OCR A Level Biology A: Microorganisms in biotechnology and immobilised enzymes (H420) – Revision Notes

These revision notes condense learning outcomes 6.2.1(e) to 6.2.1(i) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Read the study guide for full explanations and worked examples, then test yourself with the practice questions. The course hub, checklist and free diagnostics help you find weak areas.

Key definitions

Term	Meaning
Biotechnology	Industrial use of living organisms, or parts of them, to make food, drugs or other products
Aseptic technique	Methods that keep unwanted microorganisms out of a culture and your culture away from you
Broth	Liquid nutrient medium; growth shows as turbidity
Agar plate	Solid nutrient medium; each viable cell can form a colony
Serial dilution	Repeated equal dilution steps, for example tenfold, to give a countable culture
Viable count	Number of living cells able to divide, found from colony numbers
Closed culture	No nutrient added and nothing removed during growth
Zone of inhibition	Clear area around an antibiotic disc where bacteria did not grow
Immobilised enzyme	Enzyme fixed in or on an inert support so it stays in the reactor

6.2.1(e) and (f): microorganisms and food

Why use microorganisms:

• Short life cycle, so rapid production.
• Simple growth requirements and cheap or waste substrates.
• Fermenters work in any climate and take little land.
• Fewer ethical objections than using animals.
• Easy to modify genetically, so they can make new products.

Food examples: Lactobacillus makes lactic acid in yoghurt and cheese. Yeast gives carbon dioxide in bread and ethanol in brewing. Mycoprotein comes from the fungus Fusarium venenatum.

Mycoprotein advantages	Mycoprotein disadvantages
High protein and fibre, low saturated fat, no cholesterol	Texture and taste not liked by everyone
No animals; suits vegetarians	Allergy or digestive upset in some people
Cheap substrates, fast growth, little land	RNA must be reduced by heat treatment
Not affected by weather	Costly sterile fermenters; contamination risk

6.2.1(g)(i): aseptic technique and culture

Method in steps

1. Disinfect the bench.
2. Sterilise equipment and media, usually in an autoclave at about 121 °C.
3. Heat the loop in a flame until red hot, then cool it.
4. Keep lids on as much as possible; lift them only slightly.
5. Flame the neck of bottles.
6. Tape the lid in two or three places only, and invert the dish.
7. Incubate at a moderate temperature, below body temperature in school work.
8. Autoclave used cultures.

Reasons to remember: flame creates an updraught that carries contaminants away; incomplete taping lets oxygen in so anaerobic pathogens are less likely to grow; low incubation temperature reduces growth of human pathogens.

Viable count from a serial dilution

• Colonies per cm³ of plated tube = colonies / volume plated in cm³.
• Original concentration = that figure × dilution factor.
• Dilution factor for 1 cm³ into 9 cm³ repeated n times is 10^n.

6.2.1(g)(ii): batch and continuous fermentation

Feature	Batch	Continuous
Nutrients	All at start	Added steadily
Harvest	At the end	Drawn off steadily
Growth phase	Whole growth curve	Kept in exponential phase
Typical product	Secondary metabolites, such as penicillin	Primary products, such as mycoprotein
Contamination	One batch lost	Harder to avoid over a long run

Conditions to control: temperature (cooling, because respiration releases heat), pH (probes with acid or alkali), oxygen (sterile air and stirring), nutrient supply (balanced carbon and nitrogen) and sterility.

6.2.1(h)(i): growth curve and formula

Phase	What happens	Why
Lag	Little increase in number	Cells adapt, make enzymes and take up nutrient
Exponential	Number doubles every generation	Nutrient plentiful, no limiting factor
Stationary	Number steady	Division rate equals death rate; nutrient low, waste builds up, pH changes
Death	Number falls	Death exceeds division

N = N0 × 2^n, where N is the final number, N0 the starting number and n the number of generations. For exponential growth only.

• n = total time / generation time.
• To find n from numbers: n = log(N / N0) / log 2.
• Generation time = total time / n.

Worked reminder. Start 250 cells, 4 generations: N = 250 × 2⁴ = 250 × 16 = 4000.

On a log scale, the exponential phase is a straight line. A gradient of 0.30 log₁₀ units per hour means the number doubles each hour, because log₁₀ 2 = 0.301.

Reading growth data

• Name the phase from the shape: flat at the start is lag, steep rise is exponential, plateau is stationary, fall is death.
• To read a rate from a graph, draw a tangent or use two points on the straight part, and divide the change in y by the change in x. Give units, such as log₁₀ cells per cm³ per hour.
• On a log scale, a rise of 1 unit means a tenfold increase in cells.
• When a question asks you to explain a plateau, name the limiting factor and link it to the numbers or the information given.
• For a bar chart of colony counts or zone areas, compare values using figures from the chart and calculate any percentage difference as (difference / original) × 100.

6.2.1(h)(ii): investigating growth

Method in steps

1. Choose one independent variable and at least five values.
2. Keep volume of broth, starting inoculum, incubation time and strain the same.
3. Use aseptic technique and sterile controls.
4. Measure growth by colorimeter, viable count, or zone area.
5. Repeat, calculate means and compare.

Zone area = πr². A zone of 16 mm diameter has r = 8 mm and area = π × 64 = 201 mm². Using area is better than using diameter because it reflects the whole zone.

6.2.1(i): immobilised enzymes

Method	Idea	Weakness
Adsorption	Weak bonds to a surface	Enzyme can wash off
Covalent bonding	Covalent bonds to a support	Active site may be distorted
Entrapment	Held inside a gel such as alginate	Slow diffusion
Membrane separation	Behind a partially permeable membrane	Membrane may limit flow or get blocked

Advantages: reusable; product is not contaminated with enzyme; continuous process possible; often more stable to temperature and pH. Disadvantages: higher set-up cost; activity may be lower because substrate must diffuse; enzyme can be lost or fouled; contamination is harder to remove.

Examples to apply: glucose isomerase (glucose to fructose), lactase (lactose to glucose and galactose), penicillin acylase, aminoacylase, glucoamylase.

Bead worked reminder. SA:V of a sphere = 3/r. Radius 1.5 mm gives 3 / 1.5 = 2.0 mm⁻¹. Halving the radius doubles it.

Must-know distinctions

• Batch is a closed system through all four phases; continuous holds the culture in the exponential phase.
• Viable count (colonies) counts living cells; turbidity counts living and dead cells together.
• Sterilising kills all organisms; disinfecting reduces numbers on surfaces.
• Adsorption uses weak bonds; covalent bonding uses strong bonds.
• Lag phase has no big rise in number, but cells are active.

Quick self-test

1. Calculate N after 5 generations from 250 cells.
2. A culture of 1000 cells has a 15 minute generation time. How many cells after 2 hours?
3. State the dilution factor after four tenfold steps.
4. A plate of 0.1 cm³ from a 10⁻³ tube has 72 colonies. Calculate cells per cm³ of the original culture.
5. State one reason a continuous culture gives more primary product than a batch culture.
6. Give two reasons why immobilised enzymes lower costs.
7. Name the phase in which the death rate equals the division rate, and give one cause.
8. Calculate the area of a zone of inhibition of diameter 16 mm.
9. Why should used plates be autoclaved?
10. Name an immobilisation method where the enzyme may leak out, and explain why.

Answers

1. 250 × 2⁵ = 250 × 32 = 8000.
2. n = 120 / 15 = 8; N = 1000 × 256 = 2.56 × 10⁵ cells.
3. A dilution factor of 10⁴, so the tube is 10⁻⁴ of the original concentration.
4. 72 / 0.1 = 720; × 10³ = 7.2 × 10⁵ cells per cm³.
5. It stays in the exponential phase, with fresh nutrient and constant removal, so cells keep dividing.
6. The enzyme can be reused, and the product needs less purification.
7. Stationary phase; nutrient shortage, toxic waste or a change in pH limits growth.
8. π × 8² = 201 mm².
9. To kill all microorganisms, including any harmful ones, before disposal.
10. Adsorption, because the enzyme is held only by weak bonds. Entrapment can leak if the pores are large.

Where marks are usually lost

• Using days or hours where the generation time is in minutes without converting units.
• Forgetting to multiply by the dilution factor, or dividing by the volume plated.
• Describing the stationary phase as the point where the cells stop respiring or are all dead.
• Writing “to keep the culture sterile” without saying what each step stops.
• Treating a growth curve plotted on a log axis as a straight line throughout.
• Comparing zones of inhibition by diameter instead of area when asked about the effect.
• Giving “cheaper” as the only advantage of immobilised enzymes with no reason such as reuse.
• Missing the control: a sterile broth, or a disc with no antibiotic.
• Saying immobilised enzymes are always faster, when diffusion can slow them.

Official syllabus

OCR Advanced GCE in Biology A (H420), Version 4.1 (April 2026), for first assessment in 2025, published by OCR.