BCHM2072/2972
2004 THEORY of PRACTICAL PAPER
MCQs
1 - 17
Background information for questions 1 – 33.
As part of a metabolic study on the effects of the Atkin’s diet you wish to
measure the level of ketone bodies in serum and urine samples. The major
ketone bodies found in serum and urine are acetoacetate,
b-hydroxybutyrate and acetone. The relative proportions of each metabolite
are shown in the table below.
Metabolite
-hydroxybutyrate
acetoacetate
acetone
(%)
78
20
2
After surveying the literature you have found a number of methods which
can be used to measure the various ketone bodies in biological samples.
They fall into two categories; colorimetric and enzymic. You begin your
study by investigating the colorimetric methods.
Colorimetric method for ketone body analysis
(questions 1 to 17).
You have found a published method for the measurement of ketone bodies in urine
based on the reaction between nitroprusside and acetoacetate.
“Acetoacetate in the presence of 50 mM glycine reacts with 7 % nitroprusside in 0.1 M
Na2HPO4 to form a purple complex. The colour development is enhanced by 1%
lactose and can be monitored at 590 nm after a 10 min incubation at room
temperature. The complex is stable for a further 30 min at room temperature.”
You set up a standard curve for acetoacetate as per the table below.
Tube #
Nitroprusside in
Na2HPO4 (ml)
Glycine (l)
Lactose (l)
Standard
acetoacetate (l)
Water (l)
[acetoacetate]
(nmoles/tube)
Absorbance
520 nm
1
1.0
2
1.0
3
1.0
4
1.0
5
1.0
6
1.0
200
100
0
200
100
10
200
100
20
200
100
30
200
100
40
200
100
50
700
0
690
40
680
80
670
120
660
160
650
200
0
0.19
0.41
0.59
0.82
1.01
1. [glycine]
What is the concentration of the stock glycine solution used
in this assay?
A. 50 mM
B. 0.1 mM
C. 1 mM
D. 4 mM
E. 0.5 M
Final [glycine] is 50 mM
Final volume of tube is 2 ml and we used
200 ul of stock glycine solution in making
the final solution up
This is a 1/10 dilution.
If a 1/10 dilution of the stock gives
50 mM, the stock must be 500 mM
(= 0.5 M)
50 mM is 50 umol/ml
So, final tube contains 100 umol
This 100 umol came from 200 ul stock
So stock is 0.5 umol/ul = 0.5 mol/L
2. 20% lactose
You need to make up 500 ml of a 20% lactose solution for
this assay. You only have the monohydrate of lactose
available (mol. wt. 360). How much of this lactose
preparation would you need to weigh out?
A. 20.0 g
B. 21.1 g
C. 105.3 g
D. 100.0 g
E. 95.0 g
20% is 20 g/100 ml.
So if lactose wasn’t the monohydrate, we’d
need 100 g to make up 200 ml
100 g of lactose monohydrate does not
contain 100 g lactose
To get 360 g, we need 378 g of the lactose
monohydrate
To get 100 g, we need 105 g of the lactose
monohydrate
3. [acetoacetate]
What concentration of standard acetoacetate would you
use in this assay?
A. 4 mM
B. 4 μM
C. 2 mM
D. 100 μM
E. 200 μM
Look at the top standard
200 nmol in the tube
Was from 50 ul of the stock
Stock is 4 nmol/ul
= 4 umol/ml = 4 mM
Acetoacetate Standard Curve
4. Extinction coefficient
Absorbance 590 nm
1.2
y = 0.0051x - 0.0052
R2 = 0.9993
1.0
0.8
What is the approximate ε
(mM-1cm-1) for the complex
at 590 nm?
A. 0.1
B. 10
Top standard is 200 nmol/tube
C. 20
Volume of tube is 2 ml
D. 0.0051
[top standard] is 100 nmol/ml = 100 uM = 0.1 mM
E. 5
0.6
0.4
0.2
0.0
0
50
100
150
[acetoacetate] (nm ol/tube]
Absorbance of 0.1 mM solution = 1
So a 1 mM solution would give an abs of 10
Watch out for stoichiometry and pathlength
200
5. [acetoacetate]min
What is the minimum concentration of acetoacetate you
could measure in a urine sample, given there is 0.7 ml in
which to add sample and reliable measurements need an
absorbance > 0.1?
A. ~30 μM
B. ~20 μM
C. ~20 mM
D. ~30 nM
E. ~10 μM
To get an abs of 0.1 we need 20 nmol of
AcAc in the tube
This 20 nmol will be put into the tube in 0.7 ml
This has a conc = 29 nmol/ml = 29 uM
0.7 ml of a 29 uM solution gives an abs of about 0.1
6. ↓[urine]
What procedural change would enable you to measure
urine samples with a lower concentration?
A. Using cuvettes with a 0.5 cm light path
Less light goes thru cuvette. Lower absorbance.
B. Scaling the assay up by a factor of 2 (adding twice the
volume of each of the components of the assay)
Would allow 1.4 ml of urine to be added but would NEED 1.4 ml to get Abs = 0.1
C. Making the nitroprusside/phosphate stock reagent twice as
concentrated
Already in excess BUT could now add less of it, making more room for urine
D. Measuring the standard curve over twice the concentration
range
Top end is not our problem!
E. Incubating the tubes for 30 mins rather than 10 min before
measuring the absorbances
Reaction all over after 10 min
7. Scanning check
As a check on the scanning procedure enter an answer of
A to question 7 on your answer sheet.
0.7 ml of a 30 uM solution gives
abs of 0.1
1.4 ml of a 30 uM solution gives
abs of 0.1
More nmol AcAc added, but
total volume has gone up
Looking for a solution that allows more urine to be added
but which does not cause an increase in the total volume
of the tube
8. ↓ volume
You would like to scale the assay down to 1 ml final volume
(ie by a factor of 2). If the reaction was scaled down which
of the following would change?
Everyting halved… buffers, AcAc, etc
A.
The concentration range of the standard curve (μM)
Top standard is still 0.1 mM – the AcAc amount has gone down but so as the final volume
B.
The concentration range of the standard curve (nmoles/tube)
There IS less AcAc in the top standard
C.
Extinction coefficient (μM-1cm-1) of the coloured complex
E is a fundamental property of the complex. It does not change with volume/amount
D.
Absorbance range of the standard curve
Because the concentration range is the same, so must the absorbance range
E.
A&B
9. ≠ Result
Which of the following changes would invalidate the result?
A. Measuring the absorbances of both samples and
standards at 500 nm instead of 520 nm
Fine. Absorbances would be lower across the board but all ‘in proportion’
B. Using quartz cuvettes instead of plastic cuvettes
Fine. Quartz transmits visible light OK. Bit expensive though!
C. Using a pipette for both the unknown samples and
standard acetoacetate solution which consistently
delivers 5 % more than the set volume
Fine. Overestimates in the standard curve are compensated by overestimates in the test solutions
D. Using an acetoacetate preparation to make up the stock
solution which contains a 10% impurity
BAD. If the standard solution is WRONG then everything that hangs off it must be wrong. Unless we
KNOW that the standard is 10% out, then we can just plot a different value on the x-axis.
E. All of the above
10. [acetoacetate] mg/dL
You have investigated the normal range of acetoacetate in
urine and found most clinicians express the concentration
of acetoacetoate (mol wt 100) as mg/dL. You have a urine
sample with an acetoacetate concentration of 3 mM. What
is this concentration expressed as mg/dL?
3 mM is 3 mmol/L = 0.3 mmol/dL
A. 0.3
B. 3
C. 30
D. 300
E. 0.03
1 mmol is 100 mg, so 0.1 mmol is 10 mg
So 0.3 mmol is 30 mg
The following information refers to questions 11 to 13.
Having explored the standard acetoacetate reaction described above, you now try
measuring ketone bodies in a urine sample. Below are the absorbances at 590 nm
obtained from a series of urine assays using the assay described earlier.
Volume (l)
Undiluted urine sample
mM-hydroxybutyrate
(ul)
1 mM Acetone (l)
Absorbance at 590 nm
100
80
60
0.77
0.62
0.46
40
0.31
0.02
0.009
0.005
0.005
0.15
0.12
0.09
0.06
Urine samples are then classified into ranges (see table below) which can be equated
with urine dipstick measurements.
Option
[acetoacetate] range
(mM)
Grade
A
B
C
D
E
0 – 0.5
0.5 – 2.0
2.0 – 4.0
4.0 – 8.0
>8.0
trace
small
moderate
large
Very large
11.Urine category
Which category (A-E) does this urine sample fit into?
Adding more urine gives a higher absorbance… in proportion
So just consider the 100 ul sample which gives Abs of 0.77
From the std curve, this indicates that there’s about 160 nmol in the tube
160 nmol in 100 ul = 1600 nmol/ml = 1.6 umol/ml = 1.6 mM
Which is option B
12. Volume acetone
Acetone is a liquid with a density of 0.85 g/ml and a mol wt
of 58. To make up 100 mls of 1 mM acetone, what volume
of pure acetone would you need to add?
A. 14.6 ml
B. 6.8 ml
C. 6.8 μl
D. 1.18 ml
E. 118 μl
1 mM is 1 mmol/L, so 100 mls would
contain 0.1 mmol
1 mol weighs 58 g, so 1 mmol weighs 58
mg, and 0.1 mmol weighs 5.8 mg
1 ml of acetone weighs 0.85 g
So 1 g acetone has a volume of 1/0.85 = 1.176 ml
1 mg acetone has a volume of 1.176 ul
5.8 mg acetone has a volume of 6.82 ul
13. acetone:acetoacetate
What is the relative reactivity of acetone compared to
acetoacetate (%) in this colorimetric reaction?
A. 30 %
B. 50 %
C. 15 %
D. 2 %
E. 1.5%
100 ul of urine contained 160 nmol AcAc
and gave an abs of 0.77
100 ul of 1 mM acetone (ie, 0.1 umol = 100
nmol) gave an abs of 0.15
So 160 nmol would have given an abs of 0.24
0.24/0.77 is 31%
This information refers to questions 14 to 17.
This colorimetric method was then applied to a number of urine samples. The results are
presented below.
*The spike is 25 ul of the standard acetoacetate solution used to construct the standard
curve (A590 = ~0.5). The absorbances of all solutions were measured after blanking
against a reagent blank (the zero tube of the standard curve).
**Boiling urine samples for 30 sec removes acetone but does not degrade the acetoacetate
or β-hydroxybutyrate. This is done before adding the urine to the reaction.
Absorbance 590 nm
Sample
A
B
C
D
E
50 l **boiled urine
sample in assay
0.25
0.34
0.05
0.01
0.22
Abs due to AcAc
alone (not acetone)
50 l unboiled urine
sample in assay
0.4
0.35
0.05
0.01
0.23
Difference between this
and left cell = acetone
50 l unboiled urine
+ *spike
0.9
0.85
0.55
0.01
0.73
Spike should raise abs by
about 0.5 over the middle
column
14. Interference
Which sample (A-E) could contain a compound which interferes
with the reaction?
Looking for something that causes widespread low absorbances –
especially in the spike.
Option D
Absorbance 590 nm
Sample
A
B
C
D
E
50 l **boiled urine
sample in assay
0.25
0.34
0.05
0.01
0.22
50 l unboiled urine
sample in assay
0.4
0.35
0.05
0.01
0.23
50 l unboiled urine
+ *spike
0.9
0.85
0.55
0.01
0.73
15. Acetone
Which sample (A-E) contains significant amounts (>0.2
mM) of acetone?
Looking for something that has a much lower absorbance when
boiled and which still gives a good response to the spike.
Option A
Absorbance 590 nm
Sample
A
B
C
D
E
50 l **boiled urine
sample in assay
0.25
0.34
0.05
0.01
0.22
50 l unboiled urine
sample in assay
0.4
0.35
0.05
0.01
0.23
50 l unboiled urine
+ *spike
0.9
0.85
0.55
0.01
0.73
16. Acetoacetate
Which sample (A-E) contains the most acetoacetate?
Looking for a sample with a high boiled absorbance – and a
predictable response to the spike.
Option B
Absorbance 590 nm
Sample
A
B
C
D
E
50 l **boiled urine
sample in assay
0.25
0.34
0.05
0.01
0.22
50 l unboiled urine
sample in assay
0.4
0.35
0.05
0.01
0.23
50 l unboiled urine
+ *spike
0.9
0.85
0.55
0.01
0.73
17. Little acetoacetate
Which urine sample (A-E) can you CONFIDENTLY claim
contains very little acetoacetate?
Looking for something that has low boiled absorbance but behaves
properly with the spike
Option C
Absorbance 590 nm
Sample
A
B
C
D
E
50 l **boiled urine
sample in assay
0.25
0.34
0.05
0.01
0.22
50 l unboiled urine
sample in assay
0.4
0.35
0.05
0.01
0.23
50 l unboiled urine
+ *spike
0.9
0.85
0.55
0.01
0.73