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AQA A-Level Biology (Paper 1) Smart practice A mixed round aimed at your weak spots
Unit 1
Biological molecules
Monomers and polymers; carbohydrates
Monomers, polymers and carbohydrates
0%
Lipids
Lipids
0%
Proteins
General properties of proteins
0%
Many proteins are enzymes
0%
Nucleic acids are important information-carrying molecules
Structure of DNA and RNA
0%
DNA replication
0%
ATP
ATP
0%
Water; inorganic ions
Water and inorganic ions
0%
Unit 2
Cells
Cell structure
Structure of eukaryotic cells
0%
Structure of prokaryotic cells and of viruses
0%
Methods of studying cells
0%
All cells arise from other cells
All cells arise from other cells
0%
Transport across cell membranes
Membrane structure, diffusion and osmosis
0%
Active transport and co-transport
0%
Cell recognition and the immune system
Antigens, phagocytosis and T lymphocytes
0%
B lymphocytes, antibodies and vaccination
0%
HIV, monoclonal antibodies and the ELISA test
0%
Unit 3
Organisms exchange substances with their environment
Surface area to volume ratio
Surface area to volume ratio
0%
Gas exchange
Adaptations of gas exchange surfaces
0%
Human gas exchange, ventilation and lung disease
0%
Digestion and absorption
Digestion and absorption
0%
Mass transport in animals
Haemoglobin and the transport of oxygen
0%
The heart, cardiac cycle and cardiovascular disease
0%
Blood vessels and tissue fluid
0%
Mass transport in plants
Xylem and the cohesion-tension theory
0%
Phloem and translocation by mass flow
0%
Unit 4
Genetic information, variation and relationships between organisms
DNA, genes and chromosomes
DNA, genes and chromosomes
0%
DNA and protein synthesis
Genome, proteome, transcription and splicing
0%
Translation: ribosomes, tRNA and ATP
0%
Genetic diversity can arise as a result of mutation or during meiosis
Gene mutations and chromosome non-disjunction
0%
Meiosis and genetic variation
0%
Genetic diversity and adaptation
Genetic diversity and adaptation
0%
Species and taxonomy
Species and taxonomy
0%
Biodiversity within a community
Biodiversity within a community
0%
Investigating diversity
Investigating diversity
0%
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Water and inorganic ions

Practise 16 exam-style questions on Water and inorganic ions for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

  • 16 questions
  • 1–8 marks each
  • Instant marking
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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Dry seeds contain stores of starch and protein but very little water. A seed cannot germinate until it has absorbed water.

Use your knowledge of the properties of water to suggest why a seed must absorb water before it can use its stores of starch and protein for growth.

Model answer

Water is a metabolite: it is a reactant in hydrolysis reactions. The seed needs water to hydrolyse its starch into maltose and glucose, and its proteins into amino acids, by breaking glycosidic and peptide bonds. Water is also a solvent, so the hydrolytic enzymes and their substrates can move freely and collide to form enzyme-substrate complexes, and the soluble products can be transported to the growing parts of the seedling.

Explanation

Water is a metabolite because it takes part in metabolic reactions - it is a reactant in every hydrolysis reaction and a product of every condensation reaction. It is also the solvent in which metabolic reactions take place. A dry seed cannot hydrolyse its stores or move the products, which is one reason why water is needed for germination.

Mark scheme
  1. 1 mark

    Water is a metabolite / is a reactant in hydrolysis reactions

    Max 3. Insufficient: 'water is needed for reactions' alone.

  2. 1 mark

    Water is needed to hydrolyse starch to maltose/glucose / proteins to amino acids / to break glycosidic or peptide bonds

    Accept 'starch and protein are broken down by hydrolysis, which uses water'. Reject 'condensation' for the breakdown of starch or protein.

  3. 1 mark

    Water is a solvent, so enzymes and substrates (dissolve and) can move and collide / form enzyme-substrate complexes

    Accept 'metabolic reactions take place in solution'.

  4. 1 mark

    (Dissolved) glucose / amino acids can be transported (in solution) to the growing parts of the seedling

    Accept 'soluble products can diffuse to where they are used'.

Question 2

3 marks

Hydrogen ions are inorganic ions that occur in body fluids in very low concentrations. The pH of human blood plasma is normally 7.40. After several minutes of very strenuous exercise, the pH of an athlete's plasma fell to 7.25.

pH = -log₁₀[H⁺]

Calculate the hydrogen ion concentration of plasma at pH 7.40. Give your answer in mol dm⁻³, in standard form.

Calculate the percentage increase in the hydrogen ion concentration when the pH fell from 7.40 to 7.25. Give your answer to 2 significant figures.

Model answer

[H⁺] = 10 to the power -7.40 = 3.98 × 10⁻⁸ = 4.0 × 10⁻⁸ mol dm⁻³
[H⁺] at pH 7.25 = 10 to the power -7.25 = 5.62 × 10⁻⁸ mol dm⁻³
Percentage increase = (5.62 - 3.98) ÷ 3.98 × 100 = 41%

Explanation

Rearranging pH = -log₁₀[H⁺] gives [H⁺] = 10 to the power -pH, which you can calculate with the 10ˣ key on a calculator. Because the scale is logarithmic, a fall of only 0.15 pH units is a 41% rise in hydrogen ion concentration. Percentage increase = (new - original) ÷ original × 100.

Mark scheme
  1. 1 mark

    4.0 × 10⁻⁸ (mol dm⁻³); a correct answer with no working scores this mark

    Accept 3.98 × 10⁻⁸ or 3.981 × 10⁻⁸. A correct answer with no working scores the mark. Do not accept 0.00000004 (not in standard form).

  2. 1 mark

    Correct method, e.g. (5.62 × 10⁻⁸ - 3.98 × 10⁻⁸) ÷ 3.98 × 10⁻⁸ × 100; or a correct final answer of 41 with no working

    A correct answer of 41% (accept 40 to 41) with no working scores 2 marks (points 2 and 3). [H⁺] at pH 7.25 = 10 to the power -7.25 = 5.62 × 10⁻⁸ mol dm⁻³. Accept for 1 mark: 1.41 or 141 (ratio not converted to a percentage increase) OR 41.3 or 41.25 (not to 2 significant figures) OR 29 (difference divided by 5.62 × 10⁻⁸). Allow ECF from point 1.

  3. 1 mark

    41 (%) (accept 40 to 41)

    40 results from rounding both concentrations to 2 significant figures before dividing, and is accepted.

Question 3

4 marks

A student measured the water content of a sample of mushroom. The fresh mass of the sample was 12.50 g. She dried the sample in an oven at 80 °C and reweighed it every 24 hours until its mass stayed the same. The final dry mass was 1.15 g.

Calculate the percentage of the fresh mass that was water. Give your answer to 1 decimal place.

Explain why she reweighed the sample until its mass stayed the same.

Suggest why she dried the sample at 80 °C rather than at a much higher temperature.

Model answer

Mass of water = 12.50 - 1.15 = 11.35 g
Percentage water = 11.35 ÷ 12.50 × 100 = 90.8%

She reweighed until the mass stayed the same to make sure all the water had been removed, so that the mass lost was all the water in the sample.

At a much higher temperature some of the organic substances in the mushroom could burn or decompose, so some of the mass lost would not be water and the water content would be overestimated.

Explanation

Water is a major component of cells, often 70-95% of the fresh mass. Drying to constant mass shows that no more water can be lost, so the loss in mass is the total water content. Divide by the fresh (starting) mass, not the dry mass. The drying temperature must be high enough to remove the water in a reasonable time but low enough that no other substances are lost.

Mark scheme
  1. 1 mark

    Correct method, e.g. (12.50 - 1.15) ÷ 12.50 × 100; or a correct final answer of 90.8 with no working

    A correct answer of 90.8% with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 9.2 (percentage that was dry mass) OR 0.908 (not multiplied by 100) OR 90.80 (not to 1 decimal place).

  2. 1 mark

    90.8 (%)

  3. 1 mark

    To make sure all the water had evaporated / been removed (so all the mass lost is water)

    Insufficient: 'to make the result accurate' or 'to make it reliable' alone.

  4. 1 mark

    At a much higher temperature organic substances in the sample could burn / decompose / evaporate, so some of the mass lost would not be water (and the water content would be overestimated)

    Insufficient: 'the sample would burn' with no reference to losing mass that is not water. Ignore 'to save energy' and 'for safety'.

Question 4

6 marks

A large tree grows in a hot, sunny climate.

Describe and explain how the properties of water are important to the tree.

Model answer

Water molecules are held together by hydrogen bonds (cohesion), so water can be pulled up the xylem as a continuous column from the roots to the leaves.

Water has a large latent heat of vaporisation, so evaporation of water from the leaves in transpiration removes a lot of heat and cools the leaves in the hot sun, with relatively little loss of water.

Water is a solvent: mineral ions such as nitrate dissolve in it and are transported in the xylem, sucrose is carried in solution in the phloem, and metabolic reactions in the cells take place in solution.

Water is a metabolite: it is used in photosynthesis and in hydrolysis reactions, such as the breakdown of starch.

Water has a high heat capacity, so the temperature of the cells changes slowly, keeping enzymes near their optimum.

Explanation

Plan the answer property by property, and for each one give the property (with its cause where possible, e.g. hydrogen bonding) and its importance to this tree. The specification lists five properties: metabolite, solvent, high heat capacity, large latent heat of vaporisation and cohesion. The most common error is to mix up heat capacity (buffering temperature) with latent heat (cooling by evaporation).

Mark scheme
  1. 1 mark

    Cohesion / hydrogen bonds between water molecules

    Max 4 if the answer describes only two properties of water. Max 6.

  2. 1 mark

    (Cohesion) allows water to be pulled up the xylem as a continuous column

  3. 1 mark

    Large latent heat of vaporisation

  4. 1 mark

    Evaporation (transpiration) from the leaves cools them / removes a lot of heat with little loss of water

  5. 1 mark

    Water is a solvent so mineral ions (e.g. nitrate, phosphate) dissolve and are transported in the xylem

    Accept 'sucrose is transported (dissolved) in the phloem'.

  6. 1 mark

    Metabolic reactions in cells take place in solution (so molecules can move and collide)

  7. 1 mark

    Water is a metabolite / is used in hydrolysis (e.g. of starch or sucrose) / is used in photosynthesis

    Accept 'water is used in condensation and hydrolysis reactions'.

  8. 1 mark

    High (specific) heat capacity buffers temperature changes in cells (so enzymes stay near their optimum / are not denatured)

    Reject 'high heat capacity' as the reason for cooling by evaporation.

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

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