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

This guide follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology specification (Issue 2) and covers statements 4.8 to 4.13 of Topic 4, including Core Practical 8 and Core Practical 9 and the recommended additional practical on mineral deficiencies. Topic 4 belongs to the International AS (IAS) part of the course. There are no tiers.

Use it with the revision notes and the practice questions. The course page is Edexcel A Level Biology, the printable list is the Edexcel biology checklist, and the free diagnostics show where to start. The structure of fibres is covered in the plant cell structure, starch, cellulose and plant fibres guide. For exam technique across the course see the Edexcel IAL Biology exam preparation page.

What the specification asks for

Statement	You must be able to
4.8	Explain the importance of water and of nitrate, calcium and magnesium ions to plants. Recommended practical: investigate plant mineral deficiencies
4.9 (Core Practical 8)	Determine the tensile strength of plant fibres
4.10	Explain the conditions required for bacterial growth
4.11	Know that substances from plants can have antimicrobial and other therapeutic properties
4.12 (Core Practical 9)	Investigate the antimicrobial properties of plants, including aseptic techniques for safe handling of bacteria
4.13	Explain how drug testing developed from historic to modern protocols: William Withering’s digitalis soup, double blind trials, placebo and three-phased testing

4.8 Water and inorganic ions in plants

Water is the solvent for ions and sugars moving in xylem and phloem. It is a reactant in photosynthesis and in hydrolysis. Water entering vacuoles by osmosis makes cells turgid, and turgor supports non-woody tissue and drives cell expansion. Evaporation from leaves (transpiration) pulls water upwards and cools the leaf.

Ion	Used for	Deficiency symptoms
Nitrate	Nitrogen for amino acids (so proteins and enzymes), nucleotides (DNA, RNA) and chlorophyll	Stunted growth; yellow older leaves, because nitrogen is moved to young tissue
Calcium (Ca2+)	Calcium pectate in the middle lamella between cell walls; membrane stability; cell division	Poor, stunted growth; young leaves and growing tips die back or distort
Magnesium (Mg2+)	Centre of the chlorophyll molecule; activates some enzymes	Yellowing between the veins (interveinal chlorosis), usually first in older leaves

Why older leaves? Nitrate and magnesium can be moved from old to new tissue, so the older leaves show the loss first. Calcium is bound in walls and moves poorly, so new growth is affected.

Chlorosis from nitrate or magnesium shortage lowers chlorophyll content, so light absorption and photosynthesis fall. Less protein is made, so growth slows.

Recommended practical: mineral deficiencies

Grow seedlings in water culture. One solution has every required ion (the control). Others lack one ion each, such as nitrate or magnesium. Keep light, temperature, volume, aeration and starting size the same, and use several seedlings per treatment. Record height, leaf colour and leaf area, and finally dry mass (dried to constant mass, so water content does not distort the comparison).

Worked example 1. Mean dry mass per seedling after four weeks: complete solution 1.84 g; no nitrate 0.92 g; no magnesium 1.31 g; no calcium 1.12 g. Percentage reduction = (control − treatment) / control × 100.

• No nitrate: (1.84 − 0.92) / 1.84 × 100 = 50.0%
• No magnesium: (1.84 − 1.31) / 1.84 × 100 = 28.8%
• No calcium: (1.84 − 1.12) / 1.84 × 100 = 39.1%

Nitrate shortage limited growth most. Nitrogen is needed for proteins and nucleotides, so nearly every growth process is affected, not only chlorophyll.

4.9 Core Practical 8: tensile strength of plant fibres

Tensile strength is the stress a material withstands when pulled before it breaks. Stress is force divided by cross-sectional area, so units are pascals (N m⁻²).

Plant fibres are strong because sclerenchyma fibres have thick secondary walls with cellulose microfibrils lying roughly along the fibre. Hydrogen bonds hold the glucose chains within each microfibril together, and lignin adds stiffness. Pulling along the fibre loads many strong covalent bonds, which is why fibres such as sisal, hemp and flax are used in ropes and composites.

A typical method (the specification names only the aim):

1. Cut lengths of one fibre type to the same length. Measure the diameter at several points with a micrometer or a graticule under a microscope. Take the mean.
2. Clamp the fibre at the top. Attach a hanger to the lower end. Place a tray of sand below, and wear eye protection.
3. Add equal small masses gradually. Record the total mass when the fibre breaks.
4. Repeat with several fibres of each type. Discard results where the fibre broke at the clamp.

Calculate force = mass × 9.81 N kg⁻¹, area = π (d / 2)², then tensile strength = force / area.

Worked example 2. A sisal fibre of mean diameter 0.30 mm breaks under 3.1 kg.

• Force = 3.1 × 9.81 = 30.4 N
• Radius = 0.15 mm = 1.5 × 10⁻⁴ m. Area = π × (1.5 × 10⁻⁴)² = 7.07 × 10⁻⁸ m²
• Tensile strength = 30.4 / 7.07 × 10⁻⁸ = 4.3 × 10⁸ Pa (430 MPa, 2 significant figures)

Why use stress, not breaking mass? A thicker fibre carries a larger load simply because it has more material. Dividing by area allows fair comparison.

Controls and limits: same fibre length, same moisture state, same rate of loading, and a diameter that is measured rather than assumed. Fibres are not perfect cylinders and vary along their length, so repeat measurements and give a mean.

4.10 Conditions for bacterial growth

Bacteria reproduce by binary fission, so numbers double each generation: N = N₀ × 2ⁿ, where n is the number of generations.

Growth needs:

• Nutrients: a carbon source (such as glucose), a nitrogen source, mineral ions and, for some, vitamins. Nutrient agar and nutrient broth supply these.
• Water, as a solvent for metabolism.
• Suitable temperature. Enzymes control metabolism, so growth is fastest near the optimum and slows if it is lower. High temperatures denature enzymes.
• Suitable pH, again because of enzyme activity.
• Oxygen for aerobes. Obligate anaerobes are harmed by oxygen. Facultative anaerobes can grow either way.
• Freedom from inhibitors such as antibiotics and disinfectants.

Worked example 3. A culture starts with 150 cells and the generation time is 20 minutes. Number after 4 hours? 240 min / 20 min = 12 generations. N = 150 × 2¹² = 150 × 4096 = 614 400 (6.1 × 10⁵).

4.11 Plant-derived antimicrobial and therapeutic substances

Plants make chemicals that defend them against microbes and herbivores. Some are useful as medicines.

Substance	Source	Use
Digoxin and related digitalis glycosides	Foxglove (Digitalis)	Heart conditions
Quinine	Cinchona bark	Malaria
Artemisinin	Sweet wormwood (Artemisia annua)	Malaria
Salicylic acid, from which aspirin was developed	Willow bark	Pain relief, reduced inflammation
Paclitaxel	Pacific yew (Taxus brevifolia)	Cancer treatment
Morphine	Opium poppy	Pain relief
Allicin	Garlic	Antimicrobial action shown in laboratory tests

Many plant species are unstudied, which is one reason to protect plant diversity.

4.12 Core Practical 9: antimicrobial properties of plants

Aseptic technique stops unwanted microbes entering a culture and stops the culture escaping.

• Disinfect the bench. Work close to a Bunsen flame, whose rising air carries contaminants away.
• Sterilise the inoculating loop in the flame until red hot and let it cool before use.
• Lift lids only as far as needed and flame the neck of any bottle.
• Use sterile pipettes, spreaders and agar plates.
• Tape the lid with short strips, not all round, so oxygen can enter and strictly anaerobic bacteria are discouraged. Incubate the plate upside down so condensation does not fall on the agar.
• Incubate at a modest temperature, around 25 °C, not body temperature, which favours human pathogens. Use a safe laboratory strain. Do not open plates afterwards. Sterilise waste before disposal.

Method. Spread bacteria evenly over nutrient agar to make a lawn. Place sterile paper discs soaked in different plant extracts, and controls, on the lawn. Incubate. Measure the clear zone around each disc.

Controls: a disc soaked in the solvent only (negative), and a disc of a known antibiotic (positive). Standardise disc size, extract volume, extract concentration, agar depth and incubation time.

Worked example 4. Discs are 6 mm across. Extract A gives a mean zone diameter of 14 mm (including the disc), extract B 10 mm.

• A: π × 7² = 153.9 mm². Disc = π × 3² = 28.3 mm². Inhibited area = 125.7 mm², about 126 mm²
• B: π × 5² = 78.5 mm². Inhibited area = 78.5 − 28.3 = 50.3 mm²
• Ratio = 125.7 / 50.3 = 2.5

The diameter is only 1.4 times larger, but the area is 2.5 times larger. Compare areas, not diameters.

Evaluating. A larger zone suggests that a substance diffuses and stops growth, but diffusion speed varies between substances. A clear zone shows growth was prevented, which is not by itself proof that bacteria were killed.

4.13 Development of drug testing

Historic. William Withering, an eighteenth-century English physician, studied a folk remedy for dropsy (fluid retention) that contained many herbs, sometimes called digitalis soup. He identified foxglove as the active ingredient. He tried different doses and recorded effects and toxic side effects, publishing his results in 1785. His work was careful, but he had no placebo group and no blinding, so expectation and chance could influence the observations.

Modern testing. Before human trials, candidate drugs are tested on cells, tissues and computer models, and sometimes in animals. Human testing has three phases:

Phase	Participants	Main aim
1	Small number, usually healthy volunteers	Safety, side effects, a safe dose range
2	Small number of patients	Whether it works; best dose
3	Large numbers of patients	Compare with an existing treatment or placebo; confirm effectiveness and detect rarer side effects

A placebo looks the same as the drug but has no active ingredient. It allows for improvement caused by expectation. In a double blind trial, neither patient nor the staff giving treatment or assessing outcomes know who has the drug. Patients are allocated at random. This reduces bias from both sides. Ethical practice requires informed consent.

Worked example 5. 240 patients receive a drug and 240 a placebo. Improvement: 156 and 96.

• Drug: 156 / 240 × 100 = 65.0%
• Placebo: 96 / 240 × 100 = 40.0%
• Difference = 25.0 percentage points

The placebo group also improved, which is why a control is needed. A statistical test would then check whether the difference could be due to chance.

Common errors

• Calling magnesium a part of the cell wall, or calcium a part of chlorophyll. Magnesium is in chlorophyll and calcium is in calcium pectate.
• Writing “nitrogen” for the ion absorbed. Say nitrate.
• Quoting breaking mass as tensile strength without dividing by area.
• Dividing newtons by square millimetres and labelling the result pascals. Convert lengths to metres first.
• Saying bacteria “grow best at body temperature” in school work. Practical incubation is deliberately cooler.
• Describing a placebo as “a fake drug that does nothing” without linking it to the control of expectation.

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.