GENETIC ENGINEERING - EXAMPLE 27.1 : In Mendelian genetics, yellow (Y) is dominant to green (y) and round (R) is dominant to wrinkled (r). (a) What is the probability P of Rr x Rr producing wrinkled seeds? (b) What is the probability P of Yy x yy producing green seeds? (c) What is the probability that RRYy x RrYy would produce RrYy?
GENETIC ENGINEERING - ANSWER 27.1 : (a) There are 3 types of seeds produced : RR, Rr, rR, rr where RR, Rr, rR (3) are dominantly round and rr (1) is recessively wrinkled. P (wrinkled) = 1 / (3 + 1) = 1 / 4. (b) There are 2 types of seeds produced : Yy, Yy, yy, yy where Yy (2) are dominantly yellow and yy (2) are recessively green. P (green) = 2 / (2 + 2) = 1 / 2. (c) Let RR x Rr produces RR, Rr, RR, Rr, then P (Rr) = 2 / 4 = 1 / 2. Let Yy x Yy produces YY, Yy, yY, yy, then P (Yy) = 2 / 4 = 1 / 2. P (RrYy) = P (Rr) x P (Yy) = 1 / 2 x 1 / 2 = 1 / 4. 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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Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
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