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Question 87 - According to Hardy-Weinberg Equation, p x p + 2 x p x q + q x q = 1 where p = dominant allele frequency and q = recessive allele frequency. Let p + q = 1. Fraction of population has 2 copies of the p gene = p x p. Fraction of population has 2 copies of the q gene = q x q. Fraction of population has a copy of p gene and a copy of q gene = 2 x p x q. In a small town, the allele frequency is q = 0.2 for a recessive gene, the delta-32 mutation, that gives human protection from HIV infection. (a) Find the allele frequency a dominant gene, p. (b) What percent of the population has at least a copy of the gene that cause the population either immune to HIV or less susceptible to the disease?
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Answer 87 - (a) Let p + q = 1, then p = 1 - q = 1 - 0.2 = 0.8. (b) Percentage of population has 2 copies of the p gene = p x p x 100 = 0.8 x 0.8 x 100 = 64 %. Percentage of population has at least a copy of the q gene = (1 - Fraction of population has 2 copies of the p gene) x 100 = (1 - p x p) x 100 = (1 - 0.8 x 0.8) x 100 = (1 - 0.64) x 100 = 36 %. 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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POLYMER ENGINEERING - QUESTION 24.1 : The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18. The fraction of polymer chain X of 3 different monomers a, b and c in a polymer are Xa = 0.5, Xb = 0.3 and Xc = 0.2. (i) Calculate number average molecular weight by using the formula Ma Xa + Mb Xb + Mc Xc. (ii) Calculate weight average molecular weight by using the formula (Ma Xa Ma + Mb Xb Mb + Mc Xc Mc) / (Ma Xa + Mb Xb + Mc Xc). (iii) Calculate the polydispersity by using the answer in (ii) divided by answer in (i). (iv) If the molecular weight of repeat unit is 12, calculate the degree of polymerization by using the formula (Ma Xa + Mb Xb + Mc Xc) / (molecular weight of repeat unit).
Question 39 - Acetone and ethanol are separated using a distillation column with a partial condenser and partial reboiler. An equimolar, sub-cooled liquid feed enters at 100 kmol / hr and condenses 1 mole of vapor for every 6 moles of feed. The separation requires a distillate vapor that is 95 mol % acetone and bottoms liquid that is 5 mol % acetone. The reflux is returned from the condenser to the column as a saturated liquid and the operation is run at (L / V) = 1.4 * (L / V) min. Assume constant overflow conditions. (a) Feed operating line is y = [ q / (q - 1) ] x - z / (q - 1) where z = 0.5 for equimolar liquid mixture of 2 components, q = (L’ - L) / F where L’ = L + F + (F / 6) for condensation of 1 mole of vapor / 6 moles of feed. What is y = f(x)? (b) The rectifying operating line is y = (L / V) x + (D / V) (xd) where (L / V) min goes through the points A (0.95, 0.95) and B (0.53, 0.69), V = L + D. What is y = f(x)? Let xd = 0.95. (L / V) min is the slope of the 2 points A and B.
BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.
ENGINEERING ECONOMY - EXAMPLE 7.2 : In the purchase of a machine with a period n = 8.5 years, the minimum attractive rate of return, i = 12 %, the cost P = $55000, F = $4000 is the salvage, annual maintenance A = $3500. The return of the investment or equivalent uniform annual benefit is $15000. The equivalent uniform annual cost is P (A / P, i, n) + A - F (A / F, i, n). The investment is considered acceptable only when equivalent uniform annual benefit is greater than the equivalent uniform annual cost. From the compound interest table, (A / P, i = 12 %, n = 8 years) = 0.2013, (A / P, i = 12 %, n = 9 years) = 0.1877, (A / F, i = 12 %, n = 8 years) = 0.0813, (A / F, i = 12 %, n = 9 years) = 0.0677. Prove by calculations whether the investment above is acceptable.
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