Which of the the following results of the extracellular enzyme tests is FALSE? Question 3 options: A Liquid gelatin positive for gelatinase B Clear area around bacterial growth on a starch plate after iodine is added means amylase is present C Clear zone around bacterial growth on a milk plate means caseinase is present D Solid gelatin after is positive for gelatinase E No clear zone around bacterial growth on milk plate means absence of casienase

Which of the the following results of the extracellular enzyme tests is FALSE? Question 3 options: A Liquid gelatin positive for gelatinase B Clear area around bacterial growth on a starch plate after iodine is added means amylase is present C Clear zone around bacterial growth on a milk plate means caseinase is present D Solid gelatin after is positive for gelatinase E No clear zone around bacterial growth on milk plate means absence of casienase

Curren'S Math For Meds: Dosages & Sol

11th Edition

ISBN:9781305143531

Author:CURREN

Publisher:CURREN

Chapter10: Reconstitution Of Powdered Drugs

Section: Chapter Questions

Problem 17SST

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What is the result of this biochemical test in the gelatin deep? Which enzyme is this testing for? What tool is used to inoculate this medium? Gelatin positive liquid, use needle, protease,
What is the unknownn organism based on this biochemical test results? Gram Negative Oxidase Positive Catalase Positive MAC: Lactose Fermenter MSA: Mannitol Fermenter Gelatin Hydrolysis: Negative Indole Production: Negative Motility Indole Ornithine Medium: Negative MR/VP Test: Negative Nitrate Test: Negative Simmon Citrate Test: Positive Urease Test: Positive Possible Organisms: Vibrio fischeri Enterococcus faecalis Micrococcus luteus Streptococcus pneumoniae Pseudomonas aeruginosa Proteus mirabilis Acinetobacter baumannii Staphylococcus aureus Mycoplasma pneumoniae Escherichia coli Candida albicans Salmonella enterica Listeria monocytogenes
Two flasks of E. coli are grown in batch culture in the same medium (2% glucose and amino acids; no nitrate) and at the same temperature (378C). Culture #1 is well aerated. Culture #2 is anoxic. After 16 hours the following observations are made: ■ Culture #1 has a high cell density; the cells appear to be in stationary phase, and the glucose level in the medium is reduced to 1.2%. ■ Culture #2 has a low cell density; the cells appear to be in logarithmic phase, although their doubling time is prolonged (over 1 hour). The glucose level is reduced to 0.2%. Why does culture #2 have so little glucose remaining relative to culture #1, even though culture #2 displayed slower growth and has less biomass?
Can you please help me describe the colony apperance on these two stains? Thank you! Growth on Blood Agar - 48 hours Growth on Chocolate Agar Results: Growth on Blood Agar 48 hours Growth on Chocolate Agar
In the third video, glucose, glycine and maltose are tested as samples with the Benedict's reagent. Test Tube COLOR CHANGE EXPLANATION 1. Glucose 2. Glycine 3. Maltose If glucose and maltose are both sugars, why is maltose not as red in color as glucose?
Cells were incubated with chloroquine for 2 hours before stimulation with lipopolysaccharide (LPS, 1g/ml). Cell viability was measured by staining cells with 3-(4,5-dimethylthiazol-2- yl)-2,5- diphenyltetrazolium bromide. TNF in the culture supernatant and the cell lysate were measured by sandwich ELISA, using specific monoclonal antibodies according to the manufacturer's instructions. To determine the level of cell-associated cytokines, cells were washed with phosphate-buffered saline (PBS) and lysed in a buffer containing 0.5% NP-40. Standard recombinant proteins were also diluted in PBS containing the same concentrations of NP-40 as the cell lysate. Cytokine synthesized were proteolytically cleaved and released into medium. The amount of cytokines in the culture supernatant and the cell lysate were presented. Results for TNF production are shown in Figure1 (Jang et al, 2006). (Cells : PBMC: Human peripheral blood mononuclear cells, monocytes, monocytic lymphoma U-937 and leukemia…
Cells were incubated with chloroquine for 2 hours before stimulation with lipopolysaccharide (LPS, lg/ml). Cell viability was measured by staining cells with 3-(4,5-dimethylthiazol-2- yl)-2,5- diphenyltetrazolium bromide. TNF in the culture supernatant and the cell lysate were measured by sandwich ELISA, using specific monoclonal antibodies according to the manufacturer's instructions. To determine the level of cell-associated cytokines, cells were washed with phosphate-buffered saline (PBS) and lysed in a buffer containing 0.5% NP-40. Standard recombinant proteins were also diluted in PBS containing the same concentrations of NP-40 as the cell lysate. Cytokine synthesized were proteolytically cleaved and released into medium. The amount of cytokines in the culture supernatant and the cell lysate were presented. Results for TNF production are shown in Figure 1 (Jang et al, 2006). (Cells : PBMC: Human peripheral blood mononuclear cells, monocytes, monocytic lymphoma U-937 and leukemia…
Sulfur Indole Motility (SIM) Medium H2S produced, color and +/-: ______________________________ Indole present/Tryptophan hydrolysis, color and +/-: ___________________________ Motile or non-motile: _____________________________
The number of bacterial cells in a culture broth is to be determined by a culture technique. Serial dilutions were performed and a 1 mL aliquot from each dilution was mixed with warm molten agar and poured into a Petri dish. The numbers of bacterial colony forming units (CFU) after overnight incubation are shown in the table below. What is the number of colony forming units per mL of the culture broth? Choose only the most appropriate plate for your calculation. Give your answer as the number only (do not add text for the units). You may use scientific notation with the format 1.12e+6 (that is, 1.12 x 106 cfu/mL). (Note: Canvas will then display your answer a whole number.) Plate Dilution Plate 1 10 dilution Plate 2 10 dilution -5 Plate 3 10 dilution Plate 4 10 dilution Plate 5 107 dilution -8 Plate 6 100 dilution *Too many to count Number of colony forming units (CFU) TMTC* TMTC* TMTC* 867 154 18
The number of bacterial cells in a culture broth is to be determined by a culture technique. Serial dilutions were performed and a 1 mL aliquot from each dilution was mixed with warm molten agar and poured into a Petri dish. The numbers of bacterial colony forming units (CFU) after overnight incubation are shown in the table below. What is the number of colony forming units per mL of the culture broth? Choose only the most appropriate plate for your calculation. Give your answer as the number only (do not add text for the units). You may use scientific notation with the format 1.12e+6 (that is, 1.12 x 106 cfu/mL). (Note: Canvas will then display your answer a whole number.) Plate Dilution -3 Plate 1 10 dilution Plate 2 10-4 dilution Plate 3 10-5 dilution Plate 4 10 dilution Plate 5 10 dilution -8 Plate 6 100 dilution *Too many to count Number of colony forming units (CFU) TMTC* 1,957 65 21 1 0 m
Imagine you have been given a liquid culture of yeast with a starting concentration of 3.67 x 10' cells/ml and are asked to carry out the sample dilution process shown in the figure below. 100μl 100μl 100μl 100μl 100μl 0.9ml 0.9ml 0.9ml H2O H₂O 6.9ml 0.9ml H₂O H₂O H₂O Original 10-1 102 10-3 104 Culture 105 100μl 100μl 100μl Plate A Plate B Plate C a. How many colonies should have been present on Plate A in this example? - Answers must be whole numbers as partial colonies are not expected. b. Imagine you carried out the same dilution scheme shown in the figure above, but now, you do not know the concentration of the original culture. If you counted 163 colonies on Plate B, what is the concentration of cells/ml in the original culture?
The number of bacterial cells in a culture broth is to be determined by a culture technique. Serial dilutions were performed and a 0.1 mL aliquot from each dilution was spread onto Plate Count Agar (PCA). The number of bacterial colony forming units (CFU) after overnight incubation are shown as listed in the table below. What is the number of colony forming units per mL of the culture broth? Choose only the most appropriate plate for your calculation. Give your answer as the number only (do not add text for the units). You may use scientific notation with the format 1.12e+6 (that is, 1.12 x 106 cfu/mL). (Note: Canvas will then display your answer a whole number.) Plate 1 10 Plate 2 10 Plate 3 10 dilution dilution dilution Plate 4 10 dilution Plate 5 107 dilution Plate 6 10 dilution -6 *Too many to count Number of colony forming units (CFU) TMTC* TMTC* 840 28 19 1

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A	Liquid gelatin positive for gelatinase
B	Clear area around bacterial growth on a starch plate after iodine is added means amylase is present
C	Clear zone around bacterial growth on a milk plate means caseinase is present
D	Solid gelatin after is positive for gelatinase
E	No clear zone around bacterial growth on milk plate means absence of casienase