AQA A-level Biology: Water and Inorganic Ions (7402) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.1.7 Water and 3.1.8 Inorganic ions of the AQA AS and A-level Biology specification (7402 is the A-level), for first teaching 2015. Where a value is needed it is given in the question. Learn the content in the study guide and recall it with the revision notes. Course hub: AQA A-level Biology. Checklist: AQA A-level Biology checklist. More topic questions: Biological molecules practice.

Questions

1. State what is meant by water being a metabolite. Name one reaction in which water is released. [2]

2. Explain why sodium ions dissolve in water but triglycerides do not. [3]

3. A mass of 8.0 kg of body water absorbs 100 kJ of heat. Specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹.

(a) Calculate the temperature rise. [2] (b) A liquid with a specific heat capacity of 1.5 J g⁻¹ °C⁻¹ replaced the water. Calculate the new temperature rise. [1] (c) Explain why the value for water benefits enzymes in cells. [2]

4. A hiker must lose 600 kJ of heat through the skin by evaporation. Latent heat of vaporisation = 2.4 kJ g⁻¹.

(a) Calculate the mass of water that must evaporate. [2] (b) Only 80% of the sweat produced evaporates, the rest dripping off. Calculate the total mass of sweat produced. [2]

5. A leaf takes in 14 kJ of energy per hour from sunlight. In that hour it loses 5.4 g of water by evaporation. Latent heat of vaporisation = 2.4 kJ g⁻¹.

(a) Calculate the energy removed by evaporation. [1] (b) Express this as a percentage of the energy taken in. [1] (c) The stomata close. Predict the effect on leaf temperature and explain your answer. [3]

6. Explain how surface tension arises and why a small insect can stand on the surface of a pond. [3]

7. A shoot is attached to a potometer. The capillary tube has a radius of 0.40 mm. The water column moves 48 mm in 15 minutes.

(a) Calculate the volume of water taken up per minute. [3] (b) Explain how cohesion allows water to be pulled up through the xylem. [2]

8. The table shows the initial rate of an enzyme-catalysed reaction at six pH values.

pH	3	4	5	6	7	8
Rate (units)	1	8	24	40	22	5

(a) Describe the effect of pH on the rate. [2] (b) Calculate the percentage decrease in rate from the highest rate to pH 8. [2] (c) Explain the change in rate between pH 6 and pH 3. [2]

9. A red blood cell contains 2.8 × 10⁸ haemoglobin molecules.

(a) Each haemoglobin has four haem groups. Calculate how many O₂ molecules the cell holds when fully loaded. [2] (b) Explain why a diet low in iron can reduce the oxygen carried by blood. [2]

10. Glucose uptake by pieces of ileum wall was measured (nmol per mg of tissue per minute).

Condition	Uptake
Sodium ions present	42
Sodium ions absent	9
Sodium ions present, pump inhibitor added	14

(a) Calculate the percentage reduction in uptake when sodium ions are absent. [2] (b) Explain why sodium ions are needed for glucose uptake. [2] (c) Explain the result with the pump inhibitor. [2]

11. A double-stranded DNA fragment contains 15 base pairs.

(a) State the number of phosphate groups, the number of phosphodiester bonds, and the number of water molecules released as the fragment formed. [3] (b) State what happens to ATP when it is hydrolysed and what the released phosphate can do. [2]

12. A tall tree has plenty of water in the soil on a hot, sunny afternoon. Explain how the properties of water help the tree. [6]

Answers

1. A substance that takes part in, or is produced by, metabolic reactions [1]. Condensation, for example forming a peptide or glycosidic bond, releases water [1]. [2] Examiner insight: Naming a reaction type is safer than writing “respiration” alone; say what is formed or joined.

2. Sodium ions are charged, so the polar water molecules (δ− oxygen) surround them [1]. Triglycerides are non-polar, with no significant charge for water to interact with [1]. So ions dissolve while triglycerides do not [1]. [3] Examiner insight: Answers tend to gain credit when they link polarity of the solute to water’s polarity, rather than saying “oil repels water”.

3. (a) 8.0 kg = 8 000 g and 100 kJ = 100 000 J [1]. 100 000 ÷ (8 000 × 4.18) = 2.99 °C [1]. [2] (b) 100 000 ÷ (8 000 × 1.5) = 8.33 °C [1]. [1] (c) Water’s high heat capacity means a lot of energy is needed to change temperature, so temperature changes slowly [1]. Enzymes stay nearer their optimum temperature and are less likely to denature [1]. [2] Examiner insight: Unit conversion is often where working goes wrong; write the converted values down so a slip is visible.

4. (a) Mass = energy ÷ latent heat = 600 ÷ 2.4 [1] = 250 g [1]. [2] (b) Total = 250 ÷ 0.80 [1] = 313 g (312.5) [1]. [2] Examiner insight: In (b), dividing by 0.80 is the step to show; multiplying by 0.80 would give less sweat than is needed.

5. (a) 5.4 × 2.4 = 13.0 kJ (12.96) [1]. [1] (b) 12.96 ÷ 14 × 100 = 92.6% [1]. [1] (c) Leaf temperature rises [1]. Less water evaporates, so less energy is carried away by escaping molecules [1]. The energy absorbed from sunlight is no longer removed [1]. [3] Examiner insight: “Predict” needs the direction of change as well as the reason; quote the figure from (b) to support it.

6. Water molecules at the surface are attracted to neighbours by hydrogen bonds (cohesion) [1]. They are pulled inwards and sideways but not upwards, so the surface acts like a stretched skin [1]. The insect’s weight is spread over the surface and the skin supports it [1]. [3] Examiner insight: Naming cohesion and hydrogen bonds earns the first step; “surface tension” repeated without a cause is not an explanation.

7. (a) Volume = πr²l = π × 0.40² × 48 [1] = 24.1 mm³ [1]. Rate = 24.1 ÷ 15 = 1.61 mm³ min⁻¹ [1]. [3] (b) Hydrogen bonds make water molecules attract each other (cohesion) [1]. So water is pulled up as a continuous column without breaking [1]. [2] Examiner insight: Using the radius (0.40) and not a diameter is the usual trap in a volume question; state the formula first.

8. (a) Rate increases from pH 3 to a maximum of 40 units at pH 6 [1], then decreases at higher pH, to 5 at pH 8 [1]. [2] (b) (40 − 5) ÷ 40 × 100 [1] = 87.5% [1]. [2] (c) Low pH means a high H⁺ concentration, which changes the charges on R groups and disrupts ionic and hydrogen bonds [1]. The tertiary structure and active site change shape, so fewer substrate molecules fit and the enzyme denatures [1]. [2] Examiner insight: Quote figures when describing; for (c) name the bonds and the active site, not just “enzyme stops working”.

9. (a) One O₂ per haem group, so four per haemoglobin [1]. 2.8 × 10⁸ × 4 = 1.12 × 10⁹ [1]. [2] (b) Haem groups contain Fe²⁺ ions, which bind oxygen [1]. Less iron means less haemoglobin is made, so less oxygen is carried [1]. [2] Examiner insight: Linking iron to the haem group in haemoglobin is more precise than “iron carries oxygen”.

10. (a) (42 − 9) ÷ 42 × 100 [1] = 78.6% [1]. [2] (b) Sodium ions enter the cell down a gradient through a co-transporter, bringing glucose with them [1]. Glucose is therefore absorbed against its concentration gradient, which does not happen without sodium ions [1]. [2] (c) The sodium-potassium pump normally removes Na⁺ from the cell, keeping its concentration low [1]. Inhibiting it lets Na⁺ build up, the gradient is lost, and co-transport falls [1]. [2] Examiner insight: Mention the concentration gradient of sodium ions in (c); saying only “the pump stopped” is incomplete.

11. (a) 30 phosphate groups [1], 28 phosphodiester bonds (2 × 14) [1], 28 water molecules [1]. [3] (b) ATP is hydrolysed to ADP and inorganic phosphate (Pi), catalysed by ATP hydrolase [1]. The phosphate can phosphorylate other compounds, often making them more reactive [1]. [2] Examiner insight: Count bonds per strand (14 each), not nucleotides; both strands then give 28.

12. Any six from: strong cohesion holds water as a continuous column in the xylem, so it is pulled up as one chain [1]; water is a solvent, so mineral ions are carried dissolved [1]; evaporation from leaves breaks hydrogen bonds and takes energy away [1]; high latent heat means a lot of cooling for little water lost [1]; high heat capacity means cell temperature changes slowly [1]; water is a metabolite used in photosynthesis [1]. [6] Examiner insight: For six marks, each property should be paired with a role in the tree; a list of properties alone gains less.

Where marks are usually lost

• Converting kJ to J or kg to g incorrectly in energy calculations.
• Using the diameter, not the radius, in a volume calculation.
• Explaining cooling without saying that energy is carried away by escaping molecules.
• Describing pH changes as “acid” with no reference to H⁺, R groups and active site shape.
• Writing that iron “carries oxygen” with no mention of the haem group.
• Saying sodium ions “pull” glucose in without a concentration gradient.
• Counting nucleotides when the question asks for bonds.
• Quoting a property with no biological role.

Next steps

• Revise: Water and inorganic ions revision notes
• Relearn: Water and inorganic ions study guide
• Course hub: AQA A-level Biology
• Printable checklist: AQA A-level Biology checklist
• Check your gaps: all free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), first teaching 2015, sections 3.1.7 and 3.1.8.