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IB Biology 2025 · Theme B · Form and function

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

70 original IB-style questions on B1.1, written from the 2025 guide: 31 multiple-choice, 25 short-answer, 8 data-based, 3 extended-response parts, 2 drawing, 1 labelling. This subtopic has no additional higher level content, so the 21-mark practice paper below serves SL and HL alike. Hand one out as a class quiz or homework, or sit it 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

  • 31 multiple-choice
  • 25 short-answer
  • 8 data-based
  • 3 extended-response parts
  • 2 drawing
  • 1 labelling
  • 0 higher level only

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

Make your own

The practice paper

No additional higher level content in this subtopic, so one paper serves both · 21 marks
Take it online → Build a fresh paper Paper code BbB-EAAAAgAAABQAOqr8
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes21 marks

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Covers B1.1 Carbohydrates and lipids · SL
Name:
1.

Which row correctly matches a macromolecule with the monomers that are linked by condensation reactions to form it?

[1]
RowMacromoleculeMonomers
A.polysaccharideamino acids
B.polypeptidemonosaccharides
C.nucleic acidfatty acids
D.nucleic acidnucleotides
  1. Row B
  2. Row C
  3. Row D
  4. Row A
2.

Phospholipids form bilayers in water because they are amphipathic. What does amphipathic mean?

[1]
  1. Carrying a double bond in each fatty-acid tail
  2. Being equally soluble in water and in lipids
  3. Containing two phosphate groups per molecule
  4. Having both hydrophilic and hydrophobic regions
3.

Why are lipids better suited than carbohydrates for long-term energy storage in animals?

[1]
  1. Lipids store about twice as much energy per gram
  2. Lipids release energy twice as fast as carbohydrates
  3. Lipids can be broken down without using enzymes
  4. Lipids are more soluble in blood than carbohydrates
4.

A fatty acid has 18 carbon atoms and two carbon–carbon double bonds. How should it be classified?

[1]
  1. Polyunsaturated
  2. Monounsaturated
  3. Trans-saturated
  4. Saturated
5.

Cellulose and starch are both polymers of glucose, yet cellulose forms strong fibres while starch does not. Why?

[1]
  1. Cellulose is strengthened by nitrogen-containing cross-links
  2. Cellulose chains are straight and hydrogen-bond into microfibrils
  3. Starch chains are too short to form fibres of any strength
  4. Cellulose molecules are branched, so their chains interlock
6.

Oestradiol and testosterone are steroid hormones. Insulin is a protein hormone. All three are carried in the blood.

oestradiol a steroid hormone insulin a protein hormone HO OH four fused carbon rings two chains, 51 amino acids in all A chain B chain disulfide bridge
(a)Suggest why steroid hormones can pass through the plasma membrane of a target cell whereas insulin cannot.[2]
(b)Suggest why steroid hormones are usually carried in the blood attached to transport proteins.[2]
7.

Distinguish between monosaccharides, disaccharides and polysaccharides, giving an example of each.

[3]
8.

Paper is a mat of fibres made almost entirely of cellulose. Where two fibres touch, cellulose molecules at the surface of one fibre are held to those of the other. A student investigated why paper becomes weak when it gets wet. Strips 1.5 cm wide and 10 cm long were cut in the same direction from one sheet of filter paper. Each strip was clamped at both ends and pulled with a force meter, with the force increased steadily until the strip tore, and the force at which it broke was recorded. Five strips were given each of four treatments: kept dry; soaked in water for 5 min and tested at once while wet; soaked in hexane for 5 min (in a fume cupboard) and tested at once while wet; soaked in water for 5 min, then left to dry in air at 20 °C for 24 h before testing. Water molecules are polar and can form hydrogen bonds with the hydroxyl (–OH) groups of cellulose. Hexane is a non-polar liquid.

TreatmentBreaking force, strip 1 / NBreaking force, strip 2 / NBreaking force, strip 3 / NBreaking force, strip 4 / NBreaking force, strip 5 / NMean breaking force / N
Dry24.623.125.222.824.324.0
Soaked in water, tested wet1.71.41.91.51.5
Soaked in hexane, tested wet23.824.922.623.424.323.8
Soaked in water, then dried21.520.222.419.821.121.0
(a)Calculate the mean breaking force of the strips tested wet with water, as a percentage of the mean breaking force of the dry strips.[2]
(b)Using the information above, explain why the strips soaked in water broke at a much lower force than the strips soaked in hexane.[2]
(c)Another student suggested that water weakens paper because it breaks the cellulose molecules into shorter chains. Evaluate this suggestion using the data.[2]
(d)Fungi that grow on damp paper secrete enzymes that digest its cellulose to monomers. Distinguish between what happens to the cellulose when paper is soaked in water and when the cellulose is digested.[3]

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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Markscheme BbB-EAAAAgAAABQAOqr8

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. B1.1.2 [1]
  • C: nucleic acids (DNA and RNA) are polymers of nucleotides; A has the monomers of polypeptides, B has the monomers of polysaccharides, and C fails because fatty acids are not monomers of any polymer (triglycerides are not polymers);
2. B1.1.12 [1]
  • D: amphipathic = hydrophilic (phosphate head) region plus hydrophobic (hydrocarbon tail) regions; this dual character drives self-assembly: heads face water, tails hide from it, the basis of membranes;
3. B1.1.11 [1]
  • A: ~2× the energy per gram of carbohydrate, stored anhydrously (compact, less mass); glycogen is for shorter-term, faster-access storage;
4. B1.1.10 [1]
  • A: one double bond = monounsaturated; two or more = polyunsaturated; none = saturated;
5. B1.1.6 [1]
  • B: alternating orientation of β-glucose gives straight chains that pack side-by-side, cross-linked by hydrogen bonds into microfibrils of high tensile strength; α-glucose chains in starch coil instead;
6. B1.1.13
  • (a) [2 max]
    • steroids are small and non-polar/hydrophobic, so they dissolve in the hydrophobic core of the phospholipid bilayer and diffuse through;
    • insulin is a protein: large and with many polar/charged R groups on its surface;
    • so insulin is repelled by the hydrophobic core and must instead bind to a receptor on the outside of the membrane, OWTTE;
  • (b) [2 max]
    • steroids are non-polar, so they are only very slightly soluble in the watery blood plasma;
    • binding to a protein with a hydrophobic binding site keeps them in solution and allows enough to be transported;
    • bound hormone is also protected from being filtered out in the kidney or broken down in the liver, so it lasts longer in the blood, OWTTE;
7. B1.1.4 [3 max]
  • monosaccharides are single sugar units, e.g. glucose/fructose/ribose;
  • disaccharides are two units joined (by a glycosidic bond), e.g. maltose/sucrose/lactose;
  • polysaccharides are many units / polymers, e.g. starch/glycogen/cellulose;
  • (joined/split by condensation/hydrolysis), OWTTE;
8. B1.1.6, B1.1.3
  • (a) [2]
    • mean breaking force of the wet strips = 8.0 ÷ 5 = 1.6 (N);
    • 1.6 ÷ 24.0 × 100 = 6.7 (%);

    Accept 6.6 to 6.7 %. Award ECF for the second mark from an incorrect mean. Award both marks for a correct final answer with no working.

  • (b) [2]
    • where the fibres touch, they are held together by hydrogen bonds between (the –OH groups of) cellulose molecules at their surfaces;
    • water molecules form hydrogen bonds with the –OH groups of cellulose, taking the place of the hydrogen bonds between the fibres, OWTTE;
    • water molecules get between the fibres/push them apart, so the fibres separate/slide past each other at a low force;
    • hexane is non-polar, so it cannot form hydrogen bonds with the –OH groups and the bonds between the fibres remain (so the strips are almost as strong as dry strips: 23.8 N compared with 24.0 N);
    • the hexane strips show that being soaked in a liquid does not by itself weaken paper, so the weakening depends on water forming hydrogen bonds with cellulose, OWTTE;

    Do not accept 'water dissolves the cellulose' or 'water breaks down the cellulose molecules'.

  • (c) [2]
    • not supported: strips soaked in water and then dried regained most of their strength (21.0 N, about 88 % of the dry strength / about 13 times the wet strength);
    • if the cellulose chains had been broken, the strength would not return when the strips dried, OWTTE;
    • (so) water breaks only the (weak) hydrogen bonds between fibres, which form again as the strip dries, OWTTE;
    • but the re-dried strips were weaker than the dry strips (ranges 19.8 to 22.4 N and 22.8 to 25.2 N do not overlap), so the soaking changed something, for example fewer fibres touching after drying, which the data cannot identify, OWTTE;

    A statement that the suggestion is not supported without a reason from the data scores no marks.

  • (d) [3]
    • soaking breaks only hydrogen bonds between cellulose molecules/fibres, whereas digestion breaks the (covalent) glycosidic bonds between the monomers within each chain;
    • in soaking the water molecules stay whole, whereas in digestion/hydrolysis a water molecule is split/used up for each bond broken;
    • in digestion the –H and the –OH from the water are added to the monomers on either side of the broken bond, whereas nothing is added to the cellulose molecules by soaking;
    • after soaking the cellulose molecules are still unchanged polymers, whereas digestion produces monomers/beta-glucose;
    • drying reverses the effect of soaking, whereas drying does not reverse digestion (the monomers would have to be linked again by condensation reactions), OWTTE;

    Content pivot. Each point must contrast the two processes. Accept converse wording. Do not accept 'water breaks bonds' without saying which bonds.

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