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- First There Was One, Then There Were Two RESULTS Results of the plate counts for the samples you worked with. Number of Colonies for Each Dilution Plated Time Sampled 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10:40 TNTC TNTC TNTC 280 TFTC TFTC3 T FIC2 TFTC TFTC 13 TNTC TNTU TNTC TNTC I60 TNTC TNTC 1:00 TFTC q a TFTC O 1:20 11:40 12:00 TNTC TNTC87 TFTC TFTC, TFICg TNTC ITNTC ITNTC TNTC 125 TNTC TNTC TNTC TNTC 164 TFIC 13 T+TC Calculate the cfu/ml for each time sample you worked with in the space below and record those values in the table below.Determine what percentage of the culture was living (viable) and what percentage was dead (mortality). Plates Plate Dilution Volume plated No.of colonies Avg No Concentration of diluted sample Cd(cells/mL) Concentration of original sample Cu(cells/mL) 1 10^-3 10ul R1=130,R2= 110,R3=210 150 150mL 1.50*10^6 Volume of cells(mL) Volume of diluent(mL) Total dilution(D) Hemocytometer count Avg cells in 1 mm^2 area Concentration of diluted sample Cd(cells/mL) Concentration of original sample Cu(cells/mL) 4.3 0.5 0.1 grid 1= 171 , grid2 = 185 178 1.78*10^5 1.78*10^6Five mL of bacterial culture is added to 45 mL of sterile diluent. From this suspension, the following serial dilutions were made, two 1:100 and one 1:10 dilutions, and 0.1 mL is plated onto Plate Count Agar from the last dilution. After incubation, 186 colonies were counted on the plate
- Ten grams of hamburger were added to 90 mL of sterile buffer. This was mixed well in a blender. One-tenth of amL of this slurry was added to 9.9 mL of sterile buffer. After thorough mixing, this suspension was further dilutedby successive 1/100 and 1/10 dilutions. One-tenth of a mL of this final solution was plated onto Plate Count agar.After incubation 145 colonies were present. How many colony-forming units were present in the total 10 gramsample of hamburger?3. A bacterial culture was diluted and results from duplicate plates were obtained as indicated below. What was the number of colony forming units/mL of the original culture? Dilution used for plating 10-2 10-3 104 10-5 10⁰ 10-7 10-8 4. 5. Amount plated 0.1 mL 0.1 mL 0.1 mL 0.1 mL 0.1 mL 0.1 mL 0.1 mL Colony counts after incubation (Results from duplicate plates) Too many to count Too many to count 341 413 99 175 27 : 29 7:2 0:0 Ten grams of hamburger were added to 90 mL of sterile buffer. This was mixed well in a blender. One-tenth of a mL of this slurry was added to 9.9 mL of sterile buffer. After thorough mixing, this suspension was further diluted by successive 1/100 and 1/10 dilutions. One-tenth of a mL of this final solution was plated onto Plate Count agar. After incubation 145 colonies were present. How many colony-forming units were present in the total 10 gram sample of hamburger? Devise a serial dilution scheme to prepare a 10-5 dilution on a plate using the least number of…Five ml of bacterial culture is added to 45 ml of sterile diluent. From this suspension, the following serial dilutions were made, two 1:100 and one 1:10 dilutions, and 0.1 ml is plated onto Plate Count Agar from the last dilution. After incubation, 186 colonies were counted on the plate. 1. What is the dilution factor, or how much of the original sample was diluted?
- While conducting the Hydrogen Sulfide Test (SIM Agar) you open up one of the tubes and it gives off a rotten egg smell while another tube you cultured does not. What does the smell indicate? How come there was no change in color in the tube that indicated a positive test? Please explain how the color change is to occur. Think critically.1. You wish to determine viable counts on a culture of Bacillus subtilis. You begin by pipetting 1 ml of culture into 99 ml of sterile water. After mixing the dilution well you make a series of 4 further dilutions of 10-1 each. From the three most dilute samples, you prepare three pour plates using 1 ml in each. After incubation you find the plate counts of the plates are 16, 245 and 890 respectively. a) Show the dilution scheme. b) What is the estimated viable count (cells/ml) in the original culture? 2. Scientific Notation. Fill in the missing information. a) 4.5 x 109 = _______ b) 50 x 107 = _______ c) 2300 x 1010 = _______ d) 0.54 x 108 = _______Case #6: A 35-year old man with a history of intravenous drug use entered the local health clinic with complaints of a dry persistent cough, fever, malaise, and anorexia. Over the preceding 4 weeks, he had lost 15 pounds and experienced chills and sweats. A chest radiograph revealed patchy infiltrates throughout the lung fields. Because the patient had a nonproductive cough, sputum was induced and submitted for bacterial, fungal, and mycobacterial cultures, as well as examination for Pneumocystis organisms. Blood cultures and serologic tests for HIV infection were performed. The patient was found to be HIV positive. The results of all cultures were negative after 2 days of incubation.
- Which test tube is the control group? Test tube 1: 5 Hydrogen peroxide only Test tube 2: 5 Hydrogen peroxide and ground potato Test tube 3: 5 Hydrogen peroxide and diced potato Test tube 4: 5 Hydrogen peroxide And boiled potatoBelow is a diagram of the serial dilution of a culture with 1 x 10º cells. Fill out the missing information. 250 Culture Diluent 1 x 10° cells/mL 9 mL 9 mL 9.9 mL 9.9 mL 99.9 mL Volume to add to diluent Final Dilution level Theoretical count after plating 100 uLYou counted 40 colonies on a plate in your dilution series. The plate was inoculated with 1.0ml from a 10-7 test tube, what is the OCD?..