MICROBIOLOGY: CASE STUDY: Determining - Essay Example

The Effectiveness of the New Food Preservative in Cottage Cheese TEACHER                 The Effectiveness of the New Food Preservative in Cottage Cheese Plate counts were used instead of direct microscopic counts because the latter cannot distinguish between dead and living cells. Moreover, turbidity measurements were not used in bacterial counts in the experiment due to the fact that turbidity measurements cannot detect colonies of cells which number less than 10,000,000. Regarding the fairness of the test, if all variables have been carefully measured and all other extraneous variables eliminated, the test is naturally fair.

However, there also seems to be a need for the performance of the new preservative to be compared with that of an old preservative not only the control cottage cheese with no preservative at all. Without a point of comparison between the old and the new preservative, one will not be able to clearly say whether the new preservative is indeed a good or better one. Based on the results of the experiment, the only possible conclusion one can arrive at is whether the new preservative can indeed be better than nothing, which is the control sample of cottage cheese.

Now, on the effectiveness of the preservative used in the cottage cheese, the data given is simply not sufficient in establishing this. One reason is that the data on the number and log must be compared with a standard preservative or that of the control sample. Another reason is that the data for the control sample were taken after 5 hours and 29 hours, whereas those for the experimental sample were taken after 6 hours and 38 hours. One should take samples and bacterial counts at exactly the same time in order to eliminate the variable of time difference.

Moreover, data on pH is needed because “at neutral pH, such as 7.4, many preservatives do not have antimicrobial effect against Pseudomonas aeruginosa” (Abu-Ghazaleh, 2010). If, therefore, the pH of the experiment cottage cheese were neutral, then no one would be able to expect any observable effectiveness. One last thing is the temperature. The inhibition of P. aeruginosa is possible with a preservative usually only during “7 days of storage at 6, 15, and 21°C” (Bautista et al., 2003).

In the experiment, the incubation period is only 24 hours and the temperature is 25°C, which is slightly different from data from previous studies. Based on the conditions in the experiment, it seems that the bacteria cannot be eliminated at a higher temperature and a shorter time of exposure to the new preservative. All these aforementioned factors must be considered before determining the effectiveness of the new preservative.ReferencesAbu-Ghazaleh, B.M. (2010). “Inhibition of growth and caseinase production of Pseudomonas aeruginosa and Escherichia coli 28 by combination of low pH and NaCl, potassium sorbate or Thymus vulgaris extract.

” Acta Microbiology Review, 57(2), 95-108. Retrieved Oct. 11, 2011 from the National Institutes of Health: http://www.ncbi.nlm.nih.gov/pubmed/20587383Bautista, D.A., Durisin, M.D., Razavi-Rohani, S.M., Hill, A.R., and Griffiths, M.W. (2003). “Extending the shelf-life of cottage cheese using monolaurin.” Food Research International, 26(3), 203-208. Retrieved Oct. 11, 2011 from Science Direct: http://www.sciencedirect.com/science/article/pii/096399699390054M