Food Sciene problem!! Critically assess the performance of the new technician by
ID: 588767 • Letter: F
Question
Food Sciene problem!!
Critically assess the performance of the new technician by discussing the oversights and mistakes he made analyzing your assigned component.
(Try to pick out the obvious and more obscure mistakes that were made from the time the original sample was received, on through the analysis of your assigned component. Fully explain what the mistakes, oversights or problems were and what should have been done (please make sure you justify your answers). If relevant to your discussion, comment on the clarity of the report and point out where you think his explanations are ambiguous.)
Total Fat Analysis The soxhlet apparatus was used for the determination of total fat content which included both neutral and polar lipids. For nutritional labelling purposes, the results based on petroleum ether extraction are not accurate due to poor recoveries of polar lipids. To overcome these deficiencies, a more polar solvent (100% methanol) was used for total fat recovery. A second sample was removed from the refrigerator and cut into 2 equal sized portions. Each portion was placed into a metal dish, and put into a 110°C oven for 2 hours. The 2 dried samples were weighed on a top-loading balance to one decimal place (approximately 1.0 g each) and then transferred into soxhlet thimbles. Empty reflux flasks were weighed, filled 3/4 full with methanol plus two glass boiling beads, and fat was extracted during a 30 minute period. After 30 mins, the solvent was distilled off completely and the soxhlet flasks were left on the hot plates to remove the last traces of methanol. The flasks were cooled in the fume hood and weighed on the top-loading balance to one decimal place. weight flask with fat -weight of empty flask Weight of dried sample % Total Fat= × 100 The percent total lipid for the original sample was calculated from the weight of fat extracted divided by the weight of dried sample, multiplied by 100.Explanation / Answer
The technician removed the sample from the refrigerator and weighed it at different time intervals for the two samples sized. This would result in an error as the samples were not allowed to come to room temperature prior to measurement of its weight. Different levels of moisture would had been present in each case. After distiilation of methanol solvent, the last fractions were left out to dry. When the samples where left to dry in air, the amount of methanol left in sample would be different for the two samples and therefore, the %total fat content would also be different for both samples. For accuracy and precision of the measurement, samples must be handled under identical conditions at all times. This would reduce the errors.