Answer / kangchuentat

GENETIC ENGINEERING - ANSWER 27.3 : (a) Phenotypes of parents = gray wingless, black winged. Phenotypes of recombinant children = gray winged, black wingless. Number of recombinant children = gray winged + black wingless = 115 + 95 = 210. Recombination frequency between the two genes = Number of recombinant children / Total number of children = 210 / (210 + 890 + 900) = 0.105 or 10.5 % = distance = 10.5 map units. (b) Base pairs number of adenine = 180 x 0.2 = 36 = base pairs number of thymine. Base pairs number of guanine = base pairs number of cytosine = (Total number of base pairs - 2 x Base pairs number of adenine) / 2 = (180 - 2 x 36) / 2 = 54. 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.

ENVIRONMENTAL ENGINEERING - QUESTION 22.2 : 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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