: Determine the Effectiveness of a Food PreservativeBackground

Chapter 6Case History: Determine the Effectiveness of a Food PreservativeBackgroundIn 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.Control500 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 plated 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.NumberLog10.0020.3050.7060.78241.38321.512002.307002.851.00 x 10^66.006.10 x 10^66.796.10 x 10^77.79Questions1. Why were plate counts used instead of direct microscopic counts or turbidity measurements?Plate counts allow a researcher to determine 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 determine if the target organism or for example lactobacillus 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 plotted 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 initially many of the bacteria are killed which is equivalent to inoculating the culture with a lower number of bacteria.Critical Thinking 1. E. coli was incubated with aeration in a nutrient medium containing two carbon sources, and the following growth curve was made from this culture.a. Explain what happened at the time marked x. How can you tell? 4. Flask A contains yeast cells in glucose-minimal salts broth incubated at 30°C with aeration. Flask B contains yeast cells in glucose-minimal salts broth incubated at 30°C in an anaerobic jar. The yeasts are facultative anaerobes.a. Which culture produced more ATP?b. Which culture produced more alcohol?c. Which culture had the shorter generation time?d. Which culture had the greater cell mass?e. Which culture had the higher absorbance?Clinical Applications1. Assume that after washing your hands, you leave ten bacterial cells on a new bar of soap. You then decide to do a plate count of the soap after it was left in the soap dish for 24 hours. You dilute 1 g of the soap 1:10^6 and plate it on standard plated count agar. After 24 hours of incubation, there are 168 colonies. How many bacteria were in the soap? How did they get there?There were 168 x 10^6 bacteria on that 1g of soap.How did they get there?i) Were some of them there already – the soap was ‘new’ but was it sterile?ii) The ones you left there multiplied.3. The number of bacteria in saliva samples was determined by collecting the saliva, making serial dilutions, and inoculating nutrient agar by the pour plate method. The plates were incubated aerobically for 48 hours at 37°C.Bacteria per ml SalivaAfter Using Mouthwash MouthwashMouthwash 1 13.1 x 10^6 10.9 x 10^6Mouthwash 2 11.7 x 10^6 14.2 x 10^5Mouthwash 3 9.3 x 10^5 7.7 x 10^5What can you conclude from these data? Did all the bacteria present in each saliva sample grow?Chapter 7Case History: The Effect of Closure Type on PreventionMicrobial Contamination of CosmeticsData Contamination incidence (%)skin lotionPump0000000Questions1. How did the bacteria get into the products?2. Which type of closure is the best? Why do you suppose this closure works best?3. Why did the researchers test for these bacteria?Critical Thinking1. The disk-diffusion method was used to evaluate three disinfectants. The results were as follows:Disinfectant Zone of InhibitionX0 mmY5 mmZ10 mma. Which disinfectant was the most effective against the organism?b. Can you determine whether compound Y was bactericidal or bacteriostatic?Clinical Applications2. Between March 9 and April 12, five chronic peritoneal dialysis patients at one hospital became infected with Pseudomonas aeruginosa. Four patients developed peritonitis (inflammation of the abdominal cavity), and one developed a skin infection at the catheter insertion site. All patients with peritonitis had low-grade fever, cloudy peritoneal fluid, and abdominal pain. All patients had permanent indwelling peritoneal catheters, which the nurse wiped with gauze that had been soaked with an iodophor solution each time the catheter was connected to or disconnected from the machine tubing. Aliquots of the iodophor were transferred from stock bottles to small in-use bottles. Cultures from the dialysate concentrated and the internal areas of the dialysis machines were negative; iodophor from a small in-use plastic container yielded a pure culture of P. aeruginosa.What improper technique led to this infection?3. Eleven patients received injections of methylprednisolone and lidocaine to relieve the pain and inflammation of arthritis at the same orthopedic surgery office. All of them developed septic arthritis caused by Serratia marcescens. Unopened bottles of methylprednisolone from the same lot numbers tested sterile; the methylprednisolone was preserved with quat. Cotton balls were to wipe multiple-use injection vials before the medication was drawn into a disposable syringe. The site of injection on each patient was also wiped with a cotton ball. The cotton balls were soaked in benzalkonium chloride, and fresh cotton balls were added as the jar was emptied. Opened methylprednisolone containers and the jar of cotton balls contained S. marcescens. How was the infection transmitted? What part of the routine procedure caused the contamination?Chapter 8Case History: Mapping a Bacterial ChromosomeBackgroundConjunction mapping can be used to locate genes on a bacterial chromosome. In this example, the F+ strain is able to synthesize methionine, valine, leucine, and histidine. The F- strain is unable to synthesize these amino acids; therefore, they must be supplied in the growth medium. Use the results from replica-plating to answer the questions.Questions 1. On which media, if any, should the F+ be able to grow? The F-?2. Identify the prototrophs and auxotrophs. How can you tell?3. Can you map the chromosome of this species? What further information do you need?Critical Thinking3. Pseudomonas has a plasmid containing the mer operon, which includes the gene for mercuric ion Hg 2+ to the uncharged form of mercury, Hg 0.Hg 2+ is quite toxic to cells; Hg 0 is not.a. What do you suppose is the inducer for this operon? brings it into the cell. Why would a cell bring in a toxin?c. What is the value of the mer operon to Pseudomonas?Clinical Applications1. Ciprofloxacin, erythromycin and acyclovir are used to treat microbial infections. Ciprofloxacin inhibits DNA gyrase. Erythromycin binds in front of the A site on the 50S subunit of a ribosome. Acyclovir is a guanine analog.a. What steps in protein synthesis are inhibited by each drug?b. Which drug is more effective against bacteria? Why?c. Which drugs will have effects on the host’s cells? Why?is it more effective than erythromycin for treating this disease?

 
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