Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which type of epithelium lines the urinary bladder?
ATransitional epithelium (urothelium)
BSimple squamous epithelium
CSimple columnar epithelium
DPseudostratified columnar epithelium
2.Human anatomyHard
A patient has a drooping eyelid, a dilated pupil and an eye that rests "down and out". Which nerve is most likely damaged?
AAbducens nerve (VI)
BOculomotor nerve (III)
CTrochlear nerve (IV)
DOptic nerve (II)
3.Human physiologyMedium
Besides lowering glucose, insulin moves which ion into cells, an effect used to treat hyperkalaemia?
ACalcium
BPotassium
CChloride
DSodium
4.Human physiologyHard
A patient hyperventilating during a panic attack has tingling fingers. What would the arterial blood gas most likely show?
AHigh pH and high bicarbonate
BLow pH and high PaCO₂
CHigh pH and low PaCO₂
DLow pH and low bicarbonate
5.MicrobiologyMedium
How does an autoclave sterilise instruments?
AWith ultraviolet light only
BWith dry heat at about 60 °C
CWith alcohol vapour at room temperature
DWith steam under pressure at about 121 °C
6.MicrobiologyHard
Which mechanism of antibiotic resistance does a bacterium use when it pumps the drug out of the cell?
AReduced porin uptake only
BEfflux pumps
CEnzymatic inactivation
DTarget modification
7.PharmacologyMedium
Why should long-term oral corticosteroids not be stopped suddenly?
AThe immune system would immediately attack the liver and adrenal glands
BStopping causes a sudden rise in blood glucose
CThe drug forms toxic metabolites when stopped
DThe adrenal glands are suppressed and may not make enough cortisol
8.PharmacologyHard
A breastfeeding mother needs a mild analgesic. Which drug is generally considered compatible with breastfeeding?
ACodeine (as co-codamol)
BParacetamol (acetaminophen)
CPethidine (meperidine)
DAspirin
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.37, PaCO₂ 64 mmHg (8.5 kPa) and HCO₃⁻ 36 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.
pH 7.37PaCO₂ 64 mmHg (8.5 kPa)HCO₃⁻ 36 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 135, Cl⁻ 112, HCO₃⁻ 14 mmol/L and albumin 17 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 310 patients with a disease (cases) and 580 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 60b = 75c = 250d = 505
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 88% and specificity 91%. The prevalence of the disease is 9%. How many people with the disease are missed by the test (false negatives, FN)?
N = 10 000p = 9%Se = 88%Sp = 91%
FN =
3Solutions
1.Human anatomy
Answer: A · Transitional epithelium (urothelium)
Why this is the answer
Transitional epithelium can change shape from rounded to flattened cells as the bladder stretches, while staying watertight.
Answer
A · Transitional epithelium (urothelium)
2.Human anatomy
Answer: B · Oculomotor nerve (III)
Why this is the answer
The oculomotor nerve supplies most eye muscles, levator palpebrae and the pupil constrictor. Lesion leaves lateral rectus and superior oblique unopposed, so the eye goes down and out.
Answer
B · Oculomotor nerve (III)
3.Human physiology
Answer: B · Potassium
Why this is the answer
Insulin stimulates the Na⁺/K⁺-ATPase and shifts potassium into cells. Insulin with glucose is therefore used to lower dangerously high plasma potassium.
Answer
B · Potassium
4.Human physiology
Answer: C · High pH and low PaCO₂
Why this is the answer
Hyperventilation blows off CO₂ and causes acute respiratory alkalosis. Alkalosis increases calcium binding to albumin, lowering ionised calcium and causing tingling.
Answer
C · High pH and low PaCO₂
5.Microbiology
Answer: D · With steam under pressure at about 121 °C
Why this is the answer
Pressurised steam reaches about 121 °C and kills all microorganisms, including spores, typically in about 15 minutes or more depending on the load.
Answer
D · With steam under pressure at about 121 °C
6.Microbiology
Answer: B · Efflux pumps
Why this is the answer
Efflux pumps actively export antibiotics such as tetracyclines and fluoroquinolones, keeping the inside concentration too low to work.
Answer
B · Efflux pumps
7.Pharmacology
Answer: D · The adrenal glands are suppressed and may not make enough cortisol
Why this is the answer
Steroid therapy suppresses the hypothalamic–pituitary–adrenal axis. Sudden withdrawal can cause an adrenal crisis, so doses are reduced gradually.
Answer
D · The adrenal glands are suppressed and may not make enough cortisol
8.Pharmacology
Answer: B · Paracetamol (acetaminophen)
Why this is the answer
Paracetamol passes into milk in small amounts and is widely regarded as safe. Codeine can cause dangerous opioid levels in babies of ultra-rapid metaboliser mothers, aspirin carries a Reye's risk and pethidine's metabolite accumulates in infants.
Answer
B · Paracetamol (acetaminophen)
9.Arterial blood gas interpretation
Answer: fully compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.37: within 7.35–7.45
PaCO₂ 64 mmHg: high (35–45)
HCO₃⁻ 36 mmol/L: high (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: AGc = 14.75 mmol/L
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Anion gap
AG = 135 − (112 + 14) = 9
Correction for albumin
9 + 0.25·(40 − 17) = 14.75 mmol/L
Answer
AGc = 14.75 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QTc = 0.469 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 10.5 × 0.04 = 0.42 s
RR = 18 × 0.04 = 0.72 s
Fridericia’s formula
QTc = 0.42/∛0.72 = 0.42/0.8963 = 0.469 s
Answer
QTc = 0.469 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (469 ms)
12.ECG rhythm strip interpretation
Answer: atrial rate = 68 bpm
AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.
Atrial rate
P–P = 22 small boxes = 0.88 s → 1500 ÷ 22 = 68.18 ≈ 68 per minute
Ventricular rate
5 QRS complexes in 6 s × 10 = 50 per minute
Answer
atrial rate = 68 bpm
Answer
atrial rate = 68 bpm
Second-degree type I (Wenckebach): 3 P waves for every 2 QRS complexes.
13.Evidence-based practice statistics
Answer: OR = 1.62
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Odds of exposure
cases: 60/250 = 0.24
controls: 75/505 = 0.1485
Odds ratio
OR = (60·505)/(75·250) = 1.616 ≈ 1.62
Answer
OR = 1.62
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: FN = 108
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.