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MICROBIOLOGICAL ENGINEERING - QUESTION 28.3 : In the calculation of the growth of bacteria, absorbance, A in spectrophotometry is used. According to Beer-Lambert Law, A = e x l x c where A is the absorbance of the solution (no unit), l is the distance of light travels through the solution (in cm), e is the molar absorptivity or the molar extinction coefficient [ in L / (mol.cm) ]. For a particular solute and fixed path length : As / Ao = Cs / Co where Ao is the observed signal for a known concentration Co, and As is the observed signal for a sample concentration Cs. (a) For a cell concentration of 560 cells / mL, a spectrophotometre gives an absorbance reading of 1.0. A mixture of concentration 3600000 cells / mL can be diluted in several operations, with each operation having a dilution of 1:20. How many dilutions should be made so that the concentration of this mixture can be calculated within a range of A = 0.0 to 1.0. (b) In another experiment, a sample tube of 1 cm in width is used. Let A = 0.06 and e = 0.0012 ml / (cell.cm). Find the cell concentration of the sample.

MICROBIOLOGICAL ENGINEERING - QUESTION 28.3 : In the calculation of the growth of bacteria, absorban..

MICROBIOLOGICAL ENGINEERING - ANSWER 28.3 : (a) Let Ao = 1.0, Co = 560 cells / mL, Cs = 3600000 cells / mL. By using the equation As / Ao = Cs / Co, then As = Ao x Cs / Co = 1.0 x 3600000 / 560 = 6428.57. For a dilution of 1:20, first stage dilution gives A = 6428.57 / 20 = 321.43, second stage dilution gives A = 321.43 / 20 = 16.07, third stage dilution gives A = 16.07 / 20 = 0.8, which is within a range of A = 0.0 to 1.0. Total stages of dilution needed = 3. (b) By using the equation A = e x l x c, then c = A / (e x l) = 0.06 / { [ 0.0012 ml / (cell.cm) ] x 1 cm } = 50 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.

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