Echidnas are mammals that live in Australia and New Guinea. [Figure 2.1] shows an echidna. Scientists analysed the milk produced by female echidnas and identified a protein that they named EchAMP. The scientists predicted that EchAMP may have antibacterial properties. The scientists tested the effect of EchAMP on the bacterium Escherichia coli. 1. 100 E. coli cells were added to each well on a cell culture plate with 96 wells. 2. A treatment solution that contained EchAMP was added to each well on the plate. 3. A chemical that causes living E. coli cells to fluoresce was added to each well. 4. The plate was incubated at 37°C. 5. Every hour for 7 hours, the fluorescence emitted by the E. coli on the plate was recorded as a measure of E. coli population growth. 6. Steps 1-5 were repeated eight times. The scientists also carried out two control experiments. • A negative control experiment repeated the procedure (steps 1–6), but the treatment solution did not contain EchAMP. • A positive control experiment repeated the procedure (steps 1–6), but the treatment solution contained an antibiotic called bacitracin instead of EchAMP.

1. (a)Identify the independent variable in this investigation.[1]
2. (b)The scientists standardised the temperature and the initial number of E. coli cells. State two other variables that the scientists should standardise in this investigation.[2]
3. (c)(i)Explain why the scientists included the negative control experiment in their investigation.[1]
4. (c)(ii)Explain why the scientists included the positive control experiment in their investigation.[1]
5. (d)(i)Some of the results are shown in Table 2.1. Table 2.1 time/hours | mean fluorescence/arbitrary units (au) --- | --- | --- | --- | EchAMP | negative control | positive control 0 | 10 | 10 | 10 1 | 11 | 11 | 11 3 | 15 | 15 | 15 6 | 20 | 35 | 20 7 | 100 | 140 | 70 Use Table 2.1 to complete the graph in [Fig. 2.2] by: • plotting the three results at 7 hours • adding axis labels • completing the key.[2]
6. (d)(ii)Use Table 2.1 and [Fig. 2.2] to compare the effect of EchAMP and bacitracin on the population growth of E. coli.[2]
7. (e)(i)The scientists used t-tests to analyse the results. The scientists compared the fluorescence emitted by E. coli after seven hours when exposed to: • the negative control • EchAMP. State a null hypothesis for this t-test.[1]
8. (e)(ii)For E. coli exposed to the negative control, after seven hours: • the mean of nine fluorescence measurements was 140 au • the sample standard deviation was 8 au. For E. coli exposed to EchAMP, after seven hours: • the mean of nine fluorescence measurements was 100 au • the sample standard deviation was 10 au. The formula for calculating a t-test is: t = |x̄₁ - x̄₂| / sqrt(s₁²/n₁ + s₂²/n₂) key to symbols x̄ = mean s = sample standard deviation n = sample size (number of observations) Calculate a value of t for these data. Show your working and state your answer to four significant figures.[3]
9. (e)(iii)The t value calculated by the scientists was significant. After reading the scientific paper published by the scientists, a student wrote the conclusion: EchAMP would make a good treatment for bacterial infections in the human digestive system. Suggest four reasons why this conclusion might not be valid.[4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.