OCR A Level Biology A: Microorganisms in biotechnology and immobilised enzymes (H420) – 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 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. Learn the content from the study guide and revision notes first. The course hub, the checklist and the free diagnostics show the wider course.

Questions

1. State three reasons why microorganisms are used in biotechnological processes. [3]

2. Describe four precautions that make up good aseptic technique when inoculating an agar plate, and give the reason for each. [4]

3. A bacterial culture in a closed flask was sampled each hour. The log₁₀ of the viable count per cm³ was: 0 h, 3.0; 1 h, 3.0; 2 h, 3.3; 3 h, 3.9; 4 h, 4.5; 5 h, 5.1; 6 h, 5.7; 7 h, 6.0; 8 h, 6.0; 9 h, 5.7. (a) Name the phase from 0 to 1 h and describe what the cells are doing. [2] (b) Calculate the gradient from 2 h to 6 h, with units. [2] (c) Use log₁₀ 2 = 0.301 to find the generation time in minutes. [2] (d) Give one cause of the plateau from 7 to 8 h. [1]

4. A culture starts with 125 cells and has a generation time of 40 minutes. (a) Calculate the number of cells after 4 hours. [2] (b) Calculate how many hours it takes to reach 1.0 × 10⁶ cells. [2]

5. A culture was diluted tenfold four times. A volume of 0.2 cm³ of the final dilution was spread on three plates, giving 27, 31 and 29 colonies. Calculate the number of viable cells per cm³ in the original culture. [4]

6. (a) Compare batch and continuous fermentation. [4] (b) Penicillin is usually made by batch fermentation. Suggest why. [2]

7. Give two advantages and two disadvantages of mycoprotein as a food for humans. [4]

8. A student placed discs soaked in 0, 25, 50 and 100 µg of an antibiotic on a lawn of bacteria. The clear zone diameters, which include the 6 mm disc, were 6, 10, 16 and 22 mm. (a) Calculate the percentage increase in zone area between the 25 µg and 50 µg discs. [3] (b) Explain what the clear zone shows. [2] (c) Explain why the 0 µg disc was included. [2]

9. Describe two methods of immobilising an enzyme. [4]

10. Lactase is entrapped in alginate beads and used to remove lactose from milk. (a) Beads of diameter 3 mm and 5 mm are compared. Calculate the surface area to volume ratio of each, using 3/r. [2] (b) Explain which beads give the faster reaction. [2] (c) A column produced 18.0 µmol of glucose in a 6 minute run when new and 14.4 µmol in the fifth run. Calculate the percentage activity retained and suggest one cause of the loss. [2]

11. Evaluate the use of an immobilised glucose isomerase column to make fructose from glucose rather than adding free enzyme to a tank. [6]

12. Plan an investigation into the effect of temperature on the growth of yeast in broth. [8]

Answers

1. Short life cycle, so fast growth and quick production [1]. Simple, cheap growth requirements, including waste substrates [1]. Any one of: grown in any climate in a small space; fewer ethical concerns than animals; can be genetically modified to make useful products [1]. [3] Examiner insight: Answers tend to gain credit when each reason is stated as a feature of the organism; the word “cheap” alone is vague.

2. Flame the loop until red hot, to kill microorganisms already on it [1]. Keep the dish lid on or lifted only slightly, to stop airborne contaminants entering [1]. Flame the neck of bottles, to sterilise it and create an updraught away from the culture [1]. Tape the lid in two or three places and incubate inverted, so air can enter and condensation does not drip onto the agar [1]. Other valid pairs include disinfecting the bench and autoclaving used plates. [4] Examiner insight: Each precaution needs its reason. A list with no reasons leaves available marks unearned.

3. (a) Lag phase [1]. Cells are adjusting to the medium, making enzymes and taking up nutrients, with little division [1]. (b) (5.7 − 3.3) / (6 − 2) [1] = 0.6 log₁₀ cells per cm³ per hour [1]. (c) Doubling adds 0.301 to the log, so time = 0.301 / 0.6 = 0.50 h [1] = 30 minutes [1]. (d) Nutrient depletion, accumulation of toxic waste or a change in pH [1]. [7] Examiner insight: Give the units for a gradient from a log graph. Saying only “0.6” is incomplete.

4. (a) n = 240 / 40 = 6 [1]; N = 125 × 2⁶ = 8000 cells [1]. (b) 2^n = 1.0 × 10⁶ / 125 = 8000, so n = 12.97 [1]; time = 12.97 × 40 = 519 minutes = 8.6 hours [1]. [4] Examiner insight: Convert the time to the same unit as the generation time before dividing. Show the value of n so that the reader can follow each step.

5. Mean = (27 + 31 + 29) / 3 = 29 [1]. 29 / 0.2 = 145 per cm³ of the final dilution [1]. Multiply by the dilution factor 10⁴ [1]. 1.45 × 10⁶ cells per cm³ [1]. [4] Examiner insight: Both the plated volume and the dilution factor must be used. Omitting either gives a wrong power of ten.

6. (a) In batch, nutrients are added once at the start; in continuous they are added steadily [1]. Batch is harvested at the end; continuous culture is drawn off continuously [1]. Batch goes through all growth phases; continuous is held in exponential growth [1]. A contaminated batch is lost and the vessel is then cleaned; continuous runs are harder to keep sterile for long periods [1]. (b) Penicillin is a secondary metabolite made mainly as growth slows [1], so keeping the culture in exponential growth gives no gain [1]. [6] Examiner insight: Comparisons need both sides stated. Describing only batch does not earn the comparison mark.

7. Advantages, any two: high protein, low saturated fat and no cholesterol; suitable for vegetarians; cheap, fast production from waste carbohydrate [1][1]. Disadvantages, any two: some people dislike the texture or taste; allergy or digestive upset in some; RNA content must be reduced; costly sterile fermenters [1][1]. [4] Examiner insight: “Healthy” is too vague; name the feature, such as low saturated fat.

8. (a) Area at 25 µg = π × 5² = 78.5 mm² [1]; area at 50 µg = π × 8² = 201 mm² [1]; increase = (201 − 78.5) / 78.5 × 100 = 156% [1]. (b) The antibiotic diffused from the disc [1] and prevented growth or killed bacteria in the zone [1]. (c) It is a control [1]; it shows the disc and solvent alone do not stop growth, so the zones are due to the antibiotic [1]. [7] Examiner insight: Percentage change is relative to the original value. Using areas rather than diameters is the expected approach.

9. Adsorption: the enzyme sticks to the surface of a support such as a resin by weak bonds [1]; it can wash off [1]. Entrapment: the enzyme is held in a gel such as alginate [1]; substrate and product diffuse in and out through the gel [1]. Covalent bonding or membrane separation are also valid. [4] Examiner insight: A description of the method needs how the enzyme is held, not just the name.

10. (a) 3 / 1.5 = 2.0 mm⁻¹; 3 / 2.5 = 1.2 mm⁻¹ [1][1]. (b) The 3 mm beads have the larger ratio [1], so substrate diffuses to more of the enzyme and the reaction is faster [1]. (c) 14.4 / 18.0 × 100 = 80% [1]; some enzyme leaked from the beads, or beads broke [1]. [6] Examiner insight: A comparison needs a reason from the numbers, here the ratio, and not only “smaller is better”.

11. The column can be reused many times, which lowers enzyme cost [1]. The product is not contaminated with enzyme, so less purification is needed [1]. Continuous flow is possible [1]. The enzyme is often more stable to temperature and pH [1]. Against: higher set-up cost, and slower reaction if diffusion limits it [1]. Judgement, for example that the column is better for large-scale, repeated production [1]. [6] Examiner insight: An evaluation should reach a conclusion. A list of points with no judgement scores below the maximum.

12. Independent variable: at least five temperatures, for example 15 to 55 °C [1]. Same volume of sterile broth, same starting yeast concentration, same pH in every tube [1]. Aseptic technique with a sterile control [1]. Inoculate each tube and keep it in a water bath at its temperature [1]. Incubate for the same time, shaken at the same rate [1]. Measure turbidity with a colorimeter, using sterile broth to zero it, or take viable counts by serial dilution [1]. Repeat each temperature at least three times and calculate a mean [1]. Safety: use a non-pathogenic yeast and autoclave waste [1]. [8] Examiner insight: Plans earn most when each control is named and the measurement is specific. “Keep everything else the same” is too vague.

Where marks are usually lost

• Not converting time to the unit of the generation time before finding n.
• Leaving out the dilution factor or the plated volume in viable counts.
• Quoting a gradient from a log graph without units.
• Giving precautions in aseptic technique with no reason.
• Describing the stationary phase as the point where cells stop growing without naming a limiting factor.
• Comparing batch and continuous fermentation by describing only one of them.
• Comparing zones by diameter when area is needed.
• Calling a control “the same” without saying what it contains.
• Stating that immobilised enzymes always work faster, with no mention of diffusion.

Next steps

• Re-read the revision notes for any weak parts.
• Go back to the study guide for full worked examples.
• Check the course hub and the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

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