Question) A 1 mL sample of glycogen was calculated to contain 37 µmol (micromole
ID: 271683 • Letter: Q
Question
Question) A 1 mL sample of glycogen was calculated to contain 37 µmol (micromole) glucose. To 1 mL of this sample was added 2 mL of 2 M HCL. It was then hydrolysed by boiling the solution for 15 minutes. After boiling the hydrolysate was cooled and made up with H2O to a final volume of exactly 10 mL. The glucose was measured in this solution and found to have a concentration of 466 µg/mL (microgram/milliliter).
Calculate the purity of the glycogen used in the sample as
% Purity = (moles of measured glucose/ moles of calculated glucose in glycogen) *100
Show your working out such that the marker can easily understand it. Include all units. Assume the molecular mass of glucose is 180 g/mol. At each step round off your calculations to 2 decimal points. (2 mark)
Calculate the mass (mg) of glucose in the 10mL of hydrolysate. (As the 1mL of glycogen sample was made up to a final volume of 10mL, this mass of glucose was produced by the hydrolysis of the original 1mL glycogen sample) (0.5 marks)
ii) Calculate the amount (µmol) of glucose produced by the hydrolysis of the glycogen sample. (0.5 marks)
iii) Calculate the percentage purity of the glycogen sample. (0.5 marks)
iv) State the answer to iii) in a full sentence (0.5 marks).
Explanation / Answer
a.)
Calculated moles of glucose in glycogen = 37 µmol
Concentration of glucose in hydrolysate = 466 µg/mL
So mass of glucose in 10 mL of hydrosylate = 466 µg/mL * 10 mL
= 4660 µg
= 4660 * 10-3 mg
= 4.66 mg
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b.)
Calculation of moles of glucose in hydrosylate: Produced by hydrolysis of glycogen sample.
Mass of glucose = 4.66 mg = 4.66 * 10-3 g
Moles of glucose = mass of glucose / molecular mass of glucose
= 4.66 * 10-3 g
180 g/mol
= 0.02588 * 10-3 mol
= 25.88 * 10-6 µmol
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c.)
% Purity = 25.88 µmol / 37 µmol = 69.95 %
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d.)
The glycogen sample has only 69.95 % purity, remaining 30.05 % is the impurity which is present in the glycogen.
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