Question
1
How many mg (milligrams)
are in 0.8 g?
There are 1000 mg in 1
g. So in 0.8 g there are 1000 x 0.8 = 800 mg
Question
2
How many ml (milliltres)
are in 2,500 µl (microlitres)?
1000 µl = 1 ml, so 2,500
µl = 2,500/1000 ml = 2.5 ml
Question
3
How many mg (milligrams)
are in 5250 µg (micrograms)?
1000 µg = 1 mg, so 5,250
µg = 5,250/1000 mg = 5.25 mg
Question
4
How many ng (nanograms)
are in 5250 µg (micrograms)?
1 µg = 1000 ng, so 5,250
µg = 5,250 x 1000 ng = 5,250,000 ng
Question
5
Which of the following
is a CONCENTRATION and not an AMOUNT?
2ml is a volume (amount)
and 10g and 100µg are masses (amount); 100mM = 100
mmol/l and is a concentration.
Question
6
Which of the following
is an AMOUNT and not a CONCENTRATION?
10%(w/v), 35 µM
(micromolar) and 20 mmol/l (millimolar) are all concentrations. 8
mmol (millimoles) is an amount.
Question
7
The molecular weight of
the amino acid alanine is 89. How many mg are in 30 mmol (millimoles) of
alanine?
1 mol alanine = 89 g, so
1 mmol = 89 mg, so 30 mmol = 30 x 89 mg = 2670 mg
Question
8
The molecular weight of
the amino acid alanine is 89. How many g are in 45 µmol (micromoles) of
alanine?
1 mol alanine = 89 g, so
1 µmol = 89 µg = 89 x 10-6 g = 0.000089 g, so 45 µmol =
0.000089 x 45 g = 0.004005 g.
Question
9
How many µmol
(micromoles) are dissolved in 1 l of a 30 µM (micromolar) solution?
30 µM = 30 µmol/l, so in
1 litre there are 30 µmol.
Question
10
How many µmoles are in 1
ml of a 1 M solution?
1 litre of a 1 M
solution contains 1 mol, so 1 ml contains 1 mmol = 1000 µmol.
Question
11
A solution was made by
dissolving 7.687g of substance X in 720ml of water. Calculate the concentration
of X in g/litre.
If 720 ml contains 7.687
g, then 1000 ml (1 l) contains 7.687 x 1000/720 g = 10.68g. Concentration
is 10.68 g/l.
Question
12
If the molecular weight
of substance X is 212 what is the molarity of the previous solution, given that
it had a concentration of 10.68 g/l?
212 g/l = 1 M, so 10.68
g/l = (1/212) x 10.68 M = 0.05 M
Question
13
The molecular weight of
the amino acid alanine is 89. How many mg are in 250 ml of a 10 µM solution of
alanine?
1 l of a 1M solution
contains 89 g, so 250 ml of a 1 M solution contains 89 x 250/1000 = 22.25 g, so
250 ml of a 1 µM solution contains 22.25 µg, so 250 ml of a 10 µM solution
contains 222.5 µg = 0.2225 mg.
Question
14
The molecular weight of
the amino acid alanine is 89. How many µg are in 200 ml of a 0.02% (w/v)
solution of alanine?
A 0.02% (w/v) solution
contains 0.02 g/100ml, so 200 ml contains 0.04 g = 0.04 x 106 µg
= 40,000 µg. The molecular weight is not required for
the calculation.
Question
15
The molecular weight of
the amino acid alanine is 89. How many g are in 1.0 kg of a 50 ppm solid
mixture of alanine in salt ?
50 ppm (parts per
million) = 50 g per 106 g = 50 g per 103 kg =
50 x 10-3 g per kg = 0.05 g/kg. The molecular
weight is not required.
Question
16
The molecular weight of
glucose is 180. Express a blood glucose concentration of 80 mg per 100 ml as a
molarity.
80 mg/100ml = 800 mg/l =
0.8 g/l. 180 g/l = 1 M, so 0.8 g/l = 0.8/180 M = 0.0044 M = 4.4 mM.
Question
17
The molecular weight of
haemoglobin is 64,000. What is the concentration of a 5 µM solution of
haemoglobin expressed in mg/ml?
1M Hb = 64,000 g/l, so 1
µM Hb = 64,000 µg/l, so 5 µM Hb = 320,000 µg/l = 320 mg/l = 0.32 mg/ml.
Question
18
The molecular weight of
glucose is 180. 500 ml of a 10% (w/v) solution is made and 100 ml of this is
removed. What is the molar concentration of glucose in the remaining 400 ml?
10% (w/v) glucose
contains 10 g/100 ml = 100 g/l. 180 g/l is 1 M, so 100 g/l = 100/180 M = 0.555M. The
fact that 500 ml were made and 100 ml removed does not affect the concentration.
It is the same i.e. 10% (w/v), or 0.555 M, in the 500ml, the 400 ml and the
100ml.
Question
19
The labelling on a tube
of toothpaste states that it contains 0.1% (w/w) sodium fluoride (NaF). The
content is also expressed as "x" ppm fluoride ion. What is
"x"? (The molecular weight of NaF is 42 and the atomic weight of F is
19).
0.1% (w/w) is 0.1 g NaF
per 100g toothpaste, so 106 g toothpaste would contain 0.1 x 106/100
g = 1000 g NaF. This is 1000 ppm NaF, so fluoride ion (F-) is 1000 x
19/42 = 450 ppm.
Question
20
The molecular weight of
urea is 60. 5 g were dissolved in 50 ml water. 10 ml of this solution was added
to 50 ml water. What was the molarity of this final diluted solution?
5 g in 50 ml = 100g/l.
60 g/l = 1M, so 100 g/l = 100/60 M = 1.67 M. 10 ml of 1.67 M urea added to 50
ml water, so final volume = 60 ml. This is a 10/60 or 1 in 6 dilution, so
molarity of diluted solution = 1.67/6 = 0.278 M.
| Prefix |
Name
|
Which modifies an amount by
|
Examples
|
m
|
milli
|
1/1000th, i.e. by 10-3
|
mmol, mg, ml
|
µ
|
micro
|
1/1000 000th, i.e. by
10-6
|
µmol, µg, µl
|
n
|
nano
|
by 10-9
|
nmol, ng
|
p
f
k
M
|
pico
femto
kilo
mega
|
by 10-12
by 10-15
by 1000 times, i.e. by 103
by 106
|
pmol, pg
fmol, fg
kg
MPa
|
Biochemical
Calculations Test Level 2
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 µM. What was the concentration of the initial
solution of X and how many µmol/ml of X were there in the final solution?
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)
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.)
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 µM antibiotic.
How much solution do you administer?
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
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 µM, the initial concentration was 20 x 50 µM = 1000 µM
= 1 mM. 20 µM = 20 µmol/l = 0.02 µmol/ml
Answer
2
1 mM = 1 mmol/l = 1
µmol/ml. So 10-8 ml contains 10-8 µmol = 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).
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
Answer
4
Patient needs 2.5 x 40
µmol = 100 µmol = 600 x 100 µg = 60 mg antibiotic. This is in 12 ml of
a 5 mg/ml solution
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.
