Biochemical Calculations Test Level 2

The following questions are not readily attempted in the MCQ style. Do the calculations on a piece of paper then follow the [answer] link to the bottom half of the page to check the calculations.


Question 1

A 0.1ml sample of a solution of compound X was added to 4.9 ml water. The concentration of the resulting solution was 20 然. What was the concentration of the initial solution of X and how many 痠ol/ml of X were there in the final solution? .......... [answer]



Question 2

How many femtomoles (fmol) of alanine are in a liver cell if the concentration is 1.0mM and the approximate volume of the cell is 1.0 x l0-8 ml? If 1 g liver contains 109 cells, what is the %(w/w) concentration of alanine in liver? (molecular weight of alanine = 89) .......... [answer]



Question 3

A man drinks a cup of tea containing 2 teaspoonfuls of sugar (sucrose). What change might you expect in the molarity of his blood glucose? (You may need the following information: 1 teaspoonful = 5g; 1 mol sucrose = 1 mol glucose + 1 mol fructose, each with molecular weight = 180; fructose can be converted into glucose in the body; average weight of a man = 70 kg, with 20% of this weight as extracellular fluid.) .......... [answer]



Question 4

You have a stock solution of an antibiotic (molecular weight = 600) at 5 mg/ml. Your patient, with a plasma volume of 2.5 litres, needs a starting dose of 40 然 antibiotic. How much solution do you administer? .......... [answer]



Question 5

A woman has 3 g of iron in her body, of which 75% is in the form of haemoglobin. (i) how much haemoglobin is that; (ii) how much oxygen can it bind (in mols) and (iii) what volume of air would it deplete of oxygen? (You may need the following information – The molecular weight of haemoglobin is 68,000; the molecule is a tetramer with four subunits each containing one iron atom and each binding 1 molecule of oxygen. 1 mol O2 occupies 22.4 litres. Air contains 20% O2 (v/v). Atomic weight Fe = 55.8). .......... [answer]



Answer 1

Adding 0.1 ml to 4.9 ml gives a final volume of 5.0 ml. This is a dilution of 1 in 50, so if the final concentration was 20 然, the initial concentration was 20 x 50 然 = 1000 然 = 1 mM. 20 然 = 20 痠ol/l = 0.02 痠ol/ml .......... [back to question 1][go to question 2] [back to top]



Answer 2

1 mM = 1 mmol/l = 1 痠ol/ml. So 10-8 ml contains 10-8 痠ol = 10-14 mol = 10 fmol. 1 g liver contains 109 cells, each cell has 10-14 mol alanine , so 1 g liver contains 10-5 mol, so 100 g liver contains 1 mmol alanine. 1 mmol = 89 mg = 0.089g, so concentration is 0.089 g/100g = 0.089 %(w/w). .......... [back to question 2][go to question 3] [back to top]



Answer 3

10 g sucrose = 10 g glucose /fructose. Thiis is metabolically equivalent to 10 g glucose. No. of moles glucose = 10/180 = 0.055 mol. ECF = 20% of 70 kg (approx. 70 litres) = 14 litres. So, molarity = 0.055/14 M = 4 mM .......... [back to question 3][go to question 4] [back to top]



Answer 4

Patient needs 2.5 x 40 痠ol = 100 痠ol = 600 x 100 痢 = 60 mg antibiotic. This is in 12 ml of a 5 mg/ml solution. .......... [back to question 4][go to question 5] [back to top]



Answer 5

Iron in haemoglobin = 75% of 3 g = 2.25 g = 2.25/55.8 mol = 0.04 mol.

One mol Fe is associated with 1 mol haemoglobin monomer (mol. wt 17 000), so 0.04 mol Fe is associated with 0.04 mol Hb monomer i.e. 0.04 x 17 000 g = 680 g haemoglobin.

1 mol Fe binds 1 mol O2, so 0.04 mol Hb binds 0.04 mol O2. This occupies a volume of 0.04 x 22.4 litres. Since air is 20% O2, volume of air containing this is 5 x (0.04 x 22.4) = 4.5 litres. .......... [back to question 5] [back to top]



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