A capstone team is investigating the microbial load in wash-clothes used in restaurants. They enumerate their samples for total bacterial counts using pour-plate technique on PCA (Plate Count Agar) plates. The CFU counts on all PCA plates were out of countable range. However, they wanted to use the valuable data to get a rough estimate. Use their data and estimate the original bacterial concentration (plate 1 & 2 are duplicates, and inoculum volume was 1.0 mL). Tube Dilution Ratio Plate 1: CFU count Plate 2: CFU count $10^{-4}$ TNTC TNTC $10^{-5}$ TNTC TNTC $10^{-6}$ 312 308 $3.1 \times 10^8$ (estimated) Over $3.0 \times 10^5$ (estimated) Over $3.0 \times 10^8$ (estimated) $3.0 \times 10^8$ (estimated)
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- Plate 1: Dilution factor = 10^4 x 10^-5 x 10^-6 = 10^-15 - Plate 2: Dilution factor = 10^4 x 10^-5 x 10^-6 = 10^-15 Show more…
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Plate count data are used to calculate microbial population density as colony forming units (CFUs) per milliliter of culture. Culture density (CFU/mL) = number of colonies / (dilution factor) x (volume plated, in mL) Use the data below to calculate the CFU/mL in this sample of bacteria after 1 hr of growth. Important to know for these types of questions – only use colony counts that are between 10 and 300. TNTC = too numerous to count, so many colonies that you can’t distinguish them to count. Use the plates with numbers of colonies that can be counted, calculate the CFU on each one, and then the average between all of them. Looking for more assistance? Read section 6.4 in the textbook. Time: 1 h Dilution (before plating): none, 1:10 (1 x 10^-1), 1:100 (1 x 10^-2), 1:1,000 (1 x 10^-3) Volume spread on plate: 0.1 mL, 0.1 mL, 0.1 mL, 0.1 mL Number of colonies observed (replicate plates): TNTC, TNTC; 499, 518; 83, 87; 10, 8 TNTC indicates too numerous to count. * Cell density could not be calculated accurately because plates contained too many colonies to be counted, too many colonies to be counted accurately (>300), or too few colonies to be meaningful (<10)
Farhan A.
Standard Plate Count Worksheet Bacteria are often present in concentrations that are too high to easily count. Quantitative determinations of bacteria in liquid suspensions can be done using the Standard Plate Count method. This method dilutes the bacterial suspension in a series of steps, then a sample of these are added to nutrient agar plate for growth of the viable (living) cells. The resulting colonies times the dilution factor equals the original concentration of viable bacteria/mL. Plates with too few colonies (< 30) are not statistically reliable. Example: If the original bacterial suspension contained 1 million viable bacteria per milliliter (1 x 10^6 bacteria/mL), then diluting the culture by a factor of 10,000 would yield 100 colonies on a NA plate. Dilution: 1:100 1:100 Cell Conc: 10^6 b/mL -> 10^4 b/mL -> 10^2 b/mL -> 100 bacteria grow into 100 colonies Problem 1 Answer the following questions based on this dilution series. Colony count: TNTC TNTC 196 15 TNTC = Too Numerous To Count 1. What is the total dilution factor on each plate? D = E = F = G = 2. What is the concentration of viable bacteria in the original sample? 3. If the sample had started out with 6.8 x 10^5 bacteria/mL, how many colonies are expected on plate G?
Madhur L.
Table 1. Plate count results. TNTC = too numerous to count. Dilution | Colonies (1st replicate plate set) | Colonies (2nd replicate plate set) 1/2 | TNTC | TNTC 10^-1 (1/10) | 455 | 424 10^-2 (1/100) | 40 | 38 10^-3 (1/1000) | 1 | 2 10^-4 (1/10,000) | 0 | 0 7. A flask of Klebsiella bacteria has been growing overnight, and you want to determine the number of cells (CFU) per ml in your flask by viable plate count. You make a series of dilutions, and prepare two agar plates (replicates) for each dilution (1 ml of dilution is spread on each plate). Based on the results shown in the above table, what is the CFU/ml of the original culture? A. 3.9 x 10^3. B. 3.27 x 10^3. C. 4.41 x 10^3. D. 1.5 x 10^4. E. 4.0 x 10^4.
Suman K.
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