West Midlands Training Course/MSc in Clinical Biochemistry

Module 1

1. The following biochemistry results were obtained from an adult hypertensive patient

 

                       Serum                              Urine

Sodium           145 mmol/l     

Potassium       1.7 mmol/L                     Potassium  22 mmol/L

Chloride          86 mmol/L       

Bicarbonate     41 mmol/L

Urea                3.4 mmol/L

Creatinine       80 umol/L

Discuss the differential diagnosis and further appropriate biochemical investigations

(10 marks)

 

1^o Hyperaldosteronism

Cushings

 

For Hyperaldosteronism – aldosterone and renin.  Supine at 9am and Standing at 12 noon may help distinguish Conns from Bilateral Adrenal Hyperplasia

        For Cushings possible first line tests include

– Overnight dexamethasone suppression test

                        - Diurnal variation

                        - 24 hour urine free cortisol       

        Progressing to low and high dose dexamethasone suppression tests

 

1. List five causes of short stature in children.                                    (2 marks each)

Familial

IUGR

Russell-Silver Syndrome

Growth hormone deficiency

Growth hormone insensitivity syndrome

Syndromic – Turner syndrome

Skeletal dysplasia

Chronic disease  - renal disease, coeliac disease, cystic fibrosis, inflammatory bowel disease, thyroid disease

Psychosocial

Under nutrition

 

1. A GP phones to ask what further investigations are needed on a 35 year old female with a serum prolactin = 1200 mU/L.  What advice do you give?

(10 marks)

Clinical features:  oligo/amenorrhoea, infertility and galactorrhoea

Exclude macroprolactin

Causes of hyperprolactinaemia – stress, pregnancy, lactation, some drugs, pituitary

tumour, hypothyroidism, chronic renal failure, ectopic secretion.

Therefore, if ? pregnancy measure serum or urine hCG

     If ? chronic renal failure measure UE

                 If ? hypothyroidism measure FT4 & TSH

                 If ? pituitary tumour measure other pituitary hormones & cortisol  and consider referral for imaging

1. Alkaline phosphatase activity is measured using para-nitrophenyl phosphate (PNP) – molecular weight 371.  200 uL of substrate (255 mg in 5mL) are added to 3mL of diethanolamine buffer at pH 9.8 to which 25 uL of serum has already been added.

Calculate the final molar concentration of substrate in the final reaction mixture.

(10 marks)

Molarity of substrate prior to addition

255mg in 5mL = (0.255*1000)/(371*5) = 0.1374M

Molarity of substrate after addition

200uL of 0.1374M substrate added to total volume 3.225mL

Therefore final molarity = 0.1374*200×10^-3/3.225 = 8.52×10^-3 mole/L

1. A 57 year old woman known to have hypercalcaemia had the following serum biochemistry results:

Calcium 2.82 mmol/L

Phosphate 0.85 mmol/L

Alk Phos 98 mmol/L

Albumin 40 gl/L

Urea 5.7 mmol/L

Creatinine 88 umol/L

 

A  24 hour urine collected at the same time as the blood specimen gave the following results:

24 hour volume 1540 mL

Creatinine 3.7 mmol/L

Calcium 3.1 mmol/L

Calculate the calcium to creatinine clearance ratio.     (8 marks)

Calculation simplifies to:

[u. calcium (mmol/L) x s. creat (umol/L/1000)] / s.calcium (mmol/L) x u.

creat(mmol/L)

volume, minutes etc cancel each other out.

Calcium to creartinine ratio = 0.026

To exclude familial hypocalciuric hypercalcaemia, the clearance ratio should be >0.01.  Comment on your results. What further test would be helpful in deciding the cause of the hypercalcaemia?  (2 marks)

Results exclude FHH. Other common causes in someone this age could be     primary hyperparathyroidism or malignancy. Further test - plasma PTH.

Module 4

1. A 54yr old man with diabetes on insulin who drinks 35 units of alcohol per week has the following serum results:

Bilirubin                            15 mmol/L                    (2 -17)

ALT                                   101 IU/L                        (5 – 40)

Alk Phos                           129 IU/ L                       (30 –130)           

gGT                                  131 IU/L                        (5 – 50)

Albumin                           42  g/L                           (36 – 52

Iron                                  55 mmol/L                    (10 – 30)

UIBC*                                20 mmol/L                   (50 - 70)

Ferritin                             1093 ng/mL                  (15 -300)

 

*Unsaturated iron binding capacity

List three causes for a raised serum ferritin            (3 marks)

Iron overload syndromes (haemochromatosis)

Liver disease

Acute Phase Protein response (inflammation,

malignancy, infection)

Calculate the percentage transferrin saturation                      (3 marks)

 

% Saturation =       Iron /TIBC   x 100

% Saturation =       Iron/(iron +UIBC)   x 100

% Saturation =       73.3%

What is the most likely diagnosis?                        (4 marks)     

     Genetic (Primary) haemochromatosis

                 (Haemochromatosis – 3 marks)

1. Name five types of HPLC detector (2 marks each)

Mass spectrometry  e.g. Vitamin D, steroids, catecholamines, immunosuppressants, drugs of abuse

Fluorescence  e.g. Catecholamines, amino acids, pophyrins, TPMT

Electrochemical  e.g. catecholamines, 5HIAA

Ultraviolet  e.g. HBA1c, Therapeutic drugs

Refractive index e.g. Alcohols, Sugars, Fatty acids

1. A 65 year woman complains of epigastric pain and weight loss.  Fifteen years ago she was found to have anaemia due to vitamin B12 deficiency. Serum antibodies to parietal cell and intrinsic factor (IF) were negative. She had been treated with vitamin B12 injections until vitamin B12 replete and then was investigated with a Schilling test:

Following an overnight fast she was given 1 mg vitamin B12 intramuscularly and oral ⁵⁷Co & ⁵⁸Co–IF. Urine was collected for 24hours for measurement of radioisotopes and the following results were obtained:

9% of ⁵⁷Co radiolabel excreted

31% of ⁵⁸Co of radiolabel excreted

Normal response: > 30% radiolabel excreted

Abnormal response: <10% radiolabel excreted

What was the original diagnosis?                              (4 marks)

(Addison’s) Pernicious anaemia

May have to accept gastrectomy

What is the cause of the vitamin B12 deficiency in this condition?      (2 marks)

Atrophic gastritis/loss of intrinsic factor

Why was the patient given 1 mg of intramuscular vitamin B12?             (2 marks)

Blocks transcobalamin binding sites ensuring that any radiolabelled

vitamin B12 absorbed is directly excreted via kidneys

Where in the gastrointestinal tract is vitamin B12 absorbed?                        (1 mark)

Terminal ileum

(Do not accept ileum)

What is the likeliest cause of her current symptoms?            (1 mark)

Gastric cancer:  3-4 fold increased risk in PA

1. What is the Hardy-Weinberg principle?  (2 marks)

In population genetics, the Hardy–Weinberg principle states that the genotype frequencies in a population remain constant or are in equilibrium from generation to generation unless specific disturbing influences are introduced

If the observed genotype frequencies for a particular single nucleotide        polymorphism are:

AA            AB            BB

33            10            7

Calculate the expected genotype frequencies and state whether it is likely to    conform to the Hardy-Weinberg equilibrium  (8 marks)

A alleles  = (2 x 33) + 10 = 76/100 = 0.76

B alleles = (2 x 7) + 10 = 24/100 = 0.24

Expected genotypes

AA = (0.76)² = 0.578 

AB = 2 x 0.76 x 0.24 = 0.365

BB = (0.24)² = 0.058

x 50

AA = 29

AB = 18

BB = 3                                   

Observed not similar to expected.  Unlikely to conform to H-W equilibrium

1. The following results were found on an internal quality control sample

Iron (umol/L)      71, 67, 71, 66, 69, 71, 74, 68, 74, 70

Calculate the mode, mean, variance, standard deviation and coefficient of variation  (2 marks each)

Mode = 71 umol/L

Mean = 70.1 umol/L

SD = 2.7 umol/L

Variance =  SD² =  7.2 umol/L

Coefficient of variation = SD/mean x 100 = 3.8%

Module 5

1.     List 5 causes of polyuria  (2 marks each)

Nephrogenic Diabetes Insipidus (including lithium, demeclocycline hypercalcaemia, hypokalaemia, chronic renal disease)

Cranial Diabetes Insipidus

Diabetes Mellitus

Polydypsia

2.     Define the following terms:

Pharmacokinetics

Pharmacodynamics

Pharmacogenetics

Pharmacogenomics

Bioavailability  (2 marks each)

Pharmacokinetics - what the body does to drugs (the processes of absorption, distribution, metabolism and excretion).

Pharmacodynamics – what drugs do to the body (mechanisms of drug action & biochemical/physiological effects)

Pharmacogenetics – the effect of genetic variation on an individual’s response to a pharmacological agent

Pharmacogenomics – changes in gene expression in response to a pharmacological agent.  Looks across genome to identify genes that are responsible for responses to different drugs.

Bioavailability – is the dose reaching circulation divided by the dose administered

3.     A 47 year woman had recurrent and persistent duodenal ulceration despite treatment with omeprazole (proton antagonist). Two weeks after withdrawal of omperazole the following fasting results were obtained:

Plasma Gastrin:                                          1023 mU/L                (<100)

Basal gastric acid output                          11.3 mmol/H              (0-5)

Serum adjusted calcium                        2.78 mmol/L            (2.12 -2.65)

What is the most likely cause of the raised plasma gastrin?        (4 marks)

Gastrinoma

(Antral G Hyperplasia: Bonus mark)

Which cells is gastrin secreted from in healthy subjects?                (2 marks)

G cells (Stomach and duodenum)

      Why was the omeprazole stopped prior to testing?                 (2 marks)

Interferes with interpretation

Causes achlohydria and appropriate hypergastrinaemia

      What is potential relevance of the serum calcium result?                (2 marks)

       MEN type 1

4.     A subject was infused with a drug at the rate of 50umol/min until steady state plasma concentration of 250umol/L was reached.  What is the clearance of the drug?  (10 marks)

 Clearance (mL/min)  =   (U (umol/L) * V(mL/min))/P (umol/L)

      Under steady state conditions:

      Rate of excretion (U *V) = Rate of infusion = 50 umol/min

      Clearance (mL/min) = Rate of infusion (umol/min)/Plasma concentration (umol/mL)

      Plasma concentration = 250 umol/L = 0.25 umol/ml

      Clearance = 50/0.25 = 200 mL/min

 5.     The half life of serum digoxin is 38 hours.  What serum level would you expect to find 97 hours after a value of 3.9 µg/L in the absence of further medication?  (10 marks)

I = Ioe^-kt                    Where I = final concentration

                            Io = initial concentration

To find k            t = time

                            k = constant

I/Io = e^-kt

In0.5 = -kt

-0.693 = -kt

-0.693/38 = -k

k = 0.0182 hr^-1

To find serum level at 97hours

I = 3.9e^(-0.0182*97)

I = 0.67 ug/L

Module 6

1.      Briefly describe the significance of IgA deficiency in testing for celiac disease. (10 marks)

Subjects with selective IgA deficiency are at greater risk of celiac disease.       

In the diagnosis of celiac disease IgA anti tissue transglutaminase antibody and IgA     endomysial antibody are the tests of choice.  The potential for false negative results due to IgA deficiency therefore, arises.  When EMA are used total IgA should be measured on all samples to identify those with IgA deficiency.  TGA assays should be designed so that the  normal level is distinguishable from very low levels. The latter is more likely to indicate IgA deficiency  and should trigger its measurement

When IgA deficiency is found then the serological tests of choice are IgG EMA and IgG TTG.

Some workers believe that because of the association of IgA deficiency with celiac disease that a biopsy should be considered in these individuals if there are symptoms of celiac disease.

2.      A 28yr year man was admitted for removal of an ingrowing toenail. He reported to   the admitting medical officer that he had hepatitis as a child which led to the following investigations:

Hb                            15.8 g/dL                    (13.5 – 17.0)

Bilirubin                   46 mmol/L                  (2 -17)

ALT                          30 IU/L                        (5 – 40)

AST                          25 IU/L                        (5-40)

Alk Phos                   101 IU/ L                    (30 –130)           

gGT                          49 IU/L                       (5 – 50)

Albumin                   36 g/L                        (36 – 52)

Urine Dipstick                No bilirubin, increased Urobilinogen

What is the likely diagnosis?  (4 marks)

Gilbert’s syndrome                                               

(accept haemolysis)

How would confirm the diagnosis? (6 marks)

Reticulocyte count (exclude haemolysis)            3 marks

Others                                                                       

(1 mark to maximum of three for any of below)

Diagnosis of exclusion

Normal liver enzyme activity

Unconjugated hyperbilirubinaemia or Split bilirubin

Provocation tests: Nicotinic acid

                  Starvation

Genetic studies

3.              Match the following diseases, syndromes or symptoms given in part 1 with one metal given in part 2

Part1

Menkes’ Syndrome

Keshan Syndrome

Hypochromic microcytic anaemia

Acrodermatitis enteropathica

Basophilic stippling                                    (2 marks each)

Part 2.

Iron

Lead

Selenium

Copper

Zinc

Manganese

Cadmium

Menkes’ Syndrome - Copper

Keshan Syndrome - Selenium

Hypochromic microcytic anaemia - Iron

Acrodermatitis enteropathica - Zinc

Basophilic stippling            - Lead

1. Calculate the positive predictive value for IgA-tissue transglutaminase antibody in a population with a prevalence of celiac disease of 2.5% (assuming test sensitivity of 90% and specificity of 95%).  (10 marks)

            Assuming a population of 1 million

 

	Diseased	Non-Diseased	Total
Positive Test	22500(TP)	48750(FP)	71250
Negative Test	2500(FN)	926250(TN)
	25000	975000

Where TP = true positives, FP = false positives, TN = true negatives & FN = false negatives

Predictive value of a positive result = TP/(TP + FP) *100 = 22500/71250 = 31.6%

1. Your laboratory is now measuring a new analyte  - analyte Y.  You have determined that the within subject variation is 5% and between subject is 7%.  The total CV for analyte Y is 15%.  What is the:

Analytical goal for imprecision  (2 marks)

0.5*CV_I = analytical goal for imprecision   where  CV_I = within subject imprecision

0.5* 5 = 2.5%

Predicted standard deviation at 75 units of Y (2 marks)

CV_tot² = CV_A² + CV_I² + CV_G²            where CV_A = analytical imprecision

CV_G = between subject imprecision

CV_A² = CV _tot² – CV_I² -CV_G²                                         

CV_A² = 225 – 25 – 49 = 151

CV_{A =} 12.3%

%CV_A = (SD/mean) * 100

(%CV_A * mean) /100 = SD = (12.3 * 75)/100 = 9.23

Index of individuality (3 marks)

= CV_I/CV_G   = 5/7 = 0.714

Critical difference for assay assuming homogeneity of variance (3 marks)

2.77 * (CV_A²  + CV_I²)^0.5  = 2.77 *(12.3² + 5²)^0.5 = 36.7

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