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Illustrate a hypothetical graph showing the growth progression of lactic acid bacteria and coliforms in a successful burong mustasa fermentation. Remember to label your graph properly.
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- Make a hypothetical graph showing the growth progression of lactic acid bacteria and coliforms in a successful burong mustasa fermentation. Remember to put labels on the graph.1). SIM Tubes and TSI slants turn black if the organism is able to reduce iron compounds. For which organism(s) below would you expect a black reaction? Circle all that apply. Escherichia coli. Proteus vulgaris. Salmonella typhimurium. Alcaligenes faecalis. Shigella flexneri. 2). If Citrate is a sugar and fermentation generates an acid, why does the BromThymol Blue indicator in the media turn blue and not yellow? 3). Why is Salmonellosis associated with eggs? Can you get Salmonellosis from other animal sources?1. Lactococcus lactis uses fermentation as its only method of ATP production, and has SOD and catalase. How would the growth of L. lactis in oxic conditions compare to its growth in anoxic conditions? 2. Legionella pneumophila uses aerobic respiration as its only method of ATP production. How would the growth of L. pneumophila in oxic conditions compare to its growth in anoxic conditions?
- 1. List down the steps that you think will influence the fermentation and compare those steps employed in making burong mustasa. Also, note the factors that might cause variations in the consortium of LAB found in burong mustasa, and kimchi.Switchgrass is used for ethanol production. The composition of the switchgrass is 37% cellulose, 24% xylan, 3% galactan, 4% arabinan, 20% lignin, 7% extractives, and 5% ash. A dilute acid pretreatment method is applied to the switchgrass before enzymatic hydrolysis and fermentation. The pretreatment hydrolyzes 10% hexosan and 90% pentosan into monomeric sugars. Approximately 30% of the hydrolyzed pentoses further react & are decomposed to furfural. Assume that there is no decomposition of the hydrolyzed hexoses. Further Assume that lignin, extractives, and ash do not change during the pretreatment. • How much of each lignocellulosic sugar (glucose, xylose, galactose, and arabinose) is produced when pretreating 1,000 kg (dry matter) switchgrass? How much furfural is formed? • Is water consumed or produced in these pretreatment hydrolysis and dehydration reactions? How much in each?Which growth condition shown above has the largest doubling time? How would you explain the observed growth data based on information about metabolism and oxygen requirements? Definitely take into consideration the relative growth of all three growth conditions. Surprisingly, cultures grown in anerobic conditions are non-motile, whereas cultures grown in normal O2 saturation have several peritrichous flagella and swim motility. Explain briefly the mechanism of flagellar motility in terms of energy source for flagellar motility and why it may be lower in anaerobic conditions for L. monocytogenes.
- Kombucha requires two fermentation steps. The first one is partly aerobic and partly anaerobic. The aerobic growth occurs because the kombucha is allowed to brew exposed to oxygen. Where is the anaerobic growth occurring in the first fermentation step? The second fermentation step is the step in which we add the kombucha to a sealed bottle therefore I'm confused as to where it can occur before we seal it.1. Given that the total magnification for the Anabaena photomicrograph above is 400X, and the magnification of the eubacterial cells in Exercise 1 (above) is 1000X, are these Anabaena larger or smaller than the eubacteria? Approximately how much larger/smaller are they? 2. What is the mechanism of action of Crystal Violet (i.e. what does it do to the cells and how does it do it)? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. CV turns only Gram positive cells purple by interacting with the cells' peptidoglycan. Gram negative cells do not change colour. b. CV turns all cells purple by interacting with the cells' peptidoglycan. c. CV turns all cells pink by interacting with the cells' peptidoglycan. d. CV turns all cells purple by interacting with the hydrophobic layers in the cell membrane. e. CV turns all cells pink by interacting with the hydrophobic layers in the cell membrane. 3. What is the…Table 3 - Determination Tube # Potato Expt. Temp. 2c 3c 4c 5c extract (mL) 2 2 2 2 room temp 20 °℃ 40 °℃ 75 °℃ Boiling 100 °C of the optimum temperature of catechol oxidase enzyme. 1st Absorbance 0 min. at Expt. Temp. dH₂O Catechol (mL) (mL) 0 0 0 0 13 13 13 13 Start Time: 4:20 Absorbance: 0.072 Start Time: 4:25 Absorbance: 0.114 Start Time: 4:25 Absorbance: 2nd Absorbance after 10 min. in Expt Temp. Absorbance: 0.097 Time for reading: 4:30 Absorbance: 0.128 Time for reading: 4:33 Absorbance: Subtract 1st from 2nd absorbance 0.056 0.137 Time for reading: 4:35 Absorbance: 0.132 0-193 Start Time: 4:27 Time for reading: 437 Absorbance: 0-126 0.023 0.061 0.029 3rd Absorbance after 10 more min. at Room Temp. Time for reading: 4:40 Absorbance: Based on the data from Table 2 answer these questions: Q10) What is the enzyme's optimum temperature? Answer: Q11) Use the difference between 2nd and 1st absorbance to support your conclusion for the optimum temperature. 0.159 Time for reading: 4:43…
- a. What is the purpose of boiling the wort in beer preparation?b. What are hops used for?c. If fermentation of sugars to produce alcohol in wine is anaerobic,why do winemakers make sure that the early phase of yeastgrowth is aerobic?Describe how a Triple Sugar Iron agar tube would appear if the following situation was true. Please note that all items in the following list are occurring in the same tube of agar. 1. Positive for glucose fermentation 2. Negative for lactose and sucrose fermentation 3. Positive for H2S production 4. Negative for gas productionYeast cells has been cultured on glucose (Table 1). The growth data follows the Monod Equation Table 1: Growth data for yeast cells Glucose concentration (mg/L) 7 10 15 40 200 1000 Specific growth rate (h*¹) 0.066 0.088 0.12 0.294 0.330 0.364 0.419 0.451 0.2 0.304 0.340 0.458 Observed yield, Y'x/s (g/g) a) Calculate the maximum specific growth rate (max) and the substrate constant (Ks) b) Estimate the true yield Y'x/s and the maintainance coefficient ms c) Calculate the doubling time at a concentration of 15 mg/L glucose d) Calculate the substrate consumption rate (rs) assuming a substrate concentration of 40 mg/L and a biomass concentration of 100 mg/L