Question 1
Paper 1A style
Which statement describes a hydrolysis reaction?
- Two monomers are linked by a covalent bond and a water molecule is released.
- A water molecule is split and its –H and –OH are added to the products as a bond is broken.
- A polymer is broken into monomers by the removal of water.
- A monomer is oxidised to release energy for the cell.
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Answer: B [1]
Question 2
Paper 1A style
A molecular diagram shows a ring of five carbon atoms and one oxygen atom, with a sixth carbon outside the ring and an –OH group above the ring on carbon 1. Which molecule is shown, and which polymer is it a monomer of?
- Ribose, a monomer of RNA
- Alpha-glucose, a monomer of glycogen
- Beta-glucose, a monomer of cellulose
- Beta-glucose, a monomer of starch
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Answer: C [1]
Question 3
Paper 1A style
A molecule from an animal has a skeleton of four fused carbon rings, three with six carbons and one with five. Which property would allow it to enter a target cell?
- It is hydrophilic, so it passes through protein channels.
- It is amphipathic, so it forms a bilayer with the membrane.
- It is non-polar, so it diffuses through the hydrophobic core of the phospholipid bilayer.
- It is charged, so it is attracted to the phosphate heads and pulled across.
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Answer: C [1]
Question 4
Paper 1B style
The table shows the melting points of four fatty acids. Each has a chain of 18 carbon atoms. (a) Identify the monounsaturated fatty acid. [1] (b) Calculate the difference in melting point between stearic acid and linoleic acid. [1] (c) Describe the relationship between the number of C=C double bonds and melting point. [1] (d) Explain this relationship. [2] (e) Suggest why the oils stored in plant seeds contain a high proportion of unsaturated fatty acids. [1]
| Fatty acid | Number of C=C double bonds | Melting point / °C |
|---|---|---|
| Stearic acid | 0 | 69 |
| Oleic acid | 1 | 13 |
| Linoleic acid | 2 | −5 |
| Linolenic acid | 3 | −11 |
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- (a) oleic acid [1]
- (b) 74 °C [1] Accept 74 without unit; do not accept 64
- (c) more C=C double bonds, lower melting point / negative correlation [1] Accept: largest decrease between 0 and 1 double bond
- (d) double bonds cause bends / kinks in the hydrocarbon chain [1]
- (d) chains cannot pack closely together so attractions between them are weaker / less energy needed to melt [1] OWTTE
- (e) oils stay liquid at the temperatures of plant tissues / plants are not endotherms, so stores remain fluid / easy to mobilise [1]
Question 5
Paper 2A style
Outline how the structure of glycogen makes it suitable as an energy store in animals.
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- polymer of alpha-glucose [1]
- highly branched / coiled so compact, storing much glucose in a small space [1]
- large molecule so relatively insoluble / no osmotic effect / does not leave the cell [1]
- many chain ends so glucose added or removed quickly by condensation / hydrolysis [1]
- max 3
Question 6
Paper 2A style
Explain the role of glycoproteins in cell–cell recognition, using the ABO blood groups as an example.
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- glycoproteins are proteins with carbohydrate chains attached on the outer surface of the plasma membrane [1]
- carbohydrate chain acts as a marker / antigen recognised by other cells [1]
- ABO antigens on red blood cells differ in the carbohydrate chain / A and B have an extra sugar, O has neither [1]
- antibodies bind to antigens not present on the recipient's own cells, causing agglutination / so transfused blood must be matched [1]
- max 3
Question 7
Paper 2A style
A small hibernating mammal needs 1850 kJ of energy from its stores over the winter. Oxidation releases about 37 kJ g⁻¹ from triglyceride and about 17 kJ g⁻¹ from glycogen. (a) Calculate the mass of triglyceride needed. [1] (b) Calculate how many times greater the mass of glycogen needed would be. [1] (c) Suggest one further reason why storing the energy as triglyceride is an advantage to the mammal. [1]
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- (a) 1850 ÷ 37 = 50 g [1]
- (b) glycogen mass = 1850 ÷ 17 = 108.8 g; 108.8 ÷ 50 = 2.2 times [1] Accept 2.1–2.2; ECF from (a)
- (c) glycogen is stored with water, so the real difference in mass is greater / triglyceride also provides thermal insulation / insoluble so no osmotic effect [1]
Question 8
Paper 2B style
Compare and contrast carbohydrates and lipids as energy storage compounds in living organisms.
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- both contain carbon, hydrogen and oxygen / are carbon compounds [1]
- both are made by condensation and mobilised by hydrolysis [1]
- both are insoluble in their storage form so have little osmotic effect [1]
- carbohydrates are stored as starch in plants and glycogen in animals [1]
- lipids are stored as triglycerides / in adipose tissue in animals / as oils in seeds [1]
- lipids release about twice as much energy per gram when oxidised (about 37 kJ g⁻¹ compared with 17 kJ g⁻¹) [1]
- glycogen is stored with water so is heavier per unit of energy; triglycerides are stored without water [1]
- carbohydrates for short-term storage, lipids for long-term storage [1]
- glucose can be released from glycogen / starch more rapidly, from many branch ends [1]
- triglycerides also act as thermal insulation; polysaccharide stores do not [1]
- glucose is soluble and transported easily, so it is the form carried in blood rather than stored [1]
- max 6