a dairy is suspected of improperly handling milk during the pasteurization proce
ID: 78579 • Letter: A
Question
a dairy is suspected of improperly handling milk during the pasteurization process. samples of the pasteurized milk are brought to the lab for viable plate counts. a 1 ml sample of the pasteurized milk is added to 99 ml of water. 1 ml and 0.1 ml of this dilution are plated and the results are too many to count. a second dilution of 1 ml into 99 ml of water was also made with 1 ml and 0.1 ml plated. the results were too numerous to count and 320 colonies, respectively. a third dilution of 1 ml into 9 ml of water was performed and plates were made with 1 ml and 0.1 ml. the results were 290 and 27 colonies, respectively. what was the total bacterial count in the pasteurized milk?
Explanation / Answer
Answer:
Based on the given information:
first dilution: 1 ml of pasteurized milk is added to 99 ml of water, therefore:
dilution factor 1 (df 1) = total volume / volume of milk = (99+1)/(1) = 100x
When the sample from 100 fold dilution is plated then too many colonies are obtained.
Therefore second dilution is made: 1ml + 99ml water
dilution factor 2 (df 2) = total volume/volume of milk = (99+1)/(1) = 100x, still too many colonies are obtained to count (320 colonies)
Third dilution (df 3) 1ml of sample + 9ml water, dilution factor (df 3) = (9+1)/(1) = 10x
Total dilution factor = df1 x df2 x df3 = (100 x100 x 10) = 100000 or 105
Number of colonies obtained = 290 colonies (with 1ml) and 27 colonies (wth 0.1ml) plating
Therefore total bacterial count in the original undiluted pasteurized milk sample = Number of colonies x total dilution
When 1 ml sample is used, total bacterial count = 290 x 105 = 2.90 x 107CFU/ml
When 0.1ml sample is used, total bacterial count = 27 x 10 x 105 = 2.70 x 107 CFU/ml
Average bacterial count = (2.90 x 107 + 2.70 x 107 ) / 2 = 2.80 x 107 CFU/ml
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