Pearson Edexcel International A Level Biology: Culturing Microorganisms and Bacterial Growth (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 6.1 to 6.4 (culturing microorganisms, measuring growth, the bacterial growth curve and Core Practical 13) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topic 6 is International A2 (IA2) and there are no tiers. Learn the content in the study guide and revision notes first. The course hub and checklist show the wider course.

Questions

1. Agar plates inoculated with bacteria are sealed with short strips of tape and incubated upside down at 25 °C. Explain each of these three precautions. [3]

2. A broth culture of bacteria was diluted three times, each time by adding 1 cm³ to 99 cm³ of sterile diluent. A 0.2 cm³ sample of the final tube was spread on agar and gave a mean of 38 colonies. Calculate the number of colony-forming units per cm³ in the original culture. [3]

3. A yeast culture was diluted 1 in 20. A haemocytometer counted 28, 33, 30 and 33 cells in four large squares. Each large square has an area of 1 mm² and the chamber is 0.1 mm deep. Calculate the number of cells per cm³ in the original culture. [4]

4. Compare dilution plating with the use of a colorimeter to follow the growth of a bacterial culture over 12 hours. [4]

5. A colorimeter was calibrated with samples of known cell concentration.

Absorbance	0.10	0.20	0.30	0.40
Cells cm⁻³	1.0 × 10⁷	2.1 × 10⁷	3.0 × 10⁷	3.9 × 10⁷

(a) A sample reads 0.26. Estimate its cell concentration. [2] (b) A very dense sample reads 1.8. Suggest why it should be diluted and read again. [2]

6. A 20 cm³ sample of culture has a dry mass of 0.031 g at the start and 0.109 g after 6 hours. (a) Calculate the increase in dry mass in g dm⁻³. [2] (b) Calculate the mean rate of increase in g dm⁻³ h⁻¹. [1] (c) Explain why dry mass is used rather than wet mass. [1]

7. In a closed batch culture the number of viable bacterial cells changes in four phases. (a) Explain why the number of viable cells stays almost constant in the stationary phase. [2] (b) Explain why the number falls in the death phase. [2] (c) In the death phase the viable count fell from 3.6 × 10⁷ to 9.0 × 10⁶ cells cm⁻³. Calculate the percentage fall. [1] (d) Explain why the number barely changes during the lag phase. [2]

8. During exponential growth a culture rose from 4.0 × 10⁵ cells cm⁻³ at 3 h to 6.4 × 10⁷ cells cm⁻³ at 7 h. Calculate the growth rate constant, k, in generations per hour, and the generation time in minutes. [5]

9. A culture in exponential growth has k = 1.5 generations per hour. Starting from 5.0 × 10⁴ cells cm⁻³, calculate how long it takes to reach 1.0 × 10⁷ cells cm⁻³. [4]

10. Plan an investigation into the effect of glucose concentration on the rate of growth of yeast in a liquid culture. Include how you would measure growth, control variables, and take account of safety and ethical use of the organisms. [8]

11. The absorbance of a bacterial culture in nutrient broth at 25 °C was measured every hour.

Time (h)	0	1	2	3	4	5
Absorbance	0.05	0.05	0.10	0.20	0.40	0.62

(a) Identify the phase between 1 h and 4 h and justify your answer using the data. [2] (b) Assume absorbance is proportional to cell number in this range. Calculate k and the generation time. [3] (c) Suggest why the increase from 4 h to 5 h is smaller than doubling. [2] (d) State one limitation of using absorbance to follow this culture. [1]

Answers

1. Short strips of tape, not fully sealed, let oxygen enter [1], so strict anaerobes that could be harmful are not encouraged. Inverting the plate stops condensation dripping on the agar and spreading or merging the colonies [1]. A modest temperature of 25 °C does not favour human pathogens that may have entered by accident [1]. [3] Examiner insight: Each precaution needs its reason. “To stop contamination” for all three repeats one idea and gains less than a different reason for each.

2. Three 1 in 100 dilutions give a dilution factor of 10⁻⁶ [1]. 38 ÷ 0.2 = 190 CFU cm⁻³ in the final tube [1]. 190 × 10⁶ = 1.9 × 10⁸ CFU cm⁻³ [1]. [3] Examiner insight: Write the volume plated and the dilution factor in the working. Showing each step makes a slip in the dilution factor easy to spot and correct.

3. Mean = (28 + 33 + 30 + 33) ÷ 4 = 31 cells per square [1]. Volume of one square = 1 × 1 × 0.1 = 0.1 mm³ = 1 × 10⁻⁴ cm³ [1]. 31 ÷ 10⁻⁴ = 3.1 × 10⁵ cells cm⁻³ in the diluted sample [1]. Multiply by 20: 6.2 × 10⁶ cells cm⁻³ [1]. [4] Examiner insight: Unit conversion is where most working goes wrong. Show 0.1 mm³ becoming cm³, because 1000 mm³ equals 1 cm³.

4. Dilution plating counts only viable cells, whereas absorbance includes dead cells [1]. Plating needs serial dilution and a day or more of incubation, whereas a colorimeter gives an immediate reading [1]. The colorimeter can follow the same culture repeatedly without much sampling [1], but absorbance is not proportional to cell number at high density, whereas plating is valid when plates have a countable number of colonies [1]. [4] Examiner insight: A comparison needs both methods addressed on the same point. Describing each separately without a link such as “whereas” can lose credit.

5. (a) 0.26 lies between 0.20 and 0.30 [1]. 2.1 × 10⁷ + 0.6 × (3.0 × 10⁷ − 2.1 × 10⁷) = 2.6 × 10⁷ cells cm⁻³ (2.64 × 10⁷) [1]. (b) At high density, absorbance is no longer proportional to the cell number, so it would not be reliable [1]. Diluting the sample brings it into the calibrated range, and the result is then multiplied by the dilution factor [1]. [4] Examiner insight: Read from the calibration data only within the range it covers. Show the interpolation, not just the answer.

6. (a) 0.109 − 0.031 = 0.078 g in 20 cm³ [1]. × 50 = 3.9 g dm⁻³ [1]. (b) 3.9 ÷ 6 = 0.65 g dm⁻³ h⁻¹ [1]. (c) Wet mass includes a variable amount of water, so it is less reliable than dry mass [1]. [4] Examiner insight: Give the unit in every part. Mentioning that the sample is dried until the mass is constant shows why the method is reliable.

7. (a) The rate of cell division equals the rate of cell death [1], because nutrients are running out and toxic waste is building up [1]. (b) Cells die faster than they divide [1], because nutrients are exhausted and waste products are at a toxic concentration [1]. (c) (3.6 × 10⁷ − 9.0 × 10⁶) ÷ 3.6 × 10⁷ × 100 = 75% [1]. (d) Cells are adjusting to the new medium [1] by making enzymes and taking up nutrients, so few cells divide yet [1]. [7] Examiner insight: “Stationary” is explained by a balance of two rates. Writing that “growth stops” without that balance is less complete.

8. log₁₀ (6.4 × 10⁷) − log₁₀ (4.0 × 10⁵) = 7.806 − 5.602 = 2.204 [1]. n = 2.204 ÷ 0.301 = 7.32 generations [1]. Time = 7 − 3 = 4 h, so k = 7.32 ÷ 4 = 1.83 generations per hour [1]. g = 1 ÷ 1.83 = 0.546 h [1] = 33 minutes [1]. [5] Examiner insight: Show the log difference and the division by 0.301. State the unit for k. Converting hours to minutes at the end is an easy step to omit.

9. n = (log₁₀ 1.0 × 10⁷ − log₁₀ 5.0 × 10⁴) ÷ 0.301 = (7 − 4.699) ÷ 0.301 [1]. n = 7.64 generations [1]. t = n ÷ k = 7.64 ÷ 1.5 [1] = 5.1 hours [1]. [4] Examiner insight: Rearranging k = n ÷ t is the skill being tested. A check that 2 to the power 7.64 is about 200 confirms the answer is reasonable.

10. Choose several glucose concentrations, for example 0, 1, 2, 4 and 8 g dm⁻³ in sterile nutrient broth [1]. Add the same volume and density of yeast suspension to each flask, using aseptic technique [1]. Keep temperature, pH, volume of broth, flask size and aeration constant [1]. Include a flask of sterile broth without yeast as a control [1]. Measure growth by absorbance in a colorimeter at set intervals, such as every 30 minutes, blanking with sterile broth and mixing before each reading; or use a haemocytometer or dilution plating [1]. Plot a suitable measure of growth against time, find the exponential phase and calculate k for each concentration [1]. Repeat each concentration and calculate means [1]. Safety and ethics: complete a risk assessment, use an approved safe strain, incubate at a modest temperature, disinfect spills, sterilise cultures and equipment before disposal, do not release cultures, and wash hands [1]. [8] Examiner insight: A plan scores more when each controlled variable is paired with how it is held constant, and when the range of concentrations is stated with numbers. Safety needs a specific action, not “be careful”.

11. (a) Exponential phase [1], because absorbance doubles every hour (0.05, 0.10, 0.20, 0.40) [1]. (b) From 1 h to 4 h: 0.40 ÷ 0.05 = 8 = 2³, so n = 3 generations in 3 h [1]. k = 1.0 generation per hour [1]. g = 1 h (60 minutes) [1]. (c) Nutrients are becoming limiting or waste products are building up [1], so the division rate is slowing as the culture moves towards the stationary phase [1]. (d) It includes dead cells and cannot distinguish them from living cells [1]. [8] Examiner insight: Quote values from the table to justify a phase. Linking the slowing growth to a named cause and the next phase gains more than saying “it levels off”.

Where marks are usually lost

• Omitting the dilution factor or the volume plated in a colony-forming unit calculation.
• Converting 0.1 mm³ to cm³ incorrectly in haemocytometer calculations.
• Treating all the dilution steps as 1 in 10 when a question uses 1 in 100.
• Describing the lag phase as “no growth” with no mention of adaptation.
• Defining the stationary phase without the balance of division and death.
• Using stationary or lag-phase values when calculating k.
• Forgetting to divide the number of generations by the time.
• Writing aseptic steps with no reason, or listing only “keep it sterile”.
• Leaving out the control and the way of holding variables constant in a plan.

Next steps

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

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.