GENETIC ENGINEERING - EXAMPLE 27.2 : (a) Three genes, I, J and K are available. All these genes are linked with respect to one another. If the percent recombination between I and J is 8 %, that between J and K is 10 %, and that between K and I is 18 %, what is the order of the gene? (b) Twenty six genes, a, b, c, d, e, f, ... x, y and z are available. All these genes are linked with respect to one another. If the percent recombination between a and b is 3 %, between b and c is 3 %, between c and d is 3 %, ... between w and x is 3 %, between x and y is 3 %, between y and z is 3 %, then what is the percentage recombination between b and y?
GENETIC ENGINEERING - ANSWER 27.2 : (a) From I as origin, J - I = 8 %, K - J = 10 %, K - I = 18 %, then I ... 8 % ... J ... 10 % ... K. (b) There are 22 genes between b and y, separated by 3 % recombination between 2 adjacent genes. Percentage recombination between b and y = (22 + 1) x 3 % = 69 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.8 : Cash flow agreement where the sign + or - of the amounts are start negative and at a point switch to positive and stay positive is referred to has an investment. The cash flow could be denoted by the symbol of [a, b, c ...]. Consider the following two cases in a biochemical factory : (a) You pay $1 today. You receive $1.05, one year from now. (b) You pay $3 today. You receive $5, one year from today. You pay $2, two years from today. Determine if cases (a) and (b) are an investment or a business transaction with cash flow symbol. Find the interest rate r in the cash flow in both cases (a) and (b).
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FOOD ENGINEERING - QUESTION 23.1 : (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calorie, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?
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Question 48 - Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.
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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : According to Raoult's law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
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