below are links to all the short answer question papers that have been archived.

• Short Answers December 1999
• Short Answers Summer 2000
• Short Answers Spring 2000
• Short Answers Spring 2000
• Short Answers December 2000
• Short Answers Spring 2001
• Short Answers Summer 2001
• Short Answers Summer 2001
• Short Answers Spring 2002
• Short Answers Summer 2002
• Short Answers Summer 2002
• Short Answers Autumn 2002
• Short Answers Summer 2003
• Short Answers Autumn 2003
• Short Answer Spring 2004
• Short Answers Spring 2004
• Short Answers December 2004
• Short Answers Spring 2005
• Short Answers Spring 2005
• Short Answers Spring 2005
• Short Answers Winter 2005
• Short Answers December 2005
• Short Answers June 2007

                          

• Essay Questions Autumn 2008
  • Cholesterol Methods Model Answer
• Short Answer Questions Autumn 2008
  • Short Answer Questions Autumn 2008 – Model Answers

                          

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

2. Define 1 unit of enzyme activity 10 marks

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

4. 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

5. 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%.

a. Calculate the diagnostic efficiency of the test 5 marks

b. Calculate the negative predictive value of the test 5 marks

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

7. 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

8. 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

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 glucagon 2 marks each

10. 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

                          

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

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

                          

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

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

 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.

                          

Essay Questions.

Answer 2 questions only. Time allowed 1 hour 30 minutes

1. Outline the causes of diabetes insipidus. Discuss its diagnosis and investigation and the principles of its management.
2. Describe the methods for the assessment of CSF xanthochromia. What are the indications for its measurement and the precautions that need to be observed in interpreting the results?
3. Critically discuss current laboratory methods for Vitamin D and PTH measurement
4. The biochemistry laboratory has a role in the management of heart failure – critically discuss.

                          

West Midlands Training Course in Clinical Biochemistry & The University of Birmingham

1st Year Essay Paper 27th June 2008

Please answer one question from each module.

You will need to pass each module

Time allowed 3 hours

Module 1

1. Discuss the causes of hypocalcaemia.

2. Briefly describe the main causes of interference in immunoassays. State what may lead you to suspect interference and how you would investigate it.

Module 4

1. Describe the principles of mass spectrometry and outline the main applications of this technique in laboratory medicine.

2. Critically discuss the investigation of a case of suspected acute porphyria.

Module 5

1. Discuss the sources of variability in serum enzyme activity measurements.

2. A 32 year old woman presents with a blood pressure of 160/105 mmHg. Discuss the possible pathophysiology of the hypertension and the biochemical investigation required in the patient.

Module 6

1. Discuss the laboratory investigation of chronic liver disease.

2. Discuss the factors that need to be considered when determining reference intervals and how you would determine the reference interval for an analyte.