Online Submission Assignment Cover Page

Name: Wei Wu

ID#: 0819397

Date: February 6 2015

Course #: FOOD 2010

Course Name: Principles of Food Science

Instructor: Dr. Massimo Marcone

Assignment #: 1

Assignment Name: Assignment 1

Please note: If your address has changed, please complete the Change of Address Form accessible from http://www.open.uoguelph.ca/online/forms

Questions or concerns? e-mail: info@open.uoguelph.ca

# of Pages (including this one): 7

(For instructor’s use)

Grade:

Questions

1. Examine the ingredient list for Yves Jumbo Veggie Dogs. Vegetable protein and a vegetable

gum are included to help give the product its structure. Identify these ingredients and

discuss their functionality with respect to product structure. (8 marks)

The vegetable proteins found in the Yves Jumbo Veggie Dogs come from soy protein, wheat gluten

and tofu(water, soybeans, magnesium chloride). In food preparation, soy protein serves as an

emulsifying agent to encourage the mixing of ingredients and enhance the structure of the product

(Kinsella 1979). Its gelation property can also improve the firmness, pliability and texture of the veggie

dogs (Kinsella 1979). Water retention is another function of soy proteins which can increase yield by

reducing cooking shrinkage (Kinsella 1979). Wheat gluten has an elastic and sticky characteristic which

helps to improve the structure and texture of the veggie dogs (Murano 2003).

The vegetable gun included in veggie dogs is Carrageenan (Murano 2003). Carrageenan is a

polysaccharide which mainly serves as a texture agent (Murano 2003). Carrageenan is a hydrocolloid

which thickens and stabilizes the structure of the veggies dogs (Murano 2003).

These vegetable proteins and vegetable gums play important role in ensuring the structure of veggie

dogs.

References

Kinsella, J.E. (1979). Functional properties of soy proteins. Journal of the American Oil Chemists’

Society. 56(3), 242-258.

Murano, P. (2003). Understanding Food Science and Technology. Belmont: Wadsworth Publishing

Group.

2. The 2 ingredient lists presented below are derived from a product consisting of a creme

filling sandwiched between 2 cocoa cookie wafers. The 2 formulations represent different

versions of the same type of product. (Note that ingredients are listed in descending order

by weight.)

Regular product

sugar, enriched wheat flour, palm oil, cocoa, high fructose corn syrup, baking soda,

cornstarch, salt, soy lecithin (emulsifier), artificial flavour, chocolate

Reduced calorie product

maltitol, enriched wheat flour, palm oil, polydextrose, cocoa, cornstarch, glycerin, milk

protein concentrate, inulin, whey protein concentrate, soy lecithin (emulsifier), baking

soda, salt, dextrose, artificial flavour, cellulose gum, chocolate, cream, acesulfame-K,

sucralose

The lower calorie version has a modified formulation which attempts to mimic flavour

characteristics and creamy mouthfeel associated with the regular formulation. Identify 2

ingredients responsible for imparting flavour and 2 ingredients responsible for imparting

mouthfeel in the reduced calorie product. Discuss their roles in the reduced calorie

formulation in comparison with ingredients performing the same function in the regular

formulation. (12 marks)

Two ingredients responsible for imparting flavor in the reduced calorie version of the product are

maltitol and sucralose. They are substitutes for sugar and high fructose corn syrup in the regular version

of the product. Sugar and high fructose corn syrup contribute to a significant amount of calories in the

regular product. Maltitol is a polyol that has a very similar taste to sucrose (Calorie Control Council

2015). It is about 90% as sweet as sugar but it has significantly less calories, only 2.1 calories per gram

(Calorie Control Council 2015). Sugar contains 3.87 calories per gram and high fructose corn syrup

contains 2.81 calories per gram (LIVESTRONG 2013). Sucralose is a sweetener that tastes like sugar

and is also 600 times sweeter than sucrose (Calorie Control Council 2015). Hence, to produce the same

amount of sweetness, a smaller amount of sucralose can be added to the food product. Sucralose passes

through the digestive system unaltered, as it is not metabolized it has no calories, thus it can reduce

calorie intake by a large amount if it replaces sugar and high fructose corn syrup in food products

(Calorie Control Council 2015).

Two ingredients responsible for imparting mouthfeel in the reduced calorie version of the product

are polydextrose and cellulose gum. Polydextrose is able to provide thickness and texture to the food

product (Murano 2003). Besides being a sweetener, high fructose corn syrup also enhances the soft and

chewy texture of food products as it can retain moisture and resist crystallization after baking

(SweetSurprise 2015). Thus, polydextrose can compensate for the texture change due to the lack of high

fructose corn syrup in reduced calorie version of the product. As polydextrose is only partially

fermented in the colon, it contributes to 1 calorie per gram which is lower than that of high fructose corn

syrup (Calorie Control Council 2015). Cellulose gum is especially useful in making diet foods that tend

to substitute water or other liquids for fat due to its ability to bind to water (LIVESTRONG 2011).

Thus, cellulose gum can substitute high fructose corn syrup as both of them can retain water which is

important for the texture of the food product.

References

Calorie Control Council (2015). Maltitol. Available at: http://www.caloriecontrol.org/sweeteners-and-

lite/polyols/maltitol

Calorie Control Council (2015). Probiotics 3. Available at: http://www.caloriecontrol.org/health-

professional-library/prebiotics-and-probiotics/probiotics3

Calorie Control Council (2015). Sucralose. Available at: http://www.caloriecontrol.org/sweeteners-and-

lite/sugar-substitutes/sucralose#Safety

LIVESTRONG (2013). Calories Per Gram of Sucrose. Available at:

http://www.livestrong.com/article/298844-calories-per-gram-of-sucrose/

LIVESTRONG (2011). Is Cellulose Gum Harmful?. Available at:

http://www.livestrong.com/article/448293-is-cellulose-gum-harmful/

Murano, P. (2003). Understanding Food Science and Technology. Belmont: Wadsworth Publishing

Group.

SweetSurprise (2015). Why HFCS. Available at: http://sweetsurprise.com/why-use-hfcs

3. Kefir is a milk product touted to have health benefits. Explain why kefir may be slightly

alcoholic, carbonated, and have a sour flavour. What microorganisms are used to produce

kefir? (5 marks)

Kefir is a fermented milk drink that is similar to yogurt (About Food 2015). In the process of

making Kefir, a specific mixture of bacteria and yeast cultures are added to milk to start

fermentation (About Food 2015). The starter cultures are called kefir grains since they look like

small, lumpy, granules (About Food 2015). The bacteria Lactobacilus caucasius ferments lactose

in milk to form lactic acid, which explains the sour flavor of Kefir (About Food 2015). 2 yeasts,

namely Saccharomyces kefir and Torula kefir, ferment lactose into a small amount of alcohol

and carbon dioxide, which explains why kefir may be slightly alcoholic and carbonated (About

Food 2015).

References

About Food (2015). What is Kefir?. Available at:

http://foodreference.about.com/od/Dairy/a/What-Is-Kefir.htm

4. According to regulations, maple syrup must have a sugar content of 66% by weight (66

degrees Brix). Calculate the mass of maple sap required to make 500 kg of syrup. How

much water must be removed by evaporation to make 500 kg of syrup? Assume that the

sugar content of the sap is 3%. (5 marks)

X= weight of maple sap (3% sugar)Y= weight of evaporated water (no sugar)

X-Y= 500kg (66% sugar)

0.03X+0Y=0.66*500

Hence we have X=0.66*500/0.03=11000kg, thus Y=X-500=11000-500=10500kg.

So, 11000kg of maple syrup is required to make 500kg of syrup and 10500kg of water must be

removed by evaporation to make 500kg of maple syrup.

5. A bowl of potato salad was left on the picnic table for 3 hours on a hot summer day. The

microbial load was 300 CFU/g when the salad was put on the table. Would the salad be

spoiled after the 3 hour period? (Base your assessment on 1,000,000,000 CFU/g as the

threshold for spoilage.) Assume a generation time of 20 minutes under ideal growing

conditions, with no deaths, immigration or emigration. (5 marks)

3 hours = 3*60 minutes= 180 minutes

180 minutes/ (20 minutes/generation) = 9 generations

300 CFU/g * 29=153600 CFU/g

Hence the salad will not be spoiled as the microbial load would be 153600 CFU/g which is less than the threshold of 1000000000 CFU/g.

6. Strawberries are quick frozen at a rate of 6500 kg/h. The fruit enters the freezer at a

temperature of 15°C and is frozen to a final temperature of -20°C. Calculate the rate of

heat removal per hour. (5 marks)

Specific heat of strawberries above highest freezing temperature = 3.852 kJ/kg*◦C

Specific heat of strawberries below highest freezing temperature = 1.968 kJ/kg*◦C

above freezingQ1= mass*specific heat capacity*temperature change = 6500kg x 3.852 kJ/Kg*◦ C x [15 - (-0.8) ◦C]= 395600.4 kJ

Q2= mass*latent heat=6500kg*300.20 kJ/kg=1951300kJ

below freezingQ3= mass*specific heat capacity*temperature change = 6500kg*1.968kJ/kg*◦C*[-0.8 - (-20)◦C] = 245606.4 kJ

Total heat removed = Q1+Q2+Q3 = 2592506.8 kJ. Since all of this is done in 1 hour, the rate of

heat removal per hour would be 2592506.8 kJ/hr.