2. Given the results table below, create a flow chart to differentiate the organisms so you could identify an unknown isolate out of these four possibilities. Organism Urease MR TSI Glucose fermentation JM02 + A/H₂S+ A SL65 + A/A Ag DT22 K/K K YS98 + - A/A Ag

2. Given the results table below, create a flow chart to differentiate the organisms so you could identify an unknown isolate out of these four possibilities. Organism Urease MR TSI Glucose fermentation JM02 + A/H₂S+ A SL65 + A/A Ag DT22 K/K K YS98 + - A/A Ag

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Answer the following questions briefly and concisely 1.How do bacteria in a chemostat and those in a batch culture vary from one another? 2. What happens in a chemostat if the dilution rate is higher than the organism's maximum specific growth rate? 3.Does a chemostat require the use of pure cultures? 4. Why would a complicated culture media for Leuconostoc mesenteroides be simpler to make than one with a fixed chemical composition?
You have several different media onto which you inoculated eight strains of yeast (A-H). The media include a rich medium, an unsupplemented minimal medium, and minimal media each supplemented with one vitamin. Of the yeast strains, one is a prototroph and seven are auxotrophs for a vitamin. After overnight incubation, the following results were observed (tan patches represent growth): D plate 1 (A) B DE F GH plate 5 plate 4 plate 6 Which plate contains an unsupplemented minimal medium? [Select] Which plate contains a rich medium? [Select] plate 2 Which strain is a prototroph? [Select] Strain E is an auxotroph for niacin. Which plate reveals this specific auxotrophy? [ Select] plate 3 plate 7 One strain is an auxotroph for both choline and pantothenic acid. Which one is this most likely to be? [Select]
A culture consisting of 100 litres of nutrient medium containing 12 g l-1 of growth limiting substrate is inoculated with 10g of bacteria in a batch fermenter. The yield of biomass from substrate is 0.05 g g-1 . The maintenance coefficient is 2.0 g g-1 h -1 . The specific rate of product formation due to maintenance is 1.0 h-1 . The maximum specific growth rate of the organism is 0.3 h -1 . YPX= 7.7 and product formation is directly linked with metabolism. a) What is the time required to produce 15 g of biomass? !!!!! ANS IS 3.07 HRS!!!! STEPS on how to reach this please
1. Determine the inactivation rate constant (K) for a microorganism for the following treated effluent sample. The effluent temperature was 20°C. If the activation energy for the disinfection reaction is 60 kJ/mole, determine the inactivation rate constant (K) at 12°C. Assume Chick's law applies. In (N/N.) Time, min 1.5 3.9 3 2.5 5.6 7.7 9.2
Based only on your EMB and MSA results, build a dichotomous key that could differentiate between these four strains.
A lab interested in isolating Escherichia coli from food samples uses a lot of Eosin Methylene Blue (EMB) Agar. The recipe for this media is 36 g/L deionized water. How many grams of media would be required to prepare 350 plates assuming 20 mL per plate? 072 O 112 O 90 O 252
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 = _______
Below are photos of 3 bacteria, letter lableled, inoculated onto MSA plates and incubated. Note no growth on plate C C MSA (Mannitol Salt Agar) Match the descriptions below to the correct letter labeled Bacteria
of 1 stion s page 30- F2 Which of the following terms is associated with the probability of a preparation not being sterile after it has been exposed to a sterilization process? Select one: Oa. Particulate Matter level Qb. Sterility Assurance level О с. Pyrogen Load level O d. Optimal Sterilization level F3 F4 F5 Ol پیار Q Search F6 F7 F8 F9 === 8 9 F10 Time left 22:34:03 F11 Next page
Please give explanation for each step other give dislike
7. E. Coli bacteria were grown in a 5.0 L batch cell culture reactor, with the following data collected: Cell Count (per mL) 10,277,000 Inoculation Time ---> Time 11:19 am 11:35 am 11:57 pm 12:08 pm 12:45 pm 1:07 pm 1:11 pm 1:16 pm 1:19 pm 1:34 pm 1:46 pm 1:57 pm 2:09 pm 2:33 pm 10,313,000 20,984,000 29,750,000 60,202,000 210,909,000 224,750,000 227,960,000 229,216,000 229,115,000 227,260,000 204,935,000 183,745,000 146,103,000 (inoculation is when the cells are first put in the growth media) A. Draw a log10 X vs. time graph and label the phases for batch cell growth. B. Calculate the specific growth rate, μ, and mass doubling time, td, for the exponential growth phase. C. Calculate kd, the specific rate of cell mass loss due to cell death (For B & C, it will be helpful to re-plot the y-axis as In X or In (X/X.), and focus on only the data in that phase) Using spreadsheet software will be helpful in drawing this graph and analyzing data for parts B & C. Be as quantitative as possible,…
use these OD numbers to plot growth curve for E. coli K12- and estimate generation time for this culture. Table 1. Absorbance (O.D.600) measured for growing culture of E. coli K12 (time course) Incubation time O.D.600 Incubation time O.D.600 O min 3 hrs 3 hrs 30 min 4 hrs 4 hrs 30 min 5 hrs 5 hrs 30 min 0.11 1.75 30 min 0.20 1.94 1 hr 1 hr 30 min 2 hrs 2 hrs 30 min 0.27 2.24 0.51 2.48 2.31 2.19 0.65 1.30