ESTIMATION OF GLUCOSE (ORTHOTOLUIDINE METHOD)
To estimate the amount of glucose present in the given blood sample.
A solution of orthotoluidine in glacial acetic acid when treated with glucose produces a blue coloured product with an absorption maximum at about 640nm. The values obtained represent the true glucose level.
l00mg of glucose is weighed and made upto 100ml with distilled water. Concentration of glucose = 1mg/ml
10ml of stock solution is diluted to 100ml with distilled water.
Concentration of glucose = 100µg/ml
12.5 mg of thiourea and 12g of boric acid are dissolved in 50ml of distilled water by heating over a mild flame. 75ml of redistilled Orthotoluidine reagent and 375ml of analar acetic acid are mixed separately. The two solutions are mixed and the total volume is made upto 500ml with acetic acid. The reagent is kept overnight at 4°C.
0.2ml of blood sample is taken in a centrifuge tube. To this 0.3ml of 10% sodium tungstate, 0.3ml of 2 / 3N sulphuric acid and 3.2ml of distilled water are added to precipitate the proteins. It is kept aside for 10 minutes and then centrifuged at 3000 rpm for 10 min. 1 ml of the supernatant is taken for the estimation of glucose.
0.2-1.0 ml of standard glucose solutions are pipetted out into five different test tubes labelled S1- S5 with the concentration of 20 - 100µg. 1 ml of the deproteinised supernatant is pipetted out into two different test tubes labelled as T1& T2. Final volume is made upto 1ml using distilled water in all the standard tubes. 4 ml of orthotoluidine reagent is added to all the test tubes. A blank is also prepared simultaneously comprising 1ml of distilled water and 4 ml of orthotoluidine reagent. All the test tubes are heated for 20 minutes in a boiling water bath. The blue colour developed is measured at 640nm using a colorimeter.
A standard graph is drawn with optical density in Y axis vs concentration of glucose in X axis. The amount of glucose present in the given blood sample is then calculated
The optical density A of T1& T2 corresponds to B µg of glucose 1.0 ml of supernatant contains B µg of glucose
4 ml of supernatant will contain= 4 x B/1.0 µg of glucose = Z µg of glucose
0.2 ml of blood contains Zµg of glucose Therefore,
100ml of blood will contain = 100 x Z / 0.2 µg of glucose
= Cmg of glucose (1000µg=1mg)
The amount of glucose present in 100ml of the given blood sample is_____mg
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