Question { 1501 }

POLYMER ENGINEERING - QUESTION 24.3 : 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, with their respective quantities in N units having the ratio of Na : Nb : Nc = 2 : 3 : 5. (a) Find the numerical average molecular weight of the polymer by using the formula (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc). (b) Find the weighted average molecular weight of the polymer by using the formula (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc). (c) Calculate the polydispersity Q by using the answer in (b) divided by answer in (a). (d) Find the volumetric average molecular weight of the polymer by using the formula (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc). (e) Estimate the polydispersity Q by using the answer in (d) divided by answer in (b).

Answer

POLYMER ENGINEERING - ANSWER 24.3 : (a) Numerical average molecular weight = (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc) = (14 x 2 + 16 x 3 + 18 x 5) / (2 + 3 + 5) = 166 / 10 = 16.6 kg / mol. (b) Weighted average molecular weight = (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc) = (14 x 2 x 14 + 16 x 3 x 16 + 18 x 5 x 18) / (14 x 2 + 16 x 3 + 18 x 5) = 2780 / 166 = 16.747 kg / mol. (c) Polydispersity = [ answer in (b) ] / [ answer in (a) ] = (16.747 kg / mol) / (16.6 kg / mol) = 1.0089. (d) Volumetric average molecular weight = (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) = (14 x 2 x 14 x 14 + 16 x 3 x 16 x 16 + 18 x 5 x 18 x 18) / (14 x 2 x 14 + 16 x 3 x 16 + 18 x 5 x 18) = 46936 / 2780 = 16.883 kg / mol. (e) Polydispersity estimation = [ answer in (d) ] / [ answer in (b) ] = (16.883 kg / mol) / (16.747 kg / mol) = 1.0081. 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.

Question { 1609 }

PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

Answer

PETROLEUM ENGINEERING - ANSWER 25.1 : (a) Ideally (moles of nitrogen) / (moles of oxygen) = (volume of nitrogen) / (volume of oxygen) = 78.09 % V / 20.95 % V = 3.7274 where V is the volume of dry air. (Fact 1, Fact 2). (b) Moles of nitrogen = (moles of oxygen) x (volume of nitrogen) / (volume of oxygen) = 12.5 x 3.7274 = 46.5925 moles. (c) Mass of 1 mole of octane fuel = 1 mole x 114.232 g / mole = 114.232 g. (d) Mass of dry air = mass of oxygen + mass of nitrogen = 12.5 mole x 31.998 g / mole + 46.5925 mole x 28.014 g / mole = 1705.217295 g. (Fact 5) (e) Air-fuel ratio (AFR) = [ Answer in (d) ] / [ Answer in (c) ] = 1705.217295 g / 114.232 g = 14.928. (Fact 3, Fact 4) (f) Fuel-air ratio (FAR) = 1 / [ Answer in (e) ] = 1 / 14.928 = 0.067. 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.

Question { 2009 }

PETROLEUM ENGINEERING - QUESTION 25.2 : (a) The American Petroleum Institute gravity, or API gravity, is a measure of how heavy or light a petroleum liquid is compared to water. Let SG = specific gravity of petroleum liquid, and V = barrels of crude oil per metric ton. Given the formula for API gravity = 141.5 / SG - 131.5 and V = (API gravity + 131.5) / (141.5 x 0.159), find the relationship of SG as a function of V. (b) An oil barrel is about 159 litres. If a cylinder with diameter d = 50 cm and height h = 50 cm is used to contain the oil, find the volume V of the cylinder in the unit of oil barrel by using the formula V = 3.142 x d x h x d / 4. (c) First reference : 1 cubic metre = 6.2898 oil barrels. Second reference : 1 cubic metre = 6.37 oil barrels. What are the 2 factors that cause the difference in such reference data?

Answer

PETROLEUM ENGINEERING - ANSWER 25.2 : (a) Substitute API gravity = 141.5 / SG - 131.5 into V = (API gravity + 131.5) / (141.5 x 0.159) will produce V = (141.5 / SG - 131.5 + 131.5) / (141.5 x 0.159) = 1 / (0.159 SG). Then SG = 1 / (0.159 V). (b) Let V = 3.142 x d x h x d / 4 = 3.142 x 50 x 50 x 50 / 4 = 98187.5 cubic centimetres = 98.1875 litres = 98.1875 litres x (1 oil barrel / 159 litres) = 0.6175 oil barrel approximately. (c) Oil density and temperature. 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.

Question { 6259 }

Question 109 - (a) Acceptable wavefunction in quantum mechanics in the range of : negative infinity < x < positive infinity, vanishes at least at one boundary. Which of the following is the wavefunction or are the wavefunctions of acceptable theory : P = x, P = | x |, P = sin x, P = exp (-x), P = exp (-| x |)? State the reason. (b) Let linear momentum operator P = -ih d / dz. The wavefunction is S = exp (-ikz) where i x i = -1, k and h are constants. Find the linear momentum of such wavefunction by using the term P x S.

Answer

Answer 109 - (a) Acceptable wavefunctions are P = sin x as the boundaries are P = +1 and -1, and P = exp (-| x |) as the boundaries are P = 1 and 0. (b) P x S = -ih d / dz [ exp (-ikz) ] = -ik x (-ih) x exp (-ikz) = -i (-i) kh x S = -kh x S. Then P = -kh = linear momentum. 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.

Question { 1710 }

NATURAL GAS ENGINEERING - QUESTION 26.1 : (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.

Answer

NATURAL GAS ENGINEERING - ANSWER 26.1 : (a)(i) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. (ii) Range is more than 138 but less than or equal to 260 cubic feet per hour for pipe size of 1 inch. (b) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length instead of 50 feet and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. 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.

Question { 1710 }

NATURAL GAS ENGINEERING - QUESTION 26.1 : (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.

Answer

NATURAL GAS ENGINEERING - ANSWER 26.1 : (a)(i) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. (ii) Range is more than 138 but less than or equal to 260 cubic feet per hour for pipe size of 1 inch. (b) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length instead of 50 feet and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. 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.

Question { 1492 }

PETROLEUM ENGINEERING - QUESTION 25.3 : Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.

Answer

PETROLEUM ENGINEERING - ANSWER 25.3 : (a) 1 cubic metre of liquid octane has 703 kilograms or 703 kilograms x mole / (0.11423 kilograms) = 6154.25 mole. Energy needed = [ 255.68 J / (mol K) ] x (6154.25 mol) K = 1573518.64 J where K is the unit symbol for Kelvin and mol is the unit symbol for mole. (b) Density of liquid octane = 703 kilograms per cubic metre = 0.703 grams per cubic centimetre = 703 g / L. There is 0.007 mg = 0.000007 g or 0.000007 g x (1 L / 703 g) = 0.0000000099573257 L of liquid octane in 1 L of water. With negligible amount of liquid octane in 1 L of water, octane is considered insoluble in water. 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.

Question { 1400 }

NATURAL GAS ENGINEERING - QUESTION 26.2 : (a) The Hyperion sewage plant in Los Angeles burns 8 million cubic feet of natural gas per day to generate power in United States of America. If 1 metre = 3.28084 feet, then how many cubic metres of such gas is burnt per hour? (b) A reservoir of natural gas produces 50 mole % methane and 50 mole % ethane. At zero degree Celsius and one atmosphere, the density of methane gas is 0.716 g / L and the density of ethane gas is 1.3562 mg / (cubic cm). The molar mass of methane is 16.04 g / mol and molar mass of ethane is 30.07 g / mol. (i) Find the mass % of methane and ethane in the natural gas. (ii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the gases are ideal where final volume of the gas mixture is the sum of volume of the individual gases at constant temperature and pressure. (iii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the final mass of the gas mixture is the sum of mass of the individual gases. Assume the gases are ideal where mole % = volume % at constant pressure and temperature.

Answer

NATURAL GAS ENGINEERING - ANSWER 26.2 : (a) 1 foot = (1 / 3.28084) metre = 0.3048 metre, 1 cubic foot = 0.3048 x 0.3048 x 0.3048 cubic metre = 0.0283 cubic metre, 8 million cubic feet = 8 million cubic feet x (0.0283 cubic metre / cubic feet) = 0.2264 million cubic metres. If 0.2264 million cubic metres of gas is burnt per day, then 0.2264 million / 24 = 9433 cubic metres of gas is burnt per hour. (b)(i) Let 50 mole of methane and 50 mole of ethane in the gas. Mass (g) = Mole (mol) x Molar Mass (g / mol), then mass of methane = 50 x 16.04 = 802 and mass of ethane = 50 x 30.07 = 1503.5. Total mass of natural gas = 802 g + 1503.5 g = 2305.5 g, then mass % of methane = (802 / 2305.5) x 100 = 34.786 % and mass % of ethane = 100 - 34.786 = 65.214 %. (ii) Density of methane gas = 0.716 g / L, density of ethane gas = 1.3562 mg / (cubic cm) = 1.3562 mg / mL = 1.3562 g / L. Volume (V) = Mass (m) / Density (r). Then V = m / r = 0.34786 m / 0.716 + 0.65214 m / 1.3562 = 0.9667 m, average density, r = m / (0.9667 m) = 1.0344 g / L. (ii) Mass (m) = Volume (V) x Density (r). Then m = Vr = 0.5V x 0.716 + 0.5V x 1.3562 = 1.0361V, average density, r = 1.0361V / V = 1.0361 g / L. 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.

Question { 1594 }

NATURAL GAS ENGINEERING - QUESTION 26.3 : The United States of America Energy Information Administration reports the following emissions in million metric tons of carbon dioxide in the world for year 2012 : Natural gas : 6799, petroleum : 11695, coal : 13787. Coal-fired electric power generation emits around 2000 pounds of carbon dioxide for every megawatt hour generated, which is almost double the carbon dioxide released by a natural gas-fired electric plant per megawatt hour generated. If 1 metric ton = 1000 kg and 1 pound = 0.4536 kg, estimate the total energy generated by natural gas in the world for year 2012, in gigawatt hour.

Answer

NATURAL GAS ENGINEERING - ANSWER 26.3 : Mass of carbon dioxide emitted by natural gas-fired electric power generation for every megawatt hour generated = 2000 / 2 = 1000 pounds = 1000 pounds x (0.4536 kg / pound) x (metric ton / 1000 kg) = 0.4536 metric tons. Total energy generated by natural gas in the world for year 2012 = 6799 million metric tons x (megawatt hour / 0.4536 metric tons) = 14988.977 million megawatt hour = 14988977 gigawatt hour, approximately. 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.

Question { 1478 }

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?

Answer

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.

Question { 1520 }

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?

Answer

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.

Question { 1556 }

GENETIC ENGINEERING - EXAMPLE 27.3 : (a) Male with genotype GGmm and phenotype gray wingless mates with female with genotype ggMM and phenotype black winged in fruit flies. G is dominant to g in color. M is dominant to m in wing shape. If the actual distribution of the second generation of the fruit flies was as follow : 890 gray wingless, 900 black winged, 115 gray winged, 95 black wingless, calculate the recombination frequency between the two genes and distance in recombination units. Let 1 map unit = 1 % recombination. (b) A DNA molecule has 180 base pairs and 20 % adenine. How many cytosine nucleotides are present in this molecule of DNA?

Answer

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.

Question { 1512 }

GENETIC ENGINEERING - EXAMPLE 27.4 : 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 of 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?

Answer

GENETIC ENGINEERING - ANSWER 27.4 : (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.

Question { 1541 }

MICROBIOLOGICAL ENGINEERING - QUESTION 28.1 : In the calculation of the growth of bacteria, colony forming unit (CFU) in serial dilution is used. In a laboratory, viable count assay is used to estimate CFU. Formula applied is CFU / mL = (number of colonies x dilution) / (amount plated, in unit mL). Acceptable plate count is either between 20 and 200 or between 30 and 300 according to 2 different references. A wastewater sample of 200 ml is added to and mixed with 1.8 L of sterile water. Another 200 ml of the mixture is added to and mixed with 1.8 L of sterile water. (a) Calculate the dilution of first mixture and the dilution of the second mixture. (b) 100 microlitres of wastewater samples from the first mixture and the second mixture are placed separately on 2 different alga plates. The first plate has 250 colonies and the second plate has 23 colonies. Calculate the average CFU / mL.

Answer

MICROBIOLOGICAL ENGINEERING - ANSWER 28.1 : (a) Dilution = (final volume of mixture) / (initial volume of mixture). First dilution = ( 0.2 + 1.8 ) L / (0.2 L) = 10. Second dilution = first dilution x ( 0.2 + 1.8 ) L / (0.2 L) = 10 x 10 = 100. (b) 100 microlitres = 100 microlitres x 1 mL / (1000 microlitres) = 0.1 mL. CFU / mL = (number of colonies x dilution) / (amount plated, in unit mL). In the first plate, CFU / mL = 250 x 10 / 0.1 = 25000. In the second plate, CFU / mL = 23 x 100 / 0.1 = 23000. Average CFU / mL = (first plated value + second plated value) / 2 = (25000 + 23000) / 2 = 24000. 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.

Question { 1591 }

MICROBIOLOGICAL ENGINEERING - QUESTION 28.2 : A hemocytometer is a device that is used for counting cells. In an engineering experiment, 100 microlitres of cell suspension is diluted with 50 microlitres of Trypan blue dye. Only death cells appear blue in color when stained with the dye. There are 57 cells detected in a hemocytometer, where 5.3 % of them appear blue when the chamber of the meter is placed under a microscope. Each square of a chamber can contain 0.0001 mL of liquid. (a) Calculate the number of viable cells. (b) The cells occupied 5 squares. Calculate the average number of viable cells / square. (c) Calculate the dilution factor of the cell suspension by using the formula : Dilution = final volume / initial volume. (d) Calculate the concentration of viable cells / mL by using the formula : Concentration = (Average number of viable cells / square) x dilution x (square / volume).

Answer

MICROBIOLOGICAL ENGINEERING - ANSWER 28.2 : (a) Number of death cells = Total cells x % of death cells / 100 = 57 x 5.3 / 100 = 3.021 rounded to 3 cells. Number of viable cells = Total cells - number of death cells = 57 - 3 = 54 cells. (b) Average number of viable cells / square = 54 / 5 = 10.8 cells / square. (c) Dilution = final volume / initial volume = (100 + 50) microlitres / 100 microlitres = 1.5. (d) Concentration = (Average number of viable cells / square) x dilution x (square / volume) = 10.8 cells x 1.5 x (1 / 0.0001 mL) = 162000 cells / mL. 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.