Pearson Edexcel International A Level Biology: Plant Minerals, Antimicrobial Substances, Drug Testing (YBI11) – Revision Notes

These notes condense 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. They include Core Practical 8, Core Practical 9 and the recommended mineral deficiency practical. Full explanations are in the study guide, and you can test yourself with the practice questions.

Related pages: Edexcel A Level Biology hub, printable checklist, free diagnostics and the exam preparation page.

4.8 Water and ions: recall table

Item	Role in plants	If short
Water	Solvent and transport medium; reactant in photosynthesis and hydrolysis; turgor for support and cell expansion; transpiration stream; cooling	Wilting, loss of turgor, slower growth and photosynthesis
Nitrate	Nitrogen for amino acids, proteins, enzymes, nucleotides, chlorophyll	Stunted growth; older leaves pale or yellow
Calcium ions	Calcium pectate in the middle lamella; membrane function; cell division	Weak, distorted new growth; dying growing tips
Magnesium ions	Centre of chlorophyll; enzyme activator	Yellowing between veins, older leaves first

Mobile (nitrate, magnesium): old leaves show deficiency first. Poorly mobile (calcium): young tissue shows it first.

Mineral deficiency practical in steps

1. Prepare a complete nutrient solution and solutions each missing one ion.
2. Use same-sized seedlings, several per solution, with the same light, temperature and aeration.
3. Record height, leaf colour, leaf area. Finish with dry mass.
4. Compare with the complete solution (control).

Percentage reduction = (control − treatment) / control × 100.

Reminder: control 3.2 g, magnesium-free 2.0 g. Reduction = 1.2 / 3.2 × 100 = 37.5%.

Why dry mass? Fresh mass changes with water content. Dry mass measures the biological material made.

4.9 Core Practical 8: tensile strength

Quantity	Equation	Unit
Force at breaking	F = m × g (g = 9.81 N kg⁻¹)	N
Cross-sectional area	A = π (d / 2)²	m²
Tensile strength	F / A	Pa (N m⁻²)

Method in steps:

1. Same length of each fibre; measure diameter in several places; take the mean.
2. Clamp one end and hang a mass holder on the other.
3. Add equal masses gradually until the fibre breaks. Record the total.
4. Repeat for several fibres. Ignore breaks at the clamp.

Reminder: diameter 0.20 mm, breaks at 1.9 kg. F = 1.9 × 9.81 = 18.6 N. A = π × (1.0 × 10⁻⁴)² = 3.14 × 10⁻⁸ m². Strength = 18.6 / 3.14 × 10⁻⁸ = 5.9 × 10⁸ Pa.

Link to structure: cellulose microfibrils run along the fibre and are held by hydrogen bonds. Secondary thickening adds layers of wall. Lignin adds stiffness.

Controls: length, moisture state, loading rate, measured (not assumed) diameter.

4.10 Bacterial growth

Requirement	Why
Carbon and nitrogen sources, minerals, vitamins if needed	Raw materials for respiration and biosynthesis
Water	Medium for reactions
Suitable temperature	Enzymes work best near an optimum; high heat denatures them
Suitable pH	Enzyme shape and activity
Oxygen (aerobes), absence of oxygen (obligate anaerobes)	Matches respiration type
No inhibitors	Antibiotics and disinfectants prevent growth

Binary fission: N = N₀ × 2ⁿ.

Reminder: 1 cell, generation time 20 min, 3 hours. n = 180 / 20 = 9. N = 2⁹ = 512.

4.11 Plant substances

Plant	Substance	Effect or use
Foxglove	Digoxin and related digitalis glycosides	Heart conditions
Cinchona	Quinine	Malaria
Sweet wormwood	Artemisinin	Malaria
Willow	Salicylic acid (aspirin developed from it)	Pain relief
Pacific yew	Paclitaxel	Cancer
Opium poppy	Morphine	Pain relief
Garlic	Allicin	Antimicrobial in laboratory tests

Plants make these for defence. Biodiversity may hide further useful compounds.

4.12 Core Practical 9: antimicrobial plants

Aseptic technique, short list:

• Disinfect bench; work near a Bunsen flame.
• Flame the loop until red hot, then cool it.
• Open lids briefly; flame bottle necks.
• Sterile equipment only.
• Tape lid with short strips; incubate upside down at about 25 °C.
• Do not open afterwards; sterilise waste.

Method in steps:

1. Spread bacteria over nutrient agar to give a lawn.
2. Add sterile discs soaked in extracts. Include a solvent-only disc (negative control) and an antibiotic disc (positive control).
3. Incubate. Measure clear zones.
4. Compare areas: area = π (d / 2)².

Reminder: disc 6 mm, zone 16 mm across. Total area = π × 8² = 201 mm². Disc = 28.3 mm². Zone beyond the disc = 173 mm².

Variables to hold constant: disc size, extract volume and concentration, agar depth, bacterial density, incubation time and temperature.

4.13 Drug testing

Term	Meaning
Withering	Studied foxglove from a mixed herbal remedy for dropsy; varied doses; recorded benefits and toxic effects; published in 1785
Placebo	Looks identical, no active ingredient; controls for expectation
Blind	Patient does not know which treatment they have
Double blind	Neither patient nor staff who treat or assess know; reduces bias from both sides
Phase 1	Few healthy volunteers: safety, side effects, dose range
Phase 2	Small number of patients: effectiveness and dose
Phase 3	Large number of patients: compare with placebo or existing treatment

Random allocation spreads other differences between groups. Informed consent is an ethical requirement.

Reminder: 200 patients per group, improved: drug 84, placebo 60. Drug = 42.0%, placebo = 30.0%. Difference = 12.0 percentage points. A statistical test is still needed.

Evaluating the practicals

Mineral deficiencies. Seedlings differ genetically, so use several per treatment and take means. Leaf colour is a judgement, so compare against a colour chart or photograph under the same light. Deionised water and clean glassware matter, because tap water contains ions. Water culture shows that an ion is needed, but it does not copy soil, where ions bind to particles.

Tensile strength. Weak points in a fibre make it break below its true strength, and a few results can be well below the rest. Look at the spread of repeats before averaging. Adding masses too fast adds a shock load. Fibres kept in a damp store may behave differently from dry ones, so state the condition. Diameter is the largest source of error because the area depends on the diameter squared, so a small error in d gives roughly double that percentage error in area.

Antimicrobial discs. Different extracts may contain different concentrations, so equal volumes do not mean equal amounts of active substance. Compounds that dissolve in oil may not diffuse through agar, so a plant with no zone might still be antimicrobial. Measure the zone at two perpendicular diameters and take the mean. Do not treat a bigger zone as proof that a substance would work as a medicine in a person.

Command words in this topic

• Explain needs a reason joined to the fact: “magnesium is part of chlorophyll, so less light is absorbed”.
• Describe (a method) needs the order, a named control and a measurement.
• Evaluate needs a strength and a limitation, with a judgement about reliability.
• Calculate needs working and a unit, with an answer to a sensible number of significant figures.

Distinctions to know

• Nitrate (the ion absorbed) versus nitrogen (the element in the molecules made).
• Magnesium (in chlorophyll) versus calcium (in the middle lamella).
• Breaking mass versus tensile strength (force per unit area).
• Placebo (the dummy treatment) versus double blind (who knows).
• Zone of inhibition shows growth did not occur. It does not show cells were killed.

Quick self-test

1. Give two roles of water in plants.
2. Why does nitrate shortage cause yellowing of older leaves?
3. Name the compound in the middle lamella that needs calcium.
4. A fibre of diameter 0.10 mm breaks at 0.80 kg. Find its tensile strength to 2 significant figures.
5. A culture of 40 cells has a generation time of 30 minutes. How many cells after 3 hours?
6. Why is the Petri dish lid taped with short strips only?
7. Why compare areas rather than diameters of inhibition zones?
8. Why is a solvent-only disc used?
9. Give two ways a modern trial improved on Withering’s method.
10. Which phase mainly tests safety in healthy volunteers?

Answers

1. Any two of: solvent and transport; reactant in photosynthesis; gives turgor; transpiration; hydrolysis.
2. Nitrate is mobile and is moved to young leaves, so older leaves lose nitrogen needed for chlorophyll.
3. Calcium pectate.
4. F = 0.80 × 9.81 = 7.85 N. A = π × (5.0 × 10⁻⁵)² = 7.85 × 10⁻⁹ m². Strength = 1.0 × 10⁹ Pa.
5. n = 180 / 30 = 6. 40 × 2⁶ = 2560.
6. To allow oxygen in, so strictly anaerobic (and often more dangerous) bacteria are not favoured.
7. The zone is circular, so area grows with the square of the radius. Area reflects the extent of inhibition.
8. To show that the solvent alone does not stop growth.
9. Placebo control and double blinding (also random allocation or a larger sample).
10. Phase 1.

Where marks are usually lost

• Giving “nitrogen” as the ion absorbed, when the answer is nitrate.
• Saying magnesium makes chlorophyll “green” without stating it sits at the centre of the molecule.
• Forgetting to convert mm to m before calculating pascals.
• Giving breaking mass as the answer to “determine tensile strength”.
• Describing aseptic technique as “keeping things clean” without a named action.
• Not naming a control in the antimicrobial practical.
• Explaining a placebo only as a “fake drug”.
• Treating all three phases as the same, instead of stating the size and purpose of each.

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.