2004 THEORY of PRACTICAL PAPER 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. 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.

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 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 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)

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

4. Extinction coefficient What is the approximate ε (mM-1cm-1) for the complex at 590 nm? A. 0.1 B. 10 C. 20 D. 0.0051 E. 5 Top standard is 200 nmol/tube Volume of tube is 2 ml [top standard] is 100 nmol/ml = 100 uM = 0.1 mM Absorbance of 0.1 mM solution = 1 So a 1 mM solution would give an abs of 10 Watch out for stoichiometry and pathlength

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

What procedural change would enable you to measure 6. ↓[urine] What procedural change would enable you to measure urine samples with a lower concentration? Using cuvettes with a 0.5 cm light path B. Scaling the assay up by a factor of 2 (adding twice the volume of each of the components of the assay) C. Making the nitroprusside/phosphate stock reagent twice as concentrated D. Measuring the standard curve over twice the concentration range E. Incubating the tubes for 30 mins rather than 10 min before measuring the absorbances Less light goes thru cuvette. Lower absorbance. Would allow 1.4 ml of urine to be added but would NEED 1.4 ml to get Abs = 0.1 Already in excess BUT could now add less of it, making more room for urine Top end is not our problem! 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? The concentration range of the standard curve (μM) The concentration range of the standard curve (nmoles/tube) Extinction coefficient (μM-1cm-1) of the coloured complex Absorbance range of the standard curve E. A & B Everyting halved… buffers, AcAc, etc Top standard is still 0.1 mM – the AcAc amount has gone down but so as the final volume There IS less AcAc in the top standard E is a fundamental property of the complex. It does not change with volume/amount Because the concentration range is the same, so must the absorbance range

Which of the following changes would invalidate the result? Measuring the absorbances of both samples and standards at 500 nm instead of 520 nm Using quartz cuvettes instead of plastic cuvettes Using a pipette for both the unknown samples and standard acetoacetate solution which consistently delivers 5 % more than the set volume Using an acetoacetate preparation to make up the stock solution which contains a 10% impurity E. All of the above Fine. Absorbances would be lower across the board but all ‘in proportion’ Fine. Quartz transmits visible light OK. Bit expensive though! Fine. Overestimates in the standard curve are compensated by overestimates in the test solutions 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.

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? A. 0.3 B. 3 C. 30 D. 300 E. 0.03 3 mM is 3 mmol/L = 0.3 mmol/dL 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. Urine samples are then classified into ranges (see table below) which can be equated with urine dipstick measurements. Option A B C D E [acetoacetate] range (mM) 0 – 0.5 0.5 – 2.0 2.0 – 4.0 4.0 – 8.0 >8.0 Grade trace small moderate large Very large

Which category (A-E) does this urine sample fit into? 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. Abs due to AcAc alone (not acetone) Difference between this and left cell = acetone 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

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

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

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