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E. Coli Lab Report

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E. Coli Lab Report
RESULTS AND DISCUSSION E. coli and S. aureus were tested to determine effectiveness of photodynamic inactivation with the use of methylene blue as photosensitizer and blue LED as alternative light source. Table 1 shows the resulting number of colonies for each replicate of every treatment per bacteria tested. Each of the treatments was averaged to get the mean number of colonies per treatment.
All experimental setups were produced in quintuplicate, however only three of the five replicates were presented and discussed in order to attain consistency between replicates of both bacterial cultures.
Among the treatments done in E. coli (refer to Table 1), L-P- obtained the highest colony count with an average of 699 CFU/mL followed by L+P- with
…show more content…
The statistical analysis displayed by text notations in Table 1 shows that the p-value of the difference in treatment is greater than 0.05 (defined as p-value>0.05). This suggests that the difference between resulting colonies per treatment per bacteria is significant. Through this, it is determined that the photodynamic inactivation of gram-negative bacteria E. coli colonies is significantly more effective than the inactivation of gram-positive bacteria S. aureus colonies.

In table 3, the survival fraction (SF) and the growth inhibition by percentage (%) were calculated from the mean CFU miliLiter-1 of each treatment per bacteria to further determine the trend of the treatments applied. Between the two types of bacteria, the E. coli showed a more positive reaction towards Photodynamic inactivation with blue LED because the decrease in survival fraction of E. coli is 0.90 of the total colony-forming units per mL from the control treatment with the PDI30 treatment as compared to that of the S. aureus which is 0.74.

Table 3

Growth inhibition and survival fraction of different Photodynamic inactivation experimental set-ups of E. coli and S.

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