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• MRCS Part A - Sep 2019 Exam

A 24-year-old, unconscious man is admitted to Accident and Emergency. No history is available. An arterial blood gas analysis is obtained and results are as follows:

Result Normal

[H+] 80 nmol/l (pH 7.1) 35–45 nmol/l (pH 7.35–7.45)

pa(CO2) 7.0 kPa 4.7–6.0 kPa (35–54 mmHg)

pa(O2) 8.2 kPa >10.6 kPa

[HCO3–] 17.1 mmol/l 20–29 mmol/l

These results indicate which one of the following acid–base disturbances?

Metabolic acidosis with respiratory compensationMixed metabolic and respiratory acidosisRespiratory acidosisRespiratory acidosis with metabolic alkalosisUncompensated metabolic acidosis

Explanation

Mixed metabolic and respiratory acidosis The high hydrogen-ion concentration (low pH) indicates acidosis. The acidosis is combination of both metabolic and respiratory. The elevated p(CO2) indicates a respiratory component. The hydrogen-ion concentration is too low to be accounted for by a respiratory acidosis alone, there must therefore be a metabolic acidosis in addition (as the low bicarbonate concentration also indicates).

Metabolic acidosis with respiratory compensation The high hydrogen-ion concentration (low pH) and low bicarbonate concentration indicates metabolic acidosis. However, it does not show respiratory compensation as the p(CO2) is elevated. In a metabolic acidosis with respiratory compensation the p(CO2) would be low.

Respiratory acidosis The low pH indicates an acidosis and the elevated p(CO2) indicates a respiratory component to this acidosis. However, it is not only a respiratory acidosis because of the low hydrogen-ion concentration. A purely respiratory acidosis would not cause such a low hydrogen-ion concentration.

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Respiratory acidosis with metabolic alkalosis The low pH indicates an acidosis and the elevated p(CO2) indicates a respiratory component to this acidosis. However, the results do not indicate a metabolic alkalosis but rather a metabolic acidosis due to the low hydrogen-ion concentration.

Uncompensated metabolic acidosis The results show a raised p(CO2). If this was a compensated metabolic acidosis, p(CO2) would be reduced. If this was an uncompensated metabolic acidosis (a very unusual situation, as the respiratory response to a metabolic acidosis is usually a rapid one), the p(CO2) would be normal.

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A 59-year-old man is admitted to the Intensive Care Unit with a severe head injury, following a fall. An arterial blood gas (ABG) is performed which shows the following results:

Result Normal

pH plasma 7.37 7.35–7.45

pa(CO2) 6.5 kPa 4.7–6.0 kPa (35–54 mmHg)

[HCO3–] 40 mmol/l 20–29 mmol/l

This result is suggestive of which acid/base state?

Acute respiratory acidosisCompensated respiratory acidosisAcute metabolic acidosisCompensated metabolic acidosisCompensated metabolic alkalosis

Explanation

Compensated respiratory acidosis The following is a procedure for interpreting arterial blood gas results:

1. Does the pH indicate acidosis? In this case the plasma pH suggests acidosis. 2. Is the PaCO2 elevated or depressed compared with the normal range? In this case the

PaCO2 is elevated, suggesting a respiratory acidosis. 3. Is the [HCO3–] high or low compared with the normal range?

In this example the pH is normal showing a complete compensation in the face of other abnormal parameters. The pa(CO2)is raised compared with the normal range. The [HCO3–] is raised compared with the normal range. These results suggest respiratory acidosis that has been compensated resulting in a normal pH.

Acute respiratory acidosis This is incorrect as the pH is normal. Although the pa(CO2) is raised, as would be the case in acute respiratory acidosis, the pH is normal due to metabolic compensation.

Acute metabolic acidosis This is incorrect as the pH is normal, for an acidosis the pH would be reduced. In addition, for a metabolic acidosis the [HCO3–] would be reduced but in this example it is raised.

Compensated metabolic acidosis


The pH is normal and therefore this is not an acidosis. The [HCO3–] is raised as is the pa(CO2).

In a metabolic acidosis the results would show a low pH and a low [HCO3–].

Compensated metabolic alkalosis In an alkalosis the pH would be raised. In this example the pH is normal. Although the [HCO3–] is raised, as would be the case in a metabolic alkalosis the pa(CO2) is also raised, and therefore the pH remains normal.

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A 21-year-old man presents with a 2-day history of persistent vomiting and abdominal pain. His blood gas results show:

Result Normal

pH 7.44 7.36–7.44

PaCO2 7.3 kPa 4.7–6.0 kPa (35–54 mmHg)

PaO2 12.0 kPa >10.6 kPa

Base excess +12 mmol/l −2 to +2 mmol

HCO3- 38 mmol/l 20–29 mmol/l

Cl- 90 mmol/l 98–106 mmol/l

The result is suggestive of which of the following states?

Compensated metabolic alkalosisCompensated respiratory alkalosisUncompensated metabolic alkalosisRespiratory acidosisUncompensated metabolic acidosis

Explanation

Compensated metabolic alkalosis It can be seen that the striking features of this blood gas are a pH within the normal physiological range with elevated bicarbonate and base excess. The vomiting in this patient has resulted in the loss of hydrochloric acid and loss of total body H+ concentration, causing a metabolic alkalosis. This patient has compensated for this through respiratory hypoventilation and retention of CO2 (an acidic gas). Ultimately the acid–base imbalance cannot be normalised by respiratory retention of CO2, this is merely a compensatory measure and requires renal modulation of H+ and HCO3– levels.

Compensated respiratory alkalosis This is not an alkalosis as the pH is in the normal range. For an alkalosis the pH would be >7.44. Secondly, in a respiratory alkalosis the pa(CO2) would be low whereas in this example the pa(CO2) is raised.


Uncompensated metabolic alkalosis This is not an alkalosis as the pH is in the normal range. It is a metabolic alkalosis due to the loss of hydrogen ions from vomiting and diarrhoea. However, it has been compensated by hypoventilation causing an increased in pa(CO2) causing normalisation of the pH.

Respiratory acidosis This is not an acidosis as the pH is in normal range. Although the pa(CO2) value is raised, as would be the case in respiratory acidosis, the HCO3– level is also raised and therefore there is not an acidosis.

Uncompensated metabolic acidosis This is not an acidosis as the pH is in normal range. In an uncompensated metabolic acidosis we would expect to see a low HCO3

– value with a normal pa(CO2) value, which is not the case in this example.

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A 75-year-old woman presents with constant generalised abdominal pain of 12 h duration. She is a heavy smoker and has peripheral arterial disease. Examination reveals a respiratory rate of 40/minute, irregularly irregular heart rate of 150/minute, and blood pressure of 100/50 mmHg. The A&E registrar has sent an arterial blood gas.

Which blood gas result would most fit with this clinical picture and provisional diagnosis?

pH 7.30, p(O2) 10.5, p(CO2) 3.0, HCO3– 15

pH 7.30, p(O2) 9, p(CO2) 6.8, HCO3– 30

pH 7.30, p(O2) 10.5, p(CO2) 3.0, HCO3– 30

pH 7.49, p(O2) 10.5, p(CO2) 3.0, HCO3– 15

pH 7.49, p(O2) 10.5, p(CO2) 6.1, HCO3– 32

Explanation

pH 7.30, p(O2) 10.5, p(CO2) 3.0, HCO3– 15

Ischaemic gut causes a lactic acidosis. In this case the pH and bicarbonate will be low. Attempted respiratory compensation will occur leading to tachypnoea and a low p(CO2).

A primary metabolic acidosis can be due to increased acid in the form of H+ ions (either through increased production or ingestion of an acid, or by reduced renal excretion). Examples include lactic acidosis (tissue hypoperfusion, gut ischaemia), diabetic keto-acidosis and chronic renal failure. A metabolic acidosis can also be due to loss of bicarbonate ions, ie pancreatic fistula, longstanding diarrhoea, uretero-sigmoidostomy.

pH 7.30, p(O2) 9, p(CO2) 6.8, HCO3– 30

Mesenteric ischaemia will cause a metabolic acidosis due to increased levels of lactate from the muscle injury. In this example the HCO3

– level is raised, which does not fit with a lactic acidosis. We would also expect the p(CO2) to be low, while in this example the p(CO2) is elevated.

pH 7.30, p(O2) 10.5, p(CO2) 3.0, HCO3– 30

In mesenteric ischaemia we would expect to see an acute metabolic acidosis. In this example the HCO3 is raised, rather than decreased. Any cause of an acidosis will lower the


pH. In most cases, with a metabolic acidosis you expect both the bicarbonate and the p(CO2) to be low. With a respiratory acidosis the pCO2 will always be high, and the bicarbonate is usually also high.

pH 7.49, p(O2) 10.5, p(CO2) 3.0, HCO3– 15

These results show an alkalosis, rather than an acidosis as the pH >7.44. In mesenteric ischaemia we would expect to see a metabolic acidosis due to the rise in lactic acid.

pH 7.49, p(O2) 10.5, p(CO2) 6.1, HCO3– 32

These values do not fit with the clinical picture described as they show an alkalosis with raised p(CO2) and HCO3

– which is suggestive of a metabolic alkalosis.

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A 48-year old woman underwent an elective inguinal hernia repair. She is unwellpostoperatively and requires admission overnight. An arterial blood gas is taken and the results are:

Result Normal

pH 7.20 7.35–7.45

p(CO2) 3.20 4–6 kPa (35–42 mmHg)

HCO3– 20 22–26 mmol/L

Base excess −10 −2 to +2 mmol

K 2.70 3.5–5.0 mmol/l

What is the most likely diagnosis?

Acute renal failureHyperaldosteronism Diabetic ketoacidosis Hyperparathyroidism Postoperative vomiting

Explanation

Diabetic ketoacidosis In this case, the low pH and HCO3– suggest a metabolic acidosis. There is also a hypokalaemia. The low p(CO2) value suggests the patient is hyperventilating in compensation to blow off more CO2 and correct the acid–base abnormality. Metabolic acidosis is characterised by a low plasma pH and low bicarbonate ion concentration. It can be caused by the addition/failed excretion of acid or the removal of alkali from the body.

Acute renal failure Diabetic ketoacidosis and renal failure will both cause a metabolic acidosis. The hypokalaemia suggests diabetic ketoacidosis, since acute renal failure would produce hyperkalaemia. Note that unlike with other causes of metabolic acidosis, hypokalaemia is observed in diabetic ketoacidosis in the context of low total body potassium and a low


intracellular concentration of potassium. In addition to urinary losses from polyuria and volume depletion, an obligate loss of potassium from renal tubules also occurs as a cationic partner to the negatively charged ketone, β-hydroxybutyrate.

Hyperaldosteronism Hyperaldosteronism (Conn’s syndrome) will cause alkalosis, hypokalaemia and hypertension. In this example the patient has an acidosis.

Hyperparathyroidism Hyperparathyroidism would not acutely affect acid–base balance and would be an unusual diagnosis in a postoperative surgical patient.

Postoperative vomiting Vomiting causes a loss of acid and so would result in a metabolic alkalosis.

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A 62-year-old man was admitted with an exacerbation of chronic obstructive pulmonary disease. His arterial blood gases on air showed:

Result Normal

pH 7.29 7.35–7.45

p(CO2) 8.3 kPa 4.7–6 kPa

p(O2) 7.9 kPa >10.6 kPa

HCO3– 25 mmol/l 22–30 mmol/l

This patient had which one of the following?

Metabolic acidosis

Metabolic alkalosis

Mixed acidosis

Respiratory acidosis

Respiratory alkalosis

Explanation


This patient had an acidosis with a high p(CO2) and normal standard bicarbonate-respiratory acidosis. This is a common finding in acute exacerbations of chronic obstructive pulmonary disease. It is usually managed with nebulisers, steroids and antibiotics and non-invasive ventilation.

Metabolic acidosis


This patient does have an acidosis as the pH is low. However, the bicarbonate is normal and the p(CO2) is raised suggesting a respiratory cause to the acidosis rather than metabolic.

Metabolic alkalosis

This patient has an acidosis rather than an alkalosis as the pH is low. In a metabolic alkalosis we would expect the pH to be raised and the bicarbonate also raised.

Mixed acidosis

In a mixed acidosis we would expect the bicarbonate to be low and the p(CO2) to be raised. In this example the bicarbonate is within normal range.


This is not an alkalosis as the pH is low. In a respiratory alkalosis we would expect to see a low p(CO2). In this example the p(CO2) is raised.

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A 28-year-old man was admitted with status epilepticus. He was given intravenous diazepam. Arterial blood gases on 15 l/min via a reservoir bag mask showed:

Result Normal

pH 7.05 7.35–7.45

p(CO2) 8.20 kPa (61.5 mmHg) 4.7–6 kPa

p(O2) 15.33 kPa (115 mmHg) >10.6 kPa


Na 140 mmol/l 135–145 mmol/l

K 4 mmol/l 3.5–5.0 mmol/l

Cl– 98 mmol/l 98–106 mmol/l

This patient most probably had which one of the following?

Metabolic acidosis

Metabolic alkalosis

Mixed respiratory and metabolic acidosis



Explanation

Mixed respiratory and metabolic acidosis


This patient had acidosis with both a high p(CO2) and a low standard bicarbonate-mixed acidosis. The anion gap is 30 mmol/l (increased). The p(O2) is lower than expected because the patient was breathing around 70% oxygen. Does this fit with the clinical picture? Yes, he had a lactic acidosis from prolonged fitting and a respiratory acidosis from intravenous diazepam. This disturbance will return to normal with attention to either A - airway manoeuvres and oxygen, B – assisted ventilation if needed or C – treatment with fluids.

Metabolic acidosis

This pH is low which fits with an acidosis. However, it is not only a metabolic acidosis but has a respiratory component also as the p(CO2) is also raised therefore it is a mixed acidosis.

Metabolic alkalosis

This is not an alkalosis as the pH is significantly reduced. In an alkalosis we would expect to see the pH raised. In a metabolic alkalosis we would also see a raised bicarbonate whereas in this example the bicarbonate is reduced.


Components of this blood gas fit with a respiratory acidosis as the pH is low and the p(CO2) is raised. However, it is important to note the low bicarbonate also. With a low bicarbonate alongside the low pH and p(CO2) it cannot be only a respiratory acidosis but also have a metabolic component.


This is not an alkalosis as the pH is low. In a respiratory alkalosis we would expect to see a raised pH and a low p(CO2).

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A 45-year-old man with previous peptic ulcer disease was admitted with persistent vomiting. He appears dehydrated. His blood results were as follows:

Result Normal

Sodium 140 mmol/l 135–145 mmol/l

Potassium 2.5 mmol/l 3.5–5.0 mmol/l

Chloride 86 mmol/l 98–106 mmol/l

pH 7.5 7.35–7.45

p(CO2) 50 mmHg 35–45 mmHg

p(O2) 107 mmHg 10–14 Kpa (75–100 mmHg)


This patient had which one of the following?

Metabolic acidosis

Metabolic alkalosis

Mixed acidosis



Explanation

Metabolic alkalosis


This patient has an alkalosis due to a high standard bicarbonate- so a metabolic alkalosis. As vomiting results in loss of potassium and chloride, this causes a hypokalemic hypochloraemic metabolic alkalosis (with a compensatory increase in bicarbonate). Treatment is based around intravenous electrolyte replacement, with correction of the underlying cause (in this case, pyloric stenosis)

Metabolic acidosis

This is not a metabolic acidosis as the pH is raised (alkalosis) and the bicarbonate increased. In a metabolic acidosis we would expect to see a reduced pH and reduced bicarbonate.

Mixed acidosis

This is not an acidosis because the pH is raised. In a mixed acidosis we would expect to see a low pH, raised p(CO2) and low bicarbonate.


In a respiratory acidosis we would expect to see a low pH and a raised p(CO2). In this example the pH is raised (alkalosis) and the p(CO2) is close to the normal range (35 to 45 mmHg).


The pH in this example is raised which is an alkalosis. However, the cause for the alkalosis is metabolic due to the loss of potassium and chloride from vomiting causing a compensatory rise in bicarbonate. In a respiratory alkalosis we would expect to see a low p(CO2).

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You are in a pre-assessment clinic and measure the blood pressure of patients seen that day. You note the diastolic blood pressure (DBP) in a similar group of young men. This is normally distributed with a mean of 70 mmHg and a standard deviation of 10 mmHg.

What can you conclude from this data?

About 50% of the men have a DBP above 70 mmHgAbout 50% of the men have a DBP below 50mmHg All the DBPs must be less than 100 mmHg Should carry out a trial of much greater sizeThe distribution of DBP is skewed

Explanation

About 50% of the men have a DBP above 70 mmHg DBP follows a normal distribution. This means that the mean will equal the median. As such if the mean is 70, the median is also 70. The median is the middle value of the data suggesting that about 50% of the men will have a DBP above 79 mmHg.

About 50% of the men have a DBP below 50mmHg About 50% of the men will have a DBP below 70 mmHg, rather than 50 mmHg. Approximately 2.5% will have DBPs less than mean – 2 SD, approximately 2.5% will have DBPs greater than mean + 2 SD. There are very few values more than 3 SD away from the mean (about 1 in 1000 in either direction).

All the DBPs must be less than 100 mmHg This is false as although most DBPs will be less than 100 mmHg, but not all of them will be. About 95% will have DBPs between 50 and 90 mmHg.

Should carry out a trial of much greater size Larger sample sizes increase the power of the study and are therefore preferred. However, from these results this is not the main conclusion drawn.

The distribution of DBP is skewed DBP follows a normal distribution. Normally distributed data are symmetrical. The mean will equal the median.

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In a small, randomised double-blinded trial of a new treatment in acute myocardial infarction, the mortality in the treated group was half that in the control group, but the difference was not significant.

Which of the following can be concluded from this?

The treatment is uselessThere is no point in continuing to develop the treatment Reduction in mortality is so great that we should introduce the treatment We should keep adding cases to the trial until the test for the comparison of two proportions shows a significant difference We should carry out a trial of much greater size

Explanation

We should carry out a trial of much greater size The study has been inconclusive, a larger sample would be needed to identify whether the drug actually does work. The current study suggests there may be a large clinical significance to the new treatment.

The treatment is useless As the difference was not statistically significant the treatment may be useless, and the observed reduction due to random variation. However, it may actually improve survival. These results do not allow us to conclude whether the treatment is useless and warrants further larger trials to be conducted.

There is no point in continuing to develop the treatment A halving of mortality attributable to a new treatment would almost certainly be clinically significant. Although the results from this study cannot be used as they are not statistically significant, the results suggest that it is worth continuing to carry out trials on this treatment.

Reduction in mortality is so great that we should introduce the treatment The study does show a significant reduction in mortality, however the results are not statistically significant. As such the treatment should not be introduced but rather a larger trial should be conducted.

We should keep adding cases to the trial until the test for the comparison of two proportions shows a significant difference If we keep adding cases to the trial and retesting, then the probability of obtaining a falsely significant result will increase. Sequential trials need to be specially designed to take account of multiple testing

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In estimating the physiological clearance of 10 ml of an intravenous substance which has been administered at 10 mg/ml, the plasma concentration at equilibration is 15 mg/litre, the urine concentration is 150 mg/litre and the subject produces 1440 ml of urine during a 24 h collection.

What is the clearance of the substance?

1 ml/min10 ml/min0.1 ml/min100 ml/minCannot say from the information given

Explanation

10 ml/min This is the correct answer. Clearance (ml/min) is calculated using the formula (U × V)/Pwhere U = urine concentration in mg/ml, V = urine production in ml/min, P = plasma concentration in mg/ml.

• Urine concentration =150mg/litre = 0.15mg/ml = U.

• Urine production = 1440ml over 24h = 60ml per hour = 1ml/min = V

• Plasma concentration =15mg/litre = 0.015mg/ml

• Therefore: (0.15 x 1) / 0.015 = 10ml/min.

1 ml/min This is incorrect. Clearance (ml/min) is calculated using the formula (U × V)/P where U = urine concentration in mg/ml, V = urine production in ml/min, P = plasma concentration in mg/ml. This gives an answer of 10 ml/min not 1 ml/min.

0.1 ml/min This is incorrect. (U × V)/P = 10 ml/min

100 ml/min This is incorrect. (U × V)/P = 10 ml/min

Cannot say from the information given There is enough information given to calculate clearance based on the formula (U × V)/Pwhere U = urine concentration in mg/ml, V = urine production in ml/min, P = plasma concentration in mg/ml.

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A 72-year-old diabetic man with chronic renal failure has the following blood results:

Result Normal

Haemoglobin (Hb) 15.3 g/dl 13–18 g/dl

Mean corpuscular volume (MCV) 95 fl 76–96 fl

Whole cell count (WCC) 10.3 × 109/l 4–11 × 109/l

Serum corrected calcium 1.95 mmol/l 2.12–2.65 mmol/l

Phosphate 1.8 mmol/l 1.2–1.7 mmol/l

Alkaline phosphatase (ALP) 150 IU/l 30–35 IU/l

What is the underlying diagnosis?

Primary hyperparathyroidism

Secondary hyperparathyroidism

Tertiary hyperparathyroidism

Osteoporosis

Paget’s disease

Explanation

Secondary hyperparathyroidism

In chronic renal failure, secondary hyperparathyroidism is an appropriate response to low calcium levels that occur due to low calcitriol levels, which is produced in the kidneys. Phosphate is high as it cannot be cleared from the kidneys. ALP is elevated in high turnover renal osteodystrophy due to increased osteoblast activity.


Primary hyperparathyroidism

Primary hyperparathyroidism is usually caused by a tumour (parathyroid adenoma, or more rarely, parathyroid adenocarcinoma) within the parathyroid gland. It causes raised calcium, as opposed to low calcium as seen in these blood results. Bloods results typically show raised parathyroid hormone (PTH) and raised calcium and can be treated by surgical removal of one more parathyroid glands.

Tertiary hyperparathyroidism

Tertiary hyperparathyroidism refers to excessive secretion of parathyroid hormone (PTH) after a long period of secondary hyperparathyroidism. It occurs as a state of autonomous parathyroid function due to persistent parathyroid stimulation. It would result in hypercalcaemia. The blood results shown above reveal hypocalcaemia and therefore are not caused by tertiary hyperparathyroidism.

Osteoporosis

Osteoporosis is a chronic condition that results in reduced bone density. It does not however cause changes in serum levels of calcium and phosphate, and does not cause elevated ALP as is seen in these blood results.

Paget’s disease

This is incorrect. Paget’s disease of the bone is characterised by excessive breakdown and remodelling of bone. It is characterised by excessive osteoclastic and osteoblastic activity. Blood results in Paget’s disease characteristically reveal an elevated ALP with normal calcium and phosphate.

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You take a set of arterial blood gases from a patient on 28% oxygen and discover that the oxygen saturation differs significantly from the pulse oximeter reading.

What could cause inaccuracy of the pulse oximeter reading?

AcidosisAlkalosisArrhythmia

Albinism

Haemoglobin level 9g/dl

Explanation

Arrhythmia This is correct. Arrhythmias, hypotension, vasoconstriction, abnormal haemoglobin or pigments (eg bilirubin), movement, poor tissue perfusion and nail varnish can all affect pulse oximetry readings.

Acidosis The presence of acidosis does not affect pulse oximetry readings.

Alkalosis The presence of alkalosis does not affect pulse oximetry readings.

Albinism

Albinism will not affect pulse oximetry readings. There is some debate as to the effect of darker skin pigmentation may result in overestimation of arterial oxygen saturation.

Haemoglobin level 9g/dl Anaemia or polycythaemia will not affect the accuracy of the reading values, but these factors should impact the interpretation of these values. For example, an anaemic patient who has saturations of 99% will have fewer haemoglobin molecules, meaning that total oxygen content may still be significantly decreased although the saturations will not reflect this. Carboxyhaemoglobin will also cause a reading to be falsely high – pulse oximetry should be interpreted with caution in these patients.

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Consider the following data obtained from a blood gas analysis:

Result Normal

paO2 7.6 kPa >10.6 kPa

paCO2 3.2 kPa 4.7–6.0 kPa (35–54 mmHg)

pH 7.47 7.36–7.44

[HCO3–] 24 mmol/ l 20–29 mmol/l

This blood gas analysis is most consistent with which one of the following?

Compensated metabolic acidosisNormal blood gas analysisMetabolic acidosisRespiratory acidosis


Explanation


This blood gas shows respiratory alkalosis due to raised pH, with a low paO2 and low paCO2, possibly due to consolidation of lung with compensatory hyperventilation. This pattern may be seen in conditions producing a ventilation/perfusion mismatch, such as pneumonia or pulmonary embolism. If ventilation is able to increase enough to keep CO2 normal, then a respiratory acidosis will not develop.

Compensated metabolic acidosis A compensated metabolic acidosis would show decreased bicarbonate as well as a low CO2– due to respiratory compensation.

Normal blood gas analysis This is not a normal blood gas as the pH, paO2 and paCO2 are outside the normal range:

Normal

pH plasma 7.36–7.44


paCO2 4.7–6.0 kPa (35–54 mmHg)

paO2 >10.6 kPa

[HCO3–] 20–29 mmol/l

Metabolic acidosis A metabolic acidosis would have a pH of less than 7.35 and low bicarbonate.

Respiratory acidosis This is incorrect. It does not show a respiratory acidosis as the pH is not low. In a respiratory acidosis the paCO2 would be raised, whereas in this example it is low likely due to hyperventilation.

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A 26-year-old woman had a full blood count that showed:

Result Normal (women)

Red blood cell (RBC) count 3 × 106/μl 3.8–5.8 × 106/μl

Haematocrit 27% 34.9–44.5%

Haemoglobin 11 g/dl 12–16 g/dl

Mean corpuscular volume (MCV) 80–100 fl

Mean corpuscular haemoglobin concentration (MCHC) 31–37 g/dl

This patient's erythrocytes have which of the following?

Decreased MCV

Decreased MCHC

Increased cell diameter

Normal MCHC

Normal MCV

Explanation

Normal MCV

MCV is calculated from haematocrit and RBC count for example: MCV (fl) = haematocrit × 1000/RBC (106/μl) = 0.27 × 1000/3 = 90 fl, which is within the normal range.

Decreased MCV

This is incorrect as the MCV is within normal range (MCV = 80–100 fl).

Decreased MCHC


MCHC is calculated from blood haemoglobin and RBC count: MCHC (g/dl) = haemoglobin (g/dl)/haematocrit = 11/0.27 = 41 g/dl and is higher than normal.

Increased cell diameter

RBC diameter cannot be calculated from the data given, but is probably less than normal in this patient.

Normal MCHC

MCHC is calculated from blood haemoglobin and RBC count: MCHC (g/dl) = haemoglobin (g/dl)/haematocrit = 11/0.27 = 41 g/dl and is higher than normal.

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A 47-year-old patient is admitted with pancreatitis. His modified Glasgow Score is 3 based on his blood results listed:

Result Normal

White cell count (WCC) 11 × 109/l 4–11 × 109/l

pa(O2) 7.8 kPa >10.6 kPa

Urea 2.3 mmol/l 2.5–6.5 mmol/l

LDH 555 IU/l 100–190 IU/l

ALT 140 IU/l 5–30 IU/l

Glucose 11.1 mmol/l 3.5–5.4 mmol/l

Which one is the most likely remaining result?

Albumin 28 g/l

p(CO2) 5 kPa

Creatinine 115 IU/l

Haemoglobin (Hb) 8.9 mmol/mol

Potassium 5.2 mmol/l

Explanation

Albumin 28 g/l

This is correct as a low albumin (< 32 g/l) is one of the criteria used in the modified Glasgow Score done for pancreatitis on admission. A point is recorded for each criteria below. A score of 3 or more indicates a severe attack and ideally should have an intensive care unit (ITU) review/ monitored in ITU. This patient scores 3 based on pa(O2), glucose and albumin levels:


• pa(O2) < 7.9 kPa

• age > 55 years

• neutrophils: WCC > 15 × 109/l

• calcium < 2 mmol/l

• renal function: urea > 16 mmol/l

• enzymes: aspartate aminotransferase (AST)/ALT > 200 IU/l or LDH > 600 IU/l

• albumin < 32 g/l

• sugar: glucose > 10 mmol/l.

Other such scoring systems include Ranson’s criteria and APACHE.

p(CO2) 5 kPa

p(CO2) is not included in the criteria for the modified Glasgow Score for acute pancreatitis, however pa(O2) is, so an arterial blood gas is required.

Creatinine 115 IU/l

Renal function is one aspect included in the modified Glasgow Score, however it is a high urea that is recorded rather than creatinine.

Haemoglobin (Hb) 8.9 mmol/mol

Haemoglobin concentration is not included in the Modified Glasgow Scale.

Potassium 5.2 mmol/l

Potassium is not one of the clinical criteria used in the Modified Glasgow Scale.

You are analysing data from your unit comparing local recurrence rates from breast cancer over time in patients who express the gene NF1 and those expressing the gene B. The two groups (gene NF1 and gene B) are matched for other confounding variables.

Which one of the following would be the correct statistical test to use?

ANOVA test

Chi-squared test

Kaplan–Meier graphs with log-rank test

Mann-Whitney U-test

Paired student’s t-test

Explanation

Kaplan–Meier graphs with log-rank test

Kaplan–Meier curves with associated log-rank analysis provide a comparison of outcomes (in this case recurrence) between different groups over a period of time. The log-rank test assesses the statistical significance of differences between the curves.

ANOVA test Analysis of variance (ANOVA) is a statistical test used to test differences between two or more means. It would not be an appropriate test for this data set.

Chi-squared test

Chi-squared test is used for categorical data from two independent groups and therefore would not be an appropriate test in this case as the groups are matched.

Mann-Whitney U-test

This statistical test is used on non-parametric data and also assumes that the two groups are independent from one another. As such it would not be an appropriate test for this data set.

Paired student’s t-test

This statistical test is used to test the difference between population means for a pair of random samples whose differences are normally distributed. It will not help to plot events through time, such as in this case in which recurrence rates are being studied.


You are analysing data produced from your unit looking at the relationship between age and renal function. This data follows a Gaussian (normal) distribution.

Which one of the following tests is appropriate for this purpose?

Mann-Whitney U-test

Kruskal-Wallis test Pearson’s correlation coefficient Spearman’s rank correlation coefficientWilcoxson signed-rank test

Explanation

Pearson’s correlation coefficient Pearson’s correlation coefficient analyses the strength of a relationship between two continuous variables as is the case here. The other tests are non-parametric tests.

Mann-Whitney U-test

This statistical test is for use on non-parametric data and therefore would not be appropriate for this data set. It is the non-parametric equivalent of the t-test.

Kruskal-Wallis test The Kruskal–Wallis test is a non-parametric statistical test based on ranking data to compare two or more independent samples. It is similar to the Mann–Whitney U-test but can be used when there are more than two groups.

Spearman’s rank correlation coefficient This is a non-parametric measure of rank correlation which examines statistical dependence between the ranking of two variables.

Wilcoxson signed-rank test This is a non-parametric test used to compared two related samples to assess whether their population mean ranks differ. For example, it can be used to compare repeated measurements on a single sample.

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You are reading a research paper that has attempted to evaluate the effect of smoking on the incidence of pilonidal sinuses. As such they looked at two cohorts of matched patients, half of which had the disease, and then evaluated their smoking histories.

Which one is the correct description of the level of evidence this paper represents?

Level I Level II Level III Level IV Level V

Explanation

Level III This study design is a case–control study which is a form of observational research study which does not use randomisation. As such it is referred to as level III evidence. Level III evidence is divided into 3A and 3B:

3A: Systematic review (with homogeneity) of case–control studies;

3B: Individual case–control study;

3C: ‘Outcomes’ research; ecological studies.

This is an individual case–control study in which the authors are attempting to look at the patients with pilonidal disease (the cases), compared with those who did not (the controls), and how this is related to smoking.

Level I Level I evidence is highest form of evidence. This study is a case–control study, which is not ranked as the highest form of evidence. It does not include randomisation of study participants or systematic review. Level I evidence includes:

1A: Systematic reviews (with homogeneity) of randomised controlled trials;

1B: Individual randomised controlled trials (with narrow confidence interval);

1C: All or none randomised controlled trials.

Level II This is incorrect as this study design follows a case–control design. Level II research refers to:


2A: Systematic reviews (with homogeneity) of cohort studies;

2B: Individual cohort study or low quality randomised controlled trials (eg <80% follow-up);

2C: ‘Outcomes’ research; ecological studies.

Level IV Level IV research refers to case series (and poor quality cohort and case–control studies). This study does not sound like a poor quality case–control and it is not a case series and therefore it is higher quality evidence than level IV.

Level V Level V research refers to expert opinion without explicit critical appraisal, or based on physiology, bench research or ‘first principles’.

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You attended a recent regional surgical conference in which a colleague recommends to you a new surgical method of closing the abdomen using five layers of sutures. He has used it and, based on his experience, he recommends that you change your practice too.

What level of evidence is he using to inform his practice?

Level I Level II Level III Level IV Level V

Explanation

Level V Level V research refers to expert opinion without explicit critical appraisal, or based on physiology, bench research or ‘first principles’. This evidence is the type used in this scenario. The surgeon has changed his suturing practice based on ‘expert opinion’ from a course he attended that does not have any form of critical appraisal or peer review. As such it is not rigorous scientific evidence from which to change practice.

Level I Level I evidence is the highest level of evidence and can be divided into three separate subsections. Level 1A, the very highest, involves a systematic review of randomised control trials. These are designed to find an exhaustive summary of current literature regarding the research question. They are a type of literature review that follow a strict methodology to find and grade the level of evidence from each randomised control trial that has aimed to answer that research question. Level 1B involves an individual randomised control trials. These are studies which aim to reduce bias by randomly allocating subjects to receive either the treatment or control arm. Level 1C refers to all or non-randomised control trials.

Level II This is not level II evidence as no organised study has been carried out to test this new surgical approach. Level II evidence can be divided in to three sub categories. Level 2A refers to systematic reviews of cohort studies. While systematic reviews are a high level of evidence, cohort studies are not as rigorous as randomised control trials as they do not eliminate bias through randomisation and therefore they provide a lower level of evidence. Level 2B refers to either individual cohort studies or low quality randomised control trials. Level 2C refers to ecological studies. Again, this is higher level evidence than anecdotal advise from a surgical course, but not as rigorous as a randomised control trial.


Level III Level III evidence is a weaker form of evidence than I or II as it relied upon case–control studies, which are limited by several sources of bias. They are, however, still a better form of evidence than anecdotal ‘expert opinion’ which this new surgical technique is based upon.

Level IV Level IV evidence is fairly poor evidence from case series. These are published patient cases, often referring to rare diseases, that can be helpful in guiding medical practice but do not have a rigorous methodology for answering specific research questions. They are however better evidence than expert opinion.

MRCS Part A - Sep 2019 Exam

Audit is an essential part of clinical practice. You have ju

st completed one audit looking at the availability of oestrogen receptor status at the first post-operative multidisciplinary team meeting (MDT) and intend on changing practice and closing the loop.

Which one of the following statements is true regarding clinical audit?

It is a way of assessing the superiority of a new treatment which has been recommended by NICE It is an optional part of the appraisal process It is only valid as part of your appraisal folder evidence if you have completed the audit and closed the loop It is part of the process of clinical governance The availability of international guidelines as a benchmark is an essential starting point of an audit

Explanation

It is part of the process of clinical governance This is correct. Clinical governance involves seven pillars, one of which is clinical audit:

1. Service user, carer and public involvement. 2. Risk management. 3. Clinical audit. 4. Staffing and staff management. 5. Education and training. 6. Clinical effectiveness. 7. Clinical information.

It is a way of assessing the superiority of a new treatment which has been recommended by NICE This is not correct because an audit aims to study established practices, rather than new treatments. To assess superiority research would be required, rather than clinical audit. It differs from research as it does not attempt to look at new treatments but rather looks at established processes.

It is an optional part of the appraisal process Clinical audit is a mandatory part of the appraisal process as it forms one of the seven pillars of clinical governance.


It is only valid as part of your appraisal folder evidence if you have completed the audit and closed the loop Engagement with clinical audit is necessary for your appraisal folder, but it is not a requirement to have completed the audit and closed the loop. However, closing the loop on an audit does provide more usual evidence for change of clinical practice and therefore should be aimed for.

The availability of international guidelines as a benchmark is an essential starting point of an audit Guidelines and standards need to be available so that the audit has a benchmark but these can be local, national or international. The audit cycle follows a set process, which begins with identifying set criteria and standards to be reviewed as outlined below. Audit cycle:

• Identifying an issue or problem

• Setting criteria and defining standards to be reviewed

• Collecting data

• Comparing performance against criteria and standards

• Implementing change

• Repeating the audit cycle.

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You are at an interview for an ST3 appointment and are being asked a question on the different levels of evidence that you know of.

Which one of the following is an example of level II evidence-based medicine?

Expert opinion

Case series

Cohort study with a low risk of confounding or bias

Systematic review of randomised controlled trials (RCT)

Individualised RCT with a narrow confidence interval

Explanation

Cohort study with a low risk of confounding or bias

This is correct. This is regarded as level II evidence rather than level I because of the high risk of confounding or bias. In level I research (randomised control trials), subjects are randomised to either treatment or control groups thereby eliminating known bias. However, in a cohort study this is not the case and therefore bias can reduce the validity of the results. It is, however, still a higher level of research than case series as it is analytical and involves the testing of a hypothesis.

Expert opinion

Expert opinion is the lowest form of evidence-based medicine referred to as level V evidence because it is not based on any formal research methodology.

Case series This is referred to as level IV evidence. It is a better form of evidence-based medicine than expert opinion as it often involves more subjects. However, it is still referred to as descriptive research, rather than analytical research, as it does not test a hypothesis to look for cause and effect.

Systematic review of randomised controlled trials (RCT)

This option is the highest form of evidence-based medicine regarded as level 1A evidence.

Individualised RCT with a narrow confidence interval


Similarly to the previous answer this is also regarded as level I evidence. However, this is not as rigorous as a systematic review of randomised control trials and is therefore classed as level 1B

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A study comparing a new antibiotic to an old one for diverticulitis looks at the numbers of patients ‘cured’, as opposed to ‘not cured’ in each group using the chi-squared statistic. In the trial, the cure rates appeared to improve with the new antibiotic (χ2 = 4.2; P < 0.01).

Which one of the following options would you bear in mind regarding this study?

The improvement seen with the new treatment must be clinically significant

The mean survival time was significantly longer for those given the new antibiotic

Age may have been a confounding factor

The trial implies that a difference in response of 4.2 times was observed

The results may have occurred by chance 1 time in 20

Explanation

Age may have been a confounding factor

This is true because if the patients in one of the treatment groups were older and age affected cure rate then age would be a confounding factor in the analysis. A confounding factor is a variable that influences both the dependent variable and independent variable.

The improvement seen with the new treatment must be clinically significant It is important to read the answer carefully. This response is incorrect because it states ‘clinically’ significant rather than ‘statistically’ significant. Although the P-value is <0.05 (making it statistically significant) this does not imply clinical significance. Often research papers do not refer to clinical significance, this is a limitation of the usefulness of the study.

The mean survival time was significantly longer for those given the new antibiotic

This is incorrect as mean times were not compared. The chi-squared test was used to compare the proportion of people who were cured using the new antibiotic with the proportion cured on the old treatment regimen.

The trial implies that a difference in response of 4.2 times was observed

This is incorrect as it is a misinterpretation of the meaning of a chi-squared statistic. The chi-squared statistic is 4.2; by referring this to statistical tables a P-value of <0.01 is obtained. The number 4.2 has no interpretable meaning in this context.

The results may have occurred by chance 1 time in 20


This is incorrect as a P-value < 0.01 means that the difference in cure rates would have occurred by chance less than 1 time in 100 if the treatments were equally effective, not 1 time in 20.

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A study examines the systolic blood pressure and weight in a group of men aged 30–40 years. The two variables are compared using a regression method as part of this cross-sectional study and there is a positive correlation coefficient with P-value = 0.014.

Which one of the following will be true regarding this study?

Information regarding the incidence of systolic hypertension in this age group will be obtained The value of the regression coefficient is dependent on whether the weight is expressed in kilogramms or poundsThe results allow the likely systolic blood pressure of a 60-year-old man who weighs 78 kg to be predicted Weight affects systolic blood pressure Systolic blood pressure affects weight

Explanation

The value of the regression coefficient is dependent on whether the weight is expressed in kilogramms or pounds The regression coefficient (or the slope of the line) will give the average increase in systolic blood pressure for a unit increase in weight and will therefore change if weight is expressed in different units. This response is therefore true because the changes in the units of age will be completely separate from the regression equation linking systolic blood pressure and weight.

Information regarding the incidence of systolic hypertension in this age group will be obtained Systolic blood pressure and weight have been collected on 30–40-year-old men. This will give information on the prevalence of systolic hypertension in this group, but not the incidence. Prevalence is a measurement of all individuals affected by a disease at a particular time. Incidence is the number of new individuals who contract a disease during a particular time period.

The results allow the likely systolic blood pressure of a 60-year-old man who weighs 78 kg to be predicted This is incorrect as from this sample it will only be possible to make inferences about other men in the same age group.

Weight affects systolic blood pressure


The value P = 0.014 is evidence that there is a statistically significant relationship between weight and blood pressure. However, we cannot infer that this relationship is causal. It is not possible, therefore, to make this statement based on this current research.

Systolic blood pressure affects weight This is incorrect. An association has been found between weight and blood pressure but this evidence does not imply causality, ie it cannot be said that blood pressure affects weight.

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The gut transit times of two different laxatives A and B in two samples of UK patients are compared using a t-test. The study is being designed as a randomise trial.

Which one of these statements about this study is the most likely?

It is assumed that the gut transit times in the two groups are not normally distributed

A confidence interval of 95% means that 95% of the sample data lie within this interval

The null hypothesis assumes there is no difference between the two population groups

The number of patients must be the same in each group

The trial would be valid even if patients had travelled to the tropics recently

Explanation

The null hypothesis assumes there is no difference between the two population groups This is correct. The null hypothesis assumes that there is no difference in the means of the groups. Therefore when P < 0.05 the null hypothesis can be rejected and it can be stated that there is a statistical difference between the means of the groups.

It is assumed that the gut transit times in the two groups are not normally distributed To use a t-test correctly the two groups must be normally distributed.

A confidence interval of 95% means that 95% of the sample data lie within this interval

This in incorrect. A 95% confidence provides a range of values within which one can be 95% sure that the true population value lies.

The number of patients must be the same in each group

The t-test can be used to compare the means of two normally distributed samples. The samples do not have to be of the same size as long as they are both normally distributed and independent.

The trial would be valid even if patients had travelled to the tropics recently If more of one group had recently returned from the tropics and having recently been in the tropics affected gut transit time, then ‘recently been in the tropics’ would be a confounding factor and may make the results questionable.

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In a trial to test the effectiveness of a new drug against the standard treatment for improving forced expiratory volume in 1 s (FEV1) among asthmatics, the members of age-matched and sex-matched pairs are randomly allocated to the new or standard treatment. The difference (new standard) in forced vital capacity (FVC) between the pairs 1 h after treatment was 0.5 ± 0.2 (mean ± standard error), P < 0.05.

What are the chances of this difference having occurred by chance?

Less than one time in 100Less than one time in 500Less than one time in 1000Less than two in 1000

Less than five in 100

Explanation

Less than five in 100

P < 0.05 shows that the observed differences would have occurred less than five times in 100 if the new drug was no more, or less, effective than the standard (5 times in 100 = 1 time in 20).

Less than one time in 100 If this was the case then the P-value would be <0.01.

Less than one time in 500 If this was the case then the P-value would be <0.002.

Less than one time in 1000 If this was the case then the P-value would be < 0.001.

Less than two in 1000 If this was the case then the P-value would be <0.002

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The way datasets are distributed has a significant impact on the statistical tests that are required. The Gaussian distribution is used in statistical analysis to determine two limits within which an observation will fall.

Which one of the following options is correct about Gaussian distribution?

The mean value is an accurate way of assessing the central value of a data set, even when the data are skewed

The standard deviation (SD) is a measure of how accurately the calculated mean approaches the true population meanThe ANOVA test may be used to compare this set of values with one further set provided they are also normally distributedThe modal value is always equal to the mean 25% of the values will have numerical values that are smaller than the mean value minus 5.96 standard deviations

Explanation

The modal value is always equal to the mean The Gaussian distribution is more usually referred to as the normal distribution and is characterised by being ‘bell shaped’ (unimodal) and symmetrical about its central value. In a Gaussian distribution the mean, median and mode are equal. This situation is characteristic of the bell-shaped distribution.

The mean value is an accurate way of assessing the central value of a data set, even when the data are skewed

The mean is used to calculate the central value in a normal distribution. However, in skewed data (non-parametric data) the median is a more accurate calculation for assessing the central value. The median refers to the middle point of the data. It is the value that halves the data with 50% of values above the middle point and 50% below.

The standard deviation (SD) is a measure of how accurately the calculated mean approaches the true population mean The SD is a measure of the spread of the data values.

The ANOVA test may be used to compare this set of values with one further set provided they are also normally distributed Student’s t-test is an appropriate way of comparing the means of two groups of normally distributed values. When the same group of individuals is assessed on three or more occasions, the assessment times can be compared in pairs using either paired t-tests or Wilcoxon matched pairs rank-sum tests as appropriate. However the number of


comparisons needed is large if there are many groups to be compared and the risk of a type I error (false-positive result) increases rapidly. If the measure can be assumed to have a normal distribution, the assessment times can all be compared simultaneously using a repeated measures analysis of variance (ANOVA).

25% of the values will have numerical values that are smaller than the mean value minus 5.96 standard deviations The standard deviation (SD) is a measure of the spread of the data values. Here, 95% of observations lie within 1.96 SDs. Hence, 5% lie outside the interval, 2.5% less than the mean −1.96 SD and 2.5% greater than mean +1.96 SD.

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A study examines the stature of school children. The conclusion states that there is a relationship between the heights of siblings shown by the tendency of tall boys to have tall sisters (r = 0.57, P < 0.001).

In statistics, what does ‘r’ refer to?

Indicates the probability that the result is highly significant

‘r’ is the probability of having a risk factor if one has a disease

‘r’ is the risk of developing a disease for people with known exposure compared with risk of developing a disease without exposure

The symbol ‘r’ represents the correlation coefficient

When ‘r’ is less than one, a negative correlation has been demonstrated

Explanation

The symbol ‘r’ represents the correlation coefficient

The heights of boys and their sisters have been compared. The correlation coefficient has been calculated, this is denoted by ‘r’.

Indicates the probability that the result is highly significant

The ‘P’ value indicates that the correlation is highly significant. The correlation coefficient indicates how closely the data points lie to a line, it does not give any information as to the slope of the line. So, although the quoted ‘r’ and ‘P’ values indicate that as the boy’s height increases then so does his sister’s, there is no information as to how much her height increases for a given increase in his height. The statement P < 0.001 shows that the results would have been observed less than 1 time in 100 if there is no linear relationship whatsoever between the heights of boys and their sisters.

‘r’ is the probability of having a risk factor if one has a disease

This refers to an odds ratio, rather than a correlation coefficient. The odds ratio calculates the probability of an individual having a risk factor if the disease is present.

‘r’ is the risk of developing a disease for people with known exposure compared with risk of developing a disease without exposure


This refers to relative risk, rather than a correlation coefficient. Relative risk is the probability of an event occurring in an exposed group compared with the probability of an event occurring in the non-exposed group.

When ‘r’ is less than one, a negative correlation has been demonstrated

This is incorrect as a negative correlation is shown by values of ‘r’ less than 0. Values of ‘r’ less than zero indicate that one height increases as the other falls; this is known as a negative correlation.

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As part of his research protocol a medical student is looking at the correlation of height to glove size in a population of surgeons. The data are assumed to be in a normal distributionpattern.

Which one of the following tests would be most useful?

Kaplan–Meier curves with log-rank analysis

Mann–Whitney test

Kruskal–Wallis test

Pearson’s correlation co-efficient

Wilcoxon test

Explanation

Pearson’s correlation co-efficient

Glove size and height are not independent of each other. The Pearson’s correlation co-efficient measures the strength of the correlation between the variables.

Kaplan–Meier curves with log-rank analysis

The Kaplan–Meier test is used to describe and analyse comparative survival data, and the significance between curves, statistically analysed with the log-rank test.

Mann–Whitney test

This is incorrect as the data are normally distributed. The Mann–Whitney tests are non-parametric tests and are used for non-Gaussian data.

Kruskal–Wallis test


Analysis of variance (ANOVA) tests are also parametric tests. If a measure cannot be assumed to have a normal distribution, the non-parametric equivalent of the one-way ANOVA is used and this is the Kruskal–Wallis test.

Wilcoxon test

This is incorrect as the data are normally distributed. The Wilcoxon tests are non-parametric tests and are used for non-Gaussian data.

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You are in breast clinic reviewing a patient with a lesion detected via the national screening programme. You have read a recent article regarding the need to be conscious of bias in such screening programmes.

In general screening programmes for cancer are susceptible to which one of the following?

Calculation biasDetection biasInterobserver bias Lead-time biasPopulation bias

Explanation

Lead-time bias Results of screening programmes for cancer are limited by selection, length and lead-time bias. Lead-time bias occurs when screening advances the date at which diagnosis is made. This, therefore, lengthens the calculated survival time without necessarily altering the date of death. Length bias can also affect screening programmes. This is the tendency for screening to detect a disproportionate number of cancers that are slow-growing and have a better prognosis anyway.

Calculation bias This is unlikely to affect screening programmes as opposed to research studies in which particular statistical analyses are taking place and rely on particular statistical assumptions.

Detection bias Detection bias occurs when a phenomenon is more likely to be observed for a particular set of study subjects. This may lead to a false inflation of a particular phenomenon being observed in a particular group because the study authors are more likely to look for it within a set group.

Interobserver bias This occurs when the researcher subconsciously influences the experiment due to cognitive bias, in which judgement may alter how an experiment is carried out. This is unlikely to affect screening programmes.

Population bias Population bias or sampling bias arises when some members of a population are more likely to be included in a study than others, leading to systematic differences between the population being studied and the general population. This is less likely to occur in screening programmes as a large population is usually included.


You are at journal club where your colleague presents a paper on the effects of a new drug on pain control. A cross-over study has been used.

Which one of the following is an advantage of using within patient comparison?

All confounders are eliminated

More patients are needed, which strengthens the validity of the result

There may be an advantage when examining the effects of long-acting drugs

They are always free from carry-over effects

They reduce errors associated with individual differences

Explanation

They reduce errors associated with individual differences

This is true. The advantage of a cross-over study design is that the same individuals receive a sequence of different treatments therefore reducing the errors associated with individual differences. Most cross-over designs have a ‘balance’ in which all subjects should receive the same number of treatments and participate for the same period of time.

All confounders are eliminated This is incorrect. The advantage of this type of study is that the risk of confounders are reduced because each cross-over patient serves as his or her own control. However, this study design cannot eliminate all confounders.

More patients are needed, which strengthens the validity of the result

This is incorrect as fewer subjects are required as subjects act as their own controls.

There may be an advantage when examining the effects of long-acting drugs

This is incorrect as these studies are most suitable for treatments with short-term benefits. This is because a wash-out or drug-free period may be necessary between treatments to ensure that the effects of the first drug are not carried over to the other treatment period. Accordingly, this type of study is probably suitable only for treatments with short-term benefits in patients with a chronic, but relatively stable, disease as the wash-out periods need to be of adequate length between treatments and so patients may have to be in the study for a considerable period of time.


They are always free from carry-over effects

Allocating the same individual to both new and standard treatments, in random order, has the advantage that between-patient confounders are removed, however the possibility of ‘carry-over effects’ ie the effect of the first treatment remaining active partially or wholly when the second treatment is started, remains a potential confounder.

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Medical statistics deals with the analysis, interpretation and presentation of data. A false-positive result occurs if the null hypothesis is rejected when it is actually true.

What sort of error is this?

Experimental design errorObserver bias errorSelection bias errorType I errorType II error

Explanation

Type I error A type I error (false-positive result) occurs if the null hypothesis is rejected when it is actually true (eg the treatments are interpreted as having different effects when they do not).

Experimental design error An experimental design error would be likely to result in statistical bias – factors that will influence the results have not been subject to adequate control measures.

Observer bias error Observer bias occurs when for example knowledge of the type of treatment being given or received alters the perception of its effect. For example a patient knowingly given a placebo for pain relief may over-rate their pain.

Selection bias error Selection bias occurs when patients are allocated to the interventions sequentially so it is known in advance which treatment the next patient recruited will receive so a clinician may pre-select a specific patient for a particular intervention.

Type II error A type II error is a false-negative result and occurs if the null hypothesis is accepted when it is actually false (eg the treatments are interpreted as having equal effects when they are actually different).

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Quantitative measures, such as blood pressure and blood test levels, have to be summarised using two descriptive statistics – a measure of central tendency and a measure of dispersion. The arithmetic mean is such a descriptive statistic.

What do you understand by this term?

It is the average deviation

It is the most commonly occurring value

It is the nth root of the product of the observations (where n is the number of observations)

It is the sum of the observations divided by the number of observations

It is the value that divides the observations into two equal halves when they are arranged in order of increasing value

Explanation

It is the sum of the observations divided by the number of observations

The arithmetic mean is the sum of the observations divided by the number of observations

(For example: 3+1+1+2+12+1+7+2+4+1+2+4+1+3+1)/15=45/15 = 3.

It is the average deviation The average deviation is a measure of dispersion, calculated by taking the arithmetic mean of the absolute values of the deviations of the functional values from some central value, usually the mean or median.

It is the most commonly occurring value

The mode is the most commonly occurring value. For example: 3, 1, 1, 2, 12, 1, 7, 2, 4, 1, 2, 4, , 1, 3, 1 – in the above example ‘1’.

It is the nth root of the product of the observations (where n is the number of observations)

The geometric mean is the nth root of the product of the observations and is only used when the observations are positively skewed and can be assumed to have a log-normal distribution.

It is the value that divides the observations into two equal halves when they are arranged in order of increasing value


This is the median deviation rather than the arithmetic mean.

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Medical statistics deals with the analysis, interpretation and presentation of data. Being able to compare datasets for statistical differences is key for evidence-based healthcare.

Which one of the following statistical tests are appropriate for analysing parametric(normally distributed) data?

Chi-squaredKruskall–WallisMann–Whitney

Friedman's two-way ANOVA

Student’s t-test

Explanation

Student’s t-test

One-way ANOVA and Student’s t-test are both examples of statistical analysis suitable for normally distributed data. Student’s t-test is used to examine the difference between sample means.

Chi-squared This statistical test is used for non-parametric data. Observations are classified into mutually exclusive classes and the null hypothesis gives the probability that any observation falls into the corresponding classes.

Kruskall–Wallis This test is also used for non-parametric data. It is used for comparing two or more independent samples of equal or different sizes.

Mann–Whitney This statistical test is for non-parametric data to compare two sample means from the same population. It can be used as an alternative to the independent t-test when data are not normally distributed.

Friedman's two-way ANOVA

This statistical test is used for non-parametric data and aims to understand if there is an interaction between the two independent variables on the dependent variable.

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A patient presents with the following urea and electrolytes results:

Result Normal

Sodium 140 mmol/l 135–145 mmol/l

Potassium 4 mmol/l 3.5–5.0 mmol/l

Chloride 105 mmol/l 98–106 mmol/l

Bicarbonate 20 mmol/l 22–30 mmol/l

Calculate the anion gap.

19 mEq/l

5 mEq/l

10 mEq/l

30 mEq/l

0 mEq/l

Explanation

19 mEq/l

The anion gap is the difference between the primary measured cations (sodium and potassium) and primary measured anions (chloride and bicarbonate). The degree of difference between the measured cations and anions aids in determining the cause of a metabolic acidosis.

Anion gap = ([Na+] + [K+]) − ([Cl–] + [HCO3–]) (all units mmol/l).

There are many causes for decreased, normal or raised anion gap:

a decreased anion gap (< 6 mEq/l):


• hypoalbuminaemia

• plasma-cell dyscrasia

• monoclonal protein

• bromide intoxication

• normal variant

• a normal anion gap (6–12 mEq/l):

• loss of bicarbonate (ie, diarrhoea)

• recovery from diabetic ketoacidosis

• ileostomy fluid loss

• carbonic anhydrase inhibitors (acetazolamide, dorzolamide, topiramate)

• renal tubular acidosis

• arginine and lysine in parenteral nutrition

• normal variant.

an elevated anion gap (>12 mEq/l):

• methanol

• uraemia

• diabetic ketoacidosis

• propylene glycol

• isoniazid intoxication

• lactic acidosis

• ethanol ethylene glycol

• rhabdomyolysis/renal failure

5 mEq/l

The anion gap is the difference between the primary measured cations (sodium and potassium) and primary measured anions (chloride and bicarbonate).


10 mEq/l



30 mEq/l



0 mEq/l



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You are at an interview for an ST3 appointment and are being asked a question on the different levels of evidence that you know of.

Which one of the following is an example of level 2 evidence-based medicine?

Expert opinion

Case series

Cohort study with a high risk of confounding or bias

Systematic review of randomised controlled trials

Randomised controlled trial (RCT) with a low risk of bias

Explanation

Cohort study with a high risk of confounding or bias

Level 2 evidence can be divided into three subsections and include cohort studies with a high risk of confounding or bias:

• 2a Systematic reviews (with homogeneity) of cohort studies

• 2b Individual cohort study or low quality randomised controlled trials (eg <80%

follow-up)

• 2c ‘Outcomes’ research; ecological studies.

Expert opinion

Expert opinion is the lowest form of evidence classified as level 5. It is evidence that is not based on a research study design with hypothesis testing but rather anecdotal evidence that has not been tested formally. Levels of evidence are as follows:

• 1a Systematic reviews (with homogeneity) of RCTs

• 1b Individual RCTs (with narrow confidence interval)

• 1c All or none RCTs

• 2a Systematic reviews (with homogeneity) of cohort studies

• 2b Individual cohort study or low quality RCTs (eg <80% follow-up)


• 2c ‘Outcomes’ research; ecological studies

• 3a Systematic review (with homogeneity) of case–control studies

• 3b Individual case–control study

• 4 Case series (and poor quality cohort and case–control studies)

• 5 Expert opinion without explicit critical appraisal, or based on physiology, bench

research or ‘first principles’.

Case series Case series is classified as level 4 evidence along with poor quality cohort and case–control studies. These studies are classified lower in the ranking of evidence because of their level of confounding and bias reducing the validity of results.

Systematic review of randomised controlled trials

This is the highest form of evidence categorised as level 1a. This is because it is the most rigorous study design with regards to reducing bias and confounding.

Randomised controlled trial (RCT) with a low risk of bias

This is categorised as level 1b evidence. It is lower than a systematic review of RCTs but is higher quality evidence than case–control or cohort studies.

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You are taking time out of clinical practice to undertake a research fellowship, which entails managing clinical trials. You are reading through several studies to better understand clinical trials, and come across a study that looks at the use of intravenous steroids in cardiac surgery. The trial states that it is a prospective, double-blinded, randomised control trial.

Which type of bias will be reduced in this trial?

Attrition bias alone

Publication bias

Selection and observer bias

Selection bias alone

Selection, observer and funding bias

Explanation

Selection and observer bias

Selection bias will be reduced in this trial by the randomisation of participants, but observer bias will also be reduced by double-blinding. Selection bias is the bias that results from an unrepresentative sample being used in a study, and effective randomisation can reduce this discrepancy. Observer bias arises when those conducting or running the study alter its outcome; blinding both the observers and the subjects reduces this bias.

Attrition bias alone

Attrition bias comes about when there is a difference in withdrawals and dropout between study groups, which leads to incomplete outcome data.

Publication bias

Publication bias is a type of bias that occurs in published academic research, and occurs when the outcome of a study influences the decision whether to publish the data.

Selection bias alone

Observer bias, as well as selection bias, will be reduced with this methodology.


Selection, observer and funding bias

Although a prospective, double-blinded randomised controlled trial will be effective in reducing selection and observer bias there are many aspects of a trial that can be potentially affected by a study’s financial sponsor.

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A study is seeking to compare the weight of 88 patients at 6 months post sleeve gastrectomy, with the weight of 56 patients at 6 months post gastric bypass in a bariatric surgery unit 2 years ago.

What is the most suitable statistical test for this comparison?

Unpaired t-test

Paired t-test

Friedman test

Analysis of variance

Pearson correlation coefficient

Explanation

Unpaired t-test

As two separate groups are being compared at the same point in time, an unpaired t-test is used. The study is looking at a biological variable in a large sample – the data are assumed to be normally distributed.

Paired t-test

Paired t-tests are used to compare one group at two points in time, eg a patient group before and then after an intervention.

Friedman test

The Friedman test is used to compare three or more groups for statistical significance, if the data are non-parametrical ie do not follow a normal distribution.

Analysis of variance

Analysis of variance compares three or more groups for statistical significance, if the data are parametrical ie follow a normal distribution.

Pearson correlation coefficient


The Pearson correlation coefficient is a measure of strength of the linear relationship between two variables.

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A 71-year-old man with known chronic obstructive pulmonary disease is admitted to the Emergency Department with severe shortness of breath. Blood gas analysis shows:

Result Normal

arterial [H+] 55 nmol/l (pH 7.26) 35–45 nmol/l (pH 7.34–7.44)

p(CO2) 9.4 kPa 4.6–6.0 kPa

p(O2) 9.4 kPa 10.5–13.5 kPa

derived [HCO3–] 30 mmol/l 24–30 mmol/l

Which one of the following types of acid–base disturbance is most likely?

Acute respiratory acidosisChronic, compensated respiratory acidosis

Acute exacerbation of chronic respiratory acidosis

Acute respiratory alkalosis

Severe metabolic acidosis

Explanation

Acute respiratory acidosis The arterial blood glass does show an acidosis (pH 7.26) and a raised p(CO2) 9.4 kPa, which are both in keeping with acute respiratory acidosis. The marginally raised bicarbonate suggests there has not been time for complete compensation and hence the acidosis is acute.

Chronic, compensated respiratory acidosis This is not correct because the arterial blood gas is not compensated as it shows a low pH. In a compensated respiratory acidosis the pH would be normal, and the bicarbonate may be raised.

Acute exacerbation of chronic respiratory acidosis


The patient is acidotic and the high p(CO2) indicates that there is a respiratory acidosis. Were this to be chronic the bicarbonate concentration would be more elevated as metabolic compensation would be occurring. Were compensation to be complete, the patient would have a normal pH, and the bicarbonate concentration would be elevated. An acute on chronic picture would show an acidosis with a raised bicarbonate.

Acute respiratory alkalosis

The data is not compatible with respiratory alkalosis. Firstly, the pH shows acidosis, not alkalosis. Secondly, the bicarbonate is elevated, not reduced. The patient has a history of chronic obstructive pulmonary disease (COPD) and has been admitted with an acute episode of breathlessness, therefore type II respiratory failure leading to acidosis seems most likely.

Severe metabolic acidosis This is incorrect. In a metabolic acidosis we would expect to see a low bicarbonate, but in this example the bicarbonate is 30 mmol/l. In addition, the pCO2 is raised, suggesting type II respiratory failure as a cause.

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You are attempting to look at the re-admission rates following emergency appendicectomy in female patients under 35 in an attempt to change practice.

Which one of the following statements is correct regarding the principles of audit?

It evaluates the difference between two treatments before clinical application

It facilitates the evolution of guidelines for different procedures

It requires evaluation of current practice against guidelines or standards

Audit is a non-essential but important part of medical revalidationIt is not meant to influence clinical practice

Explanation

It requires evaluation of current practice against guidelines or standards

This is correct in so far as clinical audit requires evaluation of current practice against guidelines, but these guidelines do not have to be national but can be local, regional or international.

It evaluates the difference between two treatments before clinical application This is an incorrect response because clinical audit compares clinical practice against established ‘gold standards’ and identifies areas for improvement, which are usually applied locally (but not always). Audit facilitates the evolution of novel guidelines. It does not, however, facilitate new treatments (this is research).

It facilitates the evolution of guidelines for different procedures

By comparing current clinical practice against existing guidelines areas for improvement in the current medical practice can be founds. Once any changes in practice are made following audit, it is important to repeat the process to close the loop. It is an essential part of medical revalidation.

Audit is a non-essential but important part of medical revalidation This is false as audit is an essential part of medical revalidation as it is one component of clinical governance.

It is not meant to influence clinical practice The reason for carrying out a clinical audit is to influence clinical practice. Audits are carried out to aim to improve clinical practice in a particular area which is why it is vital that once an audit has been carried out its results are discussed and acted upon by the clinical team.


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