Pearson Edexcel International A Level Biology: Plant Minerals, Antimicrobial Substances, Drug Testing (YBI11) – 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 statements 4.8 to 4.13 of Topic 4 (International AS) in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology specification, Issue 2, including Core Practical 8, Core Practical 9 and the mineral deficiency practical. Use g = 9.81 N kg⁻¹ where needed. Explanations are in the study guide and the revision notes. The course page is Edexcel A Level Biology, and the checklist lists every statement.

Questions

1. Explain why a plant short of magnesium ions has yellow leaves and grows slowly. [3]

2. Seedlings were grown for four weeks in four culture solutions. Mean dry mass per seedling: complete solution 2.40 g; lacking magnesium 1.68 g; lacking nitrate 1.02 g; lacking calcium 1.44 g.

(a) Calculate the percentage reduction in dry mass caused by lack of nitrate. [2] (b) Explain why dry mass, not fresh mass, was measured. [1] (c) Seedlings lacking calcium had distorted new leaves. Suggest an explanation. [2]

3. Describe two uses of nitrate in plants and explain why nitrate deficiency shows first in older leaves. [4]

4. A fibre of mean diameter 0.25 mm breaks when the load reaches 2.4 kg. Calculate its tensile strength in Pa, giving your answer to 2 significant figures. [4]

5. Fibre A has a diameter of 0.15 mm and breaks under 1.1 kg. Fibre B has a diameter of 0.40 mm and breaks under 5.9 kg.

(a) Calculate the tensile strength of each fibre. [4] (b) Fibre B carried the larger load. Explain why this does not make it the stronger material. [1] (c) Suggest one way to make the comparison more reliable. [1]

6. Explain how the structure of a sclerenchyma fibre gives it high tensile strength. [4]

7. A bacterial culture contains 800 cells, and the generation time is 25 minutes.

(a) Calculate the number of cells after 3 hours 20 minutes, assuming no limit on growth. [3] (b) State one environmental condition, other than nutrient supply, that affects growth, and explain its effect. [2]

8. Describe how you would investigate whether extracts of garlic and mint have antimicrobial properties, including the aseptic techniques used and the controls. [7]

9. Sterile discs of diameter 6 mm were soaked in plant extract X or Y and placed on a lawn of bacteria. Three discs of each were used. Zone diameters (mm, including the disc): X 17, 18, 16; Y 12, 13, 11.

(a) Calculate the mean zone diameter for X. [1] (b) Calculate the area of clear agar around each disc for X and for Y (mean diameters), and how many times larger it is for X. [4] (c) Suggest why a disc soaked in solvent only was included. [1] (d) A student concludes that X is a better medicine than Y. Evaluate this conclusion. [1]

10. A new drug was tested in a double blind trial. 300 patients received the drug and 300 received a placebo. Improved: 135 (drug) and 90 (placebo).

(a) Calculate the percentage improving in each group and the difference. [3] (b) Explain the purpose of the placebo group. [2] (c) Explain what is meant by double blind and why it is used. [2] (d) State one reason why evidence from William Withering’s foxglove work was less reliable than this trial. [1]

11. Compare phase 1 and phase 3 of drug testing. [4]

Answers

1. Magnesium is at the centre of the chlorophyll molecule, so without it less chlorophyll is made [1], leaves go yellow [1]. Less light is absorbed, so photosynthesis and growth are slower [1]. [3] Examiner insight: “Needed for photosynthesis” is too vague. Name chlorophyll, then link to reduced light absorption.

2. (a) (2.40 − 1.02) / 2.40 × 100 [1] = 57.5% [1] (b) Fresh mass varies with water content, but dry mass measures material the plant actually made [1]. (c) Calcium is needed for calcium pectate in the middle lamella and cell walls [1]. It is poorly mobile, so new growth cannot get it from older tissue and cell walls are weak or division is disturbed [1]. [5] Examiner insight: On a “suggest” part, linking a named role of the ion to the visible symptom is what earns credit. A bare “calcium is needed for growth” is too general.

3. Any two uses from amino acids or proteins and enzymes; nucleotides for DNA, RNA or ATP; chlorophyll [2]. Nitrate is mobile and moved from old leaves to young growing tissue [1], so the older leaves lose their nitrogen first and turn yellow [1]. [4] Examiner insight: Use “nitrate” for the ion absorbed. Marks need the use stated, not just “to make nitrogen compounds”.

4. Force = 2.4 × 9.81 = 23.5 N [1]. Radius 0.125 mm = 1.25 × 10⁻⁴ m, so area = π × (1.25 × 10⁻⁴)² = 4.91 × 10⁻⁸ m² [1]. Strength = 23.5 / 4.91 × 10⁻⁸ [1] = 4.8 × 10⁸ Pa [1] [4] Examiner insight: Show the conversion from mm to m. A diameter used as a radius is a frequent slip that gives an answer four times too large or small, so check by halving.

5. (a) A: F = 1.1 × 9.81 = 10.8 N; area = π × (7.5 × 10⁻⁵)² = 1.77 × 10⁻⁸ m²; strength = 6.1 × 10⁸ Pa [2]. B: F = 57.9 N; area = π × (2.0 × 10⁻⁴)² = 1.26 × 10⁻⁷ m²; strength = 4.6 × 10⁸ Pa [2] (b) B is thicker, so more material shares the load; strength is force per unit area, and A is stronger [1]. (c) Test several fibres of each type and compare means [1]. [6] Examiner insight: “Fibre A is stronger” needs the calculated stresses next to each other, with units, to be a convincing comparison.

6. Cellulose microfibrils lie along the length of the fibre [1]. Hydrogen bonds hold the glucose chains together in each microfibril [1]. The wall is thickened by secondary thickening, so there is more material to bear the load [1]. Lignin adds stiffness and resistance to stretching [1]. [4] Examiner insight: Pair each structural feature with an effect on strength. Listing “cellulose, lignin” without a role does not show understanding.

7. (a) 200 min / 25 min = 8 generations [1]. N = 800 × 2⁸ [1] = 204 800 [1] (b) Temperature (or pH) [1] affects enzyme activity, so growth is slower away from the optimum and stops if enzymes denature [1]. [5] Examiner insight: Convert hours to minutes before dividing. Name the condition and give the enzyme-based effect.

8. Spread bacteria evenly on sterile nutrient agar to form a lawn [1]. Place sterile discs soaked in garlic extract and in mint extract on the lawn [1]. Include a solvent-only disc as a negative control [1] and an antibiotic disc as a positive control [1]. Aseptic technique, any two: disinfect bench; work near a Bunsen flame; flame the loop and cool it; lids opened briefly; sterile equipment [2]. Tape the lid with short strips, incubate upside down at about 25 °C, then measure clear zone diameters [1]. [7] Examiner insight: “Aseptic” needs named actions. A control needs saying what it shows, not just that it is there.

9. (a) (17 + 18 + 16) / 3 = 17 mm [1] (b) X: π × 8.5² = 227 mm², minus disc π × 3² = 28.3 mm², so 199 mm² [1]. Y: π × 6² = 113 mm², minus 28.3, so 84.8 mm² [1]. Ratio = 199 / 84.8 [1] = 2.3 times [1] (c) To show the solvent alone does not stop bacterial growth [1]. (d) X produced a larger zone, but diffusion rates and concentrations may differ, and a laboratory test on bacteria does not show safety or effect in people, so the conclusion is not justified [1]. [7] Examiner insight: Compare areas, not diameters. “Evaluate” asks for a judgement with a reason, not repeated data.

10. (a) Drug 135 / 300 × 100 = 45.0% [1]; placebo 90 / 300 × 100 = 30.0% [1]; difference 15.0 percentage points [1] (b) Some patients improve through expectation or natural recovery [1], so the placebo group lets you see the effect due to the drug itself [1]. (c) Neither patients nor the staff treating or assessing them know who has the drug [1]. This removes bias from patients and from observers [1]. (d) No placebo group or blinding (or no random allocation) [1]. [8] Examiner insight: For “explain the purpose”, state what the placebo group allows you to compare, not only that it is “a control”.

11. Phase 1 uses a small number of healthy volunteers, phase 3 uses large numbers of patients [1]. Phase 1 tests safety, side effects and dose range [1]. Phase 3 tests effectiveness compared with a placebo or existing treatment [1] and can detect rarer side effects [1]. [4] Examiner insight: A comparison needs both phases mentioned for the same feature, for example participants, aim and size.

Where marks are usually lost

• Answering “nitrogen” when the question is about the ion plants absorb, which is nitrate.
• Linking magnesium to the cell wall, or calcium to chlorophyll.
• Not saying why older or younger leaves show deficiency first.
• Forgetting to convert mm to m, or using diameter as radius.
• Giving breaking mass instead of force per unit area.
• Comparing zone diameters rather than areas.
• Describing aseptic technique with vague words like “clean”.
• Describing a placebo as only “a fake drug” and not what it controls.

Next steps

• Revision notes
• Study guide
• Edexcel A Level Biology course page
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology, specification Issue 2 (Pearson Education Limited), Topic 4, statements 4.8 to 4.13.