaccharides
  5. B1.1.5SL / HL Polysaccharides as energy storage compounds
  6. B1.1.6SL / HL Structure of cellulose related to its function as a structural polysaccharide in plants
  7. B1.1.7SL / HL Role of glycoproteins in cell–cell recognition
  8. B1.1.8SL / HL Hydrophobic properties of lipids
  9. B1.1.9SL / HL Formation of triglycerides and phospholipids by condensation reactions
  10. B1.1.10SL / HL Difference between saturated, monounsaturated and polyunsaturated fatty acids
  11. B1.1.11SL / HL Triglycerides in adipose tissues for energy storage and thermal insulation
  12. B1.1.12SL / HL Formation of phospholipid bilayers as a consequence of the hydrophobic and hydrophilic regions
  13. B1.1.13SL / HL Ability of non-polar steroids to pass through the phospholipid bilayer

In the bank for B1.1

  • 19 multiple-choice
  • 17 short-answer
  • 5 data-based
  • 3 extended-response part
  • 2 drawing
  • 0 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

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The practice paper

Take it on screen → Build a fresh paper Paper code BbB-EAAAAgAAABQAOqr8
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes20 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers B1.1 Carbohydrates and lipids
Name:
1.

Glycogen is more highly branched than the amylopectin of starch. What advantage does this give animals?

[1]
  1. Branched molecules are more resistant to digestion
  2. Branching makes glycogen dissolve better in the cytoplasm
  3. Many chain ends allow glucose to be released rapidly
  4. Each branch point stores extra energy in its bonds
2.

Fats, oils, waxes and steroids are all classified as lipids. Which property do they share?

[1]
  1. They are all polymers built from repeating fatty-acid monomers
  2. They are all insoluble in water but soluble in non-polar solvents
  3. They all contain glycerol and phosphate groups
  4. They all dissolve readily in blood plasma without carriers
3.

Which property of glucose makes it suitable as the main respiratory substrate transported in blood?

[1]
  1. It is insoluble, so it does not affect the osmotic balance of blood
  2. It is a large polymer that stores a great deal of energy per molecule
  3. It is soluble in water because of its many hydroxyl groups
  4. It cannot cross membranes, so it stays in the blood until needed
4.

Outline the properties of carbon atoms that allow the formation of a very large diversity of compounds in living organisms.

[3]
5.

Explain how the structure of cellulose suits its role in plant cell walls.

[3]
6.

A researcher measured the melting points of four fatty acids, each with an 18-carbon chain but differing in the number of carbon–carbon double bonds. Each measurement was repeated three times; the table shows the means with their standard errors (± SE).

Fatty acidNumber of C=C double bondsMelting point / °C (mean ± SE)
stearic acid069.4 ± 0.3
oleic acid113.2 ± 0.4
linoleic acid2−5.6 ± 0.5
linolenic acid3−11.3 ± 0.6
(a)Analyse the relationship between the number of double bonds and the melting point of these fatty acids.[2]
(b)Explain, in terms of molecular structure, why increasing the number of double bonds lowers the melting point.[2]
(c)The researcher claims the melting points of oleic and linoleic acid are significantly different. Evaluate whether the data support this claim, naming an appropriate statistical test and stating the null hypothesis it would test.[2]
(d)Suggest why the researcher used fatty acids that all had 18 carbon atoms.[1]
7.

Gram for gram, the complete oxidation of a triglyceride yields more than twice the ATP obtained from glycogen. Using the chemical structure of triglycerides, justify why lipids release more energy per gram than carbohydrates when respired. Refer to cellular respiration (C1.2) in your answer.

[4]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. B1.1.5 [1]
  • C: enzymes work at chain ends; more ends = faster mobilization of glucose (and faster storage after meals), matching animals' fast-changing energy demands;
2. B1.1.8 [1]
  • B: lipids are a chemically diverse group united by their non-polar, hydrophobic character; they are not polymers and only some contain glycerol;
3. B1.1.4 [1]
  • C: glucose is small and polar (hydroxyl groups), so it dissolves and is transported easily; it is also chemically stable and readily oxidised;
4. B1.1.1 [3 max]
  • carbon forms covalent bonds / bonds in which pairs of electrons are shared (between atoms);
  • each carbon atom can form up to four bonds (with other atoms);
  • the bonds can be single or double, or a combination of single and double bonds;
  • carbon bonds to other carbon atoms, forming chains that can be branched or unbranched;
  • carbon atoms can also form (single or multiple) rings;
  • carbon bonds to atoms of other non-metallic elements (e.g. hydrogen, oxygen, nitrogen), so molecules of many shapes and sizes are possible, OWTTE;

Do not accept 'carbon forms ionic bonds'. Accept 'strong / stable bonds' as part of the first point.

5. B1.1.6 [3 max]
  • cellulose is a polymer of β-glucose, giving straight/unbranched chains (alternate monomers inverted);
  • the straight chains lie side by side, cross-linked by (many) hydrogen bonds into bundles/(micro)fibrils;
  • fibrils have (very) high tensile strength;
  • (so) walls resist stretching/withstand turgor pressure, and cellulose is insoluble and hard to digest, OWTTE;
6. B1.1.10
  • (a) [2]
    • melting point decreases as the number of C=C double bonds increases (a negative relationship);
    • the fall is largest for the first double bond (0→1: ~56 °C) and smaller thereafter, so the relationship is not linear, OWTTE;
  • (b) [2]
    • each C=C double bond (cis) introduces a kink/bend in the chain;
    • kinked chains pack together less closely, so intermolecular attractions are weaker and less energy/lower temperature is needed to melt them, OWTTE;
  • (c) [2 max]
    • the means (13.2 vs −5.6 °C) differ by far more than the sum of their standard errors / the ±SE ranges do not overlap, so a real difference is very likely;
    • a t-test would test whether the two means differ significantly, with the null hypothesis that there is no difference between the mean melting points of the two fatty acids (any difference being due to chance), OWTTE;

    Accept 'error bars do not overlap' reasoning. Null hypothesis must state no difference.

  • (d) [1]
    • to control chain length, so that any difference in melting point is due to the number of double bonds (the independent variable) and not to chain length, OWTTE;
7. B1.1.11 [4 max]
  • fatty-acid tails are long hydrocarbon chains that are highly reduced / rich in C–H bonds;
  • oxidation of C–H bonds releases more energy than the (already partly oxidised) carbons of carbohydrate, which carry more oxygen;
  • β-oxidation of fatty acids yields many acetyl-CoA units feeding the Krebs cycle (C1.2);
  • (so) more reduced hydrogen carriers (NADH/FADH₂) are delivered to the electron transport chain, generating more ATP by oxidative phosphorylation;
  • triglycerides are also stored anhydrously (without associated water), so the energy per gram of stored mass is higher still, OWTTE;

Justify requires evidence linked to structure and to respiration; award the anhydrous-storage point only once.

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