How the agar diffusion technique measure bacteriostatic or bactericidal activity?

Answer 1

Case History: Determining the _Effectiveness of a Food Preservative Background In order to determine whether a newly synthesized chemical might be a useful food preservative, the chemical was tested for its ability to inhibit bacterial growth. Control 500 ml of cottage cheese was inoculated with 2 ml of a 24-hr culture of Pseudomonas aeruginosa and incubated at 25°C. Five hours after inoculation, a standard plate count showed there were 200 bacterial cells/ml in the cottage cheese. After 29 hours at 25°C, there were 1,000,000 cells/ml in the cottage cheese. Experiment 500 ml of cottage cheese containing the preservative was inoculated with 2 ml of a 24-hr culture of P. aeruginosa. After 6 hours of incubation at 25°C, a standard plate count was performed. There were 700 bacterial cells/ml in the cottage cheese. After 38 hours, there were 61,000,000 bacterial cells/ml in the cottage cheese. Number Log 1 0.00 2 0.30 5 0.70 6 0.78 24 1.38 32 1.51 200 2.30 700 2.85 1.00 x 106 6.00 6.10 x 106 6.79 6.10 x 107 7.79 Questions 1. Why were plate counts used instead of direct microscopic counts or turbidity measurements? Plate counts allow a researcher to determin the number of bacteria capable of reproducing from the cheese sample (viable, culturable cell count). A microscope would show all bacteria and it would be hard to determin if the target organism or for example lactobaccilus sp. were being counted, the count would be higher than the viable, culturable cell count. Turbidity is only ever a bench mark approximation of cell number and it would be almost impossible to separate the test bacteria from the cheese so most turbidity would be due to cheese. 2. How did the control cottage cheese and the experiment cottage cheese differ? Was this a fair test? There is not enough information to determine this however , from the information given the cheeses differed only in the preservative being present or not - this would be a fair test. There could however be unconsidered variables such as; were the cheeses equally lumpy, allowing equal oxygen permiation of the substrate? 3. Determine the effectiveness of the new food preservative. This is impossible without more data. The number log makes no sense as presented here. I would expect to see a lag phase followed by log. growth without enough data to plot this graph I cannot answer the question. Interestingly in this question the food preservative is classified as "new" - is it fair to compare a "new" preservative against no preservative? Would it be better to consider a trial tail using the "old" preservative as well? The additional information in this question alters the answer to question 2. 4. Does this type of test determine bacteriostatic or bactericidal activity? This would be determined by examination of the graph and comparing the ploted growth of the control and test data. However the wording of the Aim "its ability to inhibit bacterial growth" suggests were are looking for a bacteriostatic effect. The graph for a bacteriostatic preservative would show a slower gradient in log phase compared to the control as the bacteria struggle to reproduce. A graph for a bactericidal preservative would show a longer lag phase as initally many of the bacteria are killed which is equivelent to innoculating the culture with a lower number of bacteria.

Answer 2

BACTERIOSTATIC
Bacteriostatic, because it measures the presence of chemicals