West Midlands Training Course/MSc in Clinical Biochemistry

Short Answer Questions.  Answer all questions.  Time allowed 1 hour

 

1. What is the absorbance of a 1cm pathlength of acetonitrile at 200nm when the percentage transmission is 80%?                                                10 marks

A = 2-log₁₀%T                                    where     A= absorbance

A = 2-log₁₀80                                                              %T= % transmission           

A = 0.0969

 

1. Define 1 unit of enzyme activity.                                                10 marks

 The amount of enzyme that catalyses the conversion of one micromole of substrate to product in 1 minute under appropriate conditions

 3. List five biochemical abnormalities associated with rhabdomyolysis.  2 marks each

Raised CK                                                                  Hypocalcaemia

Hyperkalaemia                                                          Hyperuricaemia

Hyperphosphataemia                                                Myoglobinuria

Raised Urea/Creatinine (if renal failure)                   Raised LDH

Abnormal LFT (if liver involvement)                          Raised AST

Acidosis

           

1. A 70 kg man with severe hypomagnesaemia requires 24 mmol of magnesium intravenously.

a.     What volume of solution containing 0.48g/2mL of anhydrous magnesium sulphate would be needed ? (Atomic Weights Mg = 24, S=32)                                                               5 marks

b.     What is the final calculated osmolarity of the solution if this volume is added to 500 mL of 0.9% w/v sodium chloride?             5 marks

 

a.             24 mmol of magnesium are required

The stock solution contains 0.48g/2mL MgSO₄

                  Molecular Weight of MgSO₄ = 24 + 32 + (4 x 16) = 120

Therefore molar concentration of stock = 0.48/120 = 0.004 mol/2mL = 2 mmol/mL

            Volume containing 24 mmol of Magnesium = 24/ 2 = 12 mL

 

b.            Final volume of solution = 512 mL

500 mL of 0.9% w/v saline contains 4.5g of NaCl = 4.5/(23+35) = 0.0776 mols = 77.6 mmol NaCl

12mL of MgSO₄ contains 24 mmol

Therefore 512 mL of solution contains 24 mmol MgSO₄ and 77.6 mmol of sodium chloride

Concentration of NaCl/L = (77.6/512)*1000 = 151.56mmol/L

Concentration of MgSO₄ = (24/512)*1000 = 46.88 mmol/L

Both compounds dissociate in solution therefore the osmolarity the solution = (2 * 151.560 + (2* 46.88) = 396.9 mmol/L

 

1. A new screening test has been developed to detect a tumour with a prevalence of 1 in 5000 of the population.  It has a diagnostic sensitivity of 95% and a specificity of 97%.

1. Calculate the diagnostic efficiency of the test                 5 marks
2. Calculate the negative predictive value of the test           5 marks

 

Assuming a population of 1 million

 

	Diseased	Non-Diseased	Total
Positive Test	190 (TP)	29994 (FP)	30184
Negative Test	10 (FN)	969806 (TN)	969816
	200	999800	1000000

 

a.     Diagnostic Efficiency = (TP+TN)/(TP+TN+FN+FP) = [(190+969806)/1000000] *100 = 97%

b.     Negative Predictive Value = TN/(TN+FN) = [969903/(969903+10)]*100 = 99.999%

 

6. A 22 year old painter and decorator presented to his GP with a 2 week history of lethargy, malaise, headaches, nausea and cramping abdominal pain.  Blood results were as follows

Na 140 mmol/L

K 4.2 mmol/L

U 5.6 mmol/L

Cre 103 umol/L

Alb 40 g/L

ALT 24 U/L

ALP 88 U/L

Bili 6 umol/L

Hb 9.4 g/L

MCV 92 fl

WCC 13.2 x10⁹ L^-1

Platelets 231 x10⁹ L^-1

Basophilic stippling of erythrocytes was noted on the blood film.

a.              What is the most likely diagnosis?               4 marks

b.              What is the mechanism behind the abnormal blood results?      3 marks

c.              What management/treatment options are available?            3 marks

 

a.                                     Lead poisoning

b.            Low Hb - lead is an electropositive metal with a high affinity for sulphydryl groups and thus inhibits sulphydryl dependent enzymes such as 5-aminolaevulinic acid dehydratase and ferrochetalase which are essential for the synthesis of haem. 

Basophilic stippling – due to inhibition of pyrimidine 5’-nucleotidase

c.                                     Removal of patient from source of exposure

Use of chelating agents if blood lead sufficiently high to warrant (e.g. sodium calcium edentate, 2,3-    dimercaptosuccinic acid) to form complexes with lead preventing its binding to cell constituents and allowing it to be eliminated in urine           

     

1. A paper describing the use of a new tumour marker included the following data relating to the concentrations of histologically proven disease:-

Number of diseased patients (n) = 110

Mean concentration of tumour marker = 80 U/L

Standard error of mean = 5.5

a.     Calculate the standard deviation of the mean      5 marks

b.     If the reference interval for the tumour marker in non-diseased subjects is 10-15 U/L what can you conclude about the distribution of data in the diseased patients in this study? 5 marks

a.      SE = SD/square route of n

       SD = 5.5 * square route 110 = 57.7

 

b.    The data are not normally distributed as mean plus or minus 2SDs gives a negative value for the marker.  If normal subjects have concentrations 10-15 U/L then negative values won’t occur

 

1. One serum and one urine sample collected from a patient for the measurement of creatinine clearance were each assayed 10 times with the following results:-

Urine Creatinine      Serum Creatinine

mmol/L                               umol/L

11.1                                    102

11.5                                    108

11.9                                    107

10.9                                    112

12.1                                    103

11.2                                    105

11.8                                    109

11.7                                    114

11.6                                    110

12.1                                    100

The urine was a 24 hour collection and had a volume of 1250 mL.

What is the analytical imprecision in the creatinine clearance determination?

10 marks

 

Approach 1

%CV = (SD/mean) *100

SD = sq route of    Σ(x-x)²/n-1   where x = individual value

                                                      x = mean

However should be able to work out standard deviation on scientific calculator – suggest familiarise yourself with how to do this since it will save time           

CV overall = square route of (CV_serum² + CV_urine²)

 

                        Urine Creatinine         Serum Creatinine

                        mmol/L                  umol/L

 

Mean                  11.59                           107                          

SD                  0.415                           4.50

%CV                  3.58                           4.2

 

CV overall = square route of (4.2² + 3.58²) = 5.52%

 

Approach 2

Calculate the variable part of the creatinine clearance for each point

Creatinine clearance = UV/ST = constant * (V/S)       

where  U = urine creatinine

            S = serum  creatinine

            V = volume

            T = time

 

Urine creat mmol/L	Serum creat umol/L	Urine creat/Serum creat mmol/umol
11.1	102	0.109
11.5	108	0.106
11.9	107	0.111
10.9	112	0.097
12.1	103	0.117
11.2	105	0.107
11.8	109	0.108
11.7	114	0.103
11.6	110	0.105
12.1	100	0.121
	mean	0.108
	SD	0.0068
	%CV	6.3

 9. Match the following disease states with their biochemical/other characteristics in adults

 

a.     Carcinoid disease

b.     Phaeochromocytoma

c.     Gastrinoma

d.     VIPoma

e.     Somatostatinoma

Biochemical/Other Characteristics

B1. Fasting blood sugar of greater than 9.0mmol/L on 2 occasions, raised faecal fats

B2 Pancreatic tumour with achlorhydria

B3 Pancreatic tumour often associated with MEN1

B4 Elevated urine 3-hydroxy 4-methoxy mandelic acid output

B5 Elevated urine urine 5 –hydroxyindol acetic acid output and raised plasma chromogranin A

B6 Raised plasma glucagons            2 marks each

 

a. B5

b. B4

c. B3

d. B2

e. B1

 

1. A 54 year old man with chronic renal failure is admitted to AE with chest pain.   ECG changes are suggestive of a possible MI but are not conclusive.   Troponin T at 12 hours post onset of chest pain= 0.12 ug/L.  The renal registrar phones wanting to know if the patient has had an MI.

a.     What do you tell him?

b.     What further investigations may help?         5 marks each

 

1. TnT = 0.12 ug/L in the context of ischaemic chest pain would be consistent with MI. However, raised TnT have been reported  in CRF in the absence of MI. 
2. Definition of MI requires demonstration of change in TnT concentration so repeat TnT may help as may CK or CKMB.

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