Brewing Case Study FOOD 3270

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University of Guelph *

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3270

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Chemistry

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Dec 6, 2023

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Case Study 3 Brewing Case Study Question 1 . What is the significance of measuring the moisture content? Is the measured moisture content within acceptable limits? Ans: Moisture content is the amount of water present in a substance or material, it can affect the quality of the substance. It’s import ant to measure the moisture content during brewing because moisture content can affect the quality of product, moisture content affects the freshness and quality of ingredients involved in the process such as grains and hops, moisture content can also affect the shelf life of produce, if moisture content is too high it can lead to the spoilage such as mold formation and fungal growth on ingredient involved and if moisture content is too low the malted barley may not germinate at all. Measuring the moisture content is important to ensure consistency and quality of the final product. The measured content 5.35% falls within the acceptable limits, the acceptable range is between 4% to 6% Question 2. Convert the Specific gravity into o Plato then comment on the potential alcohol when fermented. Ans: o P = (SG 1) x 1000/4; = (1.040 1) x 1000/4 = 10 o P Potential Alcohol = ( o P/2) - 0.5. = (10/2) 0.5 = 4.5% : The potential alcohol when fermented is 4.5% Question 3. Was the distatic power, protein and FAN within acceptable limits for a Indian Pale Ale? Provide your reasoning. Ans: a) The distatic power is 100 o L, this is a high value for IPA as IPA has a recommended range 45 65 o L, this high value of distatic power could be the brewer intentions to achieve a specific flavour or character in the IPA and the value could also indicate that the beer has high alcohol content b) The protein content of 11.6% falls within the recommended range of 10 12% for IPA, this indicates that the beer may have a good balance of body and mouthfeel without impacting the foam stability and clarity of the beer negatively c) The FAN level of 200mg/l is at the higher end of the recommended range of 120 200mg/l for an IPA, this value is within the acceptable limits. This could indicate yeasts has enough nutrients to complete fermentation and produce balanced flavour in beer In conclusion the high distatic power may be unusual for an IPA, the protein content and FAN level are within the acceptable limits.
Question 4. With reference to the above pH vs time curves, calculate the AP 10 and AP 20 values for each propagation. Ans: Yeast Propagation AP 10 AP 20 A 5.3 5.2 = 0.1 5.2 3.7 = 1.5 B 5.3 4.8 = 0.5 4.8 3.1 = 1.7 C 5.3 5.3 = 0 5.3 5.3 = 0 Question 5. What caused the drop in pH in Propagation B before the introduction of glucose? Ans: The drop in pH in propagation B before the introduction of glucose could be as a result microbial activity such as growth of lactic acid and can lower the pH of the solution, the buffering capacity of the solution may also be the reason for drop in pH, if the buffering capacity of the solution is low, it can cause the pH to change rapidly. Question 6. Which yeast propagation would you take forward to prepare the pitch and why? Ans: I would use propagation A, this is because at the beginning and end of the test appropriate pH profile can be observed, the end of the fermentation had a prolonged pH of about 3.5, this matches up with our expectation
Question 7. Based on the count data, calculate the number of cells/ml (Hint, the volume of 1 mm x 1 mm square is 10 -4 ml assuming a depth of 0.1 mm). Ans. Cells/ml in propagation = (Sum of cell count*5) x dilution factor x 10,000 = (21+18+13+29+25)*5 x 100 x 10,000 = 5.30 x 10 8 cells/ml Question 8. The above is a picture of a wort sample taken prior to entering the fermentor. How would you interpret the result? Ans: Microbial growth and colonies can be observed on both plates, even on the plate with the antibiotic we can see growth, this result could indicate contamination in the wort sample or the presence oof bacteria involved in the fermentation sample, further observation of colony formed would need to be observed the determine if the microbes would be harmful to the sample. Question 9. Calculate the pitching rate for a yeast propagation containing 8 x 10 9 cells/ml, Wort o Plato 11.5 and volume of 100 liters. Ans. Pitching Rate (in million cells/mL/°P) = (Cells/mL) x (Volume of Wort in Liters) x (°P) / 1000 First, we need to convert the °Plato to specific gravity; Specific Gravity = 1 + (°Plato / (258.6 - ( (°Plato / 227.1) * 100) )) Specific Gravity = 1 + (11.5 / (258.6 - ( (11.5 / 227.1) * 100) )) = 1.045 Pitching Rate = (8 x 10^9 cells/mL) x (100 liters) x (1.045 SG) / 1000 = 836 billion cells The pitching rate for yeast propagation is 836 million cells/mL/°P. Question 10. What is the consequence of a low or high pitching rate? Ans: Low pitching rate can result in incomplete or slow fermentation, which can cause yeast to be unable to metabolize all the available sugars in the wort, this can lead to off flavors in beer, it can also increase the risk of contamination by other microbes leading to spoiled beers High pitching rate can result in rapid fermentation, which can cause yeast to produce high amounts of heat and carbon dioxide, this can lead to off flavors and thinner body in the final beer
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