Question

The blood concentration is given in mg/dL. Calculate the approximate concentrations of blood glucose in mM. The molecular weight of glucose is 180 g/mol. There are 10 dL in 1 L.

Blood glucose (mg/dL) To download image, click here In the experiment above, blood samples were taken from wild type (WT, white bar) or double knockout mice (DKO, black bar) and glucose in the blood was measured. In this experiment mice always have a supply of food pellets that they can eat any time they would like. WT and DKO mice are fed the same food in the same way.

Answer 1

Ans) From graph, you can observe that the Blood glucose concentration for wild type mice is 250 mg/dl and that of double knockout mice is around 40mg/dl.

Given that 1Litre = 10dl. Thus, 1dl = 0.1 Litre

From the definition of Molarity,

1Molar 180gram 1Litre

0.1Molar 18grams 1dl = 0.1 Litre

For wild type mice, we will first convert 250mg in grams i.e " X" 250 x10 ^-3grams 0.1 Litre

By,simple cross multiplication you get,

0.1 x 250 x 10^-3 = 18 "X"

25 x 10^-3 = 18 "X"

X = 0.00138M or 1.3888mM (1molar = 1000millimolar)

Thus,the blood glucose concentration of wild type mice is 1.3888mM

Similarly,for double knockout mice,

0.1 M 18gm 0.1 Litre

"X" 40 x 10^-3 grams 0.1Litre

By cross multiplication we get,

0.1 x 40 x 10^-3 =18 "X"

4 x 10^-3 = 18 "X"

X = 0.004/18 = 0.000222M or 0.2222mM

Thus,the blood glucose concentration of double knockout mice is approx 0.2222mM.