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
The head of the femur articulates with which structure?
AAcetabulum
BGreater trochanter
CGlenoid cavity
DObturator foramen
2.Human anatomyHard
Injury to the common fibular (peroneal) nerve at the neck of the fibula typically causes:
AWeak hip abduction
BLoss of sensation on the sole of the foot
CFoot drop
DInability to flex the knee
3.Human physiologyMedium
Why can cardiac muscle not be tetanised like skeletal muscle?
AIt is not striated
BIt has no calcium in the sarcoplasmic reticulum
CIt is controlled only by hormones
DIts long refractory period lasts almost as long as the contraction
4.Human physiologyHard
A patient with diabetic ketoacidosis breathes deeply and rapidly. What pattern is expected on the blood gas?
AHigh pH, low bicarbonate and low PaCO₂
BLow pH, normal bicarbonate and high PaCO₂
CLow pH, low bicarbonate and low PaCO₂
DLow pH, high bicarbonate and high PaCO₂
5.MicrobiologyMedium
Which bacterium causes whooping cough (pertussis)?
ACorynebacterium diphtheriae
BBordetella pertussis
CHaemophilus influenzae
DMycoplasma pneumoniae
6.MicrobiologyHard
How do retroviruses such as HIV copy their genome into the host DNA?
AReverse transcriptase makes DNA from viral RNA, and integrase inserts it
BThe viral RNA is translated into DNA by ribosomes
CHost DNA polymerase copies the RNA without help
DThe viral DNA is copied directly by host RNA polymerase
7.PharmacologyMedium
Which adverse effect of ciprofloxacin should a patient report at once?
ADarker stools than usual
BMild tiredness in the first few days
CA slightly bitter or metallic taste for a while after each dose
DPain or swelling in a tendon, such as the Achilles tendon
8.PharmacologyHard
A patient who used heroin a few hours ago is given a first dose of buprenorphine and quickly develops severe withdrawal. Why?
ABuprenorphine is a pure antagonist like naloxone with no agonist effect
BBuprenorphine causes an allergic reaction in heroin users
CBuprenorphine is a high-affinity partial agonist that displaces heroin
DBuprenorphine has no effect at opioid receptors
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.44, PaCO₂ 20 mmHg (2.7 kPa) and HCO₃⁻ 13 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.44PaCO₂ 20 mmHg (2.7 kPa)HCO₃⁻ 13 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 136, Cl⁻ 113, HCO₃⁻ 15 mmol/L and albumin 17 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = 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 60 patients with a disease (cases) and 350 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 = 37b = 98c = 23d = 252
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 89% and a specificity of 84%. It is used in a population where the prevalence is 30%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 89%Sp = 84%p = 30%
PPV = %
3Solutions
1.Human anatomy
Answer: A · Acetabulum
Why this is the answer
The acetabulum is the cup-shaped socket of the hip bone, formed by the ilium, ischium and pubis. The glenoid cavity is the shoulder's socket.
Answer
A · Acetabulum
2.Human anatomy
Answer: C · Foot drop
Why this is the answer
The common fibular nerve supplies the dorsiflexors and evertors. Damage at the fibular neck causes foot drop and a high-stepping gait. The sole is supplied by the tibial nerve.
Answer
C · Foot drop
3.Human physiology
Answer: D · Its long refractory period lasts almost as long as the contraction
Why this is the answer
The plateau of the cardiac action potential gives a long refractory period, so the muscle relaxes before it can be excited again. This allows the heart to refill.
Answer
D · Its long refractory period lasts almost as long as the contraction
Bordetella pertussis causes long coughing fits with a "whoop". It is most dangerous for young babies, which is why vaccination in pregnancy is offered in many countries.
Answer
B · Bordetella pertussis
6.Microbiology
Answer: A · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
Why this is the answer
Reverse transcriptase converts the RNA genome into DNA, which integrase inserts into the host genome as a provirus. Both enzymes are drug targets.
Answer
A · Reverse transcriptase makes DNA from viral RNA, and integrase inserts it
7.Pharmacology
Answer: D · Pain or swelling in a tendon, such as the Achilles tendon
Why this is the answer
Fluoroquinolones can cause tendinitis and tendon rupture. Patients should stop exercise and report tendon pain promptly.
Answer
D · Pain or swelling in a tendon, such as the Achilles tendon
8.Pharmacology
Answer: C · Buprenorphine is a high-affinity partial agonist that displaces heroin
Why this is the answer
Buprenorphine binds mu receptors more strongly than heroin but activates them less, so it can precipitate withdrawal. Induction is started when the patient already has mild to moderate withdrawal signs.
Answer
C · Buprenorphine is a high-affinity partial agonist that displaces heroin
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 7.44: within 7.35–7.45
PaCO₂ 20 mmHg: low (35–45)
HCO₃⁻ 13 mmol/L: low (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: AGc = 13.75 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 = 136 − (113 + 15) = 8
Correction for albumin
8 + 0.25·(40 − 17) = 13.75 mmol/L
Answer
AGc = 13.75 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QT = 0.3 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
7.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 7.5 × 0.04 s = 0.3 s
Answer
QT = 0.3 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.60 s).
12.ECG rhythm strip interpretation
Answer: atrial rate = 60 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 = 25 small boxes = 1.00 s → 1500 ÷ 25 = 60 ≈ 60 per minute
Ventricular rate
4 QRS complexes in 6 s × 10 = 40 per minute
Answer
atrial rate = 60 bpm
Answer
atrial rate = 60 bpm
Second-degree type I (Wenckebach): 4 P waves for every 3 QRS complexes.
13.Evidence-based practice statistics
Answer: OR = 4.14
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: 37/23 = 1.609
controls: 98/252 = 0.3889
Odds ratio
OR = (37·252)/(98·23) = 4.137 ≈ 4.14
Answer
OR = 4.14
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 70.4 %
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.
True and false positives (per person)
Se·p = 0.89·0.3 = 0.267
(1 − Sp)·(1 − p) = 0.16·0.7 = 0.112
Positive predictive value
PPV = 0.267/(0.267 + 0.112) = 0.7045 ≈ 70.4%
Answer
PPV = 70.4%
The lower the prevalence, the lower the PPV of the same test.